Modern Engineering Thermodynamics

8.3 Systems Undergoing Irreversible Processes 265
Solution
First, draw a sketch of the system (Figure 8.11).
The unknowns here are, after 5 s have passed, (a) T 2 ,(b)U 2 – U 1 ,
and (c) 1 (S P ) 2 ; the material is helium gas.
The solutions to the first two parts of this problem can be found
in Example 5.7 as
T ∞ = 15.0°C
0.100 m diameter
System boundary
1 (S P ) 2 = ?
a. T 2 = 32.5°C.
b. U 2 – U 1 = –0.039 kJ.
The solution to part c can be found again by the indirect method
(entropy balance) as follows. From Eq. (8.1), we have
Z
1ðS P Þ 2
= mðs 2 − s ι Þ −
Σ
dQ
T
We can assume that helium behaves as an ideal gas, then since
v 2 = v 1 = constant, we can use Eq. (7.36) to produce
act
FIGURE 8.11
Example 8.11.
h = 3.50 W/(m 2 .K)
Helium
s 2 − s 1 = c v ln T 2
+ R ln v 2
T 1 v 1
32:5 + 273:15
= ½3:123 kJ/ ðkg⋅KÞŠln 200: + 273:15 + 0
= −1:37 kJ/ ðkg⋅KÞ
and, from Example 5.7, we have
so
and
_Q = −hAðT s − T ∞ Þ = dQ
dt
dQ = −hAðT s − T ∞ Þdt
dQ
= dQ = −hA½ðT s − T ∞ Þ/T ∞ Šdt
T b T ∞
wherewehavesetT b = T ∞ (i.e., we put the system boundary slightly outside the sphere itself). Also, in Example 5.7, we
discover that
where T 1 = T s evaluated at t = 0. Then,
Z
T s = T ∞ + ðT 1 − T ∞ Þexp − hAt
mc v
Σ
dQ
T b
Now, from Example 5.7, we have the following numerical values:
act
Z 5s
T
= −hA s − T ∞
dt
0 T ∞
T
= mc 1 − T ∞
v exp − hAt
5s
T ∞
mc v 0
m = 7:46 × 10 −5 kg
c v = 3:123 kJ/kg.RK
T 1 = 200:°C = 473:15 K
T ∞ = 15:0°C = 288:15 K
and hA/(mc v ) = 0.472s –1 . Substituting these values into the preceding integration result gives
Z
dQ
= −1:35 × 10 −4 kJ/K
T b
then
1ðS P
Σ
Þ 2
= ð7:46 × 10 −5 kg
act
= 3:32 × 10 −5 kJ/K = 0:0332 J/K
Þ½−1:365 kJ/ ðkg⋅KÞŠ− ð−1:35 × 10 −4 kJ/KÞ
(Continued )