Modern Engineering Thermodynamics

664 CHAPTER 16: Compressible Fluid Flow
EXAMPLE 16.6 (Continued )
a. Assume air is a constant specific heat ideal gas with k = 1.40. Since the fluid is air, we can use either Table C.18
in Thermodynamic Tables to accompany Modern Engineering Thermodynamics with
p/p os = ð0:1013 MPaÞ/1:00 ð MPaÞ = 0:101
or Eqs. (16.12) and (16.13). Instead of interpolating in Table C.18, it is easier to solve Eq. (16.13) directly for M:
n h
io
2
1/2 M = ð
k−1
p os/pÞ ðk−1Þ/k 2 1:00 ð1:40 − 1/1:40 1/2
− 1 = − 1 = 2:15
1:40 − 1 0:101
Note that this agrees with what we would have interpolated from Table C.18.
b. Here, Eq. (16.12) can be used to give
T =
T os
1 + k−1
2 M2 =
c. Equation (16.11) can now be used to find the exit velocity as
20:0 + 273:15 K
1 + 1 − 1:40 = 152 K = −121°C
2
ð2:15Þ
2
p
V = M
ffiffiffiffiffiffiffiffiffiffiffiffi
kg c RT = ð2:15Þ ð1:40Þð1Þ
286 m 2 /s 2 .K
1/2
ð152 KÞ
= 530: m/s
d. Since the exit velocity is supersonic, the throat must be sonic; therefore, Eq. (16.19) can be used to determine the
throat pressure as
p throat = p = p os ½2/ðk + 1ÞŠ k/ðk−1Þ
2 1:4/ð1::4 − 1Þ
= ð1:00 MPaÞ
= 0:528 MPa
2:40
e. Similarly, Eq. (16.18) can be used to calculate the throat temperature as
T throat = T = T os ½2/ ðk + 1ÞŠ
= ð20:0 + 273:15 KÞð2/2:40Þ = 244K = − 29:2°C
Exercises
13. Determine the exit Mach number in Example 16.6 when the pressure in the pipe is 5.00 MPa and all the other variables
remain unchanged. Answer: M exit = 3.20.
14. Determine the exit temperature in Example 16.6 if the temperature in the pipe is 65.0°C. Assume all the other variables
remain unchanged. Answer: T exit = 176 K = –97.4°C.
15. Determine the temperature and pressure at the throat of the nozzle in Example 16.6 if the temperature and pressure
inside the pipe are 65.0°C and 5.00 MPa. Assume all the other variables remain unchanged. Answer: T* = 8.64°C and
p* = 2.64 MPa.
Example 16.6 demonstrates that a very significant temperature drop can occur inside a supersonic nozzle. If the
flowing fluid is a vapor near its saturationstate,itispossiblethatthistypeofcoolingcancausethestateto
drop through the vapor dome and produce a two-phase mixture inside the nozzle. This is a rather common
occurrence for low-temperature steam flow through a nozzle. However, the condensation process that produces
this phase change generally requires a longer time to complete than the resident time of the high-speed vapor in
the converging part of the nozzle. When this occurs, the vapor exits the nozzle’s throat in a nonequilibrium
state at a much lower temperature than the proper equilibrium saturation temperature at the exit pressure. This
nonequilibrium state is called supersaturated and is very unstable. After a sufficient time has passed, the fluid
undergoes a rapid condensation downstream from the throat due to a nucleation and growth process of the
second phase. This process is irreversible and causes the temperature of the two-phase mixture to rise to the
proper equilibrium saturation value (see Figure 16.14). The irreversible condensation process from a’ to b
shown in Figure 16.14 is sometimes called a condensation shock.