Question 1 of 6: Convergent–divergent nozzle between two reservoirs

Question 1: Convergent–divergent nozzle between two reservoirs (20 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 pressure in reservoir (b); whether the nozzle contains a normal shock; whether that shock sits in the exit plane or upstream of it; and the manometer deflection the same nozzle would show if it ran at its supersonic design point.

[Figure not reproduced: Figure 1 (redrawn). The manometer legs are tapped at the throat and in reservoir (b). Mercury stands higher in the leg connected to (b), so (b) is the lower pressure of the two. The area ratio A E /A T = 4 fixes the whole back-pressure ladder. See the official exam paper.]

Approach. Assume the throat is choked, read the throat static pressure straight off the sonic ratio (it depends on P_a alone), add the manometer head to obtain P_b, then compare P_b against the three-rung back-pressure ladder that the area ratio A_E/A_T = 4 sets; where it lands answers parts (b) and (c) with no iteration at all.

1. Fix the throat pressure from the choking assumption. If the throat is sonic the local static-to-stagnation ratio is a pure function of γ: $$\frac{P_0}{P^{*}}=\left(\frac{\gamma+1}{2}\right)^{\gamma/(\gamma-1)}=1.2^{3.5}=1.8929$$ so the throat static pressure is $$P_T=\frac{280}{1.8929}=147.9\ \text{kPa}.$$ Nothing downstream can change this number while the throat stays choked — that is what makes the manometer reading directly usable.
2. Convert the manometer deflection into the reservoir (b) pressure. With the air columns weightless compared with mercury, the head is $$\Delta P=\rho_{Hg}\,g\,h=13\,550\times 9.81\times 0.19=25\,256\ \text{Pa}=25.26\ \text{kPa}.$$ Figure 1 shows mercury standing higher in the leg tapped to reservoir (b); a manometer leg rises on the low-pressure side, so (b) is below the throat by that head: $$\boxed{P_b=P_T-\rho_{Hg}gh=147.9-25.26=122.7\ \text{kPa}}$$ This is part (a).
3. Get both exit-Mach roots from the area ratio. For isentropic quasi-one-dimensional flow, $$\frac{A}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{(\gamma+1)/2(\gamma-1)}$$ and with A_E/A_T = 4 this has the two familiar roots $$M_{E,\text{sub}}=0.1465,\qquad M_{E,\text{sup}}=2.940.$$ The subsonic root belongs to the venturi-like solution, the supersonic root to the design expansion.
4. Build the three-rung back-pressure ladder. Each rung is one isentropic evaluation. First choking, the highest back pressure that still makes the throat sonic: $$P_{b,1}=\frac{P_0}{\left(1+0.2M_{E,\text{sub}}^{2}\right)^{3.5}}=\frac{280}{1.0152}=275.8\ \text{kPa}.$$ Ideal (design) expansion, the lowest back pressure with no wave inside the nozzle: $$P_{b,3}=\frac{280}{\left(1+0.2\times 2.940^{2}\right)^{3.5}}=\frac{280}{33.57}=8.34\ \text{kPa}.$$ Shock standing exactly in the exit plane, obtained by putting the design exit state through the normal-shock static ratio: $$P_{b,2}=P_{b,3}\,\frac{2\gamma M_{E}^{2}-(\gamma-1)}{\gamma+1}=8.34\times 9.917=82.7\ \text{kPa}.$$
5. Read part (b) off the ladder. The measured value P_b = 122.7 kPa satisfies $$82.7\ \text{kPa}\;<\;P_b\;<\;275.8\ \text{kPa},$$ which is precisely the over-expanded band. The throat is choked (P_b is far below the first-choking rung, so the choking assumption of Step 1 is retro-justified), and the flow cannot reach the exit isentropically, so $$\boxed{\text{yes} - \text{a normal shock stands inside the nozzle}}$$
6. Read part (c) off the same ladder. A shock exactly in the exit plane would deliver 82.7 kPa. The measured 122.7 kPa is higher than that, meaning less supersonic expansion has been allowed before the wave, so the shock must sit $$\boxed{\text{farther upstream, inside the divergent section}}$$ Locating it is a single root-find. Let A_s be the area at the shock; because the stagnation pressure drops across the wave, the sonic area grows, A^*₂ = A_T (P₀₁/P₀₂), and the exit Mach number must be taken from A_E/A^*₂ — never from A_E/A_T. Matching the resulting exit static pressure to 122.7 kPa gives $$\frac{A_s}{A_T}=2.81\quad(A_s=25.3\ \text{cm}^{2}),\qquad M_{1}=2.570,$$ with shock strength P₂/P₁ = 7.54, stagnation-pressure recovery P₀₂/P₀₁ = 0.472 and a subsonic exit at M_E = 0.327.
7. Part (d): the deflection at the design point. At ideal expansion the back pressure equals the design exit pressure, P_b = 8.34 kPa, while the throat pressure is unchanged at 147.9 kPa because the throat is still choked. The head the manometer must now carry is therefore much larger: $$h_{\text{design}}=\frac{P_T-P_{b,3}}{\rho_{Hg}g}=\frac{(147.9-8.34)\times 10^{3}}{13\,550\times 9.81}$$ $$\boxed{h_{\text{design}}=1.050\ \text{m}=105\ \text{cm of mercury}}$$ The deflection keeps the same sense — mercury still stands higher on the (b) side — but grows by a factor of 5.5, which is the practical signature of a nozzle that has been allowed to run to its design point instead of shocking down inside the divergence.
8. Supporting number: the mass flow. A choked throat passes $$\dot m=\frac{A_TP_0}{\sqrt{T_0}}\sqrt{\frac{\gamma}{R}}\left(\frac{2}{\gamma+1}\right)^{(\gamma+1)/2(\gamma-1)}=0.527\ \text{kg/s},$$ and this value is the same for every case in the ladder, shock or no shock — a useful reminder that back pressure controls where the wave sits, not how much air flows.