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ME 336 PROBLEM
SET 3
8.5
Water at 100
o
C, quality 50% in a rigid box is heated to 110
o
C. How do the properties
(P, v, x, u and s) change? (increase, stay about the same, or decrease)
A fixed mass in a rigid box give a constant v
process. So
P goes up
(in the twophase region P = P
sat
at given T)
v stays constant.
x goes up ( we get closer to the saturated vapor state see Pv diagram)
u goes up (Q in and no work)
s goes up (Q in)
8.27
Find the entropy for the following water states and indicate each state on a Ts
diagram relative to the twophase region.
a. 250
o
C, v = 0.02 m
3
/kg
b. 250
o
C, 2000 kPa
c. –2
o
C, 100 kPa
Solution:
a)
Table B.1.1:
0.001251 = v
f
< v < v
g
= 0.05013 m
3
/kg
=>
Twophase
x =
0.02  0.001251
0.04887
= 0.38365
s = s
f
+ x s
fg
= 2.7927 + 0.38365
3.2802 =
4.05 kJ/kg K
b)
Table B.1.1:
P < P
sat
= 3973 kPa
=> superheated vapor B.1.3
s = 6.5452 kJ/kg K
c)
Table B.1.1
T < T
tripple
=0.01
o
C
so goto B.1.5
Table B.1.5:
P > P
sat
= 0.5177 kPa so compressed solid
s = –1.2369 kJ/kg K
P
v
1
2
T
s
1
2
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P
s
T
a
b
c
a
b
c
8.49
A cylinder fitted with a piston contains ammonia at 50
C, 20% quality with a
volume of 1 L. The ammonia expands slowly, and during this process heat is
transferred to maintain a constant temperature. The process continues until all the
liquid is gone. Determine the work and heat transfer for this process.
Solution:
C.V. Ammonia in the cylinder.
1
2
T
s
50 C
NH
3
o
Table B.2.1:
T
1
= 50°C,
x
1
= 0.20,
V
1
= 1 L
v
1
= 0.001777 + 0.2
0.06159 = 0.014095 m
3
/kg
s
1
= 1.5121 + 0.2
3.2493 = 2.1620 kJ/kg K
m = V
1
/v
1
= 0.001/0.014095 = 0.071 kg
v
2
= v
g
= 0.06336 m
3
/kg,
s
2
= s
g
= 4.7613 kJ/kg K
Process: T = constant to x
2
= 1.0,
P = constant = 2.033 MPa
From the constant pressure process
1
W
2
=
PdV = Pm(v
2
 v
1
) = 2033
0.071
(0.06336  0.014095) =
7.11 kJ
From the second law Eq.8.3 with constant T
1
Q
2
=
TdS = Tm(s
2
 s
1
)
= 323.2
0.071(4.7613  2.1620) =
59.65 kJ
or
1
Q
2
= m(u
2
 u
1
) +
1
W
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 Spring '09
 Borgnakke

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