Modern Engineering Thermodynamics

660 CHAPTER 16: Compressible Fluid Flow
EXAMPLE 16. 5
Find the velocity, isentropic stagnation temperature, and isentropic stagnation pressure on an aircraft flying at Mach 0.850 at
an altitude where the temperature is –20.0°C and pressure of 0.500 atm. Assume air is an ideal gas with a constant specific
heat ratio of k = 1:40 and a gas constant of R = 286 J/ðkg.KÞ:
Solution
From Eq. (16.11), we can calculate the aircraft’s velocity as
p
V = M
ffiffiffiffiffiffiffiffiffiffiffiffi
kg c RT
= ð0:850Þfð1:40Þð1Þ½286 m 2 /s ð
2 .KÞŠð−20:0 + 273:15 KÞg 1/2 = 271 m/s
The isentropic stagnation pressure and temperature can be determined from Eqs. (16.12) and (16.13) as
and
T os = T 1 + k−1
2 M2
h
= ð−20:0 + 273:15 KÞ 1 + 1:40 − 1 i
ð0:850Þ 2 = 290: K = 16:6°C
2
h
p os = p 1 + k−1 i k/ðk−1Þ
2 M2
h
= ð0:500 atmÞ 1 + 1:40 − 1 i 1:40/ð1:40−1Þ
ð0:850Þ 2
= 0:802 atm = 81:3kPa
2
Exercises
10. If the aircraft in Example 16.5 has a supersonic Mach number of 2.25 and all the other variables remain unchanged,
what would be its velocity? Answer: V = 717 m/s.
11. Determine the isentropic stagnation temperature of the aircraft in Example 16.5 if it is flying at a supersonic Mach
number of 1.30 and all the other variables remain unchanged. Answer: T os = 339 K = 65.6°C.
12. If the aircraft in Example 16.5 is flying at a hypersonic Mach number of 6.50, determine the isentropic stagnation
pressure on the aircraft, assuming all the other variables remain unchanged. Answer: p os = 1300. atm.
CRITICAL THINKING
Equations (16.12)–(16.14) are for the isentropic deceleration of an ideal gas to zero velocity (stagnation), and the resulting
isentropic stagnation pressure and temperature (p os and T os ) are valid only under these conditions. Would you expect the
actual values of the stagnation pressure and temperature (p o and T o ) for a real (nonideal) gas undergoing a real (nonisentropic)
deceleration to zero velocity to be larger or smaller than their isentropic counterparts?
Example 16.5 shows that, even at moderate subsonic velocities, there can be a considerable temperature and
pressure rise at the stagnation points of moving objects.
16.5 CONVERGING-DIVERGING FLOWS
We now investigate the effect of variations in the flow cross-sectional area on the Mach number of the flow. If we
differentiate the mass rate balance equation, _m = ρAv = constant, for an isentropic flow, we obtain
or
ρA ∂V s + ρV ∂A s + AV ∂ρ s = 0
∂V s /V + ∂A s /A + ∂ρ s /ρ = 0
∂A s /A = −∂V s /V −∂ρ s /ρ (16.15)
Next, we differentiate Eq. (16.1) for a constant isentropic stagnation enthalpy to get
∂h os = ∂h s + V ∂V s /g c = 0