Modern Engineering Thermodynamics

264 CHAPTER 8: Second Law Closed System Applications
EXAMPLE 8.10 (Continued )
Similarly,
Z
dQ
2ðS P Þ 3 = ms ð 3 − s 2 Þ−
Σ T b
In this process, we are given that dQ = KdT b , where K = 5.00 Btu/R. Therefore, 2 Q 3 = K(T b3 – T b2 )+C = –9.31 Btu, where C is
an integration constant. Then,
Therefore,
Z
Σ
Z
dQ
Tb3
= K
T b T b2
dT
b
T b
= K ln T b3
T b2
where T b2 = T 2 and T b3 = T 3 .
2ðS P Þ 3
= ms ð 3 − s 2 Þ− K ln T b3
T b2
Now, s 3 = s f 3 + x 3 s fg3 , and from Table C.7b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, at30.0psia,
we get
Then,
and
s f 3 = 0:0364 Btu/ ðlbm⋅RÞ
s fg3 = 0:2209 Btu/ ðlbm⋅RÞ
s 3 = 0:0364 + ð0:1952Þð0:2209 − 0:0364Þ = 0:07241 Btu/ ðlbm⋅RÞ
2ðS P
Þ 3
= ð0:100 lbm
= 0:976 Btu/R
Finally, the entropy production for the entire process is given by
1ðS P
Þ 3 = 1 ðS P
Þ½0:07241 − 0:2635 Btu/ ðlbm⋅RÞŠ− ð5:00 Btu/RÞln 15:38 + 459:67
120: + 459:67
Þ 2 + 1 ðS P Þ 2 = 4:00 × 10 −4 + 0:976 = 0:976 Btu/R ðto 3 significant figuresÞ
In this example, a special Q = Q(T b ) relation is introduced, which is similar to that used to describe convection heat transfer
processes. These relations, often called thermal constitutive equations, are mathematical models developed to describe specific
heat transfer mechanisms. The following exercises illustrate this concept.
Exercises
17. Find 2 (S P ) 3 in Example 8.10 when T b2 = T b3 = T b = 60.0°C, and 2 Q 3 = –9.31 Btu = constant. Answer: This is not possible
since 2 (S P ) 3 is negative for this process and violates the second law of thermodynamics.
18. Rework part d in Example 8.10 using the relation dQ = K 2 T b dT b , where K 2 = 0.001 Btu/R 2 . Note K 2 is not the same as K
in Example 8.10. Keep all other variables the same as in Example 8.10. You have to reevaluate the integral ∫dQ/T b for
this exercise. Answer: 1 (S P ) 3 = 0.0855 Btu/R.
19. Resolve part d in Example 8.10 for radiation heat transfer where dQ = K 3 T b 3 dT b , where K 3 = 6.30 × 10 –6 . Note that K 4 is
not the same as K in Example 8.10. Keep all other variables the same as in Example 8.10. You have to re-evaluate the
integral ∫dQ/T b for this exercise. Answer: 1 (S P ) 3 = 184 Btu/R.
EXAMPLE 8.11 A CONTINUATION OF EXAMPLE 5.7, WITH THE
ADDITION SHOWN IN ITALIC TYPE
A microwave antenna for a space station consists of a 0.100 m diameter rigid, hollow, steel sphere of negligible wall thickness.
During its fabrication, the sphere undergoes a heat treating operation in which it is initially filled with helium at 0.140 MPa
and 200.°C, then it is plunged into cold water at 15.0°C for exactly 5.00 s. The convective heat transfer coefficient of the sphere
in the water is 3.50 W/(m 2 · K). Neglecting any changes in kinetic or potential energy and assuming the helium behaves as an
ideal gas, determine
a. The final temperature of the helium.
b. The change in total internal energy of the helium.
c. The total entropy production in the helium.