engg 3260: thermodynamics home assignment 6 (chapter...

ENGG 3260: Thermodynamics

Home Assignment 6 (Chapter 6)

1. A heat engine with a thermal efficiency of 40 percent rejects 1000 kJ/kg of heat. How

much heat does it receive?

Answer:

a. Assumptions:

1 The plant operates steadily.

2 Heat losses from the working fluid at the

pipes and other components are negligible.

b. According to the definition of the thermal

efficiency as applied to the heat engine,

HLH

H

qqq

qw

th

thnet

which when rearranged gives

kJ/kg 1667

4.01

kJ/kg 1000

1 th

LH

qq

2. A steam power plant with a power output of 150 MW consumes coal at a rate of 60

tons/h. If the heating value of the coal is 30,000 kJ/kg, determine the overall efficiency

of this plant.

Answer:

a. Assumptions: The plant operates steadily.

b. Properties: The heating value of coal is given to be 30,000 kJ/kg.

c. The rate of heat supply to this power plant is

MW 500

kJ/h 101.8kJ/kg 30,000kg/h 60,000 9

coalHV,coal

qmQH

Then the thermal efficiency of the plant becomes

30.0% 0.300MW 500

MW 150outnet,th

HQ

W

sink

Furnace

HE

qH

qL wnet

sink

HE

60 t/h

coal

Furnace

150 MW

3. An automobile engine consumes fuel at a rate of 22 L/h and delivers 55 kW of power to

the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/ ,

determine the efficiency of this engine.

Answer:

a. Assumptions The car operates steadily.

b. Properties The heating value of the fuel is given to be 44,000 kJ/kg.

c. The mass consumption rate of the fuel is

kg/h .617)L/h 22)(kg/L 0.8()( fuelfuel V m

The rate of heat supply to the car is

kW 215.1kJ/h ,400774

)kJ/kg 44,000)(kg/h .617(

coalHV,coal

qmQH

Then the thermal efficiency of the car becomes

25.6% 0.256kW 215.1

kW 55outnet,

th

HQ

W

4. The Department of Energy projects that between the years 1995 and 2010, the United

States will need to build new power plants to generate an additional 150,000 MW of

electricity to meet the increasing demand for electric power. One possibility is to build

coal-fired power plants, which cost $1300 per kW to construct and have an efficiency of

40 percent. Another possibility is to use the clean-burning Integrated Gasification

Combined Cycle (IGCC) plants where the coal is subjected to heat and pressure to gasify

it while removing sulfur and particulate matter from it. The gaseous coal is then burned

in a gas turbine, and part of the waste heat from the exhaust gases is recovered to

generate steam for the steam turbine. Currently the construction of IGCC plants costs

about $1500 per kW, but their efficiency is about 28,000,000 kJ per ton (that is,

28,000,000 kJ of heat is released when 1 ton of coal is burned). If the IGCC plant is to

recover its cost difference from fuel savings in five years, determine what the price of

coal should be in $ per ton.

Answer:

a. Assumptions

1 Power is generated continuously by either

plant at full capacity.

2 The time value of money (interest,

inflation, etc.) is not considered.

b. Properties The heating value of the coal is

given to be 28106 kJ/ton.

sink

HE

Fuel

22 L/h

Engine

55 kW

c. For a power generation capacity of 150,000

MW, the construction costs of coal and

IGCC plants and their difference are

999

9IGCC

9coal

1030$10195$10225$differencecost on Constructi

10$225=kW)kW)($1500/ 000,000,150(coston Constructi

10$195=kW)kW)($1300/ 000,000,150(coston Constructi

The amount of electricity produced by either plant in 5 years is

kWh 106.570=h) 24365kW)(5 000,000,150( 12 tWWe

The amount of fuel needed to generate a specified amount of power can be determined

from

) valueHeating( valueHeating

or infuelin

in eee WQ

mW

QQ

W

Then the amount of coal needed to generate this much electricity by each plant and

their difference are

tons10352.0=10760.110112.2

tons10760.1kWh 1

kJ 3600

kJ/ton) 1028)(48.0(

kWh 10570.6

) valueHeating(

tons10112.2kWh 1

kJ 3600

kJ/ton) 1028)(40.0(

kWh 10570.6

) valueHeating(

999plant IGCC coal,plant coal coal,coal

9

6

12

plant IGCC coal,

9

6

12

plant coal coal,

mmm

Wm

Wm

e

e

For mcoal to pay for the construction cost difference of $30 billion, the price of coal

should be

$85.2/ton

tons10352.0

1030$differencecost on Constructicoal ofcost Unit

9

9

coalm

Therefore, the IGCC plant becomes attractive when the price of coal is above $85.2 per

ton.

5. A coal-burning steam power plant produces a net power of 300 MW with an overall

thermal efficiency of 32 percent. The actual gravimetric air-fuel ratio in the furnace is

calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg.

Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of

air flowing through the furnace.

Answer:

a. Assumptions

1 The power plant operates steadily.

2 The kinetic and potential energy changes are zero.

b. Properties The heating value of the coal is given to be 28,000 kJ/kg.

c. (a) The rate and the amount of heat inputs to the power plant are

MW 5.93732.0

MW 300

th

outnet,

in

WQ

MJ 101.8s) 360024(MJ/s) 5.937( 7inin tQQ

The amount and rate of coal consumed during this period are

kg/s 48.33s 360024

kg 10893.2

MJ/kg 28

MJ 101.8

6coal

coal

7

HV

incoal

t

mm

q

Qm

kg 102.893 6

(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is

kg/s 401.8 kg/s) 48.33(fuel) air/kg kg 12(AF)( coalair mm

6. A food department is kept at -12 by a refrigerator in an environment at 30 . The

total heat gain to the food department is estimated to be 3300 kJ/h and the heat

rejection in the condenser is 4800 kJ/h. determine the power input to the compressor,

in kW and the COP of the refrigerator.

Answer:

a. Assumptions The refrigerator operates steadily.

b. The power input is determined from

kW 0.417

kJ/h 3600

kW 1kJ/h) 1500(

kJ/h 150033004800

in LH QQW

The COP is

2.2kJ/h 1500

kJ/h 3300COP

inW

QL

12C

30C

TH

R

HQ

LQ 3300 kJ/h

4800 kJ/h

inW

7. A household refrigerator with a COP of 1.2 removes heat from the refrigerator space at

a rate of 60 kJ/min. determine (a) the electric power consumed by the refrigerator and

(b) the rate of heat transfer to the kitchen air.

Answer:


b. (a) Using the definition of the coefficient of

performance, the power input to the

refrigerator is determined to be

kW 0.83 kJ/min 051.2

kJ/min 60

COPR

innet,LQ

W

(b) The heat transfer rate to the kitchen air is determined

from the energy balance,

kJ/min 110 5060innet,WQQ LH

8. A household refrigerator that has a power input of 450 W and a COP of 2.5 is to cool five

large watermelons, 10 kg each, to 8 . If the watermelons are initially at 20 ,

determine how long it will take for the refrigerator to cool them. The watermelons can

be treated as water whose specific heat is 4.2 kJ/kg . is your answer realistic or

optimistic? Explain.

Answer:

a. Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator

through its walls, door, etc. is negligible. 3 The watermelons are the only items in

the refrigerator to be cooled.

b. Properties The specific heat of watermelons is given to be c = 4.2 kJ/kg.C.

c. The total amount of heat that needs to be removed from the watermelons is

kJ 2520C820CkJ/kg 4.2kg 105swatermelon TmcQL

The rate at which this refrigerator removes heat is

kW 1.125kW 0.452.5COP innet,R WQL

That is, this refrigerator can remove 1.125 kJ of heat

per second. Thus the time required to remove 2520 kJ

of heat is

min 37.3s 2240 kJ/s 1.125

kJ 2520

L

L

Q

Qt

This answer is optimistic since the refrigerated space will gain some heat during this

process from the surrounding air, which will increase the work load. Thus, in reality, it

will take longer to cool the watermelons.

cool space

Kitchen air

R

COP=1.2

LQ

cool space

Kitchen air

R COP = 2.5

450 W

9. When a man returns to his well-sealed house on a summer day, he finds that the house

is at 35 . He turns on the air conditioner, which cools the entire house to 20 in 30

min. if the COP of the air-conditioning system is 2.8, determine the power drawn by the

air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air

for which cv =0.72 kJ/kg.C and cp =1.0 kJ/kg.C.

Answer:

a. Assumptions

1 The air conditioner operates steadily.

2 The house is well-sealed so that no air leaks in or out during cooling.

3 Air is an ideal gas with constant specific heats at room temperature.

b. Properties The constant volume specific heat of air is given to be cv = 0.72 kJ/kg.C.

c. Since the house is well-sealed (constant volume), the total amount of heat that

needs to be removed from the house is

kJ 8640C2035CkJ/kg 0.72kg 800House

TmcQL v

This heat is removed in 30 minutes. Thus the

average rate of heat removal from the house is

kW 8.4s 6030

kJ 8640

t

QQ L

L

Using the definition of the coefficient of performance, the

power input to the air-conditioner is determined to be

kW 1.712.8

kW 4.8

COPRinnet,

LQW

10. Bananas are to be cooled from 24 to 13 at a rate of 215 kg/h by a refrigeration system.

The power input to the refrigerator is 1.4 kW. Determine the rate of cooling, in kJ/min,

and the COP of the refrigerator. The specific heat of banana above freezing is 3.35

kJ/kg.C.

Answer:


b. Properties The specific heat of banana is 3.35 kJ/kgC.

c. The rate of cooling is determined from

kJ/min 132 C )1324(C)kJ/kg 35.3(kg/min) 60/215()( 21 TTcmQ pL

The COP is

1.57kW 4.1

kW )60/132(COP

inW

QL

AC

Outside

COP = 2.8

HQ

House

3520C

11. A refrigerator is used to cool water from 23 to in a continuous manner. The heat

rejected in the condenser is 570 kJ/min and the power is 2.65 kW. Determine the rate at

which water is cooled, in L/min and the COP of the refrigerator. The specific heat of

water is 4.18 kJ/kg.C and its density is 1 kg/L.

Answer:


b. Properties The specific heat of water is 4.18 kJ/kgC and its density is 1 kg/L.

c. The rate of cooling is determined from

kW 6.85kW 65.2kW )60/570(in WQQ HL

The mass flow rate of water is

kg/s 09104.0C )523(C)kJ/kg 18.4(

kW 85.6

)()(

2121

TTc

QmTTcmQ

p

LpL

The volume flow rate is

L/min 5.46

min 1

s 60

kg/L 1

kg/s 09104.0

mV

The COP is

2.58kW 65.2

kW 85.6COP

inW

QL

12. A heat pump is used to maintain a house at a constant temperature of . The house

is losing heat to the outside air through the walls and the windows at a rate of 60,000

kJ/h while the energy generated within the house from people, lights, and appliances

amounts to 4000kJ/h. for a COP of 2.5, determine the required power input to the heat

pump.

Answer:

a. Assumptions The heat pump operates steadily.

b. The heating load of this heat pump system is the

difference between the heat lost to the outdoors and

the heat generated in the house from the people,

lights, and appliances,

, ,QH 60 000 4 000 56 000 kJ h, /

Using the definition of COP, the power input to the heat

pump is determined to be

kW6.22 kJ/h 3600

kW 1

2.5

kJ/h 56,000

COPHPinnet,

HQ

W

Outside

House

HP

COP = 2.5

HQ

60,000

kJ/h

13. Refrigeratant-134a enters the condenser of a residential heat pump at 800 kPa and 35

at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor

consumes 1.2 kW of power, determine (a) the COP of the heat pump and (b) the rate of

heat absorption from the outside air.

Answer:

a. Assumptions

1 The heat pump operates steadily.


b. Properties The enthalpies of R-134a at the condenser inlet and exit are

kJ/kg 47.950

kPa 800

kJ/kg 22.271C35

kPa 800

22

2

11

1

hx

P

hT

P

c. (a) An energy balance on the condenser gives the heat rejected in the condenser

kW 164.3kJ/kg )47.9522.271(kg/s) 018.0()( 21 hhmQH

The COP of the heat pump is

2.64kW 2.1

kW 164.3COP

inW

QH

(b) The rate of heat absorbed from the outside air

kW 1.96 2.1164.3inWQQ HL

14. Refrigeratant-134a enters the evaporator coils placed at the back of the freezer section

of a household refrigerator at 100 kPa with a quality of 20 percent and leaves at 100 kPa

and -26 . If the compressor consumes 600 W of power and the COP the refrigerator is

1.2, determine (a) the mass flow rate of the refrigerant and (b) the rate of heat rejected

to the kitchen air.

QH 800 kPa

x=0 Condenser

Evaporator

Compressor

Expansion

valve

800 kPa

35C

QL

Win

Answer:

a. Assumptions

1 The refrigerator operates steadily.


b. Properties The properties of R-134a at the evaporator inlet and exit states are

(Tables A-11 through A-13)

kJ/kg 74.234C26

kPa 100

kJ/kg 71.602.0

kPa 100

22

2

11

1

hT

P

hx

P

c. (a) The refrigeration load is

kW 72.0kW) 600.0)(2.1(COP)( in WQL

The mass flow rate of the refrigerant is determined from

kg/s 0.00414

kJ/kg )71.6074.234(

kW 72.0

12 hh

Qm L

R

(b) The rate of heat rejected from the refrigerator is

kW 1.32 60.072.0inWQQ LH

15. From a work-production perspective, which is more valuable: (a) thermal energy

reservoirs at 675 K and 325 K or (b) thermal energy reservoirs at 625 K and 275 K?

Answer:

a. Assumptions The heat engine operates steadily.

QH

100 kPa

-26C

Condenser

Evaporator

Compressor

Expansion

valve

100 kPa

x=0.2 QL

Win

b. For the maximum production of work, a heat engine

operating between the energy reservoirs would have to be

completely reversible. Then, for the first pair of reservoirs

0.519K 675

K 32511maxth,

H

L

T

T

For the second pair of reservoirs,

0.560K 625

K 27511maxth,

H

L

T

T

The second pair is then capable of producing more work for each unit of heat extracted

from the hot reservoir.

16. A heat engine operates between a source at 477 and a sink at . If heat is supplied

to the heat engine at a steady rate of 65,000 kJ/min, determine the maximum power

output of this heat engine.

Answer:

a. Assumptions The heat engine operates steadily.

b. The highest thermal efficiency a heat engine operating between two specified

temperature limits can have is the Carnot efficiency, which is determined from

60.0%or0.600K 273)(477

K 29811Cth,maxth,

H

L

T

T

Then the maximum power output of this

heat engine is determined from the

definition of thermal efficiency to be

kW 653 kJ/min 000,39kJ/min 65,0000.600thoutnet, HQW

17. A heat engine is operating on a Carnot cycle and has a thermal efficiency of 75 percent.

The waste heat from this engine is rejected to a nearby lake at 15 at a rate of 14 kW.

Determine the power output of the engine and the temperature of the source, in

Answer:

a. Assumptions

1 The heat engine operates steadily.

2 Heat losses from the working fluid at the

pipes and other components are negligible.

b. Applying the definition of the thermal

efficiency and an energy balance to the heat

HE

QH

QL Wnet

TH

TL

25°C

477°C

HE

65000 kJ/min

288 K

Source

HE

th = 75% HQ

14 kW LQ

netW

engine, the power output and the source

temperature are determined as follows:

kW 42

kW) 56)(75.0(

kW 56kW 14

175.01

thnet

th

H

H

HH

L

QW

QQQ

Q

C879

K 1152K )27315(

175.01th HHH

L TTT

T

18. A geothermal power plant uses geothermal water extracted at 150 at a rate of 210

kg/s as the heat source and produces 8000 kW of net power. The geothermal water

leaves the plant at 90 . If the environment temperature is 25 , determine (a) the

actual thermal efficiency, (b) the maximum possible thermal efficiency, and (c) the

actual rate of heat rejection from this power plant.

Answer:

a. Assumptions

1 The power plant operates steadily.


3 Steam properties are used for geothermal water.

b. Properties Using saturated liquid properties, (Table A-4)

kJ/kg 83.1040

C25

kJ/kg 04.3770

C90

kJ/kg 18.6320

C150

sinksink

sink

source

source

source,2

geo,1

source

source,1

hx

T

hx

T

hx

T

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

kW 580,53kJ/kg )04.37718.632(kg/s) 210()( geo,2geo,1geoin hhmQ

The actual thermal efficiency is

14.9%0.1493kW 580,53

kW 8000

in

outnet,

thQ

W

19. An inventor claims to have developed a heat pump that produces a 200 kW heating

effect for a 293 K heated zone while only using 75 kW of power and a heat source at 273

K. justify the validity of this claim.

Answer:


b. Applying the definition of the heat pump coefficient of

performance,

2.67kW 75

kW 200COP

innet,

HP W

QH

The maximum COP of a heat pump operating

between the same temperature limits is

7.14K) K)/(293 273(1

1

/1

1COP maxHP,

HL TT

Since the actual COP is less than the maximum COP, the claim is valid.

20. A heat pump operates on a Carnot heat pump cycle with a COP of 8.7. it keeps a space

at by consuming 4.25 kW of power. Determine the temperature of the reservoir

from which the heat is absorbed and the heating load provided by the heat pump.

Answer:


b. The temperature of the low-temperature reservoir is

K 264.6

LLLH

H TTTT

TCOP

K )299(

K 2997.8maxHP,

The heating load is

kW 37.0 LH

in

H QQ

W

QCOP

kW 25.47.8maxHP,

21. A refrigerator is to remove heat from the cooled space at rate of 300 kJ/min to maintain

its temperature at -8 . If the air surrounding the refrigerator is at 25 , determine the

minimum power input required for this refrigerator.

Answer:


b. The power input to a refrigerator will be a minimum

when the refrigerator operates in a reversible manner.

The coefficient of performance of a reversible

refrigerator depends on the temperature limits in the

cycle only, and is determined from

03.81K 2738/K 27325

1

1/

1revR,

LH TTCOP

273 K

293 K

HP

75 kW

HQ

LQ

TL

26C

TH

HP

HQ

LQ

4.25 kW

-8C

25C

R

300 kJ/min

The power input to this refrigerator is

determined from the definition of the

coefficient of performance of a refrigerator,

kW0.623 kJ/min 37.368.03

kJ/min 300

maxR,

minin,net, COP

QW L

22. A heat pump is used to maintain a house at 22 by extracting heat from the outside air

on a day when the outside air temperature is 2 . The house is estimated to lose heat at

a rate of 110,000 kJ/h, and the heat pump consumes 5 kW of electric power when

running. Is this heat pump powerful enough to do the job?

Answer:


b. The power input to a heat pump will be a minimum when the heat pump operates in

a reversible manner. The coefficient of performance of a reversible heat pump

depends on the temperature limits in the cycle only, and is determined from

14.75

K 27322/K 27321

1

/1

1COP revHP,

HL TT

The required power input to this reversible heat

pump is determined from the definition of the

coefficient of performance to be

kW 2.07

s 3600

h 1

14.75

kJ/h 110,000

COPHPminin,net,

HQW

This heat pump is powerful enough since 5 kW > 2.07 kW.

23. A Carnot refrigerator absorbs heat from a space at 15 at a rate of 16,000 kJ/h and

rejects heat to a reservoir at 36 . Determine the COP of the refrigerator, the power

input, in kW, and the rate of rejected to high-temperature reservoir, in kJ/h.

Answer:


b. The COP of the Carnot refrigerator is determined from

17.12

K )1835(

K 291maxR,

LH

L

TT

TCOP

The power input is

kW 0.195 kJ/h 0.701kJ/h 000,12

12.17maxR, ininin

L WWW

QCOP

5 kW

House

22C

HP

110,000 kJ/h

15C

36C

R

HQ

LQ 16,000 kJ/h

inW

The rate of heat rejected is

kJ/h 12,700 kJ/h 701kJ/h 000,12innet,WQQ LH

engg 3260: thermodynamics home assignment 6 (chapter...

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