Modern Engineering Thermodynamics

9.9 Unsteady State Processes in Open Systems 305
CASE STUDY 9.2. HYDRODYNAMIC FLOW SYSTEMS
A variety of hydrodynamic flow situations can be effectively
analyzed with the entropy rate balance. Consider a steady state,
steady flow, single-inlet, single-outlet system. The MERB for this system
using the definition of enthalpy, h = u + pv, and the definition
of density, ρ =1/v, can be written as
_Q − _W = _m u 2 − u 1 + ðp/ρÞ 2
− ðp/pÞ 1
+ V2 2 −
V2 1 /2gc + ðg/g c ÞðZ 2 − Z 1 Þ
= _m u 2 − u 1 − ðg/g C Þ 1 ðh L Þ 2
where, using the specific weight γ = ρg,
1 h L Þ 2
= ðpg c /γÞ 1
− ðpg c /γÞ 2
+ V1 2 − V 2 2
/2g + Z1 − Z 2 (9.52)
is the head loss between the inlet station 1 and the exit station 2. In
fluid mechanics texts, Eq. (9.52) is known as the Bernoulli equation,
named after the Swiss mathematician and hydrodynamist Daniel
Bernoulli (1700–1782). The MSRB for this system is
_S P = _m ðs 2 − s 1 Þ− _Q /T b
If the flowing fluid is a constant specific heat incompressible liquid,
ρ 1 = ρ 2 = constant
u 2 − u 1 = cT ð 2 − T 1 Þ
s 2 − s 1 = c ln T 2
T 1
and, if it is a constant specific heat ideal gas,
ρ = p/ ðRTÞ = 1/v
u 2 − u 1 = c v ðT 2 − T 1 Þ
s 2 − s 1 = c p ln T 2
T 1
− R ln p 2
p 1
Now, let us compare the rate of entropy production for two cases (i.e.,
two thermodynamic paths), adiabatic flow and isothermal flow.
Case A. SS, SF, SI, SO, aergonic, adiabatic flow 8
Incompressible, constant specific heat liquids. In this case, the
MERB gives
T 2 = T 1 + g 1 ðh L Þ 2 / ð cgc Þ
Ideal gases with constant specific heats. In this case, the MERB
gives
T 2 = T 1 + g 1 h L Þ 2 / ð cv g c Þ
and the MSRB gives
_S p
P ideal = _mc p ln 1 + g 1 ðh L Þ 2 / ð cv g c T 1 Þ − _mR 2 ln (9.54)
gas
p 1
ðadiabaticÞ
Case B. SS, SF, (SI, SO) aergonic, isothermal flow
Incompressible, constant specific heat liquids. In this case, the
MERB gives
and the MSRB gives
_S P
_Q = − _mg 1 h L Þ 2 /gc
incomp:
liquid
ðisothermalÞ
= _mg 1 h L Þ 2 /ðTb g c Þ
(9.55)
Ideal gases with constant specific heats. In this case, the MERB
again gives
and the MSRB gives
_S P ideal
gas
ðisothermalÞ
_Q = − _mg 1 h L Þ 2 /gc
= _mg 1 ðh L Þ 2 /ðTb g c Þ − _mRln p 2
(9.56)
p 1
The question is this: Which process, adiabatic or isothermal, is the
more efficient by producing less entropy? For the incompressible
liquids, we must compare Eqs. (9.53) and (9.55). Using a series
expansion for the logarithm, we find that, for all x > 0, ln(l + x) < x,
so that for T b = T l , we have
c ln 1 + g 1 ðh L Þ 2 / ð cgc T 1 Þ < g 1 ðh L Þ 2 / ð T1 g c Þ
and, consequently,
_S P
incomp:
liquid
ðadiabaticÞ
< _S P
incomp:
liquid
ðisothermalÞ
Similarly, for the ideal gases, we compare Eqs. (9.54) and (9.56) and
again find that.
_S
P ideal < _S P ideal
gas
gas
ðadiabaticÞ
ðisothermalÞ
and the MSRB gives
_S
incomp: P
liquid:
ðadiabaticÞ
= _mc ln 1 + g 1 h L Þ 2 / ð cgc T 1 Þ
(9.53)
In both of these cases, the adiabatic process produces less entropy.
This is always true in this type of comparison, because an adiabatic
system eliminates the entropy production due to heat transfer across
the system boundary.
(Continued )