Modern Engineering Thermodynamics

382 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.11 (Continued )
33. Determine an expression for the specific entropy of the material in Example 11.11 in the superheated region. Answer:
where B = B 1 – B 2 /T.
s = s 0 +
Z T
c 0 Z v
v
T dT + ∂p
= s 0 + c
∂T
v0 vdv 0 v ln T
+ R ln v
+ exp½BŠ 1 − B
2 1
T 0 v 0
T v − 1
v 0
T0
11.9 GAS TABLES
We now move from the region of the vapor dome and its immediate surroundings out into the region where
the material behaves as an “ideal gas.” However, we need to incorporate the temperature dependence of the
ideal gas’s specific heats.
When the concept of an ideal gas was introduced in Chapter 3, it was noted that ideal gases generally do not
have constant specific heats. Therefore, one of the simplest steps we can take to make the ideal gas equations
more accurate is to take into account their temperature-dependent specific heats. This is what is done for you in
Tables C.16 and C.17 in Thermodynamic Tables to accompany Modern Engineering Thermodynamics.
The specific internal energy and enthalpy values listed in thegastablesaredetermined from an integration of
Eqs. (3.37) and (3.40) by incorporating an accurate specific heat vs. temperature data curve fit as
and
Z T
u = u o + c v dT
T o
Z T
h = h o + c p dT
T o
where u o , h o ,andT o are the arbitrarily chosen reference state values, all of which are set equal to zero in these
tables.
Since the ideal gas specific entropy depends on more than just temperature, it is not listed in these tables. However,
the temperature-dependent part of the specific entropy is listed as the ϕ function, where
and, from Eq. (7.35), we have
Z T
ϕ = c p /T dT (11.34)
T o
Z T
s = s o + ðc p /TÞdT − R ln ðp/p o Þ
T o
= s o + ϕ − R ln ðp/p o Þ
(7.35)
In the gas tables, s o and T o are both arbitrarily set equal to zero, and p o is set equal to 1 atm. As always, the arbitrarily
chosen reference states for u, h, ands cancel out when we calculate the changes u 2 – u 1 , h 2 – h 1 , and s 2 – s 1 .
Such changes are calculated directly from values taken from the gas tables for u and h, but the change in entropy
is given by combining the previous equations as
or an isentropic process, s 2 = s 1 ; Eq. (11.35) then gives
s 2 − s 1 = ϕ 2 − ϕ 1 − R ln ðp 2 /p 1 Þ (11.35)
ðϕ 2 − ϕ 1 Þ/R = ln ðp 2 /p 1 Þ = ln p 2p o
= ln ðp r2 /p r1 Þ
p 1 p o