Post on 13-Mar-2020
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
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
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:
a. Assumptions The refrigerator operates steadily.
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:
a. Assumptions The refrigerator operates steadily.
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:
a. Assumptions The refrigerator operates steadily.
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.
2 The kinetic and potential energy changes are zero.
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.
2 The kinetic and potential energy changes are zero.
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.
2 The kinetic and potential energy changes are zero.
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:
a. Assumptions The heat pump operates steadily.
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:
a. Assumptions The heat pump operates steadily.
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:
a. Assumptions The refrigerator operates steadily.
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:
a. Assumptions The heat pump operates steadily.
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:
a. Assumptions The refrigerator operates steadily.
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