Energy environment and climate 2nd edition wolfson solutions manual

Energy Environment and Climate 2nd Edition Wolfson Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/energy-environment-and-climate-2nd-edition-wolfson-solutions-manual/

ENERGY, ENVIRONMENT, AND CLIMATE, Second Edition CHAPTER 8: Energy from Earth and Moon

QUESTIONS 1.) Tidal effects on the Earth’s core, and radioactivity left over from the formation of the solar system are the two fundamental sources of geothermal energy.

2.) The water may not boil (much) at the lower temperatures of a binary geothermal plant, but the working fluid should boil, and its vapor will turn turbines.

3.) Ground-source heat is mostly solar, not geothermal.

4.) The environmental impacts include depletion or contamination of water sources; land subsidence; earthquakes; emission of GHG, and sulfur and nitrogen compounds; industrialization of natural areas; and noise pollution.

5.) Two factors are: (i) loss of water due to steam extraction, and (ii) cooling of the source, or heating of the environment.

6.) Estuaries are good breeding and feeding habitat for many creatures. They are often shallow, with high surface area in vegetation as well as land, so pollutants can get trapped easily. Dams shift flows, salinity, sediment, and the motion of larger creatures.

7.) The ultimate source is gravitational interaction of the Earth–Moon system (with a little help from the Sun).

8.) No, although they might contribute substantially in a few localized regions where geothermal or ocean energy resources are concentrated.

EXERCISES 1.) Earth has surface area = 4πR 2 = 4π ( 6.38 106 m ) = 5.11014 m 2 , so 2

energy flow rate =

power 87  10−3 W  area =  5.1 1014 m 2 = 4.4  1014 W = 44 TW .

2 area m

2.) To find the high temperature TH of the heat engine, first take the geothermal gradient

dT 25 C dT 25 C over the depth dx =1 km: T = =  dx =  1 km = 25 C = 25K. dx km dx km

© 2012 by W. W. Norton & Company, Inc. — pg. 1

This is the temperature difference across the geothermal well: T = TH − TC , so the high temperature at the bottom of the well is TH = T + TC = 25 K + 295 K = 320 K . (a) A heat engine operating between these two temperatures has a maximum theoretical ΔT 25 K efficiency e = = = 7.8% .

TH 320 K (b) The total power available is 10% e Pgeothermal = 0.01 (0.078) 44 TW = 0.34 TW (from Exercise 1). available geothermal power 0.34 TW = = 2.1% total power required 16 TW

dT , so the energy flow rate per unit area is dx H dT 3.37 W  25 K  8.4  10−2 W =k = = = 84 mW/m 2 .

3 2   A dx m K  10 m  m

3.) Recall Equation 4.1: Conductive heat flow H = kA

4.) (a) The area of the city is A = R2, where 2R = 15 km: A = π ( 7.5 103 m ) = 1.77  108 m 2 .

2

The heat emitted by the cooling granite is Q = mcT, where mass m = density • volume, and the volume = A x. The mass of granite is 2700 kg m =  Ax = (1.77 108 m2 ) 4 103 m = 1.911015 kg , so the energy that could be m3  J  19 extracted is: Q = mcΔT = 1.911015 kg  840  50 K = 8.02 10 J = 80 EJ . kg·K   (b) Power = energy per time, so this geothermal resource could last for 19 eQ 0.05 (8.02 10 J ) y   a time = = = 4.0 108 s   = 12.7 y .

9 7 P 10 10 J/s  π 10 s  (c) The rate of heat flow by conduction is dT 3.37 W 25 K H =A = 1.77 108 m2 ) 3 = 1.49 107 W . ( dx m K 10 m This is far lower than the rate of power use by the city, by a factor of

P 10  109 W = = 672 , H 1.49  107 W

so the resource is not renewable under these circumstances.

5.) Recall that e =

ΔT TH − TL . = TH TH

© 2012 by W. W. Norton & Company, Inc. — pg. 2

(a) For the geothermal source, TH =150C + 273 = 423 K and ΔT 123 K ΔT = TH − TL = 423 K − 300 K = 123 K , so e = = = 29% .

TH 423 K (b) For the magma, TH =1,000C + 273 = 1273 K and ΔT 973 K ΔT = TH − TL = 1, 273 K − 300 K = 9,733 K , so e = = = 76% .

TH 1273 K The hotter source yields a more efficient heat engine.

6.) The available geothermal power to a house of area A is

Pgeothermal =

power 87 10−3 W A= 150 m 2 ) = 13 W . ( 2 area m

The geothermal floor supply could provide

Pgeothermal geothermal energy geothermal power 13 W = = = = only about 0.4% of the home's energy needs home's power needs Phome 3.5  103 W home’s needs.

7.) First, get the temperatures in Kelvin, as usual: TH = 60C + 273 = 333 K (a) Coefficient of power (COP) =

(b) Pused =

Prequired COP

=

and

T = 50C = 50 K

1 TH 333 K = = = 6.7 (so the efficiency e = 15% ) e ΔT 50 K

15 kW = 2.2 kW : This is less because Earth preheated the system.

6.7

8.) (a) The rate of heat extraction is Pth =

Pe e

=

900 MW = 12.9 GW 0.07

(b) Power = energy per time, so

time =

E 35 1018 J y   = = 2.7 109 s   = 87 y 9 7 Pth 12.9 10 J/s  π 10 s 

© 2012 by W. W. Norton & Company, Inc. — pg. 3

9.) COPfall = 4 =

heat max

and COPspring = 2.5 =

heat min

. Assuming heatmin = heatmax electricmin electric max (because the temperatures were the same in those 2 months), we can easily find the relationship between electric max /electric min : heat max = 4 electric min = heat min = 2.5 electric max electric min =

2.5 electric max = 62% electric max 4

So, your energy use in spring is 62% higher than in fall, due to ground chilling.

10.) TH =170C + 273 = 443 K and T = 170C −90C =80C = 80 K, so the efficiency is ΔT 80 K e= = = 18% .

TH 443 K 11.) (a) Following Ex.8.2, the gas furnace requires fuelfurnace 1 1 f furnace = = = = 1.18 MBtu for every MBtu of electric energy supplied. heat supplied e f 0.85 The heat pump delivers 3.5 times as much heat as it uses electricity. Taking into account the limited efficiency of the power plant supplying electricity, the heat pump’s ratio of fuel used to 1 1 heat supplied is f heat = = = 0.92 MBtu for every MBtu of electric energy COP  ee 3.5  0.31 pump supplied. f 1.18 The gas furnace uses 28% more energy than the heat pump: furnace = = 1.28 . f heat pump 0.92 (b) Now compare the costs to heat the house.Delivery of 1 MBtu of heat with the gas furnace costs f furnace  $3.85=1.18  $3.85= $4.53 .

Using Table 3.1, we can see that electricity costs over eight times more than energy from gas:

cost of electricity =

$0.11  103 kWh   0.293MW  h   = $32 per MBtu kWh  MW  h   MBtu

  so delivery of 1 MBtu of heat with the heat pump costs  f heat = 0.92  $32 = $29.71 .  pump  The heat pump costs over six times as much to heat the house, even though it uses less energy.

(Would it make sense to power this heat pump with a domestic windmill?)

© 2012 by W. W. Norton & Company, Inc. — pg. 4

12.) The kinetic energy is ½ mv2. The mass in a cube of seawater is m = V  , where the volume V = x3, so m = x3 and: 2

1 1 3  1, 030 kg  2.5 m  3 (a) Kinetic energy = x3  v = (1 m )    = 3.2 10 J = 3.2 kJ 3 2 2 m s   

(b) Rate of flow of energy =

That is the power, so the

dE dE dx E 3.2  103 J 2.5 m =  = v =  = 8  103 W dt dx dt x 1m s

power 8 103 W = = 8 103 W/m 2 .

2 area (1 m )

(Note that this is about eight times the intensity of noon sunlight—water is dense!)

ARGUE YOUR CASE 1.) As in Example 2, the heat pump delivers four times as much heat as it uses electricity.

If the power plant supplying electricity has an efficiency of 33%, however, the heat pump’s ratio 1 1 of fuel used to heat supplied is f heat = = = 0.76 .

COP  e 4  0.33 pump So, the heat pump uses more like three quarters the fuel of a conventional system all told, not one quarter.

2.) Possible responses: Pro: Every form of green energy, for which the total carbon cost is less than the carbon emissions displaced, should be developed. If geothermal meets these criteria, and if investment in geothermal is more cost effective than investment in other green energy systems, then go for it.

Con: Geothermal energy is not the solution to our energy needs. The energy quality, density, and quantity are all far too low.

© 2012 by W. W. Norton & Company, Inc. — pg. 5