Modern Engineering Thermodynamics

16.9 Shock Waves 679
6
5
4
S P /(m R )
3
2
1
0 1 2 3 4 5 6 7 8 9 10
Upstream Mach number M x
FIGURE 16.23
A plot of Eq. (16.38) for air (k = 1.4) utilizing Eqs. (16.35), (16.36), and (16.37). Note that the second law of thermodynamics requires
that _S p /ð _m R Þ ≥ 0 for all processes.
EXAMPLE 16.12
A spacecraft directional control thruster is a converging-diverging nozzle that uses high-pressure and high-temperature air.
The air enters with isentropic stagnation properties of 7.00 MPa and 2000.°C. The throat diameter is 0.0200 m and the
diameter of the exit of the diverging section is 0.100 m. Determine
a. The mass flow rate required for supersonic flow in the diverging section.
b. The Mach number, pressure, and temperature at the exit of the diverging section with this mass flow rate.
c. The outside back pressure required to produce a standing normal shock wave at the exit of the diverging section.
Solution
a. To have supersonic flow in the diverging section of a converging-diverging nozzle, the throat must have a Mach number of
unity (i.e., be choked). Therefore, the mass flow rate is the maximum value for air, in SI units, given by Eq. (16.22c), or
pffiffiffiffiffiffi
_m = _m max = 0:0404 p os A /
where A = πðD Þ 2 /4 = πð0:0200 mÞ 2 /4 = 3:14 × 10 −4 m 2 : Then,
_m = 0:0404 kg p ffiffiffi
. K /ðN .sÞ ð 7:00 × 10 6 N/m 2 Þð3:14 × 10 − 4 m 2 Þ
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
= 1:86 kg=s
2000: + 273:15 K
b. Here, A exit /A = ðD exit /D Þ 2 = ð0:100/0:0200Þ 2 = 25:0: Then, Eq. (16.23) can be inverted to find M exit , and p exit and T exit
can be found from Eqs. (16.13) and (16.12), respectively. Unless these equations are programmed into a computer, this
can be a tedious set of calculations. Table C.18 in Thermodynamic Tables to accompany Modern Engineering Thermodynamics
was created to eliminate this tedium by tabulating these equations. The preceding area ratio is a direct entry into this
table, so we use it and read
M exit = 5:00
p exit /p os = 1:89 × 10 −3 and
T exit /T os = 0:16667
Then, p exit = ð1:89 × 10 −3 Þð7:00 × 10 6 N/m 2 Þ = 13:2 kN/m 2 and
T os
T exit = ð0:16667Þð2000: + 273:15 KÞ = 378:8K
The velocity of sound at the exit is
p
c exit =
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p
kg c RT exit = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1:40Þð1Þ½286 m 2 /s ð
2 . KÞŠð378:8KÞ
= 390: m/s
then,
V exit = c exit M exit = ð390: m/sÞð5:00Þ = 1950 m/s:
(Continued )