Modern Engineering Thermodynamics

308 CHAPTER 9: Second Law Open System Applications
EXAMPLE 9.11 (Continued )
For the differential volume element dV, we use the volume of an annulus of thickness dx, ordV = 2πLx dx. Then, we can
evaluate Eq. (7.72) as
For this problem,
then
_S P
W
vis
_S P
W
vis
= 8πμLV 2 m
R 4 T
Z R
0
x 3 dx = 2πμLV2 m
T
μ = 10:1 × 10 −3 kg/ ðm.sÞ
L = 10:0m
V m = 0:500 m/s
T = 20:0°C = 293 K
= 2π ½ 10:1 × 10−3 kg/ðm.sÞŠð10:0mÞð0:500 m/sÞ 2
20:0 + 273:15 K
= 5:41 × 10 −4 W/K
In this example, we again have a very low entropy production rate. This is due, in this case, to the fact that laminar flow is a
very energy efficient type of flow. Turbulent flow (which occurs spontaneously here at higher flow velocities) is much more
dissipative and consequently is a much less energy efficient flow.
Exercises
29. If the fluid being pumped through the pipe in Example 9.11 is SAE-30 motor oil with a viscosity of 0.400 kg/(m·s)
instead of water, determine the entropy production rate due to viscosity. Keep the values of all the other variables the
same as they are in Example 9.11. Answer: _S P = 0:0214 W/K.
W-vis
30. If we increase the maximum (centerline) velocity of the water in Example 9.11 from 0.500 m/s to 3.00 m/s, determine
the new entropy production rate due to viscosity. Keep the values of all the other variables the same as they are in
Example 9.11. Answer: _S P
W-vis = 1:95 × 10−3 W/K.
31. If we increase the length of the pipe in Example 9.11 from 10.0 m to 1000. m, determine the new entropy
production rate due to viscosity. Keep the values of all the other variables the same as they are in Example 9.11.
Answer: _S P
W-vis = 5:41 × 10−3 W/K.
32. The diameter of the pipe in Example 9.11 is increased from 2.50 to 3.75 cm, but the mass flow rate is maintained
constant. Determine the new entropy production rate due to the fluid’s viscosity. Keep the values of all the variables
except V m the same as they are in Example 9.11. Answer: _S P
W-vis = 1:07 × 10−5 W/K.
SUMMARY
In this chapter, we investigate a series of open systems and carry out a second law analysis of them using the
entropy balance or the entropy rate balance. The primary purpose of this material is to stimulate your thinking
by addressing modern technologies from a second law point of view. Several new equations are introduced in
this chapter that deal with these technologies. Some of the more important equations follow.
1. The general open system entropy rate balance (SRB) equation is
Z
Σ
_q
T b act
dA +∑
inlet
+∑
in
_ms− ∑
outlet
_ms+ _S P = _S system (9.4)
and, when the system boundaries are isothermal, this equation reduces to
_Q
_ms−∑ _ms+ _S p = _S system (9.5)
T b act
out
This equation is the actual heat transfer rate, and T b is the temperature where the system boundary occurs.
When you have a steady state (SS), steady flow (SF), single inlet, single outlet (SI, SO), isothermal boundary
(IB) system, Eqn. (9.5) reduces to the modified entropy rate balance:
_Q
+ _m ðs in − s out Þ+ _S P = 0 (9.10)
T b