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Modern Engineering Thermodynamics

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384 CHAPTER 11: More Thermodynamic Relations<br />

EXAMPLE 11.12 (Continued )<br />

Now, 1 Q 2 = 0 (isentropic processes are also adiabatic), so<br />

1W 2 /m = u 1 − u 2<br />

The gas tables are to be used for the thermodynamic properties of air here, because they are more accurate than the standard<br />

constant specific heat ideal gas equations. From Table C.16a in Thermodynamic Tables to accompany <strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong>,<br />

we find that, at 60.0°F = 520. R,<br />

and<br />

u 1 = 88:62 Btu/lbm<br />

p r1 = 1:2147<br />

v r1 = 158:58<br />

For a compression ratio of 19.2 to 1, v 2 /v 1 = 1/19:2: Then, from Eq. (11.37),<br />

v r2 = v r1 ðv 2 /v 1 Þ = 158:58/19:2 = 8:26<br />

scanning down the v r column in Table C.16a, we find that v r = 8.26 at about<br />

T 2 = 1600 R = 1140°F<br />

and<br />

u 2 = 286:06 Btu/lbm<br />

Then, from Eq. (11.36),<br />

p r2 = 71:73<br />

Finally, from the preceding energy balance,<br />

p 2 = p 1 ðp r2 /p r1 Þ = ð14:7 psiaÞð71:73/1:2147Þ<br />

= 868:1 psia<br />

1W 2 /m = u 1 − u 2 = 88:62 − 286:06 = −197:44 Btu/lbm<br />

Exercises<br />

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<br />

1. Answer: T 2 = 1300. R and p 2 = 392. psia.<br />

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<br />

during the intake stroke. Answer: T 2 = 920. K, p 2 = 5.92 MPa, and 1 W 2 /m = −477 kJ/kg.<br />

36. Use the relations for an isentropic process for an ideal gas,<br />

T 2<br />

T 1<br />

= v 2<br />

v 1<br />

1− k <br />

= p 2<br />

to compute the final temperature T 2 and pressure p 2 of the compression process described in Example 11.12. Then,<br />

assuming constant specific heats, compute the work required per unit mass for the compression 1 W 2 /m = u 1 – u 2 =<br />

c v (T 1 – T 2 ). Answer: T 2 = 1236°F, p 2 = 920 psia, and 1 W 2 /m = −200. Btu/lbm.<br />

p 1<br />

k−1<br />

k<br />

11.10 COMPRESSIBILITY FACTOR AND GENERALIZED CHARTS<br />

In Chapter 3, we discussed the van der Waals, Dieterici, Berthelot, Beattie-Bridgeman, and Redlich-Kwong<br />

equations of state as possible models of nonideal, or real, gas behavior. Several of these equations are just variations<br />

on the basic ideal gas equation, pv = RT. We now introduce one of the most powerful engineering real gas<br />

equations of state: the compressibility factor.<br />

In 1880, the Dutch physicist Johannes Diderik van der Waals (1837–1923) reasoned that, if the p-v-T<br />

equation of state could be nondimensionalized, then all gases might be found to fitthesamedimensionless<br />

p-v-T equation. Further, he noted that, since every substance has a vapor dome and every vapor dome has a<br />

unique critical point (its peak), perhaps the critical point properties (p c , T c ,andv c ) couldbeusedtocreatea

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