Modern Engineering Thermodynamics

384 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.12 (Continued )
Now, 1 Q 2 = 0 (isentropic processes are also adiabatic), so
1W 2 /m = u 1 − u 2
The gas tables are to be used for the thermodynamic properties of air here, because they are more accurate than the standard
constant specific heat ideal gas equations. From Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics,
we find that, at 60.0°F = 520. R,
and
u 1 = 88:62 Btu/lbm
p r1 = 1:2147
v r1 = 158:58
For a compression ratio of 19.2 to 1, v 2 /v 1 = 1/19:2: Then, from Eq. (11.37),
v r2 = v r1 ðv 2 /v 1 Þ = 158:58/19:2 = 8:26
scanning down the v r column in Table C.16a, we find that v r = 8.26 at about
T 2 = 1600 R = 1140°F
and
u 2 = 286:06 Btu/lbm
Then, from Eq. (11.36),
p r2 = 71:73
Finally, from the preceding energy balance,
p 2 = p 1 ðp r2 /p r1 Þ = ð14:7 psiaÞð71:73/1:2147Þ
= 868:1 psia
1W 2 /m = u 1 − u 2 = 88:62 − 286:06 = −197:44 Btu/lbm
Exercises
34. Determine the final temperature and pressure in Example 11.12 if the compression ratio is 10.7 to 1 instead of 19.2 to
1. Answer: T 2 = 1300. R and p 2 = 392. psia.
35. Rework Example 11.12 for a compression ratio of 19.3 to 1 when air at 300. K and 0.100 MPa is drawn into the engine
during the intake stroke. Answer: T 2 = 920. K, p 2 = 5.92 MPa, and 1 W 2 /m = −477 kJ/kg.
36. Use the relations for an isentropic process for an ideal gas,
T 2
T 1
= v 2
v 1
1− k
= p 2
to compute the final temperature T 2 and pressure p 2 of the compression process described in Example 11.12. Then,
assuming constant specific heats, compute the work required per unit mass for the compression 1 W 2 /m = u 1 – u 2 =
c v (T 1 – T 2 ). Answer: T 2 = 1236°F, p 2 = 920 psia, and 1 W 2 /m = −200. Btu/lbm.
p 1
k−1
k
11.10 COMPRESSIBILITY FACTOR AND GENERALIZED CHARTS
In Chapter 3, we discussed the van der Waals, Dieterici, Berthelot, Beattie-Bridgeman, and Redlich-Kwong
equations of state as possible models of nonideal, or real, gas behavior. Several of these equations are just variations
on the basic ideal gas equation, pv = RT. We now introduce one of the most powerful engineering real gas
equations of state: the compressibility factor.
In 1880, the Dutch physicist Johannes Diderik van der Waals (1837–1923) reasoned that, if the p-v-T
equation of state could be nondimensionalized, then all gases might be found to fitthesamedimensionless
p-v-T equation. Further, he noted that, since every substance has a vapor dome and every vapor dome has a
unique critical point (its peak), perhaps the critical point properties (p c , T c ,andv c ) couldbeusedtocreatea