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ME 336 HW PROBLEM SET 4
9.27
A flow of 2 kg/s saturated vapor R410a at 500 kPa is heated at constant pressure
to 60
o
C. The heat is supplied by a heat pump that receives heat from the ambient
at 300 K and work input, shown in Fig. P9.27. Assume everything is reversible
and find the rate of work input.
Solution:
C.V. Heat exchanger
Continuity Eq.:
m
.
1
= m
.
2
;
Energy Eq.:
m
.
1
h
1
+ Q
.
H
= m
.
1
h
2
Table B.4.2:
h
1
= 274.33 kJ/kg,
s
1
= 1.0647 kJ/kg K
h
2
= 342.32 kJ/kg,
s
2
= 1.2959 kJ/kg K
H
Q
W
L
Q
T
L
HP
1
2
Notice we can find Q
.
H
but the temperature T
H
is not constant making it
difficult to evaluate the COP of the heat pump.
C.V. Total setup and assume everything is reversible and steady state.
Energy Eq.:
m
.
1
h
1
+ Q
.
L
+ W
.
in
= m
.
1
h
2
Entropy Eq.:
m
.
1
s
1
+ Q
.
L
/T
L
+ 0 = m
.
1
s
2
(T
L
is constant, s
gen
= 0)
Q
.
L
= m
.
1
T
L
[s
2
– s
1
] = 2
×
300 [1. 2959 – 1.0647] = 138.72 kW
W
.
in
= m
.
1
[h
2
 h
1
]  Q
.
L
= 2 (342.32 – 274.33) – 138.72 =
–
2.74 kW
9.34
A reversible steady state device receives a flow of 1 kg/s air at 400 K, 450 kPa
and the air leaves at 600 K, 100 kPa. Heat transfer of 800 kW is added from a
1000 K reservoir, 100 kW rejected at 350 K and some heat transfer takes place at
500 K. Find the heat transferred at 500 K and the rate of work produced.
Solution:
C.V. Device, single inlet and exit flows.
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View Full DocumentEnergy equation, Eq.6.12:
m
.
h
1
+ Q
.
3
 Q
.
4
+ Q
.
5
= m
.
h
2
+ W
.
Entropy equation with zero generation,
Eq.9.8:
m
.
s
1
+ Q
.
3
/T
3
 Q
.
4
/T
4
+ Q
.
5
/T
5
= m
.
s
2
1
2
T
T
500 K
3
3
4
4
Q
Q
W
Q
5
Solve for the unknown heat transfer using Table A.7.1 and Eq. 8.19 for
change in s
Q
.
5
= T
5
[s
2
– s
1
] m
.
+
T
5
T
4
Q
.
4
–
T
5
T
3
Q
.
3
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 Spring '09
 Borgnakke

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