Introduction to Renewable Energy for Engineers Kirk D. Hagen Solution Manual

Type:

Solution Manual

Resource:

Introduction to Renewable Energy for Engineers

Edition:

1st Edition

Author(s):

Kirk D. Hagen

Chapter 1 Introduction to Renewable Energy

1

1.1

Problem statement Calculate the kinetic energy of a 0.145-kg baseball with a velocity of 80 milh. Express the answer in units ofkJ. Diagram

---)o~

v = 80 mi/h

Assumptions 1. Mass and velocity are constant.

Governing equations

Calculations Converting 80 milh to rnls, we have 80 milh x

1 rnls = 35.76 rnls 2.2369 milh

KE= V2 (0.145 kg)(35.76 m/s) 2 = 92.7 J = 0.0927 kJ

Solution check No errors are detected.

Discussion A Major League pitcher can throw a fast ball 100 milh.

2

1.2

Problem statement A satellite in orbit about the earth has a photovoltaic solar panel that faces directly into the sun for 10 hours per day. How much solar energy is incident on the panel during this period of time if the panel's surface area is 5m2?

Diagram

]solar= 1366 W/m2

Assumptions 1. Solar heat flux is1366 W/m 2• Governing equations £solar = ]solar A /).(

Calculations Esolar =

(1366 W/m2)(5 m 2)(10 h X 3600 s!h)

= 2.46 X 108 J = 246 MJ

Solution check No errors are detected.

Discussion Only a small fraction of this energy is absorbed by the solar panel, so the electrical output energy would be much less than 246 MJ.

3

1.3

Problem statement The sun radiates energy at a rate of3.9 x 1026 W. How much energy has the sun radiated since the beginning of the Common Era? Diagram

-6/

I

psolar

= 3.9 X 1026 W

~

'

Assumptions 1. Sun's radiation has been constant since the beginning of the Common Era. 2. About 2016 years have passed since the beginning of the Common Era. Governing equations £solar

= psolar /).f

Calculations Esolar

= (3.9 X 1026 W)(2016 y X 365 day/y X 24 hlday X 3600 slh) =2.5 X 10 37 J

Solution check No errors are detected.

Discussion Scientists estimate that the sun has around 6 billion years oflife left before it becomes a white dwarf star. The sun's radiative power will not change significantly for a very long time.

4

1.4

Problem statement A residential gas-fired furnace is rated at 150,000 Btu/h. Convert this heating rate to units ofkW and MW. Diagram

150,000 Btu/h

Assumptions 1. Furnace rating is constant.

Governing equations N/A Calculations 150,000 Btu/h X

1w 1 kW = 44.0 kW = 0.0440 MW X 3.4121 Btulh 1000 W

Solution check No errors are detected.

Discussion Furnaces in the US, which are rated in English units of Btu/h, are sized by finding the total heat loss from the home for worst case winter conditions.

5

1.5

Problem statement If a 40-W incandescent light bulb transfers 90 percent of its power as heat instead of light, what is the rate ofheat transfer from the light bulb in units ofBtu!h?

Diagram

40W

Assumptions 1. Ninety percent of power is transferred as heat.

Governing equations NIA

Calculations 0.90 X 40 W = 36 W 36 w X 3.4121 Btu!h

123 Btu!h

w Solution check No errors are detected.

Discussion That most of the power from an incandescent light bulb is heat can be felt by touching its surface.

6

1.6

Problem statement The metabolic rate (body heat production rate) of an adult engaged in swimming is approximately 2140 Btu/h. Convert this heat production rate to units ofkW. Diagram

Assumptions 1. Heat production rate is constant.

Governing equations N/A Calculations 2140 Btulh X

1w X 1 kW = 0.627 kW 3.4121 Btulh 1000W

Solution check No errors are detected.

Discussion The human body produces heat as a function of the kind of physical activity engaged in.

For example, a person sitting in a chair produces about 356 Btulh whereas a person bicycling at 15 km/h produces about 1425 Btu/h.

7

1.7

Problem statement A coal-fired power plant steadily generates 10 MW of electrical power. In units ofkJ, how electrical energy is generated during a 24-hour period?

Diagram

/lOMW

DDDDDD Assumptions 1. Power generation is steady.

Governing equations £elect

= pelect 1'1t

Calculations Eelect

= (10 X 106 J/s) X (3600 s/h) X 24 h = 8.64 X lOll J = 8.64 X 108 kJ

Solution check No errors are detected.

Discussion Assuming that the power plant operates steadily year round, the electrical energy generated in one year is Eelect

= (8.64 X 108 kJ/day) x (365 day/y) = 3.15 x lOll kJ/y

8

1.8

Problem statement The available power in wind is proportional to turbine blade diameter squared and the cube of wind speed. For a given wind turbine, if the available power is 3 kW for a wind speed of 4 mi/h, what is the available power for a wind speed of 12 mi/h?

Diagram

D

v

Assumptions 1. Wind speeds are steady.

Governing equations P wind = c D 2 v3 ( c = proportionality constant)

Calculations P wind I_- C D2 VI 3

where vi= 4 milh, Pwind,I = 3 kW, v2 = 12 mi/h. Dividing equations, and solving for P wind, 2, we have

= (3 kW)(12 mi/h/4 mi/h) 3

9

= 81 kW Solution check No errors are detected.

Discussion The actual equation for available power in wind is

where pis air density and A is the area swept out by the turbine blades (A= rrD 2/4). Thus, we see that c = ~ p n/4.

10

1.9

Problem statement The average solar heat flux just outside the earth's atmosphere is 1366 W/m 2 • Convert this quantity to units ofBtu/h·fe. In units ofMJ, how much solar energy is intercepted by a 12 m 2 surface at this location during a time period of six hours?

Diagram

A= 12m2

/solar= 1366 W/m2

Assumptions 1. Solar heat flux is constant.

Governing equations £solar = /solar A /1t

Calculations 1366 W/m 2 X 3.4121 Btulh X

w

(1 mf = 433 Btulh·fe (3.2808 ft) 2

2 2 £solar= (1366 W/m )(12 m )(6 h X 3600 slh)

= 3.54 X 108 J = 354 MJ Solution check No errors are detected.

Discussion Within the earth's atmosphere, the solar heat flux is less than 1366 W/m2 due to scattering, absorption and other atmospheric effects.

11

1.1 0

Problem statement For a time period of30 minutes, the solar radiation incident on a photovoltaic solar panel is 825 W/m 2• Assuming that one-tenth of the solar radiation is converted to electrical energy, what is the required surface area of the panel if 1.6 kWh of electrical energy is required?

Diagram

A

/solar= 825 W/m

2

Assumptions 1. Solar radiation is constant.

Governing equations £elect= O.lJsolar A llf

Calculations Converting 1.6 kWh to J, 1.6 kWh X 3.6 X 106 J/kWh = 5.76 X 106 J Solving for A, A=

5. 76 X 106 J (0.1)(825 W/m2)(30 minx 60s/min)

= 38.8 m 2 Solution check No errors are detected.

12

Discussion A solar panel efficiency of 10 percent is conservatively low, so a panel surface area of 38.8 m 2 is large enough to supply at least 1.6 kWh of electrical energy.

13

1.11

Problem statement A biomass power plant uses wood chips for fuel. The heat of combustion (heat released by the complete burning of a given amount of fuel) of wood chips is approximately 9000 Btu/lbm. Convert this quantity to units ofkJ/kg. (Note: 1 kg= 2.2046 lbm).

Diagram

/

9000 Btu/Ibm

Assumptions 1. The wood chips combust completely.

Governing equations None (unit conversion exercise only) Calculations 9000 Btu/lbm X 2.2046 lbm/kg X 1.05506 kJ/Btu = 2.093 X 104 kJ/kg = 20.93 MJ/kg Solution check No errors are detected.

Discussion Each type of biomass fuel has a different heat of combustion value. For example, the heat of combustion of sugar cane is approximately 18 to 19 MJ/kg.

14

1.12

Problem statement The hot water from a geothermal source well steadily transfers 60 kW of heat to a working fluid that passes through a turbine. In units of Btu, how much energy is transferred to the working fluid during a 48-hour period?

Diagram

working fluid ~

60kW turbine

H,O

i

heat exchanger

source well

Assumptions 1. Heat transfer is steady.

Governing equations

E = Ql1t Calculations E = (60 kW)(48 h x 3600 s/h) = 1.04 x 107 kJ

Converting to units of Btu, 1.04 X 107 kJ X 1 Btu/1.05506 kJ = 9.83 X 106 Btu

15

Solution check No errors are detected.

Discussion As illustrated in the diagram, heat transfer between the geothermal water and the working fluid occurs in a heat exchanger, a device that facilitates the transfer of heat from one fluid to another.

16

1.13

Problem statement An underwater turbine powered by an ocean current generates 2 kW of electrical power.

This turbine is one of 24 identical devices in an ocean power plant. Assuming round the clock operation, how much energy in units of kWh does this power plant generate in a 30dayperiod?

Diagram

2kW

ocean water

•

Assumptions 1. Turbines generate power steadily around the clock.

Governing equations

E = P !J.t Calculations E = (2 kW)(30 day x 24 hiday x 3600 s/h) = 5.18 x 106 kJ

Converting to units ofkWh, 5.18 x 106 kJ x 1 kWh/3600kJ= 1440kWh(perturbine)

17

So, the total energy is 24 turbines x 1440 kWh/turbine= 3.46 x 104 kWh Solution check No errors are detected.

Discussion A more realistic assumption to make is that the turbines operate a fraction of the 30 day period. The plant capacity factor, CF, introduced in chapter 1, takes this into account.

18

1.14

Problem statement The available power in an ocean wave is proportional to the square of wave height. For a given ocean wave, if the available power is 3.8 MW for a wave height of 1.2 m, what is the available power for a wave height of 4.5 m?

Diagram

Assumptions 1. A vail able power is proportional to wave height.

Governing equations

P = c H 2 (c =proportionality constant) Calculations

P I =cHI 2 P2 =cH22 Dividing the first equation by the second equation and solving for P 2 , we obtain

= (3.8 MW)(4.5 m/1.2 m) 2 = 53.4 MW Solution check No errors are detected.

Discussion The available power increased by a factor of approximately 14.

19

1.15

Problem statement The initial cost of a wind turbine for a residential application is $14,000. If the annual energy production of the turbine is 10,500 kWh/y, and the energy cost is $0.1 0/kWh, what is the simple payback?

Diagram

10,500 kWh/y

Assumptions 1. Energy production and energy cost are constant.

Governing equations

SP=

IC (AEP)(EC)

Calculations

SP=

$14,000 (1 0,500 kWh/y)($0.1 0/kWh)

=~

20

Solution check No errors are detected.

Discussion Cost incentives, if offered by local governmental agencies, could lower the simple payback significantly.

21

1.16

Problem statement The initial cost of a micro hydro system is $1200. If the annual energy production of the system is 1800 kWh/y, and the price of electricity charged by the electric utility is $0.095/kWh, what is the simple payback?

Diagram

Assumptions 1. Energy production and energy cost are constant.

Governing equations

SP=

IC (AEP)(EC)

Calculations

SP =

$1200 (1800 kWh/y)($0.095/kWh)

= 7.02 y Solution check No errors are detected.

Discussion Micro hydro power plants are low cost renewable energy systems that can be installed in back yards or remote locations to power lights and small appliances.

22

1.17

Problem statement Rework Problem 1.15 if a loan at 4.0 percent was used to finance $5000 of the wind turbine and the annual operation and maintenance cost is $150/y.

Diagram

10,500 kWh/y

Assumptions 1. Energy production, energy cost, interest rate and maintenance costs are constant.

Governing equations

SP=

IC (AEP)(EC) - (IC)(FCR) - AOM

Calculations

SP=

IC (AEP)(EC) - (IC)(FCR) - AOM $14,000 (1 0,500 kWh/y)($0.1 0/kWh) - ($5000)(0.040/y) - $150/y

= 20.0 y

23

Solution check No errors are detected.

Discussion The simple payback in Problem 1.15 is 13.3 y. The reason that the simple payback is 20.0 yin this problem is because $5000 of the $14,000 total cost was financed at 4.0 percent interest and annual operations and maintenance cost was included.

24

1.18

Problem statement A hydroelectric plant with a power generation capacity of 12 MW costs $20 million to install. The fixed charge rate is 7.0 percent, the annual operation and maintenance cost is assumed to be 1.0 percent of the initial cost, and the levelized replacement cost is averaged over an expected 40-year lifetime. If the capacity factor for this plant is 0.38, what is the cost of energy?

Diagram

/12MW DDDDDD Assumptions 1. All costs are constant, interest rate is constant and A OM is 1.0 percent of initial cost.

Governing equations COE = (JC)(FCR) + LRC + AOM + AFC AEP

Calculations The plant has a power capacity of 12 MW, so the amount of energy that the plant can produce in one year if it runs constantly is £max= (12 MW)(8760 h/y)

= 1.051 X 105 MWhly The capacity factor is 0.38, so the actual annual energy production of the plant is AEP = (0.38)(1.051 X 105 MWhly) = 3.995 X 10

4

MWhly

For a hydroelectric plant, the annual fuel cost, AFC, is zero. The LRC and AOM are

25

LRC = ($20 X 106)/(40 y)

= $5.00 X 105/y AOM = (0.01/y)($20 X 106)

Substituting values, we obtain COE = UC)CFCR) + LRC +ADM+ AFC AEP

= ($20 X 106){0.07/y) + $5.00 X 105/y 3.995 X 104 MWh/y

= $47.56/MWh = $0.0476/kWh

Solution check No errors are detected.

Discussion Most large hydroelectric power plants can produce electrical power in the $0.040/kWh to $0.060/kWh range.

26

Chapter 2 Solar Energy

27

Section 2.2

Practice!

1. Find the declination angle and apparent solar time for Richmond, Virginia, longitude 77.4 7 o, latitude 37.53 o, on September 21 at 2:00pm.

Solution 284 8 = 23.45° sin[ N + x 360°] 365

= 23.45°sin[(264 + 284)/365 x 360°] = -0.202° D = (N -81) 360o 365

= (264- 81)(360°) 365 = 180.49°

ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D) = 9.87 sin[(2)(180.49°)]- 7.53 cos(180.49°)- 1.4 sin(180.49°)

= 7.71 min AST = LST + (4 minldeg)(LSTM- Long) + ET

= 840 min+ (4 min/deg)(75o - 77.4r) + 7.71 min = 837.8 min

28

2. A collector with a tilt angle of 50° is located in Nashville, Tennessee, longitude 86.78 o, latitude 36.17 o, and elevation 182 m. The collector faces due south, and the foreground is gravel. Find the total solar irradiance on the collector at 10:30 am on December 31.

Solution Using equations (2.1) through (2.14) in the text, we obtain the following results: D = 270.25° ET= 1.383 min AST= 644.3 min H= -18.93° /J1 =27.75° a 1 = -19.66° = 20.33 o

e

In= 858.6 W/m 2 Is= 42.87 W/m2 2 /R = 12.82 W/m !tot= 914.3 W/m

2

3. The collector in Problem 2 has a surface area of30 m2 • Find the total solar power incident on the collector at 10:30 am. Also, find the insolation and the total solar energy incident on the collector during the day.

Solution Qsolar

= ]tot Acollector

= 2.74 X 104 w = 27.4 kW Calculating the total solar irradiance at hourly intervals throughout the day and integrating under the curve using the trapezoidal rule, we obtain for the insolation H = 22.1 MJ/m2 Esolar

= H Acollector

29

Section 2.3

Practice!

1. Find the total solar irradiance for a horizontal photovoltaic solar panel in Santa Fe, New Mexico, at 9:00 am, 1:00pm and 4:00pm on October 21. Assume a foreground reflectivity ofp = 0.2.

Solution The results are summarized in the following table:

Quantity

9:00am

1:00pm

4:00pm

D (deg)

210.1

210.1

210.1

ET(min)

15.83

15.83

15.83

AST(min)

552.0

792.0

972.0

H(deg)

-42.00

18.00

63.00

/3 1 (deg)

28.18

39.62

14.02

a 1 (deg)

-48.01

23.12

64.04

e(deg)

61.82

50.38

75.97

(W/m2)

434

627

173.7

/ 5 (W/m )

2

67.1

71.8

52.3

IR (W/m2)

0

0

0

!tot (W/m2)

501

698

226

/0

30

2. The solar panel in Practice Problem 1 has a surface area of 16 m 2 • If the efficiency of the cells in the panel is 0.18, find the electrical energy generated by the panel on October 21.

Solution By plotting the total solar irradiance at hourly intervals and integrating under the curve, we obtain an insolation of H = 17.78 MJ/m2 • £elect= '1cell H A

= 51.2 MJ X

kWh 3.6MJ

= 14.2 kWh

31

3. Find the optimum tilt angle for a photovoltaic solar panel in Pittsburgh, Pennsylvania, on December 21. Also, find the insolation corresponding to this tilt angle. The panel faces due south, and the foreground is covered with snow.

Solution · By plotting the insolation as a function of panel tilt angle (see the graph below), we see that the optimum tilt angle is

25.00 24.00 23.00 22.00

121.00

~

i 20.00 ·=

0

..5 19.00 18.00 17.00 16.00 15.00 w~m~~~~~~~w~ro~w~~~~

Panel tilt ancle (de~:)

As shown in the graph, the insolation corresponding to this tilt angle is

H = 22.6 MJ/m2

32

Section 2.4

Practice!

1. A gravel roof-mounted, south-facing flat plate solar collector with a glazing surface measuring 2.5 m x 5.5 m has a tilt angle of 60 o. The working fluid is a mixture of ethylene glycol and water with a specific heat of c = 3320 Jlkg·oC and a mass flow rate of0.075 kgls.

Assuming that 55 percent of the total solar irradiance is absorbed by the working fluid and that T. = 20°C, find the temperature of the working fluid at the outlet of the collector for conditions at 12:00 noon on February 21. The collector is located in Albany, New York, longitude 73.76°, latitude 42.65° and elevation 0 m (Albany borders the Hudson River, a sea level river).

Solution The total solar irradiance for the conditions given is /tot= 1011 W/m2 • Fifty five percent of the solar irradiance is absorbed by the working fluid, so the outlet temperature of the working fluid is

T = 0.55/tot A + T me 0

•

l

= (0.55)(1011 W/m2){2.5 m x 5.5 m) + 20°C (0.075 kg/s)(3320 J/kg·°C)

= 5o.rc

33

2. The aperture diameter of a dish concentrator is 1.9 m, and the concentration ratio is 1000.

The temperature of the working fluid in the receiver is 1250°C, and the temperature of the surroundings is 25°C. If the direct solar irradiance on the dish is 680 W/m 2, find the solar power incident on the concentrator, the heat flux at the receiver and the Camot efficiency.

Solution .

Qsolar

trf)2 = JD - -

4

= 1928 w

Q"rec = JDC = (680 W/m2)(1000)

= 6.80 x 105 W/m2 TL

'17th,ideal

= 1- T. H

=1-

(25 + 273)K (1250 + 273)K

= 0.80

34

3. A parabolic trough concentrator has an aperture width and length of 1.6 m and 24m, respectively. The average direct solar irradiance on the concentrator on a given day is 590 W/m 2 • The efficiencies of the concentrator, receiver, and electrical generator are 0.90, 0.82 and 0.94, respectively. Find the maximum possible electrical output power for a working fluid temperature at the receiver of 400 o C. The temperature of the surroundings is 15 o C. Solution

= 1-

(15 + 273)K (400 + 273)K

= 0.572

= (0.572)(0.90)(0.82)(0.94)

= 0.397

= (0.397)(590 W/m2)(1.6 m x 24m) = 8991 w

35

2.1

Problem statement For Denver, Colorado on December 19 at 10:00 am, find the declination angle and apparent solar time.

Diagram (Not applicable) Assumptions 1. None Governing equations 284 8 = 23.45° sin[ N + x 360°] 365 D = (N - 81) 3 60o

365

ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D) AST = LST + (4 minldeg)(LSTM- Long) + ET Calculations 284 8 = 23.45° sin[ N + x 360°] 365 = 23.45° sin[(353 + 284)/365 x 360°]

= -23.44°

D = (N -81) 360o

365 = (353 -

81) 360°

365

36

= 268.27° ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D)

= 9.87 sin[2(268.2r)] - 7.53 cos(268.2r)- 1.5 sin(268.2r) = 2.32 min Local standard time, LST, is LST= 10 h x 60 minlh = 600 min

Denver is in the mountain time zone, so LSTM = 105 o, and the longitude of Denver is 104.88 o west.

AST = LST + (4 minldeg)(LSTM- Long) + ET

= 600 min+ (4 min/deg)(105°- 104.88°) + 2.32 min = 602.8 min Solution check No errors are detected.

Discussion Apparent solar time differs from local standard time by (602.8 min- 600 min)= 2.8 min.

37

2.2

Problem statement For Atlanta, Georgia on March 2 at 2:30pm, find the declination angle and apparent solar time.

Diagram (Not applicable) Assumptions 1. None Governing equations 284 8 = 23.45° sin[ N + x 360°] 365 D = (N -81) 360o

365

ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D) AST = LST + (4 minldeg)(LSTM- Long) + ET Calculations 284 x 360°] 8 = 23.45° sin[ N + 365 = 23.45° sin[(61 + 284)/365 x 360°]

= -7.91° D = (N- 81) 360o

365 = (61 -

81) 360° 365

= -19.276° 38

ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D)

= 9.87 sin[2(-19.276°)]- 7.53 cos(-19.276°)- 1.5 sin(-19.276°) = -12.76 min

Local standard time, LST, is

LST= 14.5 h x 60 minlh = 870 min Atlanta is in the eastern time zone, so LSTM = 75°, and the longitude of Atlanta is 84.39° west.

AST = LST + (4 minldeg)(LSTM- Long) + ET

= 870 min+ (4 min/deg)(75° - 84.39°)- 12.76 min = 819.6 min

Solution check No errors are detected.

Discussion Apparent solar time differs from local standard time by (819 .6 min - 870 min) = -50.4 min.

39

2.3

Problem statement For Albany, New York on August 20 at 11:00 am, find the declination angle and apparent solar time.

Diagram (Not applicable) Assumptions 1. None Governing equations 284 x 360°] 365

8 = 23.45° sin[N +

D = (N- 81) 360o

365

ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D) AST = LST + (4 minldeg)(LSTM- Long) + ET Calculations 284 8 = 23.45° sin[ N + x 360°] 365 = 23.45° sin[(232 + 284)/365 x 360°] = 12.10°

D = (N- 81) 360o

365 = (232- 81) 360° 365

= 148.93 ° 40

ET = 9.87 sin(2D)- 7.53 cos(D)- 1.5 sin(D) 9.87 sin[2(148.93°)]- 7.53 cos(148.93°)- 1.5 sin(148.93°)

=

= -3.05 min

Local standard time, LST, is

LST= 11 h x 60 min/h = 660 min Albany is in the eastern time zone, so LSTM = 75o, and the longitude of Albany is 73.75r west.

AST = LST + (4 minldeg)(LSTM- Long) + ET = 660 min+ (4 min/deg)(75°- 73.75r)- 3.05 min

= 661.9 min Solution check No errors are detected.

Discussion Apparent solar time differs from local standard time by (661.9 min - 660 min) = 1.92 mm.

41

2.4

Problem statement A collector, located in Columbus, Ohio has a tilt angle of60o and faces due south. Ifthe foreground is concrete, find the total solar irradiance on the collector at 3 :00 pm on January 21.

Diagram

concrete

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations

284 8 = 23.45°sin[ N + x 360°] 365 D= (N-81)360o

365 ET = 9.87 sin(2D) -7.53cos(D) -1.5sin(D) AST = LST + (4 min/ deg)( LSTM- Long) + ET H= AST-720min 4min/ deg 42

sin(/31) = cos( L) cos(8) cos( H) + sin( L) sin( 8) cos ( a 1 ) =

sin(/31) sin( L) - sin( 8) cos(/31) cos( L)

cos( B) = sin(/31) cos(/32 ) + cos(/31) sin(/32 ) cos( a 1 - a 2 ) plp0 = exp(- 0.1184z)

= Aexp(--p

I

I

S

_B_J

Po sin(/31)

DN

= CI DN [ 1+ cos(/32 ) ] 2

Calculations

284 x 360°] 8 = 23.45°sin[ N + 365 = 23.45° sin [ 21 + 284

x 360°]

365

= -20.14° D = ( N - 81) 360o 365

43

= (21 - 81) 360° 365 =-59.18°

ET = 9.87 sin(2D) -7.53cos(D) -1.5sin(D) = 9.87 sin[(2)(- 59.18 °)] - 7.53 cos(- 59.18 °) - 1.5 sin(- 59.18 °) = -11.26 min Columbus is in the eastern time zone, so LSTM = 75o. A time of 3 :00 pm gives LST = 15 h x 60 minlh = 900 min.

AST = LST + (4 min! deg)( LSTM- Long) + ET = 900 min+ (4 min/deg)(75°- 82.98°) + (-11.26 min) = 856.8 min H= AST-720min

4min! deg = 856.8 min - 720 min 4 min/deg = 34.20°

sin(,B, ) = cos( L) cos(8) cos( H) + sin( L) sin(8)

= 0.3738

[J, = sin- 1 (0.3738) = 21.95°

cos ( a, ) =

sin(/31 ) sin( L) - sin( 8) cos(/31) cos( L)

44

= sin(21.95°) sin(39.98°)- sin(-20.14°) cos(21.95 °) cos(39.98 °) = 0.8224

a 1 =COS-I (0.8224) = 34.68° cos(()) = sin(A ) cos(/32 ) + cos(/31 ) sin(/32 ) cos( a 1 - a 2 )

= 0.8474

The elevation of Columbus is z = 275 m (0.275 km), so

plp0 = exp(- 0.1184z) = exp[ -(0.1184)(0.275 km)]

= 0.9680 From Table 2.2 we see that for January 21, A= 1230 W/m2 , B = 0.142 and C = 0.058.

Hence, the direct normal irradiance is

IDN

= Aexp ( - p

B ) Po sin(A)

= (1230 W/m2) exp[(-0.9680)(0.142)] sin(21.95 °) = 851.5 W/m

2

The direct irradiance is

= (851.5 W/m2)(0.8474) = 721.6 W/m2 Scattered irradiance is 45

= (0.058)(851.5 W/rn2)[1 + cos(60°)] 2 = 37.0 W/rn

2

The foreground reflectivity for concrete is p = 0.3. The reflected irradiance is

= (851.5 W/rn2)(0.3)[0.058 + sin(21.95°)][1 - cos(60°)] 2

= 27.6 W/rn2 The total irradiance is

= (721.6 + 37.0 + 27.6) W/rn

2

= 786.2 W/rn 2 Solution check No errors are detected.

Discussion The total irradiances for a range of panel tilt angles are as follows: 2

fi2~

ltot (W/rn )

0 30 45 60 75

368 654 743 786 782

46

2.5

Problem statement Consider a south-facing collector in Springfield, Illinois. If the collector tilt angle is 70°, find the total solar irradiance on the collector at 10:30 am on November 21. The foreground is a gravel roof.

Diagram

gravel

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4.

Calculations The calculations follow the same order as those in Problem 2.4. The results are summarized below.

Constants:

Long= 89.62° L = 39.70° z = 0.170 km LSTM = 90 o (central time zone) November 21: N = 325, A= 1221 W/m2 , B = 0.149, C = 0.063 (J2 = 70° (panel tilt angle),p = 0.15 (gravel foreground) a2 = oo (panel faces due south) 47

Calculated quantities:

b= -20.44° D = 240.66° ET = 13.43 min LST= 630 min AST = 644.9 min H = -18.76°

fJi =27.38° a 1 = -19.84° e= 19.sr IoN= 888.6 W/m 2 10 = 837.3 W/m2 15 = 37.6 W/m 2 IR = 22.9 W/m2 ]tot= 897.8 W/m

2

Solution check No errors are detected.

Discussion A graph of total irradiance as a function of panel tilt angle reveals an optimum tilt angle of approximately 62 o.

1000

900

-. ;:;-

E

!II

800

u

c 700 '6

_;

..

-;;;

600

f=

500

400 0

10

W

W

40

~

60

Panel tilt ancle (dq)

48

M

00

90

2.6

Problem statement A horizontal collector on a rooftop is located in Austin, Texas. Find the total solar irradiance on the collector at 11:30 am on July 12.

Diagram

l l l l l Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

3. Given time is local standard time.

Governing equations The governing equations are the same as those in Problem 2.4.

Calculations The calculations follow the same order as those in Problem 2.4. The results are summarized below.

Constants: Long= 97.75° L = 30.25° z = 0.149 krn LSTM = 90 o (central time zone) July 12: N= 193, A= 1085 W/m2, B = 0.207, C= 0.136 /]2 = 0 o (panel tilt angle)

49

Calculated quantities:

b= 2L9r D = 110.46° ET = -5.24 min LST = 690 min (if daylight savings time was used, LST = DST- 1 h) AST = 653.8 min H = -16.56° fJ] = 73.01° a 1 = -64.75° = 16.99°

e

= 877.1 W/m2 10 = 838.9 W/m2 / 0N

Is= 119.3 W/m 2 IR = 0.0 W/m2 (because panel is horizontal) !tot= 958.1 W/m

2

Solution check No errors are detected.

Discussion A graph ofthe total solar irradiance on the panel for July 12 is shown below.

1000

900

800 i

400

>

300

200 600 700 BOO 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 Military time (h)

50

2. 7

Problem statement The collector in Problem 2.4 has a surface area of 24 m 2 • Find the total solar power incident on the collector at 1:00pm. Also, find the insolation and the total solar energy incident on the collector during the day.

Diagram

concrete

Assumptions

1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following: Qsolar

£solar

= /tot Acollector

= H Acollector

Calculations Using the governing equations in Problem 2.4, the total solar irradiance at 1:00pm is !tot= 1009.5 W/m

2

Thus, the total solar power incident on the collector at this time is

51

Qsolar = /tot Acollector

4

= 2.423 X 10

w = 24.23 kW

The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

1100 1000

-

900

"'E 800

! 700 Gl

u

I:

600

til

:ctil

500

-~ 400

... 300

iii

{:.

200 100 0 800

900

1000 1100 1200 1300 1400 1500 1600 1700 Military time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation

H = 22.8 MJ/m2 The total solar energy incident on the collector during this day is the insolation multiplied by the surface area of the collector, £solar = H Acollector

= 547.2 MJ

52

Solution check No errors are detected.

Discussion The collector cannot generate more than 54 7.2 MJ of energy during this day.

53

2.8

Problem statement The collector in Problem 2.5 has a surface area of 18m2 • Find the total solar power incident on the collector at 11 :00 am. Also, find the insolation and the total solar energy incident on the collector during the day.

Diagram

gravel

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following: Qsolar

= /tot Acollector

£solar = H Acollector

Calculations Using the governing equations in Problem 2.4, the total solar irradiance at 11:00 am is /tot= 908.4 W/m

2

Thus, the total solar power incident on the collector at this time is

54

Qsolar

= ]tot Acollector

= 1.635 X 104 w = 16.35 kW The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

1000 900

I II

800 700 600

u

;

500

~

400

] ~

300

:c

200 100 0

700

800

900 1000 1100 1200 1300 1400 1500 1600 1700 Military time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation H = 21.5 MJ/m2

The total solar energy incident on the collector during this day is the insolation multiplied by the surface area of the collector, £solar

= H Acollector

= 387.0 MJ

55

Solution check No errors are detected.

Discussion The collector cannot generate more than 387.0 MJ of energy during this day.

56

2.9

Problem statement The collector in Problem 2.6 has a surface area of 60 m 2 • Find the total solar power incident on the collector at 12:00 noon. Also, find the insolation and the total solar energy incident on the collector during the day.

Diagram

1 1 111 Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following: Qsolar

= ]tot Acollector

Esolar = H Acollector

Calculations Using the governing equations in Problem 2.4, the total solar irradiance at 12:00 noon is !tot= 983.5 W/m

2

Thus, the total solar power incident on the collector is Qsolar

= ]tot A collector

57

= 5.901 X 104 w = 59.01 kW The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

1100 1000 900 800 r;-

E 700

!.. 600 c u

•

'6

~ ;§

ii

500 400 300 200 100

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

1700

1800

1900

Milit•ry time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation

H= 29.4 MJ/m2 The total solar energy incident on the collector during this day is the insolation multiplied by the surface area of the collector, £solar

= H Acollector

= 1764 MJ

58

Solution check No errors are detected.

Discussion The collector cannot generate more than 1764 MJ of energy during this day.

59

2.10

Problem statement A manufacturing plant in San Jose, California has a roof-mounted photovoltaic solar panel that measures 16 m x 30 m. The panel is horizontal and has an efficiency of 0.18.

How much electrical energy does the panel generate on December 21?

Diagram

!

! ~

!

!

A=16mx30m

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

£solar

= H Acollector

£elect

= 11cell £solar

Calculations The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

60

550 500 450 400 1350

~

83oo c

.

] 250

;;;

~200

150 100 50 0 700

800

900

1000

1100

1200

1300

1400

1500

1600

1700

Milit•ry time(h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation

H= 10.11 MJ/m2 The total solar energy incident on the collector during this day is the insolation multiplied by the surface area of the collector, Esolar

= H Acollector = (10.11 MJ/m2)(16 m x 30m) = 4853 MJ

The total electrical energy generated by the solar panel during the day is Eelect

= IJcell Esolar = (0.18)(4853 MJ) = 873.5 MJ X 1 kWh 3.6MJ

= 242.6 kWh

61

Solution check No errors are detected.

Discussion The typical daily electrical energy consumption by a home in the US is about 30 kWh. The manufacturing plant has a higher electrical energy demand than a home, but our answer of243 kWh provides one means of comparison.

62

2.11

Problem statement A residential solar system in Montgomery, Alabama incorporates a photovoltaic solar panel with a tilt angle of60°. The roof-mounted panel, which measures 20m2, faces due south and has an efficiency of 0.16. How much electrical energy does the panel generate on October 21?

Diagram

gravel

A= 20m2

Assumptions 1. Roof is gravel.

2. Sky is clear.

3. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

£solar = H Acollector

£elect= Yfcell £solar

Calculations The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

63

1100 1000 900 BOO

1 700 ~ i 600 ~

. e

1

500

...~ 400 300 200

100 0

500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

1700

Military time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation H = 25.02 MJ/m2

The total solar energy incident on the collector during this day is the insolation multiplied by the surface area of the collector, £solar

= H Acollector

= 500.4 MJ The total electrical energy generated by the solar panel during the day is £elect

= 'lcell £solar = (0.16)(500.4 MJ)

= 80.06 MJ X 1 kWh 3.6MJ

= 22.24 kWh

64

Solution check No errors are detected.

Discussion The typical daily electrical energy consumption by a home in the US is about 30 kWh. Hence, this photovoltaic system supplies over two-thirds of this amount.

65

2.12

Problem statement A classroom building at a university in Raleigh, North Carolina has an array of photovoltaic solar panels on its roof for augmenting electrical power from the grid. The panels face due south and have a tilt angle and efficiency of 60° and 0.22, respectively.

How much electrical energy do the panels generate on December 21 for a total panel surface area of 600 m 2?

Diagram

Assumptions 1.

2.

3.

4.

Roof is gravel.

Sky is clear.

No panel blocks incident solar radiation on any other panel.

All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

£solar = H Acollector Eelect = '1cell Esolar

Calculations The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

66

1000 900 800

"'E 700 ......

s 600 II

u

1:

:0"'

500

~ .!:::

400

~

300

..

-;

200 100 0 700

800

900

1000

1100

1200

1300 1400

1500 1600

1700

Military time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation H = 22.4 MJ/m2

The total solar energy incident on the panels during this day is the insolation multiplied by the total surface area of the panels, Esolar

= H Acollector

The total electrical energy generated by the solar panels during the day is Eelect = 'lcell Esolar

= (0.22)( 1.344 X 104 MJ)

= 2957 MJ X 1 kWh 3.6MJ = 821 kWh

67

Solution check No errors are detected.

Discussion Photovoltaic solar panels on roof tops of college and university buildings are becoming more common as these institutions incorporate green policies into their operations.

68

2.13

Problem statement Derive Equation (2.18).

Diagram

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations

I S = C I DN [ 1 + cos j32 2

J

Calculations To find the optimum panel tilt angle, /J2,opt' we perform a maximization ofthe total solar irradiance, !tot· Substituting the last three equations into the first equation and taking the derivative of !tot with respect to /]2, and setting the result to zero, we obtain

69

dlrot = cos /31 cos /32 cos( a 1 - a 2 ) - sin /31 sin /32 - C sin /32 + p ( C + sin f3t) sin /32 = 0 2 2 d~ where the term / 0 N has divided out. After simplification, this relation reduces to cos A_ cos( a 1 - a 2 ) tan fJ2 = ------=--'------'---'--=-C + sinfJ1 - p (C +sin A_)

2

2

Solving for {J2, we obtain

fJ2 =tan

_1[

2cosA_ cos(a1 - a 2 ) ] C(l- p) + (2- p)sinA

which is Equation (2.18).

Solution check No errors are detected.

Discussion

It is important to note that this result applies only at an instant of time when the solar altitude angle, {J 1, and solar azimuth angle, a 1, have unique values at a given time of day.

The optimum panel tilt angle is also a function of the azimuth angle, a2, the ratio of diffuse radiation on a horizontal surface to direct normal radiation, C, and foreground reflectivity, p.

70

2.14

Problem statement Using Equation (2.18), calculate the optimum tilt angle for a solar panel in Salt Lake City, Utah at 1:00pm on January 21. The panel faces a south-southwest direction, and the foreground is snow.

Diagram

snow

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are those needed to find /] 1 (see Problem 2.4) plus the following:

_ _1[ 2cosfJ1 cos(a1 - a 2 ) ] f3:2ot-tan ,op C(l-p)+(2-p)sinfiJ Calculations The results are summarized below.

Constants:

Long= 111.88° L = 40.75° z = 1.288 km LSTM = 105 o (mountain time zone) January21: N=21, C=0.058 71

p = 0.80 (snow foreground)

a2 = +22.50° (panel faces south-southwest) Calculated quantities:

b= -20.14° D = -59.18° ET = - 11.26 min LST= 780min AST = 741.2 min H= 5.306° () = 14.66° /], = 28.91° a 1 = 5.692° Using Equation (2.18), the optimum panel tilt angle is

a

_

Plot-tan ,'P

_1[ 2cosA cos(a1 -a2 ) ] C(1-p)+(2-p)sinfJ1

=tan-! ll

2cos(28.91 °)cos(5.692 ° - 22.50°) Jl 0.058(1 - 0.80) + (2 - 0.80)sin(28.91 °)

= 70.6° Solution check No errors are detected.

Discussion The answer seems reasonable because the sun is low in the sky in January. This tilt angle applies only to 1:00 pm on January 21. The optimum tilt angle would vary throughout the day as the sun moves across the sky.

72

2.15

Problem statement A photovoltaic solar panel is to be installed on the roof of a home in Phoenix, Arizona.

The roofhas a 4 on 12 pitch, and one side of the roof faces south-southeast. For November 21, find the optimum angle ofthe solar panel with respect to the roof.

Diagram

Assumptions 1. Roof is gravel.

2. Sky is clear.

3. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

Calculations The results are summarized below.

Constants:

Long= 112.1 o L = 33.45° z = 0.331 km LSTM= 105° (mountain time zone) November 21: N = 325, C = 0.063 p = 0.15 (gravel foreground, assumed)

73

a2 = -22.50° (panel faces south-southeast) In order to find the optimum panel tilt angle for November 21, the total solar insolation as a function of tilt angle is graphed below.

22.80

22.60

22.40

'i:

~ 22.20

~ c .2

~ 22.00

..

..5

21.80

21.60

21.40 40

60

50

70

80

Panel tilt anele (dee)

As shown in the graph, the optimum tilt angle is approximately 61 o, but this angle is measured with respect to the horizontal. The roofhas a 4 on 12 pitch, which means that the angle of the roof with respect to the horizontal is tan- 1 (4/12) = 18.4 o. Thus, the optimum angle of the solar panel with respect to the roof is

Solution check No errors are detected.

Discussion As indicated in the graph, the insolation is a weak function of panel tilt angle. For example, a tilt angle of75° yields an insolation of21.5 MJ/m2, about 5 percent lower than the insolation at a tilt angle of 61 o.

74

2.16

Problem statement An office complex in San Antonio, Texas has a roof-mounted photovo1taic solar panel that measures 24 m x 36 m. The panel is horizontal and has an efficiency of 0.22. How much electrical energy does the panel generate on July 21?

Diagram

~ A~24mx36m Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

3. Given time is local standard time.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

£solar = H Acollector

£elect= 'lcell £solar

Calculations The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

75

1000 900 800

r:r 700 E

! 600 Cl

. u

c '0

500

~ 400

!

~

300 200 100 0

600 700 800 900 1000 1100 120013001400 1500 1600170018001900 2000

Military time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation

H= 29.1 MJ/m2 The total solar energy incident on the panel during this day is the insolation multiplied by the total surface area of the panel, £solar

= H Acollector = (29.1 MJ/m2)(24 m x 36m)

The total electrical energy generated by the solar panel during the day is £elect = '1cell £solar

= (0.22)( 2.514 X 104 MJ) = 5531 MJ X 1 kWh 3.6MJ = 1536 kWh

76

Solution check No errors are detected.

Discussion It is becoming more common for office buildings to incorporate photovoltaic solar panels into their roof tops to augment electrical power from the grid.

77

2.17

Problem statement A home owner in Sacramento, California wishes to install photovoltaic solar panels on his south-facing roof. The pitch and reflectivity ofthe roof are 12 on 12 and 0.2, respectively. If the efficiency of the panels is 0.18, how much electrical energy can the home owner expect to generate on December 31 ifhe mounts the panels flat on the roof?

The total surface area of the panels is 70 m2 • Diagram

p=0.2

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

Esolar = H Acollector

Eelect

= 1'/cell Esolar

Calculations The graph below, generated by calculating the total irradiance at hourly intervals, shows the total irradiance throughout the day.

78

1000 900 800 700

1 600

!

~

u

;

'6

sao

i 400

!

~ 300

•·

200 100 0

700

800

900

1000

1100

1200

1300

1400

1500

1600

1700

Military time (h)

Integrating under this curve using the trapezoidal rule, we obtain the insolation H = 20.1 MJ/m2

The total solar energy incident on the panels during this day is the insolation multiplied by the total surface area of the panels, £solar = H Acollector

= 1407 MJ The total electrical energy generated by the solar panel during the day is £elect= 'lcell £solar

= (0.18)(1407 MJ)

= 253.3 MJ X 1 kWh 3.6MJ = 70.4 kWh

79

Solution check No errors are detected.

Discussion A 12 on 12 roof pitch constitutes a 45° tilt angle for the solar panels. While this tilt angle is not optimum, the panels blend in with the roof line creating an aesthetically pleasing installation.

80

2.18

Problem statement Work Problem 2.17 for the optimum panel tilt angle for December 31.

Diagram

Assumptions 1. Sky is clear.

2. All quantities pertinent to the total solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

£solar = H Acollector

£elect= IJcell £solar

Calculations We find the optimum panel tilt angle by graphing insolation as a function of tilt angle.

As seen in the graph below, the optimum panel tilt angle is approximately 66 o.

81

21.60 21.50 21.40 21.30 21.20

::;~ 21.10

~

c 21.00 0

..

0 20.90 ·o:

..

.E

20.80 20.70 20.60 20.50 20.40

so

45

55

60

65

70

75

Panel tilt angle (dec)

The graph below, generated by calculating the total irradiance at hourly intervals for the optimum tilt angle computed above, shows the total irradiance throughout the day.

1000 900 800 N'

E 700

~ 600 II

u

c Ill .:g

500

-~ 400

... 300

Iii

~

200 100

0 700

800

900

1000

1100

1200

1300

Military time {h)

82

1400

1500

1600

1700

Integrating under this curve using the trapezoidal rule, we obtain the insolation H = 21.5 MJ/m2

The total solar energy incident on the panel during this day is the insolation multiplied by the total surface area of the panel, £solar = H Acollector

= 1505 MJ The total electrical energy generated by the solar panel during the day is £elect= 'flcell £solar

= (0.18)(1505 MJ) = 270.9 MJ X 1 kWh 3.6MJ = 75.3 kWh Solution check No errors are detected.

Discussion As shown in the first graph, solar insolation is a fairly weak function of panel tilt angle, but a maximum is clearly indicated.

83

2.19

Problem statement After cost incentives totaling $6500 by state and local governments, the initial cost of a photovoltaic system for a residence is $11,000. If the annual energy production of the solar panels is 5750 kWh/y, and the energy cost is $0.1 0/kWh, what is the simple payback?

Diagram (Not applicable) Assumptions 1. All quantities are constant.

Governing equations

SP=

IC (AEP)(EC)

Calculations The simple payback is

SP=

IC (AEP)(EC)

$11,000 (5750 kWh/y)($0.1 0/kWh)

= l2:Lx Solution check No errors are detected.

Discussion The cost incentives of $6500 were not used in the calculation because they are not part of the initial cost paid by the homeowner.

84

2.20

Problem statement Design a photovoltaic system for a residence in Tempe, Arizona. The south-facing roof has a surface area of 45m2 and a 5 on 12 pitch. State all your assumptions.

Diagram

Assumptions 1.

2.

3.

4.

5.

6.

7.

Solar panel is mounted flat on roof, giving a tilt angle of tan -I ( 5/12) = 22.6 o.

Sky is clear.

All quantities pertinent to the total solar irradiance are constant.

Foreground reflectivity is 0.2.

Solar panel efficiency is 0.18.

Solar panel occupies 85 percent of south-facing roof.

Design day is March 21.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

£solar

= H Acollector

Eeiect = 'lcell £solar

Calculations Using the assumptions above and the governing equations, we calculate the solar irradiation at hourly intervals for March 21, as shown in the graph below.

85

1100 1000 900 N"

E

800

~ 700 u

.,1: 600 .:c., 500 ... u

•!:::

-; {E.

...

400 300 200 100 700

800

900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 Military time (h)

From this graph, we obtain the solar insolation H = 26.9 MJ/m 2

The total solar energy incident on the panel during this day is the insolation multiplied by the total surface area of the panel, £solar = H Acollector

= 1029 MJ The total electrical energy generated by the solar panel during the day is £elect = '7cell £solar

= (0.18)(1029 MJ) = 185.2 MJ X 1 kWh 3.6MJ 51.4 kWh

86

March is a "transition" month in terms of solar radiation, so this result may be taken as an approximate daily electrical energy generation for any given day during the year.

Solution check No errors are detected.

Discussion To maintain a clean roofline, the solar panel was mounted flush with the roof. If the solar panel was mounted at a steep tilt angle, the insolation would be higher than the value calculated here.

87

2.21

Problem Statement A rover for gathering geological data on Mars has a photovoltaic solar panel that supplies electrical power to its sensors, motors, and other electrical components. The panel faces directly into the sun, measures 2.6 m x 1.2 m, and has an efficiency of 0.18. If the solar constant for Mars is 585 W/m 2, what is the electrical power capacity of this system?

Diagram

Assumptions 1. Mars does not have an atmosphere, so !tot= 545 W/m 2, the solar constant for Mars.

Governing equations psolar = ]tot Acollector

pelect = 1'/cell psolar

Calculations The solar power incident on the panel is psolar = ]tot Acollector

= (545 W/m2)(2.6 m x 1.2 m) = 1700W The electrical power generation of the panel is

88

pelect = Y/cell psolar

= (0.18)(1700 W)

=306W

Solution check No errors are detected.

Discussion Like Earth, Mars rotates on its axis. (One day on Mars is 24 hours, 39 minutes, 35 seconds in duration). Hence, the solar irradiation calculated here applies only to the moment when the rover's solar panel faces directly into the sun.

89

2.22

Problem statement In a domestic water heating system, an ethylene-glycol/water mixture flows through the collector at a mass flow rate of 0.065 kg/s. The glazing surface measures 3 m x 5 m, and 550 W/m 2 of solar power is absorbed by the fluid mixture. If the inlet temperature of the fluid mixture is 25°C, what is the outlet temperature?

Diagram

Assumptions 1. Fluid mixture absorbs 550 W/m2 of solar radiation.

2. Flow rate of fluid mixture is steady.

Governing equations

Q=me( I;, -I;) Calculations The specific heat of ethylene glycol/water mixture is approximately c = 3320 J/kg· °C. Solving for the outlet temperature of the fluid, we obtain

T =2_+T 0

•

me

l

90

(550 W/m 2)(3 m x 5 m) + 25°C (0.065 kg/s)(3320 J/kg·°C) 63.2°C Solution check No errors are detected.

Discussion The outlet temperature of the ethylene glycol/water mixture is 63.2°C (145.8°F), so the temperature of the water exiting the tank is limited by this temperature. Accounting for heat transfer effects, the water temperature at the tank exit is perhaps around 130°F, hot enough for domestic use, such as a dishwasher.

91

2.23

Problem statement A domestic water heating system in Los Angeles, California, incorporates a roof-mounted flat plate solar collector with a tilt angle of 60 o. The collector faces south-southeast, and the glazing surface measures 4.0 m x 6.2 m. the working fluid, which has a specific heat of c = 3150 Jlkg·°C, enters the collector at a temperature of21 oc and flows through the collector at a mass flow rate of 0.10 kg/s. If 50 percent of the total solar irradiance is absorbed by the working fluid, find the temperature of the working fluid at the outlet of the collector at 12:00 noon on April21.

Diagram

Assumptions 1.

2.

3.

4.

Sky is clear.

All quantities pertinent to the total solar irradiance are constant.

Flow rate ofworking fluid is constant.

Foreground is a gravel roof.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following: Qso/ar = ]tot A collector

92

Using the computational procedure illustrated in Problem 2.4, the total solar irradiation is ]tot= 810 W/m

2

Thus, the solar radiation incident on the collector is Qso/ar

= ]tot Acol/ector = (810 W/m2)(4.0 m x 6.2 m)

Only 50 percent of this radiation is absorbed by the working fluid, so we have

Q

~=-.-+I;

me

0.50 X 2.01 X 104 W + 21 °C (0.10 kg/s)(3150 J/kg·°C)

Solution check No errors are detected.

Discussion This temperature represents an instantaneous value for the conditions given.

93

2.24

Problem statement The direct solar irradiance on a dish concentrator is 600 W/m 2, and the dish aperture diameter is 1.80 m. The concentration ratio is 2500, the temperature of the working fluid in the receiver is 950°C, and the temperature of the surroundings is 25°C. Find the solar power incident on the concentrator, the heat flux at the receiver, and the Camot efficiency.

Diagram

In= 600 W/m 2 - - • • 0

Assumptions 1. Steady operation.

Governing equations

Q"rec =/DC TL

17th ,ideal

= 1- T, H

94

Solar power incident on the concentrator is

= 1527 w The heat flux at the receiver is Q"rec = JDC

= (600 W/m2)(2500) = 1.50 x 106 W/m2 = 1.50 MW/m2 The Camot efficiency is

TL

Tlth ,ideal

= 1- T, H

= 1 - (25 + 273) K (950 + 273) K = 0.756

Solution check No errors are detected.

Discussion In order to calculate the maximum electrical power capacity of this system, we would have to know the efficiencies of the concentrator, receiver and generator and use the relation

where Tloverall

= llth,ideal Tlcon Tlrec TJgen

95

2.25

Problem statement Consider a CSP plant with an array of 400 dish concentrators in Santa Fe, New Mexico.

The concentrators, which have an aperture diameter of2.75 m, are installed in a gravel field and have dual axis tracking. For 12:00 noon on January 21, find the total solar power incident on the concentrators and the heat flux at the receivers if the concentration ratio is 3200.

Diagram

Assumptions 1. Sky is clear.

2. All quantities pertinent to the direct solar irradiance are constant.

Governing equations The governing equations are the same as those in Problem 2.4 plus the following:

Q"rec = JDC Calculations Using the computational procedure illustrated in Problem 2.4, the direct solar irradiance lS

This value is based on a concentrator tilt angle of 56 o, the angle that maximizes the direct solar irradiance. An azimuth angle of 0 o was assumed. Thus, the solar radiation incident on one concentrator is

96

= 5999 w so the total solar radiation incident on all concentrators is (400)(5999 W) = 2.40 X 106 w = 2.40 MW The heat flux at the receivers is

Q"rec =/DC = (1010 W/m2)(3200)

Solution check No errors are detected.

Discussion Because the CSP system has dual axis tracking, the tilt and azimuth angles change throughout the day providing a maximum direct solar irradiance at all times.

97

2.26

Problem statement What is the maximum possible thermal efficiency of a concentrating solar power plant if the temperature of the working fluid at the receiver is 1400oC and the ambient air temperature is 20°C? Diagram

Assumptions 1. All parameters given in problem are constant.

Governing equations

TL

'lth ,ideal

= 1- T, H

Calculations The maximum thermal efficiency of the concentrating solar power plant is

TL

'lth ,ideal

= 1- T, H

= 1- (20+273)K (1400 + 273) K = 0.825

98

Solution check No errors are detected.

Discussion As shown in the diagram, a concentrating solar power plant is basically a heat engine that converts solar energy to electrical energy. Thermal efficiency is one efficiency of four efficiencies, concentrator, receiver and generator efficiency, that apply to concentrating solar plants.

99

2.27

Problem statement Consider a parabolic trough concentrator with an aperture width and length of 1.4 m and 26 m, respectively. During a given day, the average direct solar irradiance on the concentrator is 650 W/m2 • The efficiencies of the concentrator, receiver, and electrical generator are 0.91, 0.83, and 0.95, respectively. Ifthe temperature ofthe working fluid in the receiver is 600°C, find the maximum possible electrical output power. The temperature of the surroundings is 10°C. Diagram

10 = 650 W/m - - • • 0 2

Lap= 26m

Assumptions 1. All parameters given in problem are constant.

Governing equations

TL

lJth ,ideal

= 1- T, H

lJoverall

= lJth,ideailJcon lJrec lJgen

100

~feet = 'loverall Qeon Calculations The solar power incident on the concentrator is

= (650 W/m2)(1.4 m)(26 m)

The Camot efficiency is

TL

'lth,ideaf

= 1- T

H

= 1- (10+273)K (600 + 273) K

= 0.676 so the overall efficiency is lloverall

= 1lth,ideaf17eon llrec 'lgen = (0.676)(0.91)(0.83)(0.95)

= 0.481 Because the overall efficiency is based on Camot thermal efficiency, the maximum electrical output power is

~feet = 'loverall Qeon = (0.481)(2.37 X 10

4

W)

= 1.14 X 104 w = 11.4 kW

101

Solution check No errors are detected.

Discussion The actual thermal efficiency might be approximately 0.40, giving an overall efficiency of 0.287 and an electrical output power of 6.8 kW.

102

2.28

Problem statement A concentrating solar power plant is being designed to augment electrical power from the main grid to a small residential community. The plant is to consist of an array of parabolic dish concentrators with an aperture diameter of2.2 m. The efficiencies of the system are estimated to be: 11th = 0.30, llcon = 0.90, llrec = 0.82, llgen = 0.95

As a starting point in the design, an engineer uses 800 W1m2 as a direct solar irradiance averaged over a 12-hour period during which the sun shines on a typical day. The average electrical power consumption of a home in the United States is approximately 30 kWh per day. If the community has 1200 homes, how many dish concentrators are required to provide 25 percent of the power? Also, estimate the amount of land required for the concentrators.

Diagram

Assumptions 1. All parameters given in problem are constant.

Governing equations

~feet = 1lovera/l Qcon E elect = ~teet 1'1t Calculations The solar power incident on one concentrator is

103

= 3041 w The overall efficiency is TJovera/1

= 17th TJcon TJrec TJgen = (0.30)(0.90)(0.82)(0.95) = 0.210

The electrical output power of one dish is

~feet = TJoverall Qcon = (0.210)(3041 W) = 638.6 w so the electrical output energy of one dish for a 12-hour period is

E elect = ~feet /'1! = (638.6 W)(3600 slh x 12 h) = 2.759 X 107 J = 27.59 MJ = 27.59 MJ X 1 kWh 3.6MJ =7.66kWh Hence, the number of dishes required to provide 25 percent ofthe electricity for 1200 homes is

N = 0.25 x 1200 home x 30 kWh/day-home 7.66 kWh/day = 1175

104

To estimate the land requirement for the dishes, let's assume that each dish requires a space that is three times the aperture area of the dish so that no dish blocks the solar irradiance of adjacent dishes and to make room for maintenance. So, the total land area required is 2 Aland= 1175 X 3 X 1t(2.2 m) /4

= 1.340 x 10

4

m 2 x 1 acre 4047 m 2

= 3.31 acre Solution check No errors are detected.

Discussion If the concentrating solar power plant provided all the electrical power to the community, 4700 dishes would be needed, taking up a land area of 13.2 acre.

105

2.29

Problem statement A concentrating solar power plant with a generation-capacity of 1. 75 MW costs $2.50 million to install. The fixed charge rate is 7.5 percent, the annual operation and maintenance cost is assumed to be 1.0 percent ofthe initial cost, and the plant capacity factor is 0.20. If the levelized replacement cost is averaged over an expected 25-year lifetime, what is the cost of energy?

Diagram (Not applicable) Assumptions 1. All quantities are constant.

Governing equations

COE = (IC)(FCR) + LRC + AOM + AFC AEP Calculations The concentrating solar power plant has a power capacity of 1.75 MW, so the amount of energy that the plant can generate in one year is Emax = (1.75 MW)(8760 h/y)

= 1.533 X 104 MWhly The capacity factor is 0.20, so the actual annual energy production is AEP = (0.20)(1.533 X 104 MWhly)

= 3066 MWhly For a concentrating solar power plant, the annual fuel cost, AFC, is zero. The LRC and AOMare

LRC = ($2.50 X 106)/(25 y)

= $1.00 X 105/y

106

AOM = (0.01/y)($2.50 X 10 6) = $2.50 X 104/y Hence, the cost of energy is

COE = (IC)(FCR) + LRC + AOM + AFC AEP 6

= ($2.50 X 10 )(0.075/y)

+ $1.00 X 105/y+ $2.50 X 104/y

3066 Mwhly

= $101.92/MWh which, when converted to units of $/kWh is approximately

COE = $0.1 02/kWh Solution check No errors are detected.

Discussion The average cost of electricity in the United States is approximately $0.12/kWh, so this power plant is competitive with respect to this value.

107

Chapter 3 Wind Energy

108

Section 3.2

Practice!

1. A horizontal axis wind turbine operating at sea level has a blade diameter of 40 m. Find the maximum amount of power that the turbine can extract from the wind for wind speeds of 2 m/s and 8 m/s.

Solution

For a wind speed of2 m/s, we have

= lQ_ ~ (1.225 kg/m 3)n(40 m) 2/4(2 m/s) 3 27 2

= 3649 W = 3.65 kW For a wind speed of8 m/s, we have

= lQ_ ~ (1.225 kg/m 3)n(40 m) 2/4(8 m/s) 3 27 2

= 2.335 X 105 w = 234 kW

109

2. A small vertical axis wind turbine extracts 4.2 kW of power from a 7 m/s wind at sea level.

If the area swept out by the blades is 60 m 2, find the power coefficient for this wind turbine.

Solution

Solving for CP, we obtain

c = 2 pwind ,3max P

pAv

2(4.2 X 103 W)

= 0.33

110

Section 3.3.1 Practice!

1. AHA WT in Fargo, North Dakota, experiences winds at an average speed of 8 m/s. The power coefficient, gear box efficiency, and generator efficiency are 0.40, 0.82, and 0.94, respectively. If the turbine blade length is 26 m, find the electrical output power.

Solution Fargo, North Dakota has an elevation of z =274m, so the air density is

p = 1.225- 1.194 x w- 4 z = 1.225- (1.194 X lQ- 4)(274m) = 1.192 kg/m 3 The area swept out by the blades is

= 2124 m 2 The electrical output power is 1 3 Pelect =-n'P~V •t.

2 = 1 (0.40)(0.82)(0.94)(1.192 kg/m 3)(2124 m 2)(8 m/s) 3 2 = 1.998 X 105 w = 200 kW

111

2. Compare the electrical output power oftwo identical HAWTs subjected to the same wind speed, one operating in San Diego, California, (z = 0 m) and the other in Santa Fe, New Mexico (z = 2210 m) by calculating the output power ratio of the San Diego machine to the Santa Fe machine.

Solution P,lect ,SanDiego P,/ect ,SantaFe

f rypAv 3 ) SanDiego 3 ( f rypAv ) SantaFe

(

Other than air density, all quantities for both locations are equal. Thus, P,/ect ,SanDiego

PsanDiego

P,lect ,SantaFe

PsantaFe

1.225 kg/m3 1.225- (1.194 x w- 4)(2210m)

= 1.27

112

Section 3.3.2 Practice!

1. A H-VAWT operating in Boise, Idaho, has a rotor height and diameter of7.5 m and 6.0 m, respectively. The combined efficiency of the machine is 0.27. For a wind speed of9 rn/s, what is the electrical output power?

Solution Boise, Idaho has an elevation of z = 830 m, so the air density is p = 1.225- 1.194 X 10- 4 Z

= 1.225 - (1.194 x 10- 4)(830 m)

= 1.126 kglm 3 1 3 Pelect =-nnAv •tr

2

= 1 (0.27)(1.126 kglm 3)(7.5 m x 6.0 m)(9 rn/s) 3 2 = 4.99 X 103 w = 4.99 kW

113

2. An experimental wind machine in Seattle, Washington, has a combined efficiency of 0.18.

The machine is a V-VAWT with a rotor height and diameter of3.0 m and 7.0 m, respectively. Find the electrical output power for a wind speed of 12 m/s.

Solution We will assume that Seattle, Washington has an elevation of z = 0 m, so the air density is p = 1.225 kg/m3

The area of a V-VAWT is the V -shaped area swept out by the blades. Hence,

A= V2 (3.0 m x 7.0 m)

= 10.5 m 2 1 3 Pelect = -nl14v '/!"-'" 2

= 1 (0.18)(1.225 kg/m 3)(10.5 m 2)(12 m/s)3 2

= 2.00 X 103 w = 2.00 kW

114

3.1

Problem statement At a height of 4 m above the ground, the wind speed is 2.5 m/s. Estimate the wind speeds at heights of25 m and 50 m if the ground is covered with tall grass.

Diagram

~

v0 = 2.5 m/s

4m

~

tall grass

lllllllllllllll/1 tllllllllllllllllll/1

Assumptions 1. Friction coefficient for tall grass is a= 0.15.

Governing equations

Calculations Solving for v, the wind speed at a height of 25 m is

= (25 m/4 m) 0· 15 (2.5 m/s) = 3.29 m/s The wind speed at a height of 50 m is

=(50 m/4 m) 0.1 5 (2.5 m/s) = 3.65 m/s

115

Solution check No errors are detected.

Discussion If the ground was smooth (a = 0.1 0), the wind speeds for heights of 25 m and 50 m would be 3.00 m/s and 3.22 mls, respectively.

116

3.2

Problem statement In a large city with tall buildings, the wind speed at 8 m above the ground is 2 m/s. At what height is the wind speed approximately 5.5 m/s?

Diagram

v0 = 2 m/s )o

DD D Assumptions 1. Friction coefficient for a large city with tall buildings is a = 0.40.

Governing equations

Calculations Solving for height, H, we obtain

H~(~r H, = [(5.5 m/s)/(2 m/s)]

25

(8 m)

=lOOm

117

Solution check No errors are detected.

Discussion If the terrain consisted of a small town with tall tress (a = 0.30) instead of a large city with tall buildings, the height at which the wind speed is 5.5 rnls would be 233m.

118

3.3

Problem statement Find the elevations of Chicago, Denver, Las Vegas, Helena and Topeka, and calculate the air densities for each city.

Diagram (Not applicable) Assumptions 1. Air density varies linearly with elevation according to Equation (3.6).

2. Elevation values are taken at airports or are averages across the city.

Governing equations p = 1.225- 1.194 x 1o- 4 z

Calculations The results are summarized in the table below.

Elevation (m)

Air density (kg/m3)

Chicago

191

1.202

Denver

1648

1.028

Las Vegas

620

1.151

Helena

1181

1.084

Topeka

288

1.191

City

Solution check No errors are detected.

Discussion Power in the wind is proportional to air density, so, for example, the ratio of wind power in Denver to wind power in Chicago is (1.028/1.202) = 0.856, all other factors being equal.

119

3.4

Problem statement For the cities in Problem 3.3, find the available power in the wind for aHAWT with a blade diameter of 40 m. What is the maximum theoretical amount of power that the turbine can extract from the wind for these cities? For all calculations, use a wind speed of 13 m/s.

Diagram

J

v = 13 m/s --•~

D=40m

Assumptions 1. Wind speed is constant.

2. Air densities are those calculated in Problem 3.3.

Governing equations

Calculations All terms other than air density, p, are common for all calculations, so we have 5 pwindmax = (8.18 X 10 )

P

which yields the available power values of:

120

Chicago: Denver: Las Vegas: Helena: Topeka:

pwindmax = 983 kW

841 kW pwindmax = 942 kW p wind,max = 887 kW pwind,max = 974 kW pwindmax =

Solution check No errors are detected.

Discussion The greatest difference in wind power values is between Denver and Chicago, giving a ratio of(841 kW/983 kW) = 0.856, the same ratio found in Problem 3.3. Hence, a turbine in Denver generates only 85.6 percent of the power of the same turbine in Chicago for the same wind speed.

121

3.5

Problem statement Construct a plot of the maximum theoretical power that a wind turbine can extract from the wind per unit area as a function of wind speed. Use a wind speed range of 0 m/s to 30 rn!s, and assume a sea-level air density of 1.225 kg/m3• Diagram (Not applicable) Assumptions None Governing equations

pv.ind,rmx

= _!i_!_p~v3

27 2

Calculations A graph of maximum theoretical wind power per unit cross sectional area of turbine blade as a function of wind speed is shown below.

10000 9000

,,, E

!

~

8000 7000 6000

~ 11.

5000

E :I E

4000

. :E ·;c

3000 2000 1000 0 0

5

10

15

Wind speed (m/s)

122

20

25

30

Solution check No errors are detected.

Discussion As shown in the graph, the effect of wind speed on maximum wind power is dramatic.

Being a function of wind speed cubed, wind power is a stronger function of wind speed than any other single variable, such as air density or cross sectional area of the turbine blades.

123

3.6

Problem statement The desired power capacity of a wind turbine for a wind speed of20 m/s is 250 kW. If the power coefficient is 0.40, find the required area swept out by the turbine blades. Let p = 1.225 kg/m3 • Diagram

r/A I I I I I I I

v=20m/s - - • •

\ \ \ I I I I I I I I I I I

I I I I I

\ \ \

'-

I

Assumptions 1. Constant air density.

2. Constant wind speed.

Governing equations 1

~vind max = CP - pAv

'

3

2

Calculations Solving for the area swept out by the turbine blades, we have

2(250 X 103 W) (0.40)(1.225 kg/m3)(20 m/s) 3

124

= 127.6 m 2 Solution check No errors are detected.

Discussion The calculated area corresponds to a blade diameter of

D=/1 = [4(127.6 m 2)/nf' = 12.7 m

125

3.7

Problem statement AHA WT has a blade diameter of 18 m. For the region where the wind turbine is to be installed, wind speed never exceeds 16 m/s. Ifthe required maximum theoretical power that the turbine can extract from the wind is 350 kW, can this turbine be installed in Santa Fe, New Mexico? San Diego, California? Explain.

Diagram

350 kW

I v = 16 m/s - - • •

)

D= 18m

Assumptions 1. Constant air density.

2. Constant wind speed.

3. The power coefficient is the Betz limit.

Governing equations p = 1.225 - 1.194 X 10- 4 Z P.vind,max

= ~_!_ r• nAv 3

27 2

Calculations The elevations of Santa Fe, New Mexico and San Diego, California are 2194 m and 0 m, respectively, yielding air densities of 0.963 kg/m 3 and 1.225 kglm\ respectively.

For Santa Fe, New Mexico, the maximum theoretical power in the wind is

126

= 16(0.963 kg/m 3)n(18 m) 2/4(16 m/s) 3 27(2) = 2.97 X 105 w = 297 kW For San Diego, California, the maximum theoretical power in the wind is

= 16(1.225 kg/m 3)n(18 m) 2/4(16 m/s) 3 27(2) =3.78 X 105 W=378kW Installed in Santa Fe, New Mexico, the maximum theoretical wind power falls short of the 350 kW required. Installed in San Diego, California, however, the maximum theoretical wind power exceeds the 350 kW requirement by 28 kW. Solution check No errors are detected.

Discussion In reality, the turbine would probably not meet the requirement in San Diego, California either because the actual power coefficient is less than the Betz limit.

127

3.8

Problem statement A HA WT has a blade diameter of 40 m. The power coefficient, gear box efficiency and generator efficiency are 0.42, 0.83 and 0.91, respectively. If the wind speed is 17 rn!s, calculate the electrical output power if the turbine operates in Reno, Nevada.

Diagram

)

v = 17m/s - - • •

D=40 m

'---r-r-./

Assumptions 1. Constant air density.

2. Constant wind speed.

Governing equations

p = 1.225- 1.194 X 10- 4 Z

1

Pelect = -nf14v3 •tr'• 2 Calculations The elevation of Reno, Nevada is approximately 1300 m, so the air density is

p = 1.225- 1.194 X 10- 4 Z 128

= 1.225 - (1.194 x 10~ 4 )(1300 m)

= 1.070 kg/m3 The overall efficiency is

= (0.42)(0.83)(0.91) = 0.317 Th€ electrical output power is 1

pelect =-nnAv3 •I!-"'" 2

= V2(0.317)(1.070 kg/m3)n(40 m) 2/4(17 m/s) 3 6

= 1.05 X 10

w = 1.05 MW

Solution check No errors are detected.

Discussion If the power coefficient was equivalent to the Betz limit, the electrical output power would be 1.48 MW.

129

3.9

Problem statement A homeowner in Raleigh, North Carolina, wants to install a wind turbine in his back yard to augment electrical power from the grid. A local supplier states that the combined efficiency of their turbines is approximately 0.28. How much electrical power can this turbine generate for a wind speed of 18 m/s if the turbine blade diameter is 6 m?

Diagram

v= 18m/s - - • •

J D=6m

'----r-T---'

Assumptions 1. Constant air density.

2. Constant wind speed.

Governing equations

P = 1.225- 1.194 x 1o- 4 z 1 Pelect =-nnAv3 •tr2

Calculations The elevation of Raleigh, North Carolina is approximately 96 m, so the air density is p = 1.225- 1.194 X 10- 4 Z = 1.225 -

(1.194 x 10- 4)(96 m)

130

= 1.214 kg/m 3 The electrical output power is

1 Pelect =-nnAv3 •tr 2

= 2.80 X 104 w = 28.0 kW Solution check No errors are detected.

Discussion The wind turbine could run a refrigerator (750 W), an air conditioner (2000 W) and numerous other appliances in a typical home.

131

3.10

Problem statement A certain coastal region experiences relatively constant light winds 24 hours a day. How much electrical energy can a HAWT generate in one day in this region if the wind speed is 5 rn/s? The blade diameter is 22 m, and the combined efficiency is 0.30.

Diagram

v = 5 m/s

D=22m

Assumptions 1. Constant air density.

2. Constant wind speed.

3. Elevation is zero (coastal region).

Governing equations 1 pelect = -nnAv3 •tr• 2 £elect = pelect f:.t

Calculations The electrical output power is 1

pelect =-nnAv3 •tr· 2

132

= 12(0.30)(1.225 kg/rn 3)n(22 rn) 2/4(5 rn/s) 3 = 8731 w The electrical energy generation for one day is

= (8731 W)(24 h x 3600 slh)

= 7.54 X 105 kJ X 1kWh 3600 kJ =210kWh Solution check No errors are detected.

Discussion The average electrical energy consumption of a typical horne in the United States is about 900 kWh per month, or 30 kWh per day. Hence, this wind turbine can satisfy the electrical energy demand of (21 0 kWh)/(30 kWh) = 7 typical homes.

133

3.11

Problem statement A site for a wind farm near Spokane, Washington, as a plot of land that can accommodate 200 HA WTs. The combined efficiency of the turbine is 0.19. Using a design wind speed of 8 m/s, find the required length for the turbine blades if the required power generation capacity of the wind farm is 30 MW. Assume that all turbines are identical.

Diagram

L

v = 8 m/s

)o

Assumptions 1. Constant air density.

2. Constant wind speed.

3. All turbines are identical.

Governing equations

p = 1.225- 1.194 X 10~ 4 Z 1 Pelect = -nnAv3 2 •tr~

D=2L Calculations The elevation of Spokane, Washington is approximately 562 m, so we have

134

P = 1.225- 1.194 x w- 4 z 4

= 1.225- (1.194 X lQ- )(562m)

= 1.158 kg/m 3 Solving the first equation for the swept out area of one turbine blade, A, we obtain

A= 2~/ect 1]pv3

(0.19)(1.158 kg/m 3)(8 m/s) 3

= 2663 m 2 Combining the last two equations, we obtain the turbine blade length L = (Ain)y,

= (2663 m 21nt

= 29.1 m Solution check No errors are detected.

Discussion Neglecting the diameter of the hub, the diameter of each turbine blade is

2L = 2(29.1 m) = 58.2 m (191ft).

135

3.12

Problem statement A small HAWT in Albany, New York, has a blade diameter of 6.2 m. If the combined efficiency of the turbine is 0.21, find the wind speed required to achieve an electrical output power of6 kW. Diagram

• v

6kW

• ~ •

!

J D = 6.2 m

Assumptions 1. Constant air density.

Governing equations

p = 1.225- 1.194 x w- 4 z 1 2

3 Pelect =-nnAv "fr·

Calculations Albany, New York is on the Hudson River, a sea level river, so z = 0. Thus, the air density is p = 1.225 kg/m3

Solving the second equation for wind velocity, v, we obtain

136

V= (

=

2~;

1/3

1 )

2( 6000 W) I ] l (0.21)(1.225 kg/m )n(6.2 m) /4 3

113

2

= 11.6 m/s

Solution check No errors are detected.

Discussion A wind speed of 11.6 m/s (25.9 mi/h) would have to be maintained to achieve an electrical output power of 6 kW. Realistically, wind speeds would vary throughout the day, and may not even reach a value of 11.6 m/s. Thus, an electrical output power of 6 kW is optimistic.

137

3.13

Problem statement Based on measurements taken over a one-hour period in Lincoln, Nebraska, wind speed varies according to the data given in Table P3 .13. Find the total electrical energy generated during this period by a HAWT with a blade diameter of 20 m. The combined efficiency of the turbine is 0.25.

Table P3.13 Time period (min) 5 2 8 4 2 6 5 8 4 5 5 6

Wind speed (m/s) 6 8 12 14 11 8 5 5 7 14 15 10

Total time= 60 min

Diagram

l

v Wind speed varies according to Table P3.13

'---r-r--..J

138

D=20m

Assumptions 1. Air density is constant.

2. Wind speeds are constant over specified time periods.

Governing equations p = 1.225- 1.194 x

w- 4 z

1 3 Pelect = -nnAv ·u-·· 2 N

£elect

L

P_lect,i/).ti

i=l

Calculations The elevation of Lincoln, Nebraska is approximately 358m. Thus, air density is p = 1.225- 1.194 x

w- 4 z

= 1.225- (1.194 X 10- 4 )(358 m) = 1.182 kg/m3 The power and energy calculations are summarized in the following table:

i

!1t (s)

v (m/s)

f..etect (k W)

fietect (kWh)

1 2 3 4 5 6 7 8 9 10 11 12

300 120 480 240 120 360 300 480 240 300 300 360

6 8 12 14 11 8 5 5 7 14 15 10

10.0 23.8 80.2 127.4 61.8 23.8 5.80 5.80 15.9 127.4 156.7 46.4

0.833 0.793 10.7 8.49 2.06 2.38 0.483 0.773 1.06 10.6 13.1 4.64

139

Summing the electrical output energy values in the last column, we obtain Eelecttotal

= 55.9 kWh

Solution check No errors are detected.

Discussion This method of calculating electrical output energy of a wind turbine can be applied to any time period of interest.

140

3.14

Problem statement A rancher near Salem, Oregon, takes hourly wind speed measurements over a 24-hour period, yielding the graph shown in Figure P3.14. The specifications ofhis HAWT state that the turbine has a power coefficient of0.43, a. gear box efficiency of0.75, a generator efficiency of0.88 and a blade diameter of 15m. Using the rancher's wind speed measurements, calculate the electrical energy that the wind machine will generate during a 24-hour period.

25

1

2

3

4

5

6

7

8

9

10 11

12

Jill II

13 14 15 16

17

18 19

20 21

22

Time(h)

Figure P3.14

Diagram

J D =15m

v Wind speed varies according to Figure P3.14

'---r-.-----

141

23

24

Assumptions 1. Air density is constant.

2. Wind speeds are constant over specified time periods.

Governing equations p = 1.225- 1.194 X 10- 4 Z

1 Pelect =-n'14v3 2 •tr~

N

Eelect

L

J>_lect,i/).fi

i=l

Calculations The elevation of Salem, Oregon is approximately 47 m. Thus, air density is p = 1.225- 1.194 X 10- 4 Z

= 1.225- (1.194 x 10- 4)(47m)

= 1.219 kg/m 3 Using the same procedure outlined in Problem 3.13, we obtain the electrical output power at all time intervals. Because each time interval is one hour, the numerical values of the electrical output power (kW) and electrical output energy (kWh) at each time interval are identical.

Summing the electrical output energy values for each hour, we obtain Eelecttotal =

1782 kWh

Solution check No errors are detected.

Discussion As pointed out in Problem 3.13, this method of calculating electrical output energy of a wind turbine can be applied to any time period of interest.

142

3.15

Problem statement A wind farm near Green River, Wyoming, is designed to have 150 HAWTs, each with a combined efficiency of 0.34 and a blade diameter of 60 m. Based on wind speed measurements taken over a three-year period, average monthly wind speeds are obtained.

Use the average monthly wind speeds to estimate the annual electrical energy generation ofthe wind farm. A graph ofthe data is given in Figure P3.15.

25

20

0

:

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Figure P3.15

Diagram

l

v Wind speed varies according to Figure P3.15

'--~,.....----'

143

D=60m

Dec

Assumptions 1. Air density is constant.

2. Wind speeds are constant over specified time periods.

3. All turbines in the wind farm are identical.

Governing equations

p = 1.225- 1.194 X 10~ 4 Z 1 Pelect = -nnAv3 'It-'" 2 N

Ee/ect

L

f1.tect,if1t;

i=l

Calculations The elevation of Green River, Wyoming, is approximately 1864 m. Thus, air density is

p = 1.225- 1.194 X 10~ 4 Z 4

= 1.225- (1.194 x 10~ )(1864 m) = 1.002 kg/m 3 The calculations for one wind turbine are summarized in the following table: Month

v (m/s)

£elect (MW)

E.elect (MWh)

Jan(31) Feb (28) Mar (31) Apr (30) May (31) Jun (30) Jul(31) Aug (31) Sep (30) Oct (31) Nov (30) Dec (31)

18 20 22 16

2.81 3.85 5.13 1.97 0.641 0.482 0.351 0.351 0.482 1.63 1.97 2.37

2091 2587 3817 1418 477 347 261 261 347 1213 1418 1763

11

10 9 9 10 15 16 17

144

The days in each month are given in parentheses beside the month in the table above.

The electrical output energy for each month was calculated by multiplying the electrical output power by the number of hours in that month. For example, for January, we have Eelect =

(2.81 MW)(31 dayx 24 h/day) = 2091 MWh

Summing the values in the last column, we obtain the annual electrical output energy for one turbine, £elect= 1.60 X

104 MWh

The electrical output energy for the wind farm is £elect farm= (150)(1.60 X

4

10 MWh)

= 2.40 X 106 MWh Solution check No errors are detected.

Discussion The energy output for the wind farm is extremely large because of assumption 2, that the wind speed is constant over each month of the year.

145

3.16

Problem statement A residential community near a mountain pass with steady winds requires an electrical output power capacity of 100 MW. Planners propose to install a wind farm in the mountain pass to supply one half of this power. The diameter of the turbine blades is 50 m, and the combined efficiency of the wind machines is 0.29. If the elevation of the mountain pass is 600 m, how many wind machines are required if 16 m/s is used as a design wind speed?

Diagram

-- J

v=16m/s--)lo•

............- - '

Assumptions 1. Air density is constant.

2. Design wind speed of 16 m/s is steady.

3. All turbines in the wind farm are identical.

Governing equations

p = 1.225- 1.194 X 10- 4 Z 1 3 ~feet = 217pAv

Calculations Air density is

P = 1.22s- 1.194 x w- 4 z 146

D= 50 m

= 1.225 - (1.194 x 10- 4)(600 m) = 1.153 kg/m 3 The electrical output power of one wind turbine is 1 Pelect = -nnAv3 •II-'· 2

= 1.34 X 106 w = 1.34 MW Hence, the number of turbines required to supply half this power capacity is

N= (0.5)(100 MW)/(1.34 MW) = 37

Solution check No errors are detected.

Discussion Wind farms are sometimes located at the mouth of a canyon because canyons tend to produce a natural "venturi" effect that increases wind velocities.

147

3.17

Problem statement A homeowner wishes to have a HA WT installed on his property to offset the cost of electrical power furnished by the local electrical utility. The anticipated annual energy production of the turbine is 20,500 kWh/y, and the cost of electricity in his community is $0.085/k:Wh. If the homeowner desires a simple payback of 6 years, what is the initial cost?

Diagram (Not applicable) Assumptions 1. Energy production and energy cost are constant.

Governing equations

SP=

IC (AEP)(EC)

Calculations Solving the governing equation for initial cost, IC, we obtain

IC = (SP)(AEP)(EC) = (6 y)(20,5000kWh/y)($0.085/k:Wh) = $10,455 Solution check No errors are detected.

Discussion It is clear that initial cost varies directly with simple payback.

148

3.18

Problem statement A wind farm with an installed cost of $150 million has a power-generation capacity of 60 MW. The fixed charge rate is 6.5 percent, the plant capacity factor is 0.38 and the annual operation and maintenance is 1.25 percent of the initial cost. If the levelized replacement cost is averaged over a 25-year lifetime, what is the cost of energy?

Diagram (Not applicable) Assumptions 1. All quantities are constant.

Governing equations

COE = (IC)(FCR) + LRC + AOM + AFC AEP Calculations The wind farm has a power capacity of 60 MW, so the maximum amount of energy that the wind farm can generate in one year is Emax

= (60 MW)(8760 h/y) = 5.256 X 105 MWhly

The capacity factor is 0.38, so the actual annual energy production is

AEP = (0.38)(5.256 X 105 MWhly) = 1.997 X 105 MWhly For a wind farm, the annual fuel cost, AFC, is zero. The LRC and AOM are

LRC = ($150 X 106)/(25 y)

AOM= (0.0125/y)($150 X 106)

149

= $1.875 X 106/y Hence, the cost of energy is

COE = (IC)(FCR) + LRC + AOM + AFC AEP = ($150 X 106)(0.065/y) + $6.00 X 106/y + $1.875 X 106/y 1.997 x 105 Mwhly

= $88.26/MWh which, when converted to units of $/kWh is approximately COE = $0.088/k.Wh

Solution check No errors are detected.

Discussion The average cost of electricity in the United States is approximately $0.12/k.Wh, so this wind farm is competitive with respect to this value.

150

3.19

Problem statement The H-VAWT shown in Figure P3.19 has a combined efficiency of0.21. If the turbine operates in a coastal region, find the electrical output power for a wind speed of 18 m/s.

Diagram 4m

~I

6m

v = 18 m/s _ ____:.,..,.. )o

Figure P3 .19

Assumptions 1. Air density is constant.

2. Wind speed is constant.

Governing equations

A=DH

Calculations The presented area for the H-VA WT is

A=DH = (4 m)(6 m)

151

The electrical output power is 1

Pelect = -nnAv3 •tr 2

= 1.80 X 10

4

w = 18.0 kW

Solution check No errors are detected.

Discussion A HAWT that generates the same amount of electrical power, all other quantities being identical, would have a blade diameter of 5.53 m. This calculation is left as an exercise for the student.

152

3.20

Problem statement The Darrieus VAWT shown in Figure P3.20 has elliptical-shaped blades. The turbine operates at a sea-level location and has a combined efficiency of0.24. For a wind speed of 14 rnls, find the electrical output power.

Diagram 6m

7.5 m

v = 14 m/s --•)o

Figure P3.20 Assumptions 1. Air density is constant.

2. Wind speed is constant.

Governing equations A"' 0.65DH 1

~lect = 2_11pAv

3

Calculations The area swept out by elliptical-shaped blades is given by the approximation A"' 0.65DH

153

= (0.65)(6 m)(7.5 m) = 29.25 m 2 The electrical output power is 1 Pelect =-nnAv3 •u-" 2

= 1.18 X 104 w = 11.8 kW Solution check No errors are detected.

Discussion The area swept out by the turbine blades could be more precisely calculated by integrating the ellipse function that describes the blade's shape, as is shown in Example 3.6.

154

3.21

Problem statement A Savonius wind turbine, shown in Figure P3.21, is located in a region where the elevation is 1300 m. For a wind speed of 16 m/s and a combined efficiency of 0.20, what blade height, H, is required for an electrical output power of 45 kW? Diagram 8m

8m

H

Top view

Side view

Figure P3.21 Assumptions 1. Air density is constant.

2. Wind speed is constant.

Governing equations

A=DH 1

P,lect

= -1]pAv 3 2

Calculations Substituting the first equation into the second, and solving for H, we obtain

155

2(45 X 103 W) (0.20)(1.225 kg/m 3)(8 m)(16 m/s) 3 =11.2m Solution check No errors are detected.

Discussion The result seems reasonable given that the diameter of the turbine is 8 m. The presented area of the blades is

A=DH = (8 m)(11.2 m)

= 89.6 m 2

156

3.22

Problem statement An inventor proposes to retrofit utility poles with small Savonius wind turbines, providing some electrical power for street and parking lights. The height and diameter of the blades are 2.5 m and 1.75 m, respectively. The combined efficiency of the wind machine is 0.18. Assuming a wind speed of6 m/s, find the electrical output power ofthe turbine. Assume sea-level operation. Is this proposal viable?

Diagram

1.75 m

Assumptions 1. Air density is constant.

2. Wind speed is constant.

3. Elevation is sea level.

Governing equations

A=DH 1 3 Pelect =-nnAv · u-·· 2

157

Calculations The electrical output power is 1 3 Pelect = -nnAv ·f t-'" 2

= Yz(0.18)(1.225 kg/m 3)(1.75 m x 2.5 m)(6 m/s) 3 = 104

w

No, the proposal is not viable because the electrical output power is too low to justify the cost of the retrofit.

Solution check No errors are detected.

Discussion This problem illustrates that a well intentioned inventor should do a quick calculation before investing a lot of time and money in an impractical invention.

158

3.24

Student choice

3.25

Student choice

3.26

Student choice

3.27

Student choice

159

Chapter 5 Geothermal Energy

198

Section 5.3

Practice!

1. A dry-steam geothermal power plant operates on 300 o C steam. For an atmosphere temperature of 1 C, find the maximum possible thermal efficiency of the power plant.

r

Solution TL

'7tlz,idea/

= 1- T

H

= 1 - (17 + 273)K (300 + 273)K = 0.494

199

2. The production well of a dry-steam geothermal power plant supplies 25.0 MW to the steam that enters the turbine. The shaft of the turbine produces 6.2 MW of mechanical power, and the efficiency of the electrical generator is 0.95. Find the thermal efficiency, overall efficiency, electrical output power, and rate ofheat rejection to the atmosphere.

Solution

= 6.2 MW 25.0MW = 0.248 TJoverall = 'lth TJgen = (0.248)(0.95) = 0.236

= (0.236)(25.0 MW) =5.89MW

= 25.0 MW -6.2 MW

= 18.8 MW

200

3. The thermal and electrical generator efficiencies of a binary power plant are 0.43 and 0.88, respectively. If the electrical output power of the plant is 2.8 MW, find the rate ofheat transfer from the hot water supplied by the production well.

Solution llovera/1

= 11th 1Jgen = (0.43)(0.88) = 0.378

Q..m = P.etect 'loverall

=2.8MW 0.378 =7.40MW

201

5.1

Problem statement The heat content (internal energy) of a solid can be calculated using the relation

U=cmT where U is internal energy in units of J, m is mass in units of kg, cis specific heat in units of J/kg· °C, and Tis temperature in units of °C. The specific heat of granite is approximately 800 J/kg·°C. Find the internal energy per unit mass of200°C granite.

Diagram

Assumptions 1. All pertinent quantities are constant.

Governing equations

U=cmT Calculations The internal energy of a 1-kg mass of granite at 200°C is

U=cmT

= 1.60 X 105 J = 160 kJ

202

Solution check No errors are detected.

Discussion We could have expressed our answer in units of J/kg since the question asked to find the internal energy per unit mass. The numerical value of the answer is the same either way.

The relation used to calculate internal energy in this problem also applies to liquids, as we see in the next problem.

203

5.2

Problem statement The relation in Problem 5.1 also applies to liquids. Find the internal energy per unit mass of 190°C liquid water. Use a specific heat of 4875 Jlkg· °C. Diagram

Assumptions 1. All pertinent quantities are constant.

Governing equations

U=cmT Calculations The internal energy of a 1-kg mass of water at 190 o C is

U=cmT

=9.26 X 105 J=926kJ Solution check No errors are detected.

Discussion Again, as in Problem 5.1, we could have expressed our answer in units of Jlkg.

204

5.3

Problem statement A well-known relation in heat transfer is Fourier's law of heat conduction q" = k !1TI!1.x

where q" is heat flux in units ofW/m2, k is thermal conductivity in units ofW/m·°C, and !1T/!1.x is temperature gradient (temperature change divided by distance) in units of °C/m.

The thermal conductivity of granite is approximately 2.8 W/m·°C. Using this value fork in Fourier's law, find the heat flux at the earth's surface if the temperature difference across the earth's crust to a depth of7000 m is 300°C. Is the answer consistent with the heat flux data given in Figure 5.1?

Diagram

Ax= 7000 m ~T= 300°C

Assumptions 1. Earth's crust is made of granite.

2. All pertinent quantities are constant.

Governing equations q" = k !1TI!1.x

Calculations The heat flux is q" = k !1TI!1.x

205

= (2.8 W/m·°C)(300°C)

7000 m = 0.12 W/m2 = 120 mW/m2 Solution check No errors are detected.

Discussion Yes, the answer is consistent with the heat flux data given in Figure 5 .1. In some regions of the United States, the heat flux at the earth's surface is approximately 150 mW/m 2•

206

5.4

Problem statement A dry-steam power plant operates between the temperature limits of 15 o C and 3 80 o C. Find the maximum thermal efficiency of this power plant.

Diagram

,.

"' I /

' '

I

I

' \ \

heat engine

I

1/ ~

\ \

'' '

....... "'

~~ W out

/

low-temperature sink

Assumptions 1. Source and sink temperatures are constant.

2. Heat engine is a Camot heat engine.

Governing equations TL

llth,ideal

= 1- ~ H

Calculations The power plant is a heat engine, so the maximum thermal efficiency is given by the Camot efficiency, TL

11th ,ideal =

1- ~ H

207

= 1 - (15 + 273)K (380 + 273)K = 0.559

Solution check No errors are detected.

Discussion Various energy losses in the dry-steam power plant would result in the actual thermal efficiency of the plant being much less than the Camot efficiency.

208

5.5

Problem statement The design engineers of a binary power plant use a maximum theoretical thermal efficiency of 0.16 in the preliminary phase of the design. If the engineers assume that the temperatur of the atmosphere is 20°C, what is the required temperature of the geothermal heat source?

Diagram

/

/

....

""

'

I '1th,ideal =

0.16

I

'\\

,yw,.

I

heat engine

I \ \

I

' ' ....

/ /

""

low-temperature sink

Assumptions 1. Source and sink temperatures are constant.

2. Heat engine is a Camot heat engine.

Governing equations

-1- TL T.

'llth ,ideal -

H

Calculations Solving the equation for source temperature, TH, we obtain

209

TL

T.H-

I - 11th ,ideal = (20 + 273)K 1 - 0.16 =349K=76°C Solution check No errors are detected.

Discussion The geothermal source temperature is not high enough for a dry-steam power plant, but a binary power plant is feasible.

210

5.6

Problem statement The steam from the production well of a dry-steam power plant supplies 8.5 MW to a turbine. If the output work of the turbine is 3.9 MW, what is the thermal efficiency of the power plant?

Diagram

Q;" = 8.5 MW ,.. /

...

/

I

''\ \

I

I

heat engine

out=

I

,.. /

~ w· ~ 1

/

Qout

low-temperature sink

Assumptions 1. Heat input and work output are constant.

Governing equations

Calculations Thermal efficiency of the power plant is

=3.9MW 8.5MW

211

3 9 MW •

= 0.459 Solution check No errors are detected.

Discussion Geothermal sources are mostly unaffected by environmental factors, so the amount of heat supplied by them is nearly constant year round.

212

5.7

Problem statement The thermal and electrical generator efficiencies of a flash-steam power plant are 0.55 and 0.92, respectively. If the heat input from the steam in the production well is 16 MW, what is the electrical output power of the plant?

Diagram

Qin= 16MW /

/

17th= 0.55

/

I

llgen = 0.92

I

I

heat engine

pelect

\

I

\

'' '

/

....

/

/

low-temperature sink

Assumptions 1. Heat input is constant.

2. Efficiencies are constant.

Governing equations

17overall = 17th 17gen J>.,/ect

= 1loveral/ Qin

Calculations The overall efficiency of the power plant is

17overall = 17th 17gen 213

= (0.55)(0.92)

= 0.506 The electrical output power is

= (0.506)(16 MW)

= 8.10 MW Solution check No errors are detected.

Discussion The efficiencies of electrical generators are typically very high compared to thermal efficiencies, as shown in this problem. Thus, most of the energy lost in the conversion process is thermal, not electrical.

214

5.8

Problem statement The production well of a dry-steam power plant supplies 30 MW to the steam that enters the turbine. The efficiency of the electrical generator is 0.90, and the shaft of the turbine produces 9.5 MW of mechanical power. Find the thermal efficiency, overall efficiency, electrical output power, and the rate of heat rejection to the atmosphere.

Diagram

Qin=30MW /

/

/

1'/gen = 0.90

I

I

I

heat engine

\

I

\

' ' .....

I /

__. /

Wout=9.5 MW

Qout

low-temperature sink

Assumptions 1. Heat input power is constant.

2. Mechanical output power is constant.

3. Generator efficiency is constant.

Governing equations

TJoverall = 17th TJgen

215

Calculations Using the first equation, thermal efficiency is

=9.5 MW 30MW = 0.317 Using the second equation, overall efficiency is

= (0.317)(0.90) = 0.285 Using the third equation, electrical output power is

= (0.285)(30 MW) =8.55MW Solving the fourth equation for the rate of heat rejection to the atmosphere, we obtain

=30MW- 9.5MW =20.5 MW

216

Solution check No errors are detected.

Discussion To find the Camot efficiency, we would have to know the source and sink temperatures.

All we know is that the Camot efficiency is greater than the actual thermal efficiency of 0.317.

217

5.9

Problem statement The thermal and electrical generator efficiencies of a binary power plant are 0.40 and 0.90, respectively. Ifthe electrical output power of the plant is 3.6 MW, find the rate of heat transfer from the hot water at the production well.

Diagram

/

/

'7th= 0.40

/

I

1'/gen = 0.90

I

(

heat engine

Pelect

I

'' '

'

low-temperature sink

Assumptions 1. Efficiencies are constant.

2. Electrical output power is constant.

Governing equations

rJoverall = 11th rJgen

The overall efficiency is

= (0.40)(0.90)

218

= 3.6 MW

= 0.360 Solving the second equation for the heat input, we obtain

Q..zn = P,/ect 'loveral/

=3.6MW 0.360 = 10.0 MW Solution check No errors are detected.

Discussion Knowing the thermal efficiency and heat input, we find the work output to be

= (0.40)(1 0.0 MW) =4.0MW and the rate of heat rejection to the atmosphere to be

= 10.0 MW- 4.0 MW =6.0MW

219

5.10

Problem statement In a dry-steam power plant, 325°C superheated steam from the production well supplies 16.8 MW to the turbine. The output power of the turbine shaft is 7.0 MW, and the efficiency of the electrical generator is 0.94. Find the thermal efficiency, overall efficiency, electrical output power, rate ofheat rejection to a 20°C atmosphere, and Camot efficiency.

Diagram

Qin = 16.8 MW /

/

/

Y/gen =

I

0.94

I

I

heat engine

pelect

\

I

\

' ' .........

I

.... --

/

Wout= 7.0 MW

Qout

low-temperature sink

Assumptions 1. Heat input and work output are constant.

2. Electrical generator efficiency is constant.

Governing equations n

-

'lth -

U:,t .

Qin

'7overall = 'lth '7gen

220

TL

17th ,ideal = 1- ~ H

Calculations Thermal efficiency is

7.0MW 16.8 MW = 0.417 Overall efficiency is

17overall = 11th 17gen = (0.417)(0.94) = 0.392 Electrical output power is

= (0.392)(16.8 MW) =6.58MW The rate of heat rejection to the atmosphere is

= 16.8 MW- 7.0 MW =9.8MW Carnot efficiency is

221

TL

1Jth,ideal

= 1- T, H

= 1 - (20 + 273)K (325 + 273)K = 0.510 Solution check No errors are detected.

Discussion The actual thermal efficiency is lower than the Carnot efficiency, as demanded by the second law of thermodynamics.

222

5.11

Problem statement If the plant capacity factor for the dry-steam power plant in Problem 5.10 is 0.85, how much energy does the plant generate in one year?

Diagram

Qin = 16.8 MW /

/

/

1Jgen =

I

0.94

I

I

heat engine

pelect

\ \

I '

' ....

...--

/

I

Wout=7.0MW

low-temperature sink

Assumptions 1. Heat input and work output are constant.

2. Electrical generator efficiency is constant.

Governing equations E elect = CF ?.teet I'll

Calculations The electrical energy generated by the power plant in one year is E elect = CF ?.teet I'll

= (0.85)(6.58 x 106 W)(3.15 x 107 s) = 1.76 X 10 14 J X (1 kWh/3.6 X 106 J)

223

= 4.89 X 107 kWh Solution check No errors are detected.

Discussion An average home in the United States consumes about 11,000 kWh of electricity in one year, so this dry-steam power plant could supply electricity to (4.89 x 107 kWh/11,000 kWh)= 4445 homes

224

5.12

Problem statement In a binary power plant, 70°C water from the production well supplies 350 kW to an isobutane working fluid in the heat exchanger. The output power of the turbine shaft is 25 kW, and the efficiency of the generator is 0.95. Find the thermal efficiency, overall efficiency, electrical output power, and rate ofheat rejection to the atmosphere. If the temperature ofthe atmosphere is 22°C, find the Camot efficiency.

Diagram

Qin = 350 kW /

/

IJgen = 0.95

/

I

I

(

heat engine

\

I

\

' ' .....

I

.... /

/

Wout= 25 kW

low-temperature sink

Assumptions 1. Heat input and work output are constant.

2. Electrical generator efficiency is constant.

Governing equations

17overall = 11th 17gen

225

TL

17th,idea/

= 1- ~ H

Calculations Thermal efficiency is

=

25kW 350kW

= 0.0714 Overall efficiency is 1]overall

= 17th 1]gen = (0.0714)(0.95)

= 0.0679 Electrical output power is

= (0.0679)(350 kW) =23.8 kW The rate of heat rejection to the atmosphere is

= 350 kW - 25 kW

= 325 kW Camot efficiency is

226

11th ,ideal

= 1- T.TL H

= 1 - (22 + 273)K (70 + 273)K = 0.140 Solution check No errors are detected.

Discussion As always, the second law of thermodynamics places an upper limit on the thermal efficiency of a heat engine, and that limit for this power plant is 0.140.

227

5.13

Problem statement Consider the following parameters for a dry-steam power plant. Stearn at 280°C at the production well supplies 30 MW ofheat. The output power of the turbine shaft is 16 MW, and the temperature of the atmosphere is 20°C. By analysis, show that this power plant is impossible.

Diagram

Q;n = 30 MW ,.,.

....

/

/

.... '\

I

\

I

I

I~ wout 16 MW

heat engine

I \

=

\

' .... .....

I /

--

/

low-temperature sink

Assumptions 1. Source and sink temperatures are constant.

2. Heat input and work output are constant.

Governing equations

TL

T]th ,ideal

= 1- ~ H

Calculations The actual thermal efficiency of the power plant is

228

=16MW 30MW = 0.533 The Camot efficiency is TL

1lth,ideal

= 1- T, H

= 1 - (20 + 273)K (280 + 273)K = 0.470 The second law of thermodynamics limits the thermal efficiency of a heat engine to the Camot efficiency, which for this problem is 0.470. Because the actual efficiency is found to be 0.533, this power plant is impossible.

Solution check No errors are detected.

Discussion Had the work output been 14 MW instead of 16 MW, the actual thermal efficiency would be '1th

= 14 MW 30MW = 0.467

which is slightly lower than the Camot value, and thus permissible by the second law of thermodynamics.

229

5.14

Problem statement In a dry-steam power plant, 325 oc superheated steam from the production well supplies 20 MW to the turbine. The output power of the turbine shaft is 6.5 MW, and the efficiency of the electrical generator is 0.91. Find the thermal efficiency, overall efficiency, electrical output power, rate ofheat rejection to a 10°C atmosphere, and Carnot efficiency. If the plant capacity factor is 0.88, how much electrical energy does the plant generate in one year?

Diagram

Qin=20MW /

/

1Jgen = 0.91

/

I

I

I

heat engine

pelect

\

I

\

' ' .....

I

....

/ /

Wout=6.5 MW

low-temperature sink

Assumptions 1. Heat input and work output are constant.

2. Electrical generator efficiency is constant.

Governing equations

230

17th ,ideal

= 1- T,TL H

£elect

= CF J>.1ect /1t

Calculations Thermal efficiency is

=6.5MW 20MW = 0.325 Overall efficiency is

17overall = 17th 17gen = (0.325)(0.91) = 0.296 Electrical output power is

= (0.296)(20 MW) =5.92MW The rate of heat rejection to the atmosphere is

= 20 MW - 6.5 MW

= 13.5 MW Camot efficiency is

231

11th ,ideal

= 1- T,TL H

=1- (10+273)K (325 + 273)K = 0.527 The electrical energy generated by the power plant in one year is

= CF J>,lect /1t

£elect

= (0.88)(5.92 x 106 W)(3.15 x 107 s) = 1.64 X 10 14 J X (1 kWh/3.6 X 106 J) = 4.56 X 107 kWh Solution check No errors are detected.

Discussion Geothermal conditions are relatively unaffected by environmental factors, so plant capacity factors of geothermal power plants are typically high.

232

5.15

(Student choice)

5.16

(Student choice)

233

Chapter 6 Marine Energy

234

Section 6.4

Practice!

1. A tidal barrage has a tidal range of3.8 m, and the surface area of the basin is 475 acre. If the tidal barrage has five turbines, find the electrical output power for a tidal period of 11 h. The combined efficiency of the barrage is 0.26.

Solution p

= qgpAR

2

2T

elect

= (0.26)(9.81 m/s 2)(1025 kg/m3)(475 acre x 4046.95 m 2/acre )(3.8 m) 2 2(11 h x 3600 slh)

= 9.16 X 105 W Five turbines: 5

Pelect = 5(9.16 X 10

= 4.58 X 10

6

W)

w = 4.58 MW

235

2. The blade diameter ofthe turbines of a tidal stream generator is 5.0 m. The combined efficiency of the system is 0.35. If the velocity of the ocean current is 1.3 m/s, find the total electrical output power if the system consists of six turbines.

Solution 1 3 Pelect = -n'P~v "/, 2

= V2(0.35)(1025 kg/m3)n(5.0 m) 2/4(1.3 m/s) 3 = 7738

w

Six turbines: Pelect = 6(7738 W)

= 4.64 X 104 w = 46.4 kW

236

3. In deep ocean water, the wave period and height of an ocean wave are 5.2 sand 2.9 m, respectively, and the wave crest length is 60 m. Find the electrical output power for three waves if the overall efficiency of the system is 0.20.

Solution

= (1025 kg/m3)(9.81 m/s 2) 2(5.2 s)(2.9 m) 2 321t

~feet = l].fL = (0.20)(4.29 x 10

4

W/m)(60 m)

= 5.15 X 105 W Three waves:

= 1.54 X 106 w = 1.54 MW

237

4. Consider an OTEC plant with an ideal thermal efficiency of 0.070. If the temperature of deep ocean water is 8 o C, what is the temperature of ocean water at the surface?

Solution

T.H--

TL 1- 1Jth ,ideal

= (8 + 273)K 1 - 0.070 =302K=29°C

238

6.1

Problem statement The tidal range of a tidal barrage is 4.8 m, and the surface area of the basin is 450 acre. If the tidal barrage has 12 turbines, find the electrical output power for a tidal period of 10 h. For the combined efficiency of the barrage, let 17 = 0.27.

Diagram

A= 450 acre

~'-----barr-age

__,u

_

Assumptions 1. Tidal range, basin area and tidal period are constant.

2. Turbines are identical.

Governing equations

= 1]gpAR

p

2

2T

elect

Calculations Converting the basin area to units of m 2 , we obtain A = 450 acre x 4046.95 m 2/acre

The electrical output power of one turbine is p

_ 1]gpAR elect T 2

2

239

= (0.27)(9.81 m/s 2)(1025 kg/m3){1.82 x 106 m 2)(4.8 m) 2 2(1 0 h x 3600 s/h)

= 1.58 X 106 w = 1.58 MW so for 12 turbines, the electrical output power is Pelect =

12(1.58 MW)

= 19.0 MW

Solution check No errors are detected.

Discussion If the tidal range was just 1 m higher (5.8 m), the total electrical output power would be 27.7 MW, a 46 percent increase!

240

6.2

Problem statement The basin surface area and tidal range of a tidal barrage are 375 acre and 5.1 m, respectively. The barrage has eight turbines, each with a combined efficiency of0.35. If the tidal period is 12 h, calculate the electrical output power of the barrage.

Diagram

A= 375 acre

~'------barr-age Ill(

(turbine

_

___.u

~ ...,Ill(--

Assumptions 1. Tidal range, basin area and tidal period are constant.

2. Turbines are identical.

Governing equations

= l]gpAR

p

2

2T

elect

Calculations Converting the basin area to units of m2 , we obtain A= 375 acre x 4046.95 m 2/acre

The electrical output power of one turbine is

= l]gpAR 2

p elect

2T

241

2

3

= (0.35)(9.81 m/s )(1025 kg/m )(1.52 x 10

6

m 2 )(5.1 m) 2

2(12 h x 3600 s/h)

= 1.61 X 106 w = 1.61 MW so for eight turbines, the electrical output power is pelect = 8(1.61 MW)

= 12.9MW Solution check No errors are detected.

Discussion The graph below shows how total electrical output power of this tidal barrage varies with tidal range.

35

30

i2s

~ li

ho

. .

Q.

:I

Q.

h5

ii .!!

....jlO 5

0 0

1

2

3

4

Tidal ranee (m)

242

5

6

7

8

6.3

Problem statement Consider a tidal barrage with five turbines, each with a combined efficiency of0.40. On average, the water level in the tidal basin experiences high tide 2.6 times per day. If the surface area of the basin is 185 acre, find the electrical output power of the barrage.

Diagram A= 185 acre

barrage

Assumptions 1. Tidal basin surface area is constant.

2. Tidal period is constant.

Governing equations p elect

= 1]gpAR

2

2T

Calculations Tidal period is the number ofhours per day divided by the number of times per day that high tide is experienced. Thus, T = (24 h/day)/(2.6 high tide/day)

=9.23h The tidal basin area is A= 185 acre x 4046.95 m2/acre

243

Tidal range is not given in the problem, so we calculate the electrical output power for a range of tidal range values. The electrical output power of one turbine, as a function of R, IS

p

= 1]gpAR

2

2T

elect

= (0.40)(9.81 m/s 2)(1025 kg/m3)(7.49 x 105 m 2) R 2 2(9.23 h x 3600 s/h)

so, for five turbines, we have

The graph below shows the variation of electrical output power with tidal range.

0

0

1

2

3

4

5

Tidol r1n1• (m)

Solution check No errors are detected.

Discussion A tidal range of5 m yields an electrical output power of approximately 5.7 MW.

244

6.4

Problem statement A small tidal barrage with two turbines provides electrical power to a nearby town. The tidal range is 2.5 m, and the surface area of the basin is 60 acre. Find the electrical output power if the turbines have a combined efficiency of 0.22 and the tidal period is 11 h.

Diagram A= 60 acre

barrage

Assumptions 1. Tidal range, basin area and tidal period are constant.

2. Turbines are identical.

Governing equations p

= qgpAR

2

2T

elect

Calculations Converting the basin area to units of m 2, we obtain A = 60 acre x 4046.95 m2/acre

The electrical output power of one turbine is p

_ qgpAR T elect 2

2

245

2

3

= (0.22)(9.81 m/s )(1025 kg/m )(2.43 x 10

5

m 2 )(2.5 m) 2

2(11 h x 3600 s/h) = 4.24 X 104

w = 42.4 kW

so for two turbines, the electrical output power is Pelect

= 2(42.4 kW) = 84.8 kW

Solution check No errors are detected.

Discussion If the plant capacity factor for this tidal barrage was 0.75, the electrical energy generated in one year would be Eelect

= CF pelect f1t = (0.75)(8.48 x 104 W)(3.15 x 107 s) = 2.QQ X 10 12 J X (1 kWh/3.6 X 106 J) = 5.57 X 105 kWh

enough electrical energy to supply about 50 average homes.

246

6.5

Problem statement The required electrical output power of a tidal barrage is 45 MW. The tidal range is 4.6 m, and the tidal period is 10.4 h. If the barrage has ten turbines, each with a combined efficiency of 0.30, find the required surface area of the basin.

Diagram 45MW

/ barrage

Assumptions 1. Power production is constant.

2. Tidal range is constant.

Governing equations

p elect

= 1]gpAR

2

2T

Calculations Doing the calculation for one turbine, we set Pelect = 4.5 MW. Solving the governing equation for basin area, A, we obtain A = 2 TP,/ect

qgpR2 2(1 0.4 h x 3600 s/h)(4.5 x 106 W) (0.30)(9 .81 m/s 2)(1 025 kglm 3)(4.6 m) 2

247

6

= 5.28 x 10

m2

= 5.28 x 106 m 2 x (1 acre/4046.95 m 2) = 1305 acre

(2.04 mi 2)

Solution check No errors are detected.

Discussion Our result indicates a very large basin, about 1Yz mi on a side.

248

6.6

Problem statement The surface area of a tidal basin is 275 acre. The barrage incorporates six turbines, each with a combined efficiency of0.33. If the tidal period is 10.5 h, what is the required tidal range for an electrical output power of 5. 7 MW? Diagram 5.7MW

/

A= 275 acre

barrage

R

Assumptions 1. Power production and basin surface area are constant.

2. Turbines are identical.

Governing equations

p elect

= rygpAR

2

2T

Calculations Converting basin area from acre to m 2, we obtain A = 275 acre x 4046.95 m 2/acre

We do the analysis for one turbine, so Pelect = 5.7 MW/6 = 0.95 MW. Solving the governing equation for tidal range, R, we obtain

249

R=

=

2T~/ect

rJgpA

I 2(1 0.5 h x 3600 s/h)(0.95 x 10 W) ] l (0.33)(9.81 m/s )(1025 kg/m )(1.11 x 10 m

Yz

6

2

3

6

2

)

=4.41m Solution check No errors are detected.

Discussion Had we done the calculations using 5.7 MW, the total power generation of the barrage, we would have obtained R = 10.8 m. It is evident that to obtain the correct tidal range, we had to do the calculation of R for one turbine.

250

6. 7

Problem statement The turbine of a tidal stream generator has a blade diameter of 4.8 m, and the efficiency of the system is 0.42. Find the electrical output power for an ocean current velocity of 2.1 m/s.

Diagram

v = 2.1 m/s

D=4.8m

Assumptions 1. All pertinent quantities are constant.

Governing equations 1 3 Pelect = -n'P~V •t, 2

Calculations Electrical output power is 1 3 Pelect =-n'P~V •t, 2 = Y2(0.42)(1025 kg/m 3)n(4.8 m) 2/4(2.1 m/s) 3

= 3.61 X 104 w = 36.1 kW

251

Solution check No errors are detected.

Discussion Holding all other quantities constant, if the fluid was air instead of water, the electrical power generation of this turbine would be only 43 W! This is because the density of air is only 1.225 kg/m 3•

252

6.8

Problem statement A certain ocean current has an average velocity of 1.9 m/s. A tidal stream generator with an efficiency of 0.55 is placed in the current. Assuming the ocean current is steady, find the electrical energy generated during a three-year period. The blade diameter of the turbine is 3.8 m.

Diagram

v = 1.9 m/s

D= 3.8 m

Assumptions 1. The ocean current is steady.

2. All other pertinent quantities are constant.

Governing equations P.,lect

1 2

=- TJpAv

3

Calculations The electrical output power is 1 3 Pelect = -n'P~v •t, 2

= Y2(0.55)(1 025 kg/m3)1t(3.8 m) 2/4(1.9 m/s) 3

253

The electrical energy generated during a three-year period is

E elect = ~feet /).( = (2.19 x 10

4

W)(3)(3.15 x 107 s)

= 2.07 X 10 12 J = (2.07 X 10

12

J)(1 kWh/3 .6 X 106 J)

= 5.76 X 105 kWh Solution check No errors are detected.

Discussion The average home in the United States consumes about 11,000 kWh of electricity annually. Thus, this tidal stream generator could supply electricity to (5.76 x 105 kWh/3 y)/(11,000 kWh/home)"' 17 homely.

254

6.9

Problem statement The diameter of the turbine blades of a tidal stream generator is 4.75 m, and the efficiency of the system is 0.45. Ifthe system consists of 18 turbines, find the electrical output power for an ocean current velocity of2.6 rnls.

Diagram

v = 2.6 rnls

D=4.75m

Assumptions 1. All pertinent quantities are constant.

Governing equations

1

pelect =-nnAv3 'If/" 2

Calculations Electrical output power of one turbine is

I

pelect =-nnAv3 Z 'If/" = Yz(0.45)(1025 kg/m 3)n(4.75 m) 2/4(2.6 rnls) 3

so for 18 turbines, we have Pelect

= 18(7.18 X 104 W)

= 1.29 X 106 w = 1.29 MW 255

Solution check No errors are detected.

Discussion Holding all other quantities constant, if the fluid was air instead of water, the electrical power generation of one of these turbines would be only 86 W This is because the density of air is only 1.225 kg/m 3• See Problem 6. 7 for a similar comparison.

256

6.1 0

Problem statement The electrical power requirement of a tidal stream generator is 1. 75 MW. Each turbine in the system has a blade diameter of 4.9 mandan efficiency of0.42. For an ocean current velocity of 2.8 m/s, how many turbines are required?

Diagram

Petect = 1.75 MW (total of all turbines)

v = 2.8 m/s

D=4.9m

Assumptions 1. All pertinent quantities are constant.

2. Turbines are identical.

Governing equations 1

pelect =-nnAv3 'IY" 2

Calculations The electrical power generation of one turbine is 1

Pelect =-nnAv3 "IY" 2 3

2

= Yz(0.42)(1025 kg/m )n(4.9 m) /4(2.8 m/s)

3

= 8.91 X 104 W The number of turbines required is the total electrical power generation divided by the power generation of one turbine. Hence, we have 257

N = (1.75 X 106 W)/(8.91 X 104 W)

= 19.6 turbines (round up to 20) Solution check No errors are detected.

Discussion A total of 20 turbines would generate a total of 20(8.91 X 104 W) = 1.78 X 106 w = 1.76 MW slightly more power than required.

258

6.11

Problem statement The equation for finding available power in wind and ocean currents is identical. For the same turbine blade diameter and fluid velocity, how much more power is available in an ocean current than wind?

Diagram air: p = 1.225 kg/m3

ocean water: p = 1025 kg/m3

~

~

A

A

Assumptions 1. Fluid velocity and blade area are identical in each system 2. Sea level conditions for air density.

Governing equations

pwind =_!_ Pair A3 V 2

pv.uter =

1

2Pv.uter Av

3

Calculations Noting the cancellation of all quantities other than density, the ratio of the available power in ocean water to available power in wind is p water

= Pwater

pwind

Pair

259

Solution check No errors are detected.

Discussion Hence, there is 837 times more power in ocean water than in wind, holding all other quantities the same. Stated another way, an ocean turbine could be 837 times smaller than a wind turbine and deliver the same amount of power. See Problem 4.2 for a similar comparison of fresh water (p = 1000 kg/m 3) and air. Ocean water has a slightly higher density ( 1025 kg/m3) than fresh water because of the salt content of ocean water.

260

6.12

Problem statement A wave energy converter operates far off shore where the wave period and wave height of ocean waves are 5.2 sand 3.8 m, respectively. The wave crest length is 60 m, and the overall efficiency of the wave energy converter is 0.20. Find the electrical output power for one wave.

Diagram

L=60m

Assumptions 1. All pertinent quantities are constant.

Governing equations

P.:tect

= 17J L

Calculations Using the first equation, the energy flux is

3

2 2

= (1025 kglm )(9.81 m/s ) (5.2 s)(3.8 m)

321t

261

2

= 7.37 x 10

4

W/m

so the electrical output power is ~teet = qJ L 4

= (0.20)(7.37 x 10 5

= 8.84 X 10

W/m)(60 m)

w = 884 kW

Solution check No errors are detected.

Discussion Our result is for one wave only. The electrical output power for multiple waves is obtained by multiplying our answer by the number of waves utilized by the energy wave converter.

262

6.13

Problem statement The overall efficiency of a wave energy converter is 0.27. The system operates in ocean waters where the wave crest length is 75 m, the wave height is 2.4 m, and the wave period is 6.5 s. Find the total electrical output power for five waves.

Diagram

L=75m

Assumptions 1. All pertinent quantities are constant.

2. All waves are identical.

Governing equations

~lee/= 17JL

Calculations Using the first equation, the energy flux is

3

2 2

= (1025 kg/m )(9.81 m/s ) (6.5 s)(2.4 m)

321t

263

2

so the electrical output power for one wave is P,/ect

= 1]J L = (0.20)(3.67 x 104 W/m)(75 m)

= 5.51 X 105 W For five waves, we have Pelect

= 5(5.51 X 105 W) = 2.76 X 106 w = 2.76 MW

Solution check No errors are detected.

Discussion Efficiencies of wave energy converters are typically low, around 0.10 to 0.25. Even with low efficiencies, however, ocean waves have the potential to generate a lot of electrical power, as this problem illustrates.

264

6.14

Problem statement The required electrical power capacity of a wave energy converter is 4. 75 MW. The system is to operate in deep ocean waters where the wave period and wave height are 5.3 sand 4.8 m, respectively. If the overall efficiency of the system is 0.18, how many 75-m long waves must the system use to generate the required power?

Diagram

L=75m

Assumptions 1. All pertinent quantities are constant.

2. All waves are identical.

Governing equations

P,lect

= 77J L

Calculations Using the first equation, the energy flux is

3

2 2

= (1025 kglm )(9.81 rn/s ) (5.3 s)( m)

321t

265

2

= 1.20 x 105 W/m so the electrical output power for one wave is ~teet= 17J L

= (0.18)(1.20 x 105 W/m)(75 m) = 1.62 X 106 W The number of waves needed to generate the required electrical power is

= 2.94 (round up to ,1)

Solution check No errors are detected.

Discussion More than three waves would be needed if the waves were shorter. For example, if the waves were only 25 m long, nine waves would be needed to supply the required electrical power.

266

6.15

Problem statement A wave energy converter in the design stage has an estimated overall efficiency of 0.22.

The required electrical power capacity of the system is 7.0 MW. Find five combinations of wave height and wave period that will meet the power capacity requirement if the system is designed to exploit two waves with a wave crest length of 100 m.

Diagram

T L =100m H

Assumptions 1. All pertinent quantities are constant.

2. The two waves are identical.

Governing equations

~feet = 2ryJ L

Calculations We set Pelect = 7.0 x 106 Wand L =100m and find five combinations of wave period, T, and wave height, H, that satisfy the equations. We use a range of wave periods, T, from 4.0 s to 6.0 sand solve for the corresponding values of H. The calculations are summarized in the table below. Note that a factor of two (waves) is included in the equation for electrical output power.

267

Wave period, T ( s)

Wave height, H (m)

4.0

6.37

4.5

6.00

5.0

5.69

5.5

5.43

6.0

5.20

Solution check No errors are detected.

Discussion Obviously, there are an infinite number of combinations ofT and H that will satisfy the power requirement, but many combinations of these two quantities are not physically realistic. For example, a wave period of 1.0 s corresponds to a wave height of 12.7 m (41.7 ft).

268

6.16

Problem statement An ocean thermal energy conversion system operates between seawater temperatures of 16°C and 29°C. Find the maximum possible thermal efficiency of the system.

Diagram

, /

/

-.... '

I

I

I

' \ \

~W 1~ out I

heat engine

...

/

/

low-temperature sink

Assumptions 1. Source and sink temperatures are constant.

Governing equations TL

'lttz,idea/

= 1- ~ H

Calculations An OTEC system is a heat engine, so its efficiency is limited by the Camot efficiency.

We obtain, TL

'lttz,idea/

= 1- ~ H

= 1 - (16 + 273)K (29 + 273)K 269

= 0.0430 Solution check No errors are detected.

Discussion Efficiencies of OTEC systems are always low because of small temperature differences between warm and cool ocean water.

270

6.17

Problem statement Warm ocean water at 28oC supplies an OTEC system 500 kJ of thermal energy while 490 kJ ofheat is rejected to 19°C ocean water. Find the output work of the turbine, the actual thermal efficiency, and the Camot efficiency of the OTEC system.

Diagram

Qin = 500 kJ ;

....

/

/

'

I

I

I

' '\

heat engine

I \ \

' ' .... _

I

.... /

~W

1~

/

Qout = 490 kJ

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations

TL

'7tJz,idea/

= 1- T

H

Calculations

271

out

Solving for Wout from the first equation, we obtain

= 500 kJ - 490 kJ

= 10 kJ The actual thermal efficiency is

WQ~ut

11th=

zn

= (I 0 kJ)/( 500 kJ) = 0.020

The Camot efficiency is

TL

1lth,ideal =

1- T, H

= 1- (19+273)K (28 + 273)K = 0.0299 Solution check No errors are detected.

Discussion As demanded by the second law of thermodynamics, the actual thermal efficiency of the OTEC system is lower than the Camot efficiency.

272

6.18

Problem statement For an ideal thermal efficiency of at least 0.035, find the minimum warm ocean temperature for an OTEC power plant that rejects heat to 1rc ocean water.

Diagram

,.

.....

/

/

'

I

'

\ \

I 11th ideal= 0.035

I

,t=>w~

heat engine I

,. /

/

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations TL

1Jth,ideal

= 1- T

H

Calculations Solving the governing equation forTH, the warm ocean temperature, we obtain T.H-

TL } - 1lth,ideal

= (17 + 273)K (1 - 0.035)

273

=301 K=2rc Solution check No errors are detected.

Discussion Hence, for the OTEC power plant to have a Camot efficiency of at least 0.035, the warm ocean temperature must be at least 2rc.

274

6.19

(Student choice)

6.20

(Student choice)

6.21

(Student choice)

6.22

(Student choice)

275

Chapter7

Biomass

276

Section 7.3

Practice!

1. How much thermal energy is released by the complete burning of 600 kg of municipal solid waste? Include the latent heat of vaporization of water, and assume that at the end of the combustion process, the combustion products of the waste are returned to a temperature of 25°C. Solution

E=HHVm

= (19.9 MJ/kg)(600 kg) = 1.19 X 104 MJ = 11.9 GJ

277

2. A co-fired power plant bums a mixture of coal and a small amount of biomass at a rate of 18 kg/s. The combustion, thermal, and generator efficiencies are 0.90, 0.35, and 0.92, respectively. Find the electrical output power of the plant. If the coal-biomass mixture burns at 850°C and the temperature of the environment is l5°C, what is the maximum possible thermal efficiency of the plant?

Solution 1loveral/

= 1lcomb 17th 1Jgen = (0.90)(0.35)(0.92) = 0.290

~feet = 1loveral/HVrh

= (0.290)(24 MJ/kg)(18 kg/s) = 125 MW

1lth,idea/ =

TL 1- ~ H

= 1 - (15 + 273)K (850 + 273)K = 0.744 Note that the heating value of coal was used in the calculation, given that the amount of biomass in the mixture is small.

278

7.1

Problem statement How much thermal energy is released by the complete burning of 225 kg of Miscanthus?

Include the latent heat of vaporization of water, and assume that at the end of the combustion process, the Miscanthus combustion products are returned to a temperature of 25°C. Diagram

Miscanthus (225 kg) Assumptions 1. At the end of combustion, products are returned to 25 o C. Governing equations

E=HHVm Calculations

E=HHVm = (19.6 MJ/kg)(225 kg)

= 4410 MJ = 4.41 GJ Solution check No errors are detected.

Discussion A higher heating value (HHV) of 19.6 MJ/kg is at the upper range for Miscanthus, as shown in Table 7 .1.

279

7.2

Problem statement Find the thermal energy released by the complete combustion of350 kg of hardwood forest reside. Include the latent heat of vaporization of water, and assume that at the end of the combustion process, the hardwood combustion products are returned to a temperature of25°C. Diagram

Hardwood forest residue (350 kg) Assumptions 1. At the end of combustion, products are returned to 25 o C. Governing equations

E=HHVm Calculations

E=HHVm = (20.7 MJ/kg)(350 kg) = 7245 MJ = 7.25 GJ

Solution check No errors are detected.

Discussion As with Problem 7.1, a higher heating value (HHV) at the upper range was used in the calculation.

280

7.3

Problem statement Find the heat input to the boiler of a biomass power plant that bums 500 kg of municipal solid waste. Use the higher heating value, and assume a combustion efficiency of0.85.

Diagram Boiler

Assumptions 1. Biomass bums with higher heating value.

2. Combustion efficiency is 0.85.

Governing equations

Calculations Using the upper limit of the higher heating value (HHV) for HV, the heat input to the boiler is

= (0.85)(19.9 MJ/kg)(500 kg)

= 8458 MJ = 8.46 GJ

281

Solution check No errors are detected.

Discussion Because of various heat losses in the boiler, the value of Qin we found is not the amount ofheat imparted to the working fluid in the system.

282

7.4

Problem statement If the heat input per kilogram of fuel burned to the boiler of a biomass power plant is 16.5 MJ, what is the combustion efficiency if the fuel is corrugated paper?

Diagram Boiler

0 Qin

= 16.5 MJ/kg

Corrugated paper

Assumptions 1. All pertinent quantities are constant.

Governing equations

Calculations Letting m = 1 kg, we obtain

16.5 MJ/kg (18.5 MJ!kg)(1 kg)

= 0.892

283

Solution check No errors are detected.

Discussion The use of the upper limit on higher heating value was arbitrary. Had we used the lower limit, the combustion efficiency would be

'lcomb

=

16.5 MJ/kg (17.3 MJ/kg)(1 kg)

= 0.954.

284

7.5

Problem statement A biomass power plant bums soft wood forest residues at a rate of 1.6 kg/s. For a combustion efficiency of0.80, find the rate ofheat transfer to the boiler.

Diagram Boiler

Soft wood forest residues

Assumptions 1. All pertinent quantities are constant.

Governing equations

Calculations Solving the governing equation for rate of heat transfer, we obtain

= (0.80)(21.1 MJ/kg)(l.6 kg/s) =27.0MW

285

Solution check No errors are detected.

Discussion As with the other problems in this chapter, the upper limit of the higher heating value was used in the calculation.

286

7.6

Problem statement Find the thermal efficiency of an ideal power plant that bums biomass at a temperature of 350°C and rejects its waste heat to a 20°C environment.

Diagram

, /

.....

'

'\\ I heat engine ·~ f-----V Wout /

I

I

I

\

1 \

' .... .....

Qout

low-temperature sink

Assumptions 1. Source and sink temperatures are constant.

Governing equations TL

llth,idea/

= 1- ~ H

Calculations The thermal efficiency of the ideal power plant is TL

17th ,ideal

= 1- ~ H

= 1-

(20 + 273)K (350 + 273)K

287

= 0.530 Solution check No errors are detected.

Discussion The actual thermal efficiency of this power plant would be significantly less than the Camot value because of various energy losses in the system.

288

7. 7

Problem statement If the combustion and generator efficiencies ofthe ideal power plant in Problem 7.6 are 0.80 and 0.92, respectively, what is the overall efficiency?

Diagram

/

.....

/

/

'

'\

I

I

I \

\ \

' ' ....

/

.... /

\

·~ W

heat engine

1~

/

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations TJoverall = TJcomb 17th TJgen

Calculations The overall efficiency of the ideal power plant is

= (0.80)(0.530)(0.92) = 0.390

289

out

Solution check No errors are detected.

Discussion The actual overall efficiency would be lower than this value because the ideal (Camot) thermal efficiency was used in the calculation.

290

7.8

Problem statement A biomass power plant bums com stalks at a rate of 4.5 kg/s. If the overall efficiency of the plant is 0.28, find the electrical output power.

Diagram

/

..-

.....

/

'

I

'

I

I

heat engine

I \ \

' ' .....

':t=:> . . ._g_e_n_e_ra-to_r_ ,'- - _ r-.~

pelect

I / ..- ..-

'loverall = 0.28

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations ~feet

'loveral/

= HVrh

Calculations Solving the governing equation for electrical output power, Pelect• we obtain Peiect = 71overall HV

m

= (0.28)(18.5 MJ/kg)(4.5 kg/s)

=23.3 MW

291

Solution check No errors are detected.

Discussion Once again, the upper limit of the higher heating value was used in the calculation.

292

7.9

Problem statement The combustion and generator efficiencies of a biomass power plant are 0.88 and 0.94, respectively, and the turbine generates 20 MW of mechanical power. If the plant bums wheat straw at a rate of 6.0 kg/s, find the electrical output power, overall efficiency, and rate of heat transfer to the surroundings.

Diagram

1'/comb

/

/

"'

.....

......

''

I

I

I

heat engine

I \

''

I ......

/

'- .....

Wout=20MW

:t=>

'--------/

1'/gen

Assumptions 1. All pertinent quantities are constant.

Governing equations

= 17comb 17th 17gen

~ut

17th=-.-

Q;, ~feet

17overall

generator

/

"'

low-temperature sink

17overal/

= 0.88

= HVm 293

= 0.94

pelect

Solving the first equation for rate of heat input, we have

= (0.88)(18.9 MJ/kg)(6.0 kg/s) =99.8 MW Thermal efficiency of the plant is

n = n:ut

'lth

•

Qin = 20MW 99.8MW = 0.200

Hence, overall efficiency is

17overall = 17comb 17th TJgen = (0.88)(0.200)(0.94) = 0.165 Electrical output power is

Pe]ect = Yloveran HV m = (0.165)(18.9 MJ/kg)(6.0 kg/s) = 18.7MW Lastly, the rate of heat transfer to the environment is

=99.8MW- 20MW

294

=79.8MW Solution check No errors are detected.

Discussion As shown in the calculations, only 16.5 percent of the energy released in the combustion of the wheat straw is converted to electrical energy. The largest energy loss occurs in the conversion of thermal energy to mechanical energy in the turbine.

295

7.1 0

Problem statement In a biomass power plant, sugarcane is burned at a rate of 2.2 kg/s. The combustion, thermal, and generator efficiencies of the plant are 0.81, 0.60, and 0.93, respectively.

Find the overall efficiency, rate of heat transfer to the boiler, electrical output power, and turbine output power.

Diagram

'1comb =

;"

.....

/

I

'

I

'\

I

I

heat engine

I \ \ \

_ ' ......

I

.... "'

:t=>

/

11th= 0.60 '-g_e_n_e_ra-to_r__,l----. __ 1Jgen = 0.93

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations

(L

1lcomb

0.81

= HVrh

1]overall = 1]comb 17th 1]gen

n _ 1-i:ut •

'lth -

Qin ~feet 1loveral/ = HVrh

296

Calculations Overall efficiency is

17overall = 17comb 11th 17gen

= (0.81 )(0.60)(0.93) = 0.452

The rate of heat transfer to the boiler is

HVm n Q.in = 'lcomb

= (0.81)(19.4 MJ/kg)(2.2 kg/s) =34.6MW Electrical output power is ~lect = 'lloverai/HVm

= (0.452)(19.4 MJ/kg)(2.2 kg/s) = 19.3 MW Turbine output power is

= (0.60)(34.6 MW) =20.7MW Solution check No errors are detected.

Discussion The rate ofheat rejection to the environment is (34.6- 20.7) MW = 13.9 MW.

297

7.11

Problem statement If the Carnot efficiency of a biomass power plant is 0.55 and the temperature of the surroundings is 10°C, what must the combustion temperature of the fuel be?

Diagram

,;

,;

..... .....

I

I I

I\

'lth,ideal =

''

heat engine

\

' ' ..... .....

I

.... ,;

\

1

0.55

·~ W

~

out

/

low-temperature sink

Assumptions 1. Source and sink temperatures are constant.

Governing equations 1lth,idea/ =

TL

1- ~ H

Calculations Solving the governing equation for source temperature, TH, we obtain T. H

=

TL

1- 1Jth,ideal

= (10 + 273)K

1 - 0.55

298

= 629 K = 356°C Solution check No errors are detected.

Discussion Our result is a typical combustion temperature for some biomass fuels.

299

7.12

Problem statement In a direct-fired power plant, hard wood forest residue is burned at a rate of9.6 kg/sat a temperature of360°C. The combustion efficiency is 0.85, and the output power of the turbine shaft is 65 MW. If the efficiency of the electrical generator is 0.94, find the thermal efficiency, overall efficiency, electrical output power, rate of heat rejection to a 20°C atmosphere, and Carnot efficiency. If the plant capacity factor is 0.30, how much electrical energy does the plant generate in one year?

Diagram

1Jcomb

/

/

""

.....

'

I I

I

''

heat engine

I \ \

' ' ..... _

Wout=65 MW

:~

"'

1. All pertinent quantities are constant.

Governing equations

_

.---.,___~

1Jgen = 0.94

Assumptions

11 'lth -

__g_e_n_e_ra-to_r__

I / /

low-temperature sink

1Jcomb

= 0.85

Qin = HVrh

~ut •

Qin

300

pelect

P,lect

1lovera/l

= HVm

1lth,ideal

= 1- T,

TL

H

£elect

= CF P,/ect/1(

Calculations The rate of heat transfer to the boiler is

= (0.85)(20.7 MJ/kg)(9.6 kg/s) = 168.9 MW

Thermal efficiency is

~ut

17th

=-Q .. m

65MW 168.9MW

= 0.385 Overall efficiency is

= (0.85)(0.385)(0.94) = 0.307

Electrical output power is ~teet = TJoveral/HVfn

301

= (0.307)(20.7 MJ/kg)(9.6 kg/s) = 61.1 MW

The rate of heat rejection to the atmosphere is

= 168.9 MW- 65 MW = 104 MW Camot efficiency is

TL

1Jth,ideal

= 1- ~ H

= 1-

(20 + 273)K (360 + 273)K

= 0.537 The electrical energy generation for a one-year period is

E elect = CF ~lee/ /1t

= (0.30)(61.1 x 106 W)(3.15 x 107 s) = 5.77 X 10 14 J X (1 kWh/3.6 X 106 J)

Solution check No errors are detected.

Discussion As demanded by the second law of thermodynamics, the actual thermal efficiency is lower than the Camot efficiency.

302

7.13

Problem statement A co-fired power plant bums a mixture of coal and a small amount of municipal solid waste at a rate of 8.5 kg/s. The combustion, thermal, and generator efficiencies are 0.86, 0.40, and 0.94, respectively. Find the electrical output power of the plant. If the coalbiomass mixture bums at 800°C and the temperature of the environment is 10°C, what is the maximum possible thermal efficiency of the power plant?

Diagram

Y/comb = 0.86

/

....

.....

/

'

I

'

\

I

I

heat engine

I \

I

\ \

' ........

.... /

/

:t=>

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

2. Heating value is for pure coal.

Governing equations Qin

1Jcomb

= HVm

1Joverall = 1Jcomb 'lth 1Jgen Pelect -- 'lovera/1 Hv·m TL

'lth,ideal

= 1- ~ H

303

11th= 0.40 generator '--------/

Y/gen = 0.94

Calculations The rate of heat transfer to the boiler is

= (0.86)(24 MJ/kg)(8.5 kg/s)

= 175.4 MW

= (0.86)(0.40)(0.94) = 0.323 The electrical output power of the plant is ~feet = 1Joveral/HVm

= (0.323)(24 MJ/kg)(8.5 kg/s) =65.9MW The maximum possible thermal efficiency of the plant is 1JtJ,,ideal = 1 -

TL

T, H

= 1 - (10 + 273)K (800 + 273)K = 0.736 Solution check No errors are detected.

Discussion Again, as demanded by the second law of thermodynamics, the actual thermal efficiency is lower than the Camot efficiency.

304

7.14

Problem statement Consider a biomass power plant that operates as a Carnot heat engine between the temperature limits of 10°C and 300°C. The power plant bums black locust wood as fuel.

Assuming combustion and generator efficiencies of 0.90 and 0.95, find the electrical output power for a unit mass per second of fuel burned.

Diagram

1'/comb =

,

......

/

/

'

I

'

\

I

I

heat engine

I \ \

' ' .....

:t=:>

I

generator

-----/

/ .;

/

1'/gen =

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

2. Power plant operates as a Camot heat engine.

Governing equations TL

11th ,ideal

= 1- T

H

1loveral/

0.90

= 1lcomb 1lth,ideat1lgen

~feet = 1loveral/HVfn

305

0.95

Calculations The Carnot efficiency is '17th ,ideal = 1 -

TL

T

H

= 1- (10+273)K (300 + 273)K = 0.506

so the overall efficiency is

= (0.90)(0.506)(0.95) = 0.433 The electrical output power is ~feet = 'lloveral/HVm = (0.433)(19.9 MJ/kg)(l kg/s)

=8.62MW Solution check No errors are detected.

Discussion The higher heating value of black locust wood was used in the calculation.

306

7.15

Problem statement For the power plant in Problem 7.14, find the output power of the turbine and the rate of heat transfer to the surroundings.

Diagram

1'/comb = 0.90 ,/

.... .....

I I I

I

''

heat engine

I \

I

\

' ' ..........

.,., /

/

:t=>

generator '-------~

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

2. Power plant operates as a Camot heat engine.

Governing equations 1Jcomb

Qin = HVin

307

1'/gen = 0.95

Calculations The rate of heat transfer to the boiler is

= (0.90)(19.9 MJ/kg)(1 kg/s) = 17.9 MW so, knowing the thermal efficiency, the output power of the turbine is calculated as

= (0.506)(17.9 MW) =9.06MW The rate of heat transfer to the surroundings is

= 17.9 MW- 9.06 MW =8.84MW Solution check No errors are detected.

Discussion Keep in mind that the foregoing calculations are based on a 1 kg/s combustion rate. To calculate the actual turbine output power and rate of heat transfer to the surroundings, we would multiply our results by the actual combustion rate.

308

7.16

Problem statement An inventor proposes a biomass power plant that bums fuel at a temperature of275°C in a 20°C environment. The inventor claims that the combustion and generator efficiencies are 0.85 and 0.90, respectively. He also claims that the overall efficiency of the power plant is 0.45. Evaluate the inventor's claims.

Diagram

1'/comb = 0.85

,.,.

......

/

.....

/

I

I

I

1J overall = 0.4 5

''

heat engine

I \ \

:t=>

1

' ...... .....

...-"'

/

1'/gen = 0.90

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations

TL

17th ,ideal

= 1- ~ H

1loverall = 17comb 17th,ideal1lgen

Calculations The Camot efficiency is 17th ,ideal = 1 -

...._g_e_n_e_ra-to_r__,l---...r--

TL ~ H

309

= 1 - (20 + 273)K (275 + 273)K = 0.465 The inventor claims an overall efficiency of0.45. If the power plant has an overall efficiency of0.45, the thermal efficiency of the plant would have to be 11

7]overall

_

'lth 1lcomb 7Jgen

0.45 (0.85)(0.90)

= 0.588 This value exceeds the Camot efficiency, so the inventor's proposal violates the second law of thermodynamics. Another way to arrive at the same conclusion is to calculate the overall efficiency based on the Camot thermal efficiency, 1loverall

= 1lcomb 1lth,ideat1lgen = (0.85)(0.465)(0.90) = 0.356

Because the inventor claims an overall efficiency that exceeds the value based on the Camot thermal efficiency, the proposal violates the second law.

Solution check No errors are detected.

Discussion The problem does not provide enough information to evaluate whether the first law of thermodynamics (conservation of energy principle) is violated or not.

310

7.1 7

Problem statement Hulls, shells and prunings are burned at a rate of2.8 kg/sat a temperature of320°C in a direct-fired power plant. The combustion efficiency is 0.89, and the output power of the turbine shaft is 25 MW. If the efficiency of the electrical generator is 0.92, find the thermal efficiency, overall efficiency, electrical output power, rate ofheat rejection to a 5°C atmosphere, and Carnot efficiency. If the plant capacity factor is 0.25, how much electrical energy does the plant generate in one year?

Diagram

1'/comb

...,.

.....

/

/

'

I I

I

''

heat engine

I \

I

\

' ' .....

...,. /

/

Wout=25 MW

:t=>

generator '--------/

1'/gen

low-temperature sink

Assumptions 1. All pertinent quantities are constant.

Governing equations

Qin

'7cornb

= 0.89

= HVm

311

= 0.92

~/eel

1loverall

= HVm

1lth,idea/

= 1- ~

TL H

E elect = CF ~/eel /1t Calculations The rate of heat transfer to the boiler is

= (0.89)(20.5 MJ/kg)(2.8 kg/s) = 51.09 MW

so thermal efficiency is

25MW 51.09 MW = 0.489 Overall efficiency is TJoverall = TJcomb 17th TJgen

= (0.89)(0.489)(0.92)

= 0.400 Electrical output power is ~lect = 1loverai/HVm

312

= (0.400)(20.5 MJ/kg)(2.8 kg/s) =23.0MW The rate ofheat rejection to the atmosphere is

= 51.09 MW- 25 MW =26.1 MW Camot efficiency is '17th ,ideal = 1 -

TL

T

H

=1-

(5 + 273)K (320 + 273)K

= 0.531 The electrical energy generation for a one-year period is

E elect = CF ~feet /1t = (0.25)(23.0 x 106 W)(3.15 x 107 s) = 1.81 X 10 14 J X (1 kWh/3.6 X 106 J) = 5.03 X 107 kWh Solution check No errors are detected.

Discussion As demanded by the second law of thermodynamics, the actual thermal efficiency is lower than the Camot efficiency.

313

7.18

Problem statement A power plant that bums energy crops has a power generation capacity of 1.8 MW. The plant costs $3.0 million to install. The fixed charge rate is 7.0 percent, the annual operation and maintenance cost is assumed to be 1.0 percent of the initial cost, and the levelized replacement cost is averaged over an expected 20-year lifetime. The annual cost of energy crops is $300,000, and the capacity factor is 0.25. Find the cost of energy for the plant.

Diagram (Not applicable) Assumptions 1. All costs are constant, interest rate is constant and AOM is 1.0 percent of initial cost.

Governing equations

COE = (IC)(FCR) + LRC + AOM + AFC AEP Calculations The plant has a power generation capacity of 1.8 MW, so the amount of energy that the plant can produce in one year if it runs constantly is £max

= (1.8 MW)(8760 h/y) = 1.577 X 104 MWhly

The capacity factor is 0.25, so the actual annual energy production of the plant is

AEP = (0.25)(1.577 X 104 MWhly)

= 3.943 X 103 MWh/y The annual fuel cost, AFC, is $300,000. The LRC and AOM are

LRC= ($3.0 X 106)/(20 y)

= $1.50 X 105/y

314

AOM= (0.01/y)($3.0 X 106)

Substituting values, we obtain

COE = (IC)(FCR) + LRC + AOM + AFC AEP

= ($3.0 X 106)(0.07/y) + $1.50 X 105/y + $3.00 X 104/y + $300,000/y 3.943 X 103 MWh/y

= $175/MWh = $0.175/kWh Solution check No errors are detected.

Discussion The average cost of electrical energy in the United States is approximately $0.12/kWh, so the cost we have calculated for the biomass plant is somewhat higher than the national average.

315

7.19

(Student choice)

7.20

(Student choice)

7.21

(Student choice)

316