Problem Statement:
24
Consider a Carnot cycle executed in a closed system with air as the working fluid.
The maximum pressure in the cycle is
800.000 kPa while the maximum
temperature is
750.00 K.
If the entropy increase during the isothermal heat rejection process is
0.25 kJ/kg*K and the net work output is
100.00 kJ/kg, determine:
a) the minimum pressure in the cycle
b) the heat rejection from the cycle
c) the thermal efficiency of the cycle
d) If an actual heat engine cycle operates between the same temperature limits and produ
5200.000 kW of power for an air flow rate of
90.00 kg/s, determine
the 2nd law efficiency of this cycle.
Schematic:
Assumptions/Approximations:
Air is an ideal gas with constant specific heats
Physical Laws:
Properties/Sources:
0.29 kJ/kg*K
[Table A2]
0.718 kJ/kg*K
[Table A2]
1.005 kJ/kg*K
[Table A2]
1.400
Calculation:
350.00 K
55.467
kPa
[Book answer is in error]
3.14
23.21
kPa
87.50
kJ/kg
9000.00 kW
53.33%
w
NET
= (s
2
 s
1
)*(T
H
 T
L
)
∆
s
12
= 
∆
s
34
= Cp
AIR
*ln(T
4
/T
3
)  R
AIR
*(ln(P
4
)  ln(P
3
))
η
th
= 1  T
L
/T
H
P
1
= P
4
*(T
1
/T
4
)
(k/(k  1))
q
OUT
= T
L
*
∆
s
12
η
II
= W
ACTUAL
/W
CARNOT
R
AIR
=
Cv
AIR
=
Cp
AIR
=
k = Cp
AIR
/Cv
AIR
=
w
NET
= (s
2
 s
1
)*(T
H
 T
L
) ==> T
L
= T
4
=
P
1
= P
4
*(T
1
/T
4
)
(k/(k  1))
==> P
4
=
∆
s
12
= 
∆
s
34
= Cp
AIR
*ln(T
4
/T
3
)  R
AIR
*(ln(P
4
)  ln(P
3
)) ==> ln(P
3
)
P
3
=
q
OUT
= T
L
*
∆
s
12
=
Ẇ
CARNOT
=
*w
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 Spring '09
 n/a
 Heat engine, Qin, table a2

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