Modern Engineering Thermodynamics

9.9 Unsteady State Processes in Open Systems 299
For an incompressible isothermal liquid, this reduces to
or
u 2 − u in = cT ð 2 − T in Þ = ðpvÞ in + 1 Q 2
= 0
m 2
and, for an isothermal ideal gas, it becomes
1Q 2 = −m 2 ðpvÞ in
(9.43)
1Q 2 = −m 2 RT in = −m 2 c v T in ðk − 1Þ
The integrated entropy rate balance for this system with both S in and T b constant is
and, when m 1 = 0, this becomes
1ðS P Þ 2
= m 2 s 2 − m 1 s 1 − ðm 2 − m 1 Þs in − 1 Q 2
T b
1ðS P Þ 2 = m 2 ðs 2 − s in Þ − 1 Q 2
T b
(9.44)
For an isothermal incompressible liquid with T 2 = T in = T b = T, Eqs. (7.33), (9.43), and (9.44) give
Entropy production in isothermally filling a rigid tank with an incompressible liquid
1ðS P
Þ 2 incomp:
liquid
ðisothermalÞ
= m 2 ðpvÞ in /T (9.45)
and, for an isothermal ideal gas with p 2 = p in and T 2 = T in = T b = T,
Entropy production in isothermally filling a rigid tank with an ideal gas
1ðS P
Þ 2 ideal
gas
ðisothermalÞ
= m 2 c v ðk − 1Þ = m 2 R = p 2 V/T (9.46)
where R = c p – c v = c v (c p /c v – 1) = c v (k – 1) and m 2 = p 2 V/ðRTÞ.
Now, comparing Eqs. (9.41) and (9.45) for filling the container with the same amount of an incompressible
liquid (i.e., m 2 is the same in each case), we see that, since ln(1 + x) < x for all x >0,
1ðS P
Þ 2 incomp:
liquid
ðadiabaticÞ
< 1 ðS P
Þ 2 incomp:
liquid
ðisothermalÞ
ðfor adding the same amount of mass in each caseÞ
ð9:47Þ
but, on comparing Eqs. (9.42) and (9.46) for ideal gases, we find that we cannot add the same amount of mass
in each case because
ðm 2 Þ adiabatic
filling
= p 2 V/ ðRT 2 Þ = p 2 V/ kRT in
ð Þ = ðm 2
Þ/k isothermal
filling
(9.48)
Since
R
c p
k
= c p − c v
c v
= k − 1