Modern Engineering Thermodynamics

680 CHAPTER 16: Compressible Fluid Flow
EXAMPLE 16.12 (Continued )
c. An exit plane shock wave is illustrated by region c to d in Figure 16.23. The required back pressure p B here is equal to
the downstream isentropic stagnation pressure p osy necessary to cause a normal shock to occur in the exit plane. Then,
M x = 5.0, p x = 13.23 kN/m 2 , and T x = 378.8 K. The downstream Mach number M y can be found from Eq. (16.37), and
p y can be found from Eq. (16.36). Finally, Eq. (16.13) can be used to find p osy . However, Table C.19 is a tabular version
of these equations, and at M x = 5.0, we again have a direct entry. From this table, we find that
M y = 0:415 p osy /p osx = 0:06172
p y /p x = 29:00 p osy /p x = 32:654
T y /T x = 5:800
Here, p osx = p os = 7.00 MPa; therefore, the required back pressure p B is
alternatively,
p B = p osy = ð0:06172Þp osx = ð0:06172Þ 7:00 × 10 3 kN/m 2 = 432 kN/m 2
p B = p osy = ð32:654Þp x = ð32:654Þ 13:23 kN/m 2 = 432 kN/m 2
Exercises
32. Determine the mass flow rate required for supersonic flow in the diverging section of the converging-diverging nozzle in
Example 16.12, if the compressed air is at 10.0 MPa and 20.0°C instead of 7.00 MPa and 2000.°C. Answer: _m = 7.42 kg/s.
33. Use Table C.18 to find the Mach number at the exit of the diverging section of the converging-diverging nozzle in
Example 16.12, if the diameter at the exit of the diverging section is increased to 0.14585 m. Assume all other variables
remain unchanged. Answer: M exit = 6.00.
34. Use Table C.19 to determine the outside back pressure required to produce a standing normal shock wave at the exit of
the diverging section of the converging-diverging nozzle in Example 16.12, if the exit Mach number is 6.00 and the
upstream stagnation pressure is 7.00 MPa. Answer: p B = 208 kPa.
EXAMPLE 16.13
Air enters a converging-diverging nozzle with isentropic stagnation properties of 3.00 atm and 20.0°C and exhausts into the
atmosphere (i.e., p B = 1.00 atm). The exit to throat area ratio for the nozzle is 2.00. Determine the pressure, temperature,
and velocity at the exit.
Solution
We are given the upstream isentropic stagnation state of 3.0 atm, 20°C, and a back pressure of 1.0 atm. To find the conditions
in the exit plane, we must first determine whether or not a shock wave occurs inside the diverging section of the
nozzle. This occurs if p E < p B < p , where from Eq.(16.19),
and, from Eq. (16.13),
k/ðk−1Þ
p 2
= p os = ð3:00 atmÞð0:528Þ = 1:58 atm
k + 1
ŠM 2 k/1−k ð Þ
E
p E = p os 1 + ½ðk − 1Þ/2
Since we are given A exit /A = A E /A = 2:00, we can find M E by inverting Eq. (16.23b). However, in this case, it is again much
easier to use Table C.18 for this area ratio and read (approximately), M E = 2:20 and p E /p os = 0:09352: Then, p E = 0.093252
× (3.0 atm) = 0.281 atm. Thus, p E < p B < p here and a normal shock must occur somewhere in the diverging section of the
nozzle.
Since we now know that a shock wave occurs, we also need to know whether or not it occurs in the exit plane of the nozzle.
We could find M x from the upstream and downstream isentropic stagnation pressures from the relation,
p osy /p osx = p osy /p y py /p x ð px /p osx Þ