Modern Engineering Thermodynamics

268 CHAPTER 8: Second Law Closed System Applications
WHY IS THE ENTROPY PRODUCTION RATE IN EXAMPLE 8.12
SO SMALL?
This is due to the fact that the fin temperature T(x) is close to the environmental temperature T ∞ over most of the length of
the fin. Entropy production due to heat transfer is minimized when the temperature difference producing the heat transfer
is small. Check it out for yourself to see that _S P
Q becomes zero in the following equation when T f = T ∞ (remember,
ln(1) = 0).
_S P
Q =
p ffiffiffiffiffiffiffiffiffiffiffiffi
hPk t A
ln T f
+ T
∞
− 1
T ∞ T f
And the maximum entropy production rate occurs when the temperature difference is a maximum. This occurs in this
example when T ∞ =0KorT f = ∞. In this case,
_S P
max = lim _S P = lim
T∞!0 Tf !∞
_S P = ∞
EXAMPLE 8.13
The velocity profile in the steady isothermal laminar flow of an incompressible Newtonian fluid contained between
concentric cylinders in which the inner cylinder is rotating and the outer cylinder is stationary is given by
V = ωR2 1 R 2
2
R 2 2 − R2 1
x − x
for R 1 ≤ x ≤ R 2
where ω is the angular velocity of the inner cylinder, and x is measured radially outward. Determine the rate of entropy
production due to laminar viscous losses for SAE-40 engine oil at 20.0°C in the gap between cylinders of radii 0.0500 and
0.0510 m when the inner cylinder is rotating at 1000. rev/min. The viscosity of the oil is 0.700 N · s/m 2 and the length of
the cylinder is 0.100 m.
Solution
First, draw a sketch of the system (Figure 8.14).
Velocity profile V
R 2
R 1
x
R 2
L
R 1
ω
FIGURE 8.14
Example 8.14, system.
The unknown is the rate of entropy production due to laminar viscous losses. The material is SAE-40 engine oil at 20.0°C.
Here, we use Eq. (7.72) for the viscous dissipation of mechanical work. By differentiating the velocity formula just given,
we get
dV
dx = − ωR2 1 R 2
2
R 2 2 − R2 1
x 2 + 1