TEST II
THERMO I
3/31/04
OPEN BOOK AND NOTES
1.)
(33%) Find the entropy change s
2
 s
1
from an initial state of P
1
=
101.32 kPa, T
1
= 25°C to a final state of P
2
= 400 kPa, T
2
= 500°C for:
a) air (Gas constant R =
0.28700 kJ/(kg•K))
s
2
 s
1
= s
o
(T
2
)  s
o
(T
1
)  R ln (P
2
/P
1
)
From Table 5s
for air,
s
o
(T
2
) = s
o
(500°C) = 7.6852 kJ/(kg•K)
s
o
(T
1
) = s
o(
25°C) = 6.6999 kJ/(kg•K)
s
2
 s
1
= 7.6852  6.6999  0.28700 ln (400/101.32)
=
7.6852  6.6999  0.28700 x 1.3732 = 0.5912 kJ/(kg•K)
b) H
2
0 (do not
treat as an ideal gas.)
Using the steam tables, Tables 10s or 11s
show that state 1 is subcooled
,
state 2 is superheated
. Does not act as ideal gas!!
From Table 12s, s
2
= s(P=400 kPa, T = 500°C) = 8.1914 kJ/(kg•K)
Entropy for a subcooled liquid is independent of pressure, so we can
choose
s
1
= s
L
(T
1
), and from Table 10s,
s
1
= s
l
(25°C) = 0.3670 kJ/(kg•K)
Thus,
s
2
 s
1
= 8.1914  0.3670 = 7.8244 kJ/(kg•K)
(Note the large entropy change compared with air.
... this is because of the
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 Spring '07
 Schmidt
 Thermodynamics, Entropy, figure number, entropy change s2

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