Modern Engineering Thermodynamics

82 CHAPTER 3: Thermodynamic Properties
Figure 3.21 illustrates the temperature and pressure dependence of c p and c v for various common gases. Note
that the specific heat temperature dependence is fairly weak, and most ideal gases can be considered to have
constant specific heats over temperature ranges of a few hundred degrees.
EXAMPLE 3.7
Determine the change in specific internal energy and specific enthalpy of air as it is cooled in a closed, rigid tank from a
temperature and pressure of 240.°F, 150 psia to a temperature and pressure of 80.0°F and 14.7 psia (Figure 3.22). Assume
the air behaves as (a) a constant specific heat ideal gas and (b) as a variable specific heat ideal gas.
Solution
a. The changes in specific internal energy and specific enthalpy
of an ideal gas are given by Eqs. (3.38) and (3.42). The constant pressure and constant volume specific heats of air are
found in Table 3.7 (or Table C.13) as c p = 0.240 Btu/lbm · R and c v = 0.172 Btu/lbm · R. Then Eq. (3.38) gives
u 2 − u 1 = c v ðT 2 − T 1 Þ = ð0:172 Btu=lbm . RÞ½ð80:0 + 459:67Þ− ð240: + 459:67Þ RŠ = –27:5 Btu=lbm
and from Eq. 3.42, we have
h 2 − h 1 = c p ðT 2 − T 1 Þ = ð0:240 Btu=lbm . RÞð80:0 − 240:°FÞ = –38:4 Btu=lbm
Notice that you can use either fahrenheit or rankine values when computing the temperature difference T 2 − T 1
because the fahrenheit and rankine degree sizes are the same, only their zero points are different.
b. Values for u and h for variable specific heat air can be found in Table C.16. At T 1 = 240 + 459.67 = 700 R,
h 1 = 167.56 Btu/lbm and u 1 = 119.58 Btu/lbm; and at T 2 = 80 + 459.67 = 540. R, h 1 =129.06Btu/lbmand
u 2 = 92.04 Btu/lbm. Then the changes are
and
h 2 − h 1 = 129:06 − 167:56 = –38:5
u 2 − u 1 = 92:04 − 119:58 = –27:5,
Exercises
11. Determine the changes in specific internal energy and
specific enthalpy as air is heated at constant pressure of
0.100 MPa from 300. K to 1500. K. Assume air behaves as
(a) a constant specific heat ideal gas, and (b) as a variable
specific heat ideal gas. Answer: (a) u 2 − u 1 = 862 kJ/kg
and h 2 − h 1 = 1205 kJ/kg; (b) u 2 − u 1 = 991.38 kJ/kg and
h 2 − h 1 = 1335.8 kJ/kg. The difference in the results of (a)
and (b) is due to the large temperature difference between
the two states.
12. Determine the changes in specific internal energy and
specific enthalpy as methane is compressed at constant
temperature of 20.0°C from 0.100 MPa to 10.0 MPa.
Assume that methane behaves as a constant specific
heat ideal gas. Answer: u 2 − u 1 = h 2 − h 1 = 0. The specific
internal energy and specific enthalpy of an ideal gas
depend only on temperature, so changing just the
pressure on the gas does not alter the values of either
u or h.
State 1 Air (an ideal gas) State 2
State 1
240.°F
150. psia
FIGURE 3.22
Example 3.7.
State 2
80.0°F
14.7 psia
Normally, only low molecular mass real gases at high temperature or low pressure obey the ideal gas equation
of state with good accuracy. For real gases with complex molecular structures or real gases approaching their
saturated vapor region, more complex equations of state are required. The following equations have modifications
to the ideal gas p-v-T equation that are intended to account for observed real gas behavior.