Question 7 of 7: Question 7 (Part B, Question B3): Supersonic Wind Tunnel with a Shock in the Test Section — Sizing the Second Throat

Question 7 (Part B, Question B3): Supersonic Wind Tunnel with a Shock in the Test Section — Sizing the Second Throat (26 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The Mach numbers on both sides of the test-section shock and the stagnation temperature; the second throat area; the exit temperature and speed; and the stagnation pressure needed to drive the tunnel.

Approach. Everything except the shock is isentropic, so the two halves of the tunnel are each governed by the area–Mach relation, referenced to their own sonic areas. The shock links them: it preserves stagnation temperature and mass flow but destroys stagnation pressure, and that stagnation-pressure loss is exactly what forces the second throat to be larger than the first.

1. Part (a) — find the Mach number upstream of the shock from the area ratio. Upstream of the shock the flow is isentropic from the first throat, which is choked, so $A^{*}=A_{T1}$ and the test section runs at $$\frac{A}{A^{*}}=\frac{20}{10}=2.00 .$$ Solving the isentropic area relation $A/A^{*}=(1/M)\left[(1+0.2M^{2})/1.2\right]^{3}$ on its supersonic branch gives $$\boxed{\ M_1=2.197\ }$$ (the value 2.20 is the usual table entry, which returns $A/A^{*}=2.005$).
2. Apply the normal-shock relation for the downstream Mach number. The aid sheet gives $$M_2^{2}=\frac{M_1^{2}+\dfrac{2}{\gamma-1}}{\dfrac{2\gamma}{\gamma-1}M_1^{2}-1} =\frac{4.8277+5}{7\times4.8277-1}=\frac{9.8277}{32.794}=0.29968,$$ so $$\boxed{\ M_2=0.547\ }$$ The flow leaves the shock subsonic, as it must.
3. Recover the stagnation temperature from the measured static value. A normal shock is adiabatic, so $T_0$ is the same on both sides of it and equal to the supply value. Using the measured $T_2=280\ \text{K}$ with $M_2$, $$T_0=T_2\left(1+\frac{\gamma-1}{2}M_2^{2}\right)=280\times(1+0.2\times0.29968) =280\times1.05994,$$ giving $$\boxed{\ T_0=296.8\ \text{K}\ }$$ This single number now governs every temperature in the tunnel. As a check, the static temperature ahead of the shock is $T_1=T_0/(1+0.2M_1^{2})=151.0\ \text{K}$, so the shock raises the temperature by a factor of 1.854 and drops the speed from 541 m/s to 184 m/s.
4. Part (b) — size the second throat from the stagnation-pressure loss. The hint is the whole method: the mass flow is the same at both throats, and both are choked, so the aid-sheet choked mass flow evaluated at each throat gives $$\dot m=\sqrt{\frac{\gamma}{R}}\frac{P_{01}}{\sqrt{T_0}} \left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}A_{T1} =\sqrt{\frac{\gamma}{R}}\frac{P_{02}}{\sqrt{T_0}} \left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}A_{T2},$$ in which $T_0$ cancels because the shock is adiabatic. Hence the throats scale inversely with the stagnation pressures: $$A_{T2}=A_{T1}\frac{P_{01}}{P_{02}} .$$ The stagnation-pressure ratio across a normal shock at $M_1=2.197$ is $$\frac{P_{02}}{P_{01}}=\left[\frac{\tfrac{\gamma+1}{2}M_1^{2}} {1+\tfrac{\gamma-1}{2}M_1^{2}}\right]^{\frac{\gamma}{\gamma-1}} \left[\frac{\gamma+1}{2\gamma M_1^{2}-(\gamma-1)}\right]^{\frac{1}{\gamma-1}} =(2.9474)^{3.5}(0.18296)^{2.5}=0.6294,$$ so $$\boxed{\ A_{T2}=\frac{10}{0.6294}=15.9\ \text{cm}^{2}\ }$$ The second throat must be about 59 per cent larger than the first. Physically, the shock has degraded the flow's ability to push itself through a small opening, so the same mass now needs more area to pass sonically — this is the classic "second-throat" sizing constraint on supersonic tunnels and diffusers.
5. Part (c) — find the exit Mach number, then the exit temperature. Downstream of the shock the flow is isentropic again, but referenced to the new sonic area $A^{*}_2=A_{T2}=15.888\ \text{cm}^{2}$. The exit therefore runs at $$\frac{A_e}{A^{*}_2}=\frac{20}{15.888}=1.2588 \quad\Longrightarrow\quad M_e=1.611$$ on the supersonic branch, consistent with the question's statement that the flow exits supersonically. The exit temperature follows from the (unchanged) stagnation temperature: $$T_e=\frac{T_0}{1+\tfrac{\gamma-1}{2}M_e^{2}}=\frac{296.78}{1.5194},$$ so $$\boxed{\ T_e=195.4\ \text{K}\ }$$
6. Convert the exit Mach number into a speed. The local speed of sound is $a_e=\sqrt{\gamma RT_e}=\sqrt{1.4\times287\times195.41}=280.2\ \text{m/s}$, so $$\boxed{\ V_e=M_e\,a_e=1.611\times280.2=451\ \text{m/s}\ }$$ Note that although the exit Mach number (1.611) is lower than the test-section value (2.197), the exit is colder than the reservoir and the two effects partly offset; the exit speed of 451 m/s is well below the 541 m/s reached in the test section.
7. Part (d) — work back from the design back pressure to the supply pressure. "Design back pressure" with a supersonic exit means the tunnel is perfectly expanded: the exit static pressure equals the ambient value, $P_e=P_b=100\ \text{kPa}$. Running the isentropic relation back to the post-shock stagnation pressure, $$P_{02}=P_e\left(1+\frac{\gamma-1}{2}M_e^{2}\right)^{\frac{\gamma}{\gamma-1}} =100\times(1.5194)^{3.5}=431.8\ \text{kPa},$$ and then undoing the shock loss found in step 4, $$P_{01}=\frac{P_{02}}{0.6294}=\frac{431.8}{0.6294},$$ so $$\boxed{\ P_{01}=686\ \text{kPa}\ }$$ Equivalently $P_{01}=P_{02}(A_{T2}/A_{T1})=431.8\times1.5888=686\ \text{kPa}$, which is the same statement of constant mass flow read the other way round — a free check on both parts (b) and (d). For reference the tunnel then passes $\dot m=\sqrt{\gamma/R}\,(P_{01}/\sqrt{T_0})(1.2)^{-3}A_{T1}=1.61\ \text{kg/s}$, and the static pressures across the shock are 64.4 kPa rising to 352 kPa.