Modern Engineering Thermodynamics

298 CHAPTER 9: Second Law Open System Applications
m
h in , T in , p in
The energy rate balance analysis of the adiabatic,
aergonic system shown in Figure 9.15 is done in
Chapter 6. The result is
u 2 = h in (6.34)
Initially empty
rigid insulated tank
of volume ∀
and, in the case of an incompressible, constant specific
heat liquid, this means that the final temperature T 2 is
T 2 = T in + ðv/cÞp in (6.35)
FIGURE 9.15
The filling of an insulated, rigid container.
For an ideal gas with constant specific heats, it
means that
T 2 = c p /c v
Tin = kT in (6.36)
The entropy rate balance for this unsteady state system is
_S = dS
dt =
_ Q
T b ⎵
+ _ms in + _S P
0 ðinsulatedÞ
Integrating this equation from its initial empty state 1 to its final filled state 2 with S in equal to a constant and
solving for the entropy production gives
1ðS P Þ 2 = S 2 − S 1 − ðm 2 − m 1 Þs in = m 2 s 2 − m 1 s 1 − ðm 2 − m 1 Þs in
Now, for simplicity, assume that the container is initially evacuated, m 1 = 0, and
then, Eqs. (7.33), (7.37), and (9.40) give
1ðS P Þ 2
= m 2 ðs 2 − s in Þ (9.40)
Entropy production in adiabatically filling a rigid tank with an incompressible liquid
1ðS P Þ incomp: 2 = m
liquid
2 c ln T
2
= m 2 c ln 1 + ðpvÞ
in
T in cT in
ðadiabaticÞ
(9.41)
and
Entropy production in adiabatically filling a rigid tank with an ideal gas
1ðS P Þ 2 ideal = m 2 c p ln T 2
− m 2 R ln p 2
= m 2 c p ln k (9.42)
qas
T
in p in
ðadiabaticÞ
where, for the ideal gas case, m 2 = p 2 V / ðRT 2 Þ = p 2 V/ ðkRT in Þ, where V is the volume of the container and where
2
we use p 2 = p in .
A similar energy rate balance on the same system except now uninsulated and kept isothermal at T 1 = T in = T 2
gives
_Q + _m in h in = d ðmuÞ
dt
and multiplying this equation by dt and integrating gives
1Q 2 + ðm 2 − m 1 Þh in = m 2 u 2 − m 1 u 1
again setting m 1 = 0 for an initially evacuated container produces
u 2 = h in + 1 Q 2
m 2
= u in + ðpvÞ in
+ 1 Q 2
m 2