Modern Engineering Thermodynamics

112 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms
EXAMPLE 4.5
Figure 4.8 shows a new process in which 0.0100 kg of methane (an ideal gas) is compressed from a pressure of 0.100 MPa
and a temperature of 20.0 °C to a pressure of 10.0 MPa in a polytropic process with n = 1.35. Determine the moving boundary
work required.
Polytropic process
with n = 1.35
Methane
m 1 = 0.0100 kg
p 1 = 0.100 MPa
State 1 T 1 = 20.0°C State 2
Methane
m 2 = 0.0100 kg
p 2 = 10.0 MPa
1 W 2 =?
FIGURE 4.8
Example 4.5.
Solution
Since the methane behaves as an ideal gas and n ≠ 1, we can find the work required from Eq. (4.32):
ð 1
W 2 Þ polytropic ðn≠1Þ
= mR
1 − n ðT 2 − T 1 Þ
ideal gas
moving boundary
where the value of T 2 can be found from Eq. (4.31):
ðn−1Þ/n
p
2
10:0 MPa ð1:35 − 1Þ/1:35
T 2 = T 1 = ð20:0 + 273:15 KÞ
= 967 K = 694 °C
p 1
0:100 MPa
Using Table C.13b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics to find the value of the gas
constant for methane, R methane = 0.518 kJ/kg·K, Eq. (4.32) then gives
ð 1
W 2 Þ polytropic ðn≠1Þ
ideal gas
moving boundary
The work comes out negative, because it is being done on the system.
= ð0:0100 kgÞð0:518 kJ=kg .KÞ
ð967 − 293:15Þ = − 9:98 kJ
1 − 1:35
Exercises
10. Determine the work required in Example 4.5 if the final pressure of the methane is 0.500 MPa. Answer: −2.25 kJ.
11. If the work required in Example 4.5 is −5.00 kJ, determine the final temperature and pressure of the methane. Answer:
T 2 = 631 K, p 2 = 1.92 MPa.
12. If the gas used in Example 4.5 were air, determine the work required to compress it polytropically from 14.7 psia, 70.0°F
to 150.°F with n = 1.33. Answer: 1 W 2 = −285.1 ft·lbf
4.6.2 Rotating Shaft Work
Whenever a rotating shaft carrying a torque load crosses a system boundary, rotating shaft work is done. In this
case (see Figure 4.5b),
ðdWÞ rotating
shaft
= T ! .d θ ! (4.33)
and, for rotating shaft work,
Rotating shaft Z work
2
!
= T . d ! θ (4.34)
ð 1
W 2 Þ rotating 1
shaft
where T ! is the torque vector produced by the system on the shaft and d θ ! is its angular displacement vector.
These two vectors are in the direction of the shaft axis. Normally, thermodynamic problem statements do not
require rotating shaft work to be calculated from Eq. (4.34). The rotating shaft work is usually openly given