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Modern Engineering Thermodynamics

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680 CHAPTER 16: Compressible Fluid Flow<br />

EXAMPLE 16.12 (Continued )<br />

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<br />

the downstream isentropic stagnation pressure p osy necessary to cause a normal shock to occur in the exit plane. Then,<br />

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<br />

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<br />

of these equations, and at M x = 5.0, we again have a direct entry. From this table, we find that<br />

M y = 0:415 p osy /p osx = 0:06172<br />

p y /p x = 29:00 p osy /p x = 32:654<br />

T y /T x = 5:800<br />

Here, p osx = p os = 7.00 MPa; therefore, the required back pressure p B is<br />

alternatively,<br />

p B = p osy = ð0:06172Þp osx = ð0:06172Þ 7:00 × 10 3 kN/m 2 = 432 kN/m 2<br />

p B = p osy = ð32:654Þp x = ð32:654Þ 13:23 kN/m 2 = 432 kN/m 2<br />

Exercises<br />

32. Determine the mass flow rate required for supersonic flow in the diverging section of the converging-diverging nozzle in<br />

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.<br />

33. Use Table C.18 to find the Mach number at the exit of the diverging section of the converging-diverging nozzle in<br />

Example 16.12, if the diameter at the exit of the diverging section is increased to 0.14585 m. Assume all other variables<br />

remain unchanged. Answer: M exit = 6.00.<br />

34. Use Table C.19 to determine the outside back pressure required to produce a standing normal shock wave at the exit of<br />

the diverging section of the converging-diverging nozzle in Example 16.12, if the exit Mach number is 6.00 and the<br />

upstream stagnation pressure is 7.00 MPa. Answer: p B = 208 kPa.<br />

EXAMPLE 16.13<br />

Air enters a converging-diverging nozzle with isentropic stagnation properties of 3.00 atm and 20.0°C and exhausts into the<br />

atmosphere (i.e., p B = 1.00 atm). The exit to throat area ratio for the nozzle is 2.00. Determine the pressure, temperature,<br />

and velocity at the exit.<br />

Solution<br />

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<br />

in the exit plane, we must first determine whether or not a shock wave occurs inside the diverging section of the<br />

nozzle. This occurs if p E < p B < p , where from Eq.(16.19),<br />

and, from Eq. (16.13),<br />

k/ðk−1Þ<br />

p 2<br />

= p os = ð3:00 atmÞð0:528Þ = 1:58 atm<br />

k + 1<br />

<br />

ŠM 2 k/1−k ð Þ<br />

E<br />

p E = p os 1 + ½ðk − 1Þ/2<br />

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<br />

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<br />

× (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<br />

nozzle.<br />

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.<br />

We could find M x from the upstream and downstream isentropic stagnation pressures from the relation,<br />

<br />

p osy /p osx = p osy /p y py /p x ð px /p osx Þ

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