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Engineering

ENGG 3260

Linda Gerber

Winter

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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.
Furnace
2 Heat losses from the working fluid at the
pipes and other components are negligible.
b. According to the definition of the thermal qH
efficiency as applied to the heat engine,
HE
w netth H
q w net
qH q L qth H L
which when rearranged gives sink
q qL 1000 kJ/kg1667kJ/kg
H 1 th 10.4
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
QH mcoal HV,coal 60 t/h Furnace
9
60,000 kg/h 30,000 kJ/kg 1.810 kJ/h
500 MW
Then the thermal efficiency of the plant becomes
W
th net,ot150 MW 0.300 30.0% coal HE
QH 500 MW
150 MW
sink 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 Fuel Engine
mfuel (V )fuel (0.8 kg/L)(22 L/h) 17.6 kg/h
The rate of heat supply to the car is
55 kW
Q H mcoalHV,coal
(17.6 kg/h)(44,000 kJ/kg) 22 L/h HE
774,400 kJ/h 215.1kW
Then the thermal efficiency of the car becomes
W net,ou 55 kW 0.256 25.6% sink
th Q 215.1kW
H
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 28 10 kJ/ton. c. For a power generation capacity of 150,000
MW, the construction costs of coal and
IGCC plants and their difference are
9
Construction coscoal (150,000,000 kW)($1300/kW) = $19510
Construction cost (150,000,000 kW)($1500/kW) = $22510 9
IGCC
Construction cost difference $22510 $19510 $3010 9
The amount of electricity produced by either plant in 5 years is
12
W eWt (150,000,000 kW)(5 36524 h) = 6.57010 kWh
The amount of fuel needed to generate a specified amount of power can be determined
from
W W Q W
e Q in e or m fuel in e
Q in Heating value (Heating value)
Then the amount of coal needed to generate this much electricity by each plant and
their difference are
12
m W e 6.57010 kWh 3600 kJ 2.11210 tons
coal,coalpla(Heating value) (0.40)(2810 kJ/ton) 1kWh
W 6.57010 12kWh 3600 kJ
mcoal,IGCC plant e 1.76010 tons
(Heating value) (0.48)(2810 kJ/ton) 1kWh
9 9 9
m coal mcoal,coalplantcoal,IGCC plant112 10 1.76010 = 0.35210 tons
For m coalo pay for the construction cost difference of $30 billion, the price of coal
should be
Construction cost difference $30 109
Unit cost of coal 9 $85.2/ton
m coal 0.352 10 tons
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
W net,out300 MW
Q in 937.5MW
th 0.32
Q in Qint (937.5MJ/s)(24 3600 s) 8.110 MJ
The amount and rate of coal consumed during this period are
7
Qin 8.110 MJ 6
m coal 2.89310 kg
q HV 28 MJ/kg
6
m m coal 2.89310 kg 33.48 kg/s
coal t 243600s
(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is
m (AF)m (12 kg air/kg fuel) (33.48 kg/s) 401.8kg/s
air coal
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.
30C
T H
QH 4800 kJ/h
R Win
QL
3300 kJ/h
12C
Answer:
a. Assumptions The refrigerator operates steadily.
b. The power input is determined from
Win QH QL
4800 3300 1500 kJ/h
(1500 kJ/h 1kW 0.417kW
3600 kJ/h
The COP is
QL 3300 kJ/h
COP 2.2
W in 1500 kJ/h 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. Kitchen air
b. (a) Using the definition of the coefficient of
COP=1.2
performance, the power input to the
refrigerator is determined to be
R
QL 60 kJ/min
Wnet,in 50 kJ/min 0.83 kW QL
COP R 1.2
(b) The heat transfer rate to the kitchen air is determined cool space
from the energy balance,
Q Q W 60 50 110kJ/min
H L net,in
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
Q mcT 5 10 kg 4.2 kJ/kg C 20 8 C 2520 kJ
L watermelon s
The rate at which this refrigerator removes heat is Kitchen air
Q COP W 2.5 0.45 kW 1.125 kW
L R net,in
450 W
That is, this refrigerator can remove 1.125 kJ of heat R
per second. Thus the time required to remove 2520 kJ COP = 2.5
of heat is
t QL 2520 kJ 2240 s 37.3 min cool space
QL 1.125 kJ/s
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. 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 c v0.72 kJ/kg.C and c =1p0 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 c = 0.v2 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
Q mc T 800 kg 0.72 kJ/kg C 35 20 C 8640 kJ
L v House
This heat is removed in 30 minutes. Thus the Outside
average rate of heat removal from the house is
Q
QL 8640 kJ H
Q L 4.8 kW COP = 2.8
t 3060 s AC
Using the definition of the coefficient of performance, the
power input to the air-conditioner is determined to be 3520C
Wnet,in QL 4.8 kW 1.71kW House
COP R 2.8
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/kg C.
c. The rate of cooling is determined from
Q L mc (p 1 ) 2 (215/60 kg/min)(3.35kJ/kg C)(24 13)C 132kJ/min
The COP is
COP Q L (132 / 60) k1.57
Win 1.4 kW 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/kg C and its density is 1 kg/L.
c. The rate of cooling is determined from
Q Q W (570 /60) kW 2.65 kW 6.85 kW
L H in
The mass flow rate of water is
QL 6.85 kW
Q L mc (p 1 ) 2 m 0.09104 kg/s
cp(T1T 2 (4.18 kJ/kg C)(23 5)C
The volume flow rate is
m 0.09104 kg/s 60s
V 5.46L/min
1kg/L 1min
The COP is
COP

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