Question 1 of 6: Convergent–divergent nozzle diagnosed from one manometer reading

Question 1: Convergent–divergent nozzle diagnosed from one manometer reading (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 back pressure $P_b$ implied by the manometer; whether a normal shock exists; whether that shock sits in the exit plane or inside the divergent section; and the mercury deflection the same nozzle would show at its supersonic design point.

[Figure not reproduced: Figure 1 (redrawn). Convergent–divergent nozzle joining two large reservoirs. The U-tube spans the throat and reservoir (b); the red dashed plane is the normal shock located in part (c) at $A_s/A_T = 2.009$. See the official exam paper.]

Approach. Assume the throat is choked, which fixes $p_T$ from $P_a$ alone; add the manometer head to get $P_b$; then compare $P_b$ against the three back-pressure rungs that the area ratio alone defines — first choking, shock in the exit plane, and ideal expansion — and read the answers to (b) and (c) straight off that ladder.

1. Fix the throat pressure from the choked condition. For a convergent–divergent nozzle passing any supersonic flow at all, the throat is sonic, and the sonic static-to-stagnation ratio depends only on $\gamma$: $$\frac{p_0}{p^{*}} = \left(\frac{\gamma+1}{2}\right)^{\gamma/(\gamma-1)} = 1.2^{3.5} = 1.8929$$ so $p_T = P_a / 1.8929 = 300 / 1.8929 = 158.5\;\text{kPa}$. Note that this value is set by $P_a$ alone — nothing downstream can change it once the nozzle is choked, which is exactly why one manometer reading is enough to close the problem. The assumption is checked in step 4.
2. Convert the manometer deflection into a pressure difference. With the air density negligible against mercury, the U-tube reads the full difference between the two tapping points: $$\Delta p = \rho_{Hg}\,g\,h = 13\,550 \times 9.81 \times 0.150 = 19\,940\;\text{Pa} = 19.94\;\text{kPa}$$ The throat is the fastest station in the nozzle, so it carries the lowest static pressure anywhere in the system; the mercury therefore stands higher in the throat leg and reservoir (b) is the higher-pressure side.
3. Add the head to obtain the back pressure (part a). Reservoir (b) is large, so its pressure is the static pressure the nozzle discharges against: $$P_b = p_T + \Delta p = 158.5 + 19.94 = \boxed{178.4\;\text{kPa (absolute)}}$$
4. Build the back-pressure ladder from the area ratio (part b). With $A_E/A_T = 3.500$, the isentropic area relation $$\frac{A}{A^{*}} = \frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{(\gamma+1)/2(\gamma-1)}$$ has two roots, $M_{sub} = 0.1682$ and $M_{sup} = 2.800$. These generate the three rungs that separate every possible operating regime:

   Rung	Exit condition	$P_b$ (kPa)
   1	First choking — subsonic isentropic exit at $M_E = 0.1682$	294.1
   2	Normal shock exactly in the exit plane	99.27
   3	Ideal (design) expansion at $M_E = 2.800$	11.06

   Rung 3 is $P_a / (1+0.2 \times 2.8^{2})^{3.5} = 300/27.14 = 11.06\;\text{kPa}$, and rung 2 follows by applying the normal-shock static ratio at $M_1 = 2.800$, $$\frac{p_2}{p_1} = \frac{2\gamma M_1^{2}-(\gamma-1)}{\gamma+1} = \frac{2(1.4)(7.840)-0.4}{2.4} = 8.980 \quad\Rightarrow\quad 11.06 \times 8.980 = 99.27\;\text{kPa}$$
5. Place the measured back pressure on the ladder. Since $99.27 < 178.4 < 294.1\;\text{kPa}$, the back pressure sits strictly between rung 2 and rung 1, so $\boxed{\text{yes, a normal shock stands inside the nozzle}}$. That $P_b$ is far below the first-choking rung also retro-justifies the choked-throat assumption of step 1: the nozzle is comfortably choked, so $p_T = 158.5\;\text{kPa}$ was legitimate.
6. Answer the location question (part c). A shock in the exit plane would require $P_b = 99.27\;\text{kPa}$. The actual back pressure is much higher, and a higher back pressure pushes the shock upstream into a weaker part of the supersonic flow, so $\boxed{\text{the shock stands farther upstream, inside the divergent section}}$, not in the exit plane. Downstream of it the flow is subsonic and diffuses to the exit, where the static pressure equals $P_b$.
7. Locate the shock station. The shock sits where the subsonic diffusion downstream of it delivers exactly $P_b$ at the exit. Two facts drive the search: the stagnation pressure falls across the shock, and the sonic reference area therefore changes, $A_2^{*} = A_1^{*}\,p_{01}/p_{02}$, so the exit Mach number must be read from $A_E/A_2^{*}$ and never from $A_E/A_T$. Iterating on the shock-station area $A_s$ gives $$\frac{A_s}{A_T} = 2.009 \;(A_s = 18.08\;\text{cm}^2),\qquad M_1 = 2.202,\qquad M_2 = 0.5467$$ $$p_{02} = 0.6270\,P_a = 188.1\;\text{kPa},\qquad \frac{A_E}{A_2^{*}} = 2.194 \;\Rightarrow\; M_E = 0.2759$$ and the exit static pressure is $188.1/(1+0.2\times0.2759^2)^{3.5} = 178.4\;\text{kPa}$, which recovers $P_b$ and confirms the station. The shock therefore stands roughly a third of the way along the divergent section by area.
8. Recompute the manometer at design conditions (part d). At ideal expansion the exit runs supersonic at $M_E = 2.800$ with no shock anywhere, so the exit and hence the receiver pressure is rung 3, $P_b = 11.06\;\text{kPa}$. The throat is still choked, so $p_T$ is unchanged at 158.5 kPa, and the U-tube now spans a much larger difference in the opposite sense: $$h = \frac{p_T - P_b}{\rho_{Hg}\,g} = \frac{(158\,484 - 11\,060)}{13\,550 \times 9.81} = \frac{147\,424}{132\,926} = \boxed{1.109\;\text{m} = 110.9\;\text{cm}}$$ The sign is worth stating explicitly: in the measured condition reservoir (b) is above the throat and the mercury rises on the throat side by 15 cm, whereas at design the throat is far above the receiver and the deflection reverses, reaching 110.9 cm the other way.