05.04.2016 Views

Modern Engineering Thermodynamics

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

664 CHAPTER 16: Compressible Fluid Flow<br />

EXAMPLE 16.6 (Continued )<br />

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<br />

in Thermodynamic Tables to accompany <strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong> with<br />

p/p os = ð0:1013 MPaÞ/1:00 ð MPaÞ = 0:101<br />

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:<br />

n h<br />

io <br />

2<br />

1/2 M = ð<br />

k−1<br />

p os/pÞ ðk−1Þ/k 2 1:00 ð1:40 − 1/1:40 1/2<br />

− 1 = − 1 = 2:15<br />

1:40 − 1 0:101<br />

Note that this agrees with what we would have interpolated from Table C.18.<br />

b. Here, Eq. (16.12) can be used to give<br />

T =<br />

T os<br />

1 + k−1<br />

2 M2 =<br />

c. Equation (16.11) can now be used to find the exit velocity as<br />

20:0 + 273:15 K<br />

1 + 1 − 1:40 = 152 K = −121°C<br />

2<br />

ð2:15Þ<br />

2<br />

p<br />

V = M<br />

ffiffiffiffiffiffiffiffiffiffiffiffi <br />

kg c RT = ð2:15Þ ð1:40Þð1Þ <br />

286 m 2 /s 2 .K <br />

1/2<br />

ð152 KÞ<br />

= 530: m/s<br />

d. Since the exit velocity is supersonic, the throat must be sonic; therefore, Eq. (16.19) can be used to determine the<br />

throat pressure as<br />

p throat = p = p os ½2/ðk + 1ÞŠ k/ðk−1Þ<br />

<br />

2 1:4/ð1::4 − 1Þ<br />

= ð1:00 MPaÞ<br />

= 0:528 MPa<br />

2:40<br />

e. Similarly, Eq. (16.18) can be used to calculate the throat temperature as<br />

T throat = T = T os ½2/ ðk + 1ÞŠ<br />

= ð20:0 + 273:15 KÞð2/2:40Þ = 244K = − 29:2°C<br />

Exercises<br />

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<br />

remain unchanged. Answer: M exit = 3.20.<br />

14. Determine the exit temperature in Example 16.6 if the temperature in the pipe is 65.0°C. Assume all the other variables<br />

remain unchanged. Answer: T exit = 176 K = –97.4°C.<br />

15. Determine the temperature and pressure at the throat of the nozzle in Example 16.6 if the temperature and pressure<br />

inside the pipe are 65.0°C and 5.00 MPa. Assume all the other variables remain unchanged. Answer: T* = 8.64°C and<br />

p* = 2.64 MPa.<br />

Example 16.6 demonstrates that a very significant temperature drop can occur inside a supersonic nozzle. If the<br />

flowing fluid is a vapor near its saturationstate,itispossiblethatthistypeofcoolingcancausethestateto<br />

drop through the vapor dome and produce a two-phase mixture inside the nozzle. This is a rather common<br />

occurrence for low-temperature steam flow through a nozzle. However, the condensation process that produces<br />

this phase change generally requires a longer time to complete than the resident time of the high-speed vapor in<br />

the converging part of the nozzle. When this occurs, the vapor exits the nozzle’s throat in a nonequilibrium<br />

state at a much lower temperature than the proper equilibrium saturation temperature at the exit pressure. This<br />

nonequilibrium state is called supersaturated and is very unstable. After a sufficient time has passed, the fluid<br />

undergoes a rapid condensation downstream from the throat due to a nucleation and growth process of the<br />

second phase. This process is irreversible and causes the temperature of the two-phase mixture to rise to the<br />

proper equilibrium saturation value (see Figure 16.14). The irreversible condensation process from a’ to b<br />

shown in Figure 16.14 is sometimes called a condensation shock.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!