Question 8.2: During a nominal reentry trajectory, the Space Shuttle Orbit......
During a nominal reentry trajectory, the Space Shuttle Orbiter flies at 3964 \mathrm{ft} / \mathrm{s} at an altitude of 100,000~ \mathrm{ft}. The corresponding conditions at the stagnation point (point 2 in Fig. 8.5) are p_2=490.2~ \mathrm{lbf} / \mathrm{ft}^2 and T_2=1716.0^{\circ} \mathrm{R}. The static pressures for two nearby locations (points 3 and 4 of Fig. 8.5) are p_3=259.0 ~\mathrm{lbf} / \mathrm{ft}^2 and p_4=147.1~ \mathrm{lbf} / \mathrm{ft}^2. All three points are outside the boundary layer. What are the local static temperature, the local velocity, and the local Mach number at points 3 and 4?
At these conditions; the air can be assumed to behave approximately as a perfect gas. Furthermore, since all three points are outside the boundary layer and downstream of the bow shock wave, we will assume that the flow expands isentropically from point 2 to point 3 and then to point 4 . (Note that because the shock wave is curved, the entropy will vary through the shock layer. For a further discussion of the rotational flow downstream of a curved shock wave, refer to Chapter 12 . Thus, validity of the assumption that the expansion is isentropic should be verified for a given application.)
For an isentropic process, we can use equation (8.20b)
\frac{T^{\gamma /(\gamma-1)}}{p}=\text { constant } (8.20b)
to relate the temperature ratio to the pressure ratio between two points. Thus,
T_3=T_2\left(\frac{p_3}{p_2}\right)^{(\gamma-1) / \gamma}=(1716.0)(0.83337)=1430.1^{\circ} \mathrm{R}
Similarly,
T_4=(1716.0)(0.70899)=1216.6^{\circ} \mathrm{R}
Using the energy equation for the adiabatic flow of a perfect gas [i.e., equation (8.31)],
T_1+\frac{U_1^2}{2 c_p}=T_2+\frac{U_2^2}{2 c_p} (8.31)
and noting that U_2=0, since point 2 is a stagnation point,
\begin{aligned} U_3 & =\left[2 c_\rho\left(T_2-T_3\right)\right]^{0.5} \\ & =\left[2\left(0.2404 \frac{\mathrm{Btu}}{\mathrm{lbm} \cdot{ }^{\circ} \mathrm{R}}\right)\left(32.174 \frac{\mathrm{ft} \cdot \mathrm{lbm}}{\mathrm{lbf} \cdot \mathrm{s}^2}\right)\left(778.2 \frac{\mathrm{ft} \cdot \mathrm{lbf}}{\mathrm{Btu}}\right)\left(285.9^{\circ} \mathrm{R}\right)\right]^{0.5} \\ & =1855.2 \mathrm{ft} / \mathrm{s} \end{aligned}
Similarly,
U_4=2451.9 \mathrm{ft} / \mathrm{s}
Using equation (1.14b)
a=49.02 \sqrt{T} (1.14b)
for the speed of sound in English units,
\begin{aligned} & M_3=\frac{U_3}{a_3}=\frac{1855.2}{49.02(1430.1)^{0.5}}=1.001 \\ & M_4=\frac{U_4}{a_4}=\frac{2451.9}{49.02(1216.6)^{0.5}}=1.434 \end{aligned}
Thus, the flow accelerates from the stagnation conditions at point 2 to sonic conditions at point 3 , becoming supersonic at point 4.