Modern Engineering Thermodynamics

8.4 Diffusional Mixing 271
Exercises
26. Recompute the entropy production rate _S P in Example 8.14 when the steady state surface temperature of the board
increases to 38.0°C. Keep the values of all the other variables the same as they are in Example 8.14.
Answer: _S P
W = 1:61 × 10–4 W/K:
27. Determine the entropy production rate _S P in Example 8.14 if the current is increased to 50.0 mA. Keep the values of all
the other variables the same as they are in Example 8.14. Answer: _S P
W = 8:25 × 10–4 W/K:
28. If the circuit voltage is increased from 5.00 V to 30.0 V in Example 8.14, determine the new entropy production rate.
Keep the values of all the other variables the same as they are in Example 8.14. Answer: _S P
W = 9:90 × 10–4 W/K:
8.4 DIFFUSIONAL MIXING
Here, we wish to use the indirect method to analyze the entropy production that results from a simple diffusion
type of mixing process. Consider the rigid, insulated container shown in Figure 8.18. It contains the same substance
on both sides of the partition but generally at different pressures, temperatures, and amounts. When the
partition is removed, the material in the chambers mixes by diffusion, resulting in a final temperature T 2 and
final pressure p 2 . We are interested in the amount of entropy produced by this mixing process.
The energy balance equation for this system gives
1Q 2
0
ðadiabaticÞ
− 1 W 2
= U 2 − U 1 + KE 2 − KE 1 + PE 2 − PE 1
⎵
0 ðadiabaticÞ
0 ðstationary systemÞ
or
U 2 = m 2 u 2 = U 1 = m a u a + m b u b
where
u 2 = m au a + m b u b
m a + m b
= u b + yu ð a − u b Þ (8.9)
and
The entropy balance equation gives
y =
1 − y =
m a
m a + m b
m b
m a + m b
1ðS P Þ 2
= S 2 − S 1 − 1 Q 2
T b
0
or
ðadiabaticÞ
Entropy production in two-component mixing:
1ðS P Þ 2
= m 2 s 2 − m a s a − m b s b p
= ðm a + m b Þ½s 2 − s b + ys ð b − s a ÞŠ
(8.10)
Partition
Insulation
T a , p a
a , m a
A
T b , p b
b , m b
A
Mixing
process
T 2 , p 2
2, m 2
A
State 1 State 2
FIGURE 8.18
Mixing of single-species substances.