Thermo 8e SM Chap18

18-1

PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.

If you are a student using this Manual, you are using it without permission.

Solutions Manual for

Thermodynamics: An Engineering Approach

8th Edition

Yunus A. Çengel, Michael A. Boles

McGraw-Hill, 2015

Chapter 18

RENEWABLE ENERGY

PROPRIETARY AND CONFIDENTIAL

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18-2



Introduction

18-1C The concern over the depletion of fossil fuels and pollutant and greenhouse emissions associated by their combustion

can be tackled by essentially two methods: (i) Using renewable energy sources to replace fossil fuels. (ii) Implementing

energy efficiency practices in all aspects of energy production, distribution, and consumption so that less fuel is used while

obtaining the same useful output.

18-2C Main renewable energy sources include solar, wind, hydropower, geothermal, and biomass. Ocean, wave, and tidal

energies are also renewable sources but they are currently not economical and the technologies are still in the experimental

stage.

18-3C Although solar energy is sufficient to meet the entire energy needs of the world, currently it is not economical to do

so because of the low concentration of solar energy on earth and the high capital cost of harnessing it.

18-4C Wind is the fastest growing renewable. Hydropower represents the greatest amount of electricity production among

renewable.

18-5C We do not agree. The electricity used by the electric cars is generated somewhere else mostly by burning fuel

and thus emitting pollution. Therefore, each time an electric car consumes 1 kWh of electricity, it bears the

responsibility for the pollution s emitted as 1 kWh of electricity (plus the conversion and transmission losses) is

generated elsewhere. The electric cars can be claimed to be zero emission vehicles only when the electricity they

consume is generated by emission-free renewable resources such as hydroelectric, solar, wind, and geothermal energy.

18-3



Solar Energy

18-6C The conversion of solar energy into other useful forms of enegy can be accomplished by three conversion processes.

These are heliothermal process, heliothermal process, and helioelectrical process.

18-7C Solar energy is converted to electricity directly in solar cells while it is first converted to thermal energy in heliostats.

Heliostats are mirrors that reflect solar radiation into a single receiver. The resulting high temperature thermal energy is

converted to electricity by a heat engine.

18-8C Solar energy can be used for cooling applications in summer by absorption cooling systems but they are complex

devices involving high initial costs.

18-9C Solar radiation falls in ultraviolet, visible, and infrared regions of spectrum. Most radiation falls in near-infrared

region.

18-10C When solar radiation strikes a surface, part of it is absorbed, part of it is reflected, and the remaining part, if any, is

transmitted. The transmissivity , the reflectivity , and the absorptivity of a surface for solar energy are the fractions

of incident solar radiation transmitted, reflected, and absorbed, respectively. For an opaque surface,

0 and 1

18-11C Most solar collectors in operation today are used to produce hot water.

18-12C A thermosyphon solar water heater system operates on a natural circulation. Water flows through the system when

warm water rises into the tank as cooler water sinks. An active, closed loop solar water heater uses a pump for the

circulation of water containing antifreeze fluid.

18-13C A solar hot water collector may be equipped with an electric resistance heater to provide hot water when solar

energy is not available. A backup water heater that runs on natural gas or another fuel may also be used.

18-4



18-14C The efficiency of a flat-plate solar collector is given by

G

TTU ac

c

The efficiency of a solar collector is maximized when the temperature difference between the collector temperature and the

air temperature Tc Ta is zero. In this case, the efficiency is equal to . An unglazed collector has the highest maximum

efficiency as it has the highest value of values. This is followed by the single-glazing and double-glazing collectors.

18-15C The concentration of solar energy is low, and as a result, the temperature of hot water obtainable in a flat-plate

collector is low (usually under 80C). Hot fluid (water, steam, air, or another fluid) at much higher temperatures can be

produced using concentrating collectors by concentrating solar radiation on a smaller area. Temperatures in the receiver of a

concentrating collector can reach 400C.

18-16C The efficiency of a solar system used to produce electricity may be defined as the power produced divided by the

total solar irradiation. That is,

GA

W

Q

W

c

out

incident

outsolarth,

where Ac is the collector surface area receiving solar irradiation and G is the solar irradiation.

18-17C A promising method of power generation involves collecting and storing solar energy in large artificial lakes a few

meters deep, called solar ponds. Solar energy is absorbed by all parts of the pond, and the water temperature rises

everywhere. The top part of the pond, however, loses to the atmosphere much of the heat it absorbs, and as a result, its

temperature drops. This cool water serves as insulation for the bottom part of the pond and helps trap the energy there.

Usually, salt is planted at the bottom of the pond to prevent the rise of this hot water to the top. A power plant that uses an

organic fluid, such as alcohol, as the working fluid can be operated between the top and the bottom portions of the pond.

The main disadvantage of solar pond power plant is the low thermal efficiency.

18-18C An ocean thermal energy converter (OTEC) system uses the same principle as a solar pond system but in this case

the water at the sea or ocean surface is warmer as a result of solar energy absorption. The water at a deeper location is

cooler. Then, a heat engine can be operated that utilizes the surface warm water as heat source and deep cold water as the

heat sink. Experiments have been performed for OTEC principle but the results have not been promising due to large

installation cost and low thermal efficiency.

18-19C The current density J is defined as the current I over the cell surface area A. The current density flow from n-type

semiconductor to p-type semiconductor is denoted by Jr and called the light-induced recombination current and that from p-

type to n-type is denoted by Jo and called the dark current or reverse saturation current.

18-20C A photovoltaic system typically consists arrays, which are obtained by connecting modules and modules consists of

individual cells.

18-5



18-21C A house should be designed to receive (a) maximum solar heat in winter (to reduce heating energy consumption) in

winter-dominated climates and (b) minimum solar heat gain in summer (to reduce cooling energy consumption) in summer-

dominated climates.

18-22C Dark painted thick masonry walls called trombe walls are commonly used on south sides of passive solar homes to

absorb solar energy, store it during the day, and release it to the house during the night. The use of trombe wall reduces the

heating energy consumption of the building in winter.

12-23C (a) The spectral distribution of solar radiation beyond the earth’s atmosphere resembles the energy emitted by a

black body at 5982C, with about 39 percent in the visible region (0.4 to 0.7 m), and the 52 percent in the near infrared

region (0.7 to 3.5 m). (b) At a solar altitude of 41.8, the total energy of direct solar radiation incident at sea level on a

clear day consists of about 3 percent ultraviolet, 38 percent visible, and 59 percent infrared radiation.

12-24C A device that blocks solar radiation and thus reduces the solar heat gain is called a shading device. External shading

devices are more effective in reducing the solar heat gain since they intercept sun’s rays before they reach the glazing. The

solar heat gain through a window can be reduced by as much as 80 percent by exterior shading. Light colored shading

devices maximize the back reflection and thus minimize the solar gain. Dark colored shades, on the other hand, minimize

the back refection and thus maximize the solar heat gain.

12-25C The solar heat gain coefficient (SHGC) is defined as the fraction of incident solar radiation that enters through the

glazing. The solar heat gain of a glazing relative to the solar heat gain of a reference glazing, typically that of a standard 3

mm (1/8 in) thick double-strength clear window glass sheet whose SHGC is 0.87, is called the shading coefficient. They are

related to each other by

SHGC15.10.87

SHGC

SHGC

SHGC

glazing reference ofgain heat Solar

product ofgain heat Solar SC

ref

For single pane clear glass window, SHGC = 0.87 and SC = 1.0.

12-26C The SC (shading coefficient) of a device represents the solar heat gain relative to the solar heat gain of a reference

glazing, typically that of a standard 3 mm (1/8 in) thick double-strength clear window glass sheet whose SHGC is 0.87. The

shading coefficient of a 3-mm thick clear glass is SC = 1.0 whereas SC = 0.88 for 3-mm thick heat absorbing glass.

12-27C A window that transmits visible part of the spectrum while absorbing the infrared portion is ideally suited for

minimizing the air-conditioning load since such windows provide maximum daylighting and minimum solar heat gain. The

ordinary window glass approximates this behavior remarkably well.

18-6



12-28C A low-e coating on the inner surface of a window glass reduces both the (a) heat loss in winter and (b) heat gain in

summer. This is because the radiation heat transfer to or from the window is proportional to the emissivity of the inner

surface of the window. In winter, the window is colder and thus radiation heat loss from the room to the window is low. In

summer, the window is hotter and the radiation transfer from the window to the room is low.

18-29 The transmissivity of the glazing and the absorptivity of the absorber plate for a flat-plate solar collector are given.

The maximum efficiency of the collector is to be determined.

Properties The transmissivity of glazing and the absorptivity of absorber plate are given in the problem statement.

Analysis The efficiency of a solar collector is defined as

G

TTU

AG

TTUAAG

Q

Q acacc

)(

incident

useful

The collector efficiency is maximum when the collector temperature is equal to the air temperature ac TT and thus

0 ac TT . Therefore,

percent 77.1 7710)94.0)(82.0(max, .c

That is, the maximum efficiency of this collector is 77.1 percent.

18-30 The characteristics of a flat-plate solar collector and the rate of radiation incident are given. The collector efficiency is

to be determined.


Analysis The efficiency of this solar collector is determined from

percent 72.9

0.729

W/m750

C)2345(C) W/m3()95.0)(86.0(

2

2

G

TTU ac

c

That is, the efficiency of this collector is 72.9 percent.

18-7



18-31E The characteristics of a flat-plate solar collector and the rate of solar radiation incident are given. The collector

efficiency is to be determined.

Properties The product of transmissivity of glazing and the absorptivity of absorber plate is given in the problem statement.

Analysis (a) The efficiency of single-glazing solar collector is determined from

percent 74.8

0.748

ftBtu/h 260

F)67120(F)ftBtu/h 5.0()85.0(

2

2

G

TTU ac

c

(b) The efficiency of double-glazing solar collector is determined similarly as

percent 73.9

0.739

ftBtu/h 260

F)67120(F)ftBtu/h 3.0()80.0(

2

2

G

TTU ac

c

The efficiency of single-glazing collector is slightly greater than that of double-glazing collector.

18-32 A flat-plate solar collector is used to produce hot water. The characteristics of a flat-plate solar collector and the rate

of radiation incident are given. The temperature of hot water provided by the collector is to be determined.

Properties The product of transmissivity of glazing and the absorptivity of absorber plate is given in the problem statement.

The density of water is 1000 kg/m3 and its specific heat is 4.18 kJ/kgC.

Analysis The total rate of solar radiation incident on the collector is

W040,29) W/m880)(m 33( 22incident AGQ

The mass flow rate of water is

kg/s 105.0)L/s 60/3.6)(kg/L 1( V m

The rate of useful heat transferred to the water is determined from the definition of collector efficiency to be

W328,20 W)040,29)(70.0(incidentuseful

incident

useful QQQ

Qcc

Then the temperature of hot water provided by the collector is determined from

C64.3

out,

out,

in,out,useful

C)18C)(kJ/kg kg/s)(4.28 105.0( W328,20

)(

w

w

wwp

T

T

TTcmQ

That is, this collector is supplying water at 64.3C.

18-8



18-33E A flat-plate solar collector is considered. The collector efficiency for a given water inlet temperature and the water

temperature for a collector efficiency of zero are to be determined.


Analysis (a) The efficiency of single-glazing solar collector is determined from

percent 47.2

4720

ftBtu/h 210

F)60115(F)ftBtu/h 3.1)(92.0()97.0)(88.0)(1)(92.0(

2

2

in,

.

G

TTUFKF

awRRc

(b) Setting the collector efficiency zero in the efficiency relation gives

F198

in,

2

in,2

in,

in,

0ftBtu/h 210

F)60(F)ftBtu/h 3.1)(92.0()97.0)(88.0)(1)(92.0(

0

w

w

awRR

awRR

T

T

G

TTUFKF

G

TTUFKF

18-34 The characteristics of a concentrating solar collector are given. The collector efficiency is to be determined.

Analysis The efficiency of this solar collector is determined from

percent 87.4

0.874

) W/m520)(15(

C)20130(C) W/m4(93.0

CR 2

2

G

TTU ac

arc

That is, the efficiency of this collector is 87.4 percent.

18-35 A solar power plant utilizing parabolic trough collectors is considered. The power generated is to be determined.

Analysis The efficiency of a solar power plant is defined as the power produced divided by the total solar irradiation. That

is,

GA

W

Q

W

c

out

incident

outsolarth,

Solving for the power output,

kW 140 )kW/m )(0.700m 0(0.08)(250 22solarth,out GAW c

18-9



18-36 A solar pond power plant with specified temperatures and thermal efficiency is considered. The second-law efficiency

of the power plant power is to be determined.

Analysis The maximum thermal efficiency of this power plant is equal to

Carnot efficiency, and is determined from

1293.0K 273)(75

K 273)(3011maxth,

H

L

T

T

The second-law efficiency is the ratio of actual thermal efficiency to the

maximum thermal efficiency:

percent 27.8 278.01293.0

036.0

maxth,

thII

75C

30C

Solar pond

power plant outW

th = 3.6%

18-10



18-37 A solar-power-tower plant is considered. The power output and thermal efficiency of the plant, the annual electricity

production, and the cost of the plant are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) Using the turbine isentropic efficiency, the steam properties at the inlet and exit of the turbine are determined as

follows: (Tables A-4E, A-5E, and A-6E),

Btu/lbm 7.102943.9892.132588.02.1325

Btu/lbm 43.989psia 2

RBtu/lbm 7037.1

Btu/lbm 2.1325

F600

psia 160

211221

21

212

2

1

1

1

1

sTs

T

s

hhhhhh

hh

hss

P

s

h

T

P

Then the power output is

kW 4677

Btu/h 3412.14

kW 1Btu/h) 10(1.596

Btu/h 101.596Btu/lbm )7.10295.2lbm/h)(132 360015()(

7

721out hhmW

The thermal efficiency of this power plant is equal to power output divided by the total solar incident on the heliostats:

16.0%

1596.0

)ftBtu/h )(250ft (400,000

Btu/h 101.59622

7out

thAG

W

(b) The solar data for Houston, Texas is given in Table 18-4. The daily average solar irradiation for an entire year on a

horizontal surface is given to be 15.90 MJ/m2day. Multiplying this value with 365 days of the year gives an estimate of total

solar irradiation on the heliostat surfaces. Using the definition of the thermal efficiency,

kWh 107.189

ft 10.764

m 1

kJ 3600

kWh 1days) day)(365kJ/m 900,15)(ft 000,400)(12.0(

6

2

222

avgth,out

AGW

This is total power output from the turbine. The electrical energy output from the generator is

kWh 107.045 6 kWh) 10189.7)(98.0( 6outgenelect WW

(c) The total cost of this power plant is

7out 10$7.949kW) 4677)(kW/000,17($costUnit costPlant W

The annual revenue from selling the electricity produced is

/year10$2.315

h) 4500)(kW/11.0($kW) 4677(

hours OperatingpriceUnit revenue Annual

6

out

W

The simple payback period is then

years34.3

/year10$2.315

10$7.949

revenue Annual

costPlant periodPayback

6

7

That is, this plant pays for itself in 34 years, which is not unexpected for solar power plants.

18-11



18-38 A solar cell with a specified value of open circuit voltage is considered. The current output density, the load voltage

for maximum power output, and the maximum power output of the cell for a unit cell area are to be determined.

Analysis (a) The current output density is determined from Eq. 18-18 to be

2 A/m3.676

s

s

o

s

ooc

J

J

J

J

e

kTV

1A/m 109.1

lnJ/V 106.1

)K 298)(J/K 10381.1(V 55.0

1ln

2919

23

(b) The load voltage at which the power output is maximum is determined from Eq. 18-22 to be

V0.4737

max

23

max19

292

23

max19

max

max

V

)K 298)(J/K 10381.1(

V)J/V 106.1(1

)A/m 109.1/A/m 676.3(1

)K 298)(J/K 10381.1(

V)J/V 106.1(exp

V1

/1Vexp

kT

e

JJ

kT

e

o

oso

(c) The maximum power output of the cell for a unit cell area is determined from

2 W/m1.652

)V 4737.0(J/V) 106.1(

)K 298)(J/K 10381.1(1

)A/m 109.1)(A/m (3.676)V 4737.0(

V1

)(V

19

23

292

max

maxmax

o

os

e

kT

JJW

18-12



18-39 A solar cell with a specified value of open circuit voltage is considered. The power output and the efficiency of the

cell are to be determined.

Analysis The current output density is determined from Eq. 18-18 to be

2

2919

23

A/m 45.12

1A/m 109.3

lnJ/V 106.1

)K 300)(J/K 10381.1(V 60.0

1ln

s

s

o

s

ooc

J

J

J

J

e

kTV

The power output is

W627

1

)K 300)(J/K 10381.1(

)V 52.0(J/V) 106.1(exp)A/m 10)(3.9m 28)(V 52.0()A/m )(45.12m 28)(V 52.0(

1V

expVV

23

1929222

kT

eAJAJW o

os

The efficiency of the cell is

percent 4.62 0462.0) W/m)(485m (28

W62722th

AG

W

18-13



18-40E A solar cell with a specified value of open circuit voltage is considered. The load voltage for maximum power

output, the efficiency of the cell, and the cell area for a given power output are to be determined.

Analysis (a) The current output density is determined from Eq. 18-18 to be

2

2

2210

19

23

A/m 48.64

1

m 1

ft 764.10A/ft 1011.4

lnJ/V 106.1

K ]15.2738.1/)3275)[(J/K 10381.1(V 60.0

1ln

s

s

o

s

ooc

J

J

J

J

e

kTV

The load voltage at which the power output is maximum is determined from Eq. 18-22 to be

V0.5215

max

23

max19

2

22102

23

max19

max

max

V

K ]15.2738.1/)3275)[(J/K 10381.1(

V)J/V 106.1(1

m 1

ft 764.10A/ft 1011.4/)A/m 48.64(1

K ]15.2738.1/)3275)[(J/K 10381.1(

V)J/V 106.1(exp

V1

/1Vexp

kT

e

JJ

kT

e

o

oso

(b) The load current density JL is determined from

2

23

19

2

22102

A/m 93.50

1K ]15.2738.1/)3275)[(J/K 10381.1(

V) 56.0)(J/V 106.1(exp

m 1

ft 764.10A/ft 1011.4)A/m 48.64(

1V

exp

kT

eJJJJJ o

osjsL

The power output is

22 W/m52.28)V 56.0)(A/m 93.50( VJW L

The efficiency of the cell is

percent 4.11

0411.0

ftBtu/h 0.3171

W/m1)ftBtu/h (220

W/m28.52

2

22

2

thG

W

(c) Finally, the cell area for a power output of 500 W is

2ft 188.7

2

2

2

total

m 1

ft 764.10

W/m52.28

W500

W

WA

18-14



18-41E A solar cell with a specified value of open circuit voltage is considered. The maximum conversion efficiency of the

solar cell is to be determined.

Analysis The current output density is determined from Eq. 18-18 to be

2

2

2210

19

23

A/m 48.64

1

m 1

ft 764.10A/ft 1011.4

lnJ/V 106.1

K ]15.2738.1/)3275)[(J/K 10381.1(V 60.0

1ln

s

s

o

s

ooc

J

J

J

J

e

kTV

The load voltage at which the power output is maximum is determined from Eq. 18-22 to be

V0.5215

max

23

max19

2

22102

23

max19

max

max

V

K ]15.2738.1/)3275)[(J/K 10381.1(

V)J/V 106.1(1

m 1

ft 764.10A/ft 1011.4/)A/m 48.64(1

K ]15.2738.1/)3275)[(J/K 10381.1(

V)J/V 106.1(exp

V1

/1Vexp

kT

e

JJ

kT

e

o

oso

The maximum conversion efficiency of this solar cell is determined from

percent 4.62

04619.0

V) 5J/V)(0.521 106.1(

K ]15.2738.1/)3275)[(J/K 10381.1(1

ftBtu/h 0.3171

W/m1)ftBtu/h (220

m 1

ft 764.10A/ft 1011.4)A/m 48.64V)( (0.5215

V1

)(

19

23

2

22

2

22102

max

maxmaxcell,

o

os

e

kTG

JJV

18-42 The thermal efficiency of a solar car using solar cells is to be determined.

Analysis The thermal efficiency of the solar car is determined from

percent 7.85 0785.0) W/m)(860m (8

W54022th

AG

W

18-15



18-43 A building located near 40º N latitude has equal window areas on all four sides. The side of the building with the

highest solar heat gain in summer is to be determined.

Assumptions The shading coefficients of windows on all sides of the building are identical.

Analysis The reflective films should be installed on the side that receives the most incident solar radiation in summer since

the window areas and the shading coefficients on all four sides are identical. The incident solar radiation at different

windows in July are given to be (Table 18-3)

Month Time The daily total solar radiation incident on the surface, Wh/m2

North East South West

July Daily total 1621 4313 2552 4313

Therefore, the reflective film should be installed on the East or West windows (instead of the South windows) in order to

minimize the solar heat gain and thus the cooling load of the building.

18-16



18-44 The net annual cost savings due to installing reflective coating on the West windows of a building and the simple

payback period are to be determined.

Assumptions 1 The calculations given below are for an average year. 2

The unit costs of electricity and natural gas remain constant.

Analysis Using the daily averages for each month and noting the number

of days of each month, the total solar heat flux incident on the glazing

during summer and winter months are determined to be

Qsolar, summer = 4.2430+ 4.1631+ 3.9331+3.4830

= 482 kWh/year

Qsolar, winter = 2.9431+ 2.3330+2.0731+2.3531+3.0328+3.6231+4.0030

= 615 kWh/year

Then the decrease in the annual cooling load and the increase in the

annual heating load due to reflective film become

Cooling load decrease = Qsolar, summer Aglazing (SHGCwithout film - SHGCwith film)

= (482 kWh/year)(60 m2)(0.766-0.35)

= 12,031 kWh/year

Heating load increase = Qsolar, winter Aglazing (SHGCwithout film - SHGCwith film)

= (615 kWh/year)(60 m2)(0.766-0.35)

= 15,350 kWh/year = 523.7 therms/year

since 1 therm = 29.31 kWh. The corresponding decrease in cooling costs and increase in heating costs are

Decrease in cooling costs = (Cooling load decrease)(Unit cost of electricity)/COP

= (12,031 kWh/year)($0.15/kWh)/3.2 = $564/year

Increase in heating costs = (Heating load increase)(Unit cost of fuel)/Efficiency

= (523.7 therms/year)($0.90/therm)/0.90 = $524/year

Then the net annual cost savings due to reflective films become

Cost Savings = Decrease in cooling costs - Increase in heating costs = $564 - 524 = $40/year

The implementation cost of installing films is

Implementation Cost = ($15/m2)(60 m

2) = $900

This gives a simple payback period of

years22.5$40/year

900$

savingscost Annual

costtion Implementaperiodpayback Simple

Discussion The reflective films will pay for themselves in this case in about 23 years, which is unacceptable to most

manufacturers since they are not usually interested in any energy conservation measure which does not pay for itself within 3

years.

Air

space

Glass

Reflective

film

Sun

Transmitted

Reflected

18-17



18-45 A house located at 40ºN latitude has ordinary double pane windows. The total solar heat gain of the house at 9:00,

12:00, and 15:00 solar time in July and the total amount of solar heat gain per day for an average day in January are to be

determined.

Assumptions The calculations are performed for an average day in a given month.

Properties The shading coefficient of a double pane window with 6-mm thick glasses is SC = 0.82 (Table 18-6). The

incident radiation at different windows at different times are given as (Table 8-3)

Month Time Solar radiation incident on the surface, W/m2


July 9:00 117 701 190 114

July 12:00 138 149 395 149

July 15:00 117 114 190 701

January Daily total 446 1863 5897 1863

Analysis The solar heat gain coefficient (SHGC) of the windows is determined from Eq.12-57 to be

SHGC = 0.87SC = 0.870.82 = 0.7134

The rate of solar heat gain is determined from

incident solar,glazing

incident solar,glazinggainsolar

7134.0 qA

qASHGCQ

Then the rates of heat gain at the 4 walls at 3 different times in July become

North wall:

W334

W394

W334

) W/m(117)m 4(7134.0

) W/m(138)m 4(7134.0

) W/m(117)m 4(7134.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

East wall:

W488

W638

W3001

) W/m(114)m 6(7134.0

) W/m(149)m 6(7134.0

) W/m(701)m 6(7134.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

South wall:

W1084

W2254

W1084

) W/m(190)m 8(7134.0

) W/m(395)m 8(7134.0

) W/m(190)m 8(7134.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

West wall:

W3001

W638

W488

) W/m(701)m 6(7134.0

) W/m(149)m 6(7134.0

) W/m(114)m 6(7134.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

Similarly, the solar heat gain of the house through all of the windows in January is determined to be

January:

Double-pane

window

Solar heat

gain

Sun

18-18



Wh/day7974 day) Wh/m(1863)m 6(7134.0

Wh/day33,655 day) Wh/m(5897)m 8(7134.0

Wh/day7974 day) Wh/m(1863)m 6(7134.0

Wh/day1273 day) Wh/m(446)m 4(7134.0

22Westgain,solar

22South gain,solar

22Eastgain,solar

22North gain,solar

Q

Q

Q

Q

Therefore, for an average day in January,

kWh/day 58.9 Wh/day876,587974655,3379741273dayper gain solar Q

18-19



18-46 A house located at 40º N latitude has gray-tinted double pane windows. The total solar heat gain of the house at 9:00,

12:00, and 15:00 solar time in July and the total amount of solar heat gain per day for an average day in January are to be

determined.

Assumptions The calculations are performed for an average day in a given month.

Properties The shading coefficient of a gray-tinted double pane window with 6-mm thick glasses is SC = 0.58 (Table 18-6).

The incident radiation at different windows at different times are given as (Table 18-3)

Month Time Solar radiation incident on the surface, W/m2


July 9:00 117 701 190 114

July 12:00 138 149 395 149

July 15:00 117 114 190 701

January Daily total 446 1863 5897 1863

Analysis The solar heat gain coefficient (SHGC) of the windows is determined from Eq.11-57 to be

SHGC = 0.87SC = 0.870.58 = 0.5046

The rate of solar heat gain is determined from

incident solar,glazingincident solar,glazinggainsolar 5046.0 qAqASHGCQ

Then the rates of heat gain at the 4 walls at 3 different times in July become

North wall:

W236

W279

W236

) W/m(117)m 4(5046.0

) W/m(138)m 4(5046.0

) W/m(117)m 4(5046.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

East wall:

W345

W451

W2122

) W/m(114)m 6(5046.0

) W/m(149)m 6(5046.0

) W/m(701)m 6(5046.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

South wall:

W767

W1595

W767

) W/m(190)m 8(5046.0

) W/m(395)m 8(5046.0

) W/m(190)m 8(5046.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

West wall:

W2122

W451

W345

) W/m(701)m 6(5046.0

) W/m(149)m 6(5046.0

) W/m(114)m 6(5046.0

22 00:15 gain,solar

22 00:gain,12solar

22 00:9 gain,solar

Q

Q

Q

Similarly, the solar heat gain of the house through all of the windows in January is determined to be

January:

Double-pane

window

solarQ

Sun

Heat-

absorbing

glass

18-20



Wh/day5640 day) Wh/m(1863)m 6(5046.0

Wh/day23,805 day) Wh/m(5897)m 8(5046.0

Wh/day5640 day) Wh/m(1863)m 6(5046.0

Wh/day900 day) Wh/m(446)m 4(5046.0

22Westgain,solar

22South gain,solar

22Eastgain,solar

22North gain,solar

Q

Q

Q

Q

Therefore, for an average day in January,

kWh/day 35.895 Wh/day985,355640805,235640900dayper gain solar Q

18-21



18-47 A building at 40º N latitude has double pane heat absorbing type windows that are equipped with light colored

venetian blinds. The total solar heat gains of the building through the south windows at solar noon in April for the cases of

with and without the blinds are to be determined.

Assumptions The calculations are performed for an “average” day in April, and may vary from location to location.

Properties The shading coefficient of a double pane heat absorbing type windows is SC = 0.58 (Table 18-6). It is given to be

SC = 0.30 in the case of blinds. The solar radiation incident at a South-facing surface at 12:00 noon in April is 559 W/m2

(Table 18-3).

Analysis The solar heat gain coefficient (SHGC) of the windows without the blinds is determined from

SHGC = 0.87SC = 0.870.58 = 0.5046

Then the rate of solar heat gain through the window becomes

W21,440

) W/m)(559m 76(5046.0 22

incident solar,glazingblinds no gain,solar qASHGCQ

In the case of windows equipped with venetian blinds,

the SHGC and the rate of solar heat gain become

SHGC = 0.87SC = 0.870.30 = 0.261

Then the rate of solar heat gain through the window becomes

W11,090

) W/m)(559m 76(261.0 22

incident solar,glazingblinds no gain,solar qASHGCQ

Discussion Note that light colored venetian blinds significantly reduce the solar heat, and thus air-conditioning load in

summers.

Double-

pane

window

Heat-

absorbing

glass

Venetian

blinds

Light

colored

18-22



18-48 A house has double door type windows that are double pane with 6.4 mm of air space and aluminum frames and

spacers. It is to be determined if the house is losing more or less heat than it is gaining from the sun through an east window

in a typical day in January.

Assumptions 1 The calculations are performed for an “average” day in January. 2 Solar data at 40 latitude can also be used

for a location at 39 latitude.

Properties The shading coefficient of a double pane window with 3-mm thick clear glass is SC = 0.88 (Table 18-6). The

overall heat transfer coefficient for double door type windows that are double pane with 6.4 mm of air space and aluminum

frames and spacers is 4.55 W/m2.C. The total solar radiation incident at an East-facing surface in January during a typical

day is 1863 Wh/m2 (Table 18-3).

Analysis The solar heat gain coefficient (SHGC) of the windows is

determined from

SHGC = 0.87SC = 0.870.88 = 0.7656

Then the solar heat gain through the window per unit area

becomes

kWh 1.426 Wh1426

) Wh/m)(1863m 1(7656.0 22

ldaily tota solar,glazinggainsolar qASHGCQ

The heat loss through a unit area of the window during a 24-h period is

kWh 1.31 Wh1310

h) C(24)10)(22m 1(C) W/m55.4(

)day 1)((

22

ave ,0windowwindow windowloss, windowloss, TTAUtQQ i

Therefore, the house is loosing less heat than it is gaining through the East windows during a typical day in January.

Double-

pane

window

Sun

Solar

heat

gain

Heat

loss

22°C 10°C

18-23



18-49 A house has double door type windows that are double pane with 6.4 mm of air space and aluminum frames and

spacers. It is to be determined if the house is losing more or less heat than it is gaining from the sun through a South window

in a typical day in January.

Assumptions 1 The calculations are performed for an “average” day in January. 2 Solar data at 40 latitude can also be used

for a location at 39 latitude.

Properties The shading coefficient of a double pane window with 3-mm thick clear glass is SC = 0.88 (Table 18-6). The

overall heat transfer coefficient for double door type windows that are double pane with 6.4 mm of air space and aluminum

frames and spacers is 4.55 W/m2.C. The total solar radiation incident at a South-facing surface in January during a typical

day is 5897 Wh/m2 (Table 18-3).

Analysis The solar heat gain coefficient (SHGC) of the windows is

determined from

SHGC = 0.87SC = 0.870.88 = 0.7656

Then the solar heat gain through the window per unit area becomes

kWh 4.515 Wh4515

) Wh/m)(5897m 1(7656.0 22

ldaily tota solar,glazinggainsolar qASHGCQ

The heat loss through a unit area of the window during a 24-h period is

kWh 1.31 Wh1310

h) C(24)10)(22m 1(C) W/m55.4(

)day 1)((

22

ave ,0windowwindow windowloss, windowloss, TTAUtQQ i

Therefore, the house is loosing much less heat than it is gaining through the South windows during a typical day in January.

Double-

pane

window

Sun

Solar

heat

gain

Heat

loss

22°C 10°C

18-24



18-50E A house has 1/8-in thick single pane windows with aluminum frames on a West wall. The rate of net heat gain (or

loss) through the window at 3 PM during a typical day in January is to be determined.

Assumptions 1 The calculations are performed for an “average” day in January. 2 The frame area relative to glazing area is

small so that the glazing area can be taken to be the same as the window area.

Properties The shading coefficient of a 1/8-in thick single pane window is SC = 1.0 (Table 18-6). The overall heat transfer

coefficient for 1/8-in thick single pane windows with aluminum frames is 6.63 W/m2.C = 1.17 Btu/h.ft

2.F. The total solar

radiation incident at a West-facing surface at 3 PM in January during a typical day is 557 W/m2 = 177 Btu/h.ft

2 (Table 18-3).

Analysis The solar heat gain coefficient (SHGC) of the windows is determined from

SHGC = 0.87SC = 0.871.0 = 0.87

The window area is: 2window ft 135 ft) ft)(15 9( A

Then the rate of solar heat gain through the window at 3 PM becomes

Btu/h 789,20

)Btu/h.ft )(177ft 135(87.0 22

PM 3 solar,glazingPM 3 gain,solar

qASHGCQ

The rate of heat loss through the window at 3 PM is

Btu/h 9161

F)20)(78ft 135(F)ftBtu/h 17.1(

)(

22

0windowwindow windowloss,

TTAUQ i

The house will be gaining heat at 3 PM since the solar heat gain is larger than the heat loss. The rate of net heat gain through

the window is

Btu/h 11,630 9161789,20 windowloss,PM 3 gain,solar net QQQ

Discussion The actual heat gain will be less because of the area occupied by the window frame.

Single

glass

Sun

20°F 78°F

18-25



Wind Energy

18-51C Windmill is used for mechanical power generation (grinding grain, pumping water, etc.) but wind turbine is used for

electrical power generation, although technically both devices are turbines since they extract energy from the fluid.

18-52C Theoretically, electricity can be generated at all wind speeds but for economical power generation, the minimum

wimd speed should be about 6 m/s.

18-53C Cut-in speed is the minimum wind speed at which useful power can be generated. Rated speed is the wind speed that

delivers the rated power, usually the maximum power. Cut-out speed is the maximum wind speed at which the wind turbine

is designed to produce power. At wind speeds greater than the cut-out speed, the turbine blades are stopped by some type of

braking mechanism to avoid damage and for safety issues.

18-54C The wind power potential is given by

3

available2

1AVW

where is the density of air, A is the blade span area of the wind turbine, and V is the wind velocity. The power potential of

a wind turbine is proportional to the density of air, and the density of air is inversely proportional to air temperature through

the relation RT

P . Therefore, the location with cooler air has more wind power potential. Note that both locations are at

the same pressure since they are at the same altitude.

18-55C The wind power potential is given by

3

available2

1AVW

where is the density of air, A is the blade span area of the wind turbine, and V is the wind velocity. The power potential of

a wind turbine is proportional to the density of air and the density of air is proportional to atmospheric pressure through the

relation RT

P . Since atmospheric pressure is higher at a lower altitude, the location at higher altitude has more wind

power potential.

18-56C As a general rule of thumb, a location is considered poor for construction of wind turbines if the average wind

power density is less than about 100 W/m2, good if it is around 400 W/m

2, and great if it is greater than about 700 W/m

2.

Therefore, we recommend wind turbine installation in site B and site C only.

18-26



18-57C The overall wind turbine efficiency is defined as

3

electric

available

electricoverallwt,

2

1AV

W

W

W

where electricW is electric power output, is density of air, A is blade span area, and V is wind velocity. The overall wind

turbine efficiency is related to wind turbine efficiency as

available

electric

shaft

electric

available

shaftneratorgearbox/gewtoverallwt,

W

W

W

W

W

W

where gearbox/generator is the gearbox/generator efficiency.

18-58C The theoretical limit for wind turbine efficiency based on the second law of thermodynamics is 100 percent. This

represents the conversion of entire kinetic energy of wind into work. This would be the case only when the velocity of air at

the turbine exit is zero. This is not possible for practical reasons because air must be taken away at the turbine exit to

maintain the mass flow through the turbine. Therefore, there is a limit for wind turbine efficiency which is less than 100

percent, and this is called the Betz limit. The Betz limit is calculated to be 0.5926.

18-59C Turbine A: The given efficiency (41 percent) is possible and realistic. Turbine B: The given efficiency (59 percent)

is possible because it is less than Betz limit but is not probable. Turbine C: The given efficiency (67 percent) is not possible

because it is greater than the Betz limit.

18-60 The wind power potential of a wind turbine at a specified wind speed is to be determined.

Assumptions Wind flows steadily at the specified speed.

Properties The gas constant of air is R = 0.287 kJ/kgK.

Analysis The density of air is determined from the ideal gas relation to be

3

3kg/m 130.1

K) 296(K)/kgmkPa 287.0(

kPa) 96(

RT

P

The blade span area is

222 m 19634/m) 50(4/ DA

Then the wind power potential is

kW 468

22

32331available

/sm 1000

kJ/kg 1m/s) )(7.5m )(1963kg/m (1.130

2

1

2

1AVW

18-27



18-61 Average wind power density values are given for two locations. The average wind speed in these locations are to be

determined.

Properties The density of air is given to be = 1.18 kg/m3.

Analysis The relation for wind power density is

3

2

1WPD V

Solving for wind velocity at the two locations, we obtain

m/s 6.97 13

1323

11 )kg/m (1.182

1 W/m200

2

1WPD VVV

m/s 8.79 23

2323

22 )kg/m (1.182

1 W/m400

2

1WPD VVV

18-62 A wind turbine is to generate power with a specified wind speed. The average electric power output, the amount of

electricity produced, and the revenue generated are to be determined.



Analysis (a) The blade span area is

222 m 9.4904/m) 25(4/ DA

The wind power potential is

kW 68.92/sm 1000

kJ/kg 1m/s) )(6m )(490.9kg/m (1.3

2

1

2

122

32331available

AVW

The electric power generated is

kW 23.43 kW) (68.92)34.0(availableoverallwt,electric WW

(b) The amount of electricity produced is determined from

kWh 187,500 h) kW)(8000 43.23(hours Operatingelectricelectric WW

(c) The revenue generated is

$16,900 /kWh)kWh)($0.09 500,187(yelectricit of priceUnit Revenue electricW

18-28



18-63 A wind turbine generates a certain amount of electricity during a week period. The average wind speed during this

period is to be determined.


Analysis The total operating hours during a week period is

Operating hours = 7 24 h = 168 h

The average rate of electricity production is

kW 48.65h 168

kWh 000,11

hours Operating

electricelectric

WW

The available wind power is

kW 8.2330.28

kW 65.48

overallwt,

electricavailable

WW


222 m 12574/m) 40(4/ DA

Finally, the average wind speed is determined from the definition of available wind power to be

m/s 6.85

V

V

AVW

22

323

3available

/sm 1000

kJ/kg 1)m )(1257kg/m (1.16

2

1kW 8.233

2

1

18-29



18-64E A wind turbine is to generate power with two specified wind speeds. The amount of electricity that can be produced

by the turbine and the blade tip speed are to be determined.

Assumptions Wind flows steadily at the specified speeds.

Properties The density of air is given to be = 0.075 lbm/ft3.

Analysis The blade span area is

222 ft 880,264/ft) 185(4/ DA

The wind power potential at the wind speed of 16 ft/s is

kW 0.174/sftlbm 23,730

kW 1ft/s) )(16ft )(26,880lbm/ft (0.075

2

1

2

132

323311available,

AVW

The wind power potential at the wind speed of 24 ft/s is

kW 2.587/sftlbm 23,730

kW 1ft/s) )(24ft )(26,880lbm/ft (0.075

2

1

2

132

323322available,

AVW

The overall wind turbine efficiency at a wind speed of 16 ft/s is

2790.0)93.0)(30.0(genwt,1overall,1wt,

The overall wind turbine efficiency at a wind speed of 24 ft/s is

3255.0)93.0)(35.0(genwt,2overall,2wt,

The electric power generated at a wind speed of 16 ft/s is

kW 48.54kW) (174.0)2790.0(1available,overall,1wt,electric,1 WW

The electric power generated at a wind speed of 24 ft/s is

kW 1.191kW) (587.2)3255.0(2available,overall,2wt,electric,2 WW

The amount of electricity produced at a wind speed of 16 ft/s is

kWh 145,630h) kW)(3000 54.48(hours Operating 1electric,1electric,1 WW

The amount of electricity produced at a wind speed of 24 ft/s is

kWh ,560764h) kW)(4000 1.191(hours Operating 2electric,2electric,2 WW

The total amount of electricity produced is

kWh 910,000 560,764560,145electric,2electric,1totalelectric, WWW

Noting that the tip of blade travels a distance of D per revolution, the tip velocity of the turbine blade for a rotational speed

of n becomes

mph 99.1

ft/s 1.46667

mi/h 1

s 60

min 1min)/15(ft) 185(tip nDV

18-30



18-65E Air velocities at the inlet and exit of a turbine are given. The wind turbine efficiency is to be determined when the

frictional effects are neglected.

Assumptions Wind flows steadily at the specified speeds.

Analysis The wind turbine efficiency can be determined from Eq. 18-44 to be

percent 39.2

3916.0

1ft/s) (25ft/s 19.5

1

wt

wt

wt12

VV

18-31



Hydropower

18-66C A water pump increases the pressure of water by supplying mechanical energy. A hydroturbine converts potential

energy of water into mechanical shaft work and the pressure of water decreases across the turbine.

18-67C The maximum work that can be produced by this turbine is determined from

wmax = (P1 P2)/

where is the density of water.

18-68C We are usually more interested in the overall efficiency of turbine-generator combination rather than the turbine

efficiency because a turbine is usually packaged together with its generator and it is much easier to measure the electric

power output from the generator compared to shaft power output from the turbine.

18-69C No, we do not agree. The limitation that a machine cannot have an efficiency of 100 percent even in the absence of

irreversibilities holds for heat engines which convert heat to work. A hydraulic turbine is not a heat engine as it converts

mechanical energy into work and the upper limit for its efficiency is 100 percent.

18-70C No, we do not agree. A wind turbine converts kinetic energy of air into work and the Betz limit is due to the fact that

the velocity of air at the wind turbine exit cannot be zero to maintain the flow. A hydraulic turbine, on the other hand,

converts potential energy of water into work and the water velocities just upstream and downstream of the turbine are

essentially equal to each other. Therefore, the upper limit for the efficiency of a hydraulic turbine is 100 percent.

18-71C The overall efficiency of a hydroelectric power plant can be expressed as

pipingturbinegeneratorplant

Substituting the definition of each efficiency, we obtain

max

electric

max

piping loss, mech,

maxpiping loss, mech,max

shaft

shaft

electric

max

piping loss, mech,

piping loss, mech,max

shaft

shaft

electric

pipingturbinegeneratorplant

1)/1(

1

W

W

W

E

WEW

W

W

W

W

E

EW

W

W

W

Therefore, the overall efficiency of a hydroelectric power plant is defined as the electrical power output divided by the

maximum power potential.

18-32



18-72C There are two basic types of dynamic turbine—impulse and reaction. Comparing the two power-producing dynamic

turbines, impulse turbines require a higher head, but can operate with a smaller volume flow rate. Reaction turbines can

operate with much less head, but require a higher volume flow rate.

18-73 The power that can be produced by an ideal turbine from a large dam is to be determined.

Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant.

Properties The density of water is taken to be = 1000 kg/m3 = 1 kg/L.

Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free

surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely for an ideal

operation. Therefore, the maximum power that can be generated is equal to potential energy of the water. Noting that the

mass flow rate is V m , the maximum power is determined from

kW 15.94

22

2

grossgrossmax

/sm 1000

kJ/kg 1m) 65)(m/s L/s)(9.81 kg/L)(1500 1(

gHgHmW V

18-74 The pressures just upstream and downstream of a hydraulic turbine are given. The maximum work and the flow rate of

water for a given power are to be determined.



Analysis The maximum work per unit mass of water flow is determined from

kJ/kg 1.225

33

21max

mkPa 1

kJ 1

kg/m 1000

kPa )1001025(

PPw

The flow rate of water is

L/min 4898

min 1

s 60

kJ/kg) 225.1(kg/L) 1(

kJ/s 100

max

max

maxmaxmax

w

W

wwmW

V

V

18-33



18-75 The efficiency of a hydraulic turbine is to be determined.



Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free

surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely for an ideal



kW 300.8

/sm 1000

kJ/kg 1m) 160)(m/s (9.81

s 60

min 1L/min) 00kg/L)(11,5 1(

22

2

grossgrossmax

gHgHmW V

The turbine efficiency is determined from its definition to be

percent 83.1 831.0kW 8.300

kW 250

max

shaftturbine

W

W

18-34



18-76 A hydraulic turbine-generator unit produces power from a dam. The overall efficiency of the turbine-generator unit,

the turbine efficiency, and the power losses due to inefficiencies in the turbine and the generator are to be determined.



Analysis (a) The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the

free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely for an ideal



kW 5.750

/sm 1000

kJ/kg 1m) 75)(m/s L/s)(9.81 kg/L)(1020 1(

22

2

grossgrossmax

gHgHmW V

The overall efficiency of turbine-generator unit is determined from its definition to be

percent 84.0 8395.0kW 5.750

kW 630

max

electricgenerator-turbine

W

W

(b) The turbine efficiency is determine from

percent 87.5 8745.096.0

8395.0

generator

generator-turbine

turbine

(c) The shaft power output from the turbine is

kW 3.65696.0

kW 630

generator

electricshaft

WW

Finally, the power losses due to inefficiencies in the turbine and generator are

kW 94.2 3.6565.750shaftmaxturbineloss, WWW

kW 26.3 6303.656electricshaftgeneratorloss, WWW

18-35



18-77E The pressures just upstream and downstream of a hydraulic turbine are given. The shaft power output from the

turbine and the height of the reservoir are to be determined.


Properties The density of water is taken to be = 62.4 lbm/ft3.

Analysis (a) The maximum work per unit mass of water flow is determined from

kW 09.70Btu 0.94782

kJ 1

ftpsia 5.40395

Btu 1

lbm/ft 62.4

psia )1595(lbm/s) 280(

33

21max

PPmW

The shaft power is

kW 60.3 )kW 09.70)(86.0(maxturbineshaft WW

(b) The height of the reservoir is determined from

ft 185

32

2

max

max

/sftlbm 23,730

kW 1)ft/s 2lbm/s)(32. (280

kW 09.70

gm

Wh

ghmW

18-36



18-78 The irreversible head losses in the penstock and its inlet and those after the exit of the draft tube are given. The power

loss due to irreversible head loss, the efficiency of the piping, and the electric power output are to be determined.



Analysis (a) The power loss due to irreversible head loss is determined from

kW 4.578

22

2

pipingloss,

/sm 1000

kJ/kg 1m) 7)(m/s L/s)(9.81 /60kg/L)(4000 1(

LghE V

(b) The efficiency of the piping is determined from the head loss and gross head to be

percent 95.0 950.0m 140

m 711

grosspiping

H

hL

(c) The maximum power is determined from

kW 56.91

/sm 1000

kJ/kg 1m) 140)(m/s L/s)(9.81 /60kg/L)(4000 1(

22

2

grossgrossmax

gHgHmW V

The overall efficiency of the hydroelectric plant is

798.0)950.0)(84.0(pipinggenerator-turbineplant

The electric power output is

kW 73.1 )kW 56.91)(798.0(maxplantelectric WW

18-37



18-79 Various efficiencies of a hydroelectric power plant are given. The overall efficiency of the plant and the electric

power output are to be determined.



Analysis The overall efficiency of the hydroelectric plant is determined from

percent 82.7 827.0)98.0)(97.0)(87.0(pipinggeneratorturbineplant

The maximum power is determined from

kW 1295

/sm 1000

kJ/kg 1m) 220)(m/s L/s)(9.81 kg/L)(600 1(

22

2

grossgrossmax

gHgHmW V


kW 1071 )kW 1295)(827.0(maxplantelectric WW

18-80 A hydroelectric power plant operating 80 percent of the time is considered. The revenue that can be generated in a

year is to be determined.



Analysis The maximum power for one turbine is determined from

kW 93.80

/sm 1000

kJ/kg 1m) 150)(m/s L/s)(9.81 /60kg/L)(3300 1(

22

2

grossgrossmax

gHgHmW V


kW 84.72)kW 93.80)(90.0(maxplantelectric WW

Noting that the plant operates 80 percent of the time, the operating hours in one year is

Operating hours = 0.80 365 24 = 7008 h

The amount of electricity produced per year by one turbine is

kWh 510,500h) kW)(7008 84.72(hours Operatingelectricelectric WW

Finally, the revenue generated by 18 such turbines is

$873,000 5/kWh)kWh)($0.09 500,510)(18(yelectricit of priceUnit Revenue electricturbine Wn

18-38



Geothermal Energy

18-81C One common classification of geothermal resources is based on their temperature.

High temperature resource: T > 150C

Medium temperature resource: 90C < T < 150C

Low temperature resource: T < 90C

18-82C There are several options for utilizing the thermal energy produced from geothermal energy systems: Electricity

production, space heating and cooling, cogeneration, and heat pump. The most common use of geothermal energy is

electricity production.

18-83C The quality or work potential of geothermal resources is proportional to temperature of the source. The maximum

thermal efficiency of a geothermal power plant utilizing a source at 110C and a sink temperature of 25C is th,max = 1

(25+273)/(110+273) = 0.222 = 22.2%. The actual efficiency of the proposed power plant will be much lower than this value.

It is very unlikely that a power plant investment on this site will be feasible. It makes both thermodynamic and economic

sense to use this source for heating or cooling applications.

18-84C Approximate temperature requirements for geothermal applications are

Electricity production T > 120C

Cooling T > 95C

Heating T > 50C

18-85C Ground-source heat pumps are also called geothermal heat pumps as they utilize the heat of the earth. These systems

use higher ground temperatures in winter for heat absorption (heating mode) and cooler ground temperatures in summer for

heat rejection (cooling mode), and this is the reason for higher COPs.

18-86C The purpose of flashing process is to convert saturated or compressed liquid to a liquid-vapor mixture. The resulting

mixture is sent to turbine for power production. The flashing process is a constant-enthalpy process. Both pressure and

temperature decreases during this process.

18-87C This is a liquid geothermal resource at a relatively low temperature and a binary cycle is best suited for power

generation.

18-39



18-88E A geothermal site contains saturated mixture of geothermal water at 300ºF. The maximum thermal efficiency and the

maximum amount of power are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Water properties

will be used for geothermal fluid.

Analysis The maximum thermal efficiency can be determined from the Carnot efficiency relation using the geothermal

source and dead state temperatures:

percent 29.0

2895.0

R )460300(

R )46080(11maxth,

H

L

T

T

The enthalpy and entropy values of geothermal water at the wellhead state and dead state are obtained from steam tables:

RBtu/lbm 8566.0

Btu/lbm 31.588

35.0

F300

1

1

1

1

s

h

x

T (Table A-4E)

RBtu/lbm 09328.0

Btu/lbm 07.48

0

F80

0

0

0

0

s

h

x

T (Table A-4E)

The maximum power potential is equal to the exergy of the geothermal water available at the wellhead:

kW 13,500

Btu/s 94782.0

kW 1 RBtu/lbm)09328.08566.0(Btu/lbm)07.4831.588(lbm/s) 100(

)]([ 0100111max hhThhmXW

18-89 A residential district is to be heated by geothermal water in winter. The revenue generated due to selling geothermal

heat is to be determined.



Properties The specific heat of water at room temperature is cp = 4.18 kJ/kgC.

Analysis The rate of geothermal heat supplied to the district is determined from

kW 704,11C)5090(C)kJ/kg kg/s)(4.18 70()( 21heat TTcmQ p

The amount of heat supplied for a period of 2500 hours is

kJ 10053.1s) 3600kJ/s)(2500 704,11(hours Operating 11heatheat QQ

The revenue generated by selling the geothermal heat is

millions $1.264

6

511

heat

10264.1$

/kJ)10kJ)($1.2 10053.1(

heat geothermal of priceUnit Revenue Q

18-40



18-90E Geothermal water is to be used for space cooling. The potential revenue that can be generated by this geothermal

cooling system is to be determined.



Properties The specific heat of water at room temperature is cp = 1.0 Btu/lbmF.

Analysis The rate of heat transfer in the generator of the absorption system is

Btu/h 10580.2F)180210(F)Btu/lbm lbm/h)(1.0 000,86()( 621gen TTcmQ p

Using the definition of the COP of an absorption cooling system, the rate of cooling is determined to be

kW 3.529Btu/h 3412.14

kW 1Btu/h) 10580.2)(70.0(COP COP 6

gencool

gen

coolABS

QQ

Q

Q

If a conventional cooling system with a COP of 2.3 is used for this cooling, the power input would be

kW 1.2302.3

kW 529.3

COP COP cool

in

in

cool Q

WW

Q

The electricity consumption for a period of 2000 hours is

kWh 250,460h) kW)(2000 1.230(hours Operatinginin WW

At a discount of 20 percent, the revenue generated by selling the geothermal cooling is

$51,550

/kWh)kWh)($0.14 250,460)(8.01(

yelectricit of priceUnit )1(Revenue indiscount Wf

18-41



18-91 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of

steam through the turbine, the isentropic efficiency of the turbine, the power output from the turbine, and the thermal

efficiency of the plant are to be determined.


Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

1661.02108

09.64014.990

kJ/kg 14.990

kPa 500

kJ/kg 14.9900

C230

2

212

2

11

1

fg

f

h

hhx

hh

P

hx

T

The mass flow rate of steam through the turbine is

kg/s 38.20 kg/s) 230)(1661.0(123 mxm

(b) Turbine:

kJ/kg 7.2344)1.2392)(90.0(81.19190.0

kPa 10

kJ/kg 3.2160kPa 10

KkJ/kg 8207.6

kJ/kg 1.2748

1

kPa 500

444

4

434

4

3

3

3

3

fgf

s

hxhhx

P

hss

P

s

h

x

P

0.686

3.21601.2748

7.23441.2748

43

43

sT

hh

hh

(c) The power output from the turbine is

kW 15,410 kJ/kg)7.23448.1kJ/kg)(274 38.20()( 433outT, hhmW

(d) We use saturated liquid state at the standard temperature for dead state enthalpy

kJ/kg 83.1040

C250

0

0

h

x

T

kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in hhmE

7.6% 0.0757622,203

410,15

in

outT,

thE

W

production

well reinjection

well

separator

steam

turbine

1

condenser 2

3

4

5

6

Flash

chamber

18-42



18-92 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the

steam at the exit of the second flash chamber, the power produced from the second turbine, and the thermal efficiency of the

plant are to be determined.



1661.0kJ/kg 14.990

kPa 500

kJ/kg 14.9900

C230

212

2

11

1

xhh

P

hx

T

kg/s 80.1911661.0230

kg/s 38.20kg/s) 230)(1661.0(

316

123

mmm

mxm

kJ/kg 7.234490.0

kPa 10

kJ/kg 1.27481

kPa 500

44

4

33

3

hx

P

hx

P

kJ/kg 1.26931

kPa 150

0777.0

kPa 150

kJ/kg 09.6400

kPa 500

88

8

7

7

67

7

66

6

hx

P

x

T

hh

P

hx

P

C 111.35

(b) The mass flow rate at the lower stage of the turbine is

kg/s .9014kg/s) 80.191)(0777.0(678 mxm

The power outputs from the high and low pressure stages of the turbine are

kW 15,410kJ/kg)7.23448.1kJ/kg)(274 38.20()( 433outT1, hhmW

kW 5191 kJ/kg)7.23443.1kJ/kg)(269 .9014()( 488outT2, hhmW

(c) We use saturated liquid state at the standard temperature for the dead state enthalpy

kJ/kg 83.1040

C250

0

0

h

x

T

kW 621,203kJ/kg)83.10414kg/s)(990. 230()( 011in hhmE

10.1%

0.101621,203

5193410,15

in

outT,

thE

W

production

well

reinjection

well

separator

steam

turbine

1

condenser

2

3

4

5

6

Flash chamber

7

8

9

separator

Flash

chamber

18-43



18-93 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass

flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal

efficiencies for the binary cycle and the combined plant are to be determined.



1661.0kJ/kg 14.990

kPa 500

kJ/kg 14.9900

C230

212

2

11

1

xhh

P

hx

T

kg/s 80.19120.38230

kg/s 38.20kg/s) 230)(1661.0(

316

123

mmm

mxm

kJ/kg 7.234490.0

kPa 10

kJ/kg 1.27481

kPa 500

44

4

33

3

hx

P

hx

P

kJ/kg 04.3770

C90

kJ/kg 09.6400

kPa 500

77

7

66

6

hx

T

hx

P

The isobutane properties are obtained

from EES:

/kgm 001839.0

kJ/kg 83.270

0

kPa 400

kJ/kg 01.691C80

kPa 400

kJ/kg 05.755C145

kPa 3250

310

10

10

10

99

9

88

8

v

h

x

P

hT

P

hT

P

kJ/kg 65.27682.583.270

kJ/kg. 82.5

90.0/mkPa 1

kJ 1kPa 4003250/kgm 001819.0

/

in,1011

3

3

101110in,

p

pp

whh

PPw v

An energy balance on the heat exchanger gives

kg/s 105.46

isoiso

118iso766

kg276.65)kJ/-(755.05kg377.04)kJ/-09kg/s)(640. 81.191(

)()(

mm

hhmhhm

(b) The power outputs from the steam turbine and the binary cycle are

kW 15,410 kJ/kg)7.23448.1kJ/kg)(274 38.19()( 433steamT, hhmW

kW 6139

)kJ/kg 82.5)(kg/s 46.105(6753

kW 6753kJ/kg)01.691.05kJ/kg)(755 .46105()(

,isoisoT,binarynet,

98isoT,

inp

iso

wmWW

hhmW

(c) The thermal efficiencies of the binary cycle and the combined plant are

steam

turbine

production

well

reinjection

well

isobutane

turbine

heat exchanger

pump

BINARY

CYCLE

separator

air-cooled condenser

condenser

flash

chamber

1

2

3

4

5 6

7

8

9 1

0

1

1

18-44



kW 454,50kJ/kg)65.276.05kJ/kg)(755 .46105()( 118isobinaryin, hhmQ

12.2% 0.122454,50

6139

binaryin,

binarynet,

binaryth,Q

W

kJ/kg 83.1040

C250

0

0

h

x

T

kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in hhmE

10.6%

0.106622,203

6139410,15

in

binarynet,steamT,

plantth,E

WW

18-45



18-94 A binary geothermal cogeneration plant produces power and provides space heating. The rate of space heating

provided by the system, the mass flow rate of water used for space heating, the thermal efficiency of the power plant, and the

utilization factor for the entire cogeneration plant are to be determined.



Analysis (a) The rate of space heating provided by the system is determined from

kW 14,630 C)6585(C)kJ/kg kg/s)(4.18 175()( 76heat TTcmQ p

The mass flow rate of fresh water is

kg/s 140

w

w

pw

m

m

TTcmQ

C)5075(C)kJ/kg (4.18kW 630,14

)( 89heat

(a) The rate of heat input to the power plant is

kW 58,520C)85165(C)kJ/kg kg/s)(4.18 175()( 65in TTcmQ p

The thermal efficiency of the power plant and the utilization factor of the cogeneration plant are determined from their

definitions to be

11.8% 0.1179kW 520,58

kW 6900

in

outth

Q

W

36.8%

0.3679kW 520,58

kW 14,630kW 6900

in

heatout

Q

QWu

18-95 A binary geothermal cogeneration plant produces power and provides space heating. The amount of money than can

be made by selling geothermal heat is to be determined.



Analysis The amount of natural gas used to supply space heating at a rate of 14,630 kW is determined from

therms916,234kJ 105,500

therm1

85.0

s) 3600kJ/s)(4200 630,14(hours Operatinggas natural ofAmount

boiler

heat

Q

The revenue generated by selling the geothermal heat is

$229,000

1.3/therm) therms)($916,234)(25.01(

gas natural of priceUnit gas natural ofAmount )1(Revenue discountf

18-46



Biomass Energy

18-96C Biomass is mostly produced from agriculture and forest products and residues, energy crops, and algae. Organic

component of municipal and industrial wastes and the fuel produced from food processing waste such as used cooking oil

are also considered biomass. Despite relatively long period of times involved in growing crops and trees, they can be re-

grown by planting, and therefore biomass is considered to be a renewable energy source.

18-97C Growing of crops and trees as well as the conversion to liquid and gaseous fuels involves the consumption of energy

in the form of electricity and fossil fuels such as coal, oil, and natural gas. The consumption of fossil fuels is accompanied

by the pollutant and greenhouse emissions. Therefore, the renewability and cleanliness of biomass are not as good as other

renewables such as solar, geothermal, or wind.

18-98C In pyrolysis, biomass is exposed to high temperatures without the presence of air, causing it to decompose.

18-99C The two most common biofuels are ethanol and biodiesel.

18-100C Ethanol has a higher heating value (HHV) of 29,710 kJ/kg while the higher heating value of biodiesel is about

40,700 kJ/kg. Since the heating value of biodiesel is higher, the car using biodiesel will get more mileage.

18-101C The higher heating value of biodiesel is about 40,700 kJ/kg, which is about 9 percent less than that of petroleum

diesel (HHV = 44,800 kJ/kg). Since the heating value of petroleum diesel is higher, the car using petroleum diesel will get

more mileage. Biodiesel could increase nitrogen oxides emissions compared to petroleum diesel while significantly reducing

hydrocarbon, sulfur and carbon monoxide emissions.

18-102C Biogas usually consists of 50 to 80 percent methane (CH4) and 20 to 50 percent carbon dioxide (CO2) by volume.

It also contains small amounts of hydrogen, carbon monoxide, and nitrogen. Biogas can be produced from biological waste

such as animal manure, food waste, and agricultural waste.

18-103C An important class of biomass is produced by households as trash or garbage. This is referred to as municipal

solid waste (MSW). MSW includes mostly organic materials such as paper, food scraps, wood, and yard trimmings but

some fossil content such as plastic also exists. MSW does not include industrial, hazardous, or construction waste.

18-104C Recycling refers to recovery of useful materials such as paper, glass, plastic, and metals from the trash to use to

make new products. Composting, on the other hand, refers to storing organic waste such as food scraps and yard trimmings

under certain conditions to help it break down naturally. The resulting product can be used as a natural fertilizer.

18-47



18-105C In waste to energy (WTE) option, waste is burned directly for generating steam. This steam is run through a

turbine to generate electricity. The second option is known as landfill-gas-to-energy (LFGTE), and it involves harvesting

biogas (mostly methane) from the buried waste as it decomposes. Biogas is then used as the fuel in an internal combustion

engine or gas turbine to generate electricity.

18-106 Sugar beet roots with certain sugar content are used to produce ethanol. The amount of ethanol that can be produced

from 100 kg of sugar beet roots is to be determined.

Properties The molar masses of C, H2, and O2 are 12 kg/kmol, 2 kg/kmol, and 32 kg/kmol, respectively (Table A-1).

Analysis The molar mass of ethanol C6H12O6 and sugar C2H5OH are

MC6H12O6 = 6 12 + 6 2 + 3 32 = 180 kg/kmol

MC2H5OH = 2 12 + 3 2 + 0.5 32 = 46 kg/kmol

The mass of sugar in 100 kg of sugar beet roots is

msugar = mfsugar mbeet = (0.20)(100 kg) = 20 kg

The ratio of mass of ethanol to the mass of sugar is obtained from the chemical reaction C6H12O6 2 C2H5OH + 2 CO2 is

5111.0kg 46

kg 180)(22

C6H5OH

C6H12O6

sugar

ethanol

M

M

m

m

The mass of the ethanol produced is

kg 10.2 )5111.0)(kg 20(sugar

ethanolsugarethanol

m

mmm

18-48



18-107E Methanol is to replace gasoline in an internal combustion engine. The maximum power output from the engine with

methanol as the fuel is to be estimated and the rates of gasoline and methanol consumptions are to be determined.

Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.

Properties The molar masses of C, H2, and O2 are 12 lbm/lbmol, 2 lbm/lbmol, and 32 lbm/lbmol, respectively (Table A-1E).

The lower heating values of gasoline and ethanol are 18,490 Btu/lbm and 8620 Btu/lbm, respectively.

Analysis The balanced reaction equation with stoichiometric air is

2th2222th4 N 76.3OH 2CO 3.76NOOCH aa

The stoicihiometric coefficient ath is determined from an O2 balance:

5.1115.0 thth aa

Substituting,

222224 N 64.5OH 2CO 3.76NO5.1OCH

Therefore, the air-fuel ratio for this stoichiometric reaction is

471.6lbm 32

lbm 207.1

lbm )1611412(1

lbm )2976.4(1.5AF

fuel

air

m

m

For a given engine size (i.e., volume), the power produced is proportional to the heating value of the fuel and inversely

proportional to the air-fuel ratio. The ratio of power produced by gasoline engine to that of methanol engine is expressed as

9507.014.6

6.471

Btu/lbm 8620

Btu/lbm 18,490

AF

AF

LHV

LHV

gasoline

methanol

methanol

gasoline

methanol

gasoline

W

W

Then, the maximum power by the engine with ethanol as the fuel becomes

hp 210.40.9507

hp 200

9507.0

1gasoline

gasoline

methanolgasolinemethanol W

W

WWW

The thermal efficiency of this engine is defined as the net power output divided by the rate of heat input, which is equal to

the heat released by the combustion of fuel:

cm

W

Q

W

LHVfuel

out

in

outth

Note that the lower heating value is used in the analysis of internal combustion engines since the water in the exhaust is

normally in the vapor phase. Solving the above equation for the rates of gasoline and ethanol consumption, we obtain

lbm/min 1.15

hp 1

Btu/min 42.41

)Btu/lbm)(1 490(0.40)(18,

hp 200

LHVgasolineth

gasoline

gasolinec

Wm

lbm/min 2.59

hp 1

Btu/min 42.41

)Btu/lbm)(1 0(0.40)(862

hp 210.4

LHVmethanolth

methanolmethanol

c

Wm

18-49



18-108 A homeowner is to replace the existing natural gas furnace with a wood pellet furnace. The savings due to this

replacement are to be determined.

Assumptions 1 Combustion is complete. 2 Combustion efficiency for both furnace is 100 percent.

Analysis The unit price of useful heat supplied by the furnace in the case of natural gas, in $/MJ, is

MJ/01422.0$MJ 105.5

therm1

90.0

/therm35.1$gas of priceUnit (gas)heat of priceUnit

gasboiler,

The unit price of useful heat supplied by the furnace in the case of pellet, in $/MJ, is

MJ/009375.0$MJ/kg 20)(80.0(

/kg15.0$

HV

pellet of priceUnit (pellet)heat of priceUnit

pelletboiler,

Using these unit prices, the annual fuel cost in the case of pellet furnace would be

1978$MJ/01422.0$

MJ/009375.0$)3000($

(gas)heat of priceUnit

(pellet)heat of priceUnit cost gas Annualcostpellet Annual

Therefore, the savings due to this replacement would be

$1022 1978$3000$costpellet Annualcost gas AnnualSavings

18-50



18-109 Municipal solid waste (MSW) is burned directly in a boiler to generate saturated steam and this steam is run through

a turbine to generate power. The amount of steam and electricity generated and the revenue generated by selling the

electricity are to be determined.


Analysis The properties of water at the boiler inlet (state 1), boiler exit and turbine inlet (state 2), and turbine exit (state 3)

are obtained from steam tables (Tables A-4 through A-6)

kJ/kg 91.830

C201

1

1

h

x

T

kJ/kg 8.2328kPa 100

KkJ/kg 4302.6

kJ/kg 0.2792

1

C200

323

3

2

2

2

2

shss

P

s

h

x

T

The actual enthalpy at the turbine exit is determined from the definition of isentropic efficiency as

kJ/kg 3.23988.23280.2792

0.279285.0 3

3

32

32

h

h

hh

hh

sT

From the definition of the boiler efficiency,

kg 49,850

steam

steam

MSWMSW

12steam

in

usefulboiler

kJ/kg) kg)(18,000 000,10(

kJ/kg)91.83(2792.075.0

HV

)(

m

m

m

hhm

Q

Q

The electric power output from the turbine is

kWh 5180

kJ 3600

kWh 1kJ/kg)3.23980.2792)(95.0()( 32generatorelectric hhW

The revenue that can begenerated by selling the electricity is

$570

/kWh)kWh)($0.11 5180(

yelectricit of priceUnit Revenue electricW

18-51



Review Problems

18-110 A solar collector provides the hot water needs of a family. The annual natural gas and cost savings due to solar water

heating are to be determined.


Analysis The amount of solar water heating during a month period is

kJ/month 10003.1C)2060(C)kJ/kg 4.18kg/month)( 6000()( 621 TTmcQ p

The amount of gas that would be consumed (or saved) during the 8 month period is

therms 86.44

kJ 500,105

therm1

88.0

kJ/month 10003.1months) 8(savings Gas

11

heater

Q

The corresponding cost savings is

$116.7

)1.35/therm therms)($44.86(

heat geothermal of priceUnit savings GassavingsCost

18-111 Two concentrating collectors with given characteristics are considered. The solar irradiation rate for collector B to

have the same efficiency as collector A and the efficiency change of collector A for a change in irradiation are to be

determined.

Analysis (a) The collector efficiency of collector A is determined from

8098.0) W/m600)(7(

C)27145(C) W/m5.2(88.0

CR 2

2,

A

acAarAc

G

TTU

Using the same relation for collector B at the same collector efficiency, we obtain the necessary solar irradiation rate:

2 W/m840

B

B

B

acBarBc

G

G

G

TTU

)7(

C)27145(C) W/m5.3(88.08098.0

CR

2

,

(b) The collector efficiency of collector A for a solar irradiation of 900 W/m2 is

8332.0) W/m900)(7(

C)27145(C) W/m5.2(88.0

CR 2

2,

A

acAarAc

G

TTU

The change in the efficiency of collector A is then

percent 2.34 02341.08098.08332.0,Ac

18-52



18-112 The initial cost of a solar power plant is given. The payback period of the investment is to be determined.

Analysis The total cost of this power plant is

8out 10$1.400kW) 000,10)(kW/000,14($costUnit costPlant W

The annual revenue from selling the electricity produced is

/year10$4.500

h/year) 5000)(kW/09.0($kW) 000,10(

hours OperatingpriceUnit revenue Annual

6

out

W

The simple payback period is then

years31.1

/year10$4.500

10$1.400

revenue Annual

costPlant periodPayback

6

8

That is, this plant pays for itself in 31 years, which is not unexpected for solar power plants.

18-113 Two cities are considered for solar power production. The amount of electricity that can be produced in each city are

to be determined.

Analysis Using the average daily solar radiation values on a horizontal surface is obtained from Table 18-3 for each city as

262Flavg, kJ/m 10344.6)days day)(365kJ/m 380,17( G

262Atavg, kJ/m 10997.5)days day)(365kJ/m 430,16( G

Noting that the average thermal efficiency is 18 percent and the total collector area is 30,000 m2, the amount of electricity

that can be produced in each city would be

kWh 109.516 7

s 3600

kWh 1)kJ/m 10344.6)(m 000,30)(18.0((Fl)y electricit ofAmount 262

Flavg,th AG

kWh 108.995 7

s 3600

kWh 1)kJ/m 10997.5)(m 000,30)(18.0((At)y electricit ofAmount 262

Atavg,th AG

18-53



18-114 The initial cost of a solar photovoltaic cell system for a house is given. The payback period of this system is to be

determined.

Analysis The total cost of the photovoltaic system is

$7800 W)6000)(W/3.1($costUnit cost System out W

The current annual electricity cost to the homeowner is

$1500/year

)month/125($months) 12(

billMonthly rmonths/yea 12costy electricit Annual

The homeowner can meet approximately 80 percent of the electricity needs of the house by the solar system.

$1200/year)year/1500($)80.0(costy electricit Annual80.0savingsy electricit Annual

Then, the simple payback period is determined from

years6.5$1200/year

$7800

savingsy electricit Annual

cost SystemperiodPayback

That is, this photovoltaic system pays for itself in 6.5 years.

18-115 The ratio of power generated in two locations with different wind velocities is to be determined.


Properties The density of air is same in two locations.

Analysis The wind power potential is expressed for both locations to be

3

11available,2

1AVW

3

22available,2

1AVW

The ratio of power generated from location A and location B is then

3.3753

3

32

31

32

31

2-available

1-available

m/s) (6

m/s) (9

2

12

1

V

V

AV

AV

W

W

18-54



18-116 An investor to install 40 wind turbines with a specified average wind speed. The payback period of this investment is

to be determined.



Analysis The blade span area is

222 m 5.2544/m) 18(4/ DA

The wind power potential for one turbine is

kW 04.56/sm 1000

kJ/kg 1m/s) )(7.2m )(254.5kg/m (1.18

2

1

2

122

32331available

AVW

The electric power generated is

kW 49.18kW) (56.04)33.0(availableoverallwt,electric WW

The amount of electricity produced is determined from

kWh 0,94011h) kW)(6000 49.18(hours Operatingelectricelectric WW

The annual revenue generated by 40 turbines is

ear$332,820/y

5/kWh)kWh)($0.07 940,110)(40(

yelectricit of priceUnit revenue Annual electricturbine

Wn

The payback period of this investment is

years3.61ear$332,820/y

$1,200,000

revenue Annual

cost initial TotalperiodPayback

18-55



18-117 A school is considering installing wind turbines to meet its electricity needs. The required average velocity of wind

is to be determined.



Analysis The amount of electricity used by the school per year is determine from

kWh 090,209$0.11/kWh

ar$23,000/ye

yelectricit of priceUnit

costy electricit Annualelectric W

The average rate of electrical power is

kW 88.27h 7500

kWh 0,94011

hours Operating

electricelectric

WW


kW 93.9230.0

kW 88.27

overallwt,

electricavailable

WW


222 m 2.3144/m) 20(4/ DA

The average wind velocity is determined from available wind power relation:

m/s 7.90

V

V

AVW

22

323

3available

/sm 1000

kJ/kg 1)m )(314.2kg/m (1.2

2

1kW 93.92

2

1

18-56



18-118 A school is considering installing wind turbines to meet its electricity needs. The required average velocity of wind

is to be determined.



Analysis The amount of electricity used by the school per year is determine from

kWh 090,209$0.11/kWh

ar$23,000/ye


costy electricit Annualelectric W

The average rate of electrical power is

kW 88.27h 7500

kWh 0,94011

hours Operating

electricelectric

WW


kW 93.9230.0

kW 88.27

overallwt,

electricavailable

WW


222 m 9.7064/m) 30(4/ DA

The average wind velocity is determined from available wind power relation:

m/s 6.03

V

V

AVW

22

323

3available

/sm 1000

kJ/kg 1)m )(706.9kg/m (1.2

2

1kW 93.92

2

1

18-57



18-119 Mechanical energy of a waterfall in a village is to be converted into electricity by a hydraulic turbine. It is to be

determined if the proposed turbine can meet electricity needs of the village.

Assumptions 1 The flow is steady. 2 Water level of the waterfall remains constant.


Analysis The maximum power from the turbine is determined from

kW 578.4

/sm 1000

kJ/kg 1m) 80)(m/s L/s)(9.81 60kg/L)(350/ 1(

22

2

grossgrossmax

gHgHmW V


kW 754.3)kW 578.4)(82.0(maxgen-turbineelectric WW

Noting that the turbine will operate nonstop throughout the year, the operating hours in one year is

Operating hours = 365 24 = 8760 h

The amount of electricity that will be produced per year by the turbine is

kWh 32,885 h) kW)(8760 754.3(hours Operatingelectricproducedelectric, WW

The current amount of electricity used by the village is

kWh 38,333kWh/12.0$

4600$


yelectricit ofcost Annualconsumedelectric,W

Therefore, the proposed turbine cannot meet entire electricity needs of the village. However, this may still be a good option

as the turbine can meet 86 percent of the electricity needs.

18-58



18-120 An investor is to build a hydroelectric power plant. It is to be determined if the plant can pay for itself in 5 years.



Analysis The maximum power for one turbine is determined from

kW 8.498

/sm 1000

kJ/kg 1m) 90)(m/s L/s)(9.81 kg/L)(565 1(

22

2

grossgrossmax

gHgHmW V


kW 0.439)kW 8.498)(88.0(maxplantelectric WW

Noting that the plant operates 8200 h a year, the amount of electricity produced per year by one turbine is

kWh/year ,000600,3h/year) kW)(8200 0.439(hours Operatingelectricelectric WW

Noting that the operating and maintenance expenses of the plant are $750,000/year, the revenue generated by 10 such

turbines is

/year$3,030,000

$750,000/year)]$0.105/kWhkWh/year)( 000,600,3)(10[(

Cost M&Oy)electricit of priceUnit (Revenue electricturbine Wn

The payback period of this investment is then

years7.92/year$3,030,000

0$24,000,00

revenue Annual

costplant TotalperiodPayback

This payback period does not satisfy the investor’s criterion of 5 years.

18-59



18-121 Geothermal water is flashed into a lower pressure in a flash chamber. The temperature and the fractions of liquid and

vapor phases after the flashing process are to be determined.


Analysis The enthalpy of geothermal water at the flash chamber inlet is (Table A-4)

kJ/kg 61.8970

C2101

1

1

h

x

T

A flash chamber is a throttling device across which the enthalpy remains constant. Therefore, the properties at the chamber

outlet are (Table A-5)

1089.0

8.2085

38.67061.897kJ/kg 61.897

kPa 6002

2

2

12

2

fg

f

h

hhx

T

hh

PC158.8

The mass fractions of vapor and liquid phases are

0.1089 2vapormf x

0.8911 1089.01)1(mf 2liquid x

18-122 The vapor resulting from the flashing process is routed to a steam turbine to produce power. The power output from

the turbine is to be determined.


Analysis The enthalpy of geothermal water at the flash chamber inlet is (Table A-4)

kJ/kg 61.8970

C2101

1

1

h

x

T

A flash chamber is a throttling device across which the enthalpy remains constant. Therefore, the properties at the chamber

outlet are (Table A-5)

1089.0

8.2085

38.67061.897

C158.8

kJ/kg 61.897

kPa 6002

2

2

12

2

fg

f

h

hhx

T

hh

P

The mass flow rate of steam through the turbine is

kg/s 447.5kg/s) 50)(1089.0(123 mxm

The properties of steam at the turbine inlet and exit are (Table A-5)

kJ/kg 4.2226kPa 20

KkJ/kg 7593.6

kJ/kg 2.2756

1

kPa 600

434

4

3

3

3

3

shss

P

s

h

x

P

The power output from the turbine is

kW 2539 kJ/kg)4.2226.2kg/s)(2756 447.5)(88.0()( 433outT, sT hhmW

18-60



18-123 A single-flash geothermal power plant uses geothermal liquid water as the heat source, and produces power. The

thermal efficiency, the second-law efficiency, and the total rate of exergy destroyed in the plant are to be determined.


Analysis The properties of geothermal water at the inlet of the plant and at the dead state are obtained from steam tables to

be

kJ/kg.K 3672.0 kJ/kg, 92.104atm 1 C,25

kJ/kg.K 8418.1 kJ/kg, 18.632liquid C,150

0000

111

shPT

shT

The energy of geothermal water may be taken to be maximum heat that can be extracted from the geothermal water, and this

may be expressed as the enthalpy difference between the state of geothermal water and dead state:

kW 449,221kJ/kg )92.104(632.18kg/s) 420()( 01in hhmE

The exergy of geothermal water is

kW 888,36

kJ/kg.K)3672.08418.1)(K 27325(kJ/kg )92.104(632.18kg/s) 420(

)()( 01001in

ssThhmX

The thermal efficiency of the power plant is

7.14% 0.07135kW 449,221

kW 800,15

in

outnet,th

E

W

The exergy efficiency of the plant is the ratio of power produced to the exergy input (or expended) to the plant:

42.8% 0.4283kW 888,36

kW 800,15

in

outnet,II

X

W

The exergy destroyed in this power plant is determined from an exergy balance on the entire power plant to be

kW 21,088

dest

dest

destoutnet,in

0800,15888,36

0

X

X

XWX

18-61



18-124 The higher heating value of biogas with a given composition is to be determined.

Properties The molar masses of C, H2, and O2 are 12 kg/kmol, 2 kg/kmol, and 32 kg/kmol, respectively (Table A-1).

Analysis The mole fractions of CH4 and CO2 are equal to their volume fractions:

yCH4 = 0.65

yCO2 = 0.35

We consider 100 kmol of biogas mixture. In this mixture, there are 65 kmol of CH4 and 35 kmol of CO2. The mass of CH4

and CO2 are

kg 1540)kg 44)(kmol 35(

kg 1040)kg 16)(kmol 65(

CO2CO2CO2

CH4CH4CH4

MNm

MNm

The total mass of the mixture is

kg 258015401040CO2CH4total mmm

The mass fraction of CH4 is

4031.0kg 2580

kg 1040mf

total

CH4CH4

m

m

The heating value of CO2 is zero. Therefore, the higher heating value of biogas is

kJ/kg 22,250 )kJ/kg 200,55)(4031.0(HHVmfHHV CH4CH4biogas

18-62



FE Exam Problems

18-125 Which renewable energy sources are only used for electricity generation?

(a) Wind and solar (b) Hydro and solar (c) Solar and geothermal (d) Wind and hydro

(e) Hydro and geothermal

Answer: (d) Wind and hydro

18-126 Which renewable energy source should not be considered as the manifestation of solar energy in different forms?

(a) Wind (b) Hydro (c) Ocean wave (d) Biomass (e) Geothermal

Answer: (e) Geothermal

18-127 Solar radiation is incident on a flat-plate collector at a rate of 450 W/m2. The product of transmissivity and

absorptivity is = 0.85, and the heat loss coefficient of the collector is 4.5 W/m2C. The ambient air temperature is 10C.

The collector temperature at which the collector efficiency is zero is

(a) 95C (b) 104C (c) 112C (d) 87C (e) 73C

Answer: (a) 95C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES

screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

G=450 [W/m^2] TauAlpha=0.85 U=4.5 [W/m^2-C] T_a=10 [C] eta_c=0 eta_c=TauAlpha-U*(T_c-T_a)/G

18-63



18-128 Solar radiation is incident on a flat-plate collector at a rate of 600 W/m2. The glazing has a transmissivity of 0.85

and the absorptivity of absorber plate is 0.92. The heat loss coefficient of the collector is 3.0 W/m2C. The maximum

efficiency of this collector is

(a) 92% (b) 85% (c) 78% (d) 73% (e) 66%

Answer: (c) 78%



G=600 [W/m^2] Tau=0.85 Alpha=0.92 U=3.0 [W/m^2-C] eta_c_max=Tau*Alpha

18-129 The efficiency of a solar collector is given by G

TTU ac

c

. If the collector efficiency is plotted against the

term (Tc Ta)/G, a straight line is obtained. The slope of this line is equal to

(a) U (b) U (c) (d) (e) U/G

Answer: (b) U

18-130 A solar-power-tower plant produces 450 kW of power when solar radiation is incident at a rate of 1050 W/m2. If the

efficiency of this solar plant is 17 percent, the total collector area receiving solar radiation is

(a) 1750 m2 (b) 2090 m

2 (c) 2520 m

2 (d) 3230 m

2 (e) 3660 m

2

Answer: (c) 2520 m2



W_dot=450E3 [W] G=1050 [W/m^2] eta_th=0.17 eta_th=W_dot/(A_c*G)

18-64



18-131 In a solar pond, the water temperature is 40°C near the surface and 90°C near the bottom of the pond. The

maximum thermal efficiency a solar pond power plant can have is

(a) 13.8% (b) 17.5% (c) 25.7% (d) 32.4% (e) 55.5%

Answer: (a) 13.8%



T_L=40 [C] T_H=90 [C] eta_th_max=1-(T_L+273)/(T_H+273)

18-132 In a solar cell, the load voltage is 0.5 V and the load current density is determined to be 80 A/m2. If the solar

irradiation is 650 W/m2, the cell efficiency is

(a) 4.7% (b) 6.2% (c) 7.8% (d) 9.1% (e) 14.2%

Answer: (b) 6.2%



V=0.5 [V] J_L=80 [A/m^2] G=650 [W/m^2] eta_cell=(J_L*V)/G

18-133 The power potential of a wind turbine at a wind speed of 5 m/s is 50 kW. The power potential of the same turbine at

a velocity of 8 m/s is

(a) 80 kW (b) 128 kW (c) 180 kW (d) 205 kW (e) 242 kW

Answer: (d) 205 kW



W_dot_1=50 [kW] V1=5 [m/s] V2=8 [m/s] W_dot_2=W_dot_1*(V2^3/V1^3)

18-65



18-134 Wind is blowing through a turbine at a velocity of 7 m/s. The turbine blade diameter is 25 m and the density of air is

1.15 kg/m3. The power potential of the turbine is


Answer: (e) 97 kW



V=7 [m/s] D=25 [m] rho=1.15 [kg/m^3] A=pi*D^2/4 W_dot_available=1/2*rho*A*V^3*Convert(W, kW)

18-135 The power potential of a wind turbine at a wind speed of 5 m/s is 100 kW. The blade diameter of this turbine is 40

m. The power potential of a similar turbine with a blade diameter of 60 m at the same velocity is


Answer: (b) 225 kW



V=5 [m/s] W_dot_1=100 [kW] D1=40 [m] D2=60 [m] W_dot_2=W_dot_1*(D2^2/D1^2)

18-66



18-136 Wind is blowing through a turbine at a velocity of 9 m/s. The turbine blade diameter is 35 m. The air is at 95 kPa

and 20C. If the power output from the turbine is 115 kW, the efficiency of the turbine is

(a) 29% (b) 32% (c) 35% (d) 38% (e) 42%

Answer: (a) 29%



V=9 [m/s] D=35 [m] P=95 [kPa] T=(20+273) [K] W_dot=115 [kW] R=0.287 [kJ/kg-K] rho=P/(R*T) A=pi*D^2/4 W_dot_available=1/2*rho*A*V^3*Convert(W, kW) eta_wt=W_dot/W_dot_available

18-137 A turbine is placed at the bottom of a 70-m-high water body. Water flows through the turbine at a rate of 15 m3/s.

The power potential of the turbine is

(a) 10.3 MW (b) 8.8 MW (c) 7.6 MW (d) 7.1 MW (e) 5.9 MW

Answer (a) 10.3 MW



H=70 [m] V_dot=15 [m^3/s] g=9.81 [m/s^2] rho=1000 [kg/m^3] W_dot_ideal=rho*g*H*V_dot*Convert(W, kW)

18-67



18-138 The efficiency of a hydraulic turbine-generator unit is specified to be 85 percent. If the generator efficiency is 96

percent, the turbine efficiency is

(a) 0.816 (b) 0.850 (c) 0.862 (d) 0.885 (e) 0.960

Answer (d) 0.885



eta_turbine_gen=0.85 eta_gen=0.96 eta_turbine=eta_turbine_gen/eta_gen

18-139 A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free

surfaces upstream and downstream of the dam is 120 m. The water is supplied to the turbine at a rate of 150 kg/s. If the shaft

power output from the turbine is 155 kW, the efficiency of the turbine is

(a) 0.77 (b) 0.80 (c) 0.82 (d) 0.85 (e) 0.88

Answer (e) 0.88



h=120 [m] m_dot=150 [kg/s] W_dot_shaft=155 [kW] g=9.81 [m/s^2] DELTAE_dot=m_dot*g*h*Convert(W, kW) eta_turbine=W_dot_shaft/DELTAE_dot

18-140 The maximum thermal efficiency of a power plant using a geothermal source at 180C in an environment at 25C is

(a) 43.1% (b) 34.2% (c) 19.5% (d) 86.1% (e) 12.7%

Answer: (b) 34.2%



T_geo=180 [C] T0=25 [C] eta_th_max=1-(T0+273)/(T_geo+273)

18-68



18-141 Geothermal liquid water is available at a site at 150C at a rate of 10 kg/s in an environment at 20C. The maximum

amount of power that can be produced from this site is


Answer: (e) 955 kW



T_geo=150 [C] x_geo=0 m_dot=10 [kg/s] T0=20 [C] Fluid$='Steam_iapws' h_geo=enthalpy(Fluid$, T=T_geo, x=x_geo) s_geo=entropy(Fluid$, T=T_geo, x=x_geo) h0=enthalpy(Fluid$, T=T0, x=0) s0=entropy(Fluid$, T=T0, x=0) X_dot_geo=m_dot*(h_geo-h0-(T0+273)*(s_geo-s0)) W_dot_max=X_dot_geo

18-142 A binary geothermal power plant produces 5 MW power using a geothermal resource at 175C at a rate of 110 kg/s.

If the temperature of geothermal water is decreased to 90C in the heat exchanger after giving its heat to binary fluid, the

thermal efficiency of the binary cycle is

(a) 7.5% (b) 16.3% (c) 14.4% (d) 12.5% (e) 9.7%

Answer: (d) 12.5%



W_dot=5000 [kW] T_geo_1=175 [C] T_geo_2=90 [C] m_dot=110 [kg/s] Fluid$='Steam_iapws' h_geo_1=enthalpy(Fluid$, T=T_geo_1, x=0) h_geo_2=enthalpy(Fluid$, T=T_geo_2, x=0) Q_dot_in=m_dot*(h_geo_1-h_geo_2) eta_th=W_dot/Q_dot_in

18-69



18-143 A certain biogas consists of 75 percent methane (CH4) and 25 percent carbon dioxide (CO2) by volume. If the

higher heating value of methane is 55,200 kJ/kg, the higher heating value of this biogas is

(a) 25,100 kJ/kg (b) 28,800 kJ/kg (c) 33,900 kJ/kg (d) 41,400 kJ/kg

(e) 55,200 kJ/kg

Answer: (b) 28,800 kJ/kg



y_CH4=0.75 y_CO2=0.25 HHV_CH4=55200 [kJ/kg] MM_CH4=16 [kg/kmol] MM_CO2=44 [kg/kmol] "Consider 100 kmol of mixture" N_total=100 [kmol] N_CH4=y_CH4*N_total N_CO2=y_CO2*N_total m_CH4=N_CH4*MM_CH4 m_CO2=N_CO2*MM_CO2 m_total=m_CH4+m_CO2 mf_CH4=m_CH4/m_total mf_CO2=m_CO2/m_total HHV_biogas=mf_CH4*HHV_CH4

18-144 Consider 100 kg of sugar beet roots whose sugar content represents 15 percent of total mass. The sugar is converted

into ethanol through the reaction C6H12O6 2 C2H5OH + 2 CO2. The amount of ethanol produced is

(a) 15 kg (b) 13 kg (c) 9.8 kg (d) 7.7 kg (e) 5.5 kg

Answer: (d) 7.7 kg



m_beet=100 [kg] mf_sugar=0.15 "C6H12O6 = 2 C2H5OH + 2 CO2" MM_C6H12O6=180 [kg/kmol] MM_C2H5OH=46 [kg/kmol] m_sugar=mf_sugar*m_beet m_ethanol\m_sugar=2*MM_C2H5OH/MM_C6H12O6 m_ethanol=m_sugar*m_ethanol\m_sugar

1-145 – 18-151 Design and Essay Problems

Thermo 8e SM Chap18

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Transcript of Thermo 8e SM Chap18