Problems of Chapter 3
Solution: For perfect gas it holds h = cpT. The speed of sound is a2 = yRT. These relations are introduced into eq. (3.2):
‘У R Cp Cp cv
cp cv cp
v0/aсо the result is
Tt = Co(l + ^Ml).
From Table 4.4 we find Pr = 0.7458. The recovery factors then are riam = /Pr = 0.8636, and rtUrb = /Pr = 0.9069.
At H = 30 km the free-stream temperature is То = 226.509 K, Table 2.1. The the specific gas constant of air is R = 287.06 m2/s2K, Table B.1.
With that the speed of sound is a0 = 301.712 m/s and the flight Mach number Ыо = 3.314.
The results are Tt = 724.16 K, Trlam = 656.18 K, Trturb = 677.72 K. Problem 3.3
Simplify eq. (3.25) to the proportionality Tra = c(x/L)-n/4 = c(x/L)-0-125. Measure in Fig. 3.3 the temperature at x/L = 0.1: Tw « 1,100 K and find the constant c = 824.88 K.
Measure the temperature at a) x/L = 0.3: Tw « 994 K and b) x/L = 0.75: Tw « 880 K, and compare with the temperatures from the proportionality at these locations: a) Tw « 958 K, b) Tw « 855 K.
How do you rate the result? The temperature follows approximately the behavior (x/L)-n/4.
Simplify eq. (3.25) to the proportionality Tra = c(x/L)-n/4’6 = c(x/L)-0 0435. Measure in Fig. 3.3 the temperature at x/L = 0.1: Tw « 874 K and find the constant c « 790 K.
Measure in Fig. 3.3 the temperature at x/L = 0.75: Tw « 820 K and compare with the temperature from the proportionality at that location: Tw = 800 K. How do you rate the result? The temperature follows approximately the behavior (x/L)-n/4 6.
At H = 60.56 km the Mach number is Ыж = 15.7. The free-stream temperature is T^ « 247 K. At x/L = 0.5 the (radiation-cooled) wall temperature is Tw « 916 K.
Assume that the Reynolds number remains unchanged and that this holds also for the ratio of specific heats and the recovery factor. Assume further Tra ^ Tr and simplify eq. (3.25) to Tra ж Tr0’25. Choose Yeff = 1.3 and Pr =1 and find Tra, M^=!7/Tra, M^=i5.7 = 1.0396. Hence Tw « 952.2 K compared to originally Tw « 916 K. With the higher flight Mach number the wall temperature is larger.
Proceed like in Problem 3.5 and find Tra, M^=i4/Tra, M^=i5.7 = 0.946. Hence Tw = 866.4 K. With the lower flight Mach number the wall temperature is smaller.
From the scaling law eq. (3.34) the proportionality
The ratios are Tra, Mx, = 17/Tra, Mx,= 15.7
Tra, M^=i4/Tra, M^=i5.7 = 0.944. The agreement is good.
The Reynolds number increases with decreasing altitude. Simplify eq. (3.25) to
rp4-(2n-1) T ra
(Reref, l)1 71
It follows for both laminar (n = 0.5) and turbulent (n = 0.2) flow
Reref, L ^ ж : Tra ^ Tr,
i. e., the radiation-adiabatic temperature approaches with increasing Reynolds number the recovery temperature: radiation cooling becomes ineffective. See in this regard also Fig. 3.4.