Modern Engineering Thermodynamics

222 CHAPTER 7: Second Law of Thermodynamics and Entropy Transport and Production Mechanisms
For a stationary differential closed system at a uniform temperature T containing a pure substance that is
subjected to only a mechanical moving boundary work mode, Eq. (7.29) becomes
TdS= dU + pdV
and on dividing through by the system mass m and the absolute temperature T,
ds = du
T + p dv (7.30)
T
Since u = h − pv, this equation can also be written as
ds = dh
T − v dp (7.31)
T
InChapter3,wedefinetheconstantvolume and constant pressure specific heats for an incompressible
substance as
c v = c p = c = du
dT
Since v = constant and dv = 0 for an incompressible material, then Eq. (7.30) becomes
ðdsÞ incomp: = c dT
T
or
ðs 2 − s 1 Þ incomp: =
Z T2
T 1
c dT T
(7.32)
If the specific heat c is constant over the temperature range from T 1 to T 2 , then this equation can be integrated
to give
ðs 2 − s 1 Þ incompressible material
= c lnðT 2 /T 1 Þ (7.33)
with a constant c
In Chapter 3, we also define the constant volume and constant pressure specific heats for an ideal gas as
c v = du
dT
(3.37)
and
c p = dh
dT
(3.40)
Consequently, we can now write Eqs. (7.30) and (7.31) as
dT
ðdsÞ ideal = c v + p
T T dv = c p
gas
dT
T
− v T dp
Further, for an ideal gas, p/T = R/v and v/T = R/p, so this equation can be integrated to give
ðs 2 − s 1
Þ ideal
gas
=
=
Z 2
c v
1
Z 2
c p
1
dT
T
dT
T
+ R ln v 2
v 1
(7.34)
− R ln p 2
p 1
(7.35)
and if the specific heats are constant over the temperature range from T 1 to T 2 , then these equations become
ðs 2 − s 1 Þ ideal gas
constant
= c v ln T 2
+ R ln v 2
T 1 v 1
(7.36)
c p and c v
= c p ln T 2
T 1
− R ln p 2
p 1
(7.37)