Modern Engineering Thermodynamics

11.10 Compressibility Factor and Generalized Charts 393
Solving for T 2 gives
T 2 = 423:15 K − 14:0 kJ/kgmole
.K 304:2K
= 308:6K = 35:5°C
0:845 kJ/kg .K 44:01 kg/kgmole
Exercises
42. Determine the exit temperature in the throttle described in Example 11.15 if the inlet pressure is reduced from 20.0 MPa
to 10.0 MPa. Answer: T 2 = 117°C.
43. If the temperature of the CO 2 entering the throttle in Example 11.15 is reduced from 150.°C to 100.°C, determine the
exit temperature of the CO 2 . Answer: T 2 = −90.0°C.
44. Suppose the CO 2 in Example 11.15 is replaced with air using the same inlet and exit conditions. Determine the exit
temperature of the air from the throttle. Answer: T 2-air = 141°C.
Similarly, integration of Eq. (11.26) between a zero value reference state (p o = T o = s o = 0) and a final state at
p, v, T, and s gives
Z T
s = c p /T Z p
dT − ð∂v/∂TÞ p dp
0
0
and, for an ideal gas, we have, from Eq. (7.35),
Combining these results, we get
Now, using Eq. (11.40),
s* =
Z T
0
s = s* +
ðc p /TÞdT − R
Z p
0
Z p
0
h
R/p −ð∂v/∂T
dp/p
Þ p
i
dp
Then,
R/p −ð∂v/∂T
s* − s = R
Þ p
= R/p − ZR/p − ðRT/pÞð∂Z/∂T
h
i
= ðR/pÞ 1 − Z − Tð∂Z/∂T
Z p
0
Þ p
h
i
Tð∂Z/∂TÞ p
+ Z − 1 ðdp/pÞ
Again, nondimensionalizing with T = T c T R and p = p c p R and multiplying both sides by the molecular mass M to
remove substance dependence from the equation, we get the molar results:
s* − s = R
Z pR
0
Þ p
h
i
T R ð∂Z/∂T R Þ pR
+ Z − 1 ðdp R /p R Þ
Compressibility data have been used to integrate this equation, and the results are shown in Figure 11.11. To
find the change in specific entropy between states 1 and 2 using this figure, calculate the values of p R and T R
and obtain values for s* − s from the figure. Then, we can compute
s 2 − s 1 = ðs 2 − s 1 Þ −
ð s − sÞ 2
− s
ð − sÞ 1 ð1/MÞ (11.42)
where s 2 − s 1 = ϕ 2 − ϕ 1 − R lnðp 2 /p 1 Þ from the gas tables, or by assuming constant specific heats over the
temperature range from T 1 to T 2 and using s 2 − s 1 = c p ln(T 2 /T 1 ) – R ln(p 2 /p 1 ).