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ME 336 HOMEWORK SET 5
10.16
Find the availability of 100 kW delivered at 500 K when the ambient is 300 K.
Solution:
The availability of an amount of heat transfer equals the possible work that
can be extracted. This is the work out of a Carnot heat engine with heat
transfer to the ambient as the other reservoir. The result is from Chapter 7 as
also shown in Eq. 10.1 and Eq. 10.36
Φ
.
= W
.
rev HE
= (1 –
T
o
T
)Q
.
= (1 –
300
500
) 100 kW =
40 kW
10.20
A household refrigerator has a freezer at T
F
and a cold space at T
C
from which
energy is removed and rejected to the ambient at T
A
as shown in Fig. P10.27.
Assume that the rate of heat transfer from the cold space, Q
.
C
, is the same as from
the freezer, Q
.
F
, find an expression for the minimum power into the heat pump.
Evaluate this power when T
A
=
20
°
C, T
C
=
5
°
C, T
F
=
−
10
°
C, and Q
.
F
=
3 kW.
Solution:
C.V. Refrigerator (heat pump), Steady, no
external flows except heat transfer.
Energy Eq.:
Q
.
F
+ Q
.
c
+ W
.
= Q
.
A
(amount rejected to ambient)
Q
W
A
Q
C
REF
Q
F
Reversible gives minimum work in as from Eq. 10.1
or 10.9 on rate form.
W
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This note was uploaded on 10/20/2009 for the course MECHENG mecheng336 taught by Professor Borgnakke during the Spring '09 term at University of Michigan.
 Spring '09
 Borgnakke
 Heat Transfer

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