Question 5 of 7: Question 5 (Part B, Question B1): Supersonic Duct with a Normal Shock — Reservoir, Throat and Minimum-Area Conditions

Question 5 (Part B, Question B1): Supersonic Duct with a Normal Shock — Reservoir, Throat and Minimum-Area Conditions (29 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. (a) \(T_3\) and \(P_{o3}\); (b) the reservoir \(P_o\) and \(T_o\); (c) the static pressure, temperature and Mach number at the second throat; (d) the smallest second-throat area that still passes this flow.

Approach. Work backwards from the fully specified exit. The exit state fixes \(P_{o3}\) and, through the mass-flow relation, \(T_o\) and the sonic reference area \(A^{*}\). Comparing \(A^{*}\) with \(A_1\) reveals the subsonic Mach number behind a normal shock in the pipe; inverting the shock relation gives the supersonic Mach number ahead of it, and the shock's stagnation pressure loss recovers the reservoir pressure. The second throat then follows from \(A_T/A^{*}\).

1. Part (a) — total pressure at the exit. The exit state is fully given, so the isentropic stagnation relation applies directly: \[\begin{aligned}P_{o3}&=P_3\left(1+\tfrac{\gamma-1}{2}M_3^2\right)^{\frac{\gamma}{\gamma-1}}\\&=100\,(1+0.2(0.09))^{3.5}=100(1.018)^{3.5}\end{aligned}\] \[\boxed{P_{o3}=106.44\ \text{kPa}}\]
2. Part (a), continued — exit temperature from the mass flow. The aid sheet gives the mass-flow relation \(\dot m=\sqrt{\gamma/R}\;\dfrac{P_o}{\sqrt{T_o}}\,M\left(1+\tfrac{\gamma-1}{2}M^2\right)^{-\frac{\gamma+1}{2(\gamma-1)}}A\). Applying it at station 3, where every other quantity is now known, and solving for \(T_o\), \[\begin{aligned}\sqrt{T_o}&=\frac{\sqrt{\gamma/R}\,P_{o3}M_3(1.018)^{-3}A_3}{\dot m}\\&=\frac{(0.069843)(106\,443)(0.3)(0.94789)(15.19\times10^{-4})}{0.175}\\&=18.350,\end{aligned}\] so \(T_o=336.7\) K and, since the exit static temperature follows from the same stagnation ratio, \[\boxed{T_3=\frac{T_o}{1.018}=330.8\ \text{K}\;(57.6\,{}^{\circ}\text{C})}\]
3. Part (b) — the reservoir total temperature. The duct is insulated and no work is done, so the stagnation temperature is constant everywhere, shock included: \[\boxed{T_o=336.7\ \text{K}\;(63.6\,{}^{\circ}\text{C})}\]
4. Identify the sonic reference area downstream of the shock. The area–Mach relation for \(M_3=0.3\) is \[\frac{A_3}{A^{*}}=\frac{1}{M_3}\left[\frac{2}{\gamma+1}\left(1+\tfrac{\gamma-1}{2}M_3^{2}\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}=2.0351,\] so \(A^{*}=15.19/2.0351=7.464\ \text{cm}^{2}\). This is a key intermediate: \(A^{*}\) is smaller than the physical second throat \(A_T=7.75\ \text{cm}^{2}\), which proves the second throat is not choked and that the flow through the whole second convergent-divergent section is subsonic. The supersonic flow in the pipe must therefore already have been made subsonic, and with no friction and no area change available, only a normal shock can have done it.
5. Locate the shock and find the Mach number behind it. In the constant-area pipe the same \(A^{*}\) applies downstream of the shock, so \[\frac{A_1}{A^{*}}=\frac{10.00}{7.464}=1.3397\quad\Rightarrow\quad M_{1b}=0.500\ \ (\text{subsonic root}).\] The value lands on 0.500 to three figures, confirming the reading of the problem: the paper was designed with a normal shock in the constant-area pipe.
6. Invert the normal-shock relation for the upstream Mach number. The aid sheet gives \(M_2^2=\dfrac{M_1^2+\frac{2}{\gamma-1}}{\frac{2\gamma}{\gamma-1}M_1^2-1}\). With \(M_2=0.500\) this rearranges to \(0.25(7M_1^2-1)=M_1^2+5\), hence \(0.75M_1^2=5.25\) and \[\boxed{M_1=\sqrt{7}=2.646\ \ \text{(supersonic, ahead of the shock)}}\] The static pressure jump is \(P_2/P_1=(2\gamma M_1^2-(\gamma-1))/(\gamma+1)=19.2/2.4=8.00\), a clean design value.
7. Part (b), continued — recover the reservoir total pressure. The first nozzle is isentropic, so the reservoir stagnation pressure equals the pre-shock stagnation pressure. Across the shock, \[\begin{aligned}\frac{P_{o2}}{P_{o1}}&=\left[\frac{\frac{\gamma+1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_1^2}\right]^{\frac{\gamma}{\gamma-1}}\left[\frac{2\gamma}{\gamma+1}M_1^2-\frac{\gamma-1}{\gamma+1}\right]^{-\frac{1}{\gamma-1}}\\&=(3.500)^{3.5}(8.000)^{-2.5}=0.4434 .\end{aligned}\] Since \(P_{o2}=P_{o3}=106.44\) kPa (everything downstream of the shock is isentropic), \[\boxed{P_o=\frac{106.44}{0.4434}=240.1\ \text{kPa}}\] The shock destroys 56 per cent of the stagnation pressure — the reason supersonic diffusers are designed to swallow their shocks at the lowest possible Mach number.
8. Part (c) — conditions at the second throat. The throat sits in the subsonic, isentropic region downstream of the shock, so \[\frac{A_T}{A^{*}}=\frac{7.75}{7.464}=1.0383\quad\Rightarrow\quad\boxed{M_T=0.800\ \ (\text{subsonic root})}\] The throat is the point of minimum area and therefore of maximum velocity in this subsonic passage — it is behaving as a venturi, not as a choking throat.
9. Convert the throat Mach number to static conditions. Using \(P_{o}=106.44\) kPa and \(T_o=336.7\) K with \(1+0.2(0.64)=1.128\), \[\begin{aligned}T_T&=\frac{336.7}{1.128}=298.5\ \text{K},\\P_T&=\frac{106.44}{(1.128)^{3.5}}=\frac{106.44}{1.5243}=69.84\ \text{kPa}.\end{aligned}\] \[\begin{aligned}&\boxed{M_T=0.800}\\&\boxed{P_T=69.84\ \text{kPa},\quad T_T=298.5\ \text{K}\;(25.4\,{}^{\circ}\text{C})}\end{aligned}\]
10. Part (d) — the minimum permissible throat area. The second throat can be narrowed only until it becomes sonic; below that it cannot pass 0.175 kg/s at the prevailing stagnation state, the shock is forced upstream and the specified conditions collapse. That limit is by definition the sonic reference area computed in step 4: \[\boxed{A_{T,\min}=A^{*}=7.464\ \text{cm}^{2}}\] The installed throat of 7.75 cm² is 3.8 per cent larger, which is the design margin that keeps \(M_T=0.80\) rather than 1.00.