Solution Manual for The Physics of Energy 1st Edition Robert L. Jaffe, Washington Taylor

SOLUTION MANUAL

The Physics of Energy Solutions Manual Version 0.9 (Note: work still in progress; this version is still missing a few solutions, and may contain some errors. User beware.)

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R. L. Jaffe

W. Taylor

The Physics of Energy: Solutions manual

This solutions manual contains worked solutions for the problems in the textbook “The Physics of Energy” by Jaffe and Taylor. This is intended only for use by instructors teaching classes using the book as a text. Do not distribute this document or post it online.

These solutions have not all been thoroughly checked and some errors may be present; the users of this manual are advised to solve the problems independently themselves and compare with these solutions. Please send corrections or improvements to jaffe@mit.edu and wati@mit.edu.

Most recent revision: June 17, 2018

Part I Basic Energy Physics and Uses

11

Problems

Problems 1.1

If 83% of this were from oil, this is equivalent to

Confirm eq. (1.14) and the estimate for the rhinoceros’s kinetic energy below eq. (1.11) by explicit calculation.

0.83(1.03 × 1020 J) This is equivalent to

The mass of a proton is M p = 1.673 × 10−27 kg. When a proton and anti-proton, which has the same mass as the proton, annihilate, the energy released is equal to the energy equivalent to the total mass of the system. So, this energy is E = (M p + M p̄ )c2 = 2M p c2

0.83(1.03 × 1020 J)

1.5

= 2(1.673 × 10−27 kg)(2.998 × 108 m/s)2 = 3.007 × 10−10 J. In units of MeV, the energy is E = 3.007 × 10−10 J ×

1 MeV = 1 877 MeV. 1.602 × 10−13 J

1.2

1.3

1.6

3600 s = 3.6 × 106 J = 3.6 MJ. 1h

In a typical mid-latitude location, incident solar energy averages around 200 W/m2 over a 24-hour cycle. Compute the land area needed to supply the average person’s energy use of 200 MJ/day if solar panels convert a net 5% of the energy incident over a large area of land to useful electrical energy. Multiply by world population to get an estimate of total land area needed to supply the world energy needs from solar power under these assumptions. Compare this land area to some relevant reference areas – the Sahara Desert, your native country, etc.

The total land area needed to supply an average person’s energy use is A1 person =

200 MJ/day 106 W 1 day = 231 m2 .

(0.05)(200 W/m2 ) MJ/s 86 400 s

Earth has a total population of about 7 × 109 , so the total land area needed to supply world energy needs is Atotal = (231 m2 /person)(7 × 109 persons) ≈ 1.6 × 106 km2 .

This is approximately the size of Alaska or Iran, roughly 17% of the area of the Sahara Desert.

1.4

The US total energy consumption in 2014 was 98.0 quads. What is this quantity in joules? About 83% of US energy comes from fossil fuels. If the whole 83% came from oil, how many tons of oil equivalent (toe) would this have been? How many barrels of oil (bbl) is this equivalent to?

The US total energy consumption in joules is 98.0 quads

1.055 × 1018 J = 1.03 × 1020 J. 1 quad

[mass][distance]2 /[time]2 [distance] [energy] = = . [mass][distance] [mass][distance] [time]2

This is an acceleration, suggesting that near the surface all objects undergo the same acceleration due to gravity.

The light bulb consumes (100 W)(10 h) = (100 J/s)(10 h)

The gravitational potential energy of an object of mass m at a distance h above ground is given by E = mgh.

Use dimensional analysis to compute the units of g.

What does this result suggest about the behavior of objects near Earth’s surface?

[g] =

!2

How much energy would a 100 W light bulb consume if left on for 10 hours?

1 bbl = 1.40 × 1010 bbl.

6.118 × 109 J

Using dimensional analysis,

A 3 000 kg rhinoceros charging at 50 km/hour has a total kinetic energy of 1 000 m 1 h 1 E = mv2 = 0.5(3 000 kg) (50 km/h) 2 km 3 600 s !

1 kJ = 289 kJ. × 1 000 kg m2 /s2

1 toe = 2.04 × 109 toe.

41.868 × 109 J

The energy emitted or absorbed in chemical processes is often quoted in kilojoules per mole (abbreviated mol) of reactants, where a mole contains NA 6.022 × 1023 (Avogadro’s number) molecules (§5).

Derive the conversion factor from eV/molecule to kJ/mol.

1

1.7

eV eV 1.602 × 10−19 J 1 kJ =1 molecule molecule eV 1 000 J 6.022 × 1023 molecules kJ × = 96.47 .

1 mol mol

The energy available from one kilogram of 235 U is 82 TJ. Energy is released when each uranium nucleus splits, or fissions. (The fission process is described in §16, but you do not need to know anything about the process for this problem.) 235 grams of 235 U contain approximately Avagadro’s number (see Problem 1.6) atoms of 235 U. How many millions of electron volts (MeV) of energy are released, on average, when a 235 U nucleus fissions? How many kilograms of gasoline have the same energy content as one kilogram of 235 U? The energy released per nucleus in MeV is ! ! !

1 mol TJ 1012 J 0.235 kg kg TJ 1 mol 6.022 × 1023 atoms 1 MeV MeV 1 eV 200 × . atom 1.602 × 10−19 J 106 eV

E 82

We need the energy content of gasoline, 120 MJ/gallon(US), and the density of gasoline, 0.755 kg/L. One kg of 235 U is equivalent to M(gasoline) = 82 × 1012 J

1 gallon (US) 120 × 106 J

= 1.834 × 106 kg .

!

3.785 L gallon(US)

!

0.755 kg L

!

12

1.8

The US total electrical power consumption in 2010 was 3.9 TkWh. Utilities try to maintain a capacity that is twice the average power consumption to allow for high demand on hot summer days. What installed generating capacity does this imply?

The installed generating capacity is 2(3.9 × 1015 Wh/y)/(8 766 h/y) 890 GW .

27

Problems

Problems 2.1

which can be checked by differentiating with respect to each coordinate. We set the integration constant C = 0 so that the potential energy is zero at x = 0.

The equation of motion is m ẍ = F = −kx, and suggested solution can be written x(t) = x0 cos ωt, where x0 = (x0 , y0 , z0 ).

Differentiating twice with respect to t √we see ẍ = −ω2 x, which satisfies the equation of motion if ω = k/m.

The total energy is the sum of kinetic, Ekin = 12 m ẋ2 , and potential, V(x).

Estimate the kinetic energy of the ocean liner Queen Mary 2, with mass (displacement) 76 000 tons, moving at a cruising speed of 26 knots. Compare with the change in potential energy of the Queen Mary 2 when lifted by a three foot tide.

The knot is a nautical unit of speed, 1 knot = 0.5144 m/s, so 26 knots = 13.4 m/s. The Queen Mary 2’s kinetic energy is then Ekin =

1 2 1 2 m ẋ + kx 2 2 1 1 1 = mω2 x20 sin2 ωt + kx20 cos2 ωt = kx20 , 2 2 2

E=

1 2 mv = 0.5(76 × 106 kg) (13.4 m/s)2 = 6.82 GJ .

2

where we have used mω2 = k and sin2 + cos2 = 1.

This solution is not the most general since it requires the oscillator to be maximally stretched at t = 0. It is easy to show that x(t) = x0 cos ω(t − t0 ) is a solution for any t0 . Using the identity cos(θ − φ) = cos θ cos φ + sin θ sin φ, it is clear that this solution is a sum of sine and cosine functions.

The change in potential energy when lifted by a three foot tide is !

0.3048 m = 663 MJ , V = mgh = (76 × 106 kg)(9.81 m/s2 )(3 ft) 1 ft or about 10% of the kinetic energy.

2.2

[T] A mass m moves under the influence of a force derived from the potential V(x) = V0 cosh ax, where the properties of cosh x and other hyperbolic functions are given in Appendix B.4.2. What is the force on the mass? What is the frequency of small oscillations about the origin? As the magnitude of the oscillations grows, does their frequency increase, decrease, or stay the same?

2.4

From eq. (2.20), the potential energy relative to the surface is V(h) = −

The force on the mass is F(x) = −dV/dx = −V0 a sinh ax, with sinh x = 12 (e x − e−x ). For small x, the exponential function is approximated using a Taylor expansion by ey = 1 + y +

In a circular orbit, the gravitational force is equal to the mass times the centripetal acceleration,

so cosh y = 1+y2 /2+. . . and sinh y = y+y3 /6+. . .. For small oscillations, i.e. when ax ≪ 1, so we only need keep the first term in the expansion of sinh ax, the force is F(x) = −V0 a2 x. The equation of motion for small oscillations is

mv2 GM⊕ m . = r r2 Solving for velocity, s r GM⊕ (6.67 × 10−11 )(6 × 1024 ) v= = m/s = 7.7 km/s. r (6.37 × 106 + 3.5 × 105 )

m ẍ = −V0 a2 x , which is the equation √ for a harmonic oscillator (see eq. (2.10)) with frequency ω = a V0 /m.

This frequency is independent of magnitude of oscillations. If, however, the amplitude of the oscillation becomes large compared to 1/a, then the non-linear terms in the expansion of sinh x cannot be ignored and the equation of motion becomes !

1 m ẍ = −V0 a ax + a3 x3 + . . . , .

6

2.3

[T] An object of mass m is held near the origin by a spring, F = −kx. What is its potential energy? Show that x(t) = (x0 , y0 , z0 ) cos ωt is a possible classical motion for the mass. What is the energy of this solution? Is this the most general motion? If not, give an example of another solution.

According to eq. (2.18), the force is the negative gradient of the potential energy. Setting F = −kx = −∇V, the solution is V(x) =

1 2 kx + C , 2

GmM⊕ GmM⊕ + R⊕ + h R⊕

= (6.67 × 10−11 N m2 /kg2 ) (3.7 × 105 kg) (6 × 1024 kg)× !

1 1 × = 1.21 TJ. − 6.37 × 106 m 6.37 × 106 + 3.5 × 105 m

y2 y3 + + ..., 2 6

All of the terms in the expansion have the same sign, so as the amplitude of oscillations increases, the magnitude of the force increases compared to just a simple harmonic oscillator and the frequency of the oscillations increases.

Estimate the potential energy of the international space station (mass 370 000 kg) relative to Earth’s surface when in orbit at a height of 350 km. Compute the velocity of the space station in a circular orbit and compare the kinetic and potential energy.

The kinetic energy is equal to Ekin =

1 2 1 mv = (3.7 × 105 kg) (7.7 × 103 m/s)2 = 11 TJ. 2 2

Thus, the kinetic energy is 9.1 times the potential energy relative to Earth’s surface.

2.5

[T] Relate eq. (2.20) to eq. (2.9) and compute g in terms of G, M⊕ , R⊕ as described in the text.

The potential yielding the r−2 force law is V(r) = −GMm/r.

In the approximation that the gravitational force is constant, the potential is V(z) = mgz, where z = r − R⊕ is the height from Earth’s surface, R⊕ is Earth’s radius and z ≪ R⊕ . By definition, V(z) − V(0) = −

GM⊕ m GM⊕ m + .

R⊕ + z R⊕

Use a series expansion for 1/(R⊕ + z), (see eq. (B.63)), 1 1 z = − + ... , R⊕ + z R⊕ R2⊕ giving V(z) − V(0) =

GM⊕ m (z + O(z2 /R3⊕ )) R2⊕

28

Comparing this to V(z) = mgz, and setting V(0) = 0, we get to leading order in z/R⊕ , g=

(a) The moon orbits the earth about once a month and has an orbital radius of about 380 000 km, so its speed is v = 2πR/T 0.91 km/s. The moon has a mass of 7.3 × 1022 kg, so this corresponds to a kinetic energy of

GM⊕ (6.67 × 10−11 N m2 /kg2 )(5.97 × 1024 kg) = (6.37 × 106 m)2 R2⊕

Ekin =

= 9.81 m/s2 .

2.6

(b) Large raindrops are about 5 mm in diameter and fall at ∼9 m/s, so they have a kinetic energy of ! !3 1 4πρ d Ekin = v2 2 3 2 !

4π (2.5 × 10−3 m)3 )(9 m/s)2 0.5(1000 kg/m3 ) 3

Make a rough estimate of the maximum hydropower available from rainfall in the US state of Colorado. Look up the average yearly rainfall and average elevation of Colorado and estimate the potential energy (with respect to sea level) of all the water that falls on Colorado over a year. How does this compare with the US yearly total energy consumption?

∼ 3 mJ.

(c) Lake Powell has a volume of approximately 33 km3 and the Glen Canyon Dam has a height of Z = 220 m. Treating the lake as a box of area A and volume V = ZA, its potential energy is Z Z V = ρgA dz z = ρgAZ 2 /2 = (1/2)ρgVZ

Colorado has an average yearly precipitation of ∼0.4 m. Suppose this is typical of the whole state of Colorado. Colorado has a mean elevation of approximately 2 070 m and a total surface area of 270 000 km2 . All the precipitation falling on the state in one year then has a total potential energy of

0

∼ (0.5)(1000 kg/m3 )(9.8 m/s2 )(33 × 109 m3 )(220 m) ∼ 36 PJ .

mgh ≈ ρVgh (1000 kg/m3 )(0.4 m)(2.7 × 1011 m2 ) 2

18

× (9.8 m/s )(2070 m) 2.2 × 10

(d) The cross sectional area of a person on a bike is around 0.5 m2 with a drag coefficient around 1. The energy output is mostly to counteract air resistance. The energy lost is

J 2.2 EJ.

This is about 2% of the total energy consumed in the US each year.

2.7

Choose your favorite local mountain. Estimate how much energy it takes to hike to the top of the mountain (ignoring all the local ups and downs of a typical trail).

How does your result compare to a typical day’s food energy intake of 10 MJ, taking into account the fact that human muscles are not 100% efficient at converting energy into work?

1 ρair cd Av2 d = 0.5(1.2 kg/m3 )(0.5 m2 ) 2 × (4.2 m/s)2 (1.5 × 104 m) 79 kJ .

Assuming an efficiency of ∼25% for the human body, around 340 kJ will will have to be expended (e) Assume the runner has a cross-sectional area of approximately 0.5 m3 and a drag coefficient around 1. Good runners finish in around 10 s, so they have a speed of 10 m/s. The total energy lost to air resistance will be (1/2)cd ρair Av2 d ∼ 3 kJ. (f) Estimate that a climber and gear have a mass of approximately 90 kg. The top of Mt. Everest is 8 848 m above sea level, so the climber will have

Mount Washington is the tallest mountain in the State of New Hampshire? The peak of Mount Washington is at 1 917 meters (6 288 feet) above sea level. A typical route, however, begins at about 600 meters above sea level, so the elevation gain is roughly 1 300 m. For a hiker of 80 kg (including food, water, extra clothing, etc.), the potential energy gain on the hike is V = mgh (80 kg)(9.8 m/s2 )(1300 m) 1 MJ .

mgh (90 kg)(9.8 m/s2 )(8.9 × 103 m) 7.8 MJ of gravitational potential energy with respect to sea level.

(2.45)

At 20-30% muscle efficiency, this represents about one third to one half of the useful energy output from the typical hiker’s 10 MJ/day food energy intake. This is computed just from the elevation gain, without including local ups and downs of the trail, horizontal distance covered, or the hike down. So don’t feel bad about consuming a lot of high-calorie trail food next time you are on a strenuous hike!

2.8

1 2 mv = 0.5(7.3×1022 kg)(0.91×103 m/s)2 3.0×1028 J. 2

Use any means at your disposal (including the internet) to justify estimates of the following (an orderof-magnitude estimate is sufficient): (a) The kinetic energy of the Moon in its motion around Earth. (b) The kinetic energy of a raindrop before it hits the ground. (c) The potential energy of the water in Lake Powell (relative to the level of the Colorado river directly beneath the dam). (d) Energy needed for you to power a bicycle 15 km at 15 km/h on flat terrain. (e) The energy lost to air resistance by an Olympic athlete running a 100 m race. (f) The gravitational potential energy of a climber at the top of Mount Everest.

2.9

Verify the claim that conversion of the potential energy at the top of a 15 m hill to kinetic energy would increase the speed of the Toyota Camry from 62 to 74 mph.

A Camry has a mass of 1800 kg, so it gains 265 kJ of kinetic energy when its elevation falls by 15 m. Conservation of energy requires that 1 2 mv + mgh = constant.

2

2.10

Using this, the final velocity if it starts at 62 mi/hr is q v f = v2i + 2gh = 32.6 m/s = 73 mi/hr.

Consider a collision between two objects of different masses M, m moving in one dimension with initial speeds V, −v. In the center-of-mass frame, the total momentum is zero before and after the collision. In an elastic collision both energy and momentum are conserved. Compute the velocities of the two objects after an elastic collision in the center-of-mass frame. Show that in the limit where m/M → 0, the

29

Problems

more massive object remains at rest in the center-ofmass frame (V = 0 before and after the collision), and the less massive object simply bounces off the more massive object (like a tennis ball off a concrete wall).

and at 10 000 m has a potential energy of V = mgh = (70 000 kg)(9.8 m/s2 )(10 000 m) = 6.86 GJ .

The density of air at 10 000 m is approximately e−10/8.5 = 31% of the density at sea level ρair = 0.31(1.17 kg/m3 ) 0.36 kg/m3 .

Traveling at v = 800 km/h for d = 2 000 km at 10 000 m, the plane loses

In the center-of-mass frame the objects have velocities V ′ = V + u and v′ = v + u, where −u is the velocity of the center-of-mass viewed from the original frame. The total momentum of the two objects in the center-of-mass vanishes: ptot = M(V+u)+m(−v+u) = 0, which determines the center-of-mass velocity

1 ρair cd Av2 d 0.5(0.36 kg/m3 )(0.03) 2 × (12 m2 )(222 m/s)2 (2000 × 103 m) 6.4 GJ .

Eair =

due to air resistance. The total energy used by the plane over the flight is approximately 15.3 GJ. This is 306 MJ/passenger or 153 kJ/passenger-km. A typical car uses around 210 MJ of mechanical energy over the 340 km trip between New York and Boston.

With 2 passengers, this is 310 kJ/passenger-km, or nearly twice the energy usage of the airplane. So, before engine efficiencies are considered, automobiles use roughly twice as much energy over a long trip.

mv − MV u= m+M and m (v + V) m+M M v′ = −v + u = − (v + V) m+M V′ = V + u =

2.13

are the objects’ velocities before the collision in the center-ofmass. If energy and momentum are both conserved in the collision, the objects’ must reverse their center-of-mass velocities after the collision, m (v + V) m+M M v′f = −v′ = (v + V) . m+M

V ′f = −V ′ = −

In the limit m/M → 0, V ′ → 0 and V ′f → 0, the heavier object begins and remains at rest, while v′f = −v′ = (v + V), the lighter object bounces off.

2.11

As we explain in §34.2.1, the density of air decreases with altitude roughly as ρ ∼ ρ0 e−z/H , where z is the height above the Earth’s surface and H 8.5 km near the surface. Compute the ratio of air resistance losses for an airplane traveling at 750 km/h at an altitude of 12 000 m compared to an altitude of 2000 m. What happens to this ratio as the speed of the airplane changes?

How do automobile air resistance losses at 2000 m compare to losses at sea level?

The ball has an effective area of A = πR2 = 0.00442 m2 . For air at sea level and 25◦ C, ρair = 1.17 kg/m3 . The energy loss rate is 1 dE = − cd Aρair v3 −0.5(0.3)(0.00442 m2 ) dt 2 × (1.17 kg/m3 )(44.7 m/s)3 −69 J/s.

If the ball’s speed does not differ significantly from 100 mph, it takes (a) t = (18.4 m)/(44.7 m/s) 0.41 sec for the ball to reach the batter. The total amount of energy lost is (b) ∆E = dE dt × t −28 J. At 100 mph, the ball has a kinetic energy of

Air resistance losses are linearly proportional to the density of air, so the ratio of loses at 12 000 m compared to 2000 m is r = e−∆z/H e−10/8.5 31%. Although the losses (dEair /dt) are proportional to speed cubed, the speed cancels out in the ratio of losses at two different air densities, so the ratio is independent of the airplane’s speed. Similarly for a car at 2000 m compared to sea level r e−2/8.5 79%.

2.12

Consider an airplane with mass 70 000 kg, crosssectional area 12 m2 , and drag coefficient 0.03. Estimate the energy needed to get the plane moving at 800 km/h and lift the plane to 10 000 m, and estimate air-resistance losses for a flight of 2000 km using the formula in the previous problem. Do a rough comparison of the energy used per person to do a similar trip in an automobile, assuming that the plane carries 50 passengers and the automobile carries two people.

For simplicity, assume that acceleration occurs quickly so that air resistance during acceleration can be neglected. The plane requires a total kinetic energy of Ekin =

1 2 mv = 0.5(70 000 kg)(800 × 103 m/3600 s)2 = 1.73 GJ , 2

In the American game of baseball, a pitcher throws a baseball, which is a round sphere of diameter b = 0.075 m, a distance of 18.4 m (60.5 feet), to a batter, who tries to hit the ball as far as he can. A baseball has a mass close to 0.15 kg. A radar gun measures the speed of a baseball at the time it reaches the batter at 44.7 m/s (100 mph). The drag coefficient cd of a baseball is about 0.3. Give a semi-quantitative estimate of the speed of the ball when it left the pitcher’s hand by (a) assuming that the ball’s speed is never too different from 100 mph to compute roughly how long it takes to go from the pitcher to the batter, (b) using (a) to estimate the energy lost to air resistance, and (c) using (b) to estimate the original kinetic energy and velocity.

Ekin f =

1 2 mv 150 J. 2

(c) The initial kinetic energy of the ball is just the difference of this and the energy loss, Ekin i = 150 J + 28 J = 178 J. This corresponds to a velocity of s p 2(178 J) v = Ekin i m m/s 49 m/s 110 mph.

(0.15 kg)

2.14

Estimate the power output of an elite cyclist pedaling a bicycle on a flat road at 14 m/s. Assume all energy is lost to air resistance, the cross-sectional area of the cyclist and the bicycle is 0.4 m2 , and the drag coefficient is 0.75. Now estimate the power output of the same elite cyclist pedaling a bicycle up a hill with slope 8% at 5 m/s. Compute the air resistance assuming the

30

drag coefficient times cross-sectional area is cd A = 0.45 m2 (rider is in a less aerodynamic position). Compute the ratio of the power output to potential energy gain. Assume that the mass of the rider plus bicycle is 90 kg.

the lab frame, we add a velocity +v to everything, so that the cylinder once again is traveling at a velocity of +v and the air molecule has a velocity of +2v.

To find cd , we must now compute the rate of energy loss of the cylinder. In time dt, the front end of the cylinder passes through a volume Avdt of air, corresponding to a mass of dm = Aρvdt. This mass is accelerated from rest to a velocity of 2v. So, the column of air gains an energy of

On the flat road, the power output of the cyclist is equal to the power lost due to air resistance,

dEair =

1 P = ρcd Av3 (0.5)(1.2 kg/m3 )(0.75)(0.4 m2 )(14 m/s)3 490 W. 2

From conservation of energy, the cylinder loses an equal amount of energy, so the power output due to air resistance is ! dE 1 = −2Aρv3 = − cd Aρv3 , dt cyl 2

The power output of the cyclist on the slope is equal to the power lost due to air resistance and due to the rate of change of potential energy, P = Pair + Pg .

The contribution from air resistance is Pair =

1 ρcd Av3 = (0.5)(1.2 kg/m3 )(0.45 m2 )(5 m/s)3 = 34 W. 2

where cd = 4.

2.17

The contribution from gravitational potential energy Pg = mg

dz (90 kg)(9.8 m/s)(5 m/s) sin(arctan(8/100) 350 W dt

so the total power output of the cyclist on the slope is P = 380 W.

2.15

Compare the rate of power lost to air resistance for the following two vehicles at 60 km/h and 120 km/h: (a) General Motors EV1 with cd A 0.37 m2 , (b) Hummer H2 with cd A 2.45 m2 .

mg =

Aeff =

2mg . ρair v2∞

The effective area of the soccer ball is Aeff = cd A = cd πr2 = 0.0095 m2 . The limiting velocity is s s 2mg 2(0.43 kg)(9.8 m/s2 ) v∞ =

27 m/s . ρair Aeff (1.2 kg/m3 )(0.0095 m2 )

At 120 km/h, the power is 8 times as high, or 8.2 kW. (b). The power for the Hummer H2 scales with the cd A, so the H2 loses energy at a rate of (2.45/.37)(1.03 kW) = 6.8 kW at 60 km/h and (2.45/.37)(8.2 kW) = 54 kW at 120 km/h. In both cases, the H2 loses energy at a rate 6.6 times that of the EV1.

Suppose the cylinder is traveling at a velocity v. In the cylinder’s rest frame, the air molecules are traveling at a velocity −v perpendicular to the surface of the cylinder. Taking the limit that the cylinder is infinitely more massive than the air molecules, this is the center of mass frame. In this limit, a collision may change the direction of the air molecule but not the speed. Since the velocity of the air molecules is perpendicular to the end of the cylinder, the final velocity must be +v in the cylinder rest frame. Going back to

1 ρair Aeff v2∞ .

2

Solving for Aeff ,

dEloss 1 = ρcd Av3 dt 2 (0.5)(1.2 kg/m3 )(0.37 m2 )(16.7 m/s)3 = 1.03 kW

[T] Consider an idealized cylinder of cross-sectional area A moving along its axis through an idealized diffuse gas of air molecules with vanishing initial velocity. Assume that the air molecules are pointlike and do not interact with one another. Compute the velocity that each air molecule acquires after a collision with the cylinder, assuming that the cylinder is much more massive than the air molecules. [Hint: assume that energy is conserved in the reference frame of the moving cylinder and use the result of Problem 2.10.] Use this result to show that the drag coefficient of the cylinder in this idealized approximation is cd = 4.

One way to estimate the effective area (see eq. (2.31)) of an object is to measure its limiting velocity v∞ falling in air. Explain how this works and find the expression for Aeff as a function of m (the mass of the object), v∞ , g, and the density of air. The drag coefficient of a soccer ball (radius 11 cm, mass 0.43 kg) is cd ≈ 0.25. What is its limiting velocity?

Because the force due to air resistance increases with velocity, if an object is allowed to fall in air, there is some velocity at which the upward force due to air resistance is equal in magnitude to the downward gravitational force. When this happens, there is no net force on the object, so the object continues falling at this velocity, which is v∞ . Thus,

(a). At 60 km/h (16.67 m/s)

2.16

1 dmv2air = 2Aρv3 dt .

2

2.18

If the vehicle used as an example in this chapter accelerates to 50 km/h between each stoplight, find the maximum distance between stoplights for which the energy used to accelerate the vehicle exceeds the energy lost to air resistance. (You may assume that the time for acceleration is negligible in this calculation.) How does the result change if the vehicle travels at 100 km/h between lights? Why? The vehicle in this chapter has A = 2.7 m2 , cd = 1/3, and m = 1 800 kg. The total energy lost to friction after a distance d at a velocity v is Elost =

1 ρcd Av2 d.

2

Setting this equal to the kinetic energy and solving for d to get the maximum distance for which the kinetic energy exceeds the energy lost due to air resistance, 1 2 1 mv = ρcd Av2 d 2 2 m 1 800 kg d= 1.67 km.

ρcd A (1.2 kg/m3 )(1/3)(2.7 m2 )

Problems

The result is the same if the vehicle travels at 100 km/h. Both the kinetic energy and the energy loss are proportional to v2 so the distance at which they are equal is independent of the velocity.

2.19

Estimate the rotational kinetic energy in a spinning yoyo (a plastic toy that you can assume is a cylinder of diameter 5.7 cm and mass 52 g, which rotates at 100 Hz). Compare to the gravitational potential energy of the yo-yo at height of 0.75 m.

The moment of inertia for a cylinder of mass m, length z and radius R rotating around its axis is (see Example 2.4), Icylinder =

1 1 ρπzR4 = mR2 .

2 2

The energy of the rotating yo-yo is E = (2π s−1 /Hz)(100 Hz) 630 s−1 , so

(2.46)

1 2 2 Iω and ω = 2πν =

1 2 Iω (0.25)(0.052 kg)(0.0285 m)2 (630 s−1 )2 4.2 J .

2 (2.47) Its gravitational potential energy at a height of 0.75 m is mgh 0.38 J, almost a factor of ten smaller than its rotational kinetic energy at 10 Hz. (Most of the energy is imparted in the initial “throw,” and not from gravitational potential energy.)

Erot =

2.20

Verify the assertion (see Example 2.3) that Ekin = − 21 V for the Moon in a circular orbit around Earth. [Hint: the magnitude of the centripetal acceleration for circular motion (2.35) can be rewritten a = v2 /r.]

Assume throughout that M⊕ ≫ mmoon , so we may take Earth to be at rest. For a circular orbit, the centripetal acceleration is a = v2 /r. The acceleration due to gravity is GM⊕ /r2 where M⊕ is Earth’s mass, which is much greater than that of the moon. The gravitational potential energy of the moon is V = −GM⊕ m/r The centripetal acceleration and gravitational acceleration are identical, so

2.21

GM⊕ GM⊕ v2 = 2 or v2 = , so r r r 1 2 1 1 GM⊕ m Ekin = mv = − − = − V. 2 2 r 2

Estimate Earth’s kinetic energy of rotation (the moment of inertia of a uniform sphere is 25 MR2 ).

Earth has a mass of 5.972 × 1024 kg and a mean radius of 6 371 km. It rotates about its axis once per sidereal day (23.9345 h), giving it an angular velocity of ω = 2π/(1 sidereal day) = 7.292 × 10−5 s−1 . Its kinetic energy of rotation is MR2 2 ω = (0.2)(5.972 × 1024 kg)(6 371 × 103 m)2 5 × (7.292 × 10−5 rad/s)2 = 2.578 × 1029 J .

E=

31

60

Problems 3.1

The electric field outside a charged conducting sphere is the same as if the charge were centered at its origin. Use this fact to calculate the capacitance of a sphere of radius R, taking the second conductor to be located at infinity. What is the most charge you can store on an otherwise isolated spherical conductor of radius R without a breakdown of the surrounding air as discussed in Example 3.2? What is the maximum energy you can store on a conducting sphere of radius 1 mm?

individual point charges or infinitesimal charge elements. Since Gauss’s law holds for a point charge, it must then hold for any charge distribution.

3.3

The energy stored in the capacitor is EEM = CV 2 /2. The capacitance is C = ǫA/d and the maximum voltage is Vmax = |E|max d, so 1 ǫA 1 EEM = (|E|max d)2 = ǫAd|E|2max 2 d 2 (0.5)(8.85 × 10−10 F/m)(10 cm2 )

The capacitance is defined by C=

× (10−2 m/cm)2 (10−6 m)(3.3 × 106 V/m)2

Q .

V

For a conducting sphere of radius R and charge Q, V is just the potential of a point charge Q at a radius R, so

4.8 µJ .

3.4

C = 4πε0 R = (1.11 × 10−10 F/m) R .

Breakdown in air occurs at Emax = 3.3 × 106 V/m. The field around the sphere is highest at the radius R, where it is Emax =

Qmax .

4πε0 R2

Thus Qmax = 4πǫ0 R2 Emax = (1.11 × 10−10 F/m)(3.3 × 106 V/m) (3.7 × 10−4 C/m2 ) R2 .

For a sphere of radius 1 mm, the maximum charge that can be stored is and its capacitance is 0.111 pF . The total (maximum) energy stored by the capacitor

3.2

Q2max (3.7 × 10−10 C)2

6.2 × 10−7 J .

2C 2.2 × 10−13 F

aE = a

[T] Prove Gauss’s law from Coulomb’s law for static charge distributions by showing that the electric field of a single charge satisfies the integral form of Gauss’s law and then invoking linearity.

Due to linearity, the field of many charges or even a continuous charge distribution is identical to the sum of the fields of the

Q ǫ0 A

or

E=−

Q ẑ . ǫ0 A

The voltage difference between the upper and lower plate is given by eq. (3.7), Z d Qd V=− dx · E = Ed = , ǫ0 A 0

The integral form of Gauss’s law is given in eq. (3.10). Take a spherical surface surrounding a point charge Q. The electric field is E = Qr̂/(4πǫ0 r2 ), so integrating the field over the surface, Z Q Q = . dΩr2 ǫ0 4πǫ0 r2 The result is the same for a sphere of any radius including a sphere of infinitesimally small radius. Now, consider a sphere not enclosing the charge. In the limit that the radius is very small, the electric field is constant across the entire sphere. In this case, there is no net flux into or out of the infinitesimally small sphere, so the same integral gives a value of zero.

Any closed surface can be created by combining the surfaces of infinitesimally small spheres, as surfaces shared between the different spheres have no net contribution. Since only the infinitesimally small sphere centered on the charge has a nonzero contribution, for any closed surface S surrounding a point charge Q, I Q (3.75) dS · E = . ǫ0 S

[T] Starting from Gauss’s law and ignoring edge effects (i.e. assume that the plates are very large and the electric field is uniform and perpendicular to the plates), derive the formula for the capacitance of a parallel plate capacitor, C = ǫ0 A/d. Refer to Figure 3.8. You can assume that the electric field vanishes outside the plates defining the capacitor. Show that the energy stored in the capacitor, 21 CV 2 , can be written as the integral of ǫ0 |E|2 /2 over the region within the capacitor (as asserted in eq. (3.20)).

Ignoring edge effects, the charge Q is uniformly distributed over the plates corresponding to a surface charge density Q/A. E points in the negative ẑ direction. Apply Gauss’s law (3.10) to a surface with sides parallel to E. The top of the surface, of area a is inside the upper conducting plate, the bottom, also of area a is inside the capacitor. A slice through the surface is denoted by the dotted line in Figure 3.8. Gauss’s law gives no contribution on the top or sides,

Qmax 3.7 × 10−10 C

EEM =

How much energy can you store on a parallel plate capacitor with d = 1 µm, A = 10 cm2 , and ǫ = 100ǫ0 , assuming that the breakdown field of the dielectric is the same as for air?

(Note, the differential line element dx = +d z, so E · dx = −Edz).

From the definition Q = CV, we find C = ǫ0 A/d .

Integrating over the volume of the capacitor Z

3.5

dVǫ0 |E|2 /2 =

Q ǫ0 Ad 2 ǫ0 A

!2

=

!

1 Q2 1 d = CV 2 .

Q2 = 2 ǫ0 A 2C 2

Suppose that a capacitor with capacitance C is charged to some voltage V and then allowed to discharge through a resistance R. Write an equation governing the rate at which energy in the capacitor decreases with time due to resistive heating. Show that the solution of this equation is E EM (t) = E EM (0)e−2t/RC . You can ignore the internal resistance of the capacitor. Show that the heat produced in the resistor equals the energy originally stored in the capacitor.

61

Problems

the breakdown field of air (3 MV/m)? (It is now known that various effects cause lightning to begin at fields that are considerably smaller than the breakdown field.)

Energy conservation requires that the rate of change of the energy in the capacitor −d(Q2 /2C)/dt must equal the power dissipated in the resistor, I 2 R. Since I = Q̇, QQ̇ = −RQ̇2 or Q̇ = −Q/RC , C

The bolt lasted a time T = 5 C/(50 × 103 A) = 10−4 s. We do not know the capacitance, C. However we know both the charge on the capacitor and the energy stored. Using EEM = 12 Q2 /C and Q = CV, we can eliminate C,

since Q̇ , 0. This differential equation is Q(t) = Q0 e−t/RC , so the energy stored in the capacitor is EEM =

Q20 −2t/RC e 2C

EEM = 21 QV , or

The power dissipated in the resistor is P(t) = Q̇2 R =

Q20 RC2

V = 2EEM /Q = 2 × (5 × 108 J)/(5 C) = 200 MV.

e−2t/RC ,

and the total power dissipated in the resistor is Z ∞

dt P(t) =

0

∞ Q20 RC −2t/RC Q2 e − = 0, 2 2 2C RC 0

3.8

which is the energy originally stored in the capacitor.

3.6

The dielectrics in capacitors allow some leakage current to pass from one plate to the other. The leakage can be parameterized in terms of a leakage resistance RL . This limits the amount of time a capacitor can be used to store electromagnetic energy. The circuit diagram describing an isolated capacitor, slowly leaking charge, is therefore similar to the one analyzed in Problem 3.5. The Maxwell BCAP0310 ultracapacitor (see §37.4.2) is listed as having a capacitance of 310 F with a voltage up to 2.85 V, and a maximum leakage current of 0.45 mA when fully charged. Take this to be the current at t = 0. What is the leakage resistance of the BCAP0310? Estimate the time scale over which the charge on the capacitor falls to 1/e of its initial value.

We want the first terms in an expansion in ξ as ξ → 0 and Q → ∞ with the produce d = Qξ fixed. We need to expand the denominators to first order in ξ, |r ± ξ/2|−3 = (r2 ± r · ξ)−3/2 = r−3 (1 ± r · ξ/r2 )−3/2 !

1 3 r·ξ = 3 1∓ 2 r2 r where in the last step we have used the binomial expansion (B.66).

Next substitute into the expression for E, !

Q r·ξ E= −ξ + 3r 2 .

4πǫ0 r3 r

Q(0) −t/RC V(0) −t/RC e = e , RC R

Setting d = ξQ gives the desired result.

so the initial current is I(0) = V(0)/R. Solving for resistance, R=

2.85 V V(0) = = 6.33 kΩ.

I(0) 0.45 × 10−3 A

3.9

The time scale over which the charge falls to 1/e of its initial value is τ = RC = (6.33 kΩ)(310 F) = 1.96 × 106 s = 22.7 days.

3.7

A cloud-to-ground lightning bolt can be modeled as a parallel plate capacitor discharge, with Earth’s surface and the bottom of the cloud forming the two plates (see Example 3.2). A particular bolt of lightning passes to the ground from a cloud bottom at a height of 300 m. The bolt transfers a total charge of 5 C and a total energy of 500 MJ to the ground, with an average current of 50 kA. How long did the lightning bolt last? What was the electric field strength in the cloud–earth capacitor just before it discharged? How does this electric field compare with

[T] Consider an electric dipole composed of two charges ±Q at positions ± ξ/2. Write the exact electric field from the two charges and show that the leading term in an expansion in 1/r matches the E-field quoted in Box 3.2. [Hint: consider the binomial expansion, eq. (B.66).]

Write the field from the two charges located at r = ±ξ/2, ! r − ξ/2 r + ξ/2 Q − E= .

4πǫ0 |r − ξ/2|3 |r + ξ/2|3

Using the results of the previous problem, the current as a function of time is I(t) =

In an idealized parallel plate capacitor, the electric field is constant and equal to |E| = V/d = 200 MV/(300 m) = 0.67 MV/m. So the lightning occurred when the electric field was about a factor of 3.3/(0.67) ≈ 5 below the breakdown value for air.

If each of the batteries used in the flashlight in Example 3.3 has an internal resistance of 0.5 Ω (in series with the circuit), what fraction of power is lost to Joule heating within the batteries?

The flashlight uses two 1.5 V batteries, each with an internal resistance of 0.5Ω. The bulb has a resistance of 5Ω, so the total resistance is 6Ω. The power dissipated is P = I 2 R. The same current goes through each resistor. The total internal resistance of the batteries is 1Ω, so 1/6 of the power is lost to Joule heating within the batteries.

3.10

[T] Consider two resistors placed in series, one after the other, in an electric circuit connected to a battery with voltage V. Show that the effective resistance of the pair is R1 +R2 by using the fact that the current through both resistors is the same while voltages add. Now connect the resistors in parallel, so that the voltage across both resistors is V. Compute the total current and show that the effective resistance satisfies 1/Reff = 1/R1 + 1/R2 .

62

was added to the water for each of these. Let ηS be the efficiency for the stovetop and ηK be the efficiency for the kettle. Then

In series, the current through each resistor is the same, while the sum of the voltages across each resistor is equal to the total voltage V. So,

E = ηS PS tS = ηK PK tK

V = R1 I + R2 I = I(R1 + R2 ) = IReff ,

ηK PS tS (1132 W)(460 s) = = = 1.46. ηS P K tK (1273 W)(280 s)

where Reff = R1 + R2 . Thus the effect of the two resistors equals that of a single resistor with resistance R1 + R2 .

In parallel, the voltage drop across each resistor must be the total voltage, so

The kettle is nearly 50% more efficient at delivering energy to the water as the stovetop. This difference is likely because the heating elements of the stove are located in air rather than contained within the stove, so much of the power is lost to heating the environment.

The kettle’s heating elements are fully contained within its volume, so much less heat is lost.

V = R1 I1 = R2 I2 .

The total current I is I = I1 + I2 = V

!

1 V 1 + , = R1 R2 Reff

3.13

where 1 1 1 = + .

Reff R1 R2

3.11

The magnetic force is F = qv × B, so the work done is W = −q

An appliance that uses 1000 W of power is connected by 12 gauge (diameter 2.053 mm) copper wire to a 120 V (RMS) AC outlet. Estimate the power lost per meter, dPlost /dL, (in W/m) as resistive heating in the wire. (Remember that the wire to the appliance has two separate current-carrying wires in a single sheath.)

W = −q = −q

3.14

1000 W P = = 8.33 A. VRMS 120 V

The heating losses in the cable should be insignificant compared to the 1000 W. The power loss in the cable is given by ! dP Ω dR W 2 = IRMS = (8.33 A)2 0.0102 . = 0.708 dℓ dℓ m m

Electrical power is often used to boil water for cooking.

Here are the results of an experiment: a liter of water initially at 30 ◦ C was boiled on an electric stove top burner. The burner is rated at 6.67 A (maximum instantaneous current) and 240 VRMS . It took 7 minutes 40 seconds to reach the boiling point. The experiment was repeated using an “electric kettle.” The kettle is rated at 15 A (maximum instantaneous current again) and uses ordinary (US) line voltage of 120 VRMS . This time it took 4 minutes and 40 seconds to reach boiling. What are the power outputs and resistances of the burner and the kettle? Compare the efficiency for boiling water of the stove top burner and the kettle. To what do you attribute the difference?

The power output of the stovetop burner is PS = IRMS VRMS = √ IVRMS / 2 = 1132 W while the power output of the electric kettle is PK = 1273 W. The resistance of the burner is RS = VRMS /IRMS = 50.9Ω, while the resistance of the kettle is RK = 11.3Ω. Since the same amount of water was boiled, the same amount of energy E

dx · v × B .

Z T 0

ZT 0

dt ẋ · ( ẋ × B)

dt B · (ẋ × ẋ) = 0,

[T] Show that a charged particle moving in the xy-plane in the presence of a magnetic field B = Bẑ will move in a circle. Compute the radius of the circle and frequency of rotation in terms of the speed v of the particle. Show that adding a constant velocity uẑ in the z-direction still gives a solution with no further acceleration.

From the result of Problem 3.13 or from the fact that the force is at right angles to the velocity, the speed of the particle is constant.

An object moving at constant speed with constant perpendicular (centripetal) acceleration is undergoing circular motion. The centripetal force on the particle mv2 /R must be supplied by the magnetic force F = qvB, thus

For a reasonable cable length, this is very small compared to the power usage of the appliance.

3.12

a

where we have used eq. (B.6) and the fact that the cross product of any vector with itself vanishes.

The average current through the appliance is IRMS =

Z b

Parameterize x as x(t), where t is time, and let x(0) = a and x(T ) = b. Then dx = ẋdt (and v = ẋ) and the line integral becomes

There are 2 wires with a 2.053 mm diameter in each cable to carry the current and the resistivity of copper is ρ = 1.68 × 10−8 Ωm. The resistance per unit length of the cable is 1.68 × 10−8 Ω m Ω dR ρ ρ = 0.0102 . =2 =2 2 =2 dℓ A m πR π(0.5 × 0.002053 m)2

[T] Use the magnetic force law (3.42) and the definition Rb of work W = − a dx · F to show that magnetic forces do no work. [Hint: Consider the vector identity (B.6).]

R = mv/qB .

To compute the frequency of rotation, note that the particle completes an orbit in time T = 2πR/v = 2πm/qB so the frequency is ν = 1/T = qB/2πm.

If the particle gains a velocity component in the z direction parallel to B, there is no change in the force because the cross product of parallel vectors vanishes. The particle continues to move in a circle in the x − y plane, but with a constant velocity in the z direction.

This type of motion is in the shape of a helix.

3.15

[T] Review how magnetic fields are calculated from Ampere’s law by computing (a) the magnetic field due to a straight wire and (b) the magnetic field in the interior of a very long solenoid. The contours C1 , C2 shown in red in Figure 3.18 will help. You may use the fact that the magnetic field vanishes just outside the outer boundary of the solenoid in (b). (a). Consider the geometry in Figure 3.18(a). On a C1 , a circle of constant radius ρ, symmetry around the z-axis and mirror symmetry in the plane perpendicular to the z-axis, the magnetic

63

Problems

field has a constant magnitude and is oriented in the φ̂ direction, B = Bφ̂ for a current I in the ẑ direction. Ampere’s law then gives: I µ0 I = dx · B = 2πρB.

3.18

Thus, solving for B,

An electric motor operates at 1000 rpm with an average torque of τ = 0.1 Nm. What is its power output? If it is running on 1.2 A of current, estimate the back-EMF from the rotor.

The motor’s power output is

µ0 I µ0 ẑ × ρ̂ B= φ̂ = I .

2πρ 2π ρ (b). In this case, the curve C2 in Figure 3.18(b) applies. There can be no magnetic field in the radial direction or in the azimuthal direction due to mirror symmetry and rotational symmetry about the solenoid’s axis. Only fields along the axis can be non-zero and the field is a function of radius ρ only, B = B(ρ)ẑ.

Using Ampere’s law by integrating B around C2 , and using the fact that B vanishes outside the solenoid, I dx · B = B(ρ)l = µ0 N(l)I

The time-averaged back-EMF from the rotor is just V = P/I = 8.7 V.

3.19

C2

where l is the length of the path inside the solenoid and N(l) is the number of turns of wire enclosed by C2 . Note B is independent of radius inside the solenoid. Let n = N/L equal the number of turns per unit length in the solenoid’s windings.

Then N(l) = ln and

[T] Show that the force per unit area on the windings of an air-core solenoid from the magnetic field of the solenoid itself is of order F/A ∼ B2 /µ0 . Check that the dimensions of this expression are correct and estimate F/A in pascals if |B| = 1 T.

3.20

From eq. (3.71), the energy per unit volume in a air core solenoid is dEmagnetic /dV = |B|2 /(2µ0 ). The pressure is the rate of change of energy with volume. Since for a solenoid |B| is independent of the volume, p ∼ |B|2 /2µ0 . To check the units, use eq. (3.45) and eq. (3.52),

According to Faraday’s law E = −dΦ/dt. The flux through the current loop is Φ = AB cos θ so

[B] = [F]/[Qv] = mt−1 Q−1

E = −(−AB sin θθ̇) = wlBθ̇ sin θ

[µ0 ] = [F]/[I ] = ml/Q

−2

[B2 /µ0 ] = [p] .

in agreement with eq. (3.62).

3.21

For |B| = 1 T,

3.17

[T] In §3.4.2 we derived the EMF on a wire loop rotating in a magnetic field using the Lorentz force law to compute the forces on the mobile charges. Although Faraday’s law of induction (3.64) does not apply in a rotating reference frame, show that the same relation (3.62) follows from Faraday’s law.

[p] = [E]/[V] = ml−1 t−2 2

p=

Consider the motor described in Box 3.4. If the resistance in the wire wrapping the rotor is 1 Ω, compute the energy lost under the conditions described. What fraction of energy is lost to Joule heating in this situation?

If the current is doubled but the rotation rate is kept fixed, how do the output power, Joule heating losses, and fraction of lost energy change?

The motor draws 300 mA of current, so the power lost to Joule heating is P = I 2 R = 90 mW. The motor has an average power output of 1.8 W, so the energy lost is equal to 5% of the output energy.

If the current is doubled but everything else stays the same, the output power doubles to 3.6 W. The power lost to Joule heating increases to 360 mW. In this case, the power lost is equal to 10% of the output power.

B = µ0 nI ẑ .

3.16

1 Hz 2π s−1 = 10.5 W. 60 rpm 1 Hz

P = τω = (0.1 Nm)(1000 rpm)

1 T2 ≈ 400 000 Pa ≈ 4 atm .

8π × 10−7 N/A2

H

dS · B = 0 over any closed surface. Form a closed surface by S taking the union of two surfaces S 1 and S 2 , S = S 1 ∪ S 2 sharing the same edge, which is a closed curve C. If the normals on S 1 and S 2 were initially defined by the orientation of the curve C (e.g. by the right-hand rule), then the normal on one of the surfaces must be reversed to consistently define a normal over the whole of S. Then using eq. (3.46),

[T] Derive eq. (3.52) from eq. (3.43) and eq. (3.50).

Make sure you get the both the direction and magnitude.

The magnetic field of a wire with a current I1 along +ẑ is B=

µ0 ẑ × ρ̂ I1 2π ρ

I

S=S 1 ∪S 2

where ρ is the distance (radius) from the wire. The force on an infinitesimal segment dx = dl ẑ of a second wire parallel to another at a distance d and having current I2 along the +ẑ direction is µ0 I2 I1 dl ẑ × (ẑ × ρ̂) 2πd µ0 µ0 I2 I1 dl ẑ × φ̂ = − I2 I1 dl ρ̂ . = 2πd 2πd

dS · B = 0 =

Z

S1

Thus for any two surfaces, Z Z dS · B =

dF = I2 dl ẑ × B =

So, the force per unit length is along the radial direction and, letting F = F · ρ̂, µ0 dF =− I1 I2 . dl 2πd

R [T] Explain why the integral S dS · B that appears in eq. (3.65) is independent of the choice of surface S .

[Hint: make use of eq. (3.46).]

S1

dS · B −

S2

Z

S2

dS · B .

dS · B ,

so the choice of surface does not matter.

3.22

[T] Consider a long, hollow solenoid of volume V. Show that, ignoring end effects, its inductance is L = n2 Vµ0 , and that the magnetic energy it stores EEM = LI 2 /2 can be written in the form of eq. (3.71).

64

cross section. How many separate strings of transmission towers are needed if the transmission losses are to be kept below 5%? Assume a purely resistive load.

Suppose the solenoid has cross-sectional area A and current I. It has a length l = V/A. The magnitude of the magnetic field is B = µ0 nI

(3.76)

dB dI = µ0 n . dt dt

(3.77)

so

According to eq. (3.41) the power lost is given by PL ̺l Plost

.

2 PL AVRMS

The electromotive force from one loop is equal to change in magnetic flux through a single loop of the solenoid is Z dΦ dI E1 loop = − = −µ0 nA .

(3.78) dt dt

The total cable area is NA = 36N(750 mm2 ) = 0.027 m2 N, where N is the number of strings of transmission towers. The resistivity of aluminum is 2.82 × 10−8 Ω m. Solving for N and combining all the factors

S

There are ln total loops, so the total change in flux integrated over the entire solenoid is E = −µ0 n2 Al

dI dI = −µ0 n2 V . dt dt

N>

(3.79)

From this, the inductance is L = µ0 n2 V

3.23

1 1 1 µ0 n2 VI 2 = (µ0 nI)2 V = |B|2 V. 2 2µ0 2µ0

3.25

(3.81)

[T] Prove that the mutual inductance is a symmetric relation, M12 = M21 , by computing the energy stored when current I1 is established in loop 1 and then current I2 is established in loop 2. Then set up the same currents in the opposite order.

EP = VP = −NP (dΦ/dt)

ES = VS = −NS (dΦ/dt) .

Thus VS /VP = NS /NP . Next consider the case where the secondary circuit is open so IS =√0. Then VP = −LP I˙P and VS = −M I˙P , so VS /VP = M/LP = LS /LP . Now, close the secondary circuit through the resistor so that

P2 = L2 I2 I˙2 + M12 I1 I˙2

ES = −NS

Integrating to find the total energy (remember that I1 is being held fixed) E2 = 12 L2 I22 + M12 I1 I2 . So the total energy stored in the two circuits is

1 1 L1 I12 + L2 I22 + M21 I1 I2 .

2 2

The final configuration (the state of the system) is the same in both cases, so the energy stored must be the same, hence M12 = M21 .

3.24

Design a transmission system to carry power from wind farms in North Dakota to the state of Illinois (about 1200 km). The system should handle Illinois’s summertime electricity demand of 42 GW. Land for transmission towers is at a premium, so the system uses very high voltage (VRMS = 765 kV).

Assume that the electricity is transmitted as ordinary alternating current, although in practice three-phase power (see §38.3.1) would be used. Assume that each tower can carry 36 aluminum cables (nine lines, each consisting of four conductors separated by nonconducting spacers). The conductors are 750 mm2 in

dΦ = IS R . dt

Substituting for dΦ/dt = −VP /NP , we find IS =

1 1 EEM (1, 2) = E1 + E2 = L1 I12 + L2 I22 + M12 I1 I2 .

2 2 Next perform the same steps starting in the opposite order, first raising the current in loop 2, and then, holding I2 fixed, raising the current in loop 1. The total energy stored is

[T] Consider the transformer in Figure 3.25. Suppose the load is a resistor R and that the transformer is ideal, with M 2 = LS LP and all magnetic flux lines passing through both inductors. Show that the voltage drop √ across the resistor is VS = LS /LP VP = NS VP /NP and that the time-averaged power consumed in the resistor is P = (LS /LP )(VP2 /2R) = NS2 VP2 /2NP2 R. Since both circuits are looped by the same flux

Begin with I1 = I2 = 0. With circuit 2 open, raise the current in loop 1 to I1 . The power required is P1 = −V1 I1 with V1 = −L1 I˙1 .

The total energy required is given by E1 = 12 L1 I12 (see eq. (3.70)).

Next raise the current in loop 2 to I2 , holding the current in loop 1 constant. The power required is

EEM (2, 1) =

(42 × 109 W)(2.82 × 10−8 Ω m)(1.2 × 106 m) 1.8 .

(0.027 m2 )(7.65 × 105 V)2 (0.05)

So two strings of transmission towers would suffice.

(3.80)

The total energy held in the solenoid is EEM =

PL ̺l 2 AVRMS (Plost /PL )

NS V P .

NP R

Finally, using P = (1/2)IS2 R for an AC circuit, we obtain the desired result.

3.26

[T] Take the divergence of both sides of eq. (3.74), use Coulomb’s law on the left and current conservation on right to show that the equation is consistent.

Use Gauss’s law (3.12) and the identity of eq. (B.21), ∂ ∇ · E = µ0 ∇ · j ∂t ∂ ρ ∂ρ 0 − µ0 ǫ0 = −µ0 ∂t ǫ0 ∂t ∂ρ ∂ρ = −µ0 . −µ0 ∂t ∂t

∇ · (∇ × B) − µ0 ǫ0

77

Problems

Problems 4.1

where tan θ j = (y j+1 − y j )/∆x, so sin θ j = (y j+1 − y j )/∆l.

Substituting into the equation for (Fy ) jj+1 ,

Sound waves travel in air at roughly 340 m/s. The human ear can hear frequencies ranging from 20 Hz to 20 000 Hz. Determine the wavelengths of the corresponding sine wave modes and compare to humanscale physical systems.

(Fy ) jj+1 = k(y j+1 − y j ) .

Combining the forces on the jth mass from the masses on its left and right, mÿ j = (Fy ) j = k((y j+1 − y j ) − (y j − y j−1 )) .

The wavelength is related to the frequency by λ = v/ν, so for 20 Hz, λ = 17 m and for 20 000 Hz, λ = 17 mm. Note that both of these are scales that are easy to visualize. 17 mm is comparable to the width of a finger or, more relevant, the size of the ear.

17 m is much larger but still comparable in size to things like trees and buildings. This is in contrast to light, where the wavelength of visible light is much smaller than can be seen by the eyes.

4.2

Set m = ρ∆x, k = τ/∆x and take the limit ∆x → 0, to obtain ∆xρÿ j = (τ/∆x)(y′′j (∆x))2 .

Labeling the masses by x ≡ j∆x, we obtain eq. (4.9).

The energy stored in the array of masses and springs consists of kinetic and potential energy

A violin A-string of length L = 0.33 m with total mass 0.23 g has a fundamental frequency (for the lowest mode) of 440 Hz. Compute the tension on the string.

If the string vibrates at the fundamental frequency with maximum amplitude 2 mm, what is the energy of the vibrational motion?

The fundamental frequency is given by (eq. (4.12) and below) r r 1 π 1 τ τ ν= = .

2π L ρ 2 ML

E=

Substituting ∆m = ρ∆x, k = τ/∆x, y j+1 − y j → y′j ∆x, and R P dx (. . .), we obtain j ∆x(. . .) → Z Z ρ τ E= dx ẏ2 + y′2 + dxτ/2 .

2 2

Since the last term is a constant independent of the displacement y(x, t), we drop it.

4.4

Solving for the tension, τ = 4MLν2 = 4(0.23 × 10−3 kg)(0.33 m)(440 Hz)2 = 59 N .

To compute the total energy either integrate the energy density (4.10) over x, !

Z L Z L 1 2 1 ′2 ρẏ + τy dx E= u(x, t)dx = 2 2 0 0 # Z " ω2 1 L 2 2 2 2 ρA ω sin (ωt) sin (kx) + τA2 cos2 (ωt) cos2 (kx) dx = 2 0 τ/ρ L 2 2 L 2 2 2 2 2 = ρω A = (M/L)(2πν) A = Mπ ν A , 4 4 √ or use eq. (4.16) being careful to identify c1 = L/2A.

∂u = ρẏÿ + τy′ ẏ′ , ∂t substitute for ÿ from the wave equation, eq. (4.9), ρÿ = τy′′ . ∂ ∂S ∂u = τẏy′′ + τy′ ẏ′ = (τẏy′ ) = − ∂t ∂x ∂x

0

with S = −τẏy′ . For y = f (x − vt), y′ = f ′ and ẏ = −v f ′ , where f ′ = d f (η)/dη, where η = x − vt. Substituting into S ,

M L (2πν)2 A2 = Mπ2 ν2 A2 .

2L 2

S = −τ(−v f ′ )( f ′ ) = τv f ′2 > 0 ,

Substituting for M, ν, and A, E = 1.76 mJ.

4.3

[T] Derive the equation of motion for the string (4.9) from a microscopic model. Assume a simple model of a string as a set of masses ∆m spaced evenly on the x axis at regular intervals of ∆x, connected by springs of spring constant k = τ/∆x. Compute the leading term in the force on each mass and take the limit ∆x → 0 to get the string wave equation, where ρ = ∆m/∆x. In the same limit show that the energy density of the string is given by eq. (4.10).

Label the vertical displacement of each mass as y y j . Some relations between derivatives and the limits of finite differences are useful. As ∆x → 0, y j+1 − y j → y′j ∆x and y′j − y′j−1 → y′′j ∆x.

Apply Hooke’s law to the jth mass. The force on the jth mass in the y-direction exerted by the ( j + 1)st mass is given by q (Fy ) jj+1 = k∆l = k ∆x2 + (y j+1 − y j )2 sin θ j

[T] Show that the energy density on a string u(x, t), defined in eq. (4.10), obeys the conservation law ∂u/∂t + ∂S /∂x = 0, where S (x, t) = −τẏy′ is the energy flux, the energy per unit time passing a point x. For the traveling wave y(x, t) = f (x − vt), find u(x, t) and S (x, t) and show that energy flows to the right (for v > 0) as the wave passes a point x. Show that the total energy passing each point is equal to the total energy in the wave.

Starting from eq. (4.10)

 r 2 ZL   πx  ρ  2 dx sin2 E = ω2 A 2  L L  =

∆m X 2 k X 2 ẏ + (∆x + (y j+1 − y j )2 ) .

2 j j 2 j

which is positive, indicating a flux of energy to the right.

The integral of S over all time at fixed x is Z Z Z I= dt(−τẏy′ ) = dtτv f ′2 = τ dη( f ′ (η))2 .

4.5

The integral of u over all x at fixed time is !

Z Z ρ τ 1 1 I′ = dx ẏ2 + y′2 = dx ρv2 f ′2 + τ f ′2 = I 2 2 2 2 p where we have used v = τ/ρ. Thus the total energy flux passing a point equals the total energy in the wave.

Compute the maximum energy flux possible for electromagnetic waves in air given the constraint that the electric field cannot exceed the breakdown field described in Example 3.2.

The breakdown field of air is Emax = 3.3 × 106 V/m. In an electromagnetic wave, |B| = |E|/c. The magnitude of the energy

78

flux of an electromagnetic wave is |S| = µ1 |E × B| = µ1 c |E|2 = 0

ǫ0 c|E|2 . The maximum possible energy flux in air, then, is

0

4.9

|S|max = (8.85 × 10−12 F/m)(3.00 × 108 m/s)(3.3 × 106 V/m)2 2.89 × 1010 W/m2 .

4.6

The strongest radio stations in the US broadcast at a power of 50 kW. Assuming that the power is broadcast uniformly over the hemisphere above Earth’s surface, compute the strength of the electric field in these radio waves at a distance of 100 km.

It has been proposed that solar collectors could be deployed in space, and that the collected power could be beamed to Earth using microwaves. A potential limiting factor for this technology would be the possible hazard of human exposure to the microwave beam. One proposal involves a circular receiving array of diameter 10 km for a transmitted power of 750 MW. Compute the energy flux in this scenario and compare to the energy flux of solar radiation.

This proposal would have the array receive a power density of

The magnitude of the energy flux of an electromagnetic wave is |S| = µ1 |E × B| = µ1 c |E|2 = ǫ0 c|E|2 . The energy flux in a 0

0

hemispherical wave at distance R is |S| = W/(2πR ), so r 1 W |E| = R 2πcǫ0 s 5 × 104 W 1 = 5 10 m 2π(3.00 × 108 )(8.85 × 10−12 F/m)

4.10

[T] Derive the wave equation for B analogous to eq. (4.20).

From Maxwell’s equations (§3.6), take the time derivative of Faraday’s Law with ρ = j = 0, ∂2 B ∂E , = −∇ × ∂t ∂t2

(4.32)

and then substitute from the Ampere-Maxwell Law, 1 ∂2 B =− ∇ × (∇ × B) µ0 ǫ0 ∂t2

1 1 ∇(∇ · B) − ∇2 B = ∇2 B , =− µ0 ǫ0 µ0 ǫ0

4.8

[T,H] Consider two electromagnetic plane waves (see eq. (4.21)) one with amplitude E0a and wave vector ka and the other with amplitude E0b and wave vector kb .

These waves are said to add coherently if the average energy density u in the resulting wave is proportional to |E0a + E0b |2 or incoherently if the average energy density is proportional to |E0a |2 + |E0b |2 . Show that two electromagnetic plane waves are coherent only if they are propagating in the same direction with the same frequency and the same polarization.

A plane wave of the form of eq. (4.21) has E = E0 sin(k· x−ωt), where k̂ defines the direction of propagation and E0 defines the direction of polarization. Note that k · E0 = 0.

The energy density is given by eq. (4.22). We consider only the term (u1 ) proportional to |E|2 ; the second term can be handled analogously. For a single plane wave,

where we have used the vector identity (B.22) and ∇ · B = 0.

Suppose an electromagnetic plane wave is absorbed on a surface oriented perpendicular to the direction of propagation of the wave. Show that the pressure exerted by the radiation on the surface is prad = W/c, where W is the power absorbed per unit area. Solar radiation at the top of Earth’s atmosphere has an energy flux |S| = 1366 W/m2 . What is the pressure of solar radiation when the Sun is overhead? What is the total force on Earth exerted by solar radiation?

The pressure p is the force per unit area perpendicular to the surface, while force can be defined as the time derivative of momentum d| p|/dt. If a surface is absorbing electromagnetic radiation, the pressure then is the rate per unit area of momentum reaching the surface. So, p=

d 2 | p| 1 d2 E W = = . dtdA c dtdA c

When the sun is overhead, Earth feels a pressure of p=

1 366 W/m2 = 4.56 µPa. 3 × 108 m/s

The total force on Earth exerted by solar radiation is F = πR2⊕ P = π(6.376 × 106 m)2 (4.56 × 10−6 Pa) = 5.82 × 108 N , where we have taken an average value of R⊕ .

750 × 106 W 2.4 W/m2 . π × 108 m2

The power density of solar radiation is 1 366 W/m2 , about 500 times greater, while that of a microwave is approximately 1.5 × 104 W/m2 , about 5000 times greater. This power density is actually quite modest compared to other common sources of electromagnetic radiation.

17 mV/m

4.7

|S| =

2

hu1 i =

ǫ0 |E0 |2 4

where we have used hsin2 (. . .)i = 1/2.

First consider the case where the two waves propagate in the same direction with the same frequency, but possibly different polarizations. Then E = (Ea0 + Eb0 ) sin(k · x − ωt) .

Then hu1 i =

ǫ0 a |E + Eb0 |2 , 4 0

which is proportional to |E0a + E0b |2 if and only if Eb0 is proportional to Ea0 , i.e. they must have the same polarization. This proves that two waves with the same frequency, direction of propagation and polarization are coherent.

If the frequencies are different and/or direction of propagation are different one obtains cross terms in the computation of h|E|2 i proportional to hsin(ka · x − ωa t) sin(kb · x − ωb t)i The time average vanishes unless ωa = ωb because Z T 1 lim sin ωa t sin ωb t → 0 T →∞ T 0 unless ωa = ωb , and similarly for hsin cosi and hcos cosi. If the frea b quencies are the same but k̂ and k̂ are not proportional, the time average yields a result proportional to cos(ka − kb ) · x, which vanishes when averaged over space. So all cross terms vanish unless

79

Problems

the frequency, direction and polarization of the two waves are the same. When the cross terms vanish hu1 i = hua1 i + hub1 i, which is proportional to |E0a |2 + |E0b |2 .

4.11

The initial shape is antisymmetric about L/2, so only even n modes have non-zero coefficients. Calculating the coefficients, r ( nπx nπx L/4 2 L 4a L sin − x cos cn = L nπ L nπ L L 0 " #3L/4 nπx 4a L nπx nπx + −2a cos − sin − x cos L L nπ L L L/4 " #L    nπx 4a L nπx nπx  + 4a cos + sin − x cos   L L nπ L L 3L/4  √ (" # nπ 1 nπ nπ 2 3nπ 2a 2L sin − cos − cos + cos = nπ nπ 4 2 4 4 4 # " 3nπ 2 nπ 3 3nπ 1 nπ 2 sin − sin − cos + cos − nπ 4 nπ 4 2 4 2 4 " #) 3nπ 2 3nπ 3 3nπ + 2 cos nπ − 2 cos + − sin − 2 cos nπ + cos .

4 nπ 4 2 4

[T] Derive eq. (4.23) by taking the time derivative of eq. (4.22) and using Maxwell’s equations. [Hint: see eq. (B.23).] !

1 ∂ ∂ 1 1 u= ǫ0 |E|2 + |B|2 = ǫ0 E · Ė + B · Ḃ ∂t ∂t 2 2µ0 µ0

Substitute for Ė and Ḃ from Maxwell’s equations, !

1 1 1 ∂u = ǫ0 E · ∇× B− j − B·∇×E ∂t µ0 ǫ0 ǫ0 µ0 1 (E · ∇ × B − B · ∇ × E) − j · E µ0

=

= −∇ · S − j · E ,

There are four cases: n = 2, 10, . . ., n = 4, 12, . . ., n = 6, 14, . . ., and n = 8, 16, . . .. √ 16a 2L For n = 2, 10, . . . , cn = (nπ)2

where in the last step we have used eq. (B.23).

4.12

[T] A string of length L begins in the configuration y(x) = A[ 13 sin k1 x + 32 sin k2 x] with no initial velocity.

Write the exact time-dependent solution of the string y(x, t). Compute the contribution to the energy from each mode involved.

and for n = 6, 14, . . . ,

where yn (x) =

1 1 2 ′ 2 2 ρẏ + 2 τ(y ) .

Note that ẏ(x, 0) = 0. The energy density is u = The energy in each mode does not change with time, so take the energy density at t = 0, where ẏ = 0. Then the total energy is τπ2 A2 18L2

ZL

τπ A = 18L2

dx(cos πx/L + 4 cos 2πx/L)2

y(x, t) =

∞ X

dx(cos2 πx/L + 16 cos2 2πx/L + 8 cos πx/L cos 2πx/L)

The total energy in tension initially is

0

U=

with U1 = π2 τA2 /36L from the first mode and U2 = 4π2 τA2 /9L from the second.

At T = 0, the Fourier coefficients are:  r  3L/4 !

ZL/4 Z   4ax nπx nπx 4ax 2  dx cn = sin + sin dx 2a −   L L L L L   0

L/4

    nπx  4ax  sin + dx −4a + .   L L    3L/4 !

ZL 0

Lτ 4a τ dx (y′ )2 = 2 2 L

!2

=8

τa2 .

L

The energy in mode n is

[TH] A string of length L is initially stretched into a “zigzag” profile, with linear segments of string connecting the (x, f (x)) points (0, 0), (L/4, a), (3L/4, −a), (L, 0). Compute the Fourier series coefficients cn and the time-evolution of the string y(x, t). Compute the total energy in the tension of the initially stretched string. Compute the energy in each mode and show that the total energy agrees with the energy of the stretched string.y

ZL

2 nπx sin .

L L

8a 2π(1 + 2n)vt 2π(1 + 2n)x cos sin .

L L [(1 + 2n)π]2

(−1)n

n=0

17π2 τA2 τπ2 A2 (1 + 16) = , = 36L 36 L

4.13

r

To add the time dependence, each term gets a factor of cos(ωn t − φn ). Since the string is stationary at t = 0, φn = 0. The frequency is p ωn = nvπ/L (where v = τ/ρ), so the time evolution of the string is

0

L 2 2 Z

√ 16a 2L . (nπ)2

For all others, cn = 0. Writing this in summation notation, √ ∞ X 4a 2L y2+4n (x) , (−1)n y(x, 0) = f (x) = [(1 + 2n)π]2 n=0

No initial velocity means that ẏ(x, 0) = 0 for all x, so the time dependence requires cosine terms. Adding in the time dependence for each mode, and replacing kn = nπ/L, ! r r π 2 2π 1 τ π τ 2π sin x cos t + sin x cos t .. y(x, t) = A 3 L ρL 3 L ρ L

U=

cn = −

En =

τ 256τa2 1 ρ 2 2 ρ n2 v2 π2 2 cn = (nπ)2 c2n = . c ω = 2 n n 2 L2 2L2 Lπ2 n2

The total energy is ∞

∞

256τa2 X 1 64τa2 X 1 = Lπ2 n=0 4(1 + 2n)2 Lπ2 n=0 (1 + 2n)2 ! ∞ ∞ 1 48τa2 X 1 64τa2 X 1 − . = = 2 2 2 2 Lπ n=1 n (2n) Lπ n=1 n2

E=

The sum is equal to π2 /6, so the total energy is E = 8τa2 /L, just as we found by integrating over the initial tension of the string.

4.14

[T] What is the pressure exerted by a beam of light on a perfect mirror from which it reflects at normal (perpendicular) incidence? Generalize this to light incident at an angle θ to the normal on an imperfect mirror (which reflects a fraction r(θ) of the light incident at angle θ).

80

A perfect mirror reflects the light without altering its frequency (energy), so the pressure at normal incidence is just twice that of a perfect absorber. This doubles the change in momentum, so the pressure is 2W/c.

At an angle θ, the pressure is decreased by a factor of cos θ because only the normal force contributes to pressure. A perfect absorber would feel a pressure of (W/c) cos θ while a perfect mirror would feel a pressure of (2W/c) cos θ. An imperfect mirror reflecting a fraction r(θ) of the light and absorbing the rest would feel a total pressure of P(θ) = (2r(θ) + (1 − r(θ))

4.15

The phase velocity is v=

4.17

[T] Consider a cylindrical resistor of cross-sectional area A and length L. Assume that the electric field E and current density j are uniform within the resistor.

Prove that the integrated power transferred from elecR tromagnetic fields into the resistor d3 x j · E is equal to IV. Compute the electric and magnetic fields on the surface of the resistor, and show that the power transfer is also given by the surface integral of the Poynting vector, so that all energy dissipated in the resistor is transferred in through electric and magnetic fields.

with analogous expressions for the refracted and reflected waves.

At z = 0, continuity of the waveform requires that the sum of the incident and reflected waves equals the refracted wave, A1 cos(k1 x sin θ1 −ω1 t)+A3 cos(k3 x sin θ3 −ω3 t) = A2 cos(k2 x sin θ2 −ω2 t) .

For this functional equality to hold at all t, all the ωs must be equal, leading to k1 = k3 = v2 k2 /v1 . For the functional equality to hold at all points x along the interface, we require k1 sin θ1 = k3 sin θ3 = k2 sin θ2 . Since k1 = k3 , we have sin θ1 = sin θ3 . The only possibility distinct from the incident wave is θ1 = π − θ3 , which proves the law of specular reflection. Since k1 = (v2 /v1 )k2 , we have

1 IV E× B = − ρ̂ . µ0 2πRL

equal to the resistive power dissipated in the resistor.

The period is related to the frequency by T = 2π/ω, while the wave number is related to the wavelength by k = 2π/λ. Rewriting the dispersion relation in terms of T and λ, r p 2πg 2π = gk = .

T λ Solving for λ,

λ=

g 2 T .

2π

[T] A wave satisfying eq. (4.2) passes from one medium in which the phase velocity for all wavelengths is v1 to another medium in which the phase velocity is v2 . The incident wave gives rise to a reflected wave that returns to the original medium and a refracted wave that changes direction as it passes through the interface.

Suppose that the interface is the plane z = 0 and the incoming wave is propagating in a direction at an angle θ1 to the normal ẑ. Prove the law of specular reflection, which states that the reflected wave propagates at an angle π − θ1 with respect to the normal ẑ. Also prove Snell’s law, which states that the wave in the second medium propagates at an angle θ2 from the normal ẑ, where sin θ2 / sin θ1 = v2 /v1 . Use the fact that the wave must be continuous across the interface at z = 0.

ψ1 (x, z, t) = A1 cos(k1 x sin θ1 + k1 z cos θ1 − ω1 t)

µ0 I φ̂ , 2πR

As stated in the text, the dispersion relation relating the wave number and p angular frequency of ocean surface waves is ω = gk, where g 9.8 m/s2 . Compute the wavelength and speed of propagation (phase velocity) for ocean surface waves with periods 6 s and 12 s.

gλ .

2π

where v1 = v3 , v2 . The form of the incident wave is

so

4.16

r

ki = ωi /vi ,

because AL is the volume of the resistor.

For the second part we need the magnetic field at the surface of the wire. From eq. (3.50),

Note that the Poynting vector points inward everywhere on the surface of the cylindrical resistor, so energy in the EM fields is flowing into the resistor. The integrated (over the surface of the resistor) flux of energy gives the total power delivered to the resistor by the EM fields, Z P = | d A · S| = IV ,

g = k

Let ki , ωi , and Ai (i = 1, 2, 3) be the wave number, frequency, and amplitude of the incident, refracted, and reflected waves. The wave equation relates ki = |ki | to ωi and the phase velocity vi ,

Use cylindrical coordinates with ẑ pointing along the wire, ρ̂ pointing radially outward from the center of the wire, and φ̂, the angular unit vector with ẑ × φ̂ = −ρ̂.

For the first part: E = (V/L)ẑ and j = (I/A)ẑ, so Z Z d3 x j · E = d 3 x(IV/AL) = IV ,

S=

r

So, when T = 6 s, λ = 56 m and v = 9.4 m/s. When T = 12 s, λ = 225 m and v = 18.7 m/s.

W cos θ W = (r(θ) + 1) cos θ . c c

B=

ω = k

sin θ2 k1 v2 = = , sin θ1 k2 v1 which is Snell’s law.

4.18

A wave travels to the right on a string with constant tension τ and a mass density that slowly increases from ρ on the far left to ρ′ on the far right. The mass density changes slowly enough that its only effect is to change the speed with which the wave propagates. The waveform on the far left is A cos(kx − ωt) and on the far right is A′ cos(k′ x − ωt). Find the relation between k′ and k and then use conservation of energy to find the amplitude A′ on the far right. You may find it helpful to use the result of Problem 4.4 (S (x, t) = −τẏy′ ).

These solutions obey the wave equation with different densities ′′ on the left and right. On the left, p ρÿ = τy with y = A cos(kx − ωt), so ρω2 = τk2 , and k = ω ρ/τ. On the right, an analogous p calculation gives k′ = ω ρ′ /τ. Thus r ′ ρ k′ = k. ρ

Problems

The time averaged rate of flow of energy past a point in the far left must be the same as the rate on the far right.

S L = −τẏy′ = τωkA2 sin2 (kx − ωt)

S R = −τẏy′ = τωk′ A′2 sin2 (k′ x − ωt) .

Averaging the sin2 gives 1/2, and equating p the energy flux on left and right, we find A2 k = A′2 k′ or A′ = 4 ρ/ρ′ A.

81

101

Problems

Problems 5.1

A cylinder initially contains V0 = 1 L of argon at temperature T 0 = 0 ◦ C and pressure p0 = 1 atm. Suppose that the argon is somehow made to expand to a final volume V = 2 L in such a way that the pressure rises proportionally to the volume, finally reaching p = 2 atm. How much work has the argon done? What is its final temperature T and how much thermal energy has been added to the gas?

R

The work done by the gas can be computed from W = dV p(V).

Since (p0 , V0 ) and (p, V) are both known and the quantity of gas n does not change, we can compute n from n = p0 V0 /RT 0 and then find T using T = pV/nR. Knowing T − T 0 and the heat capacity of argon (a monatomic ideal gas) we can find the increase in thermal energy content of the gas. This plus the work done by the gas must sum to the total energy added to the gas.

First compute the work done: we are given p = p0 V/V0 , so W=

Z V

dV ′ p(V ′ ) =

V0

=

p0 V0

Z V

dV ′ V ′ =

V0

Using the ideal gas law, we can then solve for the pressure at which the density of helium is 0.260 kg/m3 , !

0.260 kg/m3 1 atm = 1.48 atm .

3 0.176 kg/m

pHe =

If the temperature increases to 20 ◦ C while volume remains constant, then the density of helium does not change, but the density of the outside air drops to ρair = (273/293)1.275 = 1.188 kg/m3 , so the mass that can be lifted is m = (ρair −ρHe )V = (1.188 kg/m3 −0.260 kg/m3 )5 740 m3 = 5 330 kg .

5.3

p0 2 V − V02 2V0

!

1.01 × 105 Pa (3 × 10−6 m6 ) = 150 J .

2 × 10−3 m3

The temperature T is obtained from the ideal gas law:

When air is inhaled, its volume remains constant and its pressure increases as it is warmed to body temperature T body = 37 ◦ C. Assuming that air behaves as an ideal gas and that it is initially at a pressure of 1 atm, what is the pressure of air in the lungs after inhalation if the air is initially (a) at room temperature, T = 20 ◦ C; (b) at the temperature of a cold winter day in Boston, T = −15 ◦ C; (c) coming from one person’s mouth to another’s during cardiopulmonary resuscitation (CPR).

For CPR, assume the air is initially at body temperature and at p = 1 atm.

As volume is assumed constant, the ideal gas law implies

T = (pV/p0 V0 )T 0 = 4(273 K) = 1092 K .

plung =

The internal (thermal) energy of argon (a monatomic, ideal gas) at temperature T is U = (3/2)nRT . We get n from n = p0 V0 /RT 0 = 0.0446 mol. So 3 nR∆T 2 3 = (0.0446 mol)(8.31 J/K mol)(1092 − 273) K = 455 J .

2

∆U =

must have been added to increase the temperature of the gas. Since the gas also did W = 150 J of work, the total thermal energy added to the gas must have been Qtot = 455 + 150 = 605 J.

5.2

Non-rigid airships known as blimps have occasionally been used for transportation. A blimp is essentially a balloon of volume V filled with helium. The blimp experiences a buoyancy force F = (ρatm − ρHe )Vg, where ρatm is the density of the surrounding air and g 9.8 m/s2 . The blimp maintains its shape because the pressure of the helium pHe is kept higher than the pressure of the surrounding atmosphere patm . A modern blimp with a volume V = 5740 m3 can lift a total mass m = 5825 kg at temperature 0 ◦ C and patm = 1 atm. Assuming both air and helium behave as ideal gases, estimate the pressure of the helium gas inside the blimp. Assuming the blimp’s volume is kept constant, how much mass can it lift at 20 ◦ C?

T lung pair , T air

In all cases T lung = 310 K. For case a) pair = 1 atm and T air = 310 293 K, hence plung = 294 atm = 1.05 atm. For case b) T air = 268 K, plung = 1.20 atm. For case c) T air = T lung , so plung = 1 atm.

5.4

[H] Everyday experience indicates that it is much easier to compress gases than liquids. This property is measured by the isothermal compressibility, β = − V1 ∂V ∂p T , the fractional change in a substance’s volume with pressure at constant temperature. What is the isothermal compressibility of an ideal gas? The smaller β is, the less work must be done to pressurize a substance. Compare the work necessary at 20 ◦ C to raise the pressure of a kilogram of air and a kilogram of water (βwater 4.59 × 10−10 Pa−1 ) from 1 atm to 104 atm.

First compute the work done to compress air, an ideal gas. βair = −

1 1 ∂ 1 1 1 ∂V = pV = −nRT = .

V ∂p T V ∂p p p V p2

Wair = −

Z Vf

Then dV p =

(d p βair V) p =

Z pf

Z pf

d pV

p0

p0

V0

= nRT

Z pf

d p/p = nRT ln(p f /p0 ) = 9.21 nRT .

p0

3

At 0 C and 1 atm (STP), ρair = 1.275 kg/m . In order to lift m = 5 825 kg, ◦

mg = (ρair − ρHe )Vg .

So the density of helium is ρHe = ρair −m/V = 1.275 kg/m3 −5 825 kg/5 740 m3 = 0.260 kg/m3 .

At any given T and p, ρHe /ρair = (4/29), the ratio of their molecular masses, so at p = 1 atm, ρHe = (4/29)1.275 kg/m3 = 0.176 kg/m3 .

The molecular weight of air is 0.029 kg/mol, so 1 kg = 1/0.029 34.5 mol, so at T = 298 K, Wair = 9.21(34.5 mol/kg)(8.31 J/mol K)(298 K) = 787 KJ/kg .

Next compute the work done to compress 1 kg water with βwater = 4.6 × 10−10 Pa−1 . The work is Wwater = −

Z Vf V0

dV p =

Z pf p0

(d p βwater V(p)) p .

102

need to estimate the average number of showers per person per year for the people of your country.)

The fractional change in the volume of the water when compressed from 1 atm to 104 atm is ∆V/V ≈ −βwater ∆p −(4.6 × 10−10 )(104 × 1.01 × 105 ) = −0.46 ,

From the example, a regular shower head averages 9.5 L/min.

The water is heated up by approximately ∆T = 30 K, and the average shower takes 8 min. Water has a heat capacity of cV = 4.18 kJ/kg K, so the total energy savings for the US (population 3 × 108 ) per year is

so the variation of V in Rthe integrationR over p cannot be ignored.

From the definition of β, dV/V = −β d p, so ln V = −βp+const, and V(p) = V0 exp(−β(p − p0 )). Substituting into the integral and ignoring p0 since p0 ≪ p f , Z pf

V0 d p p e−βp ≈ Wwater ≈ βV0 1 − (1 + βp f )e−βp f β 0

10−3 m3 /kg 1 − 1.46 e−.46 ≈ 170 kJ/kg . ≈ 4.6 × 10−10 Pa−1

∆E = cV ∆M∆T , where ∆M is the difference in the total mass of water used in showers between the normal and low-flow shower heads. ∆M = (4.8 L/min − 9.5 L/min) (8 min/shower)

The work necessary to pressurize water is 170/790 ≈ 22% of the work needed to pressurize air. Note that this ratio gets much smaller if p f is reduced.

5.5

[HT] A cylindrical tube oriented vertically on Earth’s surface, closed at the bottom and open at the top (height 100 m, cross-sectional area 1 m2 ) initially contains air at a pressure of one atmosphere and temperature 300 K. A disc of mass m plugs the cylinder, but is free to slide up and down without friction. In equilibrium, the air in the cylinder is compressed to p = 2 atm while the air outside remains at 1 atm. What is the mass of the disc?

Next, the disc is displaced slightly in the vertical (z) direction and released. Assuming that the temperature remains fixed and that dissipation can be ignored, what is the frequency with which the disc oscillates?

× (365 shower/y)(3 × 108 people)(1 kg/L) = −4.12 × 1012 kg/y.

The total energy savings is then ∆E = (4.18 kJ/kg K)(−4.12 × 1012 kg/y)(30 K) = −5.2 × 1017 J/y = −0.52 EJ/y,

which is about 0.5% of the total US yearly energy usage.

5.8

In mechanical equilibrium, the net force from the pressure of the gas inside and outside of the cylinder must balance the downwards force of gravity from the mass. That is, ∆p A = mg, or m

(p + ∆p) (V − ∆V) = pV , where ∆V = Az. Since the displacement and the resulting pressure change are small, we ignore ∆V∆p, so V∆p = p∆V, where, in terms of the z-coordinate of the oscillator, ∆V/V = z/50 m. Hence, ∆p =

z[m] (2 × 105 Pa) = 4 × 103 z[m] Pa .

50

The force on the disc then becomes mz̈ = F = −A∆p = −kz, where k = 4 × 103 N/m. This is the differential √ equation of a harmonic oscillator, with angular frequency ω = k/m 0.6 s−1 , and frequency ν = ω/2π 0.1 Hz.

5.6

How much energy does it take to heat 1 liter of soup from room temperature (20 ◦ C) to 65 ◦ C? (You may assume that the heat capacity of the soup is the same as for water.)

Water has a heat capacity of 4.18 kJ/kg K. One liter of soup, assumed to have approximately the same properties as water, has a mass of 1 kg. So, it takes 4.18 kJ to heat the soup by 1 K. Raising the temperature by 45 K would require Q = (4.18 kJ/kg K)(45 K)(1 kg/L) 188 kJ/L.

5.7

The total amount of energy stored is

(1.0 × 105 Pa)(1 m2 ) = 104 kg.

10 m/s2

Since the gas is compressed to a pressure two atmospheres and since pV is a constant at fixed T , equilibrium is reached when the height of the cylinder is 50 m.

If the disc is displaced by a small amount z, then again using pV = constant,

A “low-flow” showerhead averages 4.8 L/min. Taking other data from Example 5.3, estimate the energy savings (in J/y) if all the people in your country switched from US code to low-flow shower heads. (You will

A solar thermal power plant currently under construction will focus solar rays to heat a molten salt working fluid composed of sodium nitrate and potassium nitrate.

The molten salt is stored at a temperature of 300 ◦ C, and heated in a power tower to 550 ◦ C. The salt has a specific heat capacity of roughly 1500 J/kg K in the temperature range of interest. The system can store 6000 metric tonnes of molten salt for power generation when no solar energy is available. How much energy is stored in this system?

E = (1 500 J/kg K)(6 × 106 kg)(550 ◦ C − 300 ◦ C) = 2.25 TJ .

5.9

A new solar thermal plant being constructed in Australia will collect solar energy and store it as thermal energy, which will then be converted to electrical energy. The plant will store some of the thermal energy in graphite blocks for nighttime power distribution. According to the company constructing the plant, the plant will have a capacity of 30 MkWh/year. Graphite has a specific heat capacity of about 700 J/kg K at room temperature (see [26] for data on the temperature dependence of the specific heat of graphite). The company claims a storage capacity of 1000 kWh/tonne at 1800 ◦ C. Is this plausible? Explain. If they are planning enough storage so that they can keep up the same power output at night as in the day, roughly how much graphite do they need? Estimate the number of people whose electrical energy needs will be supplied by this power plant.

The company claims that the storage capacity of graphite is 1000 kWh/t = (103 kWh/t)(3.6 MJ/kWh)(10−3 t/kg) = 3.6 × 106 J/kg , at 1800◦ C. Room temperature is approximately 20◦ C, so this corresponds to a specific heat capacity of C=

3.6 × 106 J/kg = 2000 J/kg K , 1 780 K

103

Problems

which is approximately three times as high as the heat capacity of graphite at room temperature. This does not include any loss of energy to the environment during storage or transfer. The heat capacity of graphite shows a strong temperature dependence, increasing by a factor of three between room temperature and around 1500◦ C [26], so this does seem fairly reasonable.

The power plant will receive nearly all of its energy during the day. We assume that night covers about half the time that the plant is operating. Thus of the 30 MkWh/y generated, half can be distributed during the day and half, ≈ 15 MkWh/y, is distributed at night from stored energy. This means that the plant will need to store around 41 000 kWh/day of energy for use at night. The graphite is heated during the day and cools off at night as the plant supplies energy, so this means that the plant will need to have at least 41 000 kWh M= = 41 tonnes 1 000 kWh/t

We see that the ice is much more effective at reducing the temperature. The heat capacity per unit volume of the steel cubes is greater than that of ice, but less than that of water. The latent energy absorbed by the melting ice also significantly lowers the temperature of the water. The steel cubes, however, don’t undergo any phase transitions in the relevant temperature range, and so they are reusable, and will not dilute other drinks. They are much more massive than ice, however, and could break delicate glass if treated casually. The steel cubes would likely appeal to certain people as a luxury good.

5.11

of graphite to provide this much power. In reality, if the plant is to provide a fairly constant amount of energy throughout the year, much more graphite would likely be required since there will be large variations in the amount of incident solar energy due to seasonal variations and local weather conditions.

The daily energy needs of an average Australian is likely to be much closer to the per capita energy usage of western Europe or the US (∼0.5–1 GJ/day) than to the world average (200 MJ/day), of which about 14% is in the form of supplied electrical power (see Figure 1.2). So, the per capita electrical energy usage is 70– 140 MJ/day The plant supplies a total of 8.2 × 104 kWh/day, or 3 × 105 MJ/day. At 70 to 140 MJ per day per person, this plant will supply the energy needs of roughly 2 000–4 000 people.

5.10

The enthalpy of fusion of ice is 334 kJ/kg. Liquid water has a density of approximately 1000 kg/m3 , so the enthalpy of fusion is 334 MJ/m3 . To raise sea levels by 1 m, a total of

A start-up company is marketing steel “ice cubes” to be used in place of ordinary ice cubes. How much would a liter of water, initially at 20 ◦ C, be cooled by the addition of 10 cubes of steel, each 2.5 cm on a side, initially at −10 ◦ C? Compare your result with the effect of 10 ordinary ice cubes of the same volume the same initial temperature. Would you invest in this start-up? Take the density of steel to be 8.0 gm/cm3 and assume the heat capacities to be constants, independent of temperature.

V (0.7)(1 m)(5 × 1014 m2 ) = 3.5 × 1014 m3 of water ice must be melted. This requires E (3.5 × 1014 m3 )(334 MJ/m3 ) = 1.2 × 1023 J of energy. Humans use approximately 580 EJ of energy per year. If 1% (5.8 EJ) contributes to heat melting the ice, then it would take (1.2 × 1023 J)/(5.8 × 1018 J/y) = 21 000 years to melt this much ice.

However, if an increase in CO2 increases the amount of solar energy absorbed by the earth by 2 000 TW and 1% of this contributes to warming it would only take

The heat capacity of 10 steel cubes of volume V = (2.5 cm)3 then is

T

CS = 10(2.5 cm)3 (0.008 kg/cm3 )(0.52 kJ/kg K) 0.65 kJ/K .

One liter of water weighs 1 kg, and has a heat capacity of CW = 4.2 kJ/K. As energy is conserved and the system will come into equilibrium at a some temperature T , we have CS (T + 10) + CW (T − 20) = 0, and as the system will come into thermal equilibrium. Solving for T , we find T = 16.0 ◦ C, so the steel ice cubes cool the water by 4.0 ◦ C. The density of ice at freezing point is approximately 0.92 gm/cm3 . The heat capacity of 10 ice cubes of volume V then is CI = 10(2.5 cm)3 (0.00092 kg/cm3 )(2.05 kJ/kg K) 0.29 kJ/K .

When the ice melts at 0 ◦ C, there is a latent heat absorption of ∆Q = 10(2.5 cm)3 (0.00092 kg/cm3 )(330 kJ/kg) 47 kJ .

It is easiest to do the problem in two steps. First consider the cooling of the water by heating the ice to 0 ◦ C and melting it: CW (20 − T ) = 47 kJ + (10 K)(0.29 kJ/K) 50 kJ .

Solving for T , we find T 8.1 C. At this point we have 1 kg of water at 8 ◦ C and 0.14 kg of water at 0 ◦ C. These come into equilibrium when (8 − T ) = 0.14T , or T = 8/1.14 7 ◦ C. So the water cools by 13 ◦ C. ◦

Roughly 70% of the 5 × 1014 m2 of Earth’s surface is covered by oceans. How much energy would it take to melt enough of the ice in Greenland and Antarctica to raise sea levels 1 meter? Suppose that 1% of energy used by humans became waste heat that melts ice. How long would it take to melt this quantity of ice? If an increase in atmospheric CO2 led to a net energy flux of 2000 TW of solar energy absorbed into the Earth system of which 1% melts ice, how long would it take for sea levels to rise one meter?

1y 1.2 × 1023 × = 190 years (0.01)(2 × 1015 J/s) (3.15 × 107 s)

to melt that much ice.

5.12

Carbon dioxide sublimes at pressures below roughly 5 atm (see Figure 5.12). At a pressure of 2 atm this phase transition occurs at about −69 ◦ C with an enthalpy of sublimation of roughly 26 kJ/mol. Suppose a kilogram of solid CO2 at a temperature of −69 ◦ C is confined in a cylinder by a piston that exerts a constant pressure of 2 atm. How much heat must be added to completely convert it to gas? Assuming CO2 to be an ideal gas, how much work was done by the CO2 in the course of vaporizing? If the ambient pressure outside the cylinder is 1 atm, how much useful work was done? Finally, by how much did the internal energy of the CO2 change when it vaporized?

The molecular weight of CO2 is 0.044 kg/mol, so 1 kg = 1/0.044 = 22.7 mol. The CO2 sublimation phase transition takes place at fixed T and p. The enthalpy of sublimation for 1 kg is ∆H = (22.7 mol)(26 kJ/mol) = 590 kJ/kg. This is the thermal energy (heat) that must be added to vaporize the CO2 . The work done by the gas is p∆V. ∆V must be computed from the ideal gas

104

law, ∆NRT p (22.7 mol)(8.31 J/mol K)(273 K − 69 K) = 0.190 m3 . = 2(101 325 Pa)

∆V =

Then W = p∆V = 2(101 325 Pa)(0.19 m3 ) = 38.5 kJ. If the outside pressure is 1 atm, then half of this is useful work, Wuseful 19 kJ, and ∆U = ∆H − p∆V = (590 − 38.5) kJ/kg = 551 kJ/kg.

125

Problems

Problems 6.1

Suppose a small stoneware kiln with surface area 5 m2 sits in a room that is kept at 25 ◦ C by ventilation.

The 15 cm thick walls of the kiln are made of special ceramic insulation, which has k = 0.03 W/m K. The kiln is kept at 1300 ◦ C for many hours to fire stoneware.

The room ventilating system can supply 1.5 kW of cooling. Is this adequate? (You can assume that the walls of the kiln are thin compared to their area and ignore curvature, corners, etc.)

We compute the heat emitted (thermal power) by the kiln at temperature T = 1 300 ◦ C, assuming the room maintains its temperature of T r = 25 ◦ C, and see how this compares with the 1.5 kW of cooling available. The walls of the kiln are thin compared to its area, so we can treat is as a planar slab of surface area A = 5 m2 and thickness ℓ = 15 cm, with k = 0.03 W/m K. According to Fourier’s law, the thermal power emitted by the kiln is P = kA(T − T r )/ℓ,

Convection is over 3 000 times as important as conduction for heat transfer through air in this example.

6.4

The surface temperature of the sun with these assumptions is T=

6.5

So, the cooling ventilation system is sufficient to prevent the room from heating up.

Two rigid boards of insulating material, each with area A, have thermal conductances U1 and U2 . Suppose they are combined in series to make a single insulator of area A. What is the heat flux across this insulator as a function of ∆T ? Now suppose they are placed side-byside. What is the heat flux now?

Reff = 1/Ueff = 1/U1 + 1/U2 , or U1 U2 , U1 + U2

Q̇ =

A U1 U2 ∆T = Ueff A∆T = A∆T .

Reff U1 + U2

The expression for the heat flux is Q̇ = Ueff (2A)∆T = (U1 + U2 )A∆T , which states that the total heat flux is the sum of the fluxes through the two insulators. Note that the factor of one half in the effective thermal conductance compensates for the factor of two in the effective area.

The heat transfer coefficient h̄ for air flowing at 30 m/s over a 1 m long flat plate is measured to be 80 W/m2 K. Estimate the relative importance of heat transfer by convection and conduction for this situation.

The ratio of convection heat transfer to conduction heat transfer is given by the Nusselt number, NuL = h̄L/k. Taking k = 0.026 W/m K from Table 6.1, (80 W/m2 K)(1 m) 3 100 .

NuL = 0.026 W/K m

RQ̇ + T0 .

A

a. Concrete has k ≈ 0.9 W/m K. So R ≈ (0.3/0.9) = 0.33 Km2 /W. Thus, T eq ≈ (0.33)(400)/75 ≈ 1.8 ◦ C. b. Softwood has k ≈ 0.11 W/m K. R ≈ (0.10/0.11) = 0.91 Km2 /W. Thus, T eq ≈ (0.91)(400)/75 ≈ 4.9 ◦ C c. The fiberglass has R = 2.0 W/m K. R = (0.025/0.11 + 2.0) ≈ 2.2 Km2 /W. Thus T eq ≈ (2.2)(400)/75 ≈ 12.5 ◦ C.

Thermal resistances in parallel add as reciprocals multiplied by their areas, 2A A A = + , thus Reff R1 R2 1 Ueff = (U1 + U2 ) .

2

!1/4

Humans radiate energy at a net rate of roughly 100 W; this is essentially waste heat from various chemical processes needed for bodily functioning. Consider four humans in a roughly square hut measuring 5 m × 5 m, with a flat roof at 3 m height. The exterior walls of the hut are maintained at 0 ◦ C by the external environment.

For each of the following construction materials for the walls and ceiling, (i) compute the R-value (∆T/q) for the material, (ii) compute the equilibrium temperature in the hut when the four people are in it: (a) 0.3 m (1 foot) concrete walls/ceiling; (b) 10 cm (4") softwood walls/ceiling; and (c) 2.5 cm (1") softwood, with 0.09 m (3.5") fiberglass insulation on interior of walls and ceiling. Take thermal conductivities and/or R-values from Tables 6.1 and 6.3.

T eq =

and

6.3

P 1/4 384 × 1024 W = 4πR2 σ 4π(6.955 × 108 m)2 (5.67 × 10−8 W/m2 K4 )

P P The R value is just R = j R j = j L j /k j for insulators connected in series. Heat is conducted away from the house at a rate of Q̇ = A(T eq − T 0 )/R. The outside temperature is T 0 = 0 ◦ C, while the surface area of the hut is A = 75 m2 , ignoring the floor. At equilibrium, Q̇ is just the heat radiated from the four occupants of the hut, Q̇ = 400 W. Solving for the equilibrium temperature,

Thermal conductances are the reciprocals of thermal resistances, U j = 1/R j , for j = 1, 2. Resistances in series add, so

Ueff =

5 800 K.

P = (0.03 W/m K)(5.0 m2 )(1 300 ◦ C−298 ◦ C)/(0.15 m) 1.0 kW .

6.2

Given the Sun’s power output of 384 YW and radius 695 500 km, compute its surface temperature assuming it to be a black body with emissivity one.

6.6

Consider a building with 3200 ft2 of walls. Assume the ceiling is well-insulated and compute the energy loss through the walls based on the following materials, assuming an indoor temperature of 70 ◦ F and an outdoor temperature of 30 ◦ F: (a) walls are composed of 4" thick hard wood; (b) walls composed of 3/4" of plywood on the inside and outside, with 8" of fiberglass insulation between; and (c) same as (b) but with 18 single-pane R-1 windows (US units), each with area 0.7 m2 (remaining area is surrounded by walls as in (b)). Take thermal conductivities and/or R-values from Tables 6.1 and 6.3.

The energy loss is Q̇ = A∆T/R where A 300 m2 and ∆T = 40◦ F 22◦ C, so Q̇ 6600/R[km m2 /W]. (a). In this case, R = L/k ≈ 0.10/0.18 ≈ 0.56 Km2 /W, so Q̇ ≈ 6600/0.56 11.8 kW. (b). In this case, R = 2(0.19) + 4.6 5.0 Km2 /W, so Q̇ 1.32 kW.

126

(c). In this case, most of the wall is made of R = 5.0 K m2 /W material, while the windows are made of R = 1.0 ft ◦ F hr/BTU = 1/5.68 K m2 /W = 0.176 K m2 /W material. The windows take up 12.6 m2 , while the rest takes up 287.4 m2 of area. The rate of energy loss is !

A 287.4 12.6 Q̇ = ∆T + (22) 2.84 kW .

R 0.176 5.0

6.7

Estimate the R-value of the wall shown in the figure in Example 6.2.

Since the thermal resistance of the foam and fiberglass are given in US R-values, we use US-units throughout. Combining the two layers of fiberglass and foam with 2.5 cm of sheathing (plywood) with R ≈ 1 ft ◦ F hr/BTU and 1 cm of drywall with R ≈ 0.5 ft ◦ F hr/BTU, all in series,

Figure 6.17 (a) A section through a wall in a wood-frame

building. (b) A useful equivalent circuit diagram.

Reff = 0.5 + 13 + 10 + 13 + 1 ≈ 37.5 ft ◦ F hr/BTU

6.8

Assume that a 60 m2 wall of a house is insulated to an R-value of 5.4 (SI units), but suppose the insulation was omitted from a 1 m2 gap where only 6 cm of wood with an R-value of 0.37 remains. Show that the effective Rvalue of the wall drops to R = 4.4 leading to a 23% increase in heat loss through the wall.

with the value quoted in [28]? Use the same method to compute the U-factor of a similar window with a 1/4" argon-filled gap (measured value U = 1.87 W/m2 K).

Repeat your calculation for a 1/2" air gap (measured Ufactor l.70 W/m2 K). Can you explain why your answer agrees less well with the measured value in this case?

The insulated portion of the wall has R = 5.4 and the gap has R = 0.375. These are in parallel, so inverse R values add. The insulated portion of the wall is 59 m2 and the uninsulated portion is just 1 m2 . The resistance of the wall is given by

We compute the R-value assuming conduction alone and then use U = 1/R. In general Reff = Rglass + Rstill air + Rair gap .

59 1 60

+ 13.6 W/K .

R 5.4 0.37

For the 1/4 inch air gap

So R 60 m2 ÷ 13.6 W/K 4.4 m2 K/W, which represents a 23% increase (5.4/4.4) in the heat loss rate through the wall.

6.9

In Example 6.1, a film of still air was identified as the source of almost all of the thermal resistance of a single-pane glass window. Determine the thickness of the still air layers on both sides of the window if radiative heat transfer is ignored. It is estimated that about half the thermal conductance of the window is due to radiative heat transfer; estimate the thickness of the still air layers if this is the case.

From the example, the R-value of the glass alone is Rglass 0.0031 (SI units). The effective R-value of the window is Reff = Rglass + Rair 0.17 W/m2 K, so to this accuracy we can ignore Rglass entirely. Using k = 0.026 W/m K for air and R = L/k = 0.17 W/m2 K gives L = (0.17)(0.026) 4.4 mm. According to the discussion in the example, 80% of the still air is on the inside surface and 20% is on the outside. Thus Lair in 3.5 mm and Lair out 0.9 mm. If half the thermal conductance is due to radiative heat transfer then the R-value of the still air must be twice what we just computed. Thus a more accurate estimate of the thickness of the still air layers is 7.0 mm on the inside and 1.8 mm on the outside.

6.10

Reff (1/4) = 2(0.0031) + 0.17 +

According to [28], a double pane window with an emissivity ε = 0.05 coating and a 1/4" air gap has a measured (center of glass) U-factor of 2.33 W/m2 K. Assume that this coating is sufficient to stop all radiative heat transfer, so that the thermal resistance of this window comes from conduction in the glass, the air gap, and the still air layers on both sides of the window. Taking Rstill air ≈ 0.17 m2 K/W from Example 6.1, compute the U-factor of this window ignoring convection in the air gap. How does your answer compare

.0253 0.42 m2 K/W 4(.026)

and Ueff (1/4) = 1/R 2.4, within 4% of the observed value.

For the 1/4 in argon gap Reff (1/4 argon) = 2(0.0031) + 0.17 +

.0253 0.53 m2 K/W 4(.018)

and Ueff (1/4 argon) = 1/R 1.9, equal to the observed value within the accuracy of the calculation.

For the 1/2 inch air gap Reff (1/2) = 2(0.0031) + 0.17 +

.0253 0.66 m2 K/W 2(.026)

and Ueff (1/2) = 1/R 1.5. In this case the calculated thermal conductivity is 12% lower than the measured value. Possibly the larger measured value is due to convection within the wider air gap, as discussed in the text.

6.11

z A building wall is constructed as follows: starting from the inside the materials used are (a) 1/2" gypsum wallboard; (b) wall-cavity, 80% of which is occupied by a 3.5" fiberglass batt and 20% of which is occupied by wood studs, headers, etc.; (c) rigid foam insulation/sheathing (R = 0.7 m2 K/W); and (d) hollow-backed vinyl siding. The wall is illustrated in Figure 6.17 along with an equivalent circuit diagram.

Calculate the equivalent R-value of the wall. Do not forget the layers of still air on both inside and outside.

All the thermal resistances add in series except for the wood and fiberglass, which add in parallel to give an effective resistance Reff . Taking R-values and thermal conductivities from Tables 6.1 and 6.3, !−1 0.2 0.8 + 1.54 m2 K/W , Reff = 2.0 0.81

127

Problems

80 K with Q̇ = 10 W/m. So,

R1 = R0 exp 2πk(T 0 − T (R1 ))/Q̇ = (1 cm) exp

2π(80 K)(0.0254 W/m K) 10 W/m

= 3.6 cm.

6.13 Figure 6.18 Cross section through an insulated copper pipe.

The pipe carries fluid at 100 ◦ C. The radius of the insulation must be 3.6 times the radius of the pipe in order to keep the power loss to less than 10 W/m.

!

Ignoring convective heat transfer, estimate the change in U-factor by replacing argon by krypton in the quadruple glazed windows described in Table 6.3.

You can ignore radiative heat transfer since these are low-emissivity windows.

The contributions to the R-value from the still air, four panes of glass, and the 3/4 inch of argon or krypton add up to 3 0.0254 1.24 m2 K/W , 4 0.018 3 0.0254 2.21 m2 K/W .

Krypton : Reff = 0.17 + 4(0.0031) + 4 0.0094 Argon : Reff = 0.17 + 4(0.0031) +

where (3.5)(.0253)/0.11 = 0.81 m2 K/W is the R-value of the 3.5 inch thick softwood. Combining all the contributions

The difference in U-factors is therefore ∆U = 1/1.24 − 1/2.21 0.35 W/m2 K. The actual change is ∆Umeasured 0.29 W/m2 K.

Rtotal = Rstill air + Rdrywall + Reff + Rfoam + Rsiding = 0.17 + 0.079 + 1.54 + 0.7 + 0.11 2.6 m2 K/W

6.14

or, in US units, R 15 ft2 ◦ F hr/BTU. Note that if the wood were ignored, the calculation would give R = 3.1 m2 K/W, 18% higher than the actual value.

6.12

[H] An insulated pipe carries a hot fluid. The setup is shown in Figure 6.18. The copper pipe has radius R0 = 1 cm and carries a liquid at T 0 = 100 ◦ C. The pipe is encased in a cylindrical layer of insulation of outer radius R1 ; the insulation has been chosen to be closed-cell polyurethane spray foam with an R-value of 1.00 m2 K/W per inch of thickness. How large must R1 be so that the heat loss to the surrounding room (at 20 ◦ C) is less than 10 W/m? [Hint: First explain why the heat flux must be q = cr̂/r. You may find it useful to make the analogy between heat propagation through the insulator and current flowing outward through a cylindrical conductor. Note also that ∇ f (r) = r̂ f ′ (r).]

The conductivity of copper is very high compared to that of the insulation, so the copper itself can be neglected. Due to cylindrical symmetry, the heat flux must be in the radial direction from the copper. At thermal equilibrium, the flux of thermal energy per unit length along the pipe must be the same at all radii. For a cylinder, the area is proportional to the radius, so if Q̇ is the total energy flux per unit length, Q̇ = 2πq(r)r. Since Q̇ is independent of r, q = Q̇r̂/2πr. The flux is inversely proportional to the radius and aligned along the radial direction. Solving for T (r), Q̇r̂ = −k∇T 2πr Q̇ dT = −k 2πr dr r Q̇ ln , and at r = R1 , T (r) = T 0 − 2πk R0 q=

Q̇ R1 T 0 − T (R1 ) = ln .

2πk R0 The constant of integration is separated into into T 0 and R0 terms so that the logarithm term is 0 at the inner radius of the pipe (where T 0 = 100 ◦ C). The R value is 1.00 m2 K/W per inch, giving k = 0.0254 W/m K. We want to find the radius R1 at which T (R1 )−T 0 =

In areas where the soil routinely freezes it is essential that building foundations, conduits, and the like be buried below the frost level. Central Minnesota is a cold part of the US with yearly average temperature of T 0 ∼ 5 ◦ C and variation ∆T ∼ 20 ◦ C. Typical peaty soils in this region have thermal diffusivity a ∼ 1.0 × 10−7 m2 /s. Building regulations require foundation footings to exceed a depth of 1.5 m. Do you think this is adequate?

We make use of eqs. (6.26) and (6.27). The minimum temperature occurs when the sine-function is −1, so T min (z) = T 0 −∆T e−αz , where T 0 = 5 ◦ C, ∆T = 20 ◦ C and we want T min (1.5 m) > 0 ◦ C. First we compute s 2π α= 1.0 m−1 , 2(3.15 × 107 s)(1.0 × 10−7 m2 /s) and then evaluate, T min (1.5 m) = 5 − 20e−(1.0)(1.5) 5 − 4.5 0.5 ◦ C .

So the depth of 1.5 m seems marginally sufficient.

6.15

Consider a region with average surface temperature T 0 = 10 ◦ C, annual fluctuations of ∆T = 30 ◦ C, and surface soil with a 2.4 × 10−7 m2 /s, k 0.3 W/m K. If the local upward heat flux from geothermal sources is 80 mW/m2 , compute the depth at which the heat flux from surface fluctuations is comparable. What is the answer if the area is geothermally active and has a geothermal heat flux of 160 mW/m2 ?

The downward heat flux from surface fluctuations is proportional to the gradient of the temperature, q = k|dT/dz|, where we have suppressed the signs. The temperature gradient is dT = −α∆T e−αz (sin(ωt − αz) + cos(ωt − αz)) , dz and this has maximum magnitude√when the sine and cosine functions are equal and their sum is 2. Substituting for dT/dz and solving for z, ! q 1 . z = − ln √ α 2αk∆T

128

We compute α=

r

ω = 2a

s

2π = 0.64 m−1 .

(3.15 × 107 s)(2)(2.4 × 10−7 m2 /s)

Then substitute the other given values to find the depth z at which q = 8 × 10−2 W/m2 !

8 × 10−2 W/m2 1 7.2 m . ln √ z= 0.64 2(0.64 m−1 )(0.3 W m/K)(30 ◦ C) Doubling the heat flux to 160 mW/m2 decreases the depth at which the fluxes become equal to 6.1 m.

6.16

[T] When an object is in radiative equilibrium with its environment at temperature T , the rates at which it emits and absorbs radiant energy must be equal. Each is given by dQ0 /dt = εσT 4 A. If the object’s temperature is raised to T 1 > T , show that to first order in ∆T = T 1 − T , the object loses energy to its environment at a rate dQ/dt = 4εσ∆T T 3 A. When the temperature of the object is raised to T 1 = T + ∆T , it emits energy at a rate dQ1 = εσ(T + ∆T )4 A = εσ(T 4 + 4T 3 ∆T + . . .)A dt dQ0 + 4εσ∆T T 3 A , = dt where we have made a Taylor expansion of T 14 to first order in ∆T (see eq. (B.63)) and the omitted terms are of order (∆T/T )2 . Since the object is still absorbing energy from its environment at dQ0 /dt, then net energy radiated is dQ/dt = 4εσT 3 ∆T A.

155

Problems

Problems 7.1

Planck’s constant is very small on the scale of human affairs. Compare the angular momentum of a child’s marble spinning at 60 rpm with Planck’s constant.

Compare the energy of the same marble attached to a spring with spring constant k = 10 kg/s2 and oscillating with an amplitude of 1 cm with the ground state energy 1 ~ω of this harmonic oscillator. Quantum effects are 2 important in systems where L ≈ ~ or E ≈ ~ω. (The moment of inertia of a uniform sphere of mass m and radius R is I = 52 MR2 .)

According to eq. (7.38), the ground state energy of a particle of 2 2 mass m in a cube of side L is E1,1,1 = 3π ~2 . Substituting for L and 2mL m, Eproton =

7.4

Take the mass and radius of the marble to be M = 0.003 kg and R = 0.7 cm respectively. Then I = (2/5)(0.003 kg)(0.007 m)2 5.9 × 10−8 kg m2 and −8

2

−1

−34

L/~ = Iω/~ = (5.9 × 10 kg m )(2π s )/(1.055 × 10

3π2 (1.055 × 10−34 )2 3.9×10−12 J 24 MeV .

2(1.672 × 10−27 )(5 × 10−15 )2

Suppose an electron, constrained to move only in the vertical direction, sits on an impenetrable table in Earth’s gravitational field. Write down the (onedimensional) Schrödinger equation that determines its wavefunction as a function of height z. Use dimensional analysis to estimate its ground state energy up to a multiplicative factor. Estimate its average height above the table top, also up to a multiplicative factor.

The potential energy of the electron is mgz, where z ≥ 0 is the height above the table. The one-dimensional Schrödinger equation, eq. (7.40) reads ! ~2 d 2 + mgz ψ(z) = Eψ(z) . − 2 2m dz

2

kg m /s)

3.5 × 1027 .

The energy of a harmonic oscillator is twice its potential energy, E = kx2max = (10 kg/s2 )(0.01 m)2 = 10−3 J. The quantum ground state energy is

The ground state energy can depend only on the parameters m, g, and ~. so E = cmα ~β gγ , where c is a dimensionless constant and α, β, and γ are determined so that the result has dimensions of energy,

r 1 k 1 ~ω = ~ 2 2 m s   10 kg/s2  1   36.1 × 10−33 J = 1.055 × 10−34 2 0.003 kg 

[E] = m1 l2 t−2 = [m]α [~]β [g]γ

E0 =

7.2

= mα (ml2 t−1 )β (lt−2 )γ , which gives three equations

So E/E0 1.6 × 1029 .

[T] In one dimension, compare the plane wave solutions to the heat equation ∂T/∂t = a∂2 T/∂x2 with plane wave solutions to the free Schrödinger equation

ikx

t:

2 = β + 2γ .

For an electron, E ∼ ((9.1 × 10−31 kg)(9.8m/s2 )2 (1.05 × 10−34 J s)2 )1/3 ∼ 1 × 10−32 J. Repeating this analysis looking for a unit of length gives l ∼ (~2 /m2 g)1/3 ∼ 0.001 m for an electron. This may seem like a large distance scale; as a check, evaluate the potential energy of an electron at z = l ∼ 1 mm, V ∼ mgh ∼ (10−30 kg)(10 m/s2 )(0.001 m) ∼ 10−32 J, of the same order as the energy scale.

f (t).

7.5

f˙ = −ak2 f , 2

which has solution T (x, t) = T 0 e−ak t+ikx , which decays with t. For the Schrödinger equation, substitute ψ(x, t) = eikx f (t) and find ~k2 f, 2m 2

which has solution ψ(x, t) = ψ0 eikx−i~k /2m , which oscillates in time.

7.3

2 = 2β + γ ,

E = c(mg2 ~2 )1/3 .

Show that the heat equation has real solutions that oscillate in space and decay in time, whereas the Schrödinger equation has complex solutions that oscillate in both space and time.

i f˙ =

1 = α + β,

l:

This set of equations has solution α = 1/3, β = 2/3, and γ = 2/3, so

i∂ψ/∂t = −(~/2m)∂2 ψ/∂x2 .

Look for plain wave solution of the form T (x, t) = e Substituting into the heat equation,

m:

Nuclear forces are so strong that they keep protons and neutrons in a spherical region a few femtometers in radius (1 fm = 10−15 m). To get a crude idea of the energy scales involved in nuclear physics, model the nucleus as a cube 5 fm on a side. What is the ground state energy of a proton (mass m p 1.7 × 10−27 kg) in this box? This is the energy scale of nuclear interactions.

[T] A particle of mass m in a potential V(x) has an energy basis state with wavefunction ψ(x) = √ C/ cosh2 ( mV0 x/~). The properties of cosh x and other hyperbolic functions are summarized in Appendix B.4.2. What is the potential and what is the energy of this state? What is the value of the constant C? [Hint: ψ(x) must obey the Schrödinger equation.] ψ(x) must obey −(~2 /2m)ψ′′ = (E − V(x)) ψ. To find E and V(x) we compute ψ′′ sinh bx cosh3 bx ! −2 6 tanh2 bx ψ′′ (x) = b2 C + , 2 2 cosh bx cosh bx ψ′ (x) = −2bC

where b =

p

mV0 /~2 . Substituting into the Schrödinger equation,

(E − V(x)) ψ(x) = V0 (1 − 3 tanh2 bx) ψ(x)

from which we conclude that V(x) = 3V0 tanh2 bx and E = V0 .

156

7.6

The constant C must be chosen so that the integral of |ψ2 (x)| over all x is one: Z ∞ Z ∞ C2 1 C2 dx dy = = 1. b −∞ cosh4 y cosh4 (bx) −∞ R∞ p 4 Using −∞ dy/ cosh4 y = 4/3, gives C = 9mV0 /16~2 .

[T] A quantum particle is restricted to a onedimensional box 0 ≤ x ≤ L. It experiences no forces within the box, but cannot escape. At time t = 0, the √ √ particle is in the state 1/3ψ1 (x) + 2/3ψ2 (x), where ψi (x) are the (correctly normalized) energy basis states from eq. (7.36). (a) Compute the probability at t = 0 that the particle has energy ǫ = E1 = π2 ~2 /2mL2 . (b) Compute the probability at t = 0 that the particle is in the left half of the box (x < L/2). (c) Compute the state of the particle at time t = π~/ǫ. What is the probability at this time that the particle has energy ǫ = E1 ? What is the probability that the particle is in the left half of the box?

Then use these results to rewrite |±y i in terms of |± x i. !

E 1 1 1 ±y = √ √ (|+ x i + |− x i) ± i √ (|+ x i − |− x i) 2 2 2 1±i 1∓i = |+ x i + |− x i , .

2 2

7.8

ψ(x + δ1 ) = eiθ(δ1 ) ψ(x) ψ((x + δ1 ) + δ2 ) = eiθ(δ2 )+iθ(δ1 ) ψ(x) , and compare this with translation by δ1 + δ2 , ψ(x + δ1 + δ2 ) = eiθ(δ1 +δ2 ) ψ(x) .

Comparing these two results, θ(δ1 + δ2 ) = θ(δ1 ) + θ(δ2 ). This requires that θ depends linearly on δ. To show this, differentiate with respect to δ1 with δ2 fixed, θ′ (δ1 + δ2 ) = θ′ (δ1 ) , which requires that θ′ is a constant. Define that constant to be k, so θ(δ) ≡ kδ (using θ(0) = 0), and ψ(x) = eiθ(x) ψ(0) = eikx ψ(0). So up to an overall normalization constant ψ(x) = eikx .

7.9

ψ(x, t) = e−iE1 t/~ z1 ψ1 (x) + e−iE2 t/~ z2 ψ2 (x).

When t = π~/E1 , p p ψ(x, π~/E1 ) = − 1/3ψ1 (x) + 2/3ψ2 (x) ,

so the relative phase of the two terms has changed and the interference term in the previous result changes sign, so √ 1 8 2 P(left, t = π~/ǫ) = − 0.100 .

2 9π

7.7

[T] In the text we introduced electron spin states with spin ± 12 in the ẑ-direction, |±i, and states with spin √ ± 21 in the x̂-direction, |±x i = (|+i ± |−i)/ 2. Measuring the z-component of spin in the state |± x i gives ± 12 , each with 50% probability. The analogous states with √ spin ± 12 in the ŷ-direction are |±y i = (|+i ± i |−i)/ 2.

Show that a measurement of the spin in the x̂- or ẑdirection on these two states also yields ± 12 each with 50% probability.

The states |±y i are defined as superpositions of states with definite spin in the ẑ direction, so the probability to measure spin in the ẑ direction ± 12 is given by the absolute value of the coefficients of |±i in the expansion of |±y i, which is 1/2. To find the results of a measurement of the spin in the x̂ direction, it is necessary to express |±y i in terms of |x± i. First express |±i in terms of |x± i, √ √ |+i = (|+ x i + |− x i)/ 2 and |−i = (|+ x i − |− x i)/ 2 .

[T] Show that the set of functions ψk (x) = eikx (for real k) are the solutions to the condition ψ(x+δ) = eiθ(δ) ψ(x) (for real θ). [Hint: use sequential translations by δ1 and δ2 to show that θ must depend linearly on δ, then expand this condition about δ = 0.] Show that if k were complex, then the probability dP/dx = |ψk (x)|2 would grow without bound as x approaches either +∞ or −∞, corresponding to unphysical boundary conditions.

First note θ(0) = 0. Then following the hint, first consider a translation by δ1 followed by a translation by δ2 ,

a) WhenPthe state is written as a sum of normalized energy states, ψ = j z j ψ j , then the probability of measuring energy E j is |z j |2 , so P(E1 ) = |z1 |2 = 1/3. b The probability to be on the left half of the box is P(left) = R L/2 |ψ(x)|2 , 0 √   Z πx 2 2 2πx 2 2 πx 2πx  2 L  1  dx  sin2 + sin + sin sin P(left) = L 0 3 L 3 L 3 L L √  Z π/2   1  2 2 2 2 dy  sin2 y + sin2 2y + sin y sin 2y = π 0 3 3 3 √ 1 8 2 = + 0.900 .

2 9π c) The probability to be in an energy state does not change with time (see eq. (7.24)), so P(E1 ) = 1/3. The wavefunction of the state changes to

Since |(1 ± i)/2|2 = 1/2, the probability to measure the spin in the x̂ direction on either |+y i or |−y i yields 1/2.

7.10

[T] A particle with spin S has magnetic moment m = qgS/2Mc (see Example 7.3). Solving for the classical motion of S in a constant magnetic field B, we found that S precesses around B with angular frequency Ω = qg|B|/2Mc. We found that the quantum state of an electron |ψ(0)i = |+ x i precesses with angular frequency ω = 2ǫ/~, where ǫ is the energy of the electron in the field B. Show that ω = Ω.

For notation, etc., see Example 7.3. The energy of a quantum particle in a magnetic field is ±ǫ = ±Bm, where m = mŜ is the particle’s magnetic moment, and m = gq~/4Mc for an electron (spin-1/2). Then ω ≡ 2ǫ/~ = 2mB/~ = gqB/2Mc = Ω.

The interatomic potential for diatomic molecules like O2 and N2 can be approximated near its minimum by a simple harmonic oscillator potential 12 kx2 . Consider the classical motion of a diatomic molecule composed of two atoms of equal mass m. Show that near the minimum of the interatomic potential this classical motion is that of a single mass m/2 on a spring with spring constant k. For the O=O double bond in O2 , the spring constant near the bottom of the potential is roughly 1140 N/m. Estimate the energy gap between the ground state of the molecule and the first vibrational quantum excitation of the molecule, corresponding to the first excited state of the associated harmonic oscillator potential.

157

Problems

where ψn (x) is a solution to the one-dimensional oscillator with energy (n + 1/2)~ω. Show that the energies of the three-dimensional oscillator are (N + 3/2)~ω for N = 0, 1, 2, . . . and find the degeneracies of the first four energy levels. Do you recognize the sequence?

Standard mechanics analysis gives that the energy of a two body system can be written as the sum of the kinetic energy of the center of mass plus the energy of the relative motion Erel = 21 mr ẋ2 + V(x), where mr is the reduced mass, mr = m1 m2 /(m1 +m2 ). If m1 = m2 = m, as is the case for the O2 molecule, then mr = m/2. Together with a potential energy V(x) = 12 kx2 , the vibrations of O2 correspond to √ √ a harmonic oscillator with frequency ω = k/mr = 2k/m. The −26 mass of an oxygen atom is m 2.66 × 10 kg, so q ω = 2(1 140 N/m)/(2.66 × 10−26 kg) 2.93 × 1014 s−1 ,

A three dimensional harmonic oscillator obeys a Schrödinger equation analogous to eq. (7.43),

and the spacing between quantum excitations of the vibrational levels of O2 is

−

∆E = ~ω = (1.054 × 10−34 J s)(2.93 × 1014 s−1 )

p where |x| = x2 + y2 + z2 is the distance from the minimum of the harmonic oscillator potential. Changing to Cartesian coordinates,

3.09 × 10−20 J 0.19 eV .

7.11

[T] Consider a system of two independent harmonic oscillators, each with natural angular frequency ω. We denote states of this system by |n mi where the nonnegative integers n, m denote the excitation levels of the two oscillators. The energy of this system is just the sum of the energies of the two oscillators. What is the ground state energy of this system? How many distinct states |nmi have total energy 10 ~ω? If there are three independent oscillators, how many distinct states have total energy 92 ~ω?

For two oscillators E(n, m) = (n + m + 1)~ω, with n, m = 0, 1, 2, . . . , so the ground state energy is E(0, 0) = ~ω. The number of states with energy 10~ω, N(10), is the number of ordered pairs of non-negative integers whose sum is 9, (0, 9), (1, 8), (2, 7), . . . .

These are enumerated as the number of integers from 0 to 9, which gives N(10) = 10.

For three oscillators, the energy is E(n, m, l) = (n + m + l + 3/2)~ω, with n, m, l = 0, 1, 2, ..., so the ground state energy is E(0, 0, 0) = (3/2)~ω. So the number of states with energy E = (9/2)~ω is the number of ways that three units of excitation can be distributed over three oscillators. An easy way to visualize this is to permute three objects and two dividers on a line. Thus • • | • | describes n = 2, m = 1, l = 0. The number of distinct ways to do this is N(9/2) = 5!/(3!2!) = 10. The same result can of course be obtained by making a table of allowed values of the three integers.

7.12

7.13

Consider a macroscopic oscillator, given by a mass of m = 0.1 kg on a spring with spring constant k = 0.4 N/m. What is the natural frequency ω of this oscillator? Considering this as a quantum system, what is the spacing of the energy levels? Do you expect that we could observe the energy of the mass on the spring with sufficient precision to determine which quantum state the system is in? Classically, consider extending the spring by moving the mass a distance d = 1 cm from equilibrium, and letting go. Describe the classical motion of the mass. What is the energy of this system? Roughly what excitation number n characterizes quantum states with this energy? √ √ ω = k/m = 0.4/0.1 = 2 s−1 . The spacing between quantum levels is ~ω (1.055 × 10−34 J s)(2 s−1 ) 2 × 10−34 J. The energy of the stretched spring when the mass is at rest is E = 12 kd 2 = 1 2 −5 J. Setting E ≈ n~ω we find 2 (0.4 N/m)(0.01 m) = 2 × 10 n = (2 × 10−5 J)/(2 × 10−34 J) 1029 .

[T] Show that the solution to Schrödinger’s equation for a three-dimensional harmonic oscillator is ψlmn (x) = ψl (x)ψm (y)ψn (z),

! ~2 2 1 ∇ + k|x|2 ψ(x) = Eψ(x) , 2m 2

−

! ! ~2 ∂2 ∂2 ∂2 1 + 2 + 2 + k x2+y2+z2 ψ(x, y, z) = Eψ(x, y, z) .

2 2m ∂x ∂y ∂z 2

Since

−

! ~2 d 2 k + x2 ψl (x) = (l + 1/2)~ωψl (x) , 2 2m dx 2

direct substitution shows that the product ψl (x)ψm (y)ψn (z) obeys the three dimensional equation with E(l, m, n) = (l+m+n+3/2)~ω.

The degeneracy of the state with energy (N + 3/2)~ω is determined by the number of ways of choosing three non-negative integers, l, m, and n that add up to N. For the ground state, N = 0, the answer is trivial d(3/2) = 1; for N = 1, the sets {l, m, n} = {1, 0, 0}, {0, 1, 0}{0, 0, 1} exhaust the possibilities, so d(5/2) = 3.

For higher N it is easier to count the way that N units of excitation can be distributed among the three directions of oscillation. An easy way to visualize this is to permute N objects and two dividers on a line. Thus • • | • |• describes l = 2, m = 1, n = 1. The number of distinct ways of doing this is (N + 2)!/(N!2!) = (N + 1)(N + 2)/2.

The same result can be obtained directly by making a table of the sets of three integers that sum to N. The resulting sequence is 1, 3, 6, 10, 15, 21, ... is known as the “triangular number sequence”.

7.14

[T] Show that the usual relation E = p2 /2m arises as the leading momentum dependence of the energy in p an expansion of the relativistic relation E = m2 c4 + p2 c2 for small p ≪ mc.

First, factor out mc2 : E = mc2 (1 + p2 /m2 c2 )1/2 Then use the b(b−1)(b−2) 3 2 binomal expansion: (1 + a)b = 1 + ba + b(b−1) a + ...

2! a + 3!

E = mc2 1 + E = mc2 +

7.15

1 p2 p4 +O 4 4 2 m2 c 2 m c

p2 + ... , 2m

!!

where the terms that have been ignored as small when p ≪ mc.

[T] Find the ground state of the quantum harmonic oscillator. Start by showing that the Hamiltonian from eq. (7.43) can be “factorized” similar to (a2 − b2 ) = (a − b)(a + b) in the form ! ! !

1 d d x x H = ~ω −ℓ ℓ +1 , + + 2 dx ℓ dx ℓ √ where ℓ = ~/mω. Show that the Schrödinger equation is satisfied if (ℓ d/dx + x/ℓ)ψ0 (x) = 0. Show that 2 ψ0 (x) = Ne−ax is such a solution for the right choice of a. Find N such that ψ0 (x) is normalized. What is the energy of this state?

158

First show that the Hamiltonian given reduces to the harmonic oscillator Hamiltonian: ! ! !

1 d x x d Hψ = ~ω −ℓ + + ℓ +1 ψ 2 dx ℓ dx ℓ !

2 1 d x2 d 2 d = ~ω −ℓ − x+ 2 +1 ψ +x 2 dx dx dx2 ℓ ! d2 1 x2 ~2 1 ψ + mω2 x2 ψ . = ~ω −ℓ2 2 + 2 ψ = − 2 2m 2 dx ℓ Next, suppose (ℓd/dx + x/ℓ)ψ0 (x) = 0, then Hψ0 (x) =

1 ~ωψ0 (x) , 2

so ψ0 (x) is the harmonic oscillator ground state. To find ψ0 , it is necessary to solve the differntial equation, ψ′0 (x) = −(x/ℓ2 )ψ0 (x), whence dψ0 /ψ0 = −x dx/ℓ2 , which can be integrated to give 2 2 ln ψ0 = −x2 /2ℓ2 + constant, or ψ0 (x) = Ne−x /2ℓ R ∞.

Finally, N is determined by demanding that −∞ dx|ψ0 (x)|2 = 1, Z ∞ Z ∞ √ 2 2 2 N2 dx e−x /ℓ = ℓN 2 dye−y = ℓN 2 π = 1 , −∞

−∞

R∞

where we have used −∞ dye N = 1/

7.16

−y2

q

=

√

π. Solving for N,

r √ mω ℓ π= 4 . π~

The energy required to remove an electron from an atom is called the ionization energy (§9.3.3). Predict the energy needed to ionize the last electron from an atom with atomic number Z. (The atomic number is the charge on the atom’s nucleus (§17).)

After Z − 1 electrons have been removed, the remaining electron sees a Coulomb potential generated by the nucleus with charge Z. Its Schrödinger wavefunction therefore obeys eq. (7.50) with e2 replaced by Ze2 . The energy of this electron in its ground state is given by eq. (7.52) with e4 → Z 2 e4 , so the prediction is E(ionization) = 13.6 Z 2 eV.

7.17

[T] The polynomials appearing in eqs. (7.45)–(7.47) are known as Hermite polynomials. Compute the second, third, and fourth excited wavefunctions by assuming quadratic, cubic, and quartic polynomials and determining the coefficients by solving eq. (7.43).

Compute the associated energy values. [Hint: Rewrite eq. (7.43) in terms of the dimensionless variable ξ and explain why the solutions must be either even or odd in ξ → −ξ.]

First introduce dimensionless variables ξ = x/ℓ and ǫ = E/E0 , √ where ℓ = ~/mω and E0 = 12 ~ω. In terms of these variables, eq. (7.43) becomes ! d2 − 2 + ξ2 ψ(ξ) = ǫψ(ξ). dξ 2

Next substitute ψ(ξ) = f (ξ)e−ξ /2 , and find − f ′′ + 2ξ f ′ = (ǫ − 1) f .

Since this equation is even in ξ → −ξ, if f (ξ) is a solution, so is f (−ξ). Taking the sum and difference, it is clear that the solutions can be taken to be either even or odd in ξ.

As an example, consider a third order (odd) polynomial, f (ξ) = ξ + aξ3 . (Note that the equation for f is homogeneous in f so we can multiply by a constant so that the coefficient of the first term is one.) Substituting, −6aξ + 2ξ(1 + 3aξ2 ) = (ǫ − 1)(ξ + aξ3 ) .

This equation is satisfied for all ξ iff ǫ = 7 and a = −2/3, so the third energy state has E = (7/2)~ω and ψ(ξ) = N(ξ − 2 (2/3)ξ3 )e−ξ /2 , in agreement with eq. (7.47).

185

Problems

Problems 8.1

bits, an increase of thermodynamic entropy of

What is the information entropy of the results of flipping a biased coin 1000 times, if the coin comes up tails with probability 5/6 and heads with probability 1/6? For this weighted coin, can you find an encoding for sequential pairs of coin flips (e.g. HH, HT, etc.) in terms of sequences of bits (possibly different numbers of bits for different possible pairs) that takes fewer than 2 bits on average for each pair of flips? How many bits on average does your encoding take, and how close is it to the information-theoretic limit?

σ = −N

X j

pV (ln 2)∆σ T 5 (1.013 × 10 Pa)(0.001 m3 ) = (0.693)(9.966) 0.233 J/K .

300 K

∆S = NkB (ln 2)∆σ =

Note that the same result can be obtained from the Sackur-Tetrode formula for the entropy of an ideal gas (eq. (8.65)) using the fact that the temperature of an ideal gas does not change because Q = W = 0 in a free expansion.

8.4

Let p+ = x, then S = −kB (x ln x + (1 − x) ln(1 − x)), so at an extremum of the entropy, dp = −kB (ln x − ln(1 − x)) = 0 , dx

!

1 5 p j log2 p j = 1000 log2 6 + log2 (6/5) 650 6 6

Code pairs of flips representing the most probable pair by smallest number of bits: TT :p = 25/36 TH :p = 5/36 HT :p = 5/36 HH :p = 1/36

→

→

→

→

which gives x = 1− x or x = 1/2. It is easy to check that d 2 p/dx2 < 0 at x = 1/2 (actually for all x ∈ [0, 1]), so this stationary point is a maximum of S .

8.5

0 10 110 111

!

1 25 5 5 1 (1) + (2) + (3) + (3) 0.736.

2 36 36 36 36

j

What is the information entropy of the results of spinning a roulette wheel with 38 possible equally likely outcomes (36 numbers, 0 and 00) 10 times? How many fair coins would you need to flip to get this much information entropy?

∆S −(44.6 mol)(8.314 J/mol K) (0.78 ln(0.78) + 0.21 ln(0.21) + 0.01 ln(0.01)) 211 J/K .

The 2nd Law requires that the entropy of the universe must not decrease, so the process by which the gases are separated must, in the end, emit an amount of heat greater than or equal T ∆S into the environment. Thus the minimum energy required to separate the gases is E = T ∆S (273 K)(211 J/K) 57.5 kJ.

8.6

The probabilities are all equal, p j = 1/38, so σ = −N

X j

p j log2 p j = −10 38 ×

!

1 log2 (1/38) 52.5 .

38

Thus, for an optimal encoding, it takes on average 5.25 bits to specify which of the 38 results is obtained in a spin of the roulette wheel. It takes one bit to specify the results of a flip of a fair coin so it would take between 52 and 53 flips of a fair coin to get as much information entropy as in 10 spins of a roulette wheel.

8.3

Consider an amount of helium gas at atmospheric pressure and room temperature (you can use T = 300 K) enclosed in a 1-liter partition within a cubic meter.

The volume outside the partition containing the helium is evacuated. The helium gas is released suddenly by opening a door on the partition, and flows out to fill the volume. What is the increase in entropy of the gas?

Each helium atom, previously described by some number of bits, now requires log2 (1 m3 /l L) = log2 (1 000) 9.966 additional

j

when it is separated in to constituents with original concentrations c j . One mole of an ideal gas at STP has volume 22.4 L, so 1 m3 of air equals n = 1 m3 /(0.0224 m3 /mol) 44.6 moles. Thus

The factor of 1/2 accounts for the fact that each element of code corresponds to two flips. For 1 000 flips the total information entropy of this code is σ 736. This is better than the naive encoding (H→ 1, T→ 0), which has σ = 1 000, but not is not very close to the information-theoretic limit. An analogous three bit encoding yields σ 681.

8.2

Consider air to be a mixture of 78% nitrogen, 21% oxygen, and 1% argon. Estimate the minimum amount of energy that it takes to separate a cubic meter of air into its constituents at STP, by computing the change in entropy after the separation and using the 2nd Law. Generalizing the analysis of Example 8.1, the entropy of a gas decreases by X X c j ln c j c j ln c j = −nR ∆S = −NkB

This is an unambiguous coding (see discussion after eq. (8.6)). Its information entropy (per symbol) is σ=

[T] Prove analytically that the entropy of a two-state system (8.12) is maximized when p+ = p− = 1/2.

A heat engine operates between a temperature T + and the ambient environment at temperature 298 K. How large must T + be for the engine to have a possible efficiency of 90%? Can you find some materials that have melting points above this temperature, out of which you might build such an engine?

The 2nd Law requires η ≤ (T + − T −)T + so T + ≥ T − /(1 − η), for T − = 298 K and η = 0.9 we find T + ≥ 2 980 K. Common metals such as steel have melting points well below this temperature.

Among the chemical elements only osmium (3 306 K), rhenium (3 459 K), tantalum (3 290 K), and tungsten (3 695 K), have melting points above 2 980 K at atmospheric pressure. Carbon sublimates at a temperature around 3 800 K. Alloys involving these elements have even higher melting points.

8.7

A nuclear power plant operates at a maximum temperature of 375 ◦ C and produces 1 GW of electric power. If the waste heat is dumped into river water at a temperature of 20 ◦ C and environmental standards limit the effluent water to 30 ◦ C, what is the minimum amount of water (liters per second) that must feed through the plant to absorb the waste heat?

186

8.8

Waste heat from the power plant is absorbed by water initially at 20 ◦ C, heating it to a maximum of 30 ◦ C. For simplicity we assume that T − = 30 ◦ C. The Carnot efficiency limit on a system operating between 375 ◦ C and 30 ◦ C is ηC = (648 K − 303 K)/648 K 0.53, so the 1 GW output power plant consumes 1 GW/0.53 1.88 GW of energy. Thus thermal energy is expelled by the plant at a rate greater than or equal to 0.88 GW. Heating water from 20 ◦ C to 30 ◦ C requires q = (4.18 kJ/kg K)(10 K) 41.8 kJ/kg so the quantity of water at 20 ◦ C required per second is at least (0.88 GW)/((41.8 kJ/kg)(1 kg/L)) 21 000 L/s ≈ 21 m3 /s.

S → NkB ln 2 and the expectation value of the energy of each subsystem equals the average of the energies of the two states, so U → Nǫ/2. One might have thought that as T → ∞ all of the subsystems would be in the higher energy state. That, however, would be a state of zero entropy (and negative temperature).

8.10

In a typical step in the CPU of a computer, a 32-bit register is overwritten with the result of an operation.

The original values of the 32 bits are lost. Since the laws of physics are reversible, this information is added to the environment as entropy. For a computer running at 300 K that carries out 1011 operations per second, give a lower bound on the power requirement using the second law of thermodynamics.

From Example 8.3 and eq. (8.44), p0 = (1 − e−x ), p1 = e−x (1 − e−x ), and the vibrational heat capacity per molecule (in units of kB ) ĉvib = x2 e x /(e x − 1)2 , where x = ~ω/kB T . For hydrogen ~ω/kB T (6 000 K)/T , and direct calculation gives temperature p0 p1 ĉvib 500 K 0.999994 6.14 × 10−6 8.8 × 10−4 1 000 K 0.9975 0.00247 0.090 2 000 K 0.950 0.047 0.496 ĉvib goes from zero to one as T increases from zero to T ≫ ~ω/kB . As in the case of O2 , note that ĉvib reaches a value of order 1/2 when the occupation probability of the ground state is still ∼ 95%. Figure 5.7 shows the vibrational contribution to the heat capacity to be negligible at 500 K, small, but visible at 1 000 K, and roughly 1/2 at 2 000 K, in qualitative agreement with this computation.

With 32 bits of information lost per operation and 1011 operations per second, information entropy is created at a rate σ̇ = 3.2 × 1012 s−1 , equal to thermodynamic entropy, Ṡ = kB (ln 2)σ̇ (1.38 × 10−23 )(0.693)(3.2 × 1012 ) 3.1 × 10−11 W/K .

8.9

Since the computer is a mechanical device that does not store entropy, this entropy must be released in the form of heat into the environment, Q̇ = T Ṡ (3.1 × 10−11 )(300) 9.3 × 10−9 W.

[T] A simple system in which to study entropy and the Boltzmann distribution consists of N independent two-state subsystems coupled thermally to a reservoir at temperature T . The two states in each subsystem have energies 0 (by definition) and ǫ. Find the partition function, the internal energy, and the entropy of the total system as a function of temperature. Explain why the internal energy approaches Nǫ/2 as T → ∞.

Is this what you expected? It may help if you explain why the entropy approaches NkB ln 2 in the same limit.

The partition function Z is defined as the sum over all states of the system of e−E(state)/kB T . A state of this system is defined by assigning an energy zero or ǫPto each subsystem. This energy can be represented by E(state) = Nj=1 n j ǫ, where n j = 0 or 1. Thus Z=

X

e

{n j =0,1 ∀ j}

−

PN n ǫ/k T j=1 j B =

X

{n j =0,1 ∀ j}

  N  Y −n j ǫ/kB T    e   j=1

  N   Y  X

 −n j ǫ/kB T  −ǫ/k T N  = e .  = 1 + e B  j=1 n j =0,1

Here we have used the fact that the exponential of a sum equals a product of exponentials and the fact that a sum of products equals the product of sums. Then from eq. (8.37), U = kB T 2

d ln Z e−ǫ/kB T Nǫ = Nǫ = ǫ/k T . dT 1 + e−ǫ/kB T e B +1

Similarly S = kB ln Z + U/T gives Nǫ/T S = ǫ/k T + NkB ln(1 + e−ǫ/kB T ) e B +1 As T → ∞ the system approaches the state of maximum entropy, where each subsystem is equally likely to be in each of the two states. The entropy equals that of N flips of a fair coin,

Repeat the calculations of Example 8.3 for the H2 molecule, with ~ω/kB ∼ 6000 K. In particular, compute the probabilities that a given molecule is in the ground state or first excited state at temperatures T = 500, 1000, and 2000 K, compute the vibrational contribution to the heat capacity at each of these temperatures, and compare your results with Figure 5.7.

8.11

Consider a simple quantum system consisting of twenty independent simple harmonic oscillators each with frequency ω. The energy of this system is just the sum of the energies of the 20 oscillators. When the energy of this system is U(n) = (n + 10)~ω, show that its entropy is S (n) = kB ln((n + 19)!/(19! n!)). What is the entropy of the system in thermal equilibrium for n = 1, 2, and 3? Assume that ~ω 0.196 eV 3.14 × 10−20 J, corresponding to the vibrational excitation mode of O2 . Beginning in the state with U(1) = 11~ω, by how much does the entropy increase if the energy is increased to U(2) = 12~ω? Use this result to estimate the temperature of the system by 1/T ≡ ∆S /∆U. Repeat for the shift from U(2) to U(3), and estimate the heat capacity by comparing the change in U to the change in estimated T . Compare the heat capacity to kB to compute the effective number of degrees of freedom.

Do your results agree qualitatively with the analysis of Example 8.3?

The entropy is the S = kB ln N, where N is the number of microstates with energy U. In this case, N is the number of ways that n units of quantum excitation can be distributed over the N = 20 oscillators. This can be visualized as the number of ways to permute n objects and (N − 1) dividers on a line. There are (N + n − 1)! ways to arrange the objects and dividers, but permutations of the N − 1 dividers and the n objects separately must be divided out, hence N = (N + n − 1)!/(N − 1)! n!. Then S (1) = kB ln(20!/19! 1!) = kB ln 20 = 2.996 kB , S (2) = kB ln(21!/19! 2!) = kB ln 210 = 5.347 kB , S (3) = kB ln(22!/19! 3!) = kB ln 1540 = 7.340 kB .

187

Problems

RT where f (T 1 , T 0 ) = T 1 dT c(T )/T is finite for all T 0 and T 1 . Note 0 that the integral defining f (T 1 , T 0 ) is finite when either T 0 or T 1 R vanishes because 0 dT c(T )/T converges at its lower limit since c(0) = 0. Rearranging,

To obtain an estimate of the temperature, we use ∆S /∆U ≈ 1/T , giving U(2) − U(1) ~ω = ≈ 968 K , S (2) − S (1) 2.35kB ~ω U(3) − U(2) = ≈ 1 140 K .

T 23 = S (3) − S (2) 1.99kB

T 12 =

C0 ln(T 1 /T 0 ) = S (T 1 ) − S (T 0 ) − f (T 1 , T 0 ) .

Since by assumption S (T ) is finite for all T , the right hand side of this equation is finite for any T 0 and T 1 . The left hand side, however, diverges as either T 0 or T 1 approaches zero. Thus we conclude that C0 = 0.

To estimate the heat capacity we compute C ≈ ∆U/∆T ≈ (3.14 × 10−20 J)/170 K 1.83 × 10−22 J/K 13kB .

Since there are twenty oscillators in the system, the heat capacity per oscillator is ĉkB ≈ (13/20)kB = 0.65kB (at a temperature of approximately (1 140 K + 968 K)/2 ≈ 1 050 K). This is consistent with the analysis of the vibrational heat capacity of diatomic oxygen in Example 8.3, where it was noted at the heat capacity per molecule reaches 0.47kB at 750 K.

8.12

8.14

[T] Check the results quoted in eqs. (8.42)–(8.44) for a single harmonic oscillator. Show that as T → 0, the entropy and the heat R T capacity both vanish. Show that S (T 1 ) − S (T 0 ) = T 1 C(T )dT/T .

[T] Use the Sackur–Tetrode equation (8.64) to show that (∂S /∂U)|V = 1/T for an ideal monatomic gas.

Likewise show that (∂S /∂H)| p = 1/T , verifying eq. (8.18) and the assertion of footnote 6. ∂S ∂S /∂T |V ∂S ∂T . = = ∂U V ∂T V ∂U V ∂U/∂T |V

From eq. (8.63) and eq. (8.64)

3 ∂U = NkB ∂T V 2 ∂S 3 = NkB /T . ∂T V 2

0

From x = ~ω/kB T , d/dT = (dx/dT )d/dx = −(~ω/kB T 2 )(d/dx).

Changing variables in eq. (8.37), U ≡ hUi = −~ω

d ln Z(x) , dx

which gives ∂S /∂U|V = 1/T .

Similarly,

and taking Z from eq. (8.41), U = −~ω −

∂S /∂T | p ∂S ∂T ∂S = = , ∂H p ∂T p ∂H p ∂H/∂T | p

! !

1 d 1 1 − ln(1 − e−x ) = ~ω + , 2 dx 2 ex − 1

where H = U+pV. For an ideal gas, pV = NkB T , so H = U+NkB T and ∂H/∂T | p = (5/3)NkB . To evaluate ∂S /∂T | p , substitute V = NkB T/p from the ideal gas law into eq. (8.64), !

3 mkB T 3 + ln(NkB T/p) + ln − kB ln(N!) .

S ideal = NkB 2 2 2π~2

in agreement with eq. (8.42). Then from eq. (8.38), x x x , S = kB ln Z + U/T = kB − − ln(1 − e−x ) + + x 2 2 e −1 in agreement with eq. (8.43). Finally, C=

dU ~ω dU x2 e x =− = kB x , dT kB T 2 dx (e − 1)2

in agreement with eq. (8.44).

As T → 0, x → ∞, so 1/(e x − 1) → e−x and ln(1 − e−x ) → −e−x , where we have used the Taylor expansion, ln(1 − ǫ) = −ǫ − ǫ 2 /2 − ǫ 3 /3 − .... Combining these, we have lim S = lim kB (xe−x + e−x ) = 0 .

x→∞

x→∞

Similarly lim C = lim kB x2 e−x = 0 .

x→∞

x→∞

In both cases we have used the fact that e−x vanishes faster than any power of x as x → ∞.

Finally, since it is easier to differentiate than to integrate, we prove equivalently that dS /dT = C(T )/T , ! dS 1 ex e−x dS = −x = −x x −x x − T dT dx e −1 1 − e−x (e − 1)2 = x2

8.13

ex = C(x) . (e x − 1)2

[T] ShowR that in general the condition that S (T 1 ) − T S (T 0 ) = T 1 CV (T )dT/T and the requirement that the 0 entropy be finite at all temperatures (we cannot be infinitely ignorant about a finite system), require that C(T ) → 0 as T → 0 for all systems.

Suppose CV (T ) = C0 + c(T ), where c(0) = 0. Then S (T 1 ) − S (T 0 ) = C0 ln(T 1 /T 0 ) + f (T 1 , T 0 ) ,

8.15

From this ∂S /∂T | p = (5/2)NkB /T , which gives ∂S /∂H| p = 1/T .

[T] In §8.7.2 it was asserted that in order for entropy to be an extensive quantity, the classical partition function of a system of indistinguishable particles must be divided by N!. Consider a box evenly partitioned into two halves, each with volume V, and each containing N atoms of a monatomic ideal gas at a temperature T . Compute the entropy from eq. (8.64) omitting the last term. Now remove the partition, and compute the entropy of the resulting system of 2N particles in a volume 2V again from eq. (8.64) with the last term omitted. We have not lost any information about the system, but the entropy seems to have increased if we compute it without the Gibbs factor! Next, restore the N! factor and repeat the analysis.

First dropping the ln N!, for each half with volume V, number N, and temperature T , !

3 mkB T 3 + ln V + ln .

S (N, V, T ) = NkB 2 2 2 2π~ So for the two halves together, 2S (N, V, T ) = 2NkB

3 mkB T 3 + ln V + ln 2 2 2π~2

!

.

Computed directly, S (2N, 2V, T ) = 2NkB

!

3 mkB T 3 + ln(2V) + ln .

2 2 2 2π~

188

These two differ by S (2N, 2V, T ) − 2S (N, V, T ) = 2NkB ln 2.

Next, restore the factor −kB ln N! ≈ −NkB (ln N − 1) and recompute,

The condition for equilibrium is that the entropy of the system plus the environment is maximum, dS total = dS system + dS environment = 0 .

S ′ (N, V, T ) = S (N, V, T ) − NkB (ln N − 1) .

The entropy of the environment changes when heat is emitted by the system, dS environment = dQemitted /T . At fixed volume dQemitted = −dUsystem (see eq. (5.24)), so the condition for equilibrium is

So for the two halves together, 2S ′ (N, V, T ) = 2S (N, V, T ) − 2NkB (ln N − 1) ,

dS total = dS system − dUsystem /T = 0.

and computed directly,

Defining F = U − T S , at fixed temperature, dF = dU − T dS , so dS total = −dFsystem /T = 0 and S total is maximized when Fsystem is minimized.

At fixed pressure, dQemitted = −dHsystem (see eq. (5.36)) and the analysis leads to Gsystem being minimized at equilibrium.

A shorter though more formal proof is as follows: dF = dU − T dS at fixed T . Then at fixed V, dF = dU(1 − T ∂S /∂U|V ) and at equilibrium ∂S /∂U|V = 1/T , so dF = 0.

S ′ (2N, 2V, T ) = S (2N, 2V, T ) − 2NkB (ln(2N) − 1) .

And the we now have S ′ (2N, 2V, T ) = 2S ′ (N, V, T ) as required for S to be an extensive quantity.

8.16

[T] Compute the partition function per particle (8.61) for helium gas at NTP confined in a cube of side 1 cm.

What quantum state has the highest probability of being occupied? What is the probability that a given helium atom is in this state? Given the number of atoms in the box, what is the expected average occupation number of this state?

8.19

The ground state has with l = m = n = 1 has the largest probability of being occupied, with p0 = e−3x /zbox , where x = 2 2 ~ π /(2mL2 kB T ). Evaluate x at T = 293 K, 2

x

First consider three dimensions: From eq. (8.61) and 8.62, Zideal = (V/λ)3N3 /2 /N3 !. For a free, monatomic gas U = 3N3 kB T/2 and S is given by eq. (8.65), so !

5 3 − ln(N3 λ3 /V) F2 = U − T S = N3 kB T − NkB T 2 2

−34 2

π (1.05 × 10 ) 2.0 × 10−17 , 2(4 × 1.66 × 10−27 )(0.01)2 (1.38 × 10−23 )(293)

So e−3x 1. Using zbox =

π 3/2 4x

8.17

[T] Real crystals contain impurities, which lead to nonzero entropy at T = 0. Consider a crystal consisting of N atoms. The crystal is primarily composed of element A but contains M ≪ N atoms of an impurity B. An atom of B can substitute for an atom of A at any location in the crystal. Compute the entropy of this crystal at T = 0.

The entropy is S = kB ln N, where N is the number of microstates that describe the crystal at T = 0. There are !

N! N ≡ N= M (N − M)!M! distinct ways to distribute M impurities among N total sites. Each corresponds to a distinct microstate, so S = kB (ln N! − ln M! − ln(N − M)!)

= kB (N ln N − M ln M − (N − M) ln(N − M))

N + 1 + O(M/N) , = MkB ln M

8.18

= −N3 kB T (1 − ln(N3 λ3 /V)) .

3/2 π 7.8 × 1024 . −17 8.0 × 10

and p0 1.3 × 10−25 .

At NTP, one mole of an ideal gas occupies ∼ 24 L = 0.024 m3 , so a cube of volume 1 cm3 contains N = (1/24 000)NA ∼ 2.5×1019 helium atoms. So the average occupation number of this state is n0 = p0 N 3.3 × 10−6 , and the average occupation number of any higher energy state is smaller than this.

where Stirling’s approximation and M/N ≪ 1 have been used.

[T] Show that when a system comes into thermal equilibrium while in contact with a heat reservoir at temperature T , its free energy (Helmholtz, if V is held fixed; Gibbs if p is held fixed) is minimized.

[T] Show that the free energy of an ideal monatomic gas of N3 particles in a volume V can bepwritten as F =

−kB N3 T 1 − ln(N3 λ3 /V) , where λ = 2π~2 /mkB T is a thermal length scale. Show that a two-dimensional ideal gas of N2 particles moving in a bounded area A has free energy F2 = −kB N2 T 1 − ln(N2 λ2 /A) .

Alternatively, eq. (8.39) leads directly to the same result.

For a two dimensional gas the triple sum in eq. (8.59) is replaced by a double sum. The subsequent analysis carries through leading to Zideal = (A/λ2 )N2 /N2 ! and eq. (8.39) leads directly to the quoted result.

8.20

[H] The atoms of an ideal monatomic gas of N particles confined in a box of volume V can be adsorbed onto the surface of the box (surface area A), where they are bound to the surface with binding energy ǫ, but they can move around on the surface like an ideal two-dimensional gas. Use the results of Problem 8.19 to write the (Helmholtz) free energy of this system as a function of N3 and N2 , the numbers of gas atoms in the bulk and the surface respectively. Minimize the free energy (Problem 8.18) to find the ratio N2 /N3 when the system reaches thermal equilibrium. What happens in the limits where T → 0, T → ∞, and ~ → 0?

The effect of the binding energy is to modify the partition function of the two-dimensional gas by a factor eN2 ǫ/kB T , so the total free energy is

F(N2 ) = −kB N2 T 1 − ln(N2 λ2 /A) −kB N3 T 1 − ln(N3 λ3 /V) −N2 ǫ where N3 = N−N2 . The system is in equilibrium when ∂F/∂N2 |N = 0, or kB T ln(N2 λ2 /A) − kB T ln(N3 λ3 /V) − ǫ = 0 , which yield λA ǫ/kB T N2 = e .

N3 V As T → 0, N2 /N3 → ∞ – the effects of binding energy dominates and all the atoms√are in their lowest energy state. As T → ∞, N2 /N3 → 0 (∼ A/V T ) – the effects of entropy dominate since

Problems

there are more available states in the volume than on the surface. If Planck’s constant were zero, there would be no particles adsorbed on the surface at any temperature, thus adsorption is a quantum phenomenon(!)

189

208

of the photon gas in §22.3 shows that the heat capacity in the Debye model takes the form Z T D /T x4 ex cv /kB = 9(T/T D )3 dx x , (e − 1)2 0

Problems 9.1

As described in §21, the universe is filled with electromagnetic radiation left over from the Big Bang. This radiation has a characteristic temperature of ≈ 2.7 K. At what wavelength does the power spectrum of this radiation peak? What is the energy of a photon with this wavelength?

where x is a rescaled energy parameter. At large x, ex (ex − 1)−2 → e−x gives the Boltzmann suppression of states with increasing energy, while x4 measures the increase in the number of states available with increasing energy. Compute (approximately) the minimum temperature T/T D at which the factor x4 e−x exceeds one for some x in the range of integration.

According to the discussion after eq. (9.4), λmax = λth /2.82 and λth = hc/kB T , so λmax =

(2π)(1.056 × 10−34 J s)(3.00 × 108 m/s) ≈ 1.9 mm .

(2.82)(2.7 K)(1.38 × 10−23 J/K)

Note that this is the wavelength corresponding to the peak frequency; this is roughly 1.76 times the peak wavelength of the power spectrum per unit wavelength. For more on this, you can consult Figure 22.3.

The energy of a photon with this wavelength is

The suppression of high energy states becomes more apparent if we rewrite Z T D /T Z ∞ Z ∞ x4 e x x4 e x x4 e x 4π4 dx x − dx x = dx x −= 2 2 15 (e − 1) (e − 1) (e − 1)2 0 T D /T 0

(2π)(1.056 × 10−34 J s)(3.00 × 108 m/s) 1.9 × 10−3 m 1.05 × 10−22 J 6.55 × 10−4 eV .

E = hc/λ

9.2

A blue supergiant star may have a surface temperature as high as 50 000 K. Estimate the wavelength at which it emits maximum power. Why does the star appear blue?

When T ≪ T D the integral from T D /T is suppressed by e−T D /T , but as T increases the integrand enters regions where x4 e−x ≈ 1 and the high energy states are not suppressed. Solving the equation x4 e−x = 1 numerically yields x 8.6, so by the time T/T D ≈ 1/8, higher energy states are contributing significantly to the integral.

9.5

According to the discussion after eq. (9.4), λmax = λth /2.82 and λth = hc/kB T , so λmax =

(2π)(1.056 × 10−34 J s)(3.00 × 108 m/s) ≈ 103 nm.

(2.82)(50 000 K)(1.38 × 10−23 J/K)

This wavelength is in the X-ray part of the spectrum, far beyond range of visible light (400–750 nm). The intensity of the radiation increases as the wavelength approaches λmax , so the visible light emitted by the star is peaked at short wavelengths, i.e. in the blue.

9.3

When the spin of the electron in a hydrogen atom flips from parallel to antiparallel to the direction of the proton’s spin, a photon of energy E 9.41 × 10−25 J is emitted. What is the photon’s energy in electron volts?

What is its frequency? Its wavelength? Above roughly what temperature would you expect this transition to be thermally excited? This emission line of neutral hydrogen plays a central role in modern astronomy.

The power output of air conditioners is measured in “tons,” an ancient nomenclature dating back to the days when air was cooled by blowing it over blocks of ice.

A ton of air conditioning is defined to be the energy required to melt one ton (2000 pounds) of ice at 32 ◦ F distributed over a period of one day. What is the power equivalent (in kilowatts) of one ton of air conditioning?

2 000 lbs of ice 907 kg. The heat of fusion for water is 334 kJ/kg, so to melt one ton of ice requires (334 kJ/kg) × (907 kg) ÷ (103 J/kJ) 303 MJ of energy. One day is 86 400 seconds, and one watt is 1 J/s, so one ton of air conditioning = (303 MJ) ÷ (86,400 s) 3.51 kW.

9.6

Find the frequency of the lowest vibrational mode of a diatomic molecule in terms of the parameters of the Morse potential, eq. (9.5).

Expand VMorse (r) to second order about its minimum. First find the minimum dVMorse = 2Dα(eα(r−re ) − 1) = 0 dr has solution r = re as expected. Then

ν = E/h = 9.41 × 10−25 /6.63 × 10−34 = 1.42 GHz

d 2 VMorse = 2Dα2 dr2 r=re

λ = c/ν = 3.00 × 108 /1.42 × 109 = 0.211 m

T ∼ E/kB = 9.41 × 10−25 /1.38 × 10−23 = 0.068 K

For small r − re the energy is

The wavelength places this radiation in the short-wave region of the radio spectrum. Observation with radio telescopes enables astronomers to detect regions on neutral hydrogen atoms. This is the famous “21-cm” line of radio astronomy.

9.4

[T] The cubic growth of the heat capacity of a solid as a function of temperature continues only up to roughly one tenth of the Debye temperature associated with the shortest wavelength excitations in the solid. This may seem surprising, since this implies that at this point the heat capacity becomes sensitive to the absence of modes whose probability of excitation is on the order of e−10 . A derivation of the Debye model parallel to that

E(r, ṙ) = 21 µṙ2 + 21 2Dα2 (r − re )2 ,

9.7

where is the reduced mass of the diatomic molecule. This has the form of apharmonic oscillator (see Example 2.2). From this we read off ω = 2De /µα.

The Morse potential parameters for oxygen are given

by De = 5.211 eV, α = 2.78 Å−1 , and re = 1.207 Å [55]. Using the result from the previous problem, estimate the energy necessary to excite the oxygen molecule out of its ground state. Compare with your answer to Problem 7.10.

209

Problems

From the previous problem, ω = mass of the oxygen atom

p

2De /µα. µ is the reduced

where λ J is the wavelength of the emitted photon. Solving for λ J , ! ! ~2 (J + 1)/µre2 ~2 (J + 1)/µre2 λvib / 1 + ≈ λvib 1 − ~ωvib ~ωvib

µ = 12 (16)(1.66 × 10−27 ) = 1.33 × 10−26 kg. and De = 5.211 eV(1.602 × 10−19 J/eV) = 8.348 × 10−19 J. Then r 2(8.348 × 10−19 ) ω= 2.78 × 1010 1.33 × 10−26 = 0.311 × 1015 Hz

9.10

and ∆E = ~ω = (0.311 × 1015 )(1.055 × 10−34 ) = 3.28 × 10−20 J = 0.205 eV.

9.8

[T] Another commonly used model for the potential energy of a diatomic molecule is the Lennard–Jones (LJ) potential V(r) = ǫ((rm /r)12 − 2(rm /r)6 ). Using the results of Problem 9.6, find the choice of the parameters rm and ǫ that gives the same equilibrium separation between atoms and the same harmonic oscillator frequency as the Morse potential. Unlike the Morse potential, the depth of the LJ potential is not an independent parameter. What is the relation between the depth De , equilibrium separation re , and harmonic oscillator frequency ω of the LJ potential?

So the spacing between lines is ∆λ = (~2 (J + 1)/µre2 )/(~ωvib ).

Substituting the values for re and µ = mO mC /(mO + mC ), we find ∆λ 1.7 × 10−3 λvib = 0.0081 µ.

[T] Show that the data in Table 9.6 satisfy the definitions of Gibbs and Helmholtz free energy, (1) ∆Gr = ∆Hr − T ∆S r and (2) ∆Hr = ∆Ur + p∆Vr .

First, ∆Hr − T ∆S r = 178.3 kJ/mol − 298.15 K(160.6 J/molK) = 130.4 kJ/mol = ∆Gr .

Second: the change in volume is basically just the volume of 1 mol of CO2 at the given conditions. This is given by the ideal gas law, V = RT/p = (8.314 J/mol K)(298.15 K)/101 325 Pa = 0.02446 m3 ∆Ur + p∆Vr = 175.8 kJ/mol + (101 325 Pa)(0.02446 m3 ) = 178.3 kJ/mol = ∆Hr .

9.11

The minimum of this potential is where ! r12 r6 dV = −12ǫ m13 − m7 = 0 dr r r which is satisfied when r = rm . Expanding the potential around rm , ǫ V ′′ = 72 2 . rm Comparing it to the Morse potential, rm = re , and 72ǫ/rm2 = 2De α2 or ǫ = Dep(αre /6)2 . p ω = V ′′ /µ = 72ǫ/µrm2 , and the depth of the LJ potential is ǫ, so the relation is

[T] Suppose reaction data, ∆Hr◦ and ∆G◦r (and therefore ∆S r◦ ), on a chemical reaction are all known at a temperature T 0 and pressure p, but you want to know the reaction enthalpy and free energy at a different temperature T 1 . Suppose the heat capacities of all the reactants (at pressure p) are known throughout the range T 0 < T < T 1 . Find expressions for ∆Hr (T 1 ) and ∆Gr (T 1 ).

Express your answers in terms of ∆C p , the sum of the heat capacities of the products minus the heat capacities of the reactants, and the reaction data at T 0 . ∆C p (T ) is given by

De = µω2 re2 /72

∂Hr ∂T

∆C p (T ) =

9.9

[T] Approximating the interatomic potential as a harmonic oscillator with angular frequency ωvib , the vibrational-rotational energy levels of a diatomic molecule are given by E(n, J) = ~ωvib (n+1/2)+~2 J(J+ 1)/(2µre2 ). Here re is the equilibrium interatomic separation, µ is the reduced mass, and J = 0, 1, 2, . . . is the rotational quantum number. The CO bond in carbon monoxide has re 1.13 × 10−10 m and ~ωvib = 0.2657 eV. Ignoring rotations, what is the wavelength λvib (in microns) of the radiation absorbed when the CO molecule makes a transition from vibrational level n to n + 1? Vibrational transitions are always accompanied by a transition from rotational level J to J ± 1. The result is a sequence of equally spaced absorption lines centered at λvib . What is the spacing between these vibrational-rotational absorption lines? λvib = c/(2πωvib ) 4.666 µ.

. p

The change in enthalpy, then, is ∆Hr (T 1 ) = ∆Hr◦ +

ZT 1

dT ∆C p (T ).

T0

∆C p (T ) also determines the change in reaction entropy, ∆C p (T )dT = T dS r , so ∆S r (T 1 ) = ∆S ◦r +

ZT 1

dT ∆C p (T ).

T

T0

The changes in entropy and enthalpy are related to the Gibbs free energy by ∆Gr (T 1 ) = ∆Hr (T 1 ) − T 1 ∆S r (T 1 ) = ∆G◦r +

ZT 1

T0

9.12

E(n + 1, J + 1) − E(n, J) = ~ωvib + (~2 /2mre2 ) ((J + 1)(J + 2) − J(J + 1)) = ~ωvib + (~2 /mre2 )(J + 1) = hc/λ J

!

T1 − (T 1 − T 0 )∆S r (T 0 ). dT ∆C p (T ) 1 − T

Estimate the amount of CO2 produced per kilogram of CaCO3 in the calcination reaction (9.10). In addition to the CO2 released in the reaction, include the CO2 emitted if coal is burned (at 100% efficiency) to supply the enthalpy of reaction and the energy necessary to heat the reactants to 890 ◦ C. You can assume that the heat

210

capacity of CaCO3 is Ĉ p 0.09 kJ/mol K, and assume that the coal is pure carbon so ∆H f = 0.

9.16

Heating from 25 ◦ C to 890 ◦ C requires ∆H = (0.09 kJ/mol K)(865 K) 80 kJ/mol. Adding in ∆Hr 180 kJ/mol, gives a total thermal energy requirement of 260 kJ/mol. The molar mass of CaCO3 is approximately 40 + 12 + 3(16) = 100 g/mol. So the thermal energy requirement for calcination is ∆H ≈ 2.6 MJ/kg. The energy content of coal is approximately 30 GJ/tonne, so mass(coal) 2.6 MJ/kg(CaCO3 ) ≈ 0.087 kg(coal)/kg(CaCO3 ) mass(CaCO3 ) 30 GJ/tonne(coal) is required. Combustion of 0.087 kilograms of coal generates (44/12) × 0.087 0.32 kg(CO2 ) .

This must be added to the mole of CO2 created in the calcination reaction to give

Wood contains organic polymers such as long chains of cellulose (C6 H10 O5 )n , and is commonly used as a biofuel.15 Write a balanced equation for the complete combustion of one unit of the cellulose polymer to water and CO2 . Note that hydrogen and oxygen are already present in the ratio 2:1. Make a crude estimate of the enthalpy of combustion of cellulose by assuming that the conversion of the H and O to water does not contribute significantly, so that the energy content can be attributed entirely to the combustion of carbon. Compare your estimate to the true enthalpy of combustion of cellulose, 14.9 MJ/kg.

0.32 kg(CO2 )/kg(CaCO3 )+0.44 kg(CO2 )/kg(CaCO3 ) 0.72 kg(CO2 )/kg(CaCO3 ) .

9.13

Look up the standard enthalpies of formation and the entropies of solid ice and liquid water and verify that ice may spontaneously melt at NTP (20 ◦ C, 1 atm).

The proposal is to estimate the enthalpy of combustion of cellulose by replacing C6 H10 O5 by 6C+5H2 O, and then model the combustion of cellulose by 6C + 6O2 → 6C2 O. Since the enthalpy of formation of carbon dioxide is -394 kJ/mol, we estimate ∆Hc (cellulose) ∼ 6(−394) = −2 304 kJ/mol, or

For liquid phase, the ∆Hlf = −285.8 kJ/mol and S l = 70 J/mol K. For ice, ∆H sf = −291.8 kJ/mol and S s = 41 J/mol K. At 20 ◦ C, the Gibbs free energy of reaction for H2 O(s) → H2 O(l) is

∆Hc (cellulose) ∼ (−2 304 kJ/mol)/(0.162 kg/mol) = −14.2 MJ/kg.

∆Gr = (−285.8 − (−291.8)) kJ/mol − (293.15 K)(70 − 41 J/mol K)

To actual enthalpy of combustion of cellulose, we need f ∆H f (cellulose) = −961 kJ/mol and ∆Hl/g (H2 O) = −286/242 kJ/kg.

We compute both the HHV and LHV,

= −2.5 kJ/mol .

This is negative, so the reaction occurs spontaneously.

9.14

The convention we have used for LHV differs from the one offered by the US Department of Energy, which defines LHV as “the amount of heat released by combusting a specified quantity (initially at 25 ◦ C) and returning the temperature of the combustion products to 150 ◦ C, which assumes the latent heat of vaporization of water in the reaction products is not recovered” [53]. According to the DOE convention, the LHV of methane is 789 kJ/mol, compared with 802 kJ/mol from the convention we use in Example 9.2. Account for this difference.

The difference between the two conventions is the enthalpy ∆H ′ required to heat two moles of water vapor and one mole of CO2 from 25 ◦ C to 150 ◦ C, which can be obtained by integrating the gases heat capacities (at constant pressure). We estimate ∆H ′ by assuming the heat capacities are (roughly) constant. From Table 5.2 C p (H2 O(g)) = 38 J/mol K, and C p (CO2 (g)) = 37 J/mol K, thus ∆H ′ ≈ (125 K)(2(38 J/mol K) + (37 J/mol K)) ≈ 14 kJ/mol .

The difference between the two conventions for the LHV is 802 − 789 = 13 kJ/kg, which agrees well given the precision of our estimate.

9.15

Find the enthalpy of reaction for the two pathways of decomposition of TNT mentioned in §9.5.2.

The two modes of decomposition of TNT mentioned are: 2C6 H2 (NO2 )3 CH3 → 5H2 O + 7CO + 7C + 3N2

C6 H10 O5 + 6O2 −→ 6CO2 + 5H2 O

HHV :

∆Hc (cellulose) = 6(−394) + 5(−286) + 961

LHV :

∆Hc (cellulose) = 6(−394) + 5(−242) + 961

= −2 833 kJ/mol = −17.5 MJ/kg

= −2 613 kJ/mol = −16.1 MJ/kg So the actual result is either 13% (LHV) or 23% (HHV) higher than the estimate

9.17

Estimate the lower heating value of a typical cord of wood by first estimating the mass of the wood and then assuming that this mass is completely composed of cellulose (see Problem 9.16). Compare your answer with the standard value of 26 GJ. Real “dried” firewood (density 670 kg/m3 ) contains a significant amount of water in addition to cellulose (and lignin and hemicellulose). When it burns, the water must be driven off as

Table 9.10 Data for Problem 9.21.

Compound

∆H f (kJ/mol)

∆G f (kJ/mol)

PbO(s) PbS(s) SO2 (g)

−219.4 −98.3 −296.8

−188.6 −96.7 −300.2

2C6 H2 (NO2 )3 CH3 → 5 H2 + 12 CO + 2 C + 3 N2

The enthalpies of formation of TNT, H2 O(g), and CO are -63 kJ/mol, -242 kJ/mol, and -111 kJ/mol, respectively. The enthalpies of reaction per mol TNT are 1)

∆Hr = 21 (5(−242) + 7(−111) + 2(63)) = 931 kJ/mol

2)

∆Hr = 21 (12(−111) + 63) = 603 kJ/mol

15 Wood also contains other organic polymers such as lignin (see

§26 (Biofuels)), but the enthalpy of combustion of the other polymers is close enough to that of cellulose for the purpose of this problem.

211

Problems

vapor. Compute the fraction y of water by mass in firewood that would account for the discrepancy you found in the first part of this problem.

3

3

The oxidation reaction is 4Fe + 3O2 → 2Fe2 O3 . ∆Hr for this process is just ∆H f (Fe2 O3 ) = −824 kJ/mol(Fe2 O3 ).

There are 2 moles of iron per mole of ferric oxide, and each mole of iron has a mass of approximately 55.9 g, so the total amount of energy liberated per kilogram of iron is

3

A cord of wood is 128 ft = 3.62 m . At a density of 670 kg/m , the mass of a cord of wood is m = 2 425 kg/cord. Assuming that the wood is completely cellulose, the LHV of this mass is ∆Hc = (2 425 kg)(16.1 MJ/kg) = 39 GJ/cord, which is much greater than the standard value of 26 GJ/cord.

Next, we assume a fraction y by mass of the wood is actually water. Each kilogram of water must be heated from 25 ◦ C to 100 ◦ C and then vaporized. This requires (4.19 × 75 + 2 257) = 2 570 kJ/kg. Therefore the effective enthalpy of combustion of a cord of firewood is

1 2 (824) kJ/mol(Fe)

0.056 kg/mol(Fe)

Warming your hands by 27 ◦ C requires energy E = cV m∆T = (4.18 kJ/kg K)(0.4 kg)(27 K) = 45 kJ .

∆Hc (y) = (2 425 kg/cord)(1 − y)(16.1 MJ/kg) − 2.57 MJ/kg y = (39 − 45.2y) GJ/cord

This requires m(Fe) (45 kJ)/(7.4 MJ/kg(Fe)) 6.1 g(Fe).

9.21

To account for the standard value of 26 GJ/cord, we require y ≈ (39 − 26)/45 ≈ 29% of the mass of the wood to be water. A survey of typical water content of wood indicates that this is typical of moderately dry firewood.

9.18

Acetylene (used in welding torches) C2 H2 , sucrose (cane sugar) C12 H22 O11 , and caffeine C8 H10 O2 N4 , are all popular energy sources. Their heats of combustion are 310.6 kcal/mol, 1348.2 kcal/mol, and 1014.2 kcal/mol, respectively. Which substance has the highest specific energy density (kJ/kg)?

Acetylene has a molar mass of 0.026 kg/mol and a heat of combustion of 1300 kJ/mol, for a specific energy density of 50 MJ/kg.

Sucrose has a molar mass of 0.342 kg/mol and a heat of combustion of 5640 kJ/mol, for a specific energy density of 16.5 MJ/kg.

Caffeine has a molar mass of 0.194 kg/mol and a heat of combustion of 4240 kJ/mol, for a specific energy density of 21.9 MJ/kg.

So, acetylene has the highest specific energy density by a wide margin of 50 MJ/kg, or 50 000 kJ/kg.

9.19

Thermite is a mixture of powdered metallic aluminum and iron oxide (usually Fe2 O3 ). Although stable at room temperature, the reaction 2 Al + Fe2 O3 → Al2 O3 + 2 Fe proceeds quickly when thermite is heated to its ignition temperature. The reaction itself generates much heat. Thermite is used as an intense heat source for welding and is particularly useful because it does not need an oxygen supply. What is the energy density in megajoules per kilogram of thermite?

Fe2 O3 has a standard enthalpy of formation of ∆H f = −824 kJ/mol while Al2 O3 has a standard enthalpy of formation of ∆H f = −1676 kJ/mol. The energy per mole of Al released by this reaction is ∆Hr = −1676 − (−824) = −852 kJ/mol. The molar mass of Al is 27 g/mol and that of Fe2 O3 is 2(55.9) + 3(16) 160 g/mol, so the molar mass of thermite is m = 2(27) + 160 = 214 g/mol. So the energy content of the thermite is Ethermite (852 kJ/mol)/(0.214 kg/mol) 4.0 MJ/kg.

9.20

An inexpensive hand warmer uses an exothermic chemical reaction to produce heat: iron reacts with oxygen to form ferric oxide, Fe2 O3 . Write a balanced chemical reaction for this oxidation process. Compute the energy liberated per kilogram of iron. Assuming that your hands each weigh about 0.4 kilogram and assuming that the specific heat of your hands is the same as water, what mass of iron would be needed to warm your hands from 10 ◦ C to body temperature, ∼38 ◦ C?

) 7.4 MJ/kg(Fe) .

Show that roasting lead (2 PbS + 3 O2 → 2 PbO + 2 SO2 ) is an exothermic reaction and compute the free energy of this reaction under standard conditions. See Table 9.10 for data.

Using data from the table, the enthalpy of reaction is ∆Hr = 2(−219.4) + 2(−296.8) − 2(−98.3) = −835.8 KJ/mol , so the reaction is exothermic. The free energy of reaction is ∆Gr = 2(−188.6) + 2(−300.2) − 2(−96.7) = −784.2 KJ/mol .

9.22

Show that the method of refining tin explained in Example 9.3 will not work for alumina.

Whether or not the method works amounts to computing the reaction free energy for reducing Al2 O3 with carbon monoxide.

The reaction is Al2 O3 + 3 CO −→ 2 Al + 3 CO2 Using data from the tables in this chapter, ∆Gr = 3(−394) − 3(−137) − (−1582) = 811 kJ/mol Since the reaction free energy is positive, the reaction will not proceed at 25 ◦ C and 1 atm. To estimate the effects of temperature, we must compute the reaction entropy, ∆S r = (∆Hr − ∆Gr )/T . First compute ∆Hr = 3(−394) − 3(−111) − (−1676) = 827 kJ/mol Thus ∆S r = (827 − 811)/298 = 53.7 J/kg K. We estimate the reaction free energy as a function of temperature by assuming the enthalpy and entropy of reaction are constant, ∆Gr (T) ≈ (827 − 0.0537T) kJ/mol . ∆Gr must be negative for the reaction to be able to proceed spontanously. This does not occur until T > 15 000 K, far too high to be practical.

231

Problems

Problems 10.1

Table 10.1 Data for Carnot cycle. Data provided is in

By how much does the temperature drop in the example of adiabatic expansion in Example 10.2?

In adiabatic expansion pV γ = constant. Since V ∝ T/p for a fixed amount of an ideal gas, p(T/p)γ = constant, or solving for a relation between T 2 and T 1 , !(1−γ)/γ p1 T1 T2 = p2

black; data calculated below is in red.

Point

Temperature [K]

p [atm]

V [L]

➀ ➁ ➂ ➃

289 289 478 478

1 1.55 9 5.83

5.46 3.52 1 1.55

= (1.3)−0.4/1.4 (373 K) 346 K = 73 ◦ C ,

for the conditions of Example 10.2.

10.2

[T] Derive eq. (10.10).

Alternatively, compare the electric power flowing through a material due to the Seebeck effect to the thermal power (losses) flowing through the same material of length ∆L and cross sectional area ∆A. For the electric power

Substitute pV γ = constant = p1 V1γ into the equation for the work, Z V2 Z V2 V γ 1 dV p1 dV p = W= V V1 V1

Pelectric = V 2 /R = (|E|∆L)2 /(∆L/σ∆A) ≈ S 2 ∆T 2 σ∆A/∆L ,

p1 V1γ 1−γ = V − V11−γ 1−γ 2 !γ p1 V2 V1 p1 V1 − p2 V2 p1 V1 − . = = γ − 1 γ − 1 V2 γ−1

where we used |∇T | ≈ ∆T/∆L. The power flowing due to thermal conductivity is given by Fourier’s law, Pthermal = k|∇T |∆A ≈ k∆T ∆A/∆L , and

The second line of eq. (10.10) follow from substituting the ideal gas law, pV = NkB T .

10.3

[T] In isobaric heating (and expansion), heat is added reversibly to a gas kept at constant pressure. Find the change in volume and entropy when a sample of an ideal, monatomic gas, initially at temperature T 1 , volume V1 , and pressure p is heated isobarically to T 2 . Plot the system’s path in the pV- and S T -planes.

Z≡

From the Wiedemann-Franz-Lorenz law, k = σLT , so Z = (S 2 /L)(∆T/T ) .

10.5

From the ideal gas law, pV1 = NkB T 1 and pV2 = NkB T 2 , so V2 = (T 2 /T 1 )V1 .

To find the change in the entropy, we integrate dQreversible /T from T 1 to T 2 , using dQ| p = C p dT = 25 NkB dT , S2 − S1 =

5 NkB 2

Z T2 T1

dT 5 = NkB ln(T 2 /T 1 ) .

T 2

The path in the pV-plane is a horizontal straight line from V1 to V2 = (T 2 /T 1 )V1 at constant p. To find the path in the S T -plane, we need T (S ), which is obtained by solving the equation above for T , T = T 1 exp ((S − S 1 )/((5/2)NkB )) .

The plot resembles Figure 10.7 except that the temperature rises more slowly with entropy.

10.4

[T] Box 10.1 describes a thermoelectric generator and mentions that the materials used should have large Seebeck coefficient S , small poor thermal conductivity k, and good electrical conductivity σ. Use dimensional analysis to show that Z = σS 2 ∆T/k is the unique dimensionless figure of merit for the material for thermoelectricity. Show that if the material is a metal that obeys the Wiedemann–Franz–Lorentz law (see Box 6.1) then Z = (S 2 /L)(∆T/T ), so that the Seebeck coefficient is the critical parameter.

From their definitions: [σ] = 1/mΩ, [S ] = V/K, [k] = W/mK . Let Z = σa S b kc ∆T d , use Ω = V2 /W and find a, b, c, and d to cancel all dimensions. Solution: b = 2a, c = −a, d = a. Choose a = 1 to get quoted result.

Pelectric ≈ σS 2 ∆T/k .

Pthermal

10.6

[T] Show that the changes in entropy found for isothermal and isometric heating of an ideal gas (eqs. (10.5) and (10.12)) agree with the prediction from the Sackur– Tetrode equation (8.65).

According to eq. (8.65), S = kB N((3/2) ln T + ln V + constants), where “constants” are terms that do not vary with T or V. From this general formula, for isothermal heating, T 1 = T 2 , ∆S = kB N(ln V2 − ln V1 ). For isometric heating V1 = V2 , ∆S = (3/2)kB N(ln T 2 − ln T 1 ) = CV ln(T 2 /T 1 ).

Consider a Carnot engine cycle operating between a maximum temperature of T 3 = T 4 = 400 ◦ F and a minimum temperature of T 1 = T 2 = 60 ◦ F. Assume that the lowest pressure attained is p1 = 1 atm, the highest pressure is p3 = 9 atm, and the minimum volume of the working gas space is V3 = 1 L. Take γ = 1.4. Compute the pressure and volume at each point in the cycle, as well as the heat in, heat out, and work out. Compute the overall efficiency from these numbers and check that this agrees with theoretical Carnot efficiency.

It’s useful to tabulate the given data (see Table 10.1). The points are labeled as in Figure 10.8, where we have converted 400 ◦ F→ 478 K and 60 ◦ F→ 289 K. To find the other entries in the table we use the adiabatic gas law between points ➁ and ➂ and between ➃ and ➀, and the ideal gas equation of state between points ➀ and ➁ and between ➂ and ➃.

• First, use T V γ−1 = constant and pV γ = constant between ➁ and ➂ to find V2 = (T 3 /T 2 )1/(γ−1) V3 = 3.52 L p2 = (V3 /V2 )γ p3 = 1.55 atm .

232

• Then, find V1 from the ideal gas law pV = constant between ➀ and ➁,

Both engines achieve the Carnot efficiency limit, so η = (500 − 300)/500 = 0.4 for both. The ratio of Carnot work to Stirling work was computed in the previous problem,

V1 = (p2 /p1 )V2 = 5.46 L .

WCarnot 1 1 = 1− ln(T + /T − ) = 1− ln(5/3) = 0.53 .

WStirling (γ − 1) ln r 0.4 ln(3/0.2)

• Next, find V4 and p4 from the adiabatic gas law between ➀ and ➃,

It remains to compute the work done by either engine. The Stirling engine work is easiest, !

T+ − 1 p1 V1 ln(V3 /V1 ) WStirling = nR(T + − T − ) ln(V3 /V1 ) = T−

V4 = (T 1 /T 4 )1/(γ−1) V1 = 1.55 L p4 = (V1 /V4 ) p1 = 5.83 atm . γ

Then, with the convention that W is work done by the system and Q is heat added to the system, W{12} = Q{12} =

Z V2

dV p = nRT 1

Z V2

dV/V = p1 V1 ln(V2 /V1 )

V1

V1

= (2/3)(101 325)(3 × 10−3 ) ln(15) 550 J .

10.9

= (1 atm)(101 325 Pa/atm)(5.46 L)(0.001 m3 /L) ln(3.52/5.46) = −243 J , Z V4 dV p = p3 V3 ln(V4 /V3 ) W{34} = Q{34} = V3

= (9 atm)(101 325 Pa/atm)(1 L)(0.001 m3 /L) ln(1.55/1) = 400 J .

Finally, we compute the efficiency directly, η=

W{34} + W{12} = (400 − 243)/400 = 0.39 , Q{34}

and compare with the Carnot limit, ηCarnot = (T 3 − T 1 )/T 3 = (478 − 289)/478 = 0.39 , and note that they agree.

10.7

[T] Stirling engines generate more work per cycle than Carnot engines operating under the same conditions.

Find the ratio WCarnot /WStirling for engines run with the same compression ratio, r = V1 /V3 , and temperature ratio T + /T − . Assume the working fluid is an ideal gas.

[H] Consider the cycle proposed in Question 10.3 quantitatively. Assume that it is executed reversibly.

Show that its efficiency η relative to the efficiency ηC of a Carnot engine operating between the same temperature limits is given by !

1 r η ln r , = 1− ηC ĉ p r − 1 r−1

where ĉ p = C p /NkB is the heat capacity per molecule at constant pressure and r = V1 /V2 is the compression ratio. Plot the efficiency ratio versus r. Why is it always less than one?

Labeling the steps as described in Question 10.3, the efficiency is given by η = Wtotal /Q{23} , where Wtotal = Q{23} + Q{12} + Q{31} .

(Q is positive for heat added to the system, so Q{12} and Q{31} are both negative.) Since {12} is isothermal, p1 V1 = p2 V2 = NkB T − , and by the conditions that define the cycle, p2 V1 = NkB T + , so T + /T − = V1 /V2 = r. Then Q{23} = NkB ĉ p (T + − T − ) ,

Q{31} = NkB ĉV (T − − T + ) ,

Both engines have Carnot efficiencies so the work done in both cases is W = ηCarnot Q+ . Thus ln (V4 (Carnot)/V3 (Carnot)) WCarnot = .

WStirling ln V4 (Stirling)/V3 (Stirling)

Referring to Figures 10.8, 10.11, and 10.13, V3 (Carnot) = V3 (Stirling) = V3 , but V4 (Carnot) = V1 (T − /T + )1/(γ−1) , whereas V4 (Stirling) = V1 . Thus

V 1 ln TT +− ln V1 + γ−1 1 WCarnot 3

=1− = ln(T + /T − ) V1 WStirling (γ − 1) ln r ln V 3

10.8

Mirrors are used to concentrate sunlight and heat a molten salt mixture to 500 K. A heat engine is then used to convert the thermal energy to useful mechanical form. Compare a Carnot engine to a Stirling engine with the same operating parameters. In each case the engine operates between the maximum temperature of T + = 500 K and a minimum temperature of T − = 300 K (ambient temperature). Maximum volume for the engine in each case is 3 L, minimum volume is 0.2 L. Assume that the minimum pressure attained in the cycle is 1 atm, and the working fluid is air with γ = 1.4. Compute and compare the efficiency and net useful work per cycle for the two engines.

Q{12} = NkB T − ln(V2 /V1 ) .

Note that ĉ p = ĉV + 1. Then ĉ p (T + − T − ) − ĈV (T + − T − ) − T − ln r ĉ p (T + − T − ) ! ln r 1 1− . = ĉ p r−1

η=

Combine this with the definition of the Carnot efficiency, ηC = (T + − T − )/T + = (r − 1)/r, to obtain the quoted answer.

No figure.

This cycle cannot possibly reach the Carnot limit because heat is added at a range of temperatures that are below T + .

10.10 [T] In section §5.2.2, we gave an example of a sit-

uation where an ideal gas is heated to a temperature T in and allowed to expand a small amount against an external ambient pressure, where the external gas has temperature T out . We found that the fraction of thermal energy dU used to do useful work dW precisely realized the Carnot efficiency. Show that the expansion process described there can be completed to a closed loop giving a heat engine with Carnot efficiency by using reversible processes to bring the fluid back to the initial volume so that addition of thermal energy dU restores the initial temperature T in . [Hint: you may find it helpful to make use of a regenerator to store thermal energy at a continuous range of temperatures.]

Problems

10.11 An air conditioner run on electricity and based on

a (ideal) Carnot cycle operates between temperatures T − = 25 ◦ C and T + = 40 ◦ C. The gas in the air conditioner has γ = 1.75. The AC is designed to remove 10 000 BTU/h from a living space. How much electric power does it draw? Suppose the gas in the AC has a pressure of 2 atm at its maximum volume, which is 1 L, and the minimum volume is 0.4 L. At what rate must the AC cycle in order to accomplish its mission? [Hint: See Example 10.5.]

The Carnot limit on the CoP is T − /(T + − T − ) = 298/15 19.9.

Cooling power is Pcool = 104 BTU/h = (104 BTU/h)(1.055 kJ/BTU)/(3.6 × 103 s/h) 2 930 W .

So the required electric power is P = 2 930/19.9 147 W. To find the cycling rate, we compute the heat drawn from the cold reservoir in one cycle, Qin = W{21} = nRT − ln(V1 /V2 ) = p1 V1 ln(V1 /V2 ). We are given p1 = 2 atm and V1 = 10−3 m3 . To find V2 use the adiabatic relation, T − V2γ−1 = T + V3γ−1 , 1/(γ−1) V2 = (T + /T − ) V3 = (313/298)1.33 (0.4 × 10−3 L) = 0.43 L .

So Qin = (2 atm)(101 325 Pa/atm)(10−3 m3 ) ln(1/.43) 170 J .

The cycling rate r is given by the cooling power divided by the heat absorbed per cycle, r = Pcool /Qin 2 930 W/(170 J/cycle) 17 cycles/s .

10.12 Air conditioners are rated by their Energy Effi-

ciency Ratio, or EER, defined as the cooling power Pr (in Btu/h), divided by the total electric power input Pe (in watts), measured with T + = 95 ◦ F and T − = 80 ◦ F: EER = Pr [BTU/h]/Pe [W]. US energy efficiency standards require central air conditioners to have an EER value greater than ≈ 11.7.6 What is the CoPr of an air conditioner with EER = 12? How does it compare with the Carnot limit?

The CoP of an air conditioner is the dimensionless ratio of the cooling power to the energy input, CoP = Pcool /Pe . Since Pr is given in BTU/h, we convert to watts, Pcool = Pr (BTU/h)(1 055 J/BTU)/(3 600 s/h) = 0.293Pr W/h .

So CoP = 0.293 EER. For EER=12, CoP = 0.293(12) 3.5. The Carnot limit for T + = 95 ◦ F = 308 K and T − = 80 ◦ F = 300 K is CoP|Carnot = 300/8 = 37.5 .

So the CoP is less than 1/10th of the Carnot limit.

6 The US uses a seasonal average EER (SEER) which is only

approximately related to the EER, and requires SEER ≥ 13.

233

250

Problems 11.1

The efficiency of the Otto cycle is given by

Assume that an SI engine has the following parameters: displacement (V1 − V2 ): 2.4 L; compression ratio 9.5:1; air to fuel mass ratio 15:1; heating value of fuel 44 MJ/kg, pressure at start of compression 90 kPa, intake temperature 300 K. Compute the pressure, volume, and temperature at each of the points ➀, ➁, ➂, and ➃ in the idealized cold air standard Otto cycle.

Compute the work done per cycle, and estimate the engine efficiency in this idealized model. Compute the power at 3000 rpm.

η=1−

From eq. (11.7), r = 1 − T 1 /T 2 . T 1 is typically just the outside temperature (300 K) and T 2 is the temperature immediately before combustion.

Heat Q is only added to the engine during the isometric heating step, so the work done is just given by W = ηQ. The system is heated isometrically from T 2 to T 3 , so Q = CV (T 3 − T 2 ). From the equation for efficiency, T2 =

CV =

T3 − T2 =

W = ηCV (T 3 − T 2 ) = (0.49)(2.2 J/K)(1 800 K − 592 K) = 1.3 kJ .

If the engine runs at 5 000 rpm, then the total power output of the engine is

p2 V2 = T 1 rγ−1 = 738 K. p1 V1

Q Q Q = 5 = 4 170 K = 5 CV nR pV/T 2 2

so T 3 = 4910 K, and p3 = (T 3 /T 2 )p2 = 14.0 MPa. After adiabatic expansion, the final pressure is p4 = p3 r−γ = 0.60 MPa, and the temperature is T 4 = T 3 r−γ+1 = 2000 K. The work done per cycle is W = CV (T 3 −T 2 −T 4 +T 1 ) = 25 pV T (2470) K= 4.94 kJ. The efficiency is η = 1 − r−γ+1 = 0.59. At 3 000 rpm, the power generated by this cycle is 250 kW, which is about 335 horsepower.

11.2

The cold air standard value of γ = 1.4 was based on the heat capacity of a diatomic gas with no vibrational excitation (see §9), CV = 52 nR and C p = CV + nR. In reality, the heat capacity increases with increased temperature, and γ decreases accordingly. If we assume, however, that CV is independent of temperature through the cycle and that C p = CV + nR remains valid, what value of CV is required for γ = 1.3?

γ=

Cp nR =1+ .

CV CV

nR Solving for CV and plugging in γ = 1.3, CV = γ−1 = 10 3 nR.

11.3

An SI engine, modeled as an ideal Otto cycle, runs at a compression ratio of 9.6:1 with a maximum cylinder volume (V1 ) of 2 L and a corresponding displacement of 1.8 L. Combustion leads to a maximum temperature of T 3 = 1800 K. Using the air standard value of γ = 1.3, find the engine efficiency and the work per cycle. Compute the engine’s power at 5000 rpm.

Assume that the intake temperature and pressure are 300 K and 100 kPa. Note: you will need the value of CV from Problem 11.2.

10 10 10 p1 V1 10 (100 000 Pa)(2 L) nR = NkB = = = 2.2 J/K .

3 3 3 T1 3 300 K

So, the amount of work per cycle is

The ideal gas law tells us that

The pressure is dominated by the air, so the volume contains approximately 0.097 mol, or about 2.8 g. This corresponds to 0.19 g of fuel, so a total of Q = 8.4 kJ of energy is added per cycle. This tells us that

300 K T1 = = 592 K .

1−η 1 − 0.493

If the cycle starts at standard temperature and pressure, then

In this problem, V1 − V2 = 2.4 L while V1 /V2 = 9.5. Solving for V1 and V2 , V1 = 2.68 L and V2 = 0.28 L. The temperature starts at room temperature and pressure, T 1 = 300 K and p1 = 90 kPa. Using the cold air standard value of γ = 1.4, adiabatic compression requires that pV γ be constant. So, !γ V1 p2 = p1 = p1 (9.5)1.4 = 2.11 MPa. V2

T2 = T1

1 = 0.493 . rγ−1

P = (1.3 kJ/cycle)(5 000 cycles/min)(1 min/60 sec) = 110 kW .

11.4

[H] (Requires results from Problem 11.3) Consider the same engine as in Problem 11.3 but now run as an Atkinson cycle. The volume after expansion (V4 ) is still 2 L, but the volume before compression (V1b ) is 1.54 L. The compression ratio is still 9.6:1, so the minimum volume is V2 = 0.16 L. What is the expansion ratio? Assuming that the amount of air and fuel are both reduced by the ratio of V1b /V4 , find the work done in expansion, W{34} , the work required by compression, W{1b 2} , and the energy input from combustion Q{23} . Compute the efficiency and the engine’s power at 5000 rpm.

The expansion ratio is now 2 : 0.16 = 12.5 : 1. Since the amounts of air and fuel are both reduced by the same ratio and the compression ratio does not change, the change in temperature T 3 − T 2 remains the same (1208 K). Thus T 2 = 592 K. The heat capacity is reduced by V1b /V4 to CV = 10 3 nR = 1.7 J/K. The total heat added is now Q[23] = CV (T 3 − T 2 ) = 2050 J. The work done in isobaric compression is W[1 1b] = p1 (V1b − V1 ) = (100 kPa)(2 L − 1.54 L) − 46 J. The adiabatic compression step remains the same as in Problem 11.3. The pressure is p1b = 1 atm and the temperature is T 1b = 300 K. After the adiabatic compression step, !γ V1b p2 = p1b = 1.87 MPa. V2 Sincd T 2 = 592 K and T 3 = 1800. The pressure after ignition is p3 = (T 3 /T 2 )p2 = 5.69 MPa. The work done in expansion is   Z V4 p3 V3γ  1 1  dV γ   p V = − W[34] =  3 3  γ γ − 1 V γ−1 V γ−1  V3 V 3 4   !γ−1  V3 p3 V3  0.3 1 − =  = 3.03(1 − (1/12.5) ) = 1.62 kJ. γ−1  V4

251

Problems

The work done in the adiabatic compression, likewise, is

p2 V2 W[1b 2] = 1 − (1/r)γ−1 = −0.49 kJ. γ−1

The net work done in the cycle is

W = W[34] + W[1 1b] + W[1b 2] = (1.62 − 0.046 − 0.49)1.08 kJ so the efficiency is η=

W = 53% Q[23]

and the engine’s power at 5 000 rpm is P=W×

11.5

5 000 rpm = 90 kW. 60 s/min

(Requires results from Problem 11.1) Consider a throttled Otto cycle for an engine with the same parameters as Problem 11.1. In the throttled cycle assume that the spent fuel–air mixture is ejected at 1 atm and brought in again at 0.5 atm. Compute the work done per cycle.

Compute the associated pumping loss. Subtract from the work per cycle and compute the engine efficiency with throttling.

Since the cycle starts at only 0.5 atm (compared to 0.9 atm in Problem 11.1), the volume contains only 0.054 mol, or 1.6 g of air or 0.107 g of gasoline. This means that 4.7 kJ of energy is added per cycle. In the idealized cycle, the efficiency is 0.59, so the work done per cycle is W = 4.7 kJ × 0.59 = 2.8 kJ. In the throttled cycle, the pumping loss is equal to Wloss = (V1 − V2 )(pexhaust − pcompress ) = (0.5 atm)(2.4 L) = 120 J. With throttling, the net work per cycle is (2.8-0.12)=2.7 kJ, giving an efficiency of 0.57.

11.6

Consider the Otto and Diesel cycles shown in Figure 11.13. The parameters have been chosen so both cycles have the same heat input and the same low temperature and pressure set point. Explain why the difference between the Diesel cycle efficiency and the Otto cycle efficiency is proportional to the area of the green shaded region.

For the Diesel cycle ηD = 1 − Q[41] /Q[23] , while for the Otto cycle ηO = 1 − Q[4′ 1] /Q[2′ 3′ ] and Q[23] = Q[2′ 3′ ]R by construction.

So ηD − ηO = (Q[41] − Q[4′ 1] )/Q[23] . Since Q = T dS , the green area equals (Q[41] −Q[4′ 1] ) and thus is proportional to the difference in efficiency.

11.7

A marine diesel engine has a compression ratio r = 18 and a cutoff ratio rc = 2.5. The intake air is at p1 = 100 kPa and T 1 = 300 K. Assuming an ideal cold air standard Diesel cycle, what is the engine’s efficiency?

Use the ideal and adiabatic gas laws to determine the temperature at points ➁, ➂, and ➃. Compute the work per cycle. What is the power output of this engine if its displacement is 10 L and it runs at 2000 rpm? What is its fuel consumption per hour (assuming the energy content of diesel fuel to be 140 MJ/gal)?

The efficiency can be computed directly from eq. (11.13), η = 0.61. Since T V γ−1 is constant in adiabatic expansion, T 2 = rγ−1 T 1 T 4 = (rc /r)γ−1 T 3

11.8

and the ideal gas law gives T 3 = (V3 /V2 )T 2 = rc T 2 . From the given values of r, rc , and T 1 we find T 2 = 950 K, T 3 = 2 400 K, and T 4 = 1 080 K. The work per cycle is W = Q[23] − Q41 = nCV (γ(T 3 − T 2 ) − (T 4 − T 1 ) = (5/2)nR(1250 K). With a displacement of 10 L, so that V1 = 10.6 L, the engine contains n = p1 V1 /RT 1 = 0.42 moles of air/fuel mixture, so the work per cycle is W = (2.5)(0.42)(8.314)(1250) = 10.9 kJ/cycle. The heat added per cycle is W/η = 17.9 kJ/cycle. Running at 2 000 rpm, the power output is P = 10.9(2 000/60) = 367 kW and the fuel consumption is (P/(η(140 MJ/gal)) = 600 kW/140 MJ/gal = 4.3 × 10−3 gal/s = 15.6 gal/h.

Suppose two engines, one SI, the other CI, have the same temperature range, T 1 = 300 K and T 3 = 1800 K. Suppose the SI engine, modeled as an ideal Otto cycle, has a compression ratio of 10:1, while the CI engine, modeled as an ideal Diesel cycle, has twice the compression ratio, 20:1. Using cold air standard assumptions, what is the efficiency of each engine? How much more efficient is the diesel engine? [Hint: the fuel cutoff ratio, rc can be computed as a function of r and T 3 /T 1 .]

The efficiency of the Otto cycle is ηO = 1 − rO1−γ = 0.60.

For the diesel cycle, adiabatic compression requires that T 2 = T 1 (V1 /V2 )γ−1 = 990 K. The ideal gas law tells us that for isobaric expansion, rc = V3 /V2 = T 3 /T 2 = 1.8, so the efficiency of this diesel cycle is ! rcγ − 1 1 ηD = 1 − γ−1 = 0.655. γ(rc − 1) r The Diesel engine is about 9% more efficient than the Otto cycle.

268

Problems

12.4

Note: for a number of the problems in this chapter you will need to use steam table data from a source such as [23] or the equivalent.

12.1

The isothermal compressibility of a fluid is the relative change of volume with pressure at constant temperature β = −(1/V)∂V/∂p|T . Qualitatively describe and sketch the compressibility of water as a function of temperature from 0 ◦ C to 1000 ◦ C, at a pressure of 1 atm and at the critical pressure pc 22.06 MPa. What is the compressibility at the phase transition from liquid to vapor?

At 1 atm, water is liquid below 100 ◦ C, and has relatively small compressibility (β ∼ 10−10 Pa−1 ). At 100 ◦ C the volume changes by a finite amount with no change in pressure, so the compressibility is infinite. Once the water has vaporized it behaves approximately like an ideal gas, so β = 1/p 10−5 Pa−1 at 1 atm.

The compressibility at the critical pressure can be sketched on the basis of Figure 12.7, where it is related to the negative of the slope of the isotherms. At low temperature the water is liquid with small compressibility. As the temperature approaches the T c 647 K, the compressibility grows and becomes infinite at T c .

Above T c the compressibility drops back eventually to its ideal gas value β → 1/p.

The compressibility is always infinite at the liquid-vapor phase transition (below the critical temperature).

12.2

Consider a volume of 100 L of water, initially in liquid form at temperature T i , to which ∆U = 25 MJ of energy is added from an external reservoir at temperature T r through a thermal resistance, so that the rate of energy transfer is Q̇ = (1 kW/K)(T r − T ), where T is the instantaneous temperature of the water. The energy is added in two different ways: (a) with the water at initial temperature T i = 100 ◦ C and the external source at temperature T r = 150 ◦ C; (b) with T i = 0 ◦ C, T r = 75 ◦ C. In each case, compute the time needed to transfer ∆U and the extra entropy produced in this process.

Do the comparative results match with the expectation from the discussion of energy transfer at constant temperature in §12.1? (You may assume a constant heat capacity for liquid water, independent of temperature.) a) ta =

25 MJ ∆U

500 s . k(T r − T i ) 50 kW

The extra entropy produced equals the entropy added to the fluid minus the entropy removed from the reservoir, ∆S a = ∆U

Check that the ∆svap and ∆hvap quoted in Table 12.1 are consistent with ∆h = ∆u + p∆v and ∆s = ∆h/T .

b) The final temperature is T f = ∆U/C 60 ◦ C. The temperature as a function of time is given by

At 105 Pa,

dQ dT =C = k(T r − T ) . dt dt

∆h = 2 257.5 kJ/kg ∆u + p∆v = 2 088.2 kJ/kg + (105 )(1.6929) J/kg = 2 257.5 kJ/kg

The solution to this differential equation is

∆s = 6.0560 kJ/kg K ∆h/T = (2 257.4 kJ/kg)/(99.61 + 273.15) K = 6.056 kJ/kg K.

t=

5

Likewise, at 1.2 × 10 Pa,

∆s = 5.9368 kJ/kg K

∆S =

∆h/T = (2 243.7 kJ/kg)/(104.78 + 273.15) K = 5.9368 kJ/kg K.

a) From the table, the enthalpy per kilogram at 90 ◦ C is 377.06 kJ and at 0 ◦ Cis 0.06 kJ, so ∆H 377 kJ. b) The table does not give the specific enthalpy of liquid water at either 95 ◦ C or 100 ◦ C, however it is easy to check that the specific enthalpy is increasing at 42 kJ every 10 ◦ C in the range from 60 ◦ C to 90 ◦ C. Extrapolating we expect that 21 kJ is added in heating the water from 95 ◦ C to 100 ◦ C. A similar calculation give 20 kJ, added in heating the vapor from 100 ◦ C to 105 ◦ C. The specific enthalpy of vaporization ∆hvap = 2 257.4 kJ/kg dominates. The total enthalpy increase of 1 kg of water is ∆H 2 257 + 21 + 20 2 298 kJ. c) At this pressure, water remains a liquid up to 104.78 ◦ C, so heating a kilogram of water to 105 ◦ C requires the enthalpy of vaporization, 2 244 kJ, plus the enthalpy needed to heat liquid water from 95 ◦ C to 105 ◦ C, which we estimate to be 42 kJ from an analysis similar to part b). So the total enthalpy increase is ∆H = 2 244 + 42 2 286 kJ.

!

or

−kt/C

Solving for the time needed to reach temperature 60 ◦ C, we find tb 666 s. Total entropy transferred is

∆u + p∆v = 2 072.5 kJ/kg + (1.2 × 105 )(1.4274) J/kg = 2 243.8 kJ/kg

Using the data given in Table 12.1, estimate the enthalpy added to one kilogram of water that undergoes the following transformations: (a) from 0 ◦ C to 90 ◦ C at p = 105 Pa; (b) from 95 ◦ C to 105 ◦ C at p = 105 Pa; (c) from 95 ◦ C to 105 ◦ C at p = 1.2 × 105 Pa.

Tr − Ti C ln k Tr − T f

T (t) = T r + (T 0 − T r )e

∆h = 2 243.7 kJ/kg

12.3

!

1 1 − 8 kJ/K .

Ti Tr

Z Tf Ti

!

Tf dU ∆U ∆U ∆U − = ln 11 kJ/K − T Tr T f − Ti Ti Tr

This agrees with the expectation from the discussion in the text: method b) requires more time and is less efficient.

12.5

Data on the heat flux for laminar flow of liquid water and for pool boiling of water are shown in Figure 12.4.

Take the bulk temperature of the fluid in the case of laminar flow to be T 0 = 25 ◦ C and assume that the pool boiling process takes place at 1 atm. (a) Suppose the two methods are used to transfer heat at a rate of q = 105 W/m2 . Estimate the rate of extra entropy production (W/m2 K) for the two methods. (b) Estimate the heat flux for pool boiling when its rate of entropy production is the same as the rate of entropy production for laminar flow in part (a). a) From Figure 12.4(a) T − T 0 ≈ 10 K when q = 1 × 105 W/m2 .

Thus T ≈ 308 K and !

1 1 dS − ≈ 11 W/m2 K . =q dA dt laminar T0 T

269

Problems

From Figure 12.4(b), T − T 0 ≈ 3.5 ◦ C when q = 1 × 105 W/m2 .

Since T 0 = 100 ◦ C for boiling at 1 atm, ! dS 1 1 − ≈ 2.5 W/m2 K . =q dA dt boiling 373 376.5 b) This requires estimating and interpolating on Figure 12.4(b).

The second two data points are approximately (T − T 0 , q) = (5 K, 2 × 105 W/m2 ) and (T − T 0 , q) = (6.2 K, 3.5 × 105 W/m2 ).

These give dS /dA dt ≈ 7 W/m2 K and dS /dA dt ≈ 15 W/m2 K respectively. So we estimate that q ∼ 2.5 − 3 × 105 W/m2 , when the rate of entropy production is ∼ 11W/m2 K. Thus, under somewhat similar conditions boiling heat transfer is either more efficient at the same rate or more rapid at the same efficiency than single phase heat transfer.

12.6

An industrial freezer is designed to use ammonia (NH3 ) as a working fluid. The freezer is designed so that ammonia flowing through tubes inside the freezer at p− vaporizes at T = −40 ◦ C, drawing heat out of the interior. Outside the freezer, the ammonia vapor at T = +45 ◦ C liquefies at a pressure p+ dumping heat into the environment. Find thermodynamic data on ammonia and identify the pressures p± that enable the liquid–vapor phase transition to take place at the required temperatures.

Thermodynamic data on ammonia come from [23], where we find pvap (−40 ◦ C) = 71.7 kPa and pvap (45 ◦ C) = 1.78 MPa.

12.7

12.8

12.9

Using the data given in Table 12.1, find the deviations from the ideal gas law for water vapor just above the boiling point at p = 105 Pa. For example, does V/n = RT/p?

According to Table 12.1, at 99.51 ◦ C and p = 105 Pa, steam has a volume of 1.6959 m3 /kg. The molecular weight of water is 1.8015 × 10−2 kg/mol. So the molar volume is V/n = (1.6959 m3 /kg)(1.8015 × kg/mol) or 0.030555 m3 /mol. The ideal gas law states that V/n = RT/p = (8.3145 J)(273.15 + 99.51) K/105 Pa = 0.030985. So the deviation from the ideal gas law is (0.030985 − 0.030555)/0.030985 = 0.01388 or about 1.4%.

Revisit Question 12.2 quantitatively. Specifically, what is the pressure in the cylinder when the system comes to equilibrium after the volume has been doubled at −20 ◦ C? What is the pressure after the cylinder has been heated to 110 ◦ C? What fraction of the water in the cylinder is in vapor form at −20 ◦ C and at 110 ◦ C?

At −20 ◦ C, the vapor pressure of ice is p = 103.26 Pa. At T = 110 ◦ C, the vapor pressure of water is 143.3 kPa. We assume that the volume of ice that sublimes is small so that the cylinder contains 1 L of ice and 1 L of water vapor at p = 103.26 Pa. To find the mass of the water vapor, we apply the ideal gas law with V = 1 L, T = 253 K and p = 103.26 kPa, and find m = 8.84 × 10−7 kg of vapor. Since the total mass is ≈ 1 kg, the fraction in vapor is ≈ 9×10−7 . At 110 ◦ C, the specific volume of steam is 1.211 m3 /kg, so the density of steam is 0.826 kg/m3 . The contribution of vapor to the mass is very small and the density of liquid water is still ≈ 1 kg/L, so to good approximation, we can take the system to comprise 1 L of liquid and 1 L of vapor. Once again using the ideal gas law, we find the mass in vapor form is 0.826 × 10−3 kg, and the fraction in vapor form is approximately 8 × 10−4 .

As explained in the following chapter (§13), a steam turbine can be modeled as an isentropic expansion. The incoming steam is superheated at an initial temperature T + and pressure p+ . After expansion, the exhausted steam is at temperature T − . A high-performance turbine

cannot tolerate low-quality steam because the water droplets degrade its mechanical components. Suppose that T − = 40 ◦ C and that the turbine requires χ ≥ 0.9 throughout the expansion process. If the high-pressure system is rated to 5 MPa, what is the minimum possible initial temperature of the steam?

This problem requires the use of a steam table or online equivalent. At 40 ◦ C, a χ = 0.9 mixture has s = 7.490 kJ/kg K. At 5 MPa we search for the minimum temperature with the same specific entropy, and find that value of the specific entropy requires T + > 691 ◦ C.

12.10 Use steam table data to estimate accurately the pres-

sure at which water would boil at room temperature (20 ◦ C). Similarly estimate the temperature at which water boils on top of Mt. Everest, where the air pressure is approximately 1/3 atm?

12.11 A sample of H2 O at a pressure of 1 atm has a specific

enthalpy of 700 kJ/kg. What is its temperature? What state is it in, a sub-cooled liquid, superheated vapor, or a mixed phase? If it is in a mixed phase, what fraction is liquid?

The data in Table 12.1 includes p = 1 × 105 Pa, which is close enough to 1 atm for the accuracy of this problem. h = 700 kJ/kg is between the specific enthalpy of saturated liquid and saturated vapor (column 4 of the table, first and third row), so this H2 O is in the mixed liquid-vapor phase at the boiling point at 1 atm, which is T = 100 ◦ C. At 105 Pa and 100 ◦ C, the specific enthalpy of liquid water is 417.4 kJ/kg, while the specific enthalpy of water vapor is 2674.9 kJ/kg. Using eq. (12.3) we require that 700 = 417.4(1 − χ) + 2674.9χ = 417.4 + 2257.5χ Solving for χ, χ = 0.125, so 7/8 of the sample is liquid.

12.12 Repeat the preceding question for a sample at 10 atm

with the same specific enthalpy.

For p = 10 atm, Table 12.1 does not apply, and the data must be found elsewhere. Most steam tables list the properties of saturated liquid and vapor as a function of temperature. At p = 10 atm is one finds h = 765.1 kJ/kg. Since the sample as a lower specific enthalpy, it must be a subcooled liquid. Online tables of the properties of subcooled water (for example [23]) allow one to back out the temperature at which h = 700 kJ/kg, with the result T 167 ◦ C.

12.13 On a hot summer day in Houston, Texas, the day-

time temperature is 35 ◦ C with relative humidity φ = 50%. An air conditioning system cools the indoor air to 25 ◦ C. How much water (kg/m3 ) must be removed from this air in order to maintain a comfortable indoor humidity of 40%? Compare your answer with the result of Example 12.1.

At 35 ◦ C, pvap = 5.63 kPa, so at 50% humidity, the partial pressure of water vapor is p35 = (0.5)(5.63 kPa) = 2.82 kPa. Using the ideal gas law, n p 2.82 × 103 Pa = = 1.14 mol/m3 .

V RT (8.31 J/mol K)(308 K) At 25 ◦ C, pvap = 3.17 kPa, so at 40% humidity the partial pressure of water vapor is p25 = (0.4)(3.17 kPa) = 1.27 kPa. Again calculating n/V, we find n/V = 0.513 mol/m3 . So, 1.14 − 0.51 = 0.63 mol of water with mass m = (0.63 mol)(0.018 kg/mol) = 0.011 kg must be removed from each cubic meter of air. The result is about twice

270

the result of Example 12.1. This reflects the fact that air holds more water at high temperature (this problem) than at low temperature (the example).

296

If you took T 2 ≈ T + , then h3 = 419.4 kJ/kg, COP ≈ 6.39 (11% too high), and ṁ = 0.307 kg/s (2% too high).

Table 13.5 Data for Problem 13.3.

p (MPa)

T ( ◦ C)

χ

h (kJ/kg)

s (kJ/kg K)

0.2169 0.2169 1.0166 1.0166

−8 −8 40 40

0 1 0 1

189.3 393.9 256.4 419.4

0.9606 1.732 1.191 1.711

13.4

The Carnot limit for the temperatures (T + = 95 ◦ F, T − = 80 ◦ F) in §10 Problem 10.12 is COPcarnot 36. So we assume COPr 12 corresponding to 12 BTU/h of cooling power for 1BTU/h = (1.055 × 103 J)/(3.6 × 103 s) = 0.293 W of electric power input.

Using the definition of EER (EER = Pr [BTU/h]/Pe [W]), we find EER = 12/0.293 41, far better than the value 11.7 required by US efficiency standards.

Problems 13.1

[T] Prove that the Carnot-like vapor-compression cooling cycle {1234} in Figure 13.5(a) has a maximum CoP that equals the Carnot limit regardless of the properties of the working fluid. Likewise, prove that the efficiency of the simplest Rankine power cycle {1234} of Figure 13.12(a) equals the Carnot limit.

Any cycle in which is run reversibly and in which all heat is transferred in or out of the system at the maximum and minimum temperatures will reach the Carnot limit. Explicitly for the cycle of Figure 13.5(a), the 2nd Law gives Q[23] /T + = Q[41] /T − and the 1st Law gives W[12] − W[34] = Q[23] − Q[41] , combining these we find COP ≡ Q[41] /(W[12] − W[34] ) = T − /(T + − T − ), which is the Carnot limit.

13.2

[T] Sketch the Carnot-like VC cycle of Figure 13.5(a) in the pV-plane (superimposed on the saturation dome), and contrast the resulting shape with that of the Carnot cooling cycle described in §10.

13.5

A heat pump based on the ideal VC cycle of Figure 13.5(b) uses the refrigerant R-134a to heat a house.

The set points are T − = −8 ◦ C, T + = 40 ◦ C. What is the CoP; how does it compare with the Carnot limit?

What flow rate (ṁ) in kg/s is required to provide heat at 50 kW? Relevant thermodynamic data for R-134a is provided in Table 13.5. Note that since this is an ideal VC cycle, as opposed to the modified cycle analyzed in §13.2.5, the maximum temperature T 2 must be determined. To find T 2 , data in the superheated region are required [23]. Alternatively you can approximate T2 ≈ T+ .

For a heat pump, COP = q[23] /w[12] = (h2 − h3 )/(h2 − h1 ). From the table h1 = 393.9 kJ/kg and h3 = 256.4 kJ/kg. To obtain h2 in the superheat region, we need to find the temperature at which p = 1.0166 MPa and s = s1 = 1.732 kJ/kg K. From [23], we find T 2 43 ◦ C and h2 422.8 kJ/kg. From these we find q[23] 166.4 kJ/kg and w[12] 28.9 kJ/kg, giving COP 5.76 compared to the Carnot limit COPCarnot = T 2 /(T 2 − T − ) = 6.20. To find the mass flow rate, compute the power P = 5.0 × 104 J/s = q[23] ṁ = 1.664 × 105 J/kg ṁ .

Solving for ṁ 0.30 kg/s.

[T] Neglecting the work done by the pump (which is a good approximation), show that the efficiency of the ideal Rankine cycle is η = (h3 − h4 )/(h3 − h2 ), where h j is the specific enthalpy at the point j in the cycle.

The heat input to the cycle is q[23] = h3 − h2 , where we have used the information in Table 13.3. The net work done (ignoring pumping) is w[34] = h3 − h4 , thus the ratio η = q[23] /w[34] gives the required result.

13.6

[T] Sketch the ideal Rankine cycle on a pV-diagram including the saturation dome. Label the points ➀–➃.

As in Figure 12.7, plot the logarithm of the specific volume on the horizontal axis.

[12] is almost exactly constant volume, beginning with ➀ on the saturation dome.

[23] is constant pressure, going from subcooled to superheated vapor.

[34] is constant entropy. In the superheated region pV γ = constant in the superheated region, so p = constant × e−γ ln v between ➂ and the saturation dome close to ➃.

[41] is constant pressure.

In the carnot-like VC cycle, the isothermal steps, [41] and [23], are at constant pressure, as opposed to the gas phase Carnot cooling cycles, in which the pressure and volume change according to pV = NkB T . In the adiabatic step [12] work is done on the system and in the adiabatic step [34] the system does work. This requires lines that slope in the pV plane. The resulting cycle is loosely sketched in the figure. (Note that the volume of the saturated liquid at ➂ is negligible compared to other volumes.)

13.3

The 5-ton AC unit designed in §13.2.5 reached 33% of the Carnot limit on CoP. Look back at the definition of the “Energy Efficiency Ratio” in Problem 10.12, which employs a different temperature range than the one we specified. Assuming that the unit’s fraction of the Carnot limit remains unchanged, compute its EER. How does this compare with the current US standard of EER ≥ 11.7?

This leads to the sketch shown.

13.7

When we designed the Rankine steam cycle in §13.3.5 we ignored the circulating water’s kinetic and potential energy, having claimed earlier (see Box 13.1) that they are negligible. Using the work per unit mass done by the turbine to set the energy scale, estimate the importance of the potential energy of the fluid if the height of the boiler is 40 m? Estimate the importance of the kinetic energy if the maximum flow speed in the pipes is a few (say 5) m/s?

The potential energy of the fluid is of order gh = (9.8 m/s2 )(40 m) ≈ 400 J/kg, while the kinetic energy is of order 1 2 2 v = 12.5 J/kg. The work done per unit mass by the turbine is

297

Problems

13.10 Modify the coal plant Rankine cycle described in

Table 13.6 Data for Problem 13.9.

p (MPa)

T (K)

χ

h (kJ/kg)

s (kJ/kg K)

0.005 0.005 2.5 2.5

306.0 306.0 497.1 497.1

0 1 0 1

137.8 2561 961.9 2800.9

0.4762 8.394 2.554 6.254

1 367 kJ/kg, so the potential energy is no more than 0.03% of this and the kinetic energy is no more than 0.001% of this. Both kinetic and potential energy are negligible compared to the work done and heat exchanged.

13.8

Suppose the condenser in the 500 MWe coal power plant we analyzed in §13.3.5 is cooled by a oncethrough system, where water taken from the ocean is circulated once through the plant and then returned to the ocean. If regulations permit only a 10 ◦ C rise in the ocean water temperature, how much water (in m3 /s) must be used? (Assume that the heat capacity of seawater is the same as that of pure water. Remember that the efficiency of the plant was estimated at 38%.)

The power plant outputs (1 − η)/η × 500 = 816 MW of waste heat. The water can only rise 10 K, and has a specific heat capacity of approximately 4 kJ/kg K. The mass of water used must then be 816 MW dM = = 20 400 kg/s dt (4 kJ/kg K)(10 K)

(13.6)

which is approximately 20.4 m3 /s.

13.9

Quantitatively compare the Carnot Rankine cycle of Figure 13.12(a) with the ideal Rankine cycle of Figure 13.12(b). Compare efficiencies and the work done per cycle. All the necessary information about the ideal Rankine cycle can be found in Example 13.2.

The required information for the Carnot cycle can be obtained from Table 13.6.

The efficiency of the Carnot cycle is ηCarnot = (497.1 − 306.0)/497.1 0.38. The efficiency of the ideal Rankine cycle was computed in the text ηRankine 0.39. The work per unit mass for the Rankine cycle was also computed in the text, wRankine 1 370 kJ/kg. It remains to compute the specific work done by the Carnot cycle. wCarnot = w[34] − w[12] = h3 − h4 − h2 + h1 From the table, h2 = 961.9 kJ/kg and h3 = 2 800.9 kJ/kg. To determine the enthalpy at ➀ and ➃, we must determine the quality at those points. Using the fact that [12] and [34] are adiabatic, 0.476(1 − χ1 ) + 8.394χ1 = 2.554 → χ1 = 0.262

0.476(1 − χ4 ) + 8.394χ4 = 6.254 → χ4 = 0.730 h1 = 0.262hv + 0.738hl = 773 kJ/kg

h4 = 0.730hv + 0.270hl = 1 907 kJ/kg Thus wCarnot = 2 801 − 1 907 − 962 + 773 = 705 kJ/kg and wCarnot /wRankine 0.51. Thus, the efficiencies are almost identical, but the work per cycle of the ideal cycle is almost twice that of the Carnot cycle.)

§13.3.5 by including a regeneration cycle similar to the one shown in Figure 13.16(b). A fraction f of the steam is removed from the turbine and returned back to a second pump at a temperature T r = 130 ◦ C. Compute the fraction f needed to heat the water from the first pump to 130 ◦ C, and compute the increase in efficiency of the resulting cycle above that computed in the text.

[Hint: s5 = s3 in Figure 13.16(b).]

To compute f we need to compare the enthalpy change in the steam that goes through the regeneration cycle to the enthalpy required to heat the output of pump 2 to T r . At T r = 130 ◦ C, pvap 269 kPa. Using p5 = 269 kPa and s5 = s3 = 7.60 kJ/kg K, we can use steam tables in the superheat region to find h5 2 983 kJ/kg [23]. Saturation data at p = 269 kPa give h6 545.7 kJ/kg. We then solve f (h5 − h6 ) = (1 − f )(h6 − h2 ) , to get f 0.143. The work done is reduced by f (h5 − h4 ) 95 kJ/kg to w[34] 1 275 kJ/kg, waste heat is reduced by f (h4 − h1 ) 310 kJ/kg, and input heat is reduced by h6 − h2 405 kJ/kg (note, we assume h2 h1 ) to q[23] 3 145 kJ/kg. The efficiency of the revised cycle is roughly η′ = 1 275/3 145 40.5%, to be compared with the original efficiency of η = 1 370/3 550 38.6%. So there is an increase of efficiency of approximately (41.9 − 38.6)/38.6 ≈ 4.9%.

13.11 The wear on a steam turbine could be decreased by rais-

ing the pressure at the turbine outlet so that the quality of the steam at the outlet is one. In the ideal Rankine cycle of Example 13.2 and Figure 13.12, χ = 0.9 at the turbine outlet. Keeping the p+ , T + , and T 3 unchanged, determine p− and T − so that χ = 1 at ➃. Compare the efficiency and the power per cycle of the resulting cycle with the cycle of Example 13.2. To simplify the calculation, ignore the work done by the pump. Would you advise taking this step?

Increasing the quality at ➃ requires raising T − (and p− ) until the specific entropy at ➃ with χ4 = 1 equals the specific entropy at ➂, s3 = 7.60 kJ/kg K. Searching steam tables such as [23] for saturated steam with this specific entropy, we find T − 81.1 ◦ C, and p− 49.5 kPa. At this {T − , p− } we find h4 2646 kJ/kg and h1 340 kJ/kg. From these we compute the heat added and work done q[23] = h3 − h2 ≈ h3 − h1 3 690 − 340 = 3 250 kJ/kg ,

w[34] = h3 − h4 3 690 − 2 646 = 1 044 kJ/kg ,

where we have ignored pumping work and set h2 ≈ h1 . From this we compute an efficiency η = 1 044/3 250 0.32. So this adaptation has decreased the efficiency of the Rankine cycle by 18% (from 0.39 to 0.32) and decreased the work per kilogram per cycle by 24% (from 1 370 kJ/kg to 1 044 kJ/kg).

Raising the quality at the output of the turbine to one in an ideal Rankine cycle requires a significant sacrifice in both efficiency and work per cycle. The goal of preventing wear on the turbine must be accomplished with reheating or other more sophisticated means.

13.12 [T] Explain why the Brayton and Otto cycles look so

different in the pV-plane even though they are quite similar in the S T -plane.

The Brayton cycle uses isobaric heating and cooling steps, whereas the Otto cycle uses isometric steps. Both of these are described by exponential curves in the S T -plane (though with different constants) and so look very similar there. They look very different in the PV-plane since isobaric curves are horizontal while isometric are vertical.

298

13.13 [T] Show that the ratio of the backwork (the work nec-

13.15 [T] Verify the assertion made in §13.5.2 that the net

essary to run the compressor) to the total work done by the turbine for the ideal Brayton cycle is W{12} /W{34} = T 1 /T 4 . [Hint: use the 1st Law and the fact that both the compressor and turbine are adiabatic.] Check your result by comparing to the result stated in the text for the conditions shown in Figure 13.19.

efficiency of a CCGT is ηnet = ηB + ηR (1 − ηB ), where ηB/R is the efficiency of the Brayton/Rankine cycle.

The work done between ➂ and ➃ is given by γ pV γ = p4 V4 . Solving for V, V = V4 (p4 /p)1/γ . So,

R

pdV, where

1 1 dV = − V4 p1/γ p−1−1/γ d p = − NkB T 4 p4−1+1/γ p−1−1/γ .

4 γ γ The work, then, is

W[34] =

Z p4 p3

1 p dV = − NkB T 4 p4−1+1/γ γ

Zp4

d pp−1/γ

p3

 !1−1/γ   NkB T 4  p3  . = 1 − γ−1 p4

Likewise, the work done between ➀ and ➁ is  !1−1/γ   p2 NkB T 1   . 1 − W[12] = γ−1 p1

(13.7)

However, p2 = p3 and p1 = p4 , so

W[12] T1 = .

W[34] T4

(13.8)

13.14 [T] Calculate and plot the ratio of the efficiency of the

Brayton cycle to the efficiency of the Otto cycle with the same compression ratio (with γ = 1.4).

The Otto cycle has an efficiency of ηO = 1 −

1 = 1 − r−2/5 rγ−1

(13.9)

while the efficiency of the Brayton cycle is 1 = 1 − r−2/7 . r1−1/γ

(13.10)

ηB 1 − r−2/7 = . ηO 1 − r−2/5

(13.11)

ηB = 1 − The ratio of these two is R=

In the limit that r → 1, R → 5/7, while in the limit that r → ∞, R → 1. The ratio as a function of r for γ = 1.4 is shown in the figure.

Suppose the Brayton cycle takes in an amount of heat equal to Qin . Then it does WB = ηB Qin work and wastes Qout = (1 − ηB )Qin of heat, which is used as the input to the Rankine cycle.

The Rankine cycle does WR = ηR Qout work, so the efficiency is

WR + WB 1 ηR (1 − ηB ) + ηB Qin = ηB + ηR (1 − ηB ). = Qin Qin

Part II Energy Sources

317

Problems

Problems 14.1

14.3

A rocket fuel should have high energy density (per kilogram) because the weight of the propellant is an important consideration. A favorite fuel is liquid hydrogen plus liquid oxygen. What is the energy density (J/kg) of a stoichiometric mixture of H2 (l) and O2 (l) (assuming they combine to form water in the gaseous state)? The strong nuclear force, in principle, allows the same stoichiometric mixture of two hydrogen atoms and one oxygen atom to fuse to form an isotope of neon, 18 Ne, giving off 4.52 MeV in the process. If this reaction were the source of power for the rocket, what would be the energy density of the fuel?

Show that the combination of constants g2strong / (4πǫ0 ~c), where gstrong is defined in eq. (14.5), is dimensionless. It is taken as a dimensionless measure of the strength of an interaction. Compute this quantity for electromagnetism (g → e), where it is known as the fine structure constant and denoted by α.

From eq. (14.5),  2   gstrong   = [V(r)r] = ml3 /t2 .  4πǫ0 

Compare [~c] = (ml2 /t)(l/t) g2strong /(4πǫ0 ~c) is dimensionless α≡

The chemical reaction is

H2 (l) + (1/2)O2 (l) → H2 O(g) .

The enthalpy of vaporization of hydrogen and oxygen are both small compared to the enthalpy of formation of water, so we ignore the fact that the initial reactants are liquids and use the standard enthalpy of formation of water (Table 9.8), ∆H f (H2 O(g)) = 242 kJ/mol, or

14.4

The nuclear reaction, 2(1 H) + 16 O → 18 Ne ,

emits 4.52 MeV per nucleus of 18 Ne formed. 1 eV/nucleus = 96.5 kJ/mol (see Table C.1), so

Λ → p + e + ν̄e

The nuclear reaction releases about 2 × 106 more energy per unit mass.

Numerically 1 Egrav − (6.67 × 10−11 m3 /kg s2 )2 (1.67 × 10−27 kg)2 2 × (9.11 × 10−31 kg)3 /(1.05 × 10−34 J s)2 4.3 × 10−53 J 2.7 × 10−78 eV .

Which of the following interactions are consistent with the conservation laws for baryon number, electric charge, and the three lepton numbers? Don’t worry about energy conservation. (A bar denotes an antiparticle. See Question 14.4 for information about the Λ particle.)

τ → µ + ν̄µ + ντ

(4.5 × 1011 J/mol)/(0.018 kg/mol) 24.2 TJ/kg .

1 1 Egrav = − (GN m p me )2 me /~2 = − G2N m2p m3e /~2 .

2 2

(1.602 × 10−19 C)2 4π(8.854 × 10−12 F m)(1.054 × 10−34 J s)(2.998 × 108 m/s)

νµ + p → µ̄ + n

yielding an energy density of

Comparing Newton’s law (2.21) with Coulomb’s law (3.1), the two are related by the substitution Q1 Q2 /4πǫ0 ↔ GN Mm, so the hydrogen atom binding is modified to

e2 4πǫ0 ~c

p → n + e + νe

4.52 MeV/nucleus = (4.52 × 106 )(96.5 × 103 ) = 4.5 × 1011 J/mol ,

In principle, and in the absence of electromagnetic or any other non-gravitational interactions, an electron and a proton should form a “gravitational atom,” bound by the force of gravity. We know that the energy of the ground state of the ordinary hydrogen atom is given by eq. (14.4). Show that the ground state energy of gravitationally bound “hydrogen” is given by Egrav = − 12 G2N m2p m3e /~2 and evaluate this numerically.

ml3 /t2 , so the quantity

0.00730 1/137.0

∆h f = (242 kJ/mol)/(0.018 kg/mol) 13.4 MJ/kg .

14.2

=

p → n + ē + ν̄e

τ̄ → µ + ν̄µ + ντ

νµ + n → µ + p

Λ → n + νe + ν̄e

p → n + e + νe violates charge and electron number conservation. p → n + ē + ν̄e violates electron number conservation. τ → µ + ν̄µ + ντ conserves all. τ̄ → µ + ν̄µ + ντ violates charge and tauon number conservation. νµ + p → µ̄ + n violates muon number conservation. νµ + n → µ + p conserves all. Λ → p + e + ν̄e conserves all. Λ → n + νe + ν̄e conserves all.

14.5

Einstein’s energy–momentum relation p for a particle of mass m with momentum p is E = p2 c2 + m2 c4 . Estimate the uncertainty in the x, y, and z components of the momentum of a quark confined in a (cubic) box of side length L = 1 fm using the Heisenberg uncertainty principle ∆px,y,z & ~/L (Box 7.6). Approximate h p2 i by (∆ p)2 , then show that for a quark of mass 5 MeV/c2 , E ≈ |p|c and that the zero-point energy of three quarks is of order 900 MeV, the rest energy of a proton. ∆p x,y,z ≈ ~/L ≈ 10−34 /10−15 ≈ 10−19 kg m/s, so ∆p x,y,z c ≈ (10−19 kg m/s)(3 × 108 m/s)

≈ (3 × 10−11 J)/(1.6 × 10−19 J/eV) ≈ 200 MeV.

Comparing the kinetic (|∆p|c) and rest mass (mquark c2 ) contributions to the energy of a quark, 200 Mev ≫ 5 MeV, clearly the rest mass can be neglected. Thus, √ Equark ≈ |∆p|c ≈ 3(200 MeV) ≈ 350 MeV , and the rest energy of three confined quarks is roughly 3 × 350 ≈ 1 050 MeV, not far from the rest energy of a proton or neutron.

318

14.6

To get a sense of the potential practical relevance for energy purposes of one aspect of as-yet-undiscovered physics, consider proton decay. Making the most general assumptions about the nature of its decay, the lifetime of the proton is currently bounded by τproton > 2.1 × 1029 y. Assume that the lifetime is in fact 3 × 1029 y, and that all of the proton’s rest energy emitted in the decay can be captured and put to use. Estimate the rate of useful power output from proton decay of a cubic kilometer of hydrogen gas at NTP. Using pV = NkB T and at NTP kB T 1/40 eV, N ≈ (105 Pa)(109 m3 )(1.6 × 1019 eV/J)/(1/40 eV) ≈ 6.4 × 1034 hydrogen molecules, or 1.3 × 1035 protons. Since dN/dt = N/τ, there are roughly (1.3 × 1034 )/(3 × 1029 ) ≈ 4 × 105 decays per year, or ≈ 1.5 × 10−2 decays per second, each yielding ≈ (109 eV)/(1.6 × 1019 eV/J) ≈ 6 × 10−11 J, corresponding to a power of P ≈ (6 × 10−11 J)(1.5 × 10−2 /s) ≈ 10−12 W. This is not a useful source of power.

330

lifetime of 40 K? What is the ratio of the average number of decays of type one and type two?

Problems 15.1

The first radioactive element discovered by Polish physicist Marie Curie was radium-226. It has a halflife of 1700 years. Suppose you obtained a microgram of pure 226 Ra. How long would you have to wait before the number of decays per second from this sample fell to 104 ?

The total decay probability (see eq. (15.1)) is λ = λ1 + λ2 = 5.5 × 10−10 y−1 , and the lifetime τ = 1/λ = 1.8 × 109 y. N1 /N2 = λ1 /λ2 = 4.9/.61 = 8.0.

15.4

Radium-226 has a molar mass of 226 g/mol, so a microgram of 226 Ra is 4.42 × 10−9 mol or 2.66 × 1015 atoms. A half-life of τ1/2 = 1700 years corresponds to a lifetime of τ = τ1/2 / log 2 = τ1/2 /0.693 = 2452 y = 7.73 × 1010 s. The decay rate for N atoms is given by R=−

d N0 −t/τ N0 e−t/τ = e dt τ

where the − sign indicates that the number of atoms is decreasing as decays happen. Solving this for t, when R = 104 s−1 t = −τ ln

Rτ N0

−(7.73 × 1010 ) ln

(104 s−1 )(7.73 × 1010 s) 2.66 × 1015

9.56 × 1010 s = 3030 years

15.2

15.5

!

Uranium-238 and uranium-235 both decay by αparticle emission. Their measured half-lives are 4.5 × 109 years and 7.0 × 108 years respectively. At the present time the ratio of 235 U to 238 U in naturally occurring uranium is about 0.72%. It is believed that the uranium on Earth was formed in a supernova that took place billions of years ago. It is also believed that the ratio of 235 U/238 U produced in supernovae is roughly 1.65. Estimate how long ago the supernova that produced the uranium on Earth took place.

235

We know that r(0) = r0 = 1.65 and r(today) = rt = 0.0072, so rt 235 238 = e−t(1/τ −1/τ ) . r0 Solving for t, r

15.3

t=

ln r0t 1 1 − 238 τ τ235

= 6.5 Gy .

Potassium-40 (40 K) is a naturally occurring form of potassium whose β-decays are responsible for some of Earth’s geothermal energy (see §14.3 and §32.1.3).

40 K can decay in two different ways. The probabilities per unit time for the two decay modes are λ1 = 4.9 × 10−10 y−1 and λ2 = 6.1 × 10−11 y−1 . What is the

Consider the ground state and the first and second excited states of a simple harmonic oscillator as described in §7. Verify the statement made in the text that the probability of finding the particle outside the classically allowed region is 16% when the particle is in the ground state, and compute the corresponding probabilities for the first and second excited states.

1 dP0 2 = √ e−ξ dξ π 2 2 dP1 = √ ξ2 e−ξ dξ π 2 dP1 1 = √ (2ξ2 − 1)2 e−ξ . dξ 2 π

(Check that these probabilities integrate to one over −∞ < ξ < ∞.)

The classical turning points occur when E = V(x± ) or 12 mω2 x2± = √ (n√+ 1/2)~ω for n = 0, 1, 2. Thus ξ±0 = ±1, ξ±1 = 3 and ξ±2 = ± 5. The probability to Rbe outside the classically allowed region ∞ is given by Ptunneling = 2 ξ dξ(dP/dξ). Evaluating these integrals + numerically yields P0tunneling 0.157

,

and similarly for uranium-238. Taking the ratio of these,   N235 (t)  N0235  −t(1/τ235 −1/τ238 ) r(t) = =  238  e .

N238 (t) N0

The decays are governed by a Poisson distribution (15.3) with an average number of events λNT = 40 when T = 1 s. Thus the probability of exactly k decays is in one second is Pk = (40)k e−40 /40!. Evaluating P this, P40 0.629, P30 = 0.0185, P0 = 4.25 × 10−18 . P<40 = 39 k=0 Pk = 0.479.

The probability densities for these states are given by the square of the wavefunction (15.7). Write the squared-wavefunctions from eqs. (7.44)–(7.46) in terms of the dimensionless variable ξ = x/ℓ,

The lifetime of 238 U is τ238 = 6.49 × 109 y, while the lifetime of 235 U is τ235 = 1.01 × 109 y. If we let N0235 and N0238 be the initial numbers of uranium-235 and uranium-238 at the time of formation, then the number of 235 U atoms at time t after the supernova is given by N235 (t) = N0235 e−t/τ

A radioactive source is emitting on average 40 particles per second. In any particular one second interval, what is the probability that it emits 40 particles? 30 particles?

None? What is the probability that it emits less than 40 particles?

P1tunneling 0.112 P2tunneling 0.095

15.6

[T] Verify the barrier penetration factor obtained for the parabolic barrier in Box 15.2.

The barrier penetration factor is the exponential of −I(E), where r Z k 2 a dx 2m(V0 − x2 − E) I(E) = ~ −a 2 √ and a = 2(V0 − E)/k. Replacing V0 − E by ka2 /2 and changing integration variable to y = x/a, we get √ Z q mk 2 1 I(E) = 2 a dy 1 − y2 ~ −1 √ 2π πa2 mk = (V0 − E) , = ~ ~ω as quoted in the example.

331

Problems

p p Let k = 2mV0 /~2 and q = 2m(V0 − E)/~2 . Define the regions to the left of the barrier, within the barrier, and to the right of the barrier as I, II, and III. The Schrödinger wave functions in these three regions can be parameterized as ψI (x) = eikx + B−ikx ψII (x) = Ceqx + De−qx ψIII (x) = Aeikx

Figure 15.6 A triangular barrier.

The aim is to use continuity of ψ and ψ′ at x = 0 and x = d to solve for A and B. Continuity of ψ and ψ′ at x = 0 gives C+D=1+B

15.7

C − D = (ik/q)(1 − B) ,

[T] A particle of mass m and energy E encounters a triangular barrier of width 2w and height V0 (see Figure 15.6). Estimate the probability that it will tunnel through the barrier. How does the probability compare with that for a rectangular barrier of the same height and width?

Solving for C and D: C = z + z∗ B, D = z∗ + zB, where z = 21 (1 + ik/q). Note that for tunneling, |z| ∼ 1. Similarly matching the wave function and derivative at x = d and solving for C and D, gives C = e−qd z∗ eikd A D = eqd zeikd A Equating these two expressions for C and D give two equations relating A and B, which in turn yields,

We approximate the probability that the particle will penetrate the barrier by eq. (15.19). The potential in this problem is given by ! |x| V(x) = V 1 − w

z∗2 − z2 A = e−ikd ∗2 qd z e − z2 e−qd

z B = − ∗ 1 − e−qd A z

The turning points are the points where V(x) = E, or at

E x± = ±w 1 − V

When the barrier is “high and wide”, eqd ≫ 1 and, A = (1 − z2 /z∗2 )e−qd + O(e−3qd )

The potential is symmetric around the origin, so the integral can be done only on half of the potential: 2

Zx+ 0

x+

Z √ p dx dx 2m(V(x) − E) = 2 2mV 0

r

1−

This verifies that the probability of transmission |A|2 has the form of eq. (15.16) with a multiplicative coefficient of order one. The probability of reflection, which is proportional to |B|2 is of order 1.

The corrections to both probabilities are exponentially suppressed.

x E − V w

√ E 3/2 4w 2mV 1− = 3 V

So, the probability that the particle will tunnel through the barrier is √    8w 2mV E 3/2  1− P(E) = exp −  3~ V √ !

8w 2m(V − E) E = exp − 1− 3~ V

R For a square √ barrier of height V and width 2w, the integral is just (. . .) = 2w 2m(V − E), so the probability of tunneling through a square barrier is √ !

4w 2m(V − E) Psquare (E) = exp − ~

The probability of tunneling through the triangular barrier is significantly larger, since both the factor of 8/3 and (1 − E/V) reduce the expression inside the exponential for the triangle barrier compared to the square barrier.

15.8

B = −z/z∗ + O(e−qd )

[T,H] Compute the exact amplitude for transmission and reflection of a particle from a rectangular potential barrier of height V0 and width d by matching the wavefunction and its derivative at the boundary points x = 0, d. Note that, for a barrier that is high and wide, the leading term has the form of eq. (15.16) with a multiplicative coefficient as anticipated in Footnote 1. Show that further corrections, which arise from the exponentially increasing solution in the classically forbidden region, are exponentially suppressed.

15.9

An experimenter seeks to use a sinusoidal potential of the form studied in Example 15.2 to make a one-dimensional trap to confine a number of electrons for further study. She places one electron in the ground state of each potential well. Each electron’s energy should be 0.1 eV and the electron’s tunneling lifetime should be at least five years. Using the approximations developed in Example 15.2, estimate the minimum necessary height of the potential (W/2), the spacing d between minima in the potential, and the root mean square width of each electron’s probability distribution.

According to Example 15.2 the ground state energy E0 , the parameter W and the tunneling lifetime are related by τ∼

π~ −π πW/2E0 e e or E0

!

W 1 ∼ E0 1 + ln(E0 τ/π~) 2 π Substituting for E0 = 0.1 eV and τ & 5 y, we find W/2 & 1.7 eV. The spacing between minima in the potential is given by the condition βd = 2π, or r π~ W d = 2π/β = E0 m s −34 π(1.0 × 10 Js) 3.4(1.6 × 10−19 J)

0.1(1.6−19 J) 9.1 × 10−31 kg 1.6 × 10−8 m

332

The electron’s probability distribution is that of the ground state of the harmonic oscillator, eq. (15.10). We compute Z ∞ 1 ~ ~2 hx2 i = dx x2 |ψ0 (x)| = ℓ2 = = = (4×10−10 m)2 .

2 2mω 4mE 0 −∞ Note that the root-mean-square size of the probability distribution is only 2.5% of the distance between minima in the potential.

15.10 [T] The energy levels of the one-dimensional harmonic

oscillator are given by En = (n + 1/2)~ω with n = 0, 1, 2, . . . . Suppose two spin-1/2 fermions are trapped in a one-dimensional oscillator potential. What are the energies and degeneracies of the first three energy states of this system? (Remember that the degeneracy is the number of independent states with the same energy.)

Label the states available to a particle by (n ↑) and (n ↓), where ↑ / ↓ denotes the particle’s spin. An allowed state of two fermions must place each in a different single-particle level. Beginning at the lowest single particle level and working upward: Etot = ~ω : Etot = 2~ω : Etot = 3~ω :

(0 ↑)(0 ↓)

(0 ↑)(1 ↑), (0 ↑)(1 ↓)

(0 ↓)(1 ↑), (0 ↓)(1 ↓) (1 ↑)(1 ↓)

(0 ↑)(2 ↑), (0 ↑)(2 ↓)

(0 ↓)(2 ↑), (0 ↓)(2 ↓)

so the degeneracies are 1, 4, and 5 respectively.

15.11 [T,H] The energy levels of the three-dimensional har-

monic oscillator are given by En1 ,n2 ,n3 = (n1 + n2 + n3 + 3/2)~ω with n1,2,3 = 0, 1, . . .. Suppose three spin-1/2 fermions are trapped in a three-dimensional oscillator. What is the energy of the ground state? What is its degeneracy? (Remember that the degeneracy is the number of independent states with the same energy.)

Extra challenge: What is the energy of the first excited state? What is its degeneracy?

The ground state has energy (2(3/2) + 5/2)~ω = 11~ω/2. There are six degenerate states of the form: (000 ↑)(000 ↓)(100 l)

(000 ↑)(000 ↓)(010 l)

(000 ↑)(000 ↓)(001 l) , where l denotes the two states (. . . ↑) and (. . . ↓). The first excited state has energy (2(3/2) + 7/2)~ω = (3/2 + 2(5/2))~ω = 13~ω/2.

It has degeneracy of 42. There are six states of the form (000 ↑)(000 ↓)(110 l) plus cyclic permutations of the “110” labels; plus six states of the form (000 ↑)(000 ↓)(200 l) plus cyclic permutations of the “200” labels; plus 30 states in which one particle is in the lowest level (000 l) and two particles are in the multiplet of single particle levels of the form (100 l). There are six states of this form (cyclic permutations of “100” and the choice of ↑ or ↓), and 15 (15 = 6 × 5/2) ways of placing two particles in the six states excluding states where both particles are in the same state. Including the l degeneracy of the lowest level gives 2 × 15 = 30.

340

Problems 16.1

Given the (higher) molar heat of combustion of gasoline (§11.2.2) and knowing its molecular weight, verify that it yields about 1/2 eV per nucleon in complete combustion.

The HHV of iso-octane is 5.47 MJ/mol (§11.2.2). 96.5 kJ/mol corresponds to 1 eV/molecule, so combustion of iso-octane liberates energy E = (5.47 × 106 )/(96.5 × 103 ) 57 eV/molecule .

The atomic mass (= number of nucleons) of iso-octane is m = 8(12) + 18 = 114, so the energy per nucleon is E/m 57/114 = 0.5 eV/nucleon. If the atomic mass of the oxygen necessary for combustion is added into the calculation then (since 12.5 moles of O( 2) are required for combustion of 1 mole of C8 H18 ) m → 114+12.5(32) = 514 and the energy per nucleon drops to 57/514 0.11 eV/nucleon.

16.2

In one scenario, enrichment of 200 t of natural uranium yields 26 t enriched to 4% 235 U that is used to power a reactor for a year. How much 235 U is lost in the depleted uranium discarded during the enrichment process? The spent nuclear fuel removed after a year in the reactor contains about 0.8% 235 U. Close to 86% of the 235 U consumed actually fissions; the other 14% captures a neutron without fissioning. Verify that approximately half of the 235 U contained in the 200 t of natural uranium actually fissions during the year of running the reactor.

200 t of natural uranium contains 1.44 t of 235 U. 26 t at 4% contains 1.04 t of 235 U, so the depleted uranium contains 0.40 t or 0.40/174 = 0.23% 235 U. The spent fuel contains 0.008(26) = 0.21 t of 235 U, so 1.04 − 0.21 = 0.83 t of 235 U was consumed in the reactor. Of this (0.86)(0.83) = 0.71 t actually fissions. This amount corresponds to (0.71/1.44) = 49% of the 235 U present in the original 200 t of uranium.

16.3

Check the quoted estimate of 4 × 109 tonnes of uranium from seawater quoted in this chapter. For how many years could this amount of uranium supply 100% of the world’s energy needs at the level of 2015?

A simple, rough estimate can be made as follows. Estimate world primary energy consumption for 2015 to be 600 EJ. Natural uranium burned in “modern light-water thermal reactors” produces 1 GWye for every 200 t of natural uranium. To compare with primary energy, we divide by 033, the typical efficiency of such reactiors, to obtain an estimate of equivalent primary energy of 3 GWy per 200 t of natural uranium.

The Wikipedia data page on the abundance of the elements quotes two different estimates for the concentration of uranium in seawater: 3.2 × 10−9 and 3.3 × 10−9 kg/L. The difference is not important; to be optimistic we take te higher value. The ocean volume of 1.33 × 109 km3 contains a total mass of uranium of

m(U) = (1.33×1018 m3 )(103 L/m3 )(3.3×10−9 kg(U)/L) 4.4×109 t(U) .

Taking the estimate of 200 t(U) ⇒ 3 GWy of thermal energy, this mass of uranium could provide E = (4.4×109 t(U))/((200 t(U))/(3 GWy)) ≈ 66 PWy ≈ 2 YJ(= 2×1024 J) of primary energy. So seawater uranium has an energy content sufficient to supply world primary energy for roughly T = (2 × 1024 J)/(6 × 1020 J/y) ≈ 3 300 y .

A more sophisticated analysis would attempt to identify the manner and efficiency with which uranium could be used to supply various energy end uses.

16.4

Assume that thorium can be converted 100% into 233 U and used as a fission fuel. According to [76] world thorium reserves are roughly 6 × 106 tonnes. How long could this amount of thorium supply 100% of the world’s energy needs at the 2015 level?

Follow the same approach as in Problem 16.3. As stated in §16.2, 1.1 t of 232 Th transformed to 233 U can produce 3 GWy of primary energy (1 GWye). So the total energy content of world thorium reserves is E = (6 × 106 t(Th))/((1.1 t(Th)/(3 GWy)) ≈ 16 PWy ≈ 0.5YJ , corresponding to about 800 y of 2015 world primary energy consumption of 600 EJ/y.

363

Problems

of the nucleus and the kinetic energy of the thermal neutron in this process.)

Problems Note: for a number of the problems in this chapter you will need to use data on nuclides, which is conveniently obtained from the NNDC (Box 17.1) [70].

17.1

[T] Derive the relation quoted in eq. (17.7) between the binding energy of a neutral atom B(Z, A) and its mass excess ∆(Z, A).

From NNDC data, Q = ∆(n) + ∆(3 He) − ∆(4 He)

= 8.07 + 14.93 − 2.43 = 20.57 MeV

17.5

From eq. (17.2), B(Z, A) = ZMH + N Mn − Matom (Z, A), and from the definition of the mass excess, MH = ∆H + u Mn = ∆n + u

3

Substituting for the nuclear masses and using Z + N = A, gives the required result.

H → 3 He + e− + ν̄e .

Verify that both of these nuclear reactions are exothermic and compute the energy released in each. Check that this process is practical by looking up the fractional abundance of 6 Li in naturally occurring lithium and the lifetime of 3 H. Why should we care about the latter?

A typical fission reaction involving 235 U is 235 90 144 92 U + n(thermal) → 56 Ba + 36 Kr + 2n.

The kinetic energy of a thermal neutron is ∼1/40 eV, small enough to ignore in the energy balance. Show that the energy released in the fission is 179.5 MeV.

Q(nthermal + 6 Li → 4 He + 3 H) = 8.07 + 14.09 − 2.43 − 14.95

From the NNDC tables,

= 4.78 MeV ,

∆(235 U) = +40.92

Q(3 H → 3 He + e− + ν̄e ) = ∆(3 H) − ∆(3 He)

∆(n) = +8.07

= 14.95 − 14.93 = 0.02 MeV ,

∆(144 Ba) = −71.77

so both reactions are allowed by energy conservation. 6 Li accounts for ≈ 7.6% of naturally occurring lithium, and lithium is a fairly abundant element, so supplies of 6 Li are substantial. The lifetime of 3 H is τ ≈ 12 y. This is important because it controls the rate of production of 3 He. If R is the desired rate of production of 3 He, then Rτ is the stock of 3 H that must be maintained.

∆(90 Kr) = −74.96

where all mass excesses are in MeV. Then the energy released Q is given by, Q = 40.92 + 8.07 − (−71.77) − (−74.92) − 2(8.07) = 179.54 MeV

17.3

He is so rare that it must be produced artificially.

One way to produce it is through a sequence of reactions that can be carried out in a nuclear reactor, where thermal neutrons are quite common. The reactions are: nthermal + 6 Li → 4 He + 3 H, followed by

Matom = ∆atom + Au .

17.2

3

There is no stable nucleus with A = 5. This has profound consequences for the way stars burn and form elements. The two candidates for stable nuclei with A = 5 are 52 He and 53 Li. Look up or figure out what they decay into and show that their decays are allowed by energy conservation.

17.6

4

He is anomalously stable, so it is likely that both nuclei decay into 4 He by particle (n or p) emission. From NNDC Q(5 He →4 He + n) = 11.23 − 8.07 − 2.43 = 0.73 MeV

17.7

Q(5 Li →4 He + p) = 11.68 − 7.29 − 2.43 = 1.96 MeV .

The radius of a nucleon is about RN = 0.9 fm. Compute the volume of a nucleus with a moderately large value of A in the liquid drop model (i.e. using the SEMF) and show that roughly 2/3 of the nucleus is “empty space.”

In contrast, an atom is almost entirely empty space.

3 The volume of A nucleons is V = 4π 3 R N A, while the volume of 1/3 3 (R ) , where R0 ≈ 1.3 a nucleus with A nucleons is V(A) = 4π 0A 3 (eq. (17.10)), so V/V(A) ≈ RN /R30 ≈ (1.3/0.9)3 ≈ 0.33.

Estimate the density of a nucleus by assuming it to be a sphere of radius R0 = 1.3 A1/3 fm. What would be the radius of a nucleus with the mass of Earth?

Since both Q-values are positive, both decays are allowed.

17.4

Helium three is an extremely rare but stable isotope of helium made up of two protons and only one neutron.

Because 4 He is so tightly bound, considerable energy is given off when 3 He absorbs a neutron. This makes 3 He a very good neutron detector. The nuclear reaction involved is n + 3 He → 4 He + γ. Neutron detection is a vital aspect of homeland security in the twentyfirst century, since nuclear devices inevitably give off very-low-energy neutrons that easily leak out of almost any container. What is the energy of the γ-ray given off when a thermal neutron (with energy ∼ 1/40 eV) is absorbed by a 3 He nucleus? (You can ignore the recoil

Estimating the mass per nucleon to be 1 u, the density is ρ ≈ 1 u/(4πR30 /3) ≈ 1.66 × 10−27 kg/(4π × (1.3 × 10−15 m)3 /3) ≈ 1.8×1017 kg/m3 . M⊕ 5.97×1024 kg, so at nuclear density, Earth’s volume would be V⊕ ≈ 6 × 1024 /1.8 × 1017 ≈ 3.3 × 107 m3 ,

17.8

corresponding to a sphere of radius R⊕ ≈ (3V⊕ /4π)1/3 ≈ 200 m.

Pressure can be defined as the rate of change of energy with volume, p = ∂E/∂V. Similarly, the surface analogue of pressure, surface tension, is defined as the rate of change of energy with surface area a, σ ≡ ∂E/∂a at constant volume. Ignoring the effects of symmetry, Coulomb, and pairing energies, compute the surface tension of a nucleus as described by the empirical

364

parameters of the SEMF. The surface tension acts to keep a nucleus spherical.

Ignoring symmetry, Coulomb, and pairing energies, B(A) = ǫV A − ǫS A2/3 . The surface energy comes from the second term. a = 4πR20 A2/3 is the surface area. Then dB/da|V = −ǫa /4πR20

≈ −17 MeV/4π(1.3 fm)2 ≈ −0.8 MeV/fm2 .

17.9

Negative binding energy corresponds to positive surface tension of σ ≈ 0.8 MeV/fm2 .

Find the SEMF prediction for Zmin (A = 200) from eq. (17.20). Find the actual value of Z that gives the greatest binding energy for A = 200. How well does the SEMF do?

According to eq. (17.20), Zmin (200)

100 79.5 , 1 + (7.52 × 10−3 )(200)2/3

so the SEMF predicts that the most stable nuclide with A = 200 has either Z = 79 or Z = 80. Since 200 79 Au is an odd-odd nucleus, the paring energy strongly favors Z = 80 as the most stable A = 200 nuclide.

The mass excesses of the relatively stable nuclei with A = 200 are ∆(200 79 Au) = −27.2682 MeV

(t1/2 19 h)

∆(200 81 Tl) = −27.0474 MeV

(t1/2 26 h) ,

∆(200 80 Hg) = −29.5035 MeV

17.12 [T] The maximum value of A for which nuclei are sta-

ble against instantaneous fission can be estimated as follows. When a charged, spherical nucleus, A Z, with surface tension is deformed into an ellipsoid leaving its volume constant, it is possible to compute the change in its binding energy as a function of deformation. The result [77] is !

2 1 Z2 ∆B ≈ − ǫS A2/3 − ǫC 1/3 δ2 , (17.35) 5 5 A where δ measures the deformation (the ratio of major to minor axes is (1 + δ)3/2 ). Note the signs: the surface contribution tends to stabilize and the Coulomb contribution tends to destabilize the nucleus. Show that eq. (17.35) predicts that instantaneous fission occurs when Z 2 /A & 49. The largest A nucleus observed so far, 294 [118], has Z 2 /A = 47.4.

The stabilizing surface term dominates at small Z and A, but the surface term grows in importance until the quantity in parentheses vanishes when 2ǫS Z2 = A 49A ǫC

17.13 Explain how to read from Figure 17.8 the Q-value for

α-decay predicted by the SEMF.

(stable)

For alpha decay,

confirming that the SEMF does very well in estimating Zmin (A).

Using these mass excess values, we can also compute the binding energies, for comparison:

Q = ∆(Z, A) − ∆(Z − 2, A − 4) − ∆(4 He)

= −B(Z, A) + B(Z − 2, A − 4) + B(4 He)

1 200 B(200 79 Au) = 121∆(n) + 79∆( 1 H) − ∆( 79 Au)

= 121(8.091 MeV) + 79(7.289 MeV) − (−27.2682 MeV) = 1579.69 MeV ,

B(200 80 Hg) = 120(8.091 MeV) + 80(7.289 MeV) − (−29.5035 MeV) = 1581.14 MeV ,

dB + B(4 He) dA #! " dbmax (A) ≈ 4 b(4 He) − bmax (A) + A dA ≈ −4

B(200 , 81 Tl) = 119(8.091 MeV) + 81(7.289 MeV) − (−27.0474 MeV) = 1577.91 MeV where we have made the approximations described in the text: a) the finite difference has been replaced by a derivative; b) B(Z, A) which again confirms that Zmin = 80. has been replaced by Bmax (A).

17.10 [T] Derive eq. (17.20) by differentiating the SEMF The quantity in parentheses is the difference between the line expression for b(Z, A) with respect to Z at fixed A. labeled “4 He” and the curve in Figure 17.8, so the predicted Qvalue is four times the difference in the figure.

Only the Z dependent terms in b(Z, A) are relevant, 226

17.14 Show that the longest-lived isotope of radium,

88 Ra, is energetically allowed to decay by emission of a 14 C nucleus. This decay has been observed with a 6 probability of 3.2 × 10−9 .

b(Z, A) = ǫsym (A − 2Z)2 /A2 − ǫC Z 2 /A4/3 , and ∂b/∂Z|A = 0 implies 2ǫsym (A − 2Z) = ZA2/3 ǫC .

Solving for Z gives eq. (17.20).

17.11 [T] Check the assertion in the text that the SEMF pre-

dicts that nuclei have positive binding energy up to A ∼ 3150.

The binding energy per nucleon along the floor of the valley of stability is given by eq. (17.21). The figure shows a plot of bmax (A) versus A for 0 < A < 4 000,

5

Q(226 Ra → 214 Pb + 12 C) = 23.67 − (−.18) = 23.85 MeV, so this reaction is energetically allowed.

17.15 235 U can decay by emitting a neon nucleus. How many

such decays occur per second per mole of 235 U? For radioactive decay dN/dt = N/τ, where τ is the lifetime. The half-life of 235 U is 7 × 108 years and the fraction of decays to neon is 8 × 10−12 (from NNDC), so the effective lifetime for this decay channel is τNe 7 × 108 y/ ln 2/(8 × 10−12 ) 1.3 × 1020 y × 3 × 107 s/y 4 × 1027 s. Thus the number of decays per second per mole is dN/dt 6 × 1023 /(4 × 1027 ) 2 × 10−4 /s mol.

17.16 Radon gas 222 86 Rn is a serious environmental hazard

1000

2000

3000

4000

(see §20). It is a decay product of 238 U, which is a 92 relatively common constituent of rock. 222 86 Rn under222 goes α-decay to 218 Po. When Rn is created in the 84 86

365

Problems

foundation of a house, it diffuses out into the basement airspace. If its lifetime were very short, it would decay before getting out of the rock. If it were very long, it would diffuse away into the atmosphere. Its half-life is 3.8 days, however, just right for causing a great deal of human radiation exposure. Given this half-life, use Gamow’s formula (17.30) to estimate the Q-value for this decay and compare with the measured value. Comment on the accuracy of your estimate.

Given Z = 84, R = 1.3 (41/3 + 2181/3 ) ≈ 9.9 fm, and τ = t1/2 /.693 ≈ 5.5 d ≈ 4.7 × 105 s, we solve eq. (17.30) for Q, p

1.7Z

Q≈

and electron capture, eq. (17.34).

For β− decay eq. (17.32) gives Q = Mnucl (Z, A) − Mnucl (Z + 1, A) − me , adding and subtracting Au and Zme , = (Mnucl (Z, A) + Zme − Au) − (Mnucl (Z + 1, A) + (Z + 1)me − Au)

= ∆(Z, A) − ∆(Z + 1, A) ,

ignoring small changes in the binding energies of atomic electrons.

The calculation for electron capture is similar. For β+ -decay Q = Mnucl (Z, A) − Mnucl (Z − 1, A) − me

= (Mnucl (Z, A) + Zme − Au) − (Mnucl (Z − 1, A) + (Z − 1)me − Au) − 2me = ∆(Z, A) − ∆(Z + 1, A) − 2∆e .

√ log10 τ − log10 T 0 + 1.3 RZ 143 ≈ = 2.23 5.7 + 21 + 37.5

17.20 Ignoring small electron binding energies and the very

So we predict Q ≈ 2.232 ≈ 5.0 MeV. The observed Q-value (from NNDC) is 4 218 Q = ∆(222 86 Rn) − ∆( 84 Po) − ∆(2 He)

= 16.3736 − 8.3580 − 2.4249 = 5.5887 MeV

so the estimate based on eq. (17.30) is accurate to about 10%.

17.17 [T] Examine the derivation of Gamow’s formula for α-

decay lifetimes, and then derive the equivalent expression for decay by emission of a (small) nucleus of charge z and mass number a. Assume a/A is small enough to ignore recoil. Make sure that the z and a dependences are explicit. Compare your result to an α-decay lifetime with the same Q-value.

The general form of Gamow’s formula with the same assumptions as in the text is: r ! √ ! p az √ z a log10 τ ∼ log10 T 0 − 1.3 RZ + 1.7 Z/ Q. 8 4

And R depends weakly on a via eq. (17.28).

17.18 Check that 3 He is less tightly bound than 3 H. Explain

why 3 H decays to 3 He, and not vice versa. ∆(3 He) = 14.9312 MeV and ∆(3 H) = 14.9498 so according to eq. (17.7), B(3 He) = ∆(n) + 2∆(11 H) − ∆(3 He)

= 8.0713 + 2(7.2889) − 14.9312 = 7.7179 MeV

B(3 H) = 2∆(n) + ∆(11 H) − ∆(3 H)

= 2(8.0713) + 7.2889 − 14.9498 = 8.4817 MeV

showing that 3 H has greater binding energy than 3 He. Q-value calculations from eq. (17.34) correctly predict that 3 H decays to 3 He and not vice versa because they take account of the total energy balance including the electrons. In terms of masses and binding energies, the 3 He →3 H electron capture Q-value is 3

17.19 [T] Verify the expressions for the Q-value in β± -decay

3

Q = (2m p + mn + me − B( He)) − (m p + 2mn − B( H))

= 938.2720 + 0.5110 − 7.7179 − 939.5654 + 8.4817 = −0.0186 MeV (= ∆(3 He) − ∆(3 H)) .

Since Q < 0 the decay is forbidden. The analogous calculation for 3 H →3 He β-decay gives Q = 0.0186 MeV, so this decay is allowed.

small mass of the neutrino, show that the mass of a nucleus increases when it decays by electron capture if the Q-value of the decay is less than me c2 0.511 MeV. Verify that this is the case for the electron capture decay of the longest-lived isotope of technetium: 97 − 97 43 Tc + e →42 Mo + νe .

When a nucleus [A] [Z] decays to [A−1] [Z − 1] by electron capture, energy conservation gives M(Z, A)c2 + Zme c2 = M(Z − 1, A)c2 + (Z − 1)me c2 + Q , where we have ignored the neutrino rest mass and the small difference in electron binding energies of the two nuclei. Solving for the nuclear mass difference, M(Z − 1, A) − M(Z, A) = me c2 − Q .

The Q-value for electron capture is given by eq. (17.34). Using ∆(97 Tc) −87.2245 MeV and ∆(97 Mo) −87.5448 MeV, we find 43 42 Q 0.3202 MeV. Since Q < me c2 , the mass of the 97 42 Mo nucleus is greater than that of the 97 43 Tc nucleus.

17.21 97% of naturally occurring calcium is calcium-40,

40 Ca. This may seem surprising, since if we use the 20

semi-empirical mass formula to estimate the most stable nuclide with A = 40 we find Z ≈ 18. This suggests that 40 Ca might be unstable to electron capture, which 20 would increase its N/Z ratio. Show that 40 20 Ca cannot electron capture to 40 19 K. Show, however, that it is possible for 40 Ca to capture two electrons at the same time, 20 making 40 18 Ar. Look up the information on the lifetime of 40 Ca and comment on its value. Do we have to worry 20 about exposure to radiation from decay of the 40 Ca in 20 our bones?

40 For 40 20 Ca electron capture to 19 K,

Q = ∆(40 Ca) − ∆(40 K) = −34.8436 + 33.5354 < 0 , so this decay is forbidden by energy conservation. For 40 20 Ca double electron capture to 40 18 Ar, Q = ∆(40 Ca) − ∆(40 Ar) = −34.8436 + 35.0398 > 0 , so this decay is allowed. According to [70], the half-life for this decay is known to be greater than 3 × 1021 y 1029 s. About 1.5%

366

of the human body or 1 kg of a 70 kg person is calcium. This corresponds to ∼25 moles or 1.5 × 1025 calcium nuclei, of which almost all are 40 Ca. With a lifetime of greater than 1029 s, this results in . 1.5 × 10−4 decays per second – far less than other sources of radiation in the environment (see §20).

17.22 [H] When one of the naturally occurring radioac-

tive heavy elements like 238 U decays, a decay chain follows, leading to the creation of various unstable, radioactive descendant nuclides. Consider a sample known to be pure 238 U at time zero. Suppose that none of its decay products migrate away from the original sample. Use the fact that the lifetime of the original nuclide, in this case 238 U, is much longer than the lifetimes of any of its descendants to show that a quasiequilibrium develops in which the amounts of 238 U and all of its descendant are steadily decreasing at a rate determined by the lifetime of 238 U. Show that the ratio of the abundance of each descendant nuclide to the abundance of 238 U is very nearly constant in time, and is given by the ratio of the descendant’s lifetime to the lifetime of 238 U. Show that in this equilibrium every radioactive species contributes equally to the radioactivity (in decays per second) of the sample. [Hint: Write first-order differential equations that govern the rate of change of each nuclear species.]

Suppose the sample contains N0 (t) 238 U with lifetime τ0 , and N j (t) nuclei of daughter nuclide j with j = 1, 2, . . . m, and that the lifetime of daughter j is τ j . Then 1 dN0 = − N0 , dt τ0 dN1 1 1 = N0 − N1 , dt τ0 τ1 . . . dN j 1 1 = N j−1 − N j , dt τ j−1 τj 1 dNm = Nm−1 . dt τm−1 The solution to the first equation is N0 (t) = Ce−t/τ0 . Substituting into the second equation yields dN1 N1 C −t/τ0 + = e . dt τ1 τ0 This is an elementary differential equation with the solution (constrained to vanish at t = 0)

Cτ1 −t/τ0 e − e−t/τ1 .

N1 (t) = τ0 − τ1

Since τ0 ≫ τ1 , the “transient” term proportional to e−t/τ1 dies off quickly relative to e−t/τ0 and therefore N1 (t)

τ1 τ1 −t/τ0 Ce = N0 (t) . τ0 τ0

Proceeding in the same manner for each successive daughter in turn, we find τj N j (t) N0 (t) . τ0 The number of decays per unit time of daughter j is given by N j /τ j N0 (t)/τ0 , which is the number of decays of 238 U. Thus all nuclides in the decay chain contribute equally to the sample’s radioactivity.

17.23 In light of the results of Problem 17.22, what is the

rate of energy emission (J/kg) of a sample of 238 U that has been around long enough to come into decay equilibrium?

Since all decays contribute equally in equilibrium (Problem 17.22), the entire decay chain energy of the net reaction 238

U →206 Pb + 84 He + 6e− + 6ν̄e

is released at the rate of 238 U decay. The energy release is computed in the usual way Etotal = ∆(238 U) − ∆(206 Pb) − 8∆(4 He)

= 47.3 − (−23.8) − 8(2.4) = 51.7 MeV

(Some of this energy is lost on (anti-) neutrinos.) The energy released per second is dE dN0 N0 = Etotal = Etotal , dt dt τ where τ is the lifetime of 238 U, τ 6.45 × 109 y 2.04 × 1017 s.

Thus dE = 51.7 MeV × (1.60 × 10−13 J/MeV) dt × (1/.238 mol/kg) × (6.02 × 1023 /mol) ÷ 2.04 × 1017 s 1.0 × 10−4 W/kg

is the energy emitted per kilogram of 238 U.

17.24 The uranium isotope 234 U accounts for 0.0054% of

naturally occurring uranium. Its half-life τ1/2 (234 U) = 2.455 × 105 y is much too short for any primordial 234 U to still exist on Earth. Explain why 234 U occurs naturally and explain its fractional abundance. [Hint: Use the result of Problem 17.22.]

Assuming that the uranium has been in place long enough to achieve equilibrium with its daughters (Problem 17.22), the fractional abundance of any daughter nuclide in the 238 U decay chain is f = τ/τ0 , where τ is the daughter lifetime and τ0 is the 238 U lifetime. Using half-lives f (234 U) ≈ as observed.

2.46 × 105 5.5 × 10−5 = 0.0055% , 4.47 × 109

385

Problems

Problems 18.1

Suppose a beam of particles with flux Φ is incident on a uniform target of thickness T and density ρ (mass/volume). Assuming that the atoms in the target do not block one another (the thin-target approximation), show that the reaction rate dN/dt is given by dN/dt = IσρNA T/m, where I is the beam current (i.e., the integral of the flux over the area of the beam), m is the molecular mass of the target, NA is Avogadro’s number, and σ is the reaction cross section.

At 202.5 MeV/fission on average, the power deposited in the target is P = (5.73 × 1011 s−1 )(202.5 MeV)(1.602 × 10−19 J/ev) = 18.6 W .

18.4

First compute the number of 238 U and 235 U nuclei in a kilogram of natural uranium: 1 (6.022 × 1023 )(0.9928) = 2.51 × 1024 .238 1 N(235) = (6.022 × 1023 )(0.0072) = 1.82 × 1022 .238 N(238) =

According to eq. (18.1) dN/dt = σTOT φ, where σTOT here is the cross section of all the atoms in the path of the beam. For a section of the target of area dA, dσTOT = nT dAσ =

ρNA T dAσ , m

The number of decays per second is dN/dt = N/τ = ln 2N/τ1/2 , and the number of fissions per second is f (dN/dt), where f is the fraction of decays by spontaneous fission:

where n is the number density of atoms in the target, n = ρ NA /m.

Integrating over the area of the target, Z ρNA T ρ NA T σ I dN = σ dAΦ = dt m m

18.2

Assuming that the particles in Problem 18.1 that react are removed from the beam, use the result of that problem to show that the beam current falls exponentially I(z) = I0 e−z/λ through the target. Find an expression for λ, which is the mean free path of the particles passing through the target.

dNfission = 0.693(2.51 × 1024 )/(4.46 × 109 × 3.16 × 107 )(5.5 × 10−7 ) dt + 0.693(1.82 × 1022 )/(7.04 × 108 × 3.16 × 107 )(7.0 × 10−11 ) = 6.8 (from238 U) + 4 × 10−5 (from235 U)

So roughly 7 spontaneous fissions per second, essentially all from 238 U.

18.5

Consider a slab of the target between z and z + dz. The number of beam particles per unit time entering an area dA of the slab is dN(z)/dt = Φ(z)dA. The number leaving is dN(z + dz)/dt = Φ(z + dz)dA. The difference, dΦ dA = (Φ(z + dz) − Φ(z))dA = Φ′ dz dA must equal minus the number scattered. From Problem 18.1, dNscattered /dt = ΦσρNA dz dA/m, where we have replaced the thickness T by the thickness of the slab dz. Thus

n +A−1 [Z] →A [Z] which is given by

σρNA Φ(z) dA . m R Integrating over the area of the beam, I(z) = dA Φ(z), so

Q = ∆(n) + ∆(Z, A − 1) − ∆(Z, A) Averaging over the 5 isotopes from 234 U to 238 U

dI σρNA =− I dz m

which has the solution I(z) = I0 e

, where λ = m/(σρ NA ).

R A beam of thermal neutrons with current I ≡ dAΦ = 108 s−1 is incident on a uniform (thin) target of thickness 1 cm, consisting of 20% 235 U and 80% 238 U. See Table 18.3 and Problem 18.1. What is the fission rate? What is the thermal power being deposited in the target?

The density of uranium is ρ = 19.1 gm/cm3 . The thermal neutron fission cross section of 238 U can be neglected, so the target acts like pure 235 U with a density ρ/5. Thus dNfission = (108 s−1 )(585 × 10−24 cm2 )(19.1/5 gm/cm3 ) dt × (6.022 × 1023 mol−1 )(1 cm)/(.235 gm/mol) = 5.73 × 1011 s−1

1 (∆(92, 233) − ∆(92, 238) 5 1 = 8.071 + (36.921 − 47.310) = 5.992 MeV 5

hQi = ∆(n) +

−z/λ

dNfission = Iσρ NA T/m . dt

In §18.3.3 it is stated that a neutron in uranium is bound by ∼6 MeV. Check this by using data from [70] to compute the binding energy of a neutron averaged over the five isotopes 234 U through 238 U. The binding energy of a neutron in the nucleus A [Z] is the Qvalue of the reaction

Φ′ (z) dA = −

18.3

Given the data in Table 18.2, estimate the number of neutrons emitted by spontaneous fission per second per kilogram of naturally occurring uranium.

18.6

Verify the statements made after eq. (18.2) about the energy released for the case of thermal-neutroninduced fission of 235 U. First, use measured mass excesses to show that the asymmetric fission reaction nth + 235 U → 92 Kr + 144 Ba has a smaller Q-value than the symmetric fission into two 118 Pd nuclei. Next show that the reaction nth + 235 U → 2(116 Pd) + 4n also has a smaller Q-value than the symmetric fission into two 118 Pd nuclei. Finally, show that the SEMF underestimates the binding energy of 118 Pd and therefore overestimates the Q-value for the symmetric fission into two 118 Pd nuclei. nth +235 U → 2(118 Pd): Q = 8.071 + 40.922 − 2(−75.390) = 199.773 MeV

386

nth +235 U →92 Kr +144 Ba:

18.9

Q = 8.071 + 40.922 − (−71.767 − 68.769) = 189.529 MeV nth +235 U → 2(116 Pd) + 4n: Q = 8.071 + 40.922 − 2(−79.831) − 4(8.071) = 176.361 MeV According to the SEMF (17.19), binding energy of

118

Pd is

BSEMF (46, 118) = 986.21 MeV. This is less than the actual binding energy of

118

Pd, which is

Bactual (46, 118) = 46(7.29) + 72(8.07) − (−75.39) = 991.77 MeV. Thus the SEMF underestimates the binding energy of 118 Pd.

18.7

An example of a thermal-neutron fission reaction is nth + 235 U → 139 Ba + 94 Kr + 3n. Perform the analysis of Example 18.1 for this case.

First, the reaction, 139 94 235 92 U + nth → 56 Ba + 36 Kr + 3n .

The energy released promptly is

Qfission = ∆(235 U) + ∆(n) − ∆(139 Ba) + ∆(94 Kr) + 3∆(n)

Two sources of delayed neutrons in 235 U fission are the fission fragments 87 Br and 137 I. Find the fraction of their decays that yield a neutron, and the half-life of each. Verify that the Q-value for the β-decay with neutron emission is positive.

From [70], 87 Br has a half-life of 55.65 s and yields a neutron 2.6% of the time. That decay is 87 Br →86 Kr + n and has a Q-value of Q = −73.892 − (−83.266 + 8.071) = 1.303 MeV , which is positive.

137 I has a half-life of 24.5 s and yields a neutron in 7.14% of its β-decays. That decay is 137 I →136 Xe + n and has a Q-value of Q = −76.507 − (−86.429 + 8.071) = 1.851Ω MeV , which is positive.

18.10 [T] Verify the classical energy loss formula (18.4) for

an elastic two-body collision. Use the notation in the Figure 18.8. [Hint: Use the fact that the relationship between a velocity vector V in the target rest frame and the vector v in the center-of-mass frame is V = v+VCM , where V CM is the velocity of the center-of-mass frame.]

First relate initial velocities in the two frames: v1 = V1 − VCM and v2 = −VCM ,

= 40.922 + 8.071 − (−84.914 − 61.348 + 3(8.071))

= 171.042 MeV .

Both 139 Ba and 94 Kr decay by β-decays: 139

where all velocities are along the collision axis. Solve for the center-of-mass velocity Mv2 + mv1 = 0

Ba −−−−→ 139 La (stable) ,

−MVCM + m(V1 − VCM ) = 0

83.1 m

94

Kr −−−−→ 94 Rb −−−→ 94 Sr −−−−→ 94 Y −−−−→ 94 Zr (stable) , 2.1 ms

2.7 s

75.3 s

All these decays occur in a matter of minutes, and release additional energy, 139

Q1 = ∆(

139

Ba) − ∆(

VCM = mV1 /(M + m)

18.7 m

Then relate V1′ to the components of v′1 , V ′1 = v′1 + VCM = (v′1 sin θ, v′1 cos θ + VCM )

La) = 2.314 MeV

Q2 = ∆(94 Kr) − ∆(94 Zr) = 25.925 MeV Thus 28.24 MeV of additional energy is released (though some is lost on antineutrinos), so the total energy released in fission and subsequent β-decay is 171.04 + 28.24 = 199.28 MeV.

18.8

When 235 U fissions by absorbing a slow neutron, 93 Zr is one of the most common fission fragments. Suppose three neutrons are given off in the fission. What is the other nucleus produced in this fission reaction?

What is the total energy released in this particular fission reaction, including subsequent β-decays of the fragments?

Zr has an atomic number Z=40, so the other nucleus has Z = 92 − 40 = 52 → Te. The neutron number of the Te nucleus is 236 − 93 − 3 = 140, thus 140 Te is the other fission product. Looking for β-decays of the original fission fragments, we find that 92 Zr is stable, while 140 Te undergoes many β-decays, leading eventually to 52 140 Ce in a matter of minutes. Thus the total energy released in the 58 fission is Etotal = ∆(235 U) + ∆(n) − (∆(92 Zr) + ∆(140 Ce) + 3∆(n))

= 40.922 + 8.071 − (−88.079 − 88.460 + 3(8.071))

= 201.319 MeV

=

V1 (M sin θ, M cos θ + m) , M+m

where we have used |v′1 | = |v1 | = v1 = MV1 /(M + m) because the velocities of the colliding particles are the same before and after the collision in the center-of-mass frame. Then E′ =

m 1 mV ′2 = V 2 (M2 + 2mM cos θ + m2 ) 2 2(M + m)2 1

=E

M2 + m2 + 2mM cos θ (M + m)2

18.11 [T] Derive the expression eq. (18.7) for ξ(A) by averag-

ing over angles in the center-of-mass reference frame.

[Hint: Take E/E ′ from eq. (18.4) and remember that the measure for averaging over θ is 12 sin θdθ.]

From eq. (18.7), we write the required average as Z

E 1 π hln ′ i = dθ sin θ ln(1 + A)2 − ln(1 + A2 + 2A cos θ E 2 0 Z

1 1 dz 2 ln(1 + A) − ln(1 + A2 + 2Az = 2 −1

Evaluation of this definite integral yields the answer.

387

Problems

18.12 [T] (a) Derive the large A approximation to the loga-

rithmic energy decrement given in eq. (18.5). (b) Show that the number of collisions with a nucleus with atomic mass A required to decrease the energy of a neutron from E0 = 2 MeV to E f = 0.025 eV is N 18.2/ξ(A). (a) ξ(A) = 1 +

!

A 1 (ln(1 − 1/A) − ln(1 + 1/A)) −1+ 2 2A 2

3

4

Use ln(1 + x) → x − x2 + x3 − x4 . . .. ! !

A 1 2 2 ξ(A) −−−−→ 1 + −1+ − − + . . .

A→∞ 2 2A A 3A3

4 4 2 2 − + O(1/A3 ) = − + O(1/A3 ) A 3A2 A 3A2

−−−−→ 1 − 1 + A→∞

(b) According to eq. (18.5) hln(E j−1 )i − hln(E j )i = ξ , were E j is the neutron energy after the jth collision. Iterating this result hln(E0 )i − hln(EN )i = Nξ , so N = hln(E0 /EN )i/ξ .

For E0 = 2 MeV and EN = 0.025 eV, N = ln(80 × 106 )/ξ = 18.2/ξ .

18.13 [T] Estimate the height of the Coulomb barrier

between two protons as follows: the strong nuclear interaction potential between two protons at large separation is given by eq. (14.5). Add to this the repulsive Coulomb potential given in eq. (18.8). Show that the resulting potential has a maximum at ≈ 5.5 fm, where its value is VC ≈ 0.2 MeV. The sum of the Coulomb plus strong nuclear interaction potentials is g2strong −rb e2 V(r) = − e , 4πǫ0 r 4πǫ0 r m and g2strong

where, b 1.4 × 10 values and defining x = r/b, −15

V(r) = K

2

≈ 10e . Substituting these

! e−x 1 − 10 , x x

where K = e2 b/4πǫ0 . The maximum in this function must be determined numerically. It occurs at x0 = 3.90 (r0 ≈ 3.9(1.4 f m) = 5.6 fm) and has value V(r0 ) ≈ .205 = .205

e2 4πǫ0 b (1.602 × 10−19 C)2 4π(8.85 × 10−12 F/m)(1.4 × 10−15 m)

= .205(1.65 × 10−13 J) = 0.21 MeV

18.14 [T] Use the Maxwell–Boltzmann distribution (18.10)

to show that the mean kinetic energy of a particle in a gas at temperature T is 32 kB T .

R∞ dP hEi = 0 dE dEB E, where

! √ −E/k T 2π dPB B . = Ee 3/2 dE (πkB T )

Substitute y = E/kB T , hEi = √

2 πkB T

Z ∞ 0

√ !

2 3 π 3 dyy3/2 e−y = √ kB T = kB T 4 2 π

18.15 Show that the probability of finding a particle with

energy greater than 0.2 MeV in a gas at temperature 15 × 106 K is ∼10−69 .

We need the probability of finding a particle with energy greater than E0 = 0.2 MeV at a temperature of T = 15 × 106 K corresponding to an energy kB T = 1.25 keV. Using the Maxwell-Boltzmann distribution Z ∞ dPB P(E > 0.2 MeV) = dE . dE 0.2 MeV and substituting y = E/kB T , 2 P(E > 0.2 MeV) = √ π

Z ∞

√

ye−y

E0 /kB T

where E0 /kB T 200/1.25 = 160. The integral must be evaluated R ∞= √ numerically: 160 ye−y ∼ 4×10−69 , so (E > 0.2 MeV ∼ 3×10−69 .

18.16 The net reaction that accounts for the Sun’s power takes

four neutral hydrogen atoms and fuses them to make a neutral helium atom, 411 H→ 42 He. Compute the energy released per hydrogen atom. The Sun’s total power output is L⊙ 3.85 × 1026 W. How many protons per second are reacting in the Sun? Approximately 10% of the protons in the Sun will fuse during its stable hydrogen burning phase. Given the mass of the Sun, M⊙ 1.99 × 1030 kg, and the fact that about 75% by mass of the Sun is hydrogen (almost all the rest is helium), estimate how long the Sun’s hydrogen burning phase will last.

The energy released per hydrogen per fusion event is Q 1 = (4∆(1 H) − ∆(4 He)) = (4 × 7.289 − 2.425) = 6.68 MeV/1 H 4 4 L⊙ = (3.85 × 1026 W)/(1.6 × 10−19 J/eV) = 2.4 × 1045 eV/s Hydrogen consumed per second, L⊙ /(Q/4) = (2.45 × 1045 eV/s)/(6.7 × 106 eV/H) = 3.60 × 1038 H/s 75% of the mass of the sun is hydrogen of which 10% is consumed, and each kilogram of hydrogen contains 1/(0.001) × 6.0 × 1023 hydrogen atoms, for a total of N(1 H) = (0.075)(6.0 × 1023 H/kg)(2.0 × 1030 kg) = 9.0 × 1055 H. This yields a lifetime of T = (9.0 × 1055 H)/(3.6 × 1038 H/s) = 2.5 × 1017 s = 7.9 × 109 y , approximately 8 billion years.

18.17 The Coulomb barrier keeping two nuclei apart is given

by V(r) = Z1 Z2 e2 /(4πǫ0 r), where r is the relative separation of the two nuclei. Assume that V reaches its maximum value VC when r ∼ R1 + R2 , where R j is the radius of the jth nucleus. Using eq. (17.10), estimate the Coulomb barrier between a carbon nucleus and an α particle. Compare your results with Figure 18.9.

388

Equation (17.10) gives R(A) ≈ 1.3A1/3 , so 2

VC ∼

8e 4πǫ0 (1.3 fm)(41/3 + 121/3 )

8 × (1.6 × 10−19 C)2 = 4π(8.85 × 10−12 F/m)(3.9 × 10−15 ) m

= 4.7 × 10−13 J ∼ 3 MeV

Figure 18.9 shows a barrier of 3.43 MeV. Our calculation relies on a crude estimate of the internuclear separation and is about 15% too small.

18.18 Show that in d+t → n+ 4 He fusion reaction, the ratio of

the kinetic energy on the neutron to that on the 4 He is 4:1 if you neglect the initial kinetic energies of the reactants. Why is this a good approximation in terrestrial fusion experiments?

The energy released in dt fusion is 17.6 MeV (Table 18.7).

This is much greater than the energy necessary to overcome the dt Coulomb barrier. Thus the kinetic energies of the initial nuclei are negligible and they can be ignored. Let {m, v} ({M, V}) be the mass and speed of the neutron (α-particle). The particles are emitted back-to-back, so momentum conservation requires mv = MV or v 4V, since M = 4m. Then 1 1 E(n)/E(α) = ( mv2 )/( MV 2 ) = 4 2 2

.

18.19 Estimate the quantity of tritium (kg/year) required to

fuel a 1 GWe fusion power plant assuming 20% overall efficiency.

Convert the energy output to electron volts to compare with the energy released in the fusion reaction: E = (109 W)(3.16 × 107 s/y)/(1.6 × 10−19 J/ev) 1.7 × 1035 eV/y

Each fusion reaction yields 17.6×106 eV and consumes one tritium nucleus N(t) 5(1.7 × 1035 eV/y)(1.8 × 107 eV/t) = 4.7 × 1028 t/y , where the factor of five accounts for the efficiency. Each kilogram of tritium contains N(t) 6.0 × 1023 /0.003 2.0 × 1025 tritium nuclei. So the power plant consumes N(t) = (4.7 × 1028 t/y)/(2.0 × 1025 t/kg) 2.3 × 103 kg(t)/y which is 2.3 tonnes(tritium)/y.

18.20 Compare the energy released per nucleon for (a) fission

of 235 U; and (b) dt fusion.

235 U fission releases about 200 MeV or 200/235 0.85 MeV/nucleon. dt fusion releases 17.6 MeV and uses 5 nucleons for 17.6/5 3.5 MeV/nucleon.

18.21 Tritium is just barely unstable. How much more tightly

would tritium have to be bound in order for it to be stable? Write the nuclear reaction describing tt fusion.

How much energy is emitted in tt fusion? Unfortunately the cross section for tt fusion is much smaller than that for dt fusion, so even if tritium were stable, tt fusion would not be an interesting candidate for fusion power.

The Q-value for tritium β-decay is Q = ∆(3 H) − ∆(3 He) = 14.9498 − 14.9312 = 0.0186 MeV .

If tritium were bound by an additional 18.6 keV, the β-decay would no longer be possible, and instead 3 He would decay to 3 H by electron capture.

The tt fusion reaction would be t + t → α + 2n with an energy release of Q = 2(14.950) − (2.425 + 2(8.071)) = 11.333 MeV

418

Problems 19.1

Combining these we find k∞ = ηp f (1.84)(0.78)(0.90) = 1.30.

Starting from the Maxwell–Boltzmann distribution, eq. (18.10), for the probability of finding a particle with energy E in a gas with temperature T , show that the √ most probable speed for a particle is v̄ = 2kB T/m, but √ that the average speed of a particle is hvi = 8kB T/πm.

Verify that the most probable speed for a neutron at T = 20 ◦ C is ∼ 2200 m/s.

19.4

The relevant expressions for η, p and f are 585x 684x + 2.68(1 − x)  " #0.514    1−x 2.73  p(x, y) = exp − .51y/(1 + y) 9.3(1 − x) + 15.1x + 10.6y η(x) = 2.42

First we change variable from E to v using E = mv2 /2 and dE = mv dv, r !3/2 2 m dP 2 = e−mv /2kB T dv π kB T

f (x, y) =

19.2

Consider an infinite, homogeneous reactor fueled with uranium enriched to 3% and moderated by graphite with a ratio of graphite:uranium of 800:1. Find k∞ . To find p, you can ignore the uranium when computing hξi and in the calculation of the scattering cross section.

(You can assume ǫ = 1.)

From eq. (19.8), η = 2.42

585x 1.84 .

684x + 2.68(1 − x)

Next compute the resonance escape probability,  " #0.514    2.73 1 p = exp − 12  ξ( C) yσs (12 C)  " #0.514    2.73 1 = exp −  0.78 . .158 800(4.94) Finally, the thermal utilization factor f =

xσa (235 U) + (1 − x)σa (238 U) 0.92 xσa (235 U) + (1 − x)σa (238 U) + yσa (12 C

It is estimated that the Oklo reactor occurred when the ratio of 235 U to 238 U was 3.67%. Given the present ratio of 0.72% and the half-lives of both isotopes, estimate the time when the Oklo reactor was active.

Let U5,8 (t) be the abundance of 235 U and 238 U as functions 5,8 of time and let t1/2 be their half-lives. The abundances grow exponentially as a function of time before the present ∆t

Show from eq. (19.9) that in an infinite, homogeneous graphite-moderated reactor, the resonance escape probability is given by p(y) =

exp −17.28 (1/(9.3 + 4.9y))0.514 , where y is the ratio of moderator to fuel.

Substituting ξ(graphite) 0.158 and the scattering cross sections for 238 U and 12 C from Tables 19.1 and 19.2, gives the quoted result.

19.3

19.5

which has solutions v = 0 (a minimum in dP/dv) and v = √ 2kB T/m, a maximum.

R ∞ To find the average speed, perform the integral hvi = dv(dP/dv), which gives the quoted result.

0 For a neutron at T = 293 K, q v̄ = 2(1.381 × 10−23 J/K)(293 K)/(1.675 × 10−27 kg) 2, 200 m/s

Dividing through by number density of 238 U in the argument of the exponential in eq. (19.9), ignoring the small amount of 238 U, and substituting y = n(12 C)/n(238 U),   # "  2.73 n(238 U) 0.514  p = exp −  hξi Σs  #0.514  "   2.73 1  = exp − hξi σs (238 U) + yσs (12 C)

684x + 2.68(1 − x) 684x + 2.68(1 − x) + 0.0012y

Numerical evaluation of k∞ = ηp f for x = 0.0072 shows a maximum at y 290 where k∞ 1.18. k∞ is greater than one for 30 . y . 1 900.

To find the most probable v we differentiate with respect to v and set the result equal to zero: d dP ∝ (2v − mv3 /kB T ) = 0 , dv dv

Show that an infinite, homogeneous reactor fueled with natural uranium (0.72% 235 U) and moderated by heavy water (D2 O) can sustain a fission chain reaction. What is the optimal ratio of moderator to fuel?

∆t/t 5 1/2 U (0) 5

U5 (∆t) = 2

∆t/t 8 1/2 U (0) .

8

U8 (∆t) = 2

Defining the ratio R(t) = U5 (t)/U8 (t), and taking the logarithm,    1 1   5 − 8  ∆t = log2 (R(∆t)/R(0)) t1/2 t1/2 (1/(.704 Gy) − 1/(4.468 Gy))∆t = log2 (.0367/.0072) ∆t = 1.95 Gy

19.6

An approximate description of the Oklo uranium deposit is 90% uraninite (UO2 ) by mass, saturated with water that acted as the moderator. Assume that the Oklo deposit was homogeneously saturated with water with a ratio of 3.5 uranium nuclei per water molecule and compute the minimum enrichment (235 U/238 U) needed to sustain a fission reaction. You can ignore the moderating effect of the extra oxygen in the uraninite (it is negligible). You can also ignore the neutron-absorption cross section of the oxygen in the uraninite and of any other material in the deposit except for the water.

The question reduces to what is the minimum value of enrichment x for a water moderated reactor with y = 3.5 to obtain k∞ = 1.

The relevant equations for η, p, and f (setting ǫ = 1) are 585x 684x + 2.68(1 − x)  " #0.514    2.73 1−x  p(x, y) = exp − .92y/(1 + y) 9.3(1 − x) + 15.1x + 49.2y η(x) = 2.42

f (x, y) =

684x + 2.68(1 − x) 684x + 2.68(1 − x) + 0.664y

Numerical evaluation of k∞ = ηp f for y = 3.5 gives k∞ ≥ 1 for x ≥ 1.2%. A value of x considerably larger than this is required to make up for neutron losses. At x = 3.5% (Problem 19.5) k∞ 1.33.

419

Problems

19.7

Suppose a pressurized water reactor is loaded with 200 tons of 3% enriched uranium. The reactor has been designed to run with an average thermal-neutron flux of hΦn i = 1.5 × 1013 cm−2 s−1 for a period of four years before refueling. You can assume that Φn is held constant the whole time, ignore neutrons losses due to the finite size of the reactor, and ignore any contribution of 239 Pu fission to the reactor power. What is the thermal power output of the reactor immediately after it is started? How does the power output decrease as a function of time over the four years? Use thermal-neutron cross sections from Table 19.1.

The fuel contains 1.5 t of 235 U. At startup the fission rate is given by eq. (19.16) with N(235 U) = 1 000 kg/(0.235 kg/mol)(6.02 × 1023 nuclei/mol) = 3.8 × 1027 nuclei ,

and the power is P f = E f R f = (202.5 MeV)(684 b) × (1.5 × 1013 cm−2 s−1 )N(235 U) = 1.26 GWth

19.8

[C,T] The analysis of prompt and delayed neutrons in §19.1.6 leads to the following formula for the time dependence of the neutron density, ! ρt d exp n(t) = n0 d−ρ (d − ρ)td !! −(d − ρ)t ρ . exp − d−ρ tp Analyze the time dependence of n(t) when ρ < d and when ρ > d. For example, take t p = 2 × 10−4 s, td = 12.5 s, and d = 0.0064, and compute n(t)/n(0) for ρ = d/2 and ρ = 2d.

As shown in the figure, the ratio n(t)/n(0) grows slowly when ρ < d, but grows exponentially when d > ρ. For d = 2ρ, the ratio exceeds 1 000 when t = 2 s.

The differential equations are ṄI = −NI /τI and ṄX = NI /τI − NX /τX . Solving for NI (t) = NI0 e−t/τI and substituting into the equation for ṄX , Ṅ x = −

N0 NX + I e−t/τI . τX τI

This differential equation is solved by introducing an integrating factor, i.e by defining NX (t) = e−t/τX f (t), then f˙ = (NI0 /τI ) exp(t/τX − t/τI ), which can be integrated to give NX (t) =

NI0 τX −t/τ e I + const e−t/τX . τI − τX

Choosing the constant so that NX0 = 0, NX (t) =

NI0 τX −t/τ e I − e−t/τX . τI − τX

Substituting the lifetimes τI = 9.49 h and τX = 13.19 h, we find

NX = 3.56 NI (0) e−t/13.19 − e−t/9.49 ,

where t is measured in hours. A plot of this result over 60 hours looks qualitatively similar to Figure 19.5.

19.10 A uranium enrichment facility produces nuclear reac-

tor fuel enriched to 4% 235 U to fuel the reactor whose power to fuel consumption ratio is described by eq. (19.15). For each tonne of natural uranium, the enrichment facility produces 130 kg of fuel. What is the concentration of 235 U in the 870 kg of depleted uranium? Assume that the percentage of 235 U in spent nuclear fuel removed from the reactor is the same as in natural uranium. How many tonnes of fuel are used in a year? Show that 211 tonnes of natural uranium is required to provide the 0.9 tonnes of 235 U consumed by this reactor in one year. Repeat your analysis assuming that the enrichment facility produces 120 kg of fuel enriched to 5% and that the percentage of 235 U in the spent fuel remains the same.

The 130 kg of enriched uranium contains 0.04 × 130 = 5.2 kg of U. Since the original tonne of natural uranium contained 7.2 kg 235 of U, 2 kg remains in the 870 kg of depleted uranium, corresponding to 2/870 = 0.23%. Each tonne of fuel contains 40 kg of 235 U before use and 7.2 kg of 235 U after use, so 32.8 kg of 235 U is consumed for each tonne of fuel. Since the reactor consumes 0.9 t of 235 U, it requires 900/32.8 27.4 t of fuel. Since each tonne of natural uranium yields 130 kg of fuel, 27.4/0.13 211 t of natural uranium is required.

For the fuel enriched to 5%, the depleted uranium contains 0.14% 235 U; 21 t of fuel is required per year; and 175 t of natural uranium is required.

235

19.11 A thermal-neutron reactor is charged every year with

19.9

[C,H] When a reactor is turned off, the amount of 135 Xe decreases as it decays, but grows as 135 I decays to it.

Write a pair of differential equations that describe the time rate of change of the numbers NI and NX of 135 I and 135 Xe nuclei respectively, and solve these equations assuming that NI = NI0 and NX = NX0 at t = 0. Show (using the measured half-lives of 135 I and 135 Xe) that the resulting curve of NX (t) looks like the t > 0 section of Figure 19.5. To simplify your calculations, take NX0 = 0.

enriched uranium containing 1.6 t of 235 U. When the spent fuel is removed, it contains 400 kg of 235 U and 560 kg of 239 Pu. Assuming a conversion ratio CR = 0.6, what fraction of the reactor power was generated by fission of 239 Pu? (Ignore radiative capture of thermal neutrons in both 235 U and 239 Pu, ignore production and fission of any other actinides, and assume that the energy releases in 235 U and 239 Pu fission are identical.)

420

Let Pc and Pd be number of plutonium nuclei created and destroyed, respectively. Then CR =

Pc Pc = = 0.6 Ud + P d 1.2 + Pd

per cubic meter), and how does this compare with the density of air at room temperature and pressure?

At this temperature, bremsstrahlung can be ignored, so the second term in the denominator in eq. (19.26) can be dropped,

and Pc − Pd = 0.56 Solving for Pc = 0.96 and Pd = 0.4. Assuming equal energy from each species and ignoring radiative capture, the fraction of power from plutonium is f = Pd /(Ud +Pd ) = 0.4/(1.2+ 0.4) = 0.25.

pτE ≥

19.12 A thermal breeder reactor is fueled with a mixture of a

=

fertile nucleus A (fraction 1 − x) which breeds a fissile nucleus B (fraction x). To breed more of B it is essential that the reproduction factor η(B) in thermal-neutroninduced fission of B is greater than two. Express the maximum possible value of η(B) in terms of cross sections and the average number of fast neutrons in Bfission. Evaluate η(B)|max for the two possible breeding systems, (238 U,239 Pu) and (232 Th, 233 U). Which is the most promising for a thermal breeder reactor?

The reproduction factor η is maximized when the fuel entirely composed of the fissile nucleus, i.e. when x = 1, ηmax =

σf η0 σa

where η0 is the number of fast neutrons emitted when the fissile nucleus undergoes thermal neutron induced fission and σa/ f is the absorption/fission cross section for the fissile nucleus. The relevant data can be found in Tables 18.3 and 19.1. For a 233 U thermal breeder reactor, 531 2.50 = 2.30 , ηmax (233 U) = 576 and for a

239

Pu thermal breeder reactor ηmax (239 Pu) =

748 2.88 = 2.11 .

1016

Thus the (232 Th,233 U) reactor is more promising.

19.13 Show that a charged particle moving in the plane per-

pendicular to a constant magnetic field (eq. (19.29)) moves in a circle with Br = mv/q. What is the radius of the circle for an electron with energy 15 keV in a 10 Tesla magnetic field (typical of ITER)? What is the radius for a proton?

The magnetic force on a charged particle is given by eq. (3.45).

Since the force is at right angles to the velocity, the particle’s speed is constant, and it moves in a circle. The force must equal the centripetal force required for circular motion: Fcentripetal = mv2 /r = qvB, thus r = mv/qB. For a 15 keV electron r r p 2E 2E =c v= = c 30/511 7.3 × 107 m/s . m mc2 Thus for an electron,

(9.11 × 10−31 kg)(7.3 × 107 m/s) mv = 4.1 × 10−5 m qB (1.6 × 10−19 C)(10 T) p p For a proton v p = ve me /m p , so m p v p = me ve m p /me and r p = √ 1840 re 1.8 × 10−3 m. re =

19.14 A dt plasma fusion reactor is operating in steady state

at a temperature kB T = 15 keV and a pressure of 7 atm, where hσ f vi = 3 × 10−22 m3 /s. What is the energy confinement time τE required to satisfy the triple product criterion? What is the density of the plasma (particles

24(kB T )2 hσ f viEα

24(15 × 103 eV)2 (3 × 10−22 m3 /s)(3.5 × 106 eV)

8.2 × 105 Pa s 8.1 atm s

At a pressure of 7 atm, the confinement time must exceed 1.2 s.

The particle density is given approximately by the ideal gas law, pV = NkB T , so n=

N p 8.2 × 105 Pa = = = 3.4 × 1020 m−3 .

V kB T (1.5 × 104 eV)(1.6 × 10−19 J/eV)

The density of air at NTP is roughly 1 mole per 22.4 L or n 2.7 × 1025 m−3 . So the number density of the plasma is approximately 3.4 × 1020 /2.7 × 1025 ≈ 1.3 × 10−5 compared to air at NTP.

19.15 Suppose a fusion power plant operates at the pressure

(≈ 7 atm) and temperature (≈ 150 × 106 K) planned for ITER, and suppose the plasma volume equals that of ITER (840 m3 ) as well. Assuming that the plasma contains equal numbers of d and t nuclei, estimate how long such a fusion reactor could provide fusion power equal to 1 GW before it would have to be refueled.

Using the ideal gas law, p = 2nkB T , where nd = nt = n/2, and solving for nd = nt , nd = nt =

p 7 × 105 Pa ≈ ≈ 8×1019 m−3 4kB T 4(1.4 × 10−23 J/K)(1.5 × 108 K)

The energy release per fusion event is E f = 17.6 MeV, so the total energy available is

Etot = nd VE f ≈ (8 × 1019 m−3 )(840 m3 )(17.6 MeV)(1.6 × 10−19 J/eV) ≈ 1.3 × 1030 eV = 2 × 1011 J ,

giving a refueling time of T ≈ 2(×1011 J)/(109 J/s) ≈ 200 s, or about three minutes.

451

Problems

Problems 20.1

20.6

Show that the maximum energy transfer by a heavy particle with mass M and speed v to an electron initially at rest is 2me v2 , hence the appearance of this factor in the formula for the stopping power.

According to Figure 20.4, a 2 MeV α-particle has a range of ≈ 0.008 mm. The range scales inversely with the electron density of the material, so if we denote Rα (Z, A) as the range of an α-particle in material with nuclear charge Z and atomic mass A, then

Equation (20.2) gives the maximum energy lost by the heavy particle when M ≫ me . Energy conservation requires this to equal the energy of the electron Ee = (4me /M)( 21 Mv2 ) = 2me v2 .

20.2

Rα (Z, A) =

Estimate a lower limit on the number of collisions a 1 MeV proton must make with atomic electrons before its energy is reduced to 10 eV. Why is it reasonable to ignore the atomic binding energy of the electrons in this estimate?

Rα (Z, A) =

20.7

M ln(E0 /E) 4me

20.3

938.3 ln(105 ) 460 × 11.5 5.3 × 103 4 × 0.511

Show that a particle’s range (see §20.2.1) expressed as a function of its initial velocity R(v) is proportional to its mass and inversely proportional to the square of its charge and to the electron density of the material.

http://physics.nist.gov/cgi-bin/Star/ap_table.pl give the range of protons in water for energies 100 MeV (R = 7.72 cm) and 125 MeV (R = 11.46 cm). Interpolating linearly between the two E(10 cm) = 100+(10−7.72)/(11.46−7.72)×25 115 MeV.

20.8

The right hand side of Bethe’s formula for the stopping power (20.3) is of the form

1−β

where K includes all the other constant factors. Writing E = 21 Mv2 and integrating over the appropriate intervals, z2 (nZ) R(v) dv v / f (v ) = K M 0

so R(v) =

20.4

M Kz2 (nZ)

R

′ ′

′

dv′ v′ / f (v′ ), as required.

[T] Derive eq. (20.4) from eq. (20.3) and the approximations quoted in the text.

Problem OK, no solution written.

20.5

Show that in any material, the range of an α-particle and a proton with the same speed are the same (in the approximation that mα = 4m p ). Using Figure 20.3, find the range of a 2 MeV proton in water.

See the solution to Problem 20.3. The range is proportional to M/z2 . Since Mα 4m p and zα = 2z p , the proportionality factor is one.

A proton with energy 2 MeV has the same speed as an α-particle with energy 8 MeV, so the range in water of a 2 MeV proton is about 0.07 mm (from the figure).

[T] According to Einstein’s theory of relativity, the kinetic energy of a particle with mass m and speed v = βc is Ekin = mc2 ( √ 1 2 −1). First, verify that when

v ≪ c this reduces to the familiar result, Ekin = 21 mv2 .

Then compute β for an electron and an α-particle with 5 MeV kinetic energy.

dE = −Kz2 (nZ) f (v), dx

Z v

To a first approximation for computing stopping power, living tissue can be approximated as water. Using data on the range of protons in water, estimate the energy of a proton required to place its Bragg peak on a tumor at a depth of 10 cm inside a body.

This problem requires information available on line. Data at the NIST website

for an initial energy E0 and final energy E. Substituting the conditions of the problem, n=

ρ(water) R(water) ρ(Z, A)

The density of water/air/bone is 999/1.25/1 100 kg/m3 , so Rα (air) ≈ 999/1.25Rα (water) = 6.4 mm and Rα (bone) ≈ 999/1 100Rα (water) = 0.007 mm.

where n is the number of collisions, and we have approximated the energy loss as a continuous process. Solving this differential equation, n=

n(water)Z(water) Rα (water) n(Z, A)Z

Approximating n = ρ/m ≈ ρ/(Au), where m is the molecular mass and u is the Dalton, and taking Z/A ≈ 1/2 for all the materials, we have

Assume the maximum possible energy is transferred in each collision, then from eq. (20.2) me dE = −4 E dn Mp

Using the results of Problem 20.3, compare the range of a 2 MeV α-particle in water, air, and bone. (The density of bone is approximately 1100 kg/m3 .)

Problem OK, no solution written.

20.9

The mass attenuation coefficients (µ/ρ) for 1.5 MeV photons in concrete (ρ = 2.35 g/cm3 ) and lead (ρ=11.4 g/cm3 ) are both about 0.5 cm2 /g. How thick must concrete or lead shielding be in order to absorb 99% of these 1.5 MeV γ-rays? µ = (µ/ρ)ρ, so µ(concrete) = 1.17 cm−1 and µ(lead) = 5.7 cm−1 . From eq. (20.5) I(d)/I0 = e−µd = 0.01, then d = − ln(0.01)/µ = 4.6/µ, so d(concrete) = 4.6/1.17 = 3.9 cm, and d(lead) = 4.6/5.7 = 0.8 cm.

20.10 In §20.3.1 it is stated that activity a is often used

as measure of the mass of radionuclide in a sample m. Show that for a radionuclide with half-life t1/2 and atomic mass A, this relationship is m A t1/2 [sec] a[Ci] 8.87 × 10−17 kg.

The number of nuclei is a sample of mass m is n = m/(Au), where u = 1.661 × 10−27 kg is the dalton. The number of decays

452

per second dn/dt is given by dn/dt = n/τ = n ln 2/t1/2 = 3.7 × 1010 a[Ci]. Combining these equations,

So approximately 8 × 10−11 grams of 131 I must be absorbed per

t1/2 [sec] m = Au × (3.7 × 1010 ) a[Ci] ln 2 −17 = 8.87 × 10 t1/2 [sec] A a[Ci] kg .

ing into the conversion equation from Problem 20.10, we find

20.11 Thermal neutrons wandering through biological tissue

are most likely to be absorbed by 1 H, 12 C, or 16 O. If absorption is accompanied by emission of a single γ-ray, what are the γ-ray energies produced in these three cases?

In each case we need to evaluate the Q-value for the reaction [Z] + n →[A+1] [Z] because that is the energy carried off by the photon. Generalizing eq. (17.6)Q = ∆(Z, A) + ∆(n) − ∆(Z, A + 1).

The data can be obtained from the NNDC data site: [A]

Q(1 H) = 8.0713 + 7.2889 − 13.1357 = 2.2245 MeV

Q(12 C) = 8.0713 + 0 − 3.1250 = 4.9463 MeV

Q(16 O) = 8.0713 − 4.7370 − (−0.8087) = 4.1430 MeV

20.12 A nuclear accident has exposed a 70 kg worker to 1 Ci

of 2 MeV photons for about one minute. Assuming that all the energy of absorbed photons is deposited in the body, with uniform distribution, what absorbed dose of radiation has the worker suffered? What effective dose? What would have been his effective dose if the exposure had been concentrated entirely in his stomach (mass 0.5 kg)?

To compute the absorbed dose we need the total energy absorbed, E = (2 × 106 eV) × (60 sec) × (3.7 × 1010 sec−1 ) × (1.6 × 10−19 J/ev) 0.71 J

In a 70 kg person this corresponds to D 0.71/70 0.01 Gy. The radiation weighting factor for γ-rays is one so the effective dose is H = 10 mSv. If the dose were entirely absorbed in an organ with mass 0.5 kg it would correspond to an absorbed dose of DR,T = 0.71/0.5 = 1.4 Gy. This would be the proper measure of deterministic effects such as ARS. The risk of stochastic effects such as cancer should be measured by the effective dose, E = wT wR DR,T with wR = 1 for γ-rays, wT = 0.12 for the stomach. Thus E = 0.12 × 1.4 = .17 Sv.

20.13 131 I is a dangerous fission product because iodine is

preferentially absorbed by the thyroid gland in children. It has a half-life of about 8 days and emits a β-particle with average energy 180 keV. (Ignore the γ-radiation that accompanies this decay (see Example 20.2.)) How much mass of 131 I (per gram of thyroid tissue) must a child absorb to suffer an equivalent dose of 10 mSv to the thyroid? What is the activity per gram of this radiation dose (in curies) (see Problem 20.10)?

With a radiation weighting factor of one for β-particles, 10 mSv corresponds to 10 mGy or 0.01 J/kg. With a decay energy of 180 keV or 2.8 × 10−14 J, it takes 0.01/(2.8 × 10−14 ) = 3.5 × 1011 decays to give this dose. Assuming all the decays occur before the iodine is excreted from the body, this corresponds to a mass of M 3.5 × 1011 × 131 u × (1.66 × 10−27 ) kg/u 7.6 × 10−14 kg .

gram of thyroid tissue to provide this equivalent dose. Substitut-

a = 7.3 µCi/gm.

20.14 Rubidium-87, 87 Rb is a relatively common, very long-

lived nuclide. Look up and compare its abundance and lifetime to 40 K. How does 87 Rb decay? Why do you think it was omitted from the list of important naturally occurring radionuclei?

Both nuclei decay primarily by β-decay. There are three issues: i) the relative activity in Earth’s crust, ii) the energy of the emitted electron, iii) are there accompanying γ-rays as there are in the case of 40 K (see §20.5.4)? First, the relative activity: The ratio of the activity of 87 Rb to 40 K is proportional to the ratio of their crustal abundances times the inverse of the ratio of half-lives. The crustal abundances of Rb and K are ≈ 9 × 105 and 2 × 10−2 (by mass) respectively. Since we require abundance by number, this ratio must be reduced by a factor of 40/87. 40 K is 0.017% of naturally occurring K and 87 Rb is 27.8% of naturally occurring Rb. Their half-lives are 1.24 × 109 y(K) and 4.8 × 1010 y(Rb). Thus, a(87 Rb) 40 0.28 × (9 × 10−5 ) 1.24 × 109 ≈ 87 0.00017 × (2 × 10−2 ) 4.8 × 1010 a(40 K) ≈ 0.088

So averaged over Earth’s crust, 87 Rb contributes roughly 1/11 as much β-decay activity as 40 K. Next, electron energy: According to [70], the mean electron energy in 87 Rb β-decay is ∼82 keV, compared with ∼240 keV for 40 K β-decay, so the rubidium β-radiation is less penetrating, though in both cases the electron would be shielded by less than a centimeter of rock. Finally, γ-radiation: As stressed in §20.5.4, the γ-rays that accompany the electron capture of 40 K decay to 40 Ar are the greatest source of radiation from 40 K. A look at [70] confirms that there are no γ-rays accompanying 87 Rb β-decay, so this problem does not occur.

20.15 According to the text, roughly 10–15% of lung can-

cer deaths in the US are due to radon exposure. Use the LNT model and average yearly effective dose from radon to estimate the cancer rate from this source.

The annual average effective dose from radon exposure in the US is 2.28 mSv. Assuming the LNT model, this leads to an annual average increased risk of fatal cancer of (2.28 mSv)(0.055/Sv)≈ 10−4 per person. Integrating over the whole US population, one expects (300 × 106 )(10−4 ) ≈ 30 000 fatal cancers per year. Assuming that all radon induced cancers are lung cancer, we conclude that the annual number of lung cancer deaths in the US lies between 30 000/0.15 ≈ 200 000 and 30 000/0.10 ≈ 300 000. This is likely to be an overestimate since not all radon related cancers are lung cancer. In 2012 (the latest year for which data are available), there were approximately 157 000 lung cancer deaths in the US.

20.16 According to relativity, the clocks of an object mov-

ing withpspeed v = βc are slowed down by a factor γ ≡ 1/ 1 − β2 . The total (relativistic) energy of the object is given by mc2 γ. The mean energy of cosmic ray muons at creation is about 6 GeV. How far can such a muon travel on average before decay?

Problem OK. No solution written.

20.17 The flux of cosmic ray muons at sea level is about

one per square centimeter per minute. Given their mass stopping power of 2 MeV cm2 /gm, and their weighting factor wµ = 1, estimate an average person’s (living at sea level) equivalent dose in mSv/y from cosmic ray muons.

453

Problems

For an estimate, take a human cross sectional area as 0.1 m2 , height 1.5 m, and mass 70 kg. Then the number of muons passing through an individual per year is roughly

Use the formula from Problem 20.10 to solve for the activity as a function of t1/2 , A, and the branching fraction for spontaneous fission f . In general a[Ci] =

((1/60) × 104 /m2 s) × 3.2 × 107 s/y × 0.1 m2 ≈ 5 × 108 muons/y Taking the density of a human being to be roughly 1 gm/cm3 , each muon deposits 2×150 ≈ 300MeV in 70 kg of tissue. The equivalent dose is roughly 8 muons

H ≈ 5 × 10

y

1.13 × 1016 f A t1/2 [s]

and a[Bq] = 3.7 × 1010 a[Ci]. Substituting the data from the table, aS F (240 P) = 4.8 × 105 Bq aS F (239 P) = 6.9 Bq

MeV 1.6 × 10−19 J/eV × 300 × ≈ 0.3 mSv . muon 70 kg

aS F (235 U) = 0.0056 Bq aS F (238 U) = 6.8 Bq

which is the right order of magnitude (the average cosmic ray dose at sea level is estimated at 0.38 mSv).

20.18 Particle physics experiments are shielded from cos-

mic ray muons by placement underground, often in old mines. Cosmic ray muons have been detected at 2000 feet below ground at the Homestake Gold Mine in South Dakota. Estimate a lower limit on the energy of these muons when created high in the atmosphere.

2 000 ft ≈ 600 m. S = 2 MeV cm2 /gm, and the mean density of rock is about 2.5 − 3.0 gm/cm3 so |dE/dx| ≈ 5 − 6 MeV/cm in rock. In 600 m a muon therefore loses about 6 000 × (5 − 6) MeV = 30 − 36 GeV, so the muons that make it to a depth of 2 000 ft must have energies in excess of about 30–35 GeV.

20.19 40 K accounts for 0.0117% of naturally occurring potas-

sium and potassium accounts for about 0.2% of the human body mass. Considering only the dominant decay of 40 K, which yields a β-decay electron of average energy 0.56 MeV, compute the average annual dose of radiation from potassium inside the body of a 70 kg person. Why can the exposure due to the γ-rays accompanying the electron capture decay be ignored in this estimate?

First compute the number of 40 K nuclei in a 70 kg person: N=

0.000117 × 0.002 × 70 40 × 1.66 × 10−27 ≈ 2.5 × 1020

These give rise to an activity of dN/dt = N/τ = ln 2N/t1/2 =

.692 × (2.5 × 1020 ) ≈ 4 400 s−1 , (1.24 × 109 ) × (3.16 × 107 )

but only 90% of 40 K decays are β-decays, so dN/dt|β ≈ 4 000 s−1 .

With a radiation weighting factor of one, the equivalent dose is then H ≈ 4 000 s−1 × (.53 MeV) ×

1.6 × 10−19 J/eV ≈ 5.0 × 10−12 Sv/s 70 kg

or H≈ (5.0 × 10−12 Sv/s) × (3.16 × 107 s/y) ≈ 16 mSv/y. This is quite close to the conventional value of the DARI. The electron capture decay occurs only 10% of the time, and although the γ-ray energy is greater than the mean electron energy (1.46 MeV compared with 0.56 MeV), the attenuation length of the γ is so great that only a fraction of the γ-ray energy is absorbed by the surrounding tissue. Hence this contribution to the exposure is small compared with the β-decay contribution.

20.20 Using the data from Table 18.2, compute the sponta-

neous fission rate (in Bq/kg) for 240 Pu. Compare this with the rate in 239 Pu and in uranium enriched to 85% 235 U. (See Problem 20.10.)

The activity of uranium enriched to 85% 235 U is almost entirely due to the 238 U content, aS F (85%) = 0.15 × 6.8 = 1.0 Bq.

20.21 In §20.6.1 it was stated that approximately 21 t of SNF

are removed yearly from a 1 GWe reactor operating at 33% efficiency. Assuming that the spent fuel contains 0.9% unconsumed 235 U and that the reactor derives 1/3 of its power from burning 239 Pu created during normal operation, determine the initial enrichment of the fuel (the initial percentage 235 U).

The reactor produces 3 GW of thermal energy of which 2 GW comes from 235 U fission. The total fission energy per year is (2 × 109 J/s) × (3 × 107 s) 6 × 1016 J .

Each fission produces ∼ 200 MeV = 3.2 × 10−11 J, so (6/times1016 )/(3.2 × 10−11 1.9 × 1027 235

U nuclei fission per year. Dividing by Avogadro’s number and multiplying by the mass per mole of 235 U gives ∼ 780 kg(235 U)/y.

This corresponds to 780/21 000 = 3.7% of the total mass of the SNF. Combining with the 0.9% remaining, the original enrichment was 4.6%.

20.22 Consider a situation in which, through accident or

ill-intention, all the spent nuclear fuel from a year’s operation of a 1 GWe light-water reactor were uniformly distributed across one square kilometer of land 1000 years after the waste is sequestered. Using the information in the chapter, estimate the radiation exposure in Sv/year to an unprotected human on this land.

For simplicity, assume that the activity is dominated by 241 Am. (According to [107] the effective dose (per hour) from a planar 241 Am source of γ radiation at a distance of one meter is 1.9×10−4 mSv/h for an activity (per unit area) of 1 MBq/m2 .) How broad an area would be needed to reduce exposure below the level of background radiation from within the body (0.29 mSv/y)?

Based on this calculation, estimate the area that could be rendered “unsafe/uninhabitable for children” (>10 mSv/year) if all waste from 50 years of nuclear power (from light-water reactors) at 2 TWe were distributed uniformly on the surface 1000 years later.

According to Table 20.7 the SNF from one year of operation of a LWR initially contains ∼10 kg of 241 Am. 1 000 years later this is reduced to m(241 Am) ∼ 10 × 2−1 000/432 ∼ 2 kg. Using the formula from Problem 20.10, this corresponds to an activity of a = 250 TBq. Spread over 1 km2 = 106 m2 , the activity per unit area is 250 MBq/m2 , which generates an effective dose of

454

E = (250 MBq/m2 )(1.9 × 10−4 mSv m2 /h MBq) ∼ 50 µSv/h, or E = (50 × 10−6 Sv/h)(8 800 h/y) = 440 mSv/y. To reduce this effective dose to 0.29 mSv/y , it would have to be spread over 440/0.29 ∼1 500 km2 .

Scale this estimate up to answer the second part. 2 TWe requires 2 000 LWR of the type considered. Operating for 50 years, the SNF generated is 2000 × 50 = 105 times greater than the single-reactor, single-year case. So the dose rate if this SNF were spread over A square kilometers would be E=

(440 mSv/y)(105 reactor − years) 44 × 106 mSv/y = A[km] A[km2 ]

To reduce this dose to 10 mSv/y would require an area of ∼4.4 × 106 km2 , which is somewhat less than half of the area of the United States.

473

Problems

Problems 21.1

21.5

[T] The equation of motion of a classical harmonic oscillator subject to an external time-dependent force F(t) is m ẍ = −kx + F(t). Show that the oscillator’s energy is not conserved, but that its time dependence dE/dt can be related to the power delivered to the oscillator by the agent supplying the force.

Begin with the equation of motion and multiply by velocity, ẋ, ẋ(m ẍ + kx) = ẋF(t) , and note that the left hand side of this equation is the time derivative of the energy (kinetic plus potential) of the oscillator, ! d 1 2 1 2 dE ẋ(m ẍ + kx) = m ẋ + kx = = F(t) ẋ dt 2 2 dt and the right hand side is the power delivered by the force F(t), see eq. (2.12).

21.2

21.3

The solution method is similar to Problem 21.4.

[T] Derive the results of Box 21.1 (the last two equations) by using the Schrödinger equation for the time derivative of ψ(x) and ψ∗ (x) and integrating by parts.

[T] Confirm by explicit computations the assertions made in Example 21.1 that the Hamiltonian is invariant under translation and rotation so that total momentum and angular momentum are conserved.

∂D ∂H ∂D ∂H ∂D ∂H ∂D ∂H − + − ∂x ∂p x ∂p x ∂x ∂y ∂py ∂py ∂y ! ! ∂ ∂ ∂ ∂ = px + py +y H− x H. ∂x ∂y ∂x ∂y

Ḋ = {D, H} =

Using x = r cos φ and y = r sin φ, ! ∂x ∂ ∂y ∂ 1 ∂ ∂ ∂ = + = +y x . ∂r ∂r φ ∂x ∂r φ ∂y r ∂x ∂y Likewise

For translation, the x1 → x1 + a and x2 → x2 + a, where a is an arbitrary constant two-dimensional vector. Since da/dt = 0, p1 = m ẋ1 and p2 = m ẋ2 are unchanged, and

! ∂ 1 ∂ ∂ = + py . px ∂p p ∂p x ∂py Then

r = |x1 − x2 | → |x1 + a − (x2 + a)| = r, so V(r) is unchanged as well. Thus H is unchanged under translation.

When a vector is rotated, its length does not changed. Since p21,2 and r are the lengths of vectors, H is unchanged under a rotation.

This can be verified by explicit calculation of the relevant Poisson brackets.

21.4

Ḋ = p

because D(T ) is bounded (for a bound system). Thus h2T i = hr∂V(r)/∂ri = ahV(r)i for a power law potential.

For the Earth-moon system, a = −1. Since E = T (p) + V(r), we find E = hVi/2 = −|hVi|/2. Thus, for example, as tidal drag transfers energy from Earth’s rotation into the moon’s orbital motion, E becomes less negative and according to the virial theorem, the radius of the moon’s orbit must increase, and the moon spirals away from Earth.

From its definition, eq. (21.11), ∂L ∂H ∂L ∂H ∂L ∂H ∂L ∂H − + − . ∂x ∂p x ∂p x ∂x ∂y ∂py ∂py ∂y

Since L = xpy − yp x , we have ∂L/∂x = py , etc., so ∂H ∂H ∂H ∂H +y − px −x ∂p x ∂x ∂py ∂y ! ! ∂ ∂ ∂ ∂ −y − py H. H − px =− x ∂y ∂x ∂py ∂p x

{L, H} = py

Finally, from the relations between x, y, p x , py and r, φ, p, φ p , note that ∂ ∂x ∂ ∂y ∂ ∂ ∂ = + = −y +x ∂φ ∂φ r ∂x ∂φ r ∂y ∂x ∂y and likewise for ∂/∂pφ . Substituting into the last equation for {L, H} yields eq. (21.17).

! ∂ ∂ ∂V p2 −r −r .

H= ∂p ∂r m ∂r

In general, this does not vanish, so D is not a constant of the motion.

In a bound orbit the long term average of Ḋ is zero: Z T 1 1 Ḋ(t) = lim (D(T ) − D(0)) = 0 , hḊi = lim T →∞ T T →∞ T 0

[T] Consider motion in the xy-plane governed by a Hamiltonian H(px , py , x, y). Introduce polar coordinates x = r cos φ, y = r sin φ, px = p cos φ p , py = p sin φ p , and use the chain rule for partial derivatives, e.g. ∂H/∂φ = (∂H/∂x)(∂x/∂φ) + (∂H/∂y)(∂y/∂φ), etc., to derive eq. (21.17).

{L, H} =

[T] Consider a particle of mass m moving in two dimensions, bound in a central potential V(r), so that its Hamiltonian is H = T (p) + V(r), where T (p) = (p2x + p2y )/2m. Show that the quantity D = xpx + ypy is not a constant of the motion by computing Ḋ = {D, H}.

Compute Ḋ for the special case that V(r) is a power law, V(r) = V0 ra . Integrate your result over a time long compared to the period of this bound particle and show that 2hT i = ahV(r)i (where h. . .i denotes time averaging). This is the virial theorem of classical mechanics. Combine your result with energy conservation to express the total energy of the system in terms of hT i. What are the implications for the motion of the Moon around Earth (see eqs. (31.29) and (31.30))?

[Hint: you will need to use r∂/∂r = x∂/∂x + y∂/∂y and similar relations.]

21.6

[T] According to special relativity, the speed of light should be the same as seen by two observers moving at different relative velocities. In Newtonian physics, an observer (2) moving at a velocity v in the x-direction relative to an observer (1) labels the x-coordinate of an event by x2 = x1 − vt, where x1 is the coordinate given by the first observer, and both observers use the same time coordinate. Show that with this change of coordinates, the velocity of light traveling in the x-direction is different for the two observers. If a stationary observer sees light passing by at 3×108 m/s, and a second observer is in a rocket ship passing in the same

474

direction at 108 m/s, at what speed would the second observer see the light wave moving if the Newtonian transformation were correct?

21.7

The equation for a light ray propagating at speed c with respect to observer (1) is x1 = ct. Since x2 = x1 − vt, the equation of the light ray as seen by observer (2) would be x2 = (c − v)t = v′ t. For the values in the problem v′ = 2 × 108 m/s.

[T] The relativistic transformation law that relates the coordinates seen by an observer (2) moving at velocity v relative to an observer (1) is x2 = (cosh φ)x1 − (sinh φ)ct1 , ct2 = (cosh φ)ct1 − (sinh φ)x1 , where sinh φ/ cosh φ = v/c. (See Appendix B.4.2 for a summary of properties of the hyperbolic functions.) Note the similarity to rotation in Euclidean space by an angle θ. Confirm that under this transformation the spacetime distance (21.24) between two events is the same for both observers. Check that in the limit v ≪ c the Newtonian transformation rules described in the previous problem are reproduced.

Applying the proposed transformation law to the space time distance between two events, ∆x2 = ∆x1 cosh φ − c∆t1 sinh φ

c∆t2 = c∆t1 cosh φ − ∆x1 sinh φ .

The motion is in one dimension, so ∆y2 = ∆y1 = 0 and ∆z2 = ∆z1 = 0. Then ∆22 = ∆x22 − c2 ∆t22

= (∆x1 cosh φ − c∆t1 sinh φ)2 − (c∆t1 cosh φ − ∆x1 sinh φ)2

= (cosh2 φ − sinh2 φ)∆21 = ∆21 ,

where we have used formulas from Appendix B.4.2.

When v ≪ c, sinh φ ≈ v/c and cosh φ = 1+O(v2 /c2 ), so keeping only the leading terms when v ≪ c, x2 ≈ x1 − vt1

t2 ≈ t1 − (v/c2 )x1 ,

21.8

which reduces to the Newtonian transformation law of Problem 21.6 when v ≪ c.

[T] Consider two events seen by observer (1), who chooses her coordinates so that the first event occurs at the space-time origin xµ1 = (0, 0, 0, 0), while the second event occurs at x̃µ1 = (ct˜1 , x̃1 , 0, 0). Suppose ∆2 for the two events (see eq. (21.24)) is positive. Use the transformation defined in Problem 21.7 to show that there is a reference frame traveling at a speed v < c in which the two events occur at the same time and at different places. Such events are called causally disconnected.

Observer (2) sees the first event at the space-time origin and the µ second event at x̃2 = (ct˜2 , x̃2 , 0, 0), with ct˜2 = ct˜1 cosh φ − x̃1 sinh φ .

Observer (2) will see the two events occuring at the same time if t˜2 = 0, or tanh φ = v/c = ct˜1 / x̃1 . Since ∆2 = x̃21 − c2 t˜12 > 0 ,

|ct˜1 / x̃1 | < 1. Thus there is a velocity v = c2 t˜1 / x̃1 , with |v| < c, such that an observer moving with this velocity will see the events occurring simultaneously. The spacial separation between the events is given by | x̃2 | = ∆.

21.9

[T] Consider the circumstances described in Problem 21.8, but now suppose that ∆2 < 0. Use the transformation defined in Problem 21.7 to show that there is a reference frame traveling at a speed v < c in which the two events occur at the same place, but different times. Show that the time order of the events is the same in all reference frames related by the transformation of Problem 21.7. In this case the event at the earlier time can influence the later event.

The solution follows the same outline as the solution of Problem 21.8. The space separation between the events as seen by observer (2) is x̃2 = x̃1 cosh φ − ct˜1 sinh φ , and this can be made to vanish ( x̃2 = 0) if tanh φ = v/c = x̃1 /ct˜1 .

The fact that ∆2 < 0 guarantees that the resulting speed v is less than c. To find the time interval as seen by observer (2), use ct˜2 = ct˜1 cosh φ − x̃1 sinh φ = ct˜1 cosh φ − = ct˜1 / cosh φ ,

sinh2 φ cosh φ

!

where we have used cosh2 φ − sinh2 φ = 1. Since cosh φ > 0 for any φ, the sign of t˜2 is the same as the sign of t˜1 , the time order of the events is the same for both observers, so the notion of causality is preserved.

21.10 [T] Show that two space-time points (0, 0, 0, 0) and

(t, x = vt, 0, 0) that lie on the trajectory of an object traveling at speed v along the x-axis are separated by a distance satisfying ∆2 = 0 if and only if |v| = c. Use the relativistic invariance of ∆2 to show that the speed of light is the same in all reference frames. Show that if v > c then ∆2 > 0, so that according to Problem 21.8, the points cannot be in causal contact.

First show that ∆2 = 0 ⇒ |v| = c: ∆2 = x2 − (ct)2 = t2 (v2 − c2 ) = 0 if and only if |v| = c. Suppose that in some other frame, the light moves with velocity v2 . Then x2 = v2 t, so ∆2 = |x2 |2 − c2 t22 = t22 (|v2 |2 − c2 ) , and ∆2 = 0 requires |v2 | = c. If |v2 | > c, then ∆2 in the previous equation is greater than zero and acccording to Problem 21.8, the points cannot be in causal contact.

21.11 [T] An object moving at the speed of light along the

x-axis in the metric (21.35) of an inflating universe satisfies ẋ(t)eHt = c. Assuming x(0) = 0, compute x(t).

Show that x(t) never exceeds the value c/H. The differential equation for x(t), ẋ = ce−Ht has a solution x(t) = K − (c/H)e−Ht , where K is a constant, to be chosen so that x(0) = 0. Thus K = c/H, and

c x(t) = 1 − e−Ht .

H x(t) increases monotonically with t and reaches its maximum, xmax = c/H as t → ∞.

484

Problems 22.1

Following the same steps as in Problem 22.1, we need x1,2 for T = 3 695 K,

[C] Compute or estimate the fraction of radiated thermal energy that is in the visible range (wavelength 400– 750 nm) for the following two radiation spectra: (a) solar radiation, assumed to be perfect blackbody radiation at 5780 K; (b) EM radiation from a pot of boiling water, again assume a perfect blackbody distribution.

From eq. (22.23), the power per unit area radiated between frequencies ω1 and ω2 is P12 =

~ 4π2 c2

Z ω2

dω

ω1

0.0144 m K 5.196 (750 × 10−9 m)(3 695 K) 0.0144 m K 9.742 , x2 = (400 × 10−9 m)(3 695 K) x1 =

and numerical integration gives fvisible 0.21. So the maximum possible efficiency of the tungsten light bulb is 21%.

22.5

ω3 . e~ω/kB T − 1

Compute the frequency ωmax corresponding to the maximum power of radiation for a blackbody at temperature T = 280 K. This is roughly the average temperature of Earth’s surface. Compute the wavelength corresponding to this frequency.

Changing integration variable to x = ~ω/kB T = 2πhc/λT , and dividing by the integral over all frequencies, the fraction of the power radiated between wavelengths λ1 and λ2 is f12 =

P12 15 = 4 Ptot π

Z x2 x1

dx

= 1.03 × 1014 s−1 .

Then

[C] Compute the power radiated in the visible range from the toaster oven coil radiating with a total power of 1500 W, as described in Example 22.1.

Following the same steps as in Problem 22.1, we need x1,2 for T = 841 K, 0.0144 m K 21.9 (750 × 10−9 m)(876 K) 0.0144 m K x2 = 41.1 , (400 × 10−9 m)(876 K)

x1 =

and numerical integration gives fvisible 5.7 × 10−7 . Since the total power radiated is 1 500 W, the power radiated in the visible is Pvisible (5.7 × 10−7 )(1 500 W) 0.86 mW.

22.3

Estimate the rate of thermal radiation from a household hot-water radiator with a surface area of 1 m2 and a temperature of 80 ◦ C. The power radiated is given by the Stefan-Boltzmann law, 4

−8

2

4

ωmax 2.82(1.38 × 10−23 J/K)(280 K)/(1.055 × 10−34 J s)

x3 , ex − 1

where the integral must be done numerically.

Using hc/kB 0.0144 m K, then x j = (0.144m K)/λ j T . For solar radiation x1 = 3.322 and x2 = 6.238, and fvisible 42%. For boiling water at 100 ◦ C, x1 = 51.5 and x2 = 96.6 and fvisible 1.0 × 10−18 .

22.2

Evaluating eq. (22.14) for T = 280 K,

4

λ=

A look at Figure 9.2 indicates that this wavelength is in the infrared region of the electromagnetic spectrum.

22.6

[T] Estimate the gravitational energy in the Sun, using eqs. (22.1) and (22.2), assuming that the mass is distributed uniformly. Can you confirm the statement in the text that gravitational potential energy could only power the Sun at its present luminosity for ∼107 y? At what rate would the radius of the Sun have to be decreasing (in, say, mm/s) if its luminosity came from gravitational potential energy?

Given a sphere of uniform density ρ = 3M/4πR3 , the mass within a radius r < R is M(r) = Mr3 /R3 . The gravitational potential energy of a thin shell at radius r (mass dm = 3Mr2 dr/R3 ) is dV(r) = −GM(r)dm(r)/r. Integrating this over all radii, the total gravitational potential energy of the uniform sphere is

V=−

P = σT A = (5.670 × 10 W/m K )(353 K) (1 m ) 880 W.

22.4

The incandescent light bulb is a notoriously inefficient way to convert electric power into visible light. The tungsten filament emits blackbody radiation at a temperature that is limited by its melting point. Define the lighting efficiency as the ratio of the power emitted in the visible range (400–750 nm) to the total power emitted. What is the theoretical maximum efficiency of a tungsten light bulb, given that tungsten melts at 3422 ◦ C?

Z

GMdm =− r

ZR 0

3GM2 r4 3 GM2 . =− 5 R R6

For the sun, the total gravitational potential energy is

2

The radiator, however, also absorbs radiation from the environment at approximately 25 ◦ C, so it also absorbs heat at a rate of 448 W, for a net rate of 432 W. Taking account of the true emissivity will further reduce this value. Much of the heat transfer from the radiator occurs through conduction and convection.

c 2πc 2π(3.00 × 108 m/s) =

18.3 µm . ν ω 1.03 × 1014 s−1

V⊙ = −

3 (6.67 × 10−11 )(1.989 × 1030 )2 = −2.28 × 1041 J .

5 (6.955 × 108 )

If the contraction of the sun from an initial state with negligible potential energy powered it at its present luminosity of L⊙ 3.84 × 1026 W, the gravitational energy would last for T=

2.28 × 1041 J 5.9 × 1014 s , 3.84 × 1026 W

or approximately 2 × 107 y.

To get this luminosity from gravitational energy, the radius would need to be decreasing such that L⊙ =

|V⊙ | dR dV⊙ dR = . dR dt R dt

So, dR LR = = 1.2 × 10−6 m/s = 37 m/y. dt |V⊙ |

485

Problems

22.7

[T] Compute the energy released in each step of the solar PPI fusion chain eq. (22.5), and confirm that the total energy released matches eq. (22.6).

The relevant mass excesses for the reactions are ∆(1 H) = 7.2889 MeV, ∆(2 H) = 13.1357 MeV, ∆(3 He) = 14.9312 MeV, and ∆(4 He) = 2.4249 MeV. In the first step, the energy released is Q1 = 2∆(1 H) − ∆(2 H) = 1.4421 MeV. The second step releases Q2 = ∆(2 H) + ∆(1 H) − ∆(3 He) = 5.4934 MeV, and the third step releases Q3 = 2∆(3 He) − ∆(4 He) − 2∆(1 H) = 12.8597 MeV. To create one atom of 42 He, steps 1 and 2 must happen twice, so the total energy released per atom of 42 He is

The − sign is removed by integrating λ in the usual direction (low value to high value). So, dP dP 2πc 4π2 c2 ~ 1 = = . dAdλ dAdω λ2 λ5 e2π~c/λkB T − 1 d (dP/dAdλ) = 0, or where λmax is defined as the λ where dλ ! !

1 1 d 1 1 e1/x 1 5 0= + = − .

1/x 1/x 2 1/x dx x5 e − 1 x x e −1 x5 e − 1

with x = λkB T/2π~c. Letting y = 1/x, then, the maximum occurs where 5(1 − e−y ) = y. Solving for y, y = 1/x = 4.965. This corresponds to a wavelength of

Q = 2Q1 + 2Q2 + Q3 = 26.731 MeV .

22.8

[T] Show that the combination of the Boltzmann factor and the tunneling probability give a probability for fusion that is maximized at Emax ≈ (bkB T c /2)2/3 , as stated in §22.1.1.

The probability for fusion is proportional to the product of the Boltzmann factor and the tunneling probability, √ dP ∼ e−E/kB T c e−b/ E . dE

This probability is maximized where ! ! d dP b dP 1 0= + = − . dE dE dE kB T c 2E3/2 Solving for E, E∼

22.9

bkB T c 2

!2/3

.

[T] Show that the peak of the blackbody spectrum as a function of ω is given by eq. (22.14).

The peak wavelength of the blackbody spectrum occurs where ! d dP 0= . dω dAdω

λmax =

2.89 × 10−3 m K 2π~c = .

4.965kB T T

The frequency corresponding to this wavelength is ω(λmax ) = 2πc/λmax = 4.965kB T/~ .

This does not equal ωmax because the interval in λ, dλ varies with ω. Applying this equation, at 5780 K, λmax = 500 nm, whereas λ(ωmax ) = (4.965/2.28)λmax = 880 nm .

22.11 [T] Use dimensional analysis to show that the wave-

length scale of blackbody radiation is given by λth = hc/kB T . The average radiation energy in a cavity depends only on its volume and not on its shape when λth ≪ L, where L is a typical length scale characterizing the cavity. How low must the temperature be for the shape to matter if L ≈ 1 cm?

Planck’s constant, the speed of light, and kB are the relevant physical constants for looking at the thermodynamic properties of photons. Given a temperature scale T , we wish to find a characteristic length scale corresponding to this temperature. The energy scale associated with the temperature T is kB T , while the energy of a photon is ~ω = hc/λ. Letting these be equal and solving for λ, the wavelength scale of a photon at temperature T is

Letting x = ~ω/kB T and neglecting constant factors, this is equivalent to letting 0=

3x2 x3 e x − . e x − 1 (e x − 1)

Simplifying this, the spectrum is maximized where 3(1 − e−x ) = x , or where x = 2.82. This yields ωmax = 2.82

kB T . ~

22.10 [T] Rewrite the blackbody power spectrum as a func-

tion of λ and compute the value λmax where the power density dP/dλ is maximized. Find the frequency ω corresponding to λmax . Why is ω not equal to ωmax , where ωmax is given by eq. (22.14)? Compare λmax to 2πc/ωmax for a black body at temperature 5780 K. The blackbody power spectrum is ω3 dP ~ . = 2 2 ~ω/k T B −1 dAdω 4π c e

Letting ω = 2πc/λ,

dλ d λ2 d d = =− . dω dω dλ 2πc dλ

λth =

hc . kB T

The shape of an approximately L ≈ 1 cm cavity will matter when L ≈ khcT , or where T ≈ khcL = 1.4 K. B

B

22.12 [T] Show that the classical Rayleigh–Jeans law for

radiation follows from eq. (22.13) in the limit as ~ → 0.

Show that the total power radiated diverges in this limit (the ultraviolet catastrophe that helped lead to the discovery of quantum mechanics).

As ~ → 0, e~ω/kB T → 1 + k~ωT , so the Planck radiation law B becomes dP ~ kB T ω2 ω3 ω3 ~ = 2 2 ~ω/k T = . ≈ 2 2 B −1 dAdω 4π c e 4π c ~ω/kB T 4π2 c2

Replacing ω = 2πν gives the Rayleigh–Jeans law. After taking ~ to zero, the power radiated per unit area at frequencies below Ω is given by Z Ω dP kB T dP (Ω) = dω = Ω3 , dA dAdω 12π2 c2 0 which diverges as Ω → ∞.

22.13 [T] In §7, we showed that the energy of a particle

of mass M in an L × L × L box in the state labeled |n, m, li is En,m,l = π2 ~2 (n2 + m2 + l2 )/2ML2 . The probability of finding a particle in the state |n, m, li at temperature T is given by the Boltzmann distribution

486

P(n, m, l) ∝ exp(−En,m,l /kB T ). In a hot gas the energy levels are very close together, so E can be regarded as a continuous variable. Show that the probability of finding the particle with energy between E and E + dE at temperature T is given√by the Maxwell–Boltzmann distribution, dP/dE = 2π Ee−E/kB T /(πkB T )3/2 . [Hint: Employ the methods used to compute the spectrum of blackbody radiation, including eq. (22.20).]

First define the probability that the particle has energy less than Emax , X e−En,m,l /kB T , P(Emax ) ∝ n,m,l

where the sum is restricted to values of n, m, l such that En,m,l ≤ Emax . Note that the constant of proportionality in this expression is fixed by the condition that P(∞) = 1. As in the analysis of blackbody radiation, we treat n, l and m as continuous variables, so Z Z Z 1 dn dm dl e−En,m,l /kB T , P(Emax ) ∝ Z where again the integrals are√ over positive values of n, m, l such that En,m,l ≤ Emax . Let ρ = n2 + m2 + l2 and change integration variables according to dn dm dl =

1 4πρ2 dρ , 8

where the factor (1/8)4π comes from integrating over the angles of the vector ρp = (n, m, l). Next, note that En,m,l = π2 ~2 ρ2 /2m, √ substitute ρ = 2mE/π2 ~2 , and use ρ2 dρ ∝ EdE, to obtain Z Emax √ P(Emax ) ∝ dE Ee−E/kB T .

0

From this we can read off the probability to find the particle with energy between E and E + dE, √ dP ∝ Ee−E/kB T . dE Finally, we fix the constant of proportionality by the condition that R∞ dE(dP/dE) = 1, which gives the quoted result.

0

500

Problems 23.1

The total insolation is obtained by integrating I(t) from t− to t+ ,

The average radius of Mars’ orbit is 2.28 × 108 km.

Compute the solar constant for Mars.

Itot = I0

Zt+

dt (sin λ sin δ + cos λ cos δ cos ωt)

t−

The solar constant IMars is just IMars = I⊕

23.2

2 dMars

d⊕2

= (1 366 W/m2 )

= I0 (t+ − t− ) sin λ sin δ +

= (7.747 h)I0 = 38.1 MJ/m2 .

(1.5 × 108 km)2 = 591 W/m2 .

(2.28 × 108 km)2

Show that the insolation averaged over the year and over the entire Earth is 341 W/m2 .

23.4

The sun only shines on one side of Earth. To the sun, Earth has a cross section of πR2⊕ , while Earth has a total surface area of 4πR2⊕ .

To average over the entire surface, therefore, the solar constant is reduced by the ratio of Earth’s cross section as seen from the sun to Earth’s surface area. Thus, the insolation averaged over the entire Earth is I⊙ /4 = 341 W/m2 .

23.3

Similarly, at the winter solstice, Iwinter = 559 W/m2 .

From the geometry described in Box 23.1, and because the insolation is proportional to the the square of distance from the sun, !2 !2 Iwinter a−d 1−e R≡ = = .

Isummer a+d 1+e Solving for the eccentricity, √ √ 1− R 1 − 559/1292 e= 0.206 . √ = √ 1 + 559/1292 1+ R

23.5

!!

31 2π = 11.7◦ .

365.24

The angle of the sun is given by cos β(t) = sin λ sin δ + cos λ cos δ cos ωt ,

cos ωt± = − tan λ tan δ = −0.1393 ,

I(t) = I0 (sin λ sin δ + cos λ cos δ cos ωt) = I0 ((0.559)(0.2028) + (0.829)(0.979) cos ωt) = I0 (0.113 + 0.812 cos ωt) .

• At 6 am cos ωt = 0, so I = 0.113I0 = 154 W/m2 .

• At noon cos ωt = 1, so I = 0.925I0 = 1 264 W/m2 .

• At 3 pm cos ωt = 0.707, so I = 0.687I0 = 938 W/m2 .

Suppose a gas has an absorption cross section per molecule σ(ω) for light of frequency ω (see §18.2 for an introduction to cross sections). Show that the gas’s absorption coefficient is given by κ(ω) = σ(ω)ρ, where ρ is the number density of gas molecules.

Consider photons are traveling through a tube of cross sectional area A in a gas of molecules having the given absorption cross section and density. Then in some small distance dx there are ρAdx molecules that present an absorption area that is a fraction d f = ρσ(ω)dx of the total cross section of the tube. At every position x along the length of the tube, a fraction d f of the photons incident on a surface at x will be absorbed between x and dx. If I(x) is the number of photons that reach x, the absorption is described by dI = −I d f = −Iρσ(ω)dx or dI/I = −ρσ(ω)dx. Integrating both sides, I(x) = I0 e−ρσ(ω)x . Comparing with eq. (23.11), we see that the absorption coefficient is κ(ω) = σ(ω)ρ.

where λ = 34◦ is the latitude, ω = 2π/(24 h) and t is the time since noon. The times of sunrise and sunset t∓ (relative to noon) are determined by the condition that cos β(t± ) = 0 So sunrise and sunset occur when

which occurs when t± = ±6.536 h. The length of the day is then t+ − t− = 13.072 h. This is reasonable since the length of the day exceeds 12 hours between spring and fall equinoxes.

If I0 = 1 366 W/m2 is the insolation when the sun is directly overhead, then the insolation for δ = 11.7◦ and λ = 34◦ is

For your location, calculate the maximum instantaneous irradiance (at the top of the atmosphere) on the days of the summer and winter solstices. Now suppose Earth’s obliquity were 0◦ . How large would the eccentricity of its orbit have to be to produce the same difference in irradiance between perihelion and aphelion?

Isummer = (1366 W/m2 ) cos(42.4◦ − 23.44◦ ) = 1292 W/m2 .

The declination δ is defined to be the latitude at which the sun is directly overhead. It is given by sin δ = sin ǫ sin(2πN/365.24) where N is the number of days since the spring solstice and ǫ is the axial tilt of Earth (23.44◦ ).

δ = sin−1 sin 23.44◦ sin

!

In Cambridge, Massachusetts, the latitude is λ = 42.4◦ N. The insolation is I = I0 cos β where the irradiance is maximal at noon, where cos β = cos(λ − δ) (see eq. (23.8)). At the summer solstice, δ = ǫ = 23.44◦ , while at the winter solstice, δ = −ǫ = −23.44◦ .

At the summer solstice,

[C] Consider a location for a solar energy installation in Arizona, at latitude 34◦ . Compute the declination δ on the following dates: (i) March 20 (spring equinox), (ii) April 20. Compute the length of the day (from sunrise to sunset) at this location in Arizona on April 20.

Assuming a clear day, compute the insolation on a horizontal surface at 6 a.m., noon, and 3 p.m., and compute the total insolation for April 20.

i) On March 20th, the spring solstice, N = 0 so sin δ = 0. Thus δ = 0◦ . ii) April 20th is 31 days after spring equinox, so

2 cos λ cos δ sin ωt+ ω

23.6

The thickness of Earth’s ozone layer at any location is measured in Dobson Units (DU), where 1 DU corresponds to a gas layer of thickness 10 µm if the gas pressure were raised to 1 atm. Show that 1 DU corresponds to 2.69 × 1016 molecules/cm2 . Data on the absorption cross section for ozone are given in Figure 23.13. Normally the thickness of the ozone layer is ≈ 300 DU, but due to ozone destruction by chlorofluorocarbons it had dropped as low as ≈ 90 DU over Antarctica during the 1990s. How low

Problems

would the ozone level have to drop to allow 10−6 of the incident UV flux at λ ≈ 250 nm or 220 nm to reach sea level? (You will need to use the result of Problem 23.5.)

[C] Compute the integrated daily insolation at several different latitudes on July 30 and compare with Figure 23.6.

23.8 The absorption coefficient for crystalline silicon at 300 K for a photon of energy 1.3 eV is roughly 70 cm−1 .

Compute the thickness of silicon needed to absorb 75% of incoming photons at this energy.

23.9 Verify the estimates of global land use per terawatt in the two scenarios outlined in §23.6.

23.10 The population of the city of Cambridge in the state of Massachusetts is around 100 000. Assume that each resident of Cambridge uses energy at the rate of 1 GJ per day. If the average insolation in the area is 150 W/m2 , compute the area of land needed to supply all the energy for the residents of Cambridge assuming that the conversion efficiency of sunlight is 2%, 4%, 8%, and 16%. Compare to the land area of Cambridge, which is 16.65 km2 .

23.7

The average power consumption for the residents of the city of Cambridge is P = (105 )(109 GJ/d)/(86 400 s/d) 1.16 GW .

At a conversion efficiency of 2%, solar power provides p = (0.02)(150 W/m2 ) = 3 W/m2 . So the required area is A(2%) = (1.16 GW)/(3 W/m2 ) = 387 km2 23×A(Cambridge) and at 4%, 8%, and 16%, the required areas reduce to 193, 97, and 48 km2 , which are 11.6, 5.8, and 2.9 times the area of the city respectively.

Figure 23.13 Ozone absorption cross section in

cm2 /molecule in the Hartley Band [139].

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519

Problems

Problems 24.1

24.2

24.3

24.4

The radius of Mars’ orbit around the Sun averages roughly RMars 2.28 × 108 km. Assuming that the surface temperature is roughly uniform over the planet’s surface and constant over a Martian day, that the surface is a perfect black body, and that there is no atmosphere, estimate the average surface temperature when the planet is in radiative equilibrium.

The eccentricity of Mars’ elliptic orbit is roughly 0.0935. Estimate the range over which the solar constant and average surface temperature (in the uniform temperature blackbody approximation) vary during the year.

Suppose that the surface temperature of the Sun were to increase by 3% from the current value, assuming for simplicity that the radius of the Sun stays fixed. By how much would the solar constant on Earth increase? If, at present, Earth radiates away an amount of energy equal to the current incoming energy flux from the Sun by radiating as a black body at an effective temperature of T b 5 ◦ C, by what amount would the temperature T b have to increase to offset the increase in incident solar energy?

With no power harvested radiative equilibrium just above the absorber gives Ib = I0 + I2 and radiative balance just above the layer at T 2 gives I2 = Ig + I0 . Combining these, we obtain Ib = 2I0 + Ig .

Substituting I0 = 1000 W/m2 and Ig = σT g4 with T g = 300 K, we obtain T b 183 ◦ C.

24.6

If the temperature of the sun increased by 3%, the solar constant at Earth would increase by ≈ 12% because radiation power scales like T 4 . To adjust to this increase, Earth’s mean temperature would have to increase by 3%, from 278 K to ≈ 286 K or 13 ◦ C.

Consider an idealized flat-plate solar collector. Assume that there are no heat losses to conduction or convection. Assume that the insolation is 1000 W/m2 incident at an angle of 45◦ from the vertical. Compute the temperature of the black body in radiative equilibrium assuming that (a) the collector is horizontal, (b) the collector is tilted at 35◦ towards the Sun. (a). The intensity at normal incidence is I0 = 1000 W/m2 , and so the insolation on a collector at angle θ to the incoming radiation will be I = I0 cos √ θ. A horizontal collector is at angle θ = 45◦ , yielding I = I0 / 2 ≈ 707 W/m2 . In radiative equilibrium, the collector must radiate away at the same rate I. From the Stefan–Boltzmann Law, we then have √ σT b4 = I0 / 2 , which we can solve to find !1/4 I0 ≈ 334 K ≈ 61 ◦ C .

Tb = √ 2σ (b). When the collector is tilted at 35◦ toward the Sun, θ = 10◦ , and so we find !1/4 I0 cos(10◦ ) Tb = ≈ 363 K ≈ 90 ◦ C . σ

24.5

Figure 24.14 A double-pane flat-plate collector.

Analyze the double pane glass-covered collector described in §24.1.2 and illustrated in Figure 24.14. T g is assumed to be fixed at ≈ 300 K by conduction and convection and I0 ≈ 1000 W/m2 . Write the equations of radiative equilibrium and show that T b ≈ 180 ◦ C.

Consider a flat-plate collector covered by a single pane of glass exposed to perpendicular insolation at 1000 W/m2 . The collector operates at a net efficiency of 35% at a temperature T b 65 ◦ C. Assuming that the glass covering stays at the ambient temperature of 15 ◦ C, compute the net power radiated and non-radiative losses for the collector. Ignore incident infrared light from the atmosphere.

In the absence of non-radiative losses η = 1 − σ(T b4 − T g4 )/I0 65%. So 35% or 350 W/m2 is radiated by the collector. Since the actual efficiency is 35%, the remaining 30% or 300 W/m2 is lost through non-radiative processes.

Consider the examples of single- and double-pane glass-covered collectors described in §24.2.1. Verify the quoted results for the efficiency as a function of T b , T g , and I0 . In the case of double glazing, compute the temperature of the intermediate glass plate. Plot the efficiency as a function of T b for 25 ◦ C < T b < 100 ◦ C when T g = 280 K and I0 = 1000 W/m2 .

24.8 Redo the computation of Problem 24.7 assuming incoming radiation at I = 750 W/m2 , and fixing T b = 70 ◦ C, T g = 280 K. Show that the efficiency decreases significantly with the reduction in incoming radiation.

24.9 Compare the total (integrated daily) insolation on a flatplate collector located in Chicago, Illinois (latitude λ = 41.98◦ ) on February 1 when the collector is (a) tilted at an angle θ = λ to the south, (b) tilted at an angle θ = λ + 15◦ to the south.

24.10 Consider an idealized flat-plate solar collector with no conduction or convection losses, and with insolation 1000 W/m2 incident at an angle of 45◦ from the vertical. Assume that the collector is horizontal and is covered with a pane of glass transparent to incoming radiation but opaque to outgoing (IR) radiation.

The glass is kept at 300 K by the external environment.

Compute the temperature of the absorber in radiative 24.7

520

equilibrium. Recompute the temperature if the absorber is coated with a paint that modifies its emissivity in the IR to ε 0.2.

Radiative balance at the absorber without the paint: σT b4 = I0 cos 45◦ + σT g4 yields T b 106 ◦ C. With the paint, both the absorption and emission by the absorber are suppressed in the infrared, εσT b4 = I0 cos 45◦ + εσT g4 , which yields T b 242 ◦ C.

24.11 Consider a parabolic trough where the height of the

reflector is identical to the height of the center of the absorber, as depicted in the figure in Example 24.4.

The concentrator width is 8 m and the absorbing tube of radius 0.5 m is centered along the focal line at a height of 2 m above the bottom of the trough. Incident sunlight hits the concentrator from a direction parallel to the line of symmetry, with an intensity of 1000 W/m2 . (a) What is the effective concentration of the concentrator? (b) What is the acceptance angle within which all light hits the absorber? (c) Assume that the tube carries a fluid that is heated and removes some of the incoming energy. The tube then radiates as a black body at a temperature of 150 ◦ C. Compute the net rate at which energy is collected and transferred to the fluid, for each meter of length of the trough.

24.12 Consider a solar thermal “power tower” with a central tower of height 60 m, surrounded by an array of planar mirrors on the ground extending to a radius of 100 m around the tower. At the top of the tower there is a cylindrical absorber with its axis aligned vertically.

Each mirror is large enough so that the complete reflection of the Sun can be seen on each mirror from the center of the tower at the height of the absorber. (a) Give an upper limit on the number of mirrors that could be placed around the tower subject to these constraints. (b) How wide does the absorber have to be so that light rays coming from the center of the Sun hit the absorber, no matter which point on which mirror they are reflected from? a) The angle subtended by the mirror viewed from the center of the absorber must be greater than the angle subtended by the sun. The angular diameter of the sun’s disk is θ⊙ = 2R⊙ /d, where R⊙ 696 000 km is the sun’s radius and d 1.50 × 108 km is the mean Earth-sun separation. Thus θ 0.53◦ . This allows for a maximum of N 360/0.53 = 675 mirrors. b) All light rays from the center of the sun’s disk when viewed from Earth are essentially parallel, thus the separation between rays that strike opposite sides of a given mirror is the same at the absorber as it is at the mirror. Thus the width of the absorber must be the same as the width of the mirror, which is bounded by the circumference of the array (2π × 100 m) divided by the number of mirrors, w & 628/675 0.93 m, which is the diameter of the cylinder.

24.13 For the power tower geometry from the previous prob-

lem, assume that the absorber height is equal to its diameter, and that the diameter is that found in part (b) above. If all the light from an overhead sun that hits the circle of radius 100 m containing the reflecting mirrors were to be reflected to the absorber, what would be the effective concentration of the mirror configuration

as a solar concentrator? Assume 1000 W/m2 insolation from an overhead sun. What would be the temperature of the absorber if it was in radiative equilibrium with the reflected incoming radiation? Do you think that this is a realistic answer? Why or why not?

Let the diameter of the absorber be d and the radius of the array be R, then the concentration factor is the area of the array πR2 divided by the area of the absorber 2πRh = πd 2 , so C = R2 /d 2 .

Then in radiative equilibrium πd 2 σT 4 = CI0 . Using d 1 m from the previous problem and R = 100 m, T 3 600 K. This is unrealistic because there are no materials that could be used at this temperature.

24.14 [T] Determine the maximum concentration C of a

solar concentrator satisfying σT 4 = CI0 , where I0 is the solar constant, directly from the second law of thermodynamics and the surface temperature of the Sun. The 2nd Law requires that T < T ⊙ ≈ 5 800 K, so C = σT ⊙4 /I0 ∼ 64 000.

24.15 [T] Prove that for the compound parabolic concentra-

tor depicted in Figure 24.11 the line containing points Q and A is parallel to the axis of parabola 2. [Hint: consider the incoming ray that is reflected along QA.] 24.16 Consider a linear 2D compound parabolic concentrator built from parabolas tilted at 10◦ to the vertical, with a trough of width 3 m, and an absorber width of 0.5 m.

Compute the concentration C of the concentrator. If the incoming radiation has intensity I0 = 1000 W/m2 and the (blackbody) absorber is kept at a temperature of 100 ◦ C by circulation of a thermal fluid, compute the rate of energy transfer to the fluid. Compare this rate of energy transfer to that for a linear parabolic concentrator with the same absorber area and acceptance angle; assume that the parabolic concentrator has height equal to the focal length (as in Example 24.4), and that the absorber has an area covering the lower half of a cylinder centered on the focal line, and is kept at the same temperature of 100 ◦ C. 24.17 [T] Prove that a linear compound parabolic concentrator realizes the maximum possible concentration for a given acceptance angle θ. Suggestion: following the notation of Figure 24.11, work in a coordinate system where parabola 2 is described by the equation y = x2 /4F. Identify B and Q as the points on parabola 2 where the slopes are tan α and tan 2α, with 2α + θ = π/2. Then show that QA/BA = (1 + sin θ)/2 sin2 θ, which proves the desired result.

24.18 Consider a parabolic dish concentrator built from a dish with radius 3 m, height equal to the focal length, and an absorber at the focus with spherical shape and radius 0.1 m. Compute the concentration and acceptance angle of this concentrator, and check that the Rabl bound is satisfied.

24.19 For a concentrator with concentration C = 100, and incident sunlight at intensity I0 = 1000 W/m2 , compute the temperature T of the absorber at which solar

Problems

to electric conversion efficiency is optimized, assuming Carnot efficiency. How does your answer change if the system has an additional rate of energy loss per unit area to conduction and convection of (T − T ambient ) × 20 W/Km2 , with T ambient = 300 K?

521

550

25.10 [C] Use the photodiode equation (25.17) to compute

Problems 25.1

25.2

25.3

Show that for each unit cell in the crystal shown in Figure 25.1 there are eight atoms, all identical in the case of the diamond and four of each kind in the zincblende case. [Hint: Each atom at a corner is shared among eight cells, etc.]

Following the logic of Figure 25.6, sketch (no calculation necessary) the wavefunctions of the energy basis states at the top of the second energy band and the bottom of the third energy band in a periodic square well potential with high barriers. [Hint: First establish that these wavefunctions are both periodic in x → x + ∆.

Also remember that inside the potential wells the wavefunctions should resemble the second and third excited states of a square well potential respectively.]

Estimate the maximum possible collection efficiency ηmax collection for a germanium solar cell (E gap 0.66 eV).

R∞ 2 x x eyy−1 ηmax collection = R ∞ y3 0

ey −1

where x = Egap /kB T . By direct computation for germanium, x = 1.31 and ηmax collection 0.375.

25.4

The direct band gap in silicon is 3.4 eV. What is the maximum possible collection efficiency (for incident thermal radiation at 6000 K) for excitations over this gap?

ηmax collection =

R∞ 2 x x eyy−1 R ∞ y3 0

ey −1

where x = Egap /kB T . By direct computation for Egap = 3.4 eV, x = 6.75 and ηmax 0.074. collection

25.5

25.6

25.7

25.8

25.9

Given four materials, with band gaps 0.2 eV, 0.7 eV, 1.5 eV, and 3 eV, which two are likely to make the best double-junction solar cell? Explain your answer.

[C] Give an estimate of the collection efficiency for the best double-junction cell formed from the materials in the previous problem.

[C] Estimate the collection efficiency of a triplejunction solar cell where the first layer is an indium gallium phosphide alloy In0.53 Ga0.47 P, with band gap EIGP = 1.87 eV, the second layer is gallium arsenide, and the third layer is germanium.

[C] Numerically compute the maximum collection efficiency for a single-junction PV cell under 6000 K blackbody conditions as a function of the band gap.

Estimate the maximum possible collection efficiency and the associated band gap. Repeat for doublejunction and triple-junction cells, confirming the collection efficiencies given in the text and computing the band gaps needed in each case.

[C] Download the AM1.5 solar spectrum from [135] and numerically compute the maximum collection efficiency expected for a silicon PV cell through a computation analogous to eq. (25.11).

Iphoto /I0 for a silicon solar cell with Voc = 0.7 V. Write the power as a function of voltage and compute the fill factor, the maximum power IV attainable as a fraction of Iphoto Voc . Repeat the calculation for cells with Voc = 0.5 and 0.86 V and compute the maximum efficiency of a single-junction silicon photocell in these cases.

The maximum efficiency is given by eq. (25.26) as the product of three terms ηmax = ff ×

eVoc × ηmax collection .

Egap

We are given ηmax collection 0.49 and 0.7 eV eVoc

0.63 .

Egap 1.12 eV It remains to compute the fill factor from the given information. We need the ratio r ≡ Iphoto /I0 , which can be determined by setting the current I = 0 (“open circuit”) in eq. (25.17), r = eeVoc /kB T − 1 Given eVoc 0.7 eV and kB T 0.025 eV at T 300 K, we find r e28 , where we have dropped the subtracted factor of 1 compared to e28 .

The fill factor is defined as ( !) ) ( VI(V) 1 V ff = MaxV = MaxV 1 − eeV/kB T Voc Iphoto Voc r n o = Max x x 1 − e28(x−1) ,

where we have defined x = V/Voc . This expression must be maximized numerically, yielding ff = 0.85 at x = 0.88. Substituting this value of ff into the expression for ηmax we find ηmax = 0.85 × 0.63 × 0.49 0.26 , quite close to the value of 0.27 quoted in the text.

For Voc = 0.5 V, r = e20 and ff 0.81, thus ηmax = 0.81 ×

0.5 × 0.49 0.18 .

1.12

And for Voc = 0.85 V, r = e34 and ff = 0.87, thus ηmax = 0.87 ×

0.85 × 0.49 0.32 .

1.12

25.11 Consider a single-junction solar cell with Voc = 0.7 V

under one sun illumination conditions. Estimate the increase in Voc and the increase in overall efficiency if the incoming light is concentrated by a factor of 400.

25.12 Compute the Shockley–Queisser bound on Voc for a silicon cell with band gap 1.1 eV at temperature T = 300 K. Model the Sun as a black body at 6000 K. 25.13 Given a silicon photovoltaic solar cell with open-circuit voltage Voc = 0.7 V that operates at an overall 20% efficiency under 1000 W/m2 of normal illumination, (a) compute the current density from the collecting surface. (b) How many cells must be connected in series to make a 12 V voltage? (c) What surface area is needed to produce one ampere of current across 12 V?

Problems

25.14 Real photodiodes do not precisely obey the ideal pho-

todiode equation (25.17). A more realistic model is the non-ideal photodiode equation h i I = Iphoto − I0 eeV/mkB T − 1 ,

where m > 1 is the ideality factor. Consider a nonideal photodiode with m = 2, Voc = 0.7 V. Compute Iphoto /I0 and the fill factor. Compare the efficiency with an ideal photodiode with m = 1 and the same opencircuit voltage.

25.15 An actual solar cell has internal resistances that affect the behavior of the circuit. These can be modeled as a single shunt resistance R p in parallel with the diode, and another resistance Rs in series with the external load. Derive the modified ideal diode equation in the presence of the resistances.

25.16 Assuming a silicon PV array attains 20% efficiency over a lifetime of 25 years of use, where it is exposed to an average of 250 W/m2 , what is the total energy output of 1 m2 of PV cells (in joules). What quantity of coal would it replace (at 33% net efficiency)?

At 20% efficiency, 1 m2 of solar cells exposed to 250 W/m2 of solar power input collects E25 year (0.2)(250 W)(25 y)(3.16 × 107 s/y) 39.5 GJ .

For comparison from Appendix C, 1 tonne of coal equivalent (TCE)= 29.3 GJ, so the 1 m2 solar panel displaces (39.5 GJ)/(29.3 GJ × 0.33) 4 TCE.

25.17 The thickness of the germanium (density 5.32 gm/cm3 )

layer in a triple-junction PV is typically greater than 100 microns. If the overall cell efficiency is 40%, how many watts of PV capacity (nominal insolation of 1000 W/m2 ) could be manufactured from the world’s annual germanium production (165 t in 2014)?

Each square meter of germanium at 100µm thickness requires m(Ge) (5.32 g/cm3 )(10−2 cm)(104 cm2 ) 0.53 kg .

Therefore a year’s production (2014) of 165 × 103 kg would allow production of A = 165 × 103 kg/(0.53 kg/m2 ) 310 × 103 m2

of PV material. With an efficiency of 40% and a nominal insolation of 1000 W/m2 , the capacity of such a system would be P (3.1 × 105 m2 )(400 W/m2 ) 125 MW .

25.18 The absorption coefficient of silicon has a strong

dependence on photon energy, as shown in Figure 25.16. For simplicity, consider an idealized material, material S, similar to crystalline silicon, with an absorption coefficient of κ = 7×103 m−1 for light at any wavelength. What fraction of (normally incident) light will be absorbed by a wafer of material S that is 200 µm thick? If we could fabricate a layer of material S just 1 µm thick, what fraction of incident radiation would it absorb? The absorption coefficient of amorphous silicon is significantly higher than that of crystalline silicon. If we had another material, material A, with an

absorption coefficient κ = 1.5×106 m−1 (again independent of wavelength), what fraction would be absorbed by a 1 µm layer?

25.19 [T] Derive the Fermi–Dirac distribution (25.8). Start by considering a single electron state of energy E that is either occupied (n = 1) or not occupied (n = 0) by an electron. Now, consider coupling this two-state system to a thermal reservoir at temperature T so that not only energy but also particles can move between the smaller system and the reservoir. The entropy of the reservoir S (U, N) now depends on the total energy and total number of particles in the reservoir. Define E F = −T (∂S /∂N)|U . Use an argument similar to that used to derive the Boltzmann distribution in §8 to derive the Fermi–Dirac distribution (25.8).

25.20 [T] Compute the momentum of an electron with kinetic energy 2 eV and compare to the momenta of a photon with the same energy. Explain why such a photon cannot excite an electron in silicon without involving a phonon.

25.21 [T] Show that the right-hand side of (25.25) cannot exceed Egap and goes to this value only in the limit T → 0.

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Problems

Problems 26.1

According to Box 26.5 about 15 × 106 ha, or 1.5 × 1011 m2 are used to grow corn ethanol. If converted to STE at 3% efficiency, this would produce electric power of P = 0.03(1.5 × 1011 m2 )(200 W/m2 ) ≈ 0.91 TW. Over a year, the total electric energy produced would be ESTE ≈ (3.2 × 107 s/y)(0.91 × 1012 J/s) ≈ 28 EJ. Equivalent US energy consumption is 35(1.00) + 65(0.35) = 58 EJ. So the land dedicated to corn ethanol production could generate almost 50% of US power demand.

The enthalpy of combustion of glucose (C6 H12 O6 ) is roughly 15.6 MJ/kg. Compute the fraction of incident solar energy from 8 photons with wavelength 680 nm stored through the reaction (26.2) in a single CH2 O unit.

15.6 MJ/kg corresponds to

26.6

15.6MJ/kg = 15.6 MJ/kg(0.18 kg/mol)/(0.0965 (kJ/mol)/(eV/molecule)) = 29 eV/molecule , or 29/6=4.9 eV per CH2 O unit. Eight photons at λ = 6.8 × 10−7 m have energy 14.6 eV, so it looks like 4.9/14.6 = 33% of the energy is stored.

26.2

Taking ∆H f from the table,

Estimate the total food energy needed for the planet’s population assuming a diet of 2400 Calories/day/person. Compare to the global food production rate stated in the text.

∆Hr = 2(−278) + 2(−394) − (−1268) = −76 kJ/mol .

The standard enthalpy of combustion (HHV) of glucose/ethanol is 2805/1368 kJ/mol. Since the molecular weight of glucose is 180 gm/mol, its energy density is 2805/0.180=15.6 MJ/kg. For ethanol, molecular weight 46 gm/mol, the energy density is 1368/0.046=29.7 MJ/kg, almost twice that of glucose.

Each individual requires approximately (2 400 Cal/d)(4.2 kJ/Cal)(365 d/y) ≈ 3.7 GJ/y

26.3

of food energy. Earth’s population on Jan 1, 2016, approximately 7.4 billion, therefore consumes (7.4 × 109 )(3.7 GJ/y) ≈ 28 EJ/y of food energy. In §26.2 it was stated that 45 EJ food energy is produced per year, of which 1/3 goes to waste. This is consistent with ≈28 EJ consumed.

A feedlot cow is fed 12 kg of corn daily for 255 days, and grows from 200 kg to 500 kg in that time. Estimating the energy densities of corn and cow both at roughly 15 MJ/kg (a very rough estimate, actually more of an overestimate for the cow), compute the ratio of corn energy input to cow energy output.

26.7

E = 0.54(7 t/ha y)(29.7 GJ/t) = 112 GJ/ha y .

26.8

The fossil fuel energy input currently needed for ammonia production is roughly 36 GJ/t (§33.3.4). Estimate the energy needed to produce ammonia containing 90 Mt of nitrogen. Compare to the biomass energy of 18 EJ in 40% of all food crops that is estimated to be enabled through fertilizer inputs.

Compute the rate at which carbon is extracted from the atmosphere by a forest growing at the net rate of 4 t/ha y. Assume that the wood is purely composed of cellulose. Compare the carbon savings realized by sequestering this carbon to that realized by using corn ethanol grown on the same land as a motor fuel, according to the estimate in Box 26.6. (See Problem 26.7 for further information.)

Ammonia (NH3 ) is 14/17 = 82% nitrogen, so 90 Mt of nitrogen fixed in ammonia requires 90/(0.82)=110 Mt of ammonia. At 36 GJ/t, production of this amount of ammonia requires

Cellulose (C6 H10 O5 ) is 44.4% carbon, so a forest sequesters carbon at a rate of 0.44(4) = 1.8 t/ha y. According to Box 26.5 and Problem 26.7, energy content of corn based ethanol is produced at ∼100 GJ/ha y. According to Box ??, after including the byproducts as an offset, there is a reduction in greenhouse gases by roughly 5 g/MJ, giving

(36 × 109 GJ/t)(110 × 106 ) t = 4.0 EJ

(5 g/MJ) ∗ (100GJ/hay) = 0.5t.

of energy. This is 22% of the energy contained in the food crops produced with ammonia based fertilizer inputs.

26.5

Compute the maximum energy content of the ethanol produced from corn farmed at 10 t/ha y, assuming that the corn is 70% starch composed of polysaccharides (amylose and amylopectin) built from glucose (glucan) units of molar mass 162 g/mol.

Each glucose unit produces two molecules of ethanol. So 2(44)/162=54% of the mass of corn is transformed into ethanol.

At 7 t/ha y and with an enthalpy of combustion of ethanol of 29.7 MJ/kg, the energy content of corn-ethanol is

The cow consumes (255)(12)=3 060 kg of corn, but gains only 300 kg of mass. Assuming equal energy density in both, the ratio of corn-to-cow energy is roughly 10:1.

26.4

Look up the standard enthalpies of formation of glucose and ethanol (Table 9.8), and verify that eq. (26.4) is exothermic. Then verify that the gravimetric energy density (relative to combustion) of ethanol is approximately twice that of glucose.

If all land area in the US that is used for growing corn for ethanol were used for solar thermal electric plants operating at a gross conversion efficiency of 3%, estimate the total electrical energy yearly produced assuming an average insolation of 200 W/m2 .

Compare to US energy consumption of roughly 100 EJ, assuming 35% efficiency for 65 EJ of petroleum and coal energy used and 100% efficiency of the remainder.

26.9

Crop residues (corn stover, straw, or sugarcane bagasse) represent a potentially sustainable source of cellulosic ethanol. Assuming a production of 6 t/ha y and 270 liters of ethanol per tonne of crop residue, estimate the area of farmland necessary to sustainably provide ethanol for one million cars (using 100% ethanol) that travel on average 10 000 km/y at 40 mpg (17 km/L).

Each hectare yields 6 × 270 = 1 620 L/y of ethanol. Each car uses on average (104 km)/(17 km/L) = 590 L of ethanol. So each

572

hectare can support 1 620/590 ≈ 2.75 cars. To fuel 106 cars would require ≈ 360 000 ha = 3 600km2 .

589

Problems

potential as that of a spherical Earth plus a correction term due to centrifugal force. Compare your answer to the actual difference. Can you see how this approximation could be improved?

Problems 27.1

27.2

27.3

The absorption coefficient of water for light of wavelength 450 nm is roughly κ450 0.0001 cm−1 while for light of wavelength 750 nm it is roughly κ750 0.03 cm−1 . Estimate the depth by which 90% of light has been absorbed by the ocean in each of these wavelength ranges. Explain the color seen underwater at typical depths encountered by divers.

Assuming that one half of the energy in solar radiation absorbed at the oceans’ surface is lost to latent heat flux (evaporation), give an order-of-magnitude estimate of the rate at which water vapor enters the atmosphere from evaporation on the oceans’ surface.

Consider a location in the tropical ocean where the mixed layer has a uniform temperature of 30 ◦ C to 100 m, and the deep ocean temperature is 5 ◦ C at 1000 m. Estimate the rate of conductive heat flow downward between 100 m and 1000 m, assuming a constant gradient between these depths. At this rate, how long would it take to heat the ocean depths between 1000 m and 2000 m by 5 ◦ C?

The corrected potential is −

where r is distance from center and ρ is distance from axis of rotation. Equating at equator and pole gives a difference of δ=

3

3

3

9

27.7

27.8

C = (10 m)(10 kg/m )(4.0 × 10 J/kg K) = 4.0 × 10 J/m K .

The rate of change of temperature is then dT/dt = q/C = 0.015/(4.0 × 109 ) 3.8 × 10−12 K/s . ∆t = 5K/(3.8 × 10−12 K/s) 1.3 × 1012 s 40 000 y .

27.4

Under the same assumptions as for Problem 27.3, estimate the net rate at which energy is lost to upward thermal radiation. First, compute the rate of upward thermal radiation. Then include a downward IR radiation flux, approximated by thermal radiation from the atmosphere at an altitude of 2000 m at 20 ◦ C, and compute the net upward energy flux.

The question addresses thermal radiation alone: The outgoing IR 4 flux is qout = σT out with T out 303 K and the incoming IR flux is qin = σT in4 with T in 293 K. The net (outgoing) flux is 4 qnet = σ(T out − T in4 )

= (5.67 × 10−8 W/m2 K4 )(3034 − 2934 ) K4 60 W/m2 .

27.5

27.6

Estimate the centrifugal forces acting on a 60 kg person standing at the equator due to (a) Earth’s rotation about its axis, (b) Earth’s yearly revolution about the Sun, and (c) the Sun’s revolution about the center of the Milky Way galaxy.

Given Earth’s radius, mass, and rate of rotation, estimate the difference in radius measured along the equator from that measured to a pole in order that the apparent gravitational potential is the same everywhere on Earth’s surface by approximating the gravitational

If a pitcher throws a baseball at 44 m/s (100 mph), estimate the deviation in position, both in magnitude and direction, due to the Coriolis force when the ball reaches the batter after traveling 18.5 m, when the game is played at a latitude of (a) 30◦ N, (b) 60◦ S, or (c) on the equator. (See Problem 2.13 for more about baseball.)

Compare the gravitational force and horizontal Coriolis force on a person running at 6 m/s at latitude 45◦ . How fast would an object have to be moving for the two to be equal? g = 9.8 m/s2 , whereas ac (v) = 2ω⊕ v sin λ. With λ = π/4, and ω⊕ = 2π/(24 × 3 600)rad/s 7.27 × 10−5 rad/s, √ ac (6 m/s)/g 2(7.27×10−5 rad/s)(6 m/s)( 2/2)÷(9.8 m/s2 ) 6.3×10−5 .

2

To heat the deep water by 5 ◦ C requires

ω2 r2 10.7km.

2 × 9.8m/s2

Measured difference is about 21 km. The approximation is inadequate because the oblate shape of the Earth changes the actual gravitational potential, essentially doubling the change in radius.

According to Fourier’s Law, q = k dT/dx. k = 0.60 W/m K for seawater and dT/dx = (30 − 5)/900 = 0.028, so q = 0.015 W/m2 .

This heat flux heats a column of seawater 1 000 m in height, which has a heat capacity (Cseawater 4.0 kJ/kg K), 3

GM − ω2 ρ2 r

27.9

For this ratio to be one, v must be a factor of 1/(6.3×10−5 ) 16 000 larger, or v 6 × 16 × 103 m/s 96 km/s.

Assume that a constant wind at v = 20 knots (10.3 m/s) is blowing in a southward direction on the eastern side of an ocean basin at latitude +35◦ . The combination of the force due to the wind and the Coriolis force cause the water near the surface to drift uniformly with constant velocity. Ignore viscous drag effects and the dependence of surface currents on depth, and instead assume that the top 50 m of ocean surface moves uniformly in response to the stress exerted by the wind. If the wind stress on the ocean’s surface is estimated as a horizontal force per unit area T wind = ρcd v2 , with ρ atmospheric mass density and cd 0.0015, compute the average drift velocity (speed and direction) of the surface water such that the Coriolis force cancels the force due to the wind. Then compute the rate of mass upwelling along the western side of the continent.

As stated in the problem, the Coriolis force on the moving water exactly cancels the wind stress. Since the Coriolis force acts at right angles to the drift velocity, in order to cancel the southward directed force from the wind, the water must drift to the west.

First compute the stress due to the wind, using the air density ρ = 1.275 kg/m3 at STP, T wind = (1.275 kg/m3 )(10.3 m/s)2 (0.0013) = 0.20 N/m2 .

The Coriolis force per unit area is Fcoriolis /A = 2ρ s dvd ω⊕ sin λ, where d = 50 m, ρ s = 1 030 kg/m3 is the density of seawater, and vd is the drift speed of the water. Equating this to the wind stress,

590

vd =

0.20 N/m2 0.20 N/m2 = 2ρω⊕ d sin λ 2(1 030 kg/m3 )(7.27 × 10−5 s−1 )(50 m)(0.573)

0.05m/s .

Rate of mass upwelling is then ∆m = (0.05m/s)(50 m)(1 030 kg/m3 ) 2 500 kg/m s = 2 500 t/km s . ∆l∆t

27.10 Estimate the wind in a tropical region at latitude (30 +

y)◦ North as v = 2y x̂ m/s, where x̂ is a unit vector pointing eastward. Use the formula from Problem 27.9 to estimate the wind stress on the surface waters. Estimate the rate of southward mass transport assuming z0 = 50 m.

27.11 If an OTEC system could achieve 1/2 of Carnot efficiency, at approximately what rate could energy be extracted sustainably from ocean surface waters in the tropics? What area of ocean would need to be filled with OTEC devices at this rate of energy extraction to produce a sustained gigawatt of power? Compare to the area needed for solar thermal conversion, based on the SEGS power plant (§24).

27.12 [T] Derive the relation du/dt = d∗ u/dt + ω × u by an explicit computation in a coordinate system where ω = ωẑ as described in Box 27.1.

Define the coordinate directions x̂ and ŷ as in Example 27.2.

The wind blows in the ŷ direction (North (South) in the Northern (Southern) Hemisphere). Integrate v x and vy over 0 ≤ z < ∞. Write 1 vy = √ (cos z/z0 − sin z/z0 )e−z/z0 v s 2 1 v x = √ (cos z/z0 + sin z/z0 )e−z/z0 v s .

2 and use Z ∞ 0

dz cos ze−z =

Z ∞ 0

dz sin ze−z = 1/2

R∞ R∞ √ to obtain 0 dz vy (z) = 0 and 0 dz v x (z) = v0 z0 / 2. The mass transport is in the x̂-direction, East (West) in the Northern (Southern) Hemisphere.

27.13 [T H] When an iceberg is driven across the ocean’s sur-

face by a steady wind, it is subject to the combination of forces described in Box 27.2 plus a drag force proportional to −bv, where b is a constant. Thus Newton’s law reads m ẍ = 2mω⊕ sin λẏ − b ẋ,

mÿ = Fwind − 2mω⊕ sin λ ẋ − bẏ .

Show that the resulting motion consists of transient inertial motion plus uniform drift, ẋ = ae−γt cos ωt + v∞ cos θ,

ẏ = ae−γt sin ωt + v∞ sin θ .

Find the constants ω, γ, v∞ , and θ in terms of b, m, ω⊕ sin λ, and Fwind .

615

Problems

Problems 28.1

Choose a cubical volume of size dV = dxdydz with sides oriented perpendicular to the x, y, z axes. Forces on the two sides perpendicular to the x axis differ by f dxdA, where dA = dydz, so net force is dF/dV = f in the −x direction. Forces in the other two corrections are suppressed by a factor of order O(dx, dy, dz) and vanish in the limit. Since coordinates can be chosen arbitrarily, we can always choose local coordinates where the gradient of p is in the x̂ direction. Thus the equation in general coordinates is f p = −∇P.

28.2

f =2

[T] Assume that the pressure in a volume of air can be expressed to first order in a local coordinate system (x, y, z) as p = p0 + f x + O(x2 , y2 , xy, . . .). Show from the balance of forces on a small volume element dV that the force per unit volume is f in the −x-direction.

Using the fact that the coordinate system can be chosen arbitrarily, show that more generally f p = −∇P.

[T,H] The forces on a parcel of moving air in the Ekman layer include the pressure force f p = −∇p, the Coriolis force f hc = f v × n̂ (where f = 2Ωρ sin λ), and a frictional force directed opposite to the wind velocity f f = −α(z)v, where α(z) arises from the velocity gradient and turbulent viscosity. The coefficient function α(z) decreases with increasing height z through the Ekman layer and vanishes at the top of the layer.

Find the wind velocity by demanding that the sum of the forces on the parcel vanishes. Express v in terms of components along ∇p and n̂ × ∇p. Show that your result reproduces qualitatively the wind pattern (both in direction and magnitude) shown in Figure 28.10.

= 1.23 × 10−4 kg/m3 s If v = 30 m/s, then |∇p| = 30(1.23 × 10−4 ) Pa/m

= (36.9 × 10−4 Pa/m)/div(1.013 × 105 Pa/bar)

= 36.4 × 10−9 bar/m = 0.036 mbar/km

28.4

Suppose a gradient wind (see Box 28.1) is flowing along a circular isobar with radius R at a latitude of 45◦ . Compare the strength of the Coriolis and centrifugal forces on this flow as a function of wind speed.

At what speed (as a function of R) are the two inertial forces equal?

The magnitude of the centrifugal force (per unit volume) on a mass moving with speed v in a circle of radius R is fcent = ρv2 /R. Setting this equal to the Coriolis force, and solving for v, ρ

28.5

[T] The wind speed at any site can be written as the sum of its average hvi plus a fluctuating term that averages to zero, v(t) = hvi + δv(t). Show that the effect of the fluctuations is always to make hv3 i greater than hvi3 .

Substitute the decomposition into the average of v3 , hv3 i = hhvi3 + 3hvi2 δv + 3hviδv2 + δv3 i = hvi3 + 3hvihδv2 i > hvi3

v = v1 k̂ + v2 ˆj

28.6 f p + f hc + f f = 0 −β ĵ + + f (v1 k̂ + v2 ĵ) × n − α(v1 k̂ + v2 ĵ) = 0

f v1 − αv2 = β

f v2 + αv1 = 0

while the power density is given by ( 1 (8 m/s)3 = 326.4 W/m2 .

P = ρv3 = 0.5(1.275 kg/m3 ) (16 m/s)3 = 2.611 kW/m2 2

Solving for v1,2 and rewriting in terms of the original variables, 1 ( f n̂ × ∇p − α∇p) f 2 + α2

28.7

When α = 0 we recover geostrophic winds along isobars. When α > 0 the wind develops a component in the direction of decreasing pressure. Also the wind speed, |v| = p

What is the maximum power that could be extracted by a mechanical device of cross-sectional area 10 m2 in a steady wind at 10 m/s?

We will use the air density at STP, ρ = 1.275 kg/m3 .

Betz’z Law gives us the power density,

|∇p|

f 2 + α2

decreases as α increases. These two features reproduce the figure.

28.3

where we have used the fact that hδvi = 0.

Compute the energy density and power density in a steady wind at 8 m/s and at 16 m/s.

We will use the air density at STP, ρ = 1.275 kg/m3 .

The energy density is given by ( 1 (8 m/s)2 = 40.80 W/m2 , P = ρv2 = 0.5(1.275 kg/m3 ) (16 m/s)2 = 163.2 J/m3 2

Using k̂ × n̂ = ĵ and ĵ × n̂ = − k̂ and identifying components along ĵ and k̂,

v=

v2 = 2ρΩv sin λ R v = 2ΩR sin λ 1.0 × 10−4 R s−1

for λ = 45◦ . Thus, if R = 1 000 km, the centrifugal force exceeds the Coriolis force when v & 100 m/s.

Let ∇p = |∇p| ĵ ≡ β ĵ and let k̂ = n̂ × ˆj. Parameterize v by it’s components along ĵ and k̂,

Then

!

2π 1 (1.2 kg/m3 ) √ 24(3 600 s) 2

Evaluate the Coriolis factor f = 2Ωρ sin λ at ±45 latitude. How large a pressure gradient (in millibars/km) is necessary to sustain a geostrophic wind at 30 m/s at this latitude? ◦

Pmax =

8 3 ρv = 0.296(1.275 kg/m3 )(10 m/s)3 = 377.4 W/m2 .

27

Thus, the maximum possible power that could be extracted is Pmax = Pmax (10 m2 ) = 3 , 790 W .

616

28.8

[T] Suppose the wind speed obeys a Gaussian frequency distribution in the two components of the horizontal velocity v = (v1 , v2 ), f (v1 , v2 ) =

d2 P 1 −(v2 +v2 )/(2σ2 ) e 1 2 . = dv1 dv2 2πσ2

(28.15)

λA 5.1 m/s, kA 4.38. Estimate the maximum available wind power density and the wind power class for each site at a height of 80 m assuming a roughness length z0 ≈ 2 × 10−5 (sand). Can you explain (on the basis of geography) why the winds at Yenbu are so much more variable than at Al Qaysumah?

Show that the wind q speed distribution, f (v) = dP/dv, where v = v21 + v22 , is a Rayleigh distribution (Weibull parameter k = 2). What is the scale λ?

First the wind speeds must be scaled from 10 m to 80 m. Since λ sets the scale for the wind speed,

The Weibull distribution is a distribution in wind speed v, independent of the wind direction θv , so we change variables from {v1 , v2 } to {v, θv },

The power density is given by

λ(80 m) = λ(10 m)

P=

dv1 dv2 = v dv dθv and integrate over direction Z 2π

dP = f (v) = dθv v f (v1 , v2 ) dv 0 v −v2 /2σ2 = 2e . σ This √is the form of a Weibull distribution (28.10) with k = 2 and λ = 2σ.

28.9

[T] Show that Γ(n) = (n − 1)! for integer n.

28.10 [T] Derive the formula for Mn in eq. (28.12).

28.11 A steady wind is blowing at 8 m/s at 200 m above

ground level. Estimate the wind power at a height of 10 m above a large open cropped field. Approximately what height is needed for a turbine in a small town to reach wind flowing with the same power?

From Table 28.1, we estimate z0 ≈ 0.05 for the field and z0 ≈ 1.5 for the town. Assuming the logarithmic height variation parameterized by eq. (28.2), with zref = 200 m and vref = 8 m/s, we find the wind speed at 10 m above the field to be, ln(10/0.05) v=8 5 m/s . ln(200/0.05) To find the height z at which the wind would reach the same speed above the town, we solve 5 m/s = 8 m/s

ln(z/1.5) , ln(200/1.5)

which yields z = 34 m.

28.12 Consider building a wind turbine at a wind power class

ln(80 m/(2 × 10−5 m)) 1.16λ(10 m) . ln(10 m/(2 × 10−5 m)) 1 3 1 3 ρv̄ = ρλ Γ(1 + 3/k) 2 2

where we have substituted the Weibull expression for v̄3 . Putting these together P(80 m) =

1 ρ(1.16λ(10 m))3 Γ(1 + 3/k) .

2

With Γ(1 + 3/2.25) 1.19 and Γ(1 + 3/4.38) 0.91, and using the air density ρ = 1.17 kg/m3 at sea level and 25◦ C, we find PY (80 m) = 0.5(1.17 kg/m3 )(1.16 × 5.9 m/s)3 (1.19) 220 W/m2 ,

PA (80 m) = 0.5(1.17 kg/m3 )(1.16 × 5.1 m/s)3 (0.91) 110 W/m2 .

To assign wind power classes, we can use the original data at 10 m (since Table 28.2 refers either to 10 m or 50 m, but not to 80 m). At 10 m, we find power densities PY (10 m) = 0.5(1.17 kg/m3 )(5.9 m/s)3 (1.19) 140 W/m2 , PA (10 m) = 0.5(1.17 kg/m3 )(5.1 m/s)3 (0.91) 71 W/m2 .

Comparing with the table, we see Yenbu is Class 2 and Al Qaysumah is Class 1.

We can also compute the average wind speed at each location, given by hvi = λΓ(1 + 1/k), so hviY (10 m) = (5.9 m/s)Γ(1 + 1/2.25) 5.9(0.89) 5.3 m/s ,

hviA (10 m) = (5.1 m/s)Γ(1 + 1/4.38) 5.1(0.91) 4.6 m/s .

Note that the ratio of the power at Yenbu to the power at Al Qaysumah is ∼2:1 while the average wind speed cubed (hvi3 ) is only ∼1.5:1. The difference in power comes from the fact that v̄3 /hvi3 is considerably larger at Yenbu than at Al Qaysumah.

Yenbu is on the shore of the Red Sea whereas Al Qaysumah (airport) is several hundred kilometers inland. The variability of the wind at Yenbu could be due to a sea-land circulation associated with its proximity to the sea.

28.14 Use the data given in the wind rose Figure 28.19 to

5 site. Assume that the hub height is 50 m and the blade length is 15 m, and that the turbine operates at 50% of the maximum theoretical efficiency. Estimate the average power output of the turbine.

make a crude estimate of the average hvi and root mean cube v wind speed at Klamath Lake, Oregon. From these, estimate the Weibull parameters for this distribution. Is your result close to a Rayleigh distribution (k = 2)? Is this a promising wind power location?

According to Table 28.2, class 5 at 50 m corresponds to power density 500–600 W/m2 . At 50% of the Betz limit, this turbine captures

Taking values of v in the middle of listed ranges and ignoring the very small contribution from v > 10 m/s,

P = 0.5(16/27)(500 − 600) W/m2 = 148 − 178 W/m2 .

2

hvi ≈ 0.27(1.25) + 0.34(3) + 0.19(5) + 0.13(7) + 0.036(9)

2

≈ 3.5 m/s

With an area of AT = π(15 m) = 707 m , the average power output of the turbine should be P = I(707 m2 )(148 − 178 W/m2 ) = 105 − 126 kW.

hv i ≈ 0.27(1.25)3 + 0.34(3)3 + 0.19(5)3 + 0.13(7)3 + 0.036(9)3

eral locations in Saudi Arabia, among them Yenbu (Y) and Al Qaysumah (A). They report Weibull parameters measured at a height of 10 m: λY 5.9 m/s, kY 2.25;

√ 3 We estimate r ≡ v̄/hvi = 105/3.5 ≈ 4.7/3.5 ≈ 1.35. From Figure 28.18, we see that this value of v̄/hvi corresponds roughly k ∼ 1.6. So the wind at this site is more variable than the Rayleigh distribution would predict.

28.13 Reference [192] studies wind power prospects for sev-

3

≈ 105 m3 /s3

Problems

The mean wind speed at 10 m puts this location in wind Class 1 (Table 28.2), so this is not a promising wind power location.

28.15 The following wind frequency distribution data were

obtained in the month of January at a height of 10 m above an airport runway in Buffalo, New York. Calm: 2.0%; 0–5 kt: 5.3%; 5–12 kt: 40.1%; 12– 20 kt: 39.3%; >20 kt: 13.3% (1 knot = 0.514 m/s).

Assume that the wind above 20 kt has an average value of 25 kt. Make a crude estimate of the power in the wind at 10 m. Is this a promising location for wind power based on the January data alone?

For a crude approximation, assign a wind speed value at the middle of each bin, then compute v̄3 = hv3 i,

v̄3 ∼ (.514)3 0.053(2.5)3 + 0.40(8.5)3 + 0.39(16)3 + .13(25)3 m3 /s3 ∼ 525 m3 /s3 .

Thus, using the air density ρ = 1.275 kg/m3 at STP, hPi =

1 3 ρv̄ ∼ 0.5(1.275 kg/m3 )(525 m3 /s3 ) ∼ 335 W/m2 .

2

Referring back to Table 28.2, this power estimate is at the bottom of Class 6. So this location is promising, though January is likely to be one of the windiest months and data from the whole year must be evaluated.

617

638

Problems 29.1

Angel Falls in Venezuela, at a height of 979 m, is the highest waterfall in the world. Ignoring air resistance, how much warmer is the water at the bottom of the falls, where the kinetic energy it acquired by falling has been dissipated as heat, than it was at the top?

The water emerging from the opening with velocity v is also under atmospheric pressure. Bernoulli’s principle (29.13) gives 1 2 ρv + p0 = p0 + ρgh .

2 Thus v =

29.6

From Example 29.1, ∆Tp= v2 /2c. Falling 979 m gives the water p a velocity of v = 2gh 2(9.8 m/s2 )(979 m) 140 m/s. Thus, ∆T =

(140 m/s)2 v2

2.3 K .

2c 2(4 186 J/kg K)

(Of course the actual temperature rise will be much less because the limiting velocity of water falling in air is much less than 140 m/s; this represents the loss of most of the original potential energy to kinetic/thermal energy of air molecules.)

29.2

An entrepreneur claims to have devised a system for extracting thermal energy from the wind by cooling it.

His device lowers the temperature of the air that passes through it by 1 ◦ C. He claims to generate 800 W/m2 from air flowing at 1 m/s. Does this device conserve energy? Would you invest money in the entrepreneur’s scheme?

A stream of water emitted by a tap at velocity v0 (as in Figure 29.3) has a circular cross section with radius r0 . Use conservation of mass to predict how its radius changes as it falls.

p=

29.4

1+

r0 q

∆ρ ∆p = . ρ γp With γ ≈ 1.4, the fractional change in the density near the wings of the Boeing jet can be estimated as 8/(1.4) ≈ 6%. For the Lear jet, it is roughly 1%.

29.7

2gz/v20

Supply the missing steps in the derivation of the expression for the flow rate Q in a Venturi flow meter in Example 29.2.

Show that for the flow of a viscous fluid through a pipe, as analyzed in Box 29.1, the energy per unit mass dissipated as thermal energy by viscous forces is dQ/dm = ∆p/ρ, where ∆p is the pressure difference between the ends of the pipe. Note that the answer is independent of the viscosity and the cross-sectional shape of the pipe.

Suppose water is forced through a pipe by a pressure difference of 10 atm. How much does it heat up?

Use conservation of energy in a fashion similar to the derivation of Bernoulli’s principle. The velocity profile of the fluid does not vary along the pipe, so the kinetic energy in the fluid is the same at both ends. Therefore the work done on the fluid by the pressure must equal the energy consumed in viscous friction, which generates heat. When a volume ∆V enters/exits the pipe, the work done it by the pressure is pi/ f ∆V, so the net work done is ∆W = ∆p∆V. Thus the heat generated per unit volume is dQ/dV = ∆p and the heat generated per unit mass is

Start with Bernoulli’s law as written in the Example, and use conservation of mass, ρv1 A1 = ρv2 A2 to eliminate, say, v2 , !

1 2 A1 v 1− = g(h2 − h1 ) 2 1 A2

29.5

mg (3.5 × 105 kg)(10 m/s2 ) 8 000 Pa , = A 440 m2

which is roughly 8% of atmospheric pressure. For the Lear jet the same calculation yields 1 550 Pa, about 1.5% of atmospheric pressure.

Under adiabatic conditions,

Conservation of mass requires vπr2 = v0 πr02 . After falling a p distance z, v = v0 + 2gz. Thus r= r

2gh.

The maximum takeoff mass for a Boeing 777-300ER passenger jet is 3.5 × 105 kg. The plane’s wing area is about 440 m2 . Compute the average air pressure on the plane’s wings required to overcome the plane’s weight and lift the plane off the ground. How far does this pressure deviate from average atmospheric pressure at sea level? Repeat the calculation for a Lear jet with mass 3300 kg and wing area 21.5 m2 . Comment on the implications for the assumption that air behaves as an incompressible fluid in these situations.

Pressure is force per unit area. The force of gravity can be viewed as a downward pressure on the plane that must be counteracted by the pressure difference that generates the lift that keeps the plane in the air. For the Boeing 777-300ER,

In one second, one m3 of air passes through the device. With a heat capacity of 1.0035 kJ/kg K, lowering the temperature of air by 1 ◦ C could, in principle, yield 1 kJ/s=1 kW. This is, however, a heat engine, and is subject to the Carnot efficiency limit. Working between two temperatures close to 300 K that differ by only 1 ◦ C, the efficiency would be less than 1%, and could never yield 800 W. Don’t invest.

29.3

p

then solve for v1 , and multiply by A1 to get Q = v1 A1 as given in the Example.

dQ dm ∆p dQ = ÷ = dm dV dV ρ

Estimate the speed of water emerging from an opening at the bottom of a dam when the water in the reservoir behind the dam is held at a height h above the opening.

Compare the results of using conservation of energy directly to the application of Bernoulli’s principle.

∆p/ρ (106 Pa)/(103 kg/m3 ) 1 kJ/kg. Since the specific heat of water is c 4.2 kJ/kg K, the water heats up by less than 1/4 ◦ C.

Conservation of energy (in p elementary mechanics) requires ρgh = (1/2)ρv2 , thus v = 2gh. The water at the top of the reservoir is at rest v0 = 0 and under atmospheric pressure p0 . It has potential energy per unit mass of ρgh relative to the opening.

29.8

[T] The dependence of Reynold’s number on the parameters of fluid flow can be inferred from dimensional analysis. First show that it is not possible to construct a dimensionless number out of the viscosity η, density ρ, and speed v of a fluid flow. Next show that if a length scale K characterizing the object is

639

Problems

introduced, the dimensionless combination is that of eq. (29.23).

The dimensions of viscosity, density, and speed are [η] = m/lt, [ρ] = m/l3 , and [v] = l/t. We look for a dimensionless combination,

Finally note that the unit vector in the angular direction θ̂ = xŷ−y x̂, to obtain eq. (29.18) with Γ0 = µ0 I. Maxwell’s equations (§3.6) require that ∇ · B = 0 everywhere and ∇ × B = 0 for time independent magnetic fields in current free regions of space, so the same properties hold for v.

See the figure.

[ηa ρb vc ] = (m/lt)a (m/l3 )b (l/t)c , leading to the equations, a+b= 0 a+c = 0 −a − 3b + c = 0 .

These equations have only the trivial solution a = b = c = 0.

Including a length scale K in the combination, [ηa ρb vc K d ] = (m/lt)a (m/l3 )b (l/t)c K d , yields a+b=0 a+c=0 −a − 3b + c + d = 0 , which has the solution b = c = d = −a, so ρ v K/η ≡ Re is the only dimensionless combination.

29.9

A large wind turbine blade is 50 m long and has a maximum chord length of 4 m at a point one-third of the way out from the wind turbine hub. When rotating at its design rate, the blade sees an airstream moving at 100 m/s at the blade tip. What is the Reynold’s number of the air flow at the point of maximum chord length?

Suppose the aerodynamic properties of the blade are to be studied in a wind tunnel that can accommodate at most a blade scaled down to a length of 7.5 m. The wind tunnel operates at atmospheric pressure. To study the properties of the blade near its point of maximum chord length, what wind speed should be used in the wind tunnel? ρvK (1.17 kg/m3 )(1/3)(100 m/s)(4 m) 8.5 × 106 Re =

η 1.84 × 10−5 Pa s

To study the properties of the blade the Reynolds number must be the same in the wind tunnel as in actual use. Since the chord length has been reduced to by a factor of 7.5/50 = 0.15, the airstream velocity must be increased by a factor of 6.67, v = 220 m/s, just below 2/3 the speed of sound.

29.10 [T] Show that the flow v around a cylinder given in

eq. (29.18) has the same functional form as the magnetic field B around a long, straight wire (3.50) and therefore Maxwell’s equations imply that ∇ · v = ∇ × v = 0. (Note that the radial coordinate in the plane is denoted ρ in §3.) Then use Stoke’s theorem (B.1.5) and/or direct integration to show that the circulation around any curve that surrounds the cylinder is Γ0 .

According to eq. (3.50), Bwire =

µ0 ẑ × r̂ I .

2π r

Use r̂ = (x x̂ + yŷ)/r, ẑ × x̂ = ŷ, and ẑ × ŷ = − x̂, to obtain Bwire =

µ0 xŷ − y x̂ I .

2π r2

The integral of v around any curve C surrounding the cylinder can be written as the sum of the integral around a circle C0 , plus the integral around over the surface S ′ for which C forms one boundary and the circle forms the other boundary, Z Z Z dx · v = dx · v + dx · v C

C′ =C−C0

C0

=

Z

C0

dx · v +

Z

S′

dS · ∇ × v

The surface integral is zero since the curl of v vanishes everywhere on the surface S ′ . Thus the integral around C equals the integral around the circle.

29.11 [T,H] Suppose that in a region of steady flow the veloc-

ity of a fluid is given by v = Ω × r, where r is the vector from a fixed point in space. Describe the streamlines of this flow. Show that ∇ · v = 0, so this flow is consistent with the fluid being incompressible. Compute the circulation Γ on a circle of radius R about the center of the flow. Show that the fluid is rotating with angular velocity Ω and that the vorticity is ω = 2Ω.

Let Ω define the ẑ-axis. Then v = Ω × r = Ωẑ × (x x̂ + yŷ + zẑ) = Ω(xŷ − y x̂) = Ωrθ̂ where we have used x̂ × ŷ = ẑ, etc. The streamlines, therefore point in the direction of the polar angle θ around the Ω̂-axis.

From eq. (B.15) ! ∂ ∂ ∇·v=Ω (−y) + (x) = 0 , ∂x ∂y so the flow is incompressible. The circulation is computed by integrating on a circle, where dx = θ̂R dθ, so Z 2π Γ= dθ R2 Ω = 2πR2 Ω.

0

In a time dt, an element of fluid at a distance r from the Ω̂-axis moves in the θ̂-direction by an amount |v|dt = Ωrdt, which is the equation of circular motion with a constant angular velocity Ω.

Finally, the vorticity can be computed from eq. (B.16), ! ∂ ∂ vy − v x = 2Ω . ω = ∇ × v = 0, 0, ∂x ∂y Note that this result is consistent with Stoke’s theorem, Z dS · ω , Γ(r) = A

where S is the disc of radius R around the Ω̂-axis.

640

29.12 [T,H] In §29.4.1 it is asserted that eq. (29.28)

describes the steady, circulation-free flow of an incompressible ideal fluid around a cylinder. To confirm this, verify (a) that va → v∞ x̂ as r → ∞; (b) that no fluid flows into or out of the cylinder; (c) that the circulation is zero around any circle of radius r > R; and (d) that ∇ × va = 0 (zero vorticity) and ∇ · v∞ = 0 (incompressibility) everywhere outside the cylinder. [Hint: For part (d), note that va can be written as the gradient of a scalar function, va = ∇Φ, where Φ = v∞ x̂ · r + v∞ R2

x̂ · r . r2

Also make use of the identities eq. (B.18) and eq. (B.20), and the facts that ∇(1/r p ) = −pr̂/r p+1 and ∇2 (1/r p ) = p2 /r p+2 . The identity ∇2 ( f g) = f ∇2 g + g∇2 f + 2∇ f · ∇g may be useful as well.] a) Since the second term in the first form of eq. (29.28) falls like 1/r2 , this result is obvious. b) The radial component of va vanishes when r = R. c) The circulation is obtained by integrating the tangential component of va on a circle of radius r, ! Z 2π R2 sin θ = 0. Γ(r) = −v∞ 1 + 2 r 0 Finally for part d), first take the gradient of Φ using ∇(x̂ · r) = x̂ and ∇(1/r2 ) = −2r̂/r3 . !!

1 1 ∇Φ = v∞ x̂ + v∞ R2 2 ∇(x̂ · r) + x̂ · r ∇ 2 r r = v∞ x̂ + v∞

R2 ( x̂ − 2(x̂ · r̂)r̂) , r2

as asserted. Identity eq. (B.20) then gives ∇ × va = ∇ × ∇Φ = 0.

To show ∇ · va = 0, first use the identity eq. (B.18) and then the identities given in the hint, ! !!

1 1 ∇ · va = ∇2 Φ = v∞ R2 x̂ · r ∇2 2 + 2∇(x̂ · r) · ∇ 2 r r !! −2r̂ 2 4 x̂ · r = v∞ R + 2 x̂ · = 0, r4 r3 where we have used ∇2 (x̂ · r) = 0.

29.13 [T] Investigate the location of the stagnation points for

fluid flow around a cylinder as described in §29.4.1.

Show that for small Γ there are two stagnation points on the cylinder and find their location. What is the critical value of Γ/Rv∞ for which there is only one stagnation point? For larger values of Γ are there stagnation points in the flow? If so, where are they?

Express the components of the velocity field of eq. (29.28) in terms of the polar coordinates r and θ. Check that vr = 0 on the surface of the cylinder. For r = R, then, any value of θ for which vθ = 0 is a stagnation point. The second to last equation in the box gives the necessary decomposition of v, vr (r, θ) = v∞ cos θ(1 − R2 /r2 ) vθ (r, θ) =

Γ − v∞ sin θ(1 + R2 /r2 ) 2πr

Note that vr = 0 at r = R as required. Setting r = R, the condition for a stagnation point is vθ (R, θ) =

Γ − 2v∞ sin θ = 0 .

2πR

which gives sin θ = Γ/4πRv∞ . This has a two solutions θ1 = sin−1 (Γ/4πRv∞ ) and θ2 = π − sin−1 (Γ/4πRv∞ ) , as long as Γ ≤ 4πRv∞ . Both stagnation points are below the midline of the cylinder because Γ < 0 for the flow described in the text.

They coalesce at θ = −π/2 when Γcritical /Rv∞ = 4π.

If Γ > Γcritical there are no stagnation points on the cylinder. The radial component of the velocity vr never vanishes for r > R, so there are no stagnation points in the flow under these conditions.

29.14 A modern medium size commercial jet plane has a

wing span of L = 30 m and a mean chord length of K = 4 m. Its lift coefficient is cl = 0.83 at its cruising angle of attack α ∼ 2◦ . What is the lift force generated by the airplane wings when cruising at v 800 km/h?

While cruising, the engines exert a thrust of ∼80 kN. Estimate the wings’ drag coefficient cd and the lift to drag ratio.

From eq. (29.32), and taking the density of air to be 1.17 kg/m3 , fl = (1/2)cl ρv2 K (0.5)(0.83)(1.17 kg/m3 )(222 m/s)2 (4 m) 9.6 × 104 N/m .

So the total lift force if Fl = fl L 2.9 × 106 N. We equate the thrust to the drag force on the wings alone: Fd = (cd /cl )Fl ∼ 8 × 104 N. Substituting for Fl and cl , we find cd ∼ (8 × 104 )(0.83)/(2.9 × 106 ) 0.023, so cl /cd 36.

655

Problems

with ρ0 = 1.17 kg/m3 , Pd = 3 MW, vr = 12 m/s, and AT = π(45 m)2 . This gives CP = 0.47. The Weibull parameter λ must be scaled from 10 m to 105 m, with a roughness length z0 = 0.01 corresponding to rough pasture,

Problems 30.1

A variable-speed HAWT has a power coefficient of 0.45 at its rated wind speed vr and maintains its rated power for wind speeds between this value and its cutout speed of 25 m/s. If the rotor diameter is 100 m, and the rated power is 3 MW, what is vr ?

λ=

According to eq. (30.18) v3 =

30.2

2P 2(3 MW) = 1 400 (m/s)3 . ρ0 ACP (1.2 kg/m3 )(π × 2 500 m2 )(0.45)

The average power is then computed by carrying out the integral of eq. (30.19) with f (v) given by eq. (28.10), with CP = 0.47 for vin ≤ v ≤ vr and CP (v) = CP (vr )v3r /v3 for vr ≤ v ≤ vout . The result (from a numerical calculation) is an average power of P = 1.3 MW. Over a year, this gives a total of 41 TJ, or 11 × 103 MWh.

So v 11 m/s.

[T] Verify the results of eq. (30.7), and evaluate them at the Betz limit.

A wind pump based on the “American farm windmill” design (see Figures 30.2(b) and 30.7) with rotor diameter 4.5 m and tip-speed ratio λ ∼ 1 claims to be able to pump 2400 L/h of water from a depth of 30 m when the wind speed is 4 m/s. Do you think this is possible?

First solve eq. (30.6) for v2 = (1−a)v1 . The substitute for both v1 and v2 in eq. (30.5), giving v4 = (1−2a)v1 . From mass conservation (30.1) v1 A1 = v2 AT , which we solve for A1 = (1 − a)AT . Similarly v4 A4 = v1 A1 , and substituting for v4 yields A4 = (1−a)AT /(1−2a).

Substituting for v4 and Av = AT v2 into eq. (30.4) yields the formula for P quoted in eq. (30.7).

The power required to pump this volume of water is P = (dm/dt)gh = (2 400 kg/3 600 s)(9.8 m/s2 )(30 m) 200 W. The power coefficient of the wind pump is limited to be less than roughly CP 0.15 (see Figure 30.7(b)) due to its low tip-speed ratio. Its maximum possible power (taking the density of air to be 1.17 kg/m3 ) is then

30.5

30.6

[T] Derive eq. (30.15) by equating the normal force per unit length on a blade element given in eq. (30.12) with the thrust per unit length obtained from the axialmomentum analysis (30.14).

Equating the two expressions for f N ,

1 Pmax = ρ0 v3 ACP 2 = (0.5)(1.17 kg/m3 )(4 m/s)3 (π × (2.25 m)2 )(0.15) 90 W .

1 ρ0 KBv2 (1 − a)2 cl cos φ ρ0 v2 8πra(1 − a) = 2 2 sin2 φ 8πra tan φ sin φ = KB(1 − a)cl

This does not seem to be possible.

30.3

The maximum power of a Savonius wind turbine (see Example 30.2) is claimed to be P = 0.36 hrv3 , where P is in watts, and the height h and radius r of the Savonius rotor, and the wind velocity v, are all in SI units. What is its maximum power coefficient as a fraction of the Betz limit?

Substituting for tan φ from eq. (30.11), leads directly to the quoted result for K/R.

30.7

The maximum power is Pmax = (1/2)CP ρ0 v3 A. Here A = 2hr, so Pmax (0.5)(1.2 kg/m3 )CP 2hr v3 = (1.2CP )hrv3 . So CP = 0.36/1.2 = 0.3 (in agreement with Figure 30.7). Since 0.3 (1/2)(16/27), this Savonius turbine reaches 1/2 of the Betz limit.

30.4

ln(105 m/(0.01 m)) 9 m/s 12 m/s . ln(10 m/(0.01 m))

[C] Wind frequency data obtained at a height of 10 meters at a particular location is summarized by a Weibull distribution with λ = 9 m/s and k = 1.3.

The land use is best characterized as “rough pasture.”

Your wind power company is considering placing a variable-speed wind turbine with a hub height of 105 m, a rotor diameter of 90 m, and a rated power of 3 MW at this site. The cut-in, rated, and cut-out speeds are vin = 3 m/s, vr = 12 m/s, and vout = 25 m/s respectively. Calculate the wind turbine’s power coefficient at its rated wind speed. Assume a constant power coefficient between vin and the design speed, and assume that the turbine produces its rated power between the rated speed and vout . How much total energy (in MWh) would you estimate this turbine can produce per year?

The power coefficient can be calculated from Pd =

1 ρ0 v3r AT CP , 2

[T] Axial-momentum analysis led to a power coefficient C P = 4a(1 − a)2 for a HAWT (eq. (30.7)).

Compute the power coefficient another way: start from the tangential force per unit length f T in the simple model that led to eq. (30.12); implement the axialmomentum theory constraint (30.15); find the torque unit length dτ/dr and compute the power P = Rper R dr Ω(dτ/dr). Show that the result is the same as the 0 result of the axial-momentum analysis (30.7).

Combining the expression for f T from eq. (30.12) with the constraint (30.15) yields, fT ≡ | f T | =

4π ρ0 v3 a(1 − a)2 . Ω

The torque per unit length, dτ/dr = r fT , times the angular velocity Ω, gives the power per unit length that turns the turbine (see eq. (2.44)). Integrating over r gives the total power P=

Z R 0

dr Ω

dτ = 4πρ0 v3 a(1−a)2 dr

Z R 0

dr r =

1 ρ0 v3 4a(1−a)2 AT 2

in agreement with eq. (30.7).

30.8

[T] The axial-momentum theory approximation to blade design breaks down when r/R becomes too small. Use eqs. (30.11) and (30.15) to show that the condition KB < 2πr limits the angle φ to sin φ tan φ < cl (1 − a)/(4a). For the choice of parameters in Figure 30.10 what is the smallest value of r/R allowed by this condition?

656

Solving for KB (using eq. (30.11)) yields eq. (30.15) modified by the factor (1 + (cd /cl ) tan φ),

From eq. (30.15), KB = 8πav sin φ/(Ωcl ) < 2πr ,

30.9

K 8πa sin φ = .

R λcl B 1 + (cd /cl ) tan φ

and from eq. (30.11), r = v(1−a)/(Ω tan φ). Substituting for r yields the required result. For a = 1/3 and cl = 1, sin φ tan φ < 1/2 and the maximum allowed value of φ is 39◦ = 0.22π radians. From the figure, this corresponds to r/R ≈ 0.15.

To compute the power we want to integrate P dr Ω(dτ/dr). We substitute for KB in the equation for f T , 0

RR

[T] Model the blades of a HAWT rotor as rods of equal length and mass, equally spaced around a circle, rotating with angular frequency Ω. Compute the moment of inertia of the rotor about the vertical axis. Show that the moment of inertia is independent of time as long as there are three or more blades.

 βΩr   1 − (1−a)v  1 dτ  = r fT = rρ0 v3 8πa(1 − a)2   β(1−a)v dr 2 1+ Ωr

where β = cd /cl and we have substituted from eq. (30.11) for tan φ = (1 − a)v/Ωr. The power is obtained by integrating over r, 1 ρ0 v3 4a(1 − a)2 AT (1 + ∆) where 2 Z 1 1 − βy/ξ −1 ∆=2 dy y 1 + βξ/y 0 P=

The moment of inertia of a rod inclined at an angle Ωt to a vertical axis is to I0 cos2 (Ωt), where I0 is a time independent constant.

If there are N such rods equally spaced around the axis, then the total moment of inertia is equal to I(N) = I0

N−1 X

where y = r/R and ξ = (1 − a)/λ. ∆ vanishes when cd = 0; it is the correction we seek. Keeping only terms of first order in β,

cos2 (Ωt + 2πn/N) .

∆ = −2β(ξ + 1/3ξ)

n=0

Noting that we can write

Substituting into the expression for P, and reverting to the original variables give the answer quoted. For the parameter values given β = 0.01 and ξ = 2/30, so

1 + cos(2Ωt + 4πn/N) , cos (Ωt + 2πn/N) = 2 2

∆ = −(0.01)(2)(2/30 + 5) ≈ −.10

this sum becomes   N−1 N−1 X X  I0  1 + cos(2Ωt + 4πn/N) N + = I(N) = I0 cos(2Ωt + 4πn/N) .

2 2 n=0

So drag reduces the power of the turbine by roughly 10%.

n=0

We can then evaluate the remaining sum be writing the cosines as the real parts of complex exponentials, N−1 X n=0

cos(2Ωt+4πn/N) =

N−1 X n=0

=

h

Re e

i(2Ωt+4πn/N)

i

N−1    cos(2Ωt) ,  X    i(2Ωt+4πn/N)    =   = Re  e 2 cos(2Ωt) ,      0 , n=0

Thus, for N = 1, I(1) = I0 cos2 (Ωt), and for N = 2, I(2) = 2I0 cos2 (Ωt), but for N ≥ 3, Ωt drops out of the sum, yielding I(N) = NI0 /2.

30.10 [H] Use blade-element theory to compute the correc-

tion to the axial-momentum theory expression for the power (30.7) to first order in cd /cl for an optimized blade. First, recompute f T and f N including the drag force and show that the right-hand side of eq. (30.15) is modified by a factor (1 + (cd /cl ) tan φ)−1 . Then compute the power as in Problem 30.7, keeping only terms to first order in cd /cl . How big is this correction for a turbine with cd /cl = 0.01, λ = 10, and a = 1/3?

From the geometry of Figure 30.9, 1 f T = − ρ0 KBw2 (cl sin φ − cd cos φ) ẑ 2 ! cl cd cos φ 1 ẑ − = − ρ0 KBv2 (1 − a)2 2 2 sin φ sin φ 1 f N = ρ0 KBw2 (cl cos φ + cd sin φ) x̂ 2 ! cd cl cos φ 1 + x̂ . = ρ0 KBv2 (1 − a)2 2 2 sin φ sin φ Equating the normal force per unit length to the thrust force computed from axial-momentum theory, 1 1 KBcl dT cd = ρ0 v2 4a(1 − a)2πr = ρ0 v2 (1 − a)2 tan φ 1+ dr 2 2 sin φ tan φ cl

!

N = 1, N = 2, N ≥ 3.

685

Problems

Problems 31.1

31.2

Compute the pressure at a depth Z below the surface in a reservoir behind a hydroelectric dam. Compute the work done by a volume of water ∆V as it passes from this pressure on one side of a turbine to essentially zero pressure on the other side. Show that this analysis yields the same formula (31.2) for the power output as the energy analysis presented in §31.1.1.

Compare land use of hydropower to solar power by computing the ratio of power output to reservoir surface area for the Three Gorges Dam (maximum capacity 22 500 MW, reservoir area 1084 km2 ) and the Hoover Dam (maximum capacity 2080 MW, reservoir area 680 km2 ), and comparing to the power density of a typical large-scale solar thermal or solar photovoltaic power plant in the desert.

31.6

p p By dimensional analysis, ω = σk3 /ρ. v p = ω/k = σk/ρ, vg = dω/dk = 3ω/2k = 3v p /2. Since vg > v p , crests move backwards through the packet in contrast to gravity waves, where they move forward. Surface tension should dominate at large k (small wavelength) and gravity should dominate at large k.

31.7

Defining power density (per unit area) as power capacity divided by reservoir size, we find P/A = (2.25 × 1010 W)/(1.084 × 9 10 m2 ) 20.8 W/m2 for the Three Gorges dam and P/A = (2.08 × 109 W)/(6.8 × 108 m2 ) 3.1 W/m2 for Hoover dam.

For comparison, the gross power density of an STE or PV plant in a desert (gross conversion efficiency ∼3%, insolation 300 W/m2 ) is roughly 10 W/m2 (§23.6 and §24.4). Thus hydropower and solar power have similar land use requirements.

31.3

E=

P = Evg =

31.8

If surface tension dominates, then a liquid will form a sphere since that is the shape with the least surface to volume ratio. If gravity dominates then the liquid will spread out in a sheet at the lowest possible height. The length scale l is an indicator of the transition from one behavior to the other, consistent with the observed behavior of water on a porcelain surface. Note that the interactions with the surface material can be decisive: a hydrophilic surface has reduced surface tension and will cause water to spread out even in small droplets. A hydrophobic surface will cause water to “bead up” even in larger drops.

[T] Show that the function a cosh(k(z + h)) sin(kx − ωt) ψ(x, t) = k sinh kh

describes a (plane) gravity wave propagating in the x̂direction in a uniform ocean of depth h. In particular, show that ψ(x, t) satisfies eq. (31.7) and the boundary condition that sz = 0 at z = −h. (See Appendix B.4.2 for properties of the hyperbolic functions cosh x and sinh x, and in particular, for their expansions near x = 0.) Show that for the limiting cases kh ≪ 1 and kh ≫ 1 this correctly reproduces eq. (31.22) and eq. (31.9).

For waves of fixed period, graph H/H0 , λ/λ0 as functions of depth, where H0 , λ0 are the wave height and wavelength in the limit of infinite depth.

[σ] = [energy/area] = m/t2 , [ρ] = m/l3 , and [g] = l/t2 , so p p [l] = [ σ/ρg] = (m/t2 )(l3 /m)(t2 /l) = l

For water, at 25 ◦ Cσ 76 × 10−3 J/m2 , so q l(water) = (76 × 10−3 J/m2 )/((1 000 kg/m3 )(9.8 m/s2 ) 2.8 mm .

1 T 2 2 ρa2 g2 = ρa g = 45.9 kW/m 4ω 8π

When the wave moves into shallow water, the power is going to be approximately the same. The group velocity, however, becomes p p vg = gh so the energy density is E = P/ gh which is 10.4 kJ/m2 at 2 m depth and 14.7 kJ/m2 at 1 m depth.

The energy density is stillpgiven by E = (1/2)ρga2 so, solving for the wave amplitude, a = 2E/ρg which is 1.46 m at 2 m depth and 1.73 m at 1 m depth.

80 MW .

31.5

1 ρga2 = 4900 J/m2 2

and the power is (eq. (31.20))

Estimate the power output possible for a hydroelectric dam with hydraulic head Z = 30 m, flow rate Q = 300 m3 /s, and turbine efficiency 90%.

[T] Show that the displacement s given by (31.9) satisfies the constraints eqs. (31.4) and (31.5). p Show that the quantity l = σ/ρg has the dimensions of length, where the surface tension σ has units of energy per unit area. Look up the surface tension of water and evaluate l for water. When a small drop of water is placed on a porcelain surface it forms a round bead, but as the drop gets bigger, it flattens out. Explain what l has to do with this. Does the value of l agree with your experience?

Consider a deep-water wave with amplitude a = 1 m and period 12 s. Compute the energy density and power. When the wave moves into shallow water, what is the height and energy density of the wave at 2 m and at 1 m depth?

The energy density of the deep water wave is

P = ρgZǫQ = (1 000 kg/m3 )(10 m/s2 )(30 m)(0.90)(300 m3 /s)

31.4

[T] Use dimensional analysis to determine the dispersion relation ω(k) for surface tension waves (up to a multiplicative constant), which can only depend on the surface tension σ, the density ρ, and the wave number k. Find the phase velocity and the group velocity, and show that vg = 3v p /2. Describe how you could distinguish visually between gravity waves and surface waves by observing the way that individual wave crests move through a wave packet.

31.9

[T] The function ψ(x, t) =

a cosh(k(z + h)) cos kx cos ωt k sinh kh

is a standing wave solution analogous to eq. (31.10) for a surface gravity in an ocean of depth h. Following the derivation in the text, construct the kinetic and potential energy per unit area and find the dispersion relation for arbitrary h. Show that your solution has the correct p limit for deep and shallow water. (Answer: ω = gk tanh(hk), where tanh x = sinh x/ cosh x.)

686

Show that the power density in this wave is the same as in a deep-water wave, eq. (31.20).

31.10 A tsunami is a very long-wavelength wave produced by a sudden movement in the sea floor over a distance of hundreds of kilometers. The wavelengths in a tsunami are generally much greater than ocean depth so that a tsunami can be described accurately in the shallow wave approximation, and is non-dispersive.

Consider a tsunami described by a wave with approximate wavelength 100 km and amplitude 1 m. Compute the velocity and period of this wave and the power density per unit length of shoreline perpendicular to the incident wave. (Note, you may approximate ocean depth as h ∼ 3.5 km. Also you will need the power density in the shallow wave approximation, see Problem 31.9.)

The velocity of the tsunami is q p v = gh = (9.8 m/s2 )(3.5 km) 190 m/s

Earth’s oceans would remain fixed in an Earth-based reference system. Give an estimate of the tidal displacement ∆ in this situation by computing the gravitational potential along a line passing through the centers of the Moon and Earth and comparing to the potential at a second point also on Earth’s equator but perpendicular to this line. [Hint: Expand the potentials in R/r and ∆/R (in the notation of eq. (31.25)) and discard the term linear in R that is associated with the Moon’s gravitational force acting on the full Earth.]

Use the notation of eq. (31.25). The gravitational potential at a point on Earth’s surface is the sum of contributions from Earth and the moon. Along a line joining the centers, at a height ∆ above Earth’s surface, Vk (∆) = −

where we have expanded to first order in ∆/R and R/r. The gravitational potential at a point on Earth’s surface perpendicular to the line joining the centers is V⊥ = −

or 670 km/h (410 mph). The period is

T = 2π/ω = λ/v 100 km/(670 km/h) 9 minutes.

1 ρga2 v 930 kW/m.

2

This result was found using the density of ocean water, 1 030kg/m3 . If you use the density of fresh water, 1 000kg/m3 , you will instead find P 907 kW/m.

31.11 Derive the statements made about the power density of

1 m and 10 m deep ocean swells in §31.2.4.

31.12 In the JONSWAP model, the energy transferred to the

moving ocean surface waters is proportional to the distance over which wind travels (fetch, F). For a steady wind at 10 m/s compute the rate of energy transfer per unit area to the ocean’s surface. What fraction of the energy of the wind in the 100 meters over the ocean’s surface is transferred to the ocean with a fetch of 200 km?

According to the equation for the energy density E in Box 31.3, dE/dF = 1.67 × 10−7 ρw2 , thus the rate of transfer of energy per unit area is dE dF E = = 1.67 × 10−7 ρw3 . dt dF dt Substituting w = 10 m/s and ρ = 103 kg/m3 , we find dE/dt 0.17 W/m2 .

Over a fetch of 200 km, E (1.67 × 10−7 )(103 kg/m3 )(10 m/s)2 (2 × 105 m) 3.3 kJ/m2 .

The energy per unit area in the wind in the 100 meters above the ocean’s surface is Ewind

1 3 ρw h = 0.5(1.3 kg/m3 )(10 m/s)2 )(100 m) 6.5 kJ/m2 2

So the ratio E/Ewind 3.3/6.5 0.5.

31.13 [H] Imagine that Earth rotated about its axis once

per month and that the Moon moved in a circular orbit above Earth’s equator. Under these tidally locked conditions, the Moon and the resulting tidal bulge in

Gm GM Gm GM − √ ≈− − .

R R r r 2 + R2

Equating these two potentials and solving for ∆,

The power density per unit length along the wave crest is P = Ev =

Gm GM GM Gm Gm GM − ≈− + 2 ∆− − 2 R, R+∆ r−R R r R r

∆=

m R3 .

M r2

Substituting numerical values from eq. (31.25), ∆ 21 m.

31.14 Show that energy loss from Earth’s rotational kinetic

energy at a rate of 3.9 TW (including dissipation and energy transferred to lunar orbit) corresponds to an increase in length of day of 2.3 ms/century. You may take Earth’s moment of inertia about its rotational axis to be I = 8.0 × 1037 kg m2 .

For rotational motion of Earth, Ekin = 12 Iω2 , where ω = 2π/T is Earth’s angular velocity and T = 24 h is Earth’s rotational period.

Solving for T , √ 2π2 I T= , 1/2 Ekin So dT T dEkin T 3 dEkin =− = 2 . dt 2Ekin dt 4π I dt Substituting numerical data, 1 6.45 × 1014 s3 dT =− 2 (3.9 × 1012 W) dt 4π 8.0 × 1037 kg m2

= (8.0 × 10−13 s/s)(3.16 × 109 s/century) 2.5 ms/century

31.15 Is tidal power a renewable resource? Assuming the

rate of energy loss due to the tides remains constant at 3.9 TW, estimate the time it will take for Earth to lose 10% of its rotational energy to tidal losses. If humans were to extract additional tidal energy at the rate of 100 MW in such a way that friction increases in the system and this additional energy was also lost by Earth, how much sooner would Earth lose 10% of its rotational energy? (See Problem 31.14 for Earth’s moment of inertia.)

687

Problems

While technically not a renewable resource, the rate of energy loss due to tidal dissipation is so slow that it will take hundreds of millions of years for Earth to lose 10% of its rotational energy.

Using the moment of inertia from the previous problem and Ekin = (1/2)Iω2 , Ekin = 2π2 I/T 2 2.1 × 1029 J and Ėkin = 3.9 × 1012 J/s, so

The total energy contained in water in the lagoon when the lagoon is held at the high tide level and the outside tide is low is Z h 1 E= dxρgAx = ρgAh2 400GJ .

2 0

Extracting this entire energy at each low and high tide gives

∆t = (0.1Ekin )/Ėkin (2.1×1028 )/(3.9×1012 ) = 5.4×1015 s = 171 My. Increasing the rate of energy loss by (∆Ėkin )/Ėkin = 100 MW/3.9 TW = 2.6 × 10−5 will decrease the time for loss of 10% of rotational energy by (2.6 × 10−5 )(171 × 106 y) 4 500 y.

P = 4E/(24.8h) 18MW .

Pumping in an additional 1m of water at high tide costs an additional energy

31.16 Consider building a tidal barrage along a straight sec-

tion of beach in California. Assume that the average tidal range is 3 m, with two high tides a day. Assume that the tidal basin is 20 m wide and extends for 1 km along the coast. Assume that all available energy from both incoming and outgoing tides is extracted. Compute the average power output of this installation. Do you think this installation would be worth building?

The amount of energy that can be extracted from either filling or emptying the basin is 1 E = ρgAh2 .

2 With two high tides per day, the basin can be filled and emptied twice a day, so the total power is P=

2ρgAh2 = 42.1 kW .

1 day

This is not very much power for an area taking up 1 km of beach, so probably this installation would not be worth building.

31.17 A basin under consideration for a tidal barrage power

installation has an area that grows quadratically with height z, A(z) = A0 z2 /h2 , where h is the difference between high and low tide. Calculate the maximum power that could be extracted (a) if the full basin is emptied at low tide, and (b) if the empty basin is filled at high tide. a) dE = −dz(h − z)A(h − z)ρg, where z is the distance below the top of the filled basin. So Z h dz(h − z)3 /h2 E = ρgA0 0

=

1 A0 gρh2 4

b) dE = dz(h − z)A(z)ρg, so E = ρgA0 =

E1 =

1 ρgA(1 m)2 25GJ .

2

The total energy available for h = 5 m is then E5 =

1 ρgA(5 m)2 625GJ .

2

The same is true when water is pumped out at low tide, so net energy which can be extracted at each change of tide is E5 − E1 600 GJ. This gives a power of P 27 MW .

31.19 When a tidal barrage follows the water release profile

shown in Figure 31.27, it only captures a fraction of the energy (31.35) available in the water stored at high tide. Note that for simplicity the discharge is indicated continuing until the outside and inside water levels are equal. Show that this fraction is given by f (t1 , t2 ) =

! (sin ωt2 − sin ωt1 ) 1 sin2 ωt2 − 2(1 − cos ωt2 ) , 4 ω(t2 − t1 )

where ω = 2π/T and T 12 h. For the parameters shown in the figure, where ωt1 = π/4 and ωt2 = 5π/4, show that f 0.5.

At time t, a mass of water dm = gρAdz is released with a hydraulic head of z(t) − h(t) where z(t) is the height of the water within the tidal basin and h(t) is the height of the (tidal) water outside the basin, so Z t2 dt gρA(dz/dt)dt.

E= t1

From Figure 31.27, z(t) − h(t) = (h/2) − α(t − t1 ) − (h/2) cos ωt , where α is the slope of the release line determined by z(t2 )−h(t2 ) = 0, so h(1 − cos ωt2 ) . α= 2(t2 − t1 ) Substituting the above equations into the integral,

Z h 0

dz(h − z)z2 /h2

1 A0 gρh2 12

The barrage can extract only 1/3 as much power when filling as when emptying.

31.18 Consider a tidal barrage constructed to enclose a lagoon

of surface area 5 km2 , with typical tidal range of 4 m.

What is the maximum possible average power output of such a barrage? How much would the net output be increased if an additional 1 m of water were pumped in at every high tide and pumped out at every low tide?

(Remember to subtract energy used for pumping.)

E = gρAα

! h α h (t2 − t1 ) − (t2 − t1 )2 − (sin ωt2 − sin ωt1 ) 2 2 2ω

(sin ωt2 − sin ωt1 ) 1 1 gρAh2 (1 − cos ωt2 ) (1 + cos ωt2 ) − 4 2 ω(t2 − t1 ) ! !

1 (sin ωt2 − sin ωt1 ) 2 1 2 = gρAh sin ωt2 − 2(1 − cos ωt2 ) 2 4 ω(t2 − t1 ) =

We evaluate this when ωt1 = π/4 and ωt2 = 5π/4, √  !  √  1  1 2 1 )(− 2/π) E = gρAh2  − 2(1 + 2 4 2 2 √  !  !

1  1 2(1 + 2)  1 1 2 = gρAh2  +  0.51 gρAh .

2 4 2 π 2

!

688

31.20 Consider an idealized tidal reservoir of surface area

A = 30 km2 that fills and drains through a narrow channel of width w = 60 m and depth d = 20 m. Assume that the tidal range is h = 2.5 m. For simplicity, assume that the water flows uniformly throughout the channel and varies sinusoidally in time, with a 12 hour period, at a velocity v(t) = v0 sin ωt, where ω = 4π/24 hours.

Compute the maximum speed v0 needed for the reservoir to drain and fill completely on each tidal cycle and the power density of the water passing through the channel. Compute the total tidal stream energy passing through the channel daily. Consider installing four tidal stream turbines in the channel. Assume that they operate at 20% of the Betz limit, and that each has two rotors of radius 8 m. Estimate the average power output of the turbines. Compare this result with the energy extracted by a tidal barrage that utilizes both the ebb and flood tide with 50% efficiency.

The reservoir fills over the time 0 ≤ t ≤ π/ω = T = 6 h, during which a volume V = hA must pass through the channel of area hw, Z π/ω V = hA = dw dt v0 sin ωt = 2dwv0 /ω , 0

which we solve for v0 =

hAω πhA π(2.5 m)(3 × 107 m2 ) 4.5 m/s . = = 2dw 2dwT 2(1 200 m2 )(6 × 3 600 s)

The power density is P=

1 3 ρv = 0.5(1 000 kg/m3 )(4.5 m/s)3 sin3 ωt = 47 sin3 ωt kW/m3 .

2

The power averaged over a single tide is hPi = hwhPi where !

Z 1 π hPi hsin3 ωti(47 kW/m2 ) = dθ sin3 θ (47 kW/m2 ) π 0 = (4/3π)(47 kW/m2 ) 20 kW/m2 .

So hPi = (1 200 m2 )(20 kW/m2 ) 24 MW is the average power potential of the tidal stream energy. The total energy flowing through the channel per day is E = (24 MW)(8.64×104 s) = 2.1 TJ per day.

The turbines cover a total area of 4(πr2 ) 800 m2 . At 20% of the Betz limit, they extract average power 0.20 ×

16 × 800 hPi 1.9 MW .

27

Using E = 12 ρgAh2 , the energy available per tide is 1 × (1000 kg/m3 ) × (9.8 m/s2 ) × (2.5 m)2 × (30 km2 ) 920 GJ 2 With 4 tides/day the total energy available is 3.7 TJ, or an average power of 43 MW, at 50% we would get 21 MW.

31.21 Estimate the total power of the Gulf Stream in the

Florida Straits, given that the mass flux is roughly 30 Sv. Take the speed of the flow to be 4 m/s.

The mass flux is given by the volume flux times the water density Fm (1025 kg/m3 ) × 30 Sv 3.1 × 1010 kg/s .

The rate of kinetic energy flow is then P=

1 Fm v2 250 GW .

2

713

Problems

For 232 Th, from the data in Table 20.6, E = 40.6 MeV, t1/2 = 14.0 × 109 y, m = 0.232 kg, so

Problems 32.1

32.2

Use Newton’s law to compute the gravitational acceleration g(r) as a function of radius r within a sphere of total mass M that is uniformly distributed through the volume. Compute g(r) in a simple model of Earth with a core of radius 3500 km with density 1.2 × 104 kg/m3 surrounded by a mantle of depth 3000 km with density 5 × 103 kg/m3 . Graph this function and compare with Figure 32.3(b).

Make a very rough estimate of the thermal energy content of Earth, assuming that the core has radius 3480 km, temperature 4000 K, density 11 000 kg/m3 and heat capacity 800 J/K kg, and that the rest of the planet, dominated by the mantle, has radius 6370 km, temperature 2000 K, density 4500 kg/m3 and heat capacity 1200 J/K kg. Numbers for a more detailed model of the planetary interior can be found in the appendices to [212].

Q̇Th = 0.693

(6.02 × 1023 )(40.6 MeV)(1.60 × 10−19 J/eV) (0.232 kg)(14 × 109 y)

= 26.5 µW/kg .

The contributions of uranium and 40 K are computed similarly.

32.4

Assume that a region of continental crust has a typical surface heat flux of 65 mW/m2 , crustal density of 2750 kg/m3 , and typical crustal abundances of radioactive nuclides (as given in Table 20.6). If 60% of the surface heat flux comes from radioactivity in the upper crust, to what depth does this typical crustal composition extend?

The rate of heat production (per kilogram) of crust, given by eq. (32.4) is Q̇ = 98(2.7 × 10−6 + 26.5(9.6 × 10−6 ) + (3.5 × 10−3 )(2 × 10−2 ) 5.9 × 10−10 W/kg .

The thermal energy content of a sphere of density ρ = 11 000 kg/m3 , specific heat capacity c = 800 J/K kg, radius R = 3.48 × 106 m, and temperature T = 4000 K is 4π 3 ρR cT 3 4π (11 000 kg/m3 )(3.48 × 106 m)3 (800 J/K kg)(4000 K) 6.2 × 1030 J , = 3 32.5

U=

or (5.9 × 10−10 W/kg)(2.75 kg/m3 ) = 1.62 × 10−6 W/m3 throughout the crust of thickness h, yielding a heat flux at the surface of (1.62 × 10−6 )h[m] W/m2 . If this accounts for 60% of the heat flow of 65 mW/m2 , then h = (0.6)(65 × 10−3 W/m2 )/(1.62 × 10−6 W/m3 ) 24 × 103 m .

Consider a house in a northern location that needs a total of 100 GJ of thermal energy over a winter season from November through March. Assume that the house has a ground source heat pump that operates with a 4π ′ ′3 CoP of 3. The ground around the house has reasonable ′ 3 ′ ′ U = ρ (R − R )c T 3 moisture content, a heat capacity of 3 MJ/m3 K, and a 4π 30 conductivity of 1.5 W/m K. Ground tempera= (4500 kg/m3 )((6.37 × 106 m)3 − (3.48 × 106 m)3 )(1200 J/K kg)(2000 K)thermal 9.8 × 10 J. 3 tures average 10 ◦ C at 6 m. Compute (a) the volume of The total thermal energy is the sum, Utot = 1.6 × 1031 J. For ground whose temperature must be reduced by 8 K in comparison, annual primary human energy consumption is approxorder to provide thermal energy to supply the season’s imately 6 × 1020 J, so Earth’s stored thermal energy corresponds to heating energy needs for the house, and (b) the area 26 billion years of human primary energy use. of a horizontal piping field needed for the rate of therCheck eq. (32.4) by estimating the radioactive heat promal energy replacement through ground conduction to duction rate using data from Table 20.6. Take the fracmatch the average rate of power needed over the winter, tional abundances of 235 U, 40 K to be 0.72%, 0.012%.

40 assuming a gradient near the pipes of 8 K/m.

Take the average energy release in K decay to be and the thermal energy content of a shell of inner radius R and outer radius R′ = 6.37 × 106 m, density ρ′ = 4500 kg/m3 , specific heat capacity c′ = 1200 J/K kg, and temperature T ′ = 2000 K is

32.3

0.71 MeV.

a) The heat removed from the environment, Q− is given by Q− = Q+ − W and CoP = Q+ /W = 3, so Q− = (2/3)Q+ 67 GJ. Then

The power emitted by N j atoms of a radioactive isotope j with half-life t1/2 j , which emits energy E j in each decay, is

Q− = cV ∆T V = (3 MJ/m3 K)(8 K)V = (24 MJ/m3 )V = 67 GJ ,

Ej dE = 0.693 Nj . dt t1/2 j Here we take E j to be the total energy emitted in the decay chain leading to a stable isotope and we assume that the half-life of the original nuclide dominates the half-life for the entire chain. There are N j = NA /m j atoms per kilogram, where m j is the atom’s molecular mass (in kilograms) and NA is Avogadro’s number. Thus the power emitted per kilogram is NA E j .

Q̇ j = 0.693 t1/2 j m j The total power emitted in a kilogram of rock is Q̇ = where the sum is over all radioactive isotopes.

P

j Q̇ j C j ,

where cV is the heat capacity per unit volume. Solving for V yields V 2 800 m3 . b) The power needed is P = 67 GJ/(150 d) 5 170 W. The heat flux in the vicinity of the piping is q = κ(dT/dz) where κ = 1.5 W/m K and dT/dz = 8 K/m. Assuming that the fluxes from above and below the piping are equal, P = 2qA = 2(12 W/m2 )A = 5 170 W, which yields A 215 m2 .

32.6

Compute the drawdown coefficient (32.11) in a simple model where the geothermal reservoir is a large cylindrical volume with height h much smaller than the radius r (see Figure 32.17). Assume that the pressure is constant at pw in a small cylinder of the radius of

714

the wellbore rw , and that flow is radial and vertically uniform outside that volume. Show that η CD = ln (r/rw ) .

2πKhρ

32.8

[Hint: Use conservation of mass to show that |u| ∝ 1/r, and note that ∇ f (r) = r̂ f ′ (r).]

The mass flow rate through a cylinder of radius r, ṁ(r), is given

The density and vapor pressure of water at 200 ◦ C are ρ = 864 kg/m3 and pvap = 1.55 MPa respectively. The difference in pressure between the top and bottom of the wellbore is

by ṁ(r) = 2πrhρu(r) ,

∆p = ρgh = (864 kg/m3 )(9.8 m/s2 )(1 500 m) 12.70 MPa.

and conservation of mass in steady state requires ṁ(r) to be independent of r. Thus

When the wellhead pressure ptop = pw − ∆p = pvap , the water will flash at the wellhead. Solving for pw ,

ṁ r̂ . u(r) = − 2πρh r According to Darcy’s law (32.9), u = −(K/η)∇p = −(K/η)r̂p′ (r).

Combining these two equations, dp ηṁ 1 = , dr 2πρhK r and integrating both sides from rw to r, pr − pw =

η ln(r/rw )ṁ , 2πKhρ

which is the required result.

32.7

Consider a geothermal reservoir (see Figure 32.17) at a depth of 1.5 km, modeled as a cylindrical volume of radius r = 5 km and height h = 500 m, with pressure at the periphery of pr 15 MPa, pressure at the bottom of the well of pw 12 MPa, and permeability of K 50 mD. Use the answer to Problem 32.6 to estimate the mass flow rate through a wellbore of radius rw 0.125 m. Assume that the fluid is water at roughly 180 ◦ C, where the viscosity is η 0.15 cP.

pw = 1.55 + 12.70 = 14.25 MPa .

32.9

For a geothermal well that has been running for some time, pressure at the bottom of the well is pw 15 MPa at a depth of 2 km. Assume the fluid is water at 280 ◦ C. Estimate the location of the flash horizon. (You may neglect the contribution to the pressure from the weight of the vapor above the flash horizon.)

The density and vapor pressure of water at 280 ◦ C are ρ = 750 kg/m3 and pvap = 6.42 MPa. Assuming that the pressure in the wellbore drops linearly with increasing height, then the pressure at a height h above the bottom of the well is p(h) = pw − ρgh .

Setting ph = pvap and inserting numerical values, pw − pvap ρg (15 − 6.42) MPa 1.17 km . = (750 kg/m3 )(9.8 m/s2 )

h=

So the flash horizon occurs at a depth of 0.83 km.

32.10 [T] Assume that enthalpy is a linear function h(s, T ) =

The mass flow rate ṁ is given by ṁ = (pr − pw )/CD = 3 MPa/CD Using the expression for CD from Problem 32.6, CD =

Given a geothermal reservoir containing water at temperature T = 200 ◦ C at a depth of 1.5 km, use steam tables to determine the density and vapor pressure of the water. Assuming that pressure in the wellbore decreases according to the overlying weight of water, at what well pressure Pw will the flash occur in the wellbore?

(1.5 × 10−4 kg/m s) ln(5000/0.15) 2π((50 × 10−3 )(9.87 × 10−13 m2 ))(500 m)(1 000 kg/m3 )

10 000 m−1 s−1

as + bT . Consider a single-flash cycle, assuming that the saturated liquid and vapor curves between T − and T + form an isosceles trapezoid, i.e. sliquid (T + ) − sliquid (T − ) = svapor (T − ) − svapor (T + ) .

Prove that the optimal separator temperature is T s = (T + + T − )/2 .

So 6

2

4

−1 −1

ṁ = (3 × 10 kg/m s )/(10 m s ) 300 kg/s .

32.11 Consider the efficiency computed in the example

single-flash cycle in Example 32.5. Use thermodynamic data on water to show that running the separator at T = 140 ◦ C or T = 160 ◦ C will decrease the efficiency of the ideal cycle.

The necessary thermodynamic data for water (in addition to the data given in the Example) are:

Figure 32.17 Thermal reservoir for Problems 32.6 and 32.7.

T

hliquid

hvapor

sliquid

svapor

160

675

2 757

1.942

6.747

140

589

2 733

1.739

6.928

715

Problems

The units are the same as in the Example. For both cases h6 − h1 = 1 086 − 209 = 877 kJ/kg. For T s = 160 ◦ C, the quality is given by h1 = 1 086 = 2 757χ + 675(1 − χ) which gives χ(160 ◦ C) = 0.197. Assuming the turbine is isentropic (s5′ = s4 ), we find h5′ = 2 162 kJ/kg and χ5′ = 0.820. Combining these in eq. (32.14), we find η(160 ◦ C) = 0.197

(2 757 − 2 162) 0.134 = 13.4% , 877

which is less than the efficiency computed at 150 ◦ C in the Example. The Baumann factor further reduces the efficiency by (1 + χ5′ )/2 = 0.91, to (0.91)(13.4) = 12.2%.

The analogous calculation for T s = 140 ◦ C gives η(140 ◦ C) = 0.232

(2 733 − 2 221) 0.135 = 13.5% , 877

also less than η(150 ◦ C). In this case χ5′ = 0.84 so the Baumann

factor is 0.92, reducing the efficiency to (0.92)(13.5) = 12.4%.

32.12 Compute the (ideal) efficiency of a double-flash plant

with the same conditions as the example single-flash plant described in Example 32.5. The cycle is labeled in Figure 32.15. You will need to choose temperatures for the two separators. Try using T s1 = 183.3 ◦ C, T s2 = 116.7 ◦ C. Why do you think this would be better than, for example, T s1 = 150 ◦ C, T s2 = 100 ◦ C?

The required thermodynamic data on water (specific enthalpy in kJ/kg and entropy in kJ/kg K) is

T

hliquid (T )

hvapor (T )

sliquid (T )

svapor (T )

250

1 085.8

2 800.9

2.7935

6.0721

183.13

776.90

2 779.9

2.1695

6.5593

116.7

489.8

2 701.1

1.4922

7.1643

50

209.34

2 591.3

0.70381

8.0748

χ2 (h4 − h5 ) + χtotal (h8 − h9 ) ≈ 0.17 (h1 − h10 )

The choice of the temperatures should be such that the separator temperatures are not too close to either T + or T − or to one another.

This is similar to the single flash plant where the optimal T s is approximately the mean of T + and T − . If the four temperatures, T + , T − , T s1 and T s2 are roughly equally spaced the efficiency will be maximized.

32.13 Estimate the energy extracted from 1 km3 of granite if

cooled from 300 ◦ C to 200 ◦ C. Assume that a power plant runs between the (time-varying) temperature T g of the granite and an ambient temperature of 20 ◦ C and reaches 1/2 of the Carnot efficiency at each T g . For how long could a 100 MW power plant run off this energy?

For granite, c p = 790 J/kg K, and ρ = 2.7 × 103 kg/m3 . The total thermal energy that can be extracted from 1 km3 of granite if cooled from 300◦ C to 200◦ C is Q = c p ρ ∆T = 213.3 PJ/km3 .

The granite has temperatures from T 1 = 300◦ C= 573.15 K to T 0 = 200◦ C= 473.15 K. The Carnot efficiency of running a power plant between these temperatures and ambient temperature T − = 293.15, can be estimated as an average of the Carnot efficiencies at T 0 and T 1 !

1 T0 − T− T1 − T− η= + 0.435.

2 T0 T1 Running a power plant at half this efficiency, a P = 100 MW power plant could run for T = (η/2)(Q/P) 14.7 years.

For a more precise answer, we compute the electrical energy (work) produced by the power plant at half the Carnot efficiency as a function of T as the granite cools, Z T1 cp ρ T1 T − T− 1 T 1 − T 0 − T − log dT = W = cp ρ 46.7 PJ 2 T 2 T0 T0 and the power plant could run for T =

At points 1, 3, 4, 6, 8, 10, and 11 we have h1 = hliquid (T + )

h3 = hliquid (T s1 )

h4 = hvapor (T s1 )

h6 = hliquid (T s2 )

h8 = hvapor (T s2 )

h10 = hliquid (T − )

h11 = hvapor (T − )

η=

Following the steps for single-flash hydrothermal plant ➀→➁ is isenthalpic h1 = h2 We can use this to find the quality at ➁: χ(T s1 )h4 + (1 − χ(T s1 ))h3 = h2 → χ(T s1 ) = 0.154 Similarly h3 = h7 and χ(T s2 )h8 + (1 − χ(T s2 ))h6 = h7 → χ(T s2 ) = 0.13 We can estimate the turbine steps as isentropic steps so s4 = s5 and s8 = s9 . To estimate χ5 we use χ5 s8 + (1 − χ5 )s6 = s5 = s4 → χ5 = 0.893 Since s8 = s9 we find χ(T − ) = 0.877. We can now find the enthalpy at points ➄ and ➈ : h5 = 2464.49 KJ/Kg h9 = 2298.32 KJ/Kg The second flash means that the quality before the second turbine is χtotal = 0.893 × 0.154 + 0.13 × (1 − 0.154) = 0.248. We can now calculate the efficiency of the full process:

W 14.8 years, P

in very good agreement with our previous estimation.

749

Problems

mining operation in one year by one person. (Make reasonable assumptions and approximations.)

Problems 33.1

Estimate the energy content of a coal seam measuring 1 km2 by 1 m deep. Estimate the electrical power output of a coal plant burning this fuel over one year, and the area required for a solar thermal plant in a high insolation location to produce a comparable power output.

Take the density of coal to be ρ = 1300 kg/m3 .

Using 20 GJ/t, 1 y = 3.16 × 107 s, and assuming 33% efficiency, the power averaged over a year from one person’s output is P ≈ (104 t)(20 × 109 GJ/t)(1/3)/(3.16 × 107 s) ≈ 2.1 MWe. A one meter-squared solar panel operating at 15% efficiency with daily average insolation 300 W/m3 produces power at an average rate of 45 We. It would require N ≈ 2.1 × 106 /45 ≈ 47 000 such panels to produce the same (average) power.

An offshore wind turbine operating in an (average) wind class 6 location with a hub height above 50 m sees, on average ∼ 700 W/m2 wind power. Assuming a power coefficient of 0.4, a wind turbine of blade radius R would produce electric power at

Assume that the energy content of the coal is 20 GJ/t. The energy content of the coal is E ≈ (106 m3 )(1.3 t/m3 )(20 × 109 GJ/t) ≈ 26 PJ .

At, say, 33% efficiency, over a year this amount of coal would produce power at an average rate of 15

P ≈ (26 × 10

33.2

P ≈ (0.4)(700 W/m2 )(πR2 ) .

7

Setting P = 2.1 MW and solving we find R ≈ 50 m. So a single wind turbine of blade length ≈ 50 m would, on average, supply the same power.

J)(1/3)/(3.15 × 10 s) ≈ 275 MWe

Assuming a STE power plant operates at a gross efficiency of 3% and assuming daily average insolation of ≈ 300 W/m2 . The power plant produces P ≈ 9A We, where A is the area in meterssquared. Comparing to the coal power plant, we find A ≈ 30 km2 .

33.5

A rough measure of the enthalpy of combustion of a hydrocarbon can be obtained by assuming that all CC bonds contribute the same energy to the combustion process when broken, and that the same is true for all C-H bonds. This is particularly appropriate for alkanes and cycloalkanes, where all C-C bonds are roughly equivalent single bonds. Using this simple model, write a general formula for the combustion energy of an alkane with n carbon atoms and m hydrogen atoms in terms of two variables characterizing the energies of the two bond types. Fit for the two variables by matching with the energy of methane and dodecane, and compare the intermediate energies to those of Table 33.1.

U=µ

am + b(2n − m/2) a′ m + 2bn ≡ . m + 12n m + 12n

m (125.8 MJ/kg) + n (384.8 MJ/kg) . m + 12n

This gives a reasonable fit for the other linear hydrocarbons.

33.3

33.4

Estimate how many solar panels or wind turbines are needed to produce the same amount of electrical energy as the 10 kt of coal that can be extracted in a longwall

3 3 1 X 1 X 2 δi j ukk )2 + K u , (33.13) 3 k=1 2 i=1 ii

P For case (a), ui j = δi j u and 3k=1 ukk = 3u, so the first term in eq. (33.13) vanishes. This leaves U = (3K/2)u2 , which depends only on K. For case (b),

When computing efficiencies, European power plants sometimes use the LHV of a fuel, while for US power plants the HHV is standard. Consider a power plant burning bituminous coal with HHV/LHV of 31.1/30 MJ/kg. If the power plant’s efficiency is 35% when computed from the HHV, what is its efficiency when computed from the LHV? The same power is produced using “less” fuel energy when calculated in LHV, so the apparent efficiency of the plant will increase by the factor 31.1/30.0 1.037, so the LHV efficiency is (1.037)(35) 36.3%.

(ui j −

where µ is the shear modulus, K is the bulk modulus, and δi j = 1 if i = j and 0 otherwise. (a) Show that for uniform hydrostatic compression, where u11 = u22 = u33 = u, and all other components vanish, the energy of compression depends only on K. (b) Show that for Pwave compression, where u11 = u sin(kx) and all other components vanish, the energy of compression is proportional to the elastic modulus χ = K + 4µ/3. For a further introduction to the theory of elasticity, see [242].

Fitting to methane and dodecane, we have |∆H(Cn Hm )| ≈

3 X

i, j=1

The number of C-H bonds is m and the number of C-H bonds is 2n − m/2. The enthalpy of combustion is then modeled as |∆H(Cn Hm )| ≈

[T] The deformation in a solid can be described by a vector field u(x) = (u1 , u2 , u3 ) describing the displacement of the solid away from the equilibrium point x. This is similar to the displacement vector s used to analyze waves in §31.2. For small deformations, the strain tensor σ is a 3 × 3 matrix with elements σi j = (∂ui /∂x j + ∂u j /∂xi )/2. For example, a rod compressed uniformly to 99% of its original length in the x-direction would have σ11 = −0.01. The energy of compression is given by

1 U = µ (u11 − (1/3)u11 )2 + (−(1/3u11 ))2 + (−(1/3u11 ))2 + Ku211 2 2 1 = µu211 + Ku211 3 2 1 = (K + (4/3)µ)u2 sin2 (kx) .

2

33.6

Compute the fraction of Carnot efficiency realized by a coal plant operating at 600 ◦ C and 45% efficiency (assume ambient temperature of 300 K). If another plant operates at the same fraction of Carnot efficiency at a temperature of 700 ◦ C, what is the thermal efficiency of the second plant?

750

ηCarnot = 873−300 = 0.66, so the plant operates at 0.45/0.66 = 873 68% of Carnot efficiency. For a maximum temperature of 700 ◦ C, ηCarnot = 973−300 = 0.69, so if the plant still reaches 68% of the 973 Carnot limit, its thermal efficiency is (0.68)(0.69)=46.9%.

33.7

An underground mass of salt may form a dome trapping a petroleum reservoir. Compute the gravitational anomaly (deviation of gravitational acceleration) produced by an underground salt mass, where for simplicity we assume that the salt (density 2220 kg/m3 ) forms a sphere of radius 0.5 km centered 0.5 km beneath the surface, and compute the gravitational anomaly by determining the difference from a corresponding mass of granite (density 2640 kg/m3 ). Express your answer in mGal = 10−5 m/s2 . (Gravity data can be measured to a precision of ∼ mGal.) The change in gravitational acceleration is equivalent to the value of G∆M/r2 on the surface of a sphere of radius r = 500 m, with a mass of ∆M = (4π/3)∆ρr3 , so 4π Gr∆ρ 3 4π = (6.67 × 10−11 m3 /kg s2 )(500 m)(2220 − 2640) kg/m3 3 = 5.87 × 10−5 m/s2 = 5.87 mGal

∆g =

33.8

[T] Use the Arrhenius equation in Box 33.5 to show that a reaction rate doubles under a temperature shift of ∆T (kB T 2 ln 2)/E, assuming E ≫ kB T . = e−E/kB T j , j = 1, 2, so the

For two different temperatures r j ratio of rates is r2 = e−E(1/T 2 −1/T 1 )/kB r1 If r2 /r1 = 2, then ln 2 =

∆T E kB T (T + ∆T )

where T 1 = T and T 2 = T + ∆T . Rearranging terms, ! ∆T kB T ∆T = 1+ ln 2 .

T E T Since kB T ≪ E, ∆T/T ≪ 1 and the second term in parentheses can be ignored, leaving ∆T = kB T 2 ln 2/E.

33.9

Using an activation energy E = 250 kJ/mol, compute the change in rate of petroleum generation when temperature increases from 130 ◦ C to 430 ◦ C. If petroleum generation through such a process takes 10 million years at 130 ◦ C, estimate the time it would take in the laboratory at 430 ◦ C. For two different temperatures r j = e−E/kB T j , j = 1, 2, so the ratio of rates is r2 = e−E(1/T 2 −1/T 1 )/kB . r1 First, convert E/kB to temperature: E = 250 kJ/mol = (250 kJ/mol)(96.45 eV mol/kJ molecule) = (250)(96.45 eV)(1.602 × 10−19 J/eV) 4.15 × 10−19 J .

So E/kB = (4.15 × 10−19 J)/(1.28 × 10−23 J/K) 3.00 × 104 K. Next compute (1/T 2 − 1/T 1 ) = (1/703 − 1/403) = −0.00106

So 4 r2 e(0.00106)(3×10 ) 1013.8 r1

The reaction that takes 107 y in nature at 130 ◦ C, would take 107 y/1013.8 10−6.8 y 5 s in the lab at 430 ◦ C.

33.10 Bitumen extracted from tar sands in Athabasca,

Canada, has an API gravity of ◦ API 8. Compute the mass of one liter of such bitumen and compare to that of a liter of light crude at ◦ API 35.

From eq. (33.6), ρ = (141.5 kg/L)/(◦ API + 131.5) Substituting ◦

API = 8 for Athabasca bitumen, ρbitumen 1.01 kg/L, and ◦ API =

35 for light crude, ρlight crude 0.85kg/L.

33.11 Compute and compare the amount of carbon released

per GJ of delivered energy by (a) a coal plant operating at 35% efficiency, (b) a natural gas combined cycle plant operating at 60% efficiency, and (c) an automobile engine running on gasoline at 25% efficiency.

From Table 33.2, the carbon intensity of coal is ∼25 kg(C)/GJ. At 35% efficiency this gives 25/0.35 ∼ 0.70 kg(C)/GJe. For natural gas in a CCGT plant, the analogous calculation give 15/0.60 ∼ 0.25 kg(C)/GJe. Finally, the energy density of gasoline is ∼47 MJ/kg and gasoline (assumed to be octane) is 96/114 = 84% carbon, so the carbon intensity of burning gasoline at 25% efficiency is (1000/47 kg(gasoline)/GJ)(0.84 kg(carbon)/kg(gasoline))/0.25 ∼ 0.70 kg(C)/GJ .

So the carbon footprint of a gasoline automobile engine at 25% efficiency is roughly the same as that of a coal power plant. A CCGT power plant has roughly 1/3 the carbon footprint of either.

33.12 Compare the energy density of oil and geothermal fluid

at 250 ◦ C pumped from oil/geothermal wells at a rate of 30 L/s. To make a fair comparison, assume that the oil is transformed into useful mechanical energy at 20% net efficiency, and that the geothermal fluid emerges from the well as saturated liquid and is converted into electricity in a single-flash geothermal power plant as described in Example 32.5.

Estimate the energy density in oil (density ≈ 0.85 kg/L, energy density ≈ 43 MJ/kg) Poil = (30 L/s)(.85 kg/L)(43 MJ/kg)(0.2) 220 MW For the geothermal fluid in the notation of Example 32.5, the available energy density is h1 − h6 = 1086 kJ/kg − 209 kJ/kg = 877 KJ/kg. At a pumping rate of 30 L/s, density 0.95 kg/L, and used with efficiency 13.6%, the electric power output is Pgeothermal = (30 L/s)(0.95 kg/L)(877 kJ/kg)(0.136) 3.4 MWe So the useful energy content of the oil is ∼65 times greater than the geothermal fluid.

Part III Energy System Issues and Externalities

781

Problems

(layer 2) and the other 1/2 passes through both atmospheric layers and escapes to space. Set up the three linear equations for the temperatures T s , T 1 , T 2 of the surface and the two atmospheric layers. What is the temperature at the surface? [Hint: to simplify the algebra you may want to first solve for the values τi = σT i 4 , for i = s, 1, 2.]

Problems 34.1

Given the following approximations for planetary albedos and given approximate radii of their orbits, estimate the average effective temperature of the planets Mars and Neptune. (The effective temperature is the temperature at which the planet radiates thermal radiation into space, which is not necessarily equal to the surface temperature). You may assume that the planets have circular orbits. Mars: albedo 0.16, rorbit 2.28 × 108 km; Neptune: albedo 0.29, rorbit 4.5 × 109 km.

We can treat the problem following the method in Example 34.2. Looking at the radiation balance for the whole system we find 1 1 hIin i = σT 24 + σT s4 + σT 14 2 2

Insolation falls proportional to the distance from the sun squared, so for a planet with an orbital distance r, the insolation is I = (r⊕ /r)2 I⊕ , where I⊕ = 1366 W/m2 . So the insolation at Mars and Neptune is

Now we consider only the surface and the layer of the atmosphere above it at T 2 . For this subsystem the radiative balance gives hIin i + σT 14 = σT 24 +

IMars (1.50 × 108 /2.28 × 108 )2 (1366) 590 W/m2 ,

If we now consider the subsystem that has only the surface we find

INeptune (1.50 × 108 /4.5 × 109 )2 (1366) 1.5 W/m2 ,

hIin i +

8 where we have used the radius of Earth’s orbit, √ r⊕ 1.50 × 10 km.

Including the planets’ albedos, we find T = 4 a(I/4)/σ, so

T Mars T Neptune

34.2

p 4

p 4

1 σT 4 + σT 24 = σT s4 2 1

Solving the linear system for τi = σT i4 gives

(0.84)(590/4)/(5.67 × 10−8 ) 216 K ,

τ s = 15τ1 ,

(0.71)(1.5/4)/(5.67 × 10−8 ) 46 K .

τ2 = 4τ1 ,

10.5τ1 = hIin i

so T 1 = 148.3, T 2 = 209.8, T s = 291.9 K

Show that an n-layer atmosphere of the type described Example 34.2 gives a surface temperature T s =

in√ 4 n + 1 × 255 K.

34.4

Generalizing the analysis of Example 34.2 we impose radiative balance first at the outermost layer, σT 14 = hIin i ,

then at the nth layer,

We have described hydrostatic equilibrium in §34.2.1 under the approximation that the atmosphere lies above a planar surface. Compute the pressure in hydrostatic equilibrium as a function of height in an atmosphere around a spherical planet, and compare to the Boltzmann distribution.

We follow the logic in parallel with the planar analysis in the text, assuming again that the atmosphere is an ideal gas composed of identical molecules of mass m. The vertical change in pressure at radius r is GMρ d p = − 2 dr , r

4 σT n4 = hIin i + σT n−1 ,

and finally at the surface, σT s4 = hIin i + σT n4 .

where M is the Earth mass. We then have

The first two equations are solved by σT n4 = nhIin i, which substituted into the last equation gives

dp = −

√

34.3

1 σT 4 2 s

p 4 4 σT s4 = (n + 1)hIin i or T s = (n + 1)hIin i/σ = n + 1 255 K .

so

Consider a slight improvement on the two-layer toy model of Earth’s atmosphere shown in the figure in Example 34.2. Assume that there are two atmospheric layers, each transparent to incoming solar radiation with incoming solar radiation intensity I = hIin i 0.84I⊙ /4. Assume that all outgoing infrared radiation (from the surface and from each atmospheric layer) consists of two equal parts. One part is in a frequency range that is completely absorbed by each atmospheric layer, and the second part is in a frequency range that passes through all atmospheric layers. Thus, for example, 1/2 of the IR radiation emitted by the surface is absorbed completely by the lower atmosphere layer

giving

GMm pdr , kB T r 2

dp dr GMm =− 2 p r kB T

p(r) = p0 e−V(r)/kB T ,

ρ(r) = ρ0 e−V(r)/kB T ,

where V(r) = −GMm/r is the potential energy of a molecule

of mass m at radius r, matching with the expectation from the Boltzmann distribution.

34.5

The bond energy of the double bond in the O2 molecule is 5.06 eV. Verify that light must have wavelength less than 246 nm in order to break this bond. Assuming sunlight has a blackbody spectrum at T = 6000 K, what fraction of the photons in the sunlight incident on the upper atmosphere can dissociate O2 ? [Hint: see eqs. (22.23) and (22.24).] Repeat this calculation for

782

falls exponentially κ(λ) = κ0C CO2 e−c(λ−λ0 ) , beyond some λ0 in the right wing of the 15 µm absorption peak.

Here κ0 is a constant and CCO2 is the concentration of atmospheric CO2 . Show that the range of wavelengths over which absorption is saturated grows proportional to the logarithm of CCO2 .

the triple bond in N2 , which has a bond energy of 9.61 eV. Compare the relative importance of O2 and N2 as absorbers of ultraviolet light in Earth’s upper atmosphere.

From the Planck relation hν = E and νλ = c we find λ = (2π~c)/E = (1.24 × 10−6 eV m)/E. For O2 , λO2 1.24 × −6 10 /5.06 245 nm, and for N2 , λN2 1.24 × 10−6 /9.61 129 nm. Both are well into the ultraviolet part of the electromagnetic spectrum.

The effectiveness of either gas as an absorber of sunlight depends on the number of photons in the incident light with energy high enough to break the bond. The number of photons per unit frequency per unit area is given by !

1 1 dP ω2 , = 2 2 ~ω/k T ~ω dωdA 4π c e B − 1 where we have used eq. (22.23). The fraction of the photons in sunlight with energy greater than E0 = ~ω0 is given by the integral of this equation from ω0 to infinity divided by the integral from ω = 0 to infinity, Z ∞ y2 f (E0 ) = 0.416 dy y , e −1 y0 R∞ where y0 = E0 /kB T and 0 dy y2 /(ey − 1) 1/0.416. For T = 6 000 K, kB T 0.5 eV, so y0 5.06/0.5 10.1 for O2 and y0 9.61/0.5 19.2 for N2 . Then performing the integrals numerically,

The region of saturated absorption extends until λ = λ∗ where κ(λ∗ ) = κ∗ . Substituting for κ(λ∗ ) and solving for λ∗ , ∗

κ0 CCO2 e−c(λ −λ0 ) = κ∗ ∗ κ0 ec(λ −λ0 ) = ∗ CCO2 κ

1 κ0 ln ∗ + ln CCO2 , λ∗ = λ0 + c κ

which grows logarithmically with the concentration of CO2 .

34.8

f (O2 ) 2.74 × 10−3

f (N2 ) 1.35 × 10−6 .

So sunlight contains 2.74×10−3 /1.35×10−6 2025 times as many photons capable of breaking the O2 bond as are capable of breaking the N2 bond.

34.6

Use the value of atmospheric pressure at sea level to estimate the total mass of the atmosphere. Estimate the number of molecules in the atmosphere and check for agreement with recent numbers for the total mass of CO2 in the atmosphere and parts per million by volume.

For (a), we compute with some precision that if T 4 = 240/σ then T = 255.069 K, λ0 3.7637. With a = b = 1/2, y = 1K, we have T 4 + 6T 2 y2 + y4 = 240/σ gives T = 255.0.63, and −F0 = 4σ(T 3 + 3T y2 )∆T 3.7636, giving a slightly smaller λ0 .

The equation of (hydrostatic) equilibrium (34.5), d p = −gρdz, can be integrated from the bottom of the atmosphere, where p = p0 = 1 atm = 1.013 × 105 Pa. to the top, where p = 0.

Z ∞ p0 = g ρdz = gσ

and we wish to show g(T, y) = T 3 + 3T y2 < T 03 for positive T, y.

To show this we compute

where σ is the mass per unit area of atmosphere above Earth’s surface. The total mass of the atmosphere is then

f 3 −g4 = 6T 10 y2 +57T 8 y4 +144T 6 y6 +30T 4 y8 +18T 2 y10 +y12 > 0 .

For (b), we have in the case a = b = 1/2, f (T, y) = T 4 + 6T 2 y2 + y4 = T 04 ,

0

M(atm) = A⊕ σ = 4πR2⊕ p0 /g

4π(6.38 × 106 )2 (1.013 × 105 )/9.81 5.29 × 1018 kg .

At 608 ppmm (see Box 34.3), we find M(CO2 ) (6.08 × 10−4 )(5.29 × 1018 ) 3.22 × 1015 kg and the mass of carbon is obtained by multiplying by 12/44, M(C) =

12 (3.22 × 1015 ) = 877 × 1012 kg = 877 Gt , 44

in rough agreement with the number quoted in Box 34.3.

34.7

Show that radiation at a distribution of temperatures decreases the estimate (34.34) of the uniform temperature radiative response λ0 by proving the result in the case where the radiation comes from two different temperatures. Assume I = aσ(T − y)4 + bσ(T + y)4 = 240 W/m2 , for fixed values of a + b = 1 and y, and repeat the computation leading to the estimate (34.34). (a) Show that the distributed estimate decreases for the specific values (a = b = 1/2, y = 1 K); (b) prove analytically that the distributed estimate decreases for any values of a, b, y [Hint: start with a = b = 1/2]; (c) argue that if the result for λ0 decreases in magnitude when the radiation is distributed between two temperatures, the same is true when the radiation is distributed between any number of different temperatures for fixed outgoing total radiation flux.

Prove that radiative forcing grows logarithmically with CO2 concentration under the circumstances illustrated in Figure 34.13. Specifically, assume that for wavelengths in the vicinity of the 15 µm absorption peak, the CO2 absorption coefficient κ(λ) exceeds the value κ∗ at which absorption saturates. Further, assume that κ(λ)

34.9

Earth’s albedo is currently estimated to be a⊕ ≈ 0.30.

By what percentage would a⊕ have to increase to offset the estimated increase in forcing eq. (34.36) due to doubling of CO2 since pre-industrial levels?

A change in Earth’s albedo by y% changes the albedo by ∆a⊕ = 0.3y, causing a decrease in radiative forcing, ∆F = −∆a⊕ (I⊕ /4), (RF) with I⊕ = 1 366 W/m2 . Setting |∆F| equal to F2× = 3.7 W/m2 , we find 0.3y(1 366/4) W/m2 = 3.7 W/m2 , or y 3.6%.

34.10 Assume that ice albedo feedback gives a feedback

parameter λ = 0.5 W/m2 ◦ C. Estimate the corresponding addition to the change in temperature under a doubling of atmospheric CO2 in the absence of other

Problems

feedbacks. Assume that water vapor and the lapse rate feedback together contribute a feedback parameter λ = 1 W/m2 ◦ C. Estimate the temperature change with this feedback alone, and compare to the combined temperature change when both feedbacks are included.

(RF) Taking F2× = 3.7 W/m2 and λ0 = −3.2 W/m2 ◦ C, we find in the absence of feedbacks, ∆T 2× = 3.7/3.2 1.16 ◦ C. With ice albedo feedback −3.2 → −2.7 and ∆T 2× → 3.7/2.7 1.37 ◦ C. So ice albedo feedback raises T 2× by (1.37 − 1.16) 0.21 ◦ C. Water vapor and lapse rate feedback of λwvl ≈ 1.0 W/m2 ◦ C gives T 2× 3.7/(3.2−1.0) 1.68 ◦ C, so water vapor and lapse rate feedbacks raise the temperature by 0.52 ◦ C. Together, the two sources of feedback produce T 2× 3.7/(3.2 − 1.5) 2.18 ◦ C, raising the temperature by 1.02 ◦ C over the no-feedback estimate. Thus, the two feedbacks individually raise the temperature by 0.21 + 0.52 = 0.73 ◦ C, but together they raise the temperature by an additional 1.02 − 0.73 = 0.29 ◦ C.

34.11 Using the simplified model of Earth’s atmosphere as

in Example 34.1, with no absorption, no meridional energy transport, and an albedo of 0.16, use the yearly average integrated insolation computed in §23.3 to compute the temperature in radiative equilibrium on the equator and at the poles.

From Figure 23.6, the hours of perpendicular sunshine average about 7.3 h/d at the equator and about 3.0 h/d at the pole. This corresponds to yearly average insolation of Iequator = 0.84(1 366)(7.3/24) 350 W/m2 and Ipole = 0.84(1 366)(3/24) 140 W/m2 . Following the argument in Example 34.1, we find q p 4 T equator = 4 Iequator /σ = 350/5.67 × 10−8 280 K = 7 ◦ C , q p 4 T pole = 4 Ipole /σ = 143/5.67 × 10−8 223 K = −50 ◦ C .

34.12 Assume that the linearized analysis of climate leading

to (34.46) is correct, and F2× 3.7 W/m2 . Assume also that climate models correctly predict that quadrupling of CO2 levels would give a temperature shift of ∆T 4× 6.4 ◦ C. Use a quadratic function to fit the change in temperature as a function of the logarithm of the multiplicative increase µ in atmospheric CO2 levels to estimate the change in temperature with a doubling of pre-industrial carbon dioxide levels.

We have ∆T (µ) = 2.2 ◦ C log2 µ + β(log2 µ)2 , where the linear term is fixed from (34.47). For µ = 4, we have 4.4 + 4β 6.4, so β = 0.5, and ∆T (2) = 2.7 ◦ C.

783

815

Problems

8.5 W/m2 weaker than present. This corresponds to radiation forcing of

Problems 35.1

Compute the insolation at 65◦ latitude on the summer solstice for tilt of 22.05◦ and compare to the value at 24.5◦ .

∆F = −(1 − a)(∆I⊙ )/4 −0.7(8.5/4) = −1.5 W/m2 . ∆F is negative corresponding to a decrease in net downward radiation at the tropopause. Positive radiative forcing from CO2 is related to CO2 concentration by eq. (34.35),

The task is to integrate I0 cos(β(t)) from sunrise to sunset Z T Z T I = I0 dt cos(β(t)) = 2I0 dt cos(β(t)) ,

∆FCO2 = F2× log2 (MCO2 /M∗ ) ,

0

−T

with F2× 3.7 W/m2 . Setting ∆F = +1.5 W/m2 , and solving for MCO2 ,

where T is the time of sunset with t = 0 at solar noon.

At summer solstice angle of incidence, β(t) is given by eq. (23.9) with δ = ǫ

MCO2 21.5/3.7 M∗ 1.32(310 ppm) 410 ppm ,

cos(β(t)) = sin λ sin ǫ + cos λ cos ǫ cos(ωt) , with ω = 2π/(24 h). So

which happens to be very close to the present value of MCO2 .

35.4

I(ǫ) = 2I0 T sin λ sin ǫ + cos λ cos ǫ

! sin(ωT ) . ω

The time of sunset is determined by cos(β(T )) = 0, or T=

24 h cos−1 (− tan λ tan ǫ) 2π

The radiative forcing resulting from a 10% increase in the solar

constant is F = 0.1(1 − 0.3) 1366 W/m2 /4 ≈ 23.9 W/m2 . Using only the uniform temperature response, this gives a temperature change of

Substituting for λ = 65◦ and ǫ = 22.05◦ , we find T = 10.02 h, and for ǫ = 24.5◦ , T = 11.18 h. Substituting into the equation above for I(ǫ), we find

∆T = −

I(22.05◦ ) = 8.30 kWh/m2 d and I(24.5◦ ) = 9.03 kWh/m2 d ,

Variation in sunspot activity leads to variation in the solar constant by roughly 1–2 W/m2 with a roughly 11 year cycle. Assuming an albedo of a = 0.3, estimate the radiative forcing arising from an increase in the solar constant by 2 W/m2 . Using the IPCC mean result for feedback, what increase in surface temperature would this cause?

∆T = σc F ≈ 0.6 ◦ C/ W/m2 23.9 W/m2 ≈ 14.3 ◦ C ,

and if we use the IPCC5 value σc ≈ 0.9 ◦ C/ W/m2 , we find

35.5

The radiative forcing would be (1 − α) × δI/4 0.35 W/m2 .

Using a feedback of f /λ0 0.83◦ Cm2 /W, would give an increase in surface temperature of about 0.29◦ C. Since this phenomenon oscillates rapidly (11 year cycle) it should not significantly affect

∆T ≈ 0.9 ◦ C/ W/m2 23.9 W/m2 ≈ 21.5 ◦ C .

Compute the time to melt 3000 m thickness of ice if there is an extra 5% of 200 W/m2 average daily insolation for four months out of the year beyond the insolation that would give an ablation rate matching the accumulation rate from fresh snowfall.

The latent heat of melting for water is h = 334 KJ/kg, and the density of ice is ρice = 916.7 kg/m3 , so the heat per square meters needed to melt a depth of z = 3000 m of ice is

long-term climate. Note, however, that a long-term suppression of sunspots, such as the Maunder Minimum from 1645-1715, may

Q = ρice h z ≃ 9.17 × 1011 J/m2 .

lower the average temperature over a period of decades; it is believed that this kind of decreased sunspot activity played a role

Given that the net average insolation is 5% of 200 W/m2 , so P = 10 W/m2 , the time needed to melt the given amount of ice is

in the Little Ice Age in Europe from about 1300-1850.

35.3

Since the formation of the solar system 4.6 billion years ago, the net solar luminosity has increased roughly 40% from its initial value. Assuming that luminosity has increased at a constant rate, estimate the solar constant 100 million years ago, during the Cretaceous period. Assuming current climate conditions and an albedo of a = 0.3, what radiative forcing would be needed to keep Earth in its current state (as of say 1950) if the Sun had the luminosity of the Cretaceous period? What level of atmospheric CO2 would achieve this radiative forcing?

4.6 billion years ago the solar constant was I = 1366/1.4 973 W/m2 , and assuming linear increase over 4.6 billion years, during the Cretaceous the solar constant was (1366 − 973)/46

1 23.9 W/m2 F≈ ≈ 7.5 ◦ C . λ0 3.2 W/m2 ◦ C

If we instead use the linear feedback climate sensitivity parameter, σc ≈ 0.6 ◦ C/ W/m2 , we find

a difference of about 8%.

35.2

In another billion years solar luminosity will increase by roughly another 10%. Estimate the resulting radiative forcing and change in terrestrial surface temperature.

t=3

35.6

Q ≃ 27.5 × 1010 sec ≃ 8738 years.

P

Compute the time for the oceanic mixed layer (350 M km2 × 200 m) to warm by 3.2 K, using 50% of the energy coming from a change in radiative forcing of 3.7 W/m2 . Advanced version: assume that radiative forcing is proportional to T − T eq as the surface warms and describe the time-history of the surface water by writing and solving a differential equation.

The oceans represent 70% of earth surface. Thus, the rate of heat input to an average m2 of ocean surface is I=

1 × 3.7 W/m2 × 1/0.7 2.64 W/m2 2

816

To warm a column of water 200 m deep by 1 m2 area by 3.2◦ C requires E = 3.2 × 4.19 MJ/m3 × 200 m3 2.68 GJ So the time required is E/I 32.25 years. Indeed, the lag time for the climate to adjust to external forcing of this type is generally measured in decades.

35.7

Reproduce the graphs in Example 35.1 by performing the numerical calculations on a computer. Find a way of improving the assumptions and redo the calculation.

35.8

Assume that 1% of a net radiative forcing of 3.7 W/m2 worldwide goes to melting ice over land. Estimate the rate of sea level rise from this melting.

The total rate of energy added per decade is roughly 0.01 × (3.15 × 108 s/decade) × (3.7 W/m2 ) × (5 × 1014 m2 ) 5.8ZJ It takes 334 MJ to melt a cubic m of water, so each decade the volume of ice melted is 17.45 × 1012 m3 Distributed over 3.5 × 1014 m2 of ocean area, this gives a sea level rise of about 5 cm/decade.

35.9

(a) Assume that the rate of carbon emission at 10 Gt/year increases at a constant rate to a maximum of 20 Gt/year in 2050, and then decreases at the same rate. Assume that ocean and land biomass together absorb excess CO2 at a constant rate of −5 Gt/year.

In what year will atmospheric CO2 levels peak? What will be the level of atmospheric CO2 in that year? (b) Repeat the analysis under the more optimistic assumption that ocean and land biomass together absorb 2.5% of the excess in atmospheric carbon above 600 Gt each year, assuming for simplicity that atmospheric carbon is 800 Gt in 2010.

Assuming the rate of carbon emission now at r2011 = 10 Gt/year increases at a constant rate to a maximum of r2050 = 20 Gt/year in 2050 means that this rate is increasing every year by ∆r = 0.25 Gt/year. This is also the rate at which this rate decreases after 2050. Atmospheric CO2 concentrations will only begin to drop when the rate at which CO2 is removed by natural systems exceeds the anthropogenic excess in CO2 . The removal rate is given in our problem to be the constant value rremoval = −5 Gt/year. The peak of CO2 levels will then be reached at year t such that r2050 − (t − 2050) ∆r + rremoval = 0,

=⇒ t = 2050 + 60 = 2110.

To find out the amount Q of of atmospheric CO2 in 2110, we need to integrate the rate of emission from now, where the value is Q0 = 800 Gt to then. The rate of emission is given by the following equations dQ = r(t) = r2010 + (t − 2010) ∆r + rremoval , dt dQ = r(t) = r2050 + (t − 2050) ∆r + rremoval , dt

2010 < t < 2050 t > 2050

so that Q(t) = Q0 + (t − 2010) (r2010 + rremoval )

t2 − 20102 − 2010 (t − 2010) ∆r, 2010 < t < 2050 + 2 Q(t) = Q(2050) + (t − 2050) (r2050 + rremoval ) t2 − 20502

− − 2050 (t − 2050) ∆r, t > 2050 2

Plugging in numbers, we find Q(2050) = 1200 Gt and Q(2110) = 1650 Gt. We now repeat the analysis assuming that rremoval (t) = 0.025 (Q(t) − 600) = 0.025Q(t) − 15. The equation for Q(t) with 2011 < t < 2050 is dQ = −0.025Q(t) + r2010 + 15 + (t − 2010)∆r. dt Since the rate of emission now depends on the total amount of atmospheric CO2 , we cannot repeat the same procedure we used in part a, but we need to solve our equations for Q(t), first. A simple way to do that is to solve the discrete equation Qt = Qt−1 + r(t) − s(t), where r(t) is the rate of emission of CO2 and s(t) is the absolute value of the rate of absorption. An example of a Mathematica code that realizes it, produces a table whose elements are a year and the corresponding amount of atmospheric CO2 , and find the the peak year and the corresponding amount of atmospheric CO2 is the following: I n [ 1 ] : = ( ∗ c o mp u t e s s o l u t i o n f o r c a r b o n max given simple assumptions ∗) carbon [ y e a r _ ] := carbon [ y e a r ] = I f [ y e a r == 2 0 1 0 , 8 0 0 , carbon [ year − 1] + o u t p u t [ year − 1] − ( carbon [ year − 1] − 600)∗0.025] o u t p u t [ y e a r _ ] : = 20 − Abs [ y e a r − 2 0 5 0 ] ∗ 0 . 2 5 tab = Table [{ year , carbon [ year ] } , { year , 2010 , 2 2 0 0 } ]; So rt [ tab , # 2 [ [ 2 ] ] < # 1 [ [ 2 ] ] &][[1 ]] Out [ 4 ] = { 2 0 7 4 , 1 1 6 3 . 8 5 } A more precise answer involves solving the differential equation for dQ/dt directly. This is a linear non homogenous differential equation, like the one we encountered in pset 6. The solution is the same and is given by Q(t) = e−0.025(t−2010) Z t

′ × dt′ r2010 + 15 + ∆r(t′ − 2010) e0.025(t −2010) + Q(2010) 2010

= e−0.025(t−2010) Z t−2010

′ × dt′ r2010 + 15 + ∆r t′ e0.025 t + Q(2010) 0

e0.025(t−2010) − 1 =e−0.025(t−2010) (r2010 + 15) 0.025

(t − 2010) e0.025 (t−2010) e0.25(t−2010) − 1 − + Q(2010) + ∆r 2 0.025 0.025 1 − e−0.025(t−2010) =(r2010 + 15) 0.025 t − 2010 1 − e−0.025(t−2010) + Q(2010) e−0.025(t−2010) , + ∆r − 0.025 0.0252

which gives Q(2050)=1073.6 Gt and for t > 2050

1 − e−0.025(t−2050) 0.025 t − 2050 1 − e−0.025(t−2050) + Q(2050) e−0.025(t−2050) . − − ∆r 0.025 0.0252

Q(t) =(r2050 + 15)

817

Problems

The peak of CO2 level is reached when dQ(t) = 0. dt This equation can be solved numerically to give t = 2074. At that time, the amount of CO2 will be Q(2074) = 1161 Gt. Another way to solve this problem is to use Mathematica to solve the differential equation. An example of a Mathematica code that solve part b of the problem (it is straightforward to adapt the code to solve the easier case in part a) is included in the following.

We use here the command NDSolve to solve numerically the differential equation. This code will produce, as an output, the time at which atmospheric CO2 peaks and the amount of atmospheric CO2 at this peak.

In [ 1 ] : = t0 = 2010; t1 = 2050; r 0 = 1 0 ( ∗ Gt / y e a r ∗ ) ; r 1 = 2 0 ( ∗ Gt / y e a r ∗ ) ; (∗ c o n s t a n t r a t e of carbon emission , Gt / y e a r ^2 : ∗ ) d e l t a r = ( r1 − r0 ) / ( t1 − t0 ) / / N; Q0 = 8 0 0 ( ∗ Gt ∗ ) ; q0 ( ∗ Gt ∗)= NDSolve [ { y ’ [ t ] == −0.025 y [ t ] + 15 + r 1 − Abs [ t − 2 0 5 0 ] d e l t a r , y [ 2 0 1 0 ] == Q0 } , y , { t , 2 0 1 0 , 2 2 0 0 } ] ; T = F i n d R o o t [ E v a l u a t e [ y ’ [ t ] / . q0 ] [ [ 1 ] ] , {t , 2100}][[1 , 2]] ; {T , E v a l u a t e [ y [ T ] / . q0 ] [ [ 1 ] ] }

plant operating at 30% efficiency? How large a population (in the US or in your country) could have its carbon footprint eliminated by this DAC plant assuming the plant to be powered by a non-carbon source?

Use the results of Problem 35.11 to redo this estimate if the coal power plant itself powers the DAC facility.

Compute the mass of CO2 removed from the air per unit time per unit length, Ṁ/L. The molar volume at STP (273 K and 105 Pa is 22.7 L/mol, so the molar volume at NTP (293 K and 1 atm) is (293 K/273 K)(105 Pa/1 atm)(22.7 L/mol) = 0.0240 m3 /mol , so Ṁ/L =

(0.5)(2 m/s)(400 × 10−6 )(0.044 kg/mol)(100 m) 0.073 kg/s m .

0.0240 m3 /mol

At an efficiency of 30%, the power plant consumes coal at the rate Ṁ(coal) =

and therefore produces CO2 at the rate Ṁ(CO2 ) =

35.10 Estimate the contribution to global sea level rise over

a century if Greenland’s glaciers continue to melt at a rate of 290 Gt/y.

The surface area of Earth’s oceans is approximately A = 360 × 106 km2 , so the height of the layer of water produced by 100 years of ice melt is h=

(100 y)(290 Gt/y)(1 m3 /t) 8 cm .

360 × 1012 m2

35.11 Using the methods developed in Box 35.4.2, find a

lower limit on the energy per kilogram of carbon for removing carbon directly from ambient air. Compute the energy cost of removing all 10 Gt/year of current carbon emissions and compare to total electric production worldwide.

35.12 A concern with plans for capture of CO2 directly from ambient air (DAC), is the sheer volume of material that must be processed. One design requires fans to blow air at 2 m/s through 2.8 m thickness of absorber and other materials and aims to absorb 50% of the CO2 from the airstream. Assume a rectangular installation of length L, height 100 m (and thickness 2.8 m).

How large must L be in order for this DAC plant to remove the CO2 produced by a 1 GWe coal-fired power

44 (114 kg/s) 420 kg/s .

12

Dividing the rate of production by the rate of removal per unit length we get

Out [ 8 ] = { 2 0 7 3 . 8 7 , 1 1 6 1 . 3 3 } To reach 20 Gt/year in 2050, an extra 0.253 Gt must be emitted each year starting with 10 Gt in 2012. The atmospheric CO2 level will peak in the year in which the total CO2 added is equal to the total CO2 removed. After 2050, the total net CO2 added is 15 − 0.263i Gt/year where i is the number of years since 2050. The 15 is the 20 Gt added minus the 5 absorbed by the ocean and land biomass. This is 0 when i = 57, so no net CO2 is added in 2107.

Thus, the atmospheric CO2 level peaks in 2107.

The total CO2 added by 2106 is 810 Gt above the current level.

There is currently about 800 Gt of CO2 in the atmosphere, so the total CO2 will be around 1610 Gt at that time.

1 GW = 114 kg/s , (0.3)(29.3 GJ/t)

L

420 kg/s 5920 m = 5.7 km .

0.073 kg/s m

The average US power consumption per person is ∼ 1 GJ/d, so a 1 GWe power plant can supply daily energy to about N=

(1 GW)(86 400 s/d) = 86 400 people .

1 GJ

Clearly a ∼ 6 km long structure 100 m high and 2.8 m thick required to separate the CO2 generated to supply power to . 90 000 people is not practical. This problem is independent of considerations such as the amount of power and the chemical processes required to run such an installation.

850

electric power at 13% (1st law) efficiency. The electric power in turn runs a heat pump with a 2nd law efficiency of 33% that operates between T and an outside temperature T 0 . Further, the waste heat from the geothermal power plant in the form of saturated water at 50 ◦ C is circulated through a heat exchanger at 90% (1st law) efficiency to provide additional space heating. What is the minimum ambient temperature T 0 for which the second system outperforms the first?

[Hint: You will need to look up the specific enthalpy of saturated (liquid) water at various temperatures.]

Problems 36.1

36.2

Estimate the thermodynamic limit on the 1st law efficiency of a device that turns heat radiation from a wood fire into electricity.

In §13.3.5 we specified a realistic Rankine cycle working between T + = 600 ◦ C and T − = 36 ◦ C. We found its efficiency to be 38%. What is its 2nd law efficiency?

The maximum efficiency of a heat engine operating between these temperatures is η|Carnot = (600 − 36)/(273 + 600) 0.65, so the 2nd law efficiency of this cycle is ǫ 0.38/0.65 59%.

36.3 36.4

Verify eqs. (36.14)–(36.17).

A heating system uses thermal energy from a reservoir at temperature T + to run a heat engine that in turn powers a mechanical heat pump that takes thermal energy from an environment at temperature T − and delivers it for space heating at temperature T . Define a suitable 1st law efficiency for this system. First assume that the heat engine dumps its waste heat into the environment at T − . Show that the limit on the 1st law efficiency of this system is the product of the Carnot limits on the efficiency of the heat engine and the CoP of the mechanical heat pump. Next, assume that the heat engine dumps its waste heat into the interior space at temperature T .

Compute the limit on the 1st law efficiency of the system and show that it is not the product of the Carnot limits on the efficiency of the heat engine and the CoP of the mechanical heat pump.

The 1st law efficiency for this system is the ratio of the heat delivered for space heating Q over the thermal energy removed from the high temperature reservoir Q+ ,

At 250 ◦ C, the specific enthalpy of saturated water is h+ = 1 086 kJ/kg. We are not told the pressure of the water at the low temperature end of the heat exchanger, but enthalpy of liquid water depends only weakly on pressure, so we take the enthalpy of saturated water at 20 ◦ C, h = 84 kJ/kg. Thus option one transfers ∆h1 = 0.90(1 086 − 84) 902 kJ/kg of thermal energy as space heating.

Option 2 provides W = 0.13(1 086) 141 kJ/kg of electrical energy that runs a heat pump with a first law efficiency of η = 0.33/(1 − T 0 /T ). The net space heat delivered by the heat pump is ∆h2 (heat pump) = (0.33)(141)/(1 − T0 /T) = 47/(1 − T0 /T) kJ/kg.

The enthalpy of the saturated water at 50 ◦ C is h− = 209 kJ/kg, so the heat exchanger transfers ∆h2 (heat exchanger) = 0.9(209 − 84) 113 kJ/kg .

So the total amount of thermal energy available for space heating in option 2 is ∆h2 = 113 + 47/(1 − T 0 /T ) kJ/kg . ∆h2 increases as T 0 → T , so there is a minimum temperature above which ∆h2 always exceed ∆h1 . Setting ∆h2 equal to ∆h1 and solving for T 0 , we find T 0 = 3 ◦ C.

36.6

η = Q/Q+ .

The limit on the 1st law efficiency of this system is obtained by assuming that both the heat engine and the heat pump perform at their Carnot limits.

In the first case the heat engine supplies work W1 = (1 − T − /T + )Q+ to the mechanical heat pump, which in turn transfers heat Q = W1 /(1 − T − /T ) for space heating. The limit on the 1st law efficiency is therefore η1 ≤

1 − T − /T + .

1 − T − /T

In the second case W2 = (1 − T/T + )Q+ and Q = W2 /(1 − T − /T ). In addition, however, the waste heat from the engine, Q′ = (T/T + )Q+ , can also be used for space heating. The limit on the 1st law efficiency is

Taking T + = 400 ◦ C and T − = T = 40 ◦ C as in §36.3, from eq. (36.19) we find the 2nd law efficiency of the heat pump, ǫheat pump = (0.40)(0.55)(1−313 K/673 K)|∆Hc /∆Gc | = 0.12|∆Hc /∆Gc | ,

1 − T/T + T 1 − T − /T + η2 ≤ (Q + Q )/Q+ = + = .

1 − T − /T T+ 1 − T − /T

and similarly for the furnace,

′

36.5

ǫfurnace = (0.95)(1 − T 0 /313 K)|∆Hc /∆Gc | .

The limits on η1 and η2 are the same, but the limit on η2 is not the product of the limit of the efficiency of the engine (1 − T/T + ) times the limit on the CoP of the heat pump 1/(1 − T − /T ).

A geothermal resource outputs saturated (liquid) water at 250 ◦ C (as in Example 32.5). Consider two alternative ways to use this resource to deliver space heating at T = 20 ◦ C. First, the hot water is circulated through a heat exchanger that transfers its thermal energy with 90% (1st law) efficiency to the space to be heated.

Second, the geothermal resource is used to generate

A homeowner is trying to decide whether to heat with a furnace rated at 95% efficiency or by an electrically powered heat pump. She lives in a town where electricity is produced by a coal-fired power plant that claims to operate with 1st law efficiency that is 55% of the Carnot limit. The heat pump’s CoP is advertised to be 40% of the Carnot limit. Take the values of T + , T − , and T from §36.3. For what range of outside temperatures T 0 would the 2nd law efficiency of the furnace be greater than that of the heat pump?

Setting the two efficiencies equal and solving for T 0 yields T 0 313(1 − 0.12/0.95) 273 K .

Thus the heat pump is more efficient than the furnace when the outside temperature is above approximately 0 ◦ C.

36.7

Reproduce the results in Example 36.4 using eq. (36.25) rather than by the direct calculation of available work that was used in the example. [Hint: you can use the Sackur–Tetrode formula (8.65) for the entropy of an

851

Problems

ideal gas, expressed as a function of temperature and pressure.]

where we have used the 1st Law to set ∆U = Q at constant volume and the 2nd Law to set dS = CV dT/T . T = T 0 + Q/CV is the final temperature of the gas. The total exergy change is then

1. Compressed air: for an ideal gas, U is a function of T alone, so U − U0 = 0. From the ideal gas law V − V0 = nRT 0 (1/p − 1/p0 ) and from eq. (8.65) S − S 0 = nR(ln p0 − ln p), so

∆B = −Belectrical +B f −Bi = −T 0 (S f −S i ) = −T 0 CV ln(1+Q/(CV T 0 ))

B = U − U0 + p0 (V − V0 ) − T 0 (S − S 0 ) = nRT 0 (p0 /p − 1) − nRT 0 ln(p0 /p) ,

in agreement with Box 36.1.

2. Vacuum: U = U0 = 0, S = S 0 = 0, and V − V0 = V, so from eq. (36.25), B = p0 V. 3. Cold R T water: U − U0 = c(T c − T 0 ), V − V0 = 0, and S − S 0 = − T 0 dT c/T = −c ln(T 0 /T c ), so c

B = c(T c − T 0 ) + cT 0 ln(T 0 /T c )

36.8

36.9

Systems that are colder than their environment have positive exergy. Suppose a quantity of monatomic gas originally at ambient temperature and pressure (T 0 , p0 ) is adiabatically and reversibly expanded until its temperature has dropped to T. Compute its exergy from eq. (36.25) and show that it is positive. How can this gas do useful work?

What is the internal energy and the exergy of a one ton block of stone (c = 0.8 kJ/kg K) at 100 ◦ C relative to an environment at 20 ◦ C? What would be the mass and internal energy of a block of stone with the same exergy at 60 ◦ C? This problem illustrates exergy as a measure of the quality of stored energy.

The internal energy (relative to 20 ◦ C) is U − U0 = mc∆T = (1 000 kg)(0.8kJ/kg K)(80 K) = 64 MJ .

To compute the exergy we use eq. (36.25). We need the entropy difference Z T mc S − S0 = dT ′ ′ = mc ln(T/T 0 ) , T T0 giving B = U − U0 − T 0 (S − S 0 ) = mc (∆T − T 0 ln(1 + ∆T/T 0 ))

= (800 kJ/K)(80 K − (293K) ln(1 + 80/293) 7.4 MJ .

Note how small the exergy is compared to the internal energy.

The exergy per unit mass of a block of stone at 60 ◦ C is B/m = (0.8 kJ/kg K)(40 K−(293 K) ln(1+40/293)) 2.00 kJ/kg .

So to have the same exergy, the stone would have to have a mass of m = 7.4 MJ/(2.00 kJ/kg) 3.7 t and its internal energy relative to 20 ◦ C would be U − U0 = (3 700 kg)(0.8kJ/kg K)(40 K) 118 MJ.

36.10 Suppose a quantity of gas, held in a fixed volume V and

initially at the ambient temperature T 0 , is heated by a resistor to temperature T . Let Q be the heat delivered by the resistor and let CV be the (assumed constant) heat capacity of the gas at constant volume. How much exergy is destroyed in this process? Show that when T − T 0 ≪ T 0 essentially all of the initial exergy is destroyed.

The exergy of the electrical energy before it was converted into heat was Belectrical = Q. According to eq. (36.25), the change in exergy of the gas was Z T CV dT ′ ′ B f − Bi = U f − Ui − T 0 (S f − S i ) = Q − T 0 T T0

When T −T 0 ≪ T 0 the logarithm can be expanded using ln(1+ x) = x + O(x2 ), yielding ∆B = −Q + O(Q2 /CV T 0 ).

36.11 In §5.2.2 we described a simple system in which a

piston is used to extract energy from a hot gas with efficiency (5.20). Show that the energy extracted precisely corresponds to the exergy of the heated gas.

36.12 [H] A colloquial measure of power used in the air

conditioning industry is the ton of air conditioning = 3.517 kW (see §9, Problem 9.5). It is the average power required to melt one ton of ice in one day. Check this number. If melting ice was actually used to provide cooling, it would not only be allowed to melt but would also be allowed to warm to room temperature. Compute the 2nd law efficiency of this method of air conditioning a home with indoor temperature 27 ◦ C and outdoor temperature 37 ◦ C. To do this you must find the maximum reversible work that can be done by a ton of ice under these conditions.

One ton of ice is 907.2 kg, with a latent heat of melting of 333.6 kJ/kg, so melting one ton of ice extracts ∆H f = (907.2 kg)(333.6 kJ/kg) = 302.6 MJ from the room. If this occurs over one day, it corresponds to an instantaneous power rate of P = 302.6 MJ/86 400 s) = 3.503 kW. The most efficient strategy to use the exergy in the ice to cool the room is to use thermal energy from the room to run a reversible heat engine between room temperature and the melting temperature of the ice, and after the ice is melted, continue to run the engine warming the water to room temperature. The work output of the engine is the used to run a reversible air-conditioner to remove further heat from the room. Let T = 27 ◦ C be the room temperature, T + = 37 ◦ C the outside temperature, T c = 0 ◦ C the melting temperature of ice, and T − be the temperature of the water as it warms.

We compute in two steps. First melt the ice. Heat Q1 removed from the room by a reversible heat engine yields work W = (1 − T c /T )Q1 and thermal energy ∆H f = (T c /T )Q1 added to the ice.

Solving for W = (T/T c − 1)∆H f . This work runs an air conditioner which extracts Q2 = W/(T + /T − 1). The total heat removed from the room is !

T T/T c − 1 Q = Q1 + Q2 = + ∆H f T c T + /T − 1 T + /T c − 1 = 4.07∆H f = T + /T − 1

=

Second, warm the ice to room temperature. Let C = (902.7 kg)(4.186 kJ/kg K) 3.80 MJ/K be the heat capacity of a ton of water. To warm the ton of water to room temperature, the reversible heat engine must discard Q− = C(T − T c ) = 80 MJ into the water. The analysis of Box 36.1 give maximum work extracted from warming the water, W ′ = C(T ln(T/T c ) − T + T c ) , and the heat engine extracts Q′1 = W ′ +Q− = CT ln(T/T c ) from the room. Finally, the air conditioner uses the work to remove further heat Q′2 = W ′ /(T + /T − 1). The The total heat removed as the water warms is Q′ = Q′1 + Q′2 =

C (T + ln(T/T c ) + −T + T c ) = 255 MJ T + /T − 1

for the temperatures in the problem.

852

The total cooling power is Qtotal = 1 232 MJ + 255 MJ = 1 487 MJ, which gives a 2nd law efficiency of ǫ 302.6/1 487 .204.

36.13 [H] Consider an isothermal and isobaric combustion

system that operates as follows: a stream of fuel and air originally at T 0 and p0 absorbs thermal energy reversibly from a regenerator reaching temperature T .

It then enters a combustion chamber at temperature T and pressure p0 , where the fuel is combusted yielding enthalpy of combustion |∆Hc (T )| per mole. This energy drives a reversible heat engine with T + = T and T − = T 0 . The combustion products leave the combustion chamber at temperature T and cool to temperature T 0 transferring some of their energy to the regenerator used to heat the incoming fuel–air stream. Assume that the (molar) heat capacity of the products exceeds the heat capacity of the reactants ( f (t) = Ĉproducts (t) − Ĉreactants (t) > 0 for all t between T 0 and T ), so some residual energy in the products can also be used to run a reversible heat engine as the products cool. Assume also that |∆Hc (T 0 )| > |∆Gc (T 0 )|, where ∆Gc (T ) is the Gibbs free energy of combustion. Show that the exergy (maximum useful work) of the fuel combusted in this system is given by B(T ) = −∆Gc (T 0 ) + ∆Gc (T ) .

Since ∆Gc (T ) must be negative for the combustion reaction to occur, B(T ) reaches its maximum at T M where ∆Gc (T M ) = 0.

The useful work B(T ) has two sources, the heat of combustion extracted at T , extracted reversibly by an engine running at Carnot efficiency, W1 (T ) = |∆Hc (T )|(1 − T 0 /T ) , and the work that can be extracted reversibly as the products of reaction cool, Z T0 dt f (t)(1 − T 0 /T ) , W2 (T ) = − T

from eq. (36.23). Combining these B = |∆Hc (T )|(1 − T 0 /T ) +

Z T T0

Since combustion of H2 is simply the formation of water, we find from Table 9.8, ∆Hc (T 0 ) = −242 kJ/mol and ∆Gc (T 0 ) = −229 kJ/mol. Since ∆G(T ) = ∆H(T ) − T ∆S (T ), ∆S c (T 0 ) = (−242 + 229)/298 −0.044 kJ/mol K. Since ∆S c is negative, ∆Gc increases with temperature and eventually becomes positive at some temperature T M . Above that temperature, the combustion reaction will not proceed. Setting ∆Gc (T M ) = ∆Hc (T 0 ) − T M ∆S c (T 0 ) = 0 we find T M = −∆Hc (T 0 )/∆S c (T 0 ) = (242/0.044) 5 500 K .

At this (impractically) high temperature, the useful work performed by the engine (operating at Carnot efficiency) is Wuseful = |∆Hc (T 0 )|(1 − T 0 /T M ) = |∆Gc (T 0 )| .

36.15 According to the US EIA, 1350 Mt(CO2 ) were emit-

ted from combustion of coal and 530 Mt(CO2 ) were emitted from combustion of natural gas for electricity production in 2015. Using the data in §36.5.1 show that improving the efficiency of coal and natural gas power plants to 45% and 60% respectively would abate roughly 500 Mt/y of CO2 emissions.

Data from §36.5.1: For coal (1/3)13.3 EJ 4.5 EJ was produced from combustion of coal at 33% efficiency in 2015. For natural gas 4.5 EJ was produced at 43% average efficiency The emissions per unit of primary energy consumption for coal was therefore 1350 Mt 100 Mt(CO2 )/EJ .

4.5 EJ/0.33 Improving the efficiency of coal power plants from 33 to 45% would reduce primary energy consumption from coal from 4.5/0.33 13.5 EJ to 4.5/0.45 10.0 EJ. The resulting reduction in CO2 emissions would be ∆coal = (13.5 EJ − 10.0 EJ)(100 Mt(CO2 )/EJ) 350 Mt(CO2 ) .

The emissions per unit of primary energy consumption for natural gas is presently

dt f (t)(1 − T 0 /T )

From the definitions of enthalpy and entropy, Z T ∆Hc (T ) = ∆Hc (T 0 ) + dt f (t) T0

∆S c (T ) = ∆S c (T 0 ) +

hydrogen is combusted isothermally at temperature T in an engine that runs at the Carnot limit in an environment at T 0 . Assume for simplicity that ∆Hc and ∆S c are independent of temperature. Show that the maximum amount of work that can be extracted from such an engine is in fact Wisothermal (T M ) = |∆Gc (T 0 )| and find the temperature T M at which the engine must run to reach this limit.

Z T

dt f (t)/t.

T0

If we replace both integrals over t in B(T ) with the differences of combustion enthalpies and entropies, the required result follows immediately.

530 Mt 51 Mt(CO2 )/EJ , 4.5 EJ/0.43 Improving the efficiency of natural gas power plants to 60% would reduce primary energy consumption from natural gas from 4.5/.43 10.5 EJ to 4.5/.6 7.5 EJ and the resulting reduction in CO2 emissions would be ∆gas = (10.5 EJ − 7.5 EJ)(51 Mt(CO2 )/EJ) 150 Mt(CO2 ) .

Combining the two leads to a total reduction of approximately ∆total = 350 + 150 500 Mt(CO2 )/y .

36.14 Look up the molar enthalpy of combustion ∆Hc (T 0 )

36.16 Given T and ∆T for Boston, show that the heating

and Gibbs free energy of combustion ∆Gc (T 0 ) for hydrogen under standard conditions. Notice that |∆Hc (T 0 )| > |∆Gc (T 0 )|, suggesting that one might be able to extract more than |∆Gc | from combustion of hydrogen at high temperature, contradicting the claim that the exergy of a fuel is |∆Gc (T 0 )|. Suppose that the

season is on average ≈ 260 days long and evaluate eq. (36.44) to obtain the average number of heating degree days in Boston. Then choose a city for which monthly average temperature data are available, compute T , ∆T , and Dcity . Compare your answer with the (commonly available) actual value for Dcity .

853

Problems

According to eq. (36.43) the half heating season is determined by the day on which T l (τ/2) = T s , or cos(ωτ/2) = (T Boston − T s )/∆T Boston

= (10.7 − 18.3)/12.4 = −0.613

so ωτ/2 = cos−1 (−.613) = 2.23; with ω = 2π/365,

The heat runs at the lower thermostat setting for t0 = ttotal − ∆t = 8(3 600) − 8 100 = 20 700 s, providing heat Q1 =

and reheating requires Q2 = CH (T s − T s′ ) = 20 MJ. So the total heating energy supplied over the 8 hours is Qtotal = Q1 + Q2 = 68.2 MJ. Had the thermostat not been set back, the heat delivered would have been

τBoston (2.23)(365)/π 260 .

As an example, choose St. Paul, Minnesota. Data online give T = 8.4 ◦ C, ∆T = 16.3 ◦ C, and D(T s ) 4 376 averaged over 1873 to 2013. Using the same method as we applied to Boston, τSt. Paul 260. (The heating season on the two locations is the same.) And from eq. (??), T 0,St. Paul 2.5. Substituting into eq. (??), DSt. Paul (T s ) 4 100, which differs from the measured value by about 6%.

36.17 Show that ≈ 1000 L of home heating oil with energy

density of about 40 MJ/L are required to provide heat for our typical dwelling (AH = 100 m2 , C H = 10 MJ/K, RH = 0.7 m2 K/W) in Boston. Assume that the (1st law) efficiency of the oil furnace is 95%.

QF =

AH (# of degree days) RH 100 m2 (3 100 K d)(8.64 × 104 s/d) 0.7 K m2 /W

38 GJ The fuel provides (0.95)(40 MJ/L) 38 MJ/L, so the volume of fuel required is V = (38 GJ)/(38 MJ/L) 1 000 L.

36.18 [C] Perform a numerical estimate of the savings in

heating energy for the residence modeled in §36.5.3 in Boston when the thermostat is turned off for eight hours, assuming that the outdoor temperature over the day varies sinusoidally with a mean temperature of 0 ◦ C and a maximum variation of ±5 ◦ C. You may assume the interior temperature setpoint is T s = 18.3 ◦ C and that the thermostat is turned off over the coldest eight hours of the day.

36.19 A homeowner turns down the thermostat from T s =

18.3 ◦ C to T s′ = 16.3 ◦ C for ttotal = 8 h every night.

Including the effects of reheating, compute the fraction of heating energy saved compared with leaving the thermostat set at 18.3 ◦ C throughout. Assume that the outside temperature remains fixed at 0 ◦ C. First calculate the time ∆t that the heat is off while the dwelling cools by 2 ◦ C. Then compute the energy saved at the low thermostat setting, and finally compute the energy required for reheating. Evaluate your result for the reference dwelling with AH = 100 m2 , RH = 0.7 m2 K/W, and C H = 10 MJ/K. How significant are the effects of reheating?

First compute the time required for the house to cool. From Example 36.4 ∆t =

!

T s − Tl RH CH ln ′ 8 100 s AH T s − Tl

AH ′ (T − T l )t0 = 48.2 MJ , RH s

Q0 =

AH (T s − T l )ttotal = 75.3 MJ .

RH

So the savings was 75.3 − 68.2 = 7.1 MJ or about 10%. If reheating was not included the savings would have been 75.3 − 48.2 = 27.1 MJ or about 36%.

36.20 A frequent traveler, who lives alone in Washington

DC (T 13.2 ◦ C, ∆T 12 ◦ C), is away from home every other week throughout the year. Normally she keeps her thermostat at 20 ◦ C, and turns it down to 10 ◦ C when she is away. Estimate her fractional savings in heating energy first ignoring the need to reheat her home upon return. Then estimate the correction for reheating. Take AH = 100 m2 and C H = 10 MJ/K, and assume that she has insulated her home so that RH = 2.1 m2 K/W. 36.21 A rural cottage sits on pilings with a breezeway underneath. It has insulated wooden walls (Awalls = 50 m2 , Rwalls = 3 m2 K/W), a plywood floor (Afloor = 35 m2 ) 2.5 cm thick, and a simple peaked slate roof (slope 30◦ ) covering the floor area. The slates are roughly 2 cm thick, laid over plywood sheathing 2.5 cm thick.

The cottage has four double pane glazed windows ( 14 " air gap), each 1 m2 . Estimate the effective R value for the dwelling as a whole. Now imagine doubling the thermal resistance of any component, walls, floor, or roof. Recompute the effective R-value and decide which would be most effective for energy conservation.

We need R-values for the floor and roof. From Table 6.3, 2.5 cm ( 1 in) of wood has R 0.25 m2 K/W. Slate adds so little to the insulating value that we ignore it. Thus R 0.25 m2 K/W for 35 m2 of floor and 35/ cos(30◦ ) 40 m2 of roof. We take the R-value of the windows from the same table, to be R = 1/U = 1/3.1 0.3 m2 K/W. Combining all contributions in parallel, 125 4 46 35 40 = + + + = 328 W/m2 K , Reff 0.3 3.0 0.25 0.25 which gives Reff 0.38 m2 /W. Doubling the thermal resistance of any component halves its contribution to the sum in this equation. The biggest effect comes from the roof, which reduces R125 eff

to 248 W/m2 K and increases Reff to 0.50. Heating costs are proportional to 1/Reff so the decease is from (1/.38)x = 2.7x to (1/.5)x = 2.0x, for a decrease of (2.7 − 2.0)/2.7 = 26%.

36.22 Verify the claim that the yearly heating requirement of

the dwelling described in §36.5.3 drops from ≈ 40 GJ/y to ≈ 28 GJ/y when the 15 m2 of windows are changed from single to double glazing.

The 15 m2 of windows and the remaining 85 m2 of exterior surface (“walls”) add thermal resistances in parallel (see eq. (6.13)).

When the effective thermal resistance of the (single glazed) windows is 0.16 K m2 /W, the effective thermal resistance of the whole

854

dwelling is 0.7 K m2 /W. Thus 100 85 15 = + .

0.7 Rwalls 0.16 Solving, we find Rwalls = 1.73. Then changing the thermal resistance of the windows from 0.16 to 0.29, we find, 100 85 15 = + , Reff 1.73 0.29 which gives Reff = 1.0 K m2 /W. The yearly heat required scales proportional to 1/Reff (see eq. (36.39)), so QF → Q′F = (0.7/1.0)QF = 0.7(40) = 28 GJ.

36.23 The cottage described in the previous problem sits in

a location with average mean temperature T = 8 ◦ C, and maximum average annual variation, ∆T = 16 ◦ C. Estimate the number of heating degree days at this location. If you solved the previous problem, calculate the annual heating energy requirement of the cottage before any improvements.

Integrating eq. (36.44), we find  s     T − T  365  ∆T 2 − (T − T )2 ∆T   DD = T ) arccos + (T − π  ∆T  ∆T 2

and subsituting for T = 18.3 ◦ C, T = 8 ◦ C, and ∆T = 16 ◦ C, DD 4 140 ◦ C-days. From eq. (36.39) we find QF =

125 m2 (4 140 K/d)(86 400 s/d) 118 GJ 0.38 m2 K/W

36.24 Check the claimed rating and actual energy produc-

tion of the solar photovoltaic system described in §36.5.3.

36.25 Lithium is the source of tritium, which is the primary “fuel” for a dt fusion reactor. (The deuterium, which is abundant and relatively cheap, can be considered as analogous to the air required for combustion of a fossil fuel.) Compute the energy density (J/kg) of lithium with respect to dt fusion. First, ignore the difference between isotopes 6 Li and 7 Li, assume that one lithium atom produces one tritium atom at no energy cost, and use the fact that 17.6 MeV is emitted in dt fusion. Next improve your estimate by noting that 7.6% of naturally occuring Li is 6 Li and 92.4% is 7 Li. Note that the reactions eq. (18.14) and eq. (18.15) are exothermic and endothermic respectively, thereby allowing you to estimate the energy required (or emitted) to convert lithium to tritium. Show that this correction to the energy density of naturally occurring lithium is ≈ 10% and that the result agrees with the number quoted in Table 36.3.

The atomic mass of naturally occurring lithium is 6.94, so 1 kg(Li) = 1/0.00694 144 moles of lithium, which yields 144 moles of tritium. Fusion of this amount of tritium produces energy E = 144(6.023 × 1023 )(17.6 MeV) 1.53 × 1033 eV 246 TJ .

So the energy density of lithium with respect to dt fusion is 246 TJ/kg.

Taking account of the isotopic differences: 1 kg of lithium → 76 gm 6 Li and 924 gm 7 Li with atomic masses very close to 6 and 7 respectively. So 1 kg of lithium contains 12.7 moles of 6 Li and 132 moles of 7 Li. The production of tritium from 6 Li has a Q value of 4.78 MeV (energy emitted) and from 7 Li Q = −2.47

(energy required). So the energy required to produce 144.7 moles of tritium from naturally occurring lithium is E =(6.023 × 1023 )(1.609 × 10−19 J/eV)

× 132(2.47 × 106 eV) − 12.7(4.78 × 106 eV) 26 TJ

and fusion of 144.7 moles of tritium yields 247 TJ/kg, so the energy available from the lithium is reduced by 26/247 10.5% to ≈ 220 TJ/kg(Li).

879

Problems

would require N = ((20/12.5) × 103 )(500/1 000)(70 cars) = 56 000 cars.

Problems 37.1

A typical lithium-ion battery for a laptop has a storage capacity of 200 kJ and a mass of 0.5 kg. Compute the number of such batteries required and total mass in order to store 30% of US daily electric energy consumption of 12 PJ.

37.5

N = 12 PJ/(200 kJ/battery) = 6 × 1010 batteries with a mass of 30 Mt.

37.2

after the work done by/on the atmosphere is subtracted.

Here r = pH /p0 and γ = 1.4 is the adiabatic index of air.

Consider the scope of pumped hydro storage needed to store and release 12 PJ of energy per day. Compare to the total US hydropower output. If the full 12 PJ were stored using local modest-sized UPH reservoirs of volume 20 000 m3 at a depth of 100 m, how many such reservoirs would be needed?

A volume V0 is compressed from p0 to volume VH at pressure pH . The energy stored equals to the work done on the gas. For adiabatic compression pV γ = p0 V0γ = pH VHγ Wad = −

US hydropower use in 2008 was 280 TWh 1 EJ, or 2.8 PJ/d. So 12 PJ is about 4.3 times US daily hydropower output. Each reservoir has the capacity to store E (20 000 m3 )(1 000 kg/m3 )(10 m/s2 )(100 m) 2 × 1010 J, so N 12 × 1015 /2 × 1010 60 000 such reservoirs would be required to store 4.3 times US daily hydropower output.

37.3

The world’s largest pumped hydro storage facility, located in Bath County, Virginia, has a generating capacity of about 3 GW. When generating power, the turbine flow rate is roughly 850 m3 /s. Estimate the difference in elevation between the upper reservoir and the turbines. The area of the upper reservoir is 110 ha and its surface level fluctuates by approximately 30 m during operation. Estimate the total energy storage capacity of the facility.

dV (p − p0 ) Z VH V0

dV/V γ + p0 (VH − V0 )

pH V H 1 − (VH /V0 )γ−1 + p0 (VH − V0 ) γ−1

!

1 1 − (p0 /pH )(γ−1)/γ + p0 /pH − (p0 /pH )(γ−1)/γ = pH V H γ−1 !

γ 1 − (1/r)(γ−1)/γ − (1 − 1/r) = pH V H γ−1

37.6

Consider using the McIntosh salt mine described in Example 37.2 for adiabatic storage. Compute the energy stored if a volume of air initially at p0 = 1 atm is compressed to pH = 75 atm, and compare to the energy stored if the air were compressed isothermally. Take γ = 1.4 for air. Compute the temperature of the air at pressure pH assuming that the initial temperature was 300 K. Substituting into the result from Problem 37.5, pH = 75 atm, VH = 270 000 m3 , r = 75, and γ = 1.4,

gives h 360 m. The total capacity is V = (30 m)(110 × 104 m2 ) = 3.3 × 107 m3 or m = 3.3 × 1010 kg. Then the total stored potential energy is U = (3.3 × 1010 kg)(10 m/s2 )(360 m) 118 TJ 33 GWh.

E = 12.5 MWh = 12.5 × 106 × 3 600 = 45 × 109 J. Since E = mgh, m = (45 × 109 J)/(10 m/s2 × 500 m) = 9 000 t. At 130 t/car, about 70 railroad cars would be required. To scale up to 20 GWh

V0

=

3 GW = (8.5 × 105 kg/s)(9.8 m/s2 )h

A company is proposing to build an advanced rail energy storage system: excess electric energy will power engines that will transport a heavy mass up a slope. During periods of peak demand, the process will be reversed, converting the gravitational potential energy of the mass back into electric power. The company proposes to build a first commercial-scale facility that will store 12.5 MWh using a height difference of 500 m. The railroad cars have a loaded mass of 130 tonnes. How many railroad cars are required for the proposed installation? The company suggests a grid-scale storage facility might have a capacity of 16– 24 GWh. For a height difference of 1000 m, how many of these railroad cars would be needed?

Z VH

= −pH VHγ

Ignoring inefficiencies (which should be included in a more realistic estimate) and ignoring the change in height during the drawdown,

37.4

[T] Show that the energy stored when air is compressed adiabatically from pressure p0 to pressure pH in a final volume VH = V is given by !

γ (adiabatic) 1−(1/r)(γ−1)/γ −(1−1/r) Estored = pH V γ−1

(adiabatic) Estored = 2.05 TJ(

1.4 (1 − (1/75)2/7 − (1 − 1/75)) 3.1 TJ . .4

In the isothermal case, (isothermal) Estored =−

Z VH V0

= −pH VH

dV (p − p0 ) Z VH V0

dV/V + p0 (VH − V0 )

= pH VH (ln(pH /p0 ) − (1 − p0 /pH )) , which evaluates numerically to (isothermal) Estored 6.8 T J .

In the adiabatic case, the final temperature obeys T ∝ p(γ−1)/γ , thus T H = 300 K(75)2/7 = 1 030 K.

37.7

Confirm explicitly that the work (37.2) done in isothermal compression of a CAES system with volume V, and low, high, and ambient atmospheric pressures pL , pH , p0 is equivalent to the exergy (36.25) of the system.

880

First compute the exergy of a volume V of air compressed isothermally to pH . Since the process is isothermal, U = U0 . Since p0 V0 = pH VH for isothermal compression, ! p0 −1 . p0 (V − V0 ) = V pH pH At fixed temperature, the Sackur-Tetrode entropy (8.65) is proportional to NkB ln V + constants, so T 0 (S − S 0 ) = NkB T 0 ln(V/V0 ) = V pH ln(p0 /pH ) .

converted to electrical power with 33% efficiency and that the salt ends up at a temperature of 250 ◦ C after transferring its energy to the working fluid that drives the turbine. How big does the plant need to be, and how many kilotonnes of molten salt are needed per square kilometer of the power plant?

To gather 100 MW at 3% efficiency requires insolation of 3.33 GW. At 250 W/m2 average insolation, the required area is

Combining these terms we find BH = U − U0 + p0 (V − V0 ) − T 0 (S − S 0 ) ! p0 = V pH − 1 + V pH ln(pH /p0 ) . pH

37.8

In 24 hours the plant requires Ethermal = (100 MW)(24 × 3 600 s)/0.33 = 26 TJ

The computation for BL is analogous with pH → pL throughout. Subtracting, we obtain B = BH − BL = W isothermal in eq. (37.2).

of stored thermal energy, or 2 TJ/km2 . The molten salt stores ∆E = (1 500 J/kg K)(450 K − 250 K) = 3.0 × 105 J/kg. So the mass of salt required is

[H] Verify the computations in Example 37.2 describing the McIntosh CAES plant. In particular, use eq. (37.2) to compute the energy stored through isothermal compression; verify that 1.4 TJ of work would be done if the stored air were allowed to expand adiabatically as described in Example 37.2 and verify the temperatures of the air left in the mine and the air released into the environment after adiabatic expansion.

m = (2 TJ/km2 )/(300 kJ/kg) 6.7 kt/km2 .

First, substituting into eq. (37.2), Wiso = (270 000 m3 )(101 325 Pa)(75 ln 75−45 ln 45−75+45) 3.4 TJ. Next, consider the gas stored at PH = 75 atm in volume V = 270 000 m3 . Upon discharge, a volume V1 is expanded to V at PL and the remainder V − V1 = V2 is expanded to p0 . V1 can be determined by the adiabatic relation V1 pγH = V pγL , so V1 = (1/r1/γ )V where r = pH /pL . This gives V1 = 187 000 m3 , V2 = 83 000 m3 .

We can then apply the result of Problem 37.5, computing the difference between adiabatic storage at volume V at compression ratios 75 and 45, giving 3.06 − 1.65 1.4 TJ. More detail: We use the same methods as applied in the solution to Problem 37.5, we find W1 =

pH V ((1/r)1/γ − 1/r) − p0 V(1 − (1/r)1/γ ) γ−1

The remaining air which occupies volume V2 at pH is expanded to p0 , and the same methods give W2 =

p H V2 (1 − (1/r′ )(γ−1)/γ ) − p0 V2 (r′1/γ − 1) , ] γ−1

where r′ = pH /p0 and V2 = V − V1 = V(1 − (1/r)γ/(γ−1) ). Substituting V = 270 000 m3 , pH = 75 atm, pL = 45 atm, p0 = 1 atm, so V1 = 187 000 m3 , V2 = 83 000 m3 , r = 75/45 = 1.67, and r′ = 75, we find W1 = .47 TJ, and W2 = .94 TJ, for a total of W 1.4 TJ. The temperature changes can be obtained from the adiabatic relation T ∝ p(γ−1)γ , so for the air expanded to pL = 45 atm, T = (45/75)2/7 (300 K) 260 K, and for the air expanded to p0 = 1 atm, T = (1/75)2/7 (300 K) 87 K.

37.9

A = (3.33 GW)/(250 W/m2 ) 13 km2 .

Estimate the specifications for a solar thermal energy plant and storage system. The plant should output a steady 100 MW, 24/7. Assume that the plant is built in a desert area with an average 250 W/m2 insolation. Assume a gross conversion efficiency of 3%.

Tanks of molten salt at 450 ◦ C (specific heat capacity 1.5 kJ/kg K) should be able to store 24 hours of power.

You can assume that thermal energy from the salt is

37.10 In a lithium-iodine (LiI) battery, a solid lithium anode is

surrounded by a polymer impregnated with iodine. The net reaction, Li + 21 I2 →LiI, has a standard reaction free energy of ∆G0 = −266.9 kJ/mol and a standard reaction enthalpy of ∆H 0 = −270.08 kJ/mol. Estimate the cell voltage E0 , the upper limits on the thermodynamic efficiency η, and the upper limit on the specific energy εeffective of a LiI battery.

Lithium is singly ionized in the reaction, so from eq. (37.7) with n = 1, E0 cell = −(−266.9 kJ/mol)/(9.648 × 104 C/mol) 2.77 V. The thermodynamic limit on the cell efficiency η = ∆G0 /∆H 0 = 266.9/270.1 0.988. The molecular mass of LiI is 0.134 kg/mol, so from eq. (37.5), the upper limit on the effective energy density is εeffective = (266.9 kJ/mol)(0.134 kg/mol) 1.99 MJ/kg .

37.11 Compute the free energy of reaction −∆G 0 , ideal

cell voltage E0 , maximum possible effective energy density εeffective , and thermodynamic efficiency limit η = ∆G0 /∆H 0 for an alkaline dry-cell battery (37.11) given the enthalpies and free energies of formation in Table 37.2.

From eq. (37.10), the net reaction is

Zn + 2MnO2 → ZnO + Mn2 O3 so 0 = −972 − 348 + 2(521) = −278 kJ/mol ∆Hreaction

∆G0reaction = −890 − 318 + 2(466) = −276 kJ/mol

These give E0cell = −(−276 kJ/mol)/(2 × 9.648 × 104 C/mol) = 1.43 V, η = 276/278 = 0.993, and εeffective = (276 kJ/mol)/(0.239 kg/mol) 1.15 MJ/kg.

37.12 Compute an upper bound on the energy density of a

lead-acid battery in which the overall chemical process is Pb + PbO2 + 2H2 SO4 → 2PbSO4 + 2H2 O

(37.23)

and for each such reaction two electrons pass around the external circuit, assuming that the only material needed is that involved in the reaction. The standard

881

Problems

Table 37.2 Reaction data for Problem 37.11.

Compound

−∆H 0f (kJ/mol)

−∆G0f (kJ/mol)

ZnO MnO2 Mn2 O3

348 521 972

318 466 890

compare with the quoted figure of 5.6 MJ/L? Compare the total energy stored in hydrogen compressed to 700 atm with the energy density of gasoline.

The work required to compress an ideal gas isothermally from an ambient pressure p0 to pH is W = pH V (ln pH /p0 − 1 + p0 /pH ) For pH = 700 atm,

cell potential for this battery is 2.04 V. The actual energy density is significantly lower than the upper bound due to the need for additional water, as well as other battery components, for operation of the battery.

From eq. (37.7), ∆G0 = −nFE0cell = −2(9.648 × 104 C/mol)(2.04 V) −394 kJ/mol. The molecular mass of the reactants is m = 2(207.2)+6(16.0)+2(32.1)+4(1.0) 579 gm/mol, so εeffective = (394 kJ/mol)/(0.579 kg/mol) 0.68 kJ/kg .

37.13 Confirm the 2nd law bound η = 83% quoted in the text

on the efficiency of a hydrogen fuel cell.

The reaction is formation of liquid water, for which ∆G0 = −237.1 kJ/mol and ∆H 0 = −285.8 kJ/mol, so η = ∆G0 /∆H 0 = 237.1/285.8 0.83 .

37.14 Compute the 2nd law bound on the efficiency of a

methanol fuel cell.

The net reaction is

CH3 OH(l) +

3 O2 → CO2 (g) + 2H2 O(l) 2

The enthalpies and free energies of formation are Compound

−∆H 0f (kJ/mol)

−∆G0f (kJ/mol)

CH3 OH(l) H2 O(g) CO2 (g)

239 286 394

166 237 394

So the standard enthalpy and free energy of reaction are ∆H 0 = 2(−286) + (−394) − (−239) = −727 kJ/mol

∆G0 = 2(−237) + (−394) − (−166) = −705 kJ/mol ,

and efficiency bound is η = ∆G0 /∆H 0 = 0.97.

37.15 Estimate the relative range in highway miles of a

passenger vehicle with comparable size and drag coefficient to the Camry described in §2, assuming it is operating using (a) a 40 L gasoline tank and 25% efficient engine, (b) a 150 kg lithium-ion battery, and (c) a compressed hydrogen fuel tank at 700 atm with volume 60 L and a hydrogen fuel cell with twice the energy loss required by the 2nd Law. 37.16 According to Figure 37.7, the energy density of hydrogen compressed to 700 atm is approximately 5.6 MJ/L. Assuming hydrogen behaves as an ideal gas throughout, estimate the energy density of hydrogen compressed to 700 atm by adding the enthalpy of combustion and the energy required to compress the hydrogen isothermally starting at 1atm. How does your answer

W/V = (700)(101 325 Pa)(ln(700) − 1)(0.001 L/m3 ) = 0.39 MJ/L From the ideal gas law, 1 L of hydrogen at 700 atm contains about 31 moles or 0.062 kg of H2 . With an enthalpy of combustion of 140 MJ/kg, this amounts to energy density 8.68 MJ/kg. So the sum of the energy added by compression and the enthalpy of combustion give a total energy density of 8.68 + 0.39 9.1 MJ/kg. This is over 60% higher than the actual result indicating that the ideal gas law is badly violated at very high pressure. The energy density of gasoline is approximately 30 MJ/L, a factor of ∼5 greater than the energy density of compressed hydrogen.

37.17 Compute the speed of a fragment of metal released

from the rim of the example flywheel described in the box in Example 37.4. Compare to the speed of a bullet fired by a handgun. v = ωr, r = 0.25 m and ω ≈ 5 800 s−1 , so v ≈ 1 450 m/s.

The initial speed of a bullet fired by a handgun is typically below 400 m/s. v is more comparable to the muzzle speed of a bullet fired by a high power rifle.

37.18 For the conditions described in Example 37.4 com-

pute the moment of inertia of the flywheel, its angular momentum, and kinetic energy of rotation. Assuming that the car loses kinetic energy only to wind resistance (with a drag coefficient of cd = 1/3) and that the flywheel’s energy can be converted to kinetic energy of forward motion at 90% efficiency, find how long the flywheel could keep the car moving at 100 km/h.

Finally, show that left/right imbalance of approximately ±6% in the weight on the tires is sufficient to supply the torque needed when the car tilts forward 10◦ in 0.36 s.

For a solid disk, the moment of inertia I = MR2 /2 = 5 kg(.25 m)2 /2 0.156 kg m2 .

At an angular velocity of ω 5 800 s−1 , the angular momentum is L = Iω = (5 800 s−1 )(0.156 kg m2 ) 910 kg m2 /s. The flywheel’s kinetic energy is Ekin = Iω2 /2 = 2.64 MJ. The vehicle loses energy to wind resistance at the rate given by eq. (2.27). The speed is held fixed at v = 100 km/h 28 m/s, the cross sectional area of the vehicle is A = (1.4 m)(1.6 m) = 2.24 m2 , so dEair 1 1 = cd Aρv3 = (1/3)(2.24 m2 )(1.28 kg/m3 )(28 m/s)3 10.3 kJ/s . dt 2 2 Thus the flywheel could keep the vehicle in motion at 100 km/h for T = Ekin /(dEair /dt) = (2.63×106 J)/(10.3×103 J/s) ∼ 4.3 minutes .

As described in Box 37.4, the change in angular momentum produced by at 10◦ forward tilt in 0.36 s exerts a torque of τ 450 Nm about a horizontal axis perpendicular to the car’s direction of motion. A counterbalancing torque can be produced by shifting the load on the car’s tires. Normally the ground exerts a force mg/2 on each pair of tires (left and right) to counteract the force of gravity. Suppose that the alignment of the car shifts so that one side experiences a force mg(1 − ǫ)/2 and the other experiences a force

882

mg(1 + ǫ)/2. These forces produce a force-couple and generate a torque τ′ =

mg w mg w 1 (1 + ǫ) − (1 − ǫ) = mgwǫ , 2 2 2 2 2

where w is the width of the car. For a car of mass 1 800 kg and width 1.6 m, in order to produce a counterbalancing torque τ′ = 450 Nm, ǫ = 2τ/mgw = 2(450 Nm)/(1 800 kg)(10 m/s2 )(0.8 m) 0.063.

37.19 For the Maxwell BCAP0310 ultracapacitor described

in Example 37.5, the capacitance is 310 F, mass is 0.06 kg, maximum voltage is 2.85 V, and internal resistance is r = 2.2 mΩ. Consider connecting the ultracapacitor to an external load with resistance RL . Compute the maximum possible instantaneous power delivered to the load when the capacitor is fully charged by optimizing over RL .

The internal and load resistance add in series so the total resistance is R = RL + r, and the current in the circuit is I = V/(RL + r).

The power delivered to the load is P = RL I 2 =

RL V2 .

(RL + r)2

To maximize with respect to RL , differential by RL , set equal to zero and solve for RL , with the result RL = r. The maximum power is then Pmax = rV 2 /(2r)2 = V 2 /4r. For V = 2.85 V and r = 2.2mΩ, Pmax 0.92 kW.

37.20 Suppose that the Maxwell BCAP0310 ultracapacitor

described in Example 37.5 used “conventional” capacitor technology. Specifically, assume that it can be described essentially as a parallel plate capacitor of area A, plate separation d, and dielectric constant ǫ = kǫ0 rolled into a spiral to fit into a cylinder of radius R = 16.5 mm and height H = 60 mm. First compute the plate separation d in terms of k. Then estimate the insulator/dielectric’s breakdown voltage in terms of k, assuming that the breakdown voltage sets the limit V ≤ 2.85 V quoted in the box. Ordinary dielectrics have k in the range 1–100. Are the distance scales and dielectric properties of the BCAP0310 consistent with your expectations for a conventional capacitor?

The volume can then be expressed as V = πR2 H ∼ Ad. We parameterize the capacitance in terms of A, d, and an effective dielectric constant, k, C ∼ kǫ0 A/d. These two equations allow us to solve for d in terms of R, H, C, and k, all of which are known except for k p √ d ∼ R ǫ0 kπH/C ∼ 1.23 k nm .

At a separation √ d(k), 2.85 V corresponds to an electric field of Emax ≈ 2, 200/ k MV/m.

1 nm is about ten times the diameter of an atom, a remarkably small scale for device structure. Ordinary insulators have breakdown voltages of hundreds of MV/m or less. Both the separation d and the breakdown voltage surpass the expectations for conventional materials, so we expect that the BCAP0310 uses novel technologies.

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Problems

inductor with inductance 1.5µH. What size capacitor should you choose for your oscillator?

The circuit is described by eq. (38.1) with R = V(t) = 0, LQ̈ +

Figure 38.27 For Problem 38.2.

which is the equation of a harmonic oscillator. The general √ solution is of the form Q(t) = Q0 cos(ωt + φ), where ω = 1/ LC. We use the conditions that Q(0) = CV and Q̇(0) = I(0) = 0 to set φ = 0 and Q0 = CV. Then the current I(t) = Q̇(t) is given by r C V sin ωt , I(t) = − L

Problems 38.1

Thomas Edison’s first DC power plant on Pearl St. in New York City produced roughly 100 kW at 110 V (DC), which was distributed over two copper wires spanning distances of order 1 km. Suppose resistive transmission losses were limited to 5%. Estimate the diameter of the copper wires necessary to distribute this electric power. Compare to the wire diameter if the same conditions were served by a single-phase 20 kV AC distribution line.

where the minus sign indicates that the charge on the capacitor is decreasing. For ν = 880 kHz, LC = 1/ω2 = 1/((2π)(880×103 ))2 = 3.27 × 10−14 , and if L = 1.5 µH, C = (3.27 × 10−14 )/(1.5 × 10−6 ) 0.022 µF.

38.4

The relevant formula is eq. (3.41), which relates the fractional power lost to the load power, voltage, resistivity, and the length and cross sectional area of the wires. Solving for the area, A=

PL ρl .

V 2 (Plost /PL )

Here ρ is the resistivity of copper (see Table 3.1), ρ = 1.68 × 10−8 Ω m.

For the DC case, (105 W)(1.68 × 10−8 Ω m)(2 × 103 m) = 5.6 × 10−3 m2 (110 V)2 (0.05) √ This corresponds to a wire diameter of d = 4A/π = 8.4 cm!

For the AC case (at VRMS = 20 kV), the only difference is to replace 110 V by 2 × 104 V. This reduces the wire diameter by a factor of 110/20 000, giving a diameter of 0.046 cm.

38.3

U(t) =

[T] Suppose a capacitor C, initially charged to a voltage V, is connected to an inductor L. Show that the current in the resulting LC circuit √oscillates, I(t) = I0 sin ωt.

Find I0 and show ω = 1/ LC. Sketch the current in the circuit and the voltage across the capacitor as functions of time. An (idealized) circuit like this without resistance would slowly radiate energy away as electromagnetic radiation with frequency ν = ω/2π. Suppose you wanted to construct an oscillator to produce radiation with a frequency of ν = 880 kHz. You have an

L 2 1 2 Q̇ + Q .

2 2C

√ Substituting Q(t) = Q(t) cos ωt, where ω = 1/ LC, from the solution to Problem 38.3, and keeping only the leading term in Q,

[T] Suppose a mass m is connected to a rigid support by a spring with spring constant k. A frictional force −b ẋ is applied by a damper that provides viscous friction, and an external force F(t) may also be applied (see Figure 38.27). Show that Newton’s laws give an equation of motion for the mass that is precisely analogous to the equation for an LRC circuit, eq. (38.1). Identify the electromagnetic analogs of m, k, b, and F(t).

Newton’s laws give m ẍ = −kx − b ẋ + F(t). If we identify the variable x(t) with the charge Q(t), then with the identifications m ↔ L, b ↔ R, k ↔ 1/C, and F(t) ↔ V(t), Newton’s law for a damped harmonic oscillator maps onto the equation for an LRC series circuit.

[T] Suppose a small resistance R is added in series to the LC circuit of Problem 38.3. Assume that Q(t) has the form Q(t) = Q(t) cos ωt, where Q(t) is a slowly changing function that would √ be constant if there were no resistance and ω = 1/ LC was determined in Problem 38.3. Compute the energy stored in this circuit U(t) as a function of Q(t) in the approximation that ˙ ≪ ωQ. Then compute the average (over a cycle Q of oscillation) rate of energy loss in the resistor U̇, also in terms of Q. Combine these results to show that U̇ = −RhUi/L and therefore U̇ = U0 e−Rt/L , when R is small.

The energy stored in the circuit is

A=

38.2

1 Q = 0, C

U(t) ≈

1 2 Q .

2C

Evaluating I 2 R in the same fashion, R 2 I 2 R ≈ Rω2 Q sin2 ωt = 2 U sin2 ωt .

L

38.5

Equating dU/dt to −I 2 R as required by eq. (38.1), U̇ = −2(R/L)U sin2 ωt. Averaging over a period of oscillation, hsin2 ωti = 1/2, so dU/dt = −RhUi/L, as required.

[T] Consider an AC circuit with only a capacitor of capacitance C, driven by an AC voltage V(t) = V0 cos ωt. Compute the average h|P|i of the absolute value of the rate of energy transfer back and forth from the power source to the capacitor. Compare to the apparent (= reactive) power for this system.

The charge on the capacitor is Q(t) = CV = CV0 cos ωt. The capacitor goes from no charge to fully charged (Q = CV0 ) in time π/2ω, so the average rate of energy transfer is the average power h|P|i =

1 Q2 2ω = CV02 ω .

2C π π

This can also be computed from h|IV|i. The apparent power, however, is Papparent = VRMS IRMS = V0 I0 /2 = CV02 ω/2 , so the two disagree by a factor of 2/π = 2h| cos ωt sin ωt|i.

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38.6

38.7

Repeat the analysis that led to eq. (3.40) for a load with impedance described by a power factor cos φ.

Show that the fraction of power lost compared to real power consumed in the load is approximately given by eq. (38.12), when resistive losses are much smaller than power consumed in the load.

Fluorescent light bulbs have significant impedance.

Measurements on a compact fluorescent bulb with luminosity equivalent to a 40 W incandescent indicate that it consumes 9 W real power. Measurements of RMS current and voltage, however, indicated that it drew an apparent power of 14 VA. What is the bulb’s power factor? If the RMS voltage is 110 V, what is the bulb’s resistance? What is its inductance?

The power factor cos φ is defined by the ratio of the real power to the apparent power, so cos φ = 9/14 = 0.64. Given the RMS voltage and the apparent power, the RMS current is IRMS = 14 VA/110 V 0.127 A. The real power is Preal = 2 IRMS R, so R = 9 W/(0.127 A)2 560 Ω. The reactive power is Papparent sin φ = (14 W)(0.77) 10.8 VAR, and the reactance X = 2 ωL is given by X = Papparent /IRMS 10.8 W/(0.127 A)2 667 Ω.

Since ω = 2π(60) s−1 377 s−1 , L = 667 Ω/377 s−1 1.77 H.

38.8

[H] A 150 kV (RMS) power supply is providing 300 MW of (real) power at 60 Hz to a load with power factor cos φ = 0.8. What capacitance would you have to install in parallel with the load to restore the power factor to one? Given a large number of individual 310 µF capacitors rated to a maximum of, say, 1100 VRMS , how would you construct the needed capacitance by grouping them in parallel and in series?

If the power factor is 0.8, then the reactive power drawn by the load is Preactive = sin φVapparent = tan φVreal = (0.6/0.8)(300) = 225 MW. Capacitors in parallel supply reactive power Preactive = −V 2 /|Z| (the minus sign indicates that capacitors add rather than draw reactive power), where Z = −i/ωC. Solving for C, Preactive 225 × 106 W C= = = 2.6 × 10−5 F = 25 µF ωV 2 (2π)(60)(150 × 103 V)2

If you gang n p capacitors in parallel the capacitance is Cgang = n p C, but these groups must be put in series in order to reduce the voltage across each group to 1 100 V. This requires n s = 150 000/1 100 136 groups in series, and reduces the capacitance of the whole device to Ctot = n p C/n s = n p (310 µF)/(136) = 25µF Solving for n p (25)(136)/310 11. So to achieve the necessary capacitive reactance it is necessary to array 136 groups of capacitors in series, each group containing ≈ 11 of the supplied 310 µF capacitors. No wonder that “capacitor banks” used to compensate for small power factors are large objects.

38.9

[T] Show that the stator magnetic field of a three-phase generator has a constant magnitude and rotates with constant angular frequency. You can assume that the magnetic fields produced by the stator windings A, B, and C, are BA = êA sin ωt, BB = êB sin(ωt − 2π/3), and BC = êC sin(ωt − 4π/3), where êA,B,C are unit vectors rotated by 120◦ with respect to one another. Note: if you are comfortable with manipulating complex numbers, this may be the simplest approach to take, though the problem can be solved without complex numbers.

38.10 [T] A three-phase generator delivers voltages VA (t) =

V0 cos ωt, VB (t) = V0 cos(ωt + 2π/3), and VC (t) = V0 cos(ωt − 2π/3). First, assume that all three phases are connected to identical, purely resistive loads. Show that the total power delivered is 3V02 /2R, independent of time. Next, assume that the loads have impedance |Z|eiφ , and show that the total power remains time independent. What is the role of the power factor, cos φ?

For the resistive case, the instantaneous power delivered to each load is P(t) = I 2 (t)R, where I(t) = V(t)/R, so the total power is P(t) =

3V02 V02 2 cos ωt + cos2 (ωt − 2π/3) + cos2 (ωt + 2π/3) = , R 2R

where the quantity in parentheses evaluates to 3/2 independent of t. (Use the identity cos2 x = (1 + cos 2x)/2 for a start.) When the resistance is replaced by impedance the power is still I 2 (t)R, but IA (t) = (V0 /|Z|) cos(ωt − φ), etc. so

V02 2 cos (ωt − φ) + cos2 (ωt − φ − 2π/3) + cos2 (ωt − φ + 2π/3) R 3V 2 3V02 R = cos2 φ 0 , = 2R 2|Z|2

P(t) =

So the real power delivered to the load drops by cos2 φ.

38.11 The rating of each generator at the Three Gorges Dam

(TGD) in China is quoted as 778 MVA of apparent power (see §38.2.3) produced at 20 kV and 50 Hz. Given that the quoted real power is 700 MW, what power factor has been assumed? The TGD generator rotors turn at 75 rpm. Assuming a three-phase stator winding, as in Figure 38.9, what is the pole structure of the rotors? What is the current in the stator windings?

The ratio of real to apparent power defines the power factor, so cos φ = 700/778 = 0.900. The rotor frequency is given by ν = ν0 /n, where ν0 is the line frequency and 2n is the number of poles on the rotor. So 2n = 2ν0 /ν = 2(60)(50)/75 = 80 poles.

Each of the three phase produces 778/3 = 260 MVA at VRMS = 20 kV, so IRMS = (260 × 106 J/s)/(2 × 104 V) 13 000 A .

38.12 [TH] Suppose a synchronous generator is supplying

power at (RMS) voltage V to a load characterized by impedance Z0 = R0 + iX0 = z0 eiφ0 , when an inductive load with negligible resistance Z1 = iωL is added in parallel with the original load. Show that the real power delivered by the generator does not change. What is the change in reactive power? Similarly show that if a purely resistive load Z1 = R1 is added in parallel to the original load, the reactive power in the circuit does not change and compute the change in the real power.

When the two circuit elements are put in parallel, the voltage across each of them must be V, so the current through Z0 remains V/Z0 no matter what is added in parallel. Thus neither the real nor reactive power drawn by the impedance Z0 changes. If an inductance is added in parallel, the current through the inductance is V/(iωL). This is 90◦ out of phase with the voltage, so it draws no real power, but it adds a new reactive power ∆Preactive = V 2 /ωL. If a resistance is added in parallel, the current through it is V/R,

913

Problems

Figure 38.28 For Problem 38.13.

which is in phase with the voltage, so it draws real power V 2 /R and the reactive power remains that which is drawn by the original impedance.

38.13 [T] Suppose, as in Figure 38.28, a generator is supply-

ing (time-averaged) power PL to a purely resistive load √ R. The voltage at the generator is V1 (t) = 2V1 cos ωt.

The power is carried to the load along a single conductor with inductance L and negligible resistance. V2 (t) is the voltage at the end of the conductor, where it connects to the load. This is an example of an LR circuit and can be analyzed using complex impedance as described in §38.2.2. Show that the power delivered to the load is PL = V12 cos2 φ/R and that V2 (t) = √ 2V2 cos(ωt − φ) with V2 = V1 cos φ, where the phase difference φ between the voltages satisfies tan φ = ωL/R. Combine these results to show that PL = V1 V2 sin φ/ωL, thereby deriving eq. (38.16).

From the figure, I = V1 /Z where Z = R + iωL, or in terms of time dependent, real quantities, √ 2V1 I(t) = cos(ωt − φ) , |Z| where tan φ = ωL/R (or sin φ/ωL = cos φ/R) The voltage V2 (t) is given by V2 (t) = I(t)R =

√ R √ 2V1 cos(ωt − φ) = 2V2 cos(ωt − φ) |Z|

where V2 = V1 cos φ, and the power delivered to the load PL is PL = hI 2 (t)iR = V12 R/|Z|2 = V1 V2 sin φ/(ωL) , where we have used R/|Z| = cos φ and ωL/|Z| = sin φ.

38.14 A 200 kV transmission line is being designed to carry

power 400 km from one power plant to another as part of a transmission network. The line has inductive reactance per unit length ωL′ = 0.44 Ω/km. Assuming that the maximum phase difference that can be tolerated between the sending and receiving ends of the line is 25◦ , what is the maximum power that can be sent on this line?

P12 = V1 V2 sin δ12 /X12 . Here X12 = (0.44 Ω/km)(400km) = 176 Ω, sin δ12 < 0.42 and V1 ≈ V2 = 2 × 105 V, so P12 < (2 × 105 V)2 (0.42)/(176 Ω) 96 MW .

38.15 A 345 kV transmission line consisting of three

conductors has a resistance per unit length of R′ =

R/l = 0.0351 Ω/km, and an inductive reactance per unit length of ωL′ = ωL/l = 0.371 Ω/km, as described in §38.4.1. Assess the utility of this line for transmission of three-phase power over a distance of 200 km.

Assume that the line delivers (real) power PL = 350 MW to a purely resistive load. Estimate the fraction of power lost in transmission and the effect on the power factor of line impedance.

The total line resistance is R = (0.0351Ω/km)(200 km) = 7Ω.

The line reactance is X = (0.371Ω/km)(200 km) = 74Ω. The power delivered to the load is PL = RL I 2 = RL V 2 /Z 2 , where Z = p (RL + R)2 + X 2 . These allow us to solve for the load resistance RL , toZ 2 = (RL + R)2 + X 2 = (V 2 /PL )RL = αRL .

Rearranging this equation and dropping R2 compared to 2RRL , yields a quadratic equation for RL , R2L − (α − 2R)RL + X 2 = 0 so

p 1 RL = (α − 2R + (α − 2R)2 − 4X 2 .

2

Substituting values we find RL = 308 Ω, and p Z = (308 + 7)2 + 742 = 324 Ω

Then the power factor is given by cos φ = (R + RL )/Z = 0.97.

The fraction of power lost to line resistance is Plost /(Plost + PL ) = R/(R + RL ) = 7/315 = 2.2%.

38.16 [H] Suppose a power plant is supplying 300 MW of real

power over a 150 kV (RMS) transmission line to a load that has an impedance characterized by a phase φ = 26◦ (corresponding to a power factor of cos φ = 0.9 lagging). The (three-phase) power is supplied over lines with inductive reactance and resistance per unit length ωL′ = ωL/l = 0.49 Ω/km and R′ = R/l = 0.11 Ω/km respectively. Assume that an equal fraction of the load is supplied by each phase. Compute the power factor seen at the busbar of the generator as a function of the length l of the line. How long a transmission line would reduce the power factor to 0.75? [Hint: First compute the impedance of the load.]

The real power at the load is Preal = V 2 Rload /|Z|2load , with Rload /|Z|load = cos φ = 0.9. Substituting for Z and solving for R, Rload = V 2 cos2 φ/Preal = (150 × 103 )2 (0.81)/(300 × 106 ) 61Ω , and Xload = tan φRload = 30Ω. Now add in the line impedance Z = (61 + 0.11l) Ω + (30 + 0.49l)i Ω. The power factor seen by the generator is cos φ′ = Re(Z)/|Z|. If cos φ′ = 0.75, then tan φ′ = 0.88, so that 30 + 0.49l tan φ′ = = 0.88 .

61 + 0.11l Solving for l = 60 km.

38.17 A load consists of a large number N of identical resis-

tors R (e.g. incandescent lights, toasters,. . . ) in parallel.

Suppose they are powered by an AC generator with RMS voltage V. Compare the line losses if (a) the load is connected to the generator by single-phase power, or (b) the load is divided in three and each group is connected (in a wye configuration) to a single phase of three-phase power. Assume that all the conductors have

914

resistance R0 and ignore line inductance. Finally, suppose the N resistors are connected in parallel to a DC generator. Compare the line losses if they are powered at the same level as the previous two cases.

The power delivered to the load is P = VRMS IRMS , where IRMS is the current in a circuit, and the losses per line are Plost = 2 rIRMS , where r is the line resistance. For single-phase AC there is one circuit and two lines (feed and return), so the lost power 2 is Plost (1 − phase) = 2rP2 /VRMS . For three-phase power there are three circuits each delivering 1/3 of the power, so IRMS = P/(3VRMS ). Three-phase power requires three lines (assuming no return needed), so Plost (3 − phase) = 3rP2 /(3VRMS )2 = Plost (1 − phase)/6 .

The DC case has the same lost power as the single-phase AC case.