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Modern Engineering Thermodynamics

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394 CHAPTER 11: More Thermodynamic Relations<br />

70<br />

65<br />

60<br />

55<br />

0.75<br />

T R<br />

0.50<br />

0.55<br />

0.60<br />

, kJ/(kgmole . K)<br />

s*–s<br />

T c<br />

Entropy correction,<br />

50<br />

45<br />

40<br />

35<br />

30<br />

25<br />

20<br />

0.80<br />

Saturated liquid<br />

0.85<br />

0.90<br />

0.92<br />

0.94<br />

0.96<br />

0.98<br />

0.94<br />

0.96<br />

0.98<br />

1.00<br />

1.04<br />

1.02<br />

1.06 1.08<br />

0.65<br />

0.70<br />

0.75<br />

0.80<br />

0.85<br />

0.92 0.90<br />

1.10 1.15<br />

0.94<br />

0.98<br />

1.02<br />

1.06<br />

15<br />

10<br />

5<br />

Saturated gas<br />

0.90<br />

1.10<br />

1.40<br />

0.95<br />

1.04 1.00<br />

1.20<br />

1.60<br />

1.20<br />

1.30<br />

1.40 1.50<br />

1.60<br />

1.80 2.00<br />

2.50<br />

3.00<br />

0<br />

0.1 0.2 0.3 0.4 0.5 1.0 2.0 3.0 4.05.0 10 20 30<br />

Reduced pressure, p R = p/p c<br />

FIGURE 11.11<br />

Generalized chart for the entropy correction. Note: To convert this figure to English units use the factor 1 Btu/(lbmole ·R = 4.1865 kJ/<br />

(kgmole ·K). (Source: Van Wylen, G. J., Sonntag, R. E., 1973. Fundamentals of Classical Thermodynamic. Wiley, New York. Copyright ©<br />

John Wiley & Sons. Reprinted by permission of John Wiley & Sons.)<br />

EXAMPLE 11.16<br />

Ethylene (C 2 H 4 ) gas is to be isothermally compressed from 150. psia to 15.0 × 10 3 psia at 80.0°F. Using the compressibility<br />

charts, determine<br />

a. The change in specific enthalpy.<br />

b. The change in specific internal energy.<br />

c. The change in specific entropy of the ethylene.<br />

Solution<br />

a. The change in specific enthalpy of the ethylene is given by Eq. (11.41) as<br />

<br />

h 2 − h 1 = ðh 2 − h 1 Þ − h <br />

− h<br />

T c<br />

−<br />

2<br />

<br />

h − h<br />

T c<br />

<br />

1<br />

<br />

Tc<br />

M

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