Question 2 of 6: Convergent–divergent nozzle between two reservoirs, with a mercury manometer

Question 2: Convergent–divergent nozzle between two reservoirs, with a mercury manometer (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.

The flow is inviscid and adiabatic, so it is isentropic everywhere except across any shock.

Find. The back-reservoir pressure $P_b$ implied by the manometer, whether a normal shock exists, where that shock stands, and the manometer deflection that ideal supersonic (design) expansion would produce.

Approach. Assume the throat is choked, read the throat static pressure from the critical ratio, add the manometer head to obtain $P_b$, then compare $P_b$ with the three limiting back pressures that the area ratio $A_E/A_T$ defines — first choking, shock in the exit plane, and perfect expansion — to place the operating point. Finally locate the shock by matching the exit static pressure to $P_b$.

1. Part (a) — assume the throat is choked and fix the throat static pressure. A convergent–divergent nozzle with any appreciable pressure drop reaches $M=1$ at the throat; the assumption is checked in step 4. At $M=1$ the isentropic ratio is $$\frac{P_0}{P}=\left(1+\frac{\gamma-1}{2}M^{2}\right)^{\gamma/(\gamma-1)} =\left(\frac{\gamma+1}{2}\right)^{\gamma/(\gamma-1)}=1.8929,$$ so with $P_0=P_a=300$ kPa the throat static pressure is $$P_T=\frac{300}{1.8929}=158.5\ \text{kPa}.$$
2. Convert the mercury deflection into a pressure difference. The manometer legs connect the throat tap to reservoir (b), and the air columns are neglected against mercury, so the whole deflection is carried by the mercury: $$\Delta P=\rho_{\text{Hg}}\,g\,h=(13\,550)(9.81)(0.150)=19\,939\ \text{Pa}=19.94\ \text{kPa}.$$
3. Add the head to the throat pressure. The throat of a choked nozzle is the lowest-pressure station in the whole system, so mercury stands higher on the throat leg and reservoir (b) is the higher pressure of the pair. Hence $$\boxed{\;P_b=P_T+\Delta P=158.5+19.94=178.4\ \text{kPa}\;}$$ Note the estimate is insensitive to the shock: whatever happens downstream of the throat, the throat pressure is fixed by $P_a$ alone once the nozzle is choked.
4. Part (b) — build the three limiting back pressures for the area ratio. With $A_E/A_T=31.5/9=3.5$, the isentropic area relation $$\frac{A}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1} \left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{(\gamma+1)/2(\gamma-1)}=3.5$$ has two roots, $M_E=0.1682$ on the subsonic branch and $M_E=2.800$ on the supersonic branch. They generate the following limits, each an exit static pressure that must equal $P_b$ when the corresponding regime is exactly realised.
5. Evaluate the limits. Wholly subsonic, just-choked operation gives $P_b=P_a/(1+0.2M_E^{2})^{3.5}=300/1.0200=294.1$ kPa — the highest back pressure that still chokes the throat. Perfect (design) expansion gives $P_b=300/(1+0.2\times 2.800^{2})^{3.5}=11.06$ kPa. A normal shock standing exactly in the exit plane raises the design pressure by the shock strength at $M_1=2.800$, $$\frac{P_2}{P_1}=\frac{2\gamma M_1^{2}-(\gamma-1)}{\gamma+1} =\frac{2(1.4)(7.84)-0.4}{2.4}=8.980,$$ so $P_b=11.06\times 8.980=99.27$ kPa. Since $$99.27\ \text{kPa}<P_b=178.4\ \text{kPa}<294.1\ \text{kPa},$$ the operating point lies inside the shock band, and $$\boxed{\;\text{yes — a normal shock stands in the flow}\;}$$ The same inequality confirms the step-1 assumption, because $P_b$ is well below the 294.1 kPa first-choking limit.
6. Part (c) — decide between the exit plane and a station upstream of it. The exit-plane shock is the strongest shock the nozzle can hold and therefore corresponds to the lowest back pressure of the band, 99.27 kPa. Raising the back pressure above that value pushes the shock upstream into a region of smaller area and lower upstream Mach number, where the shock is weaker. Here $P_b=178.4$ kPa is nearly twice the exit-plane value, so $$\boxed{\;\text{the shock stands farther upstream, inside the divergent section}\;}$$
7. Locate the shock station. Let the shock sit where the area is $A_s$. The pre-shock Mach number follows from $A_s/A_T$ on the supersonic branch, the post-shock Mach number from $$M_2^{2}=\frac{M_1^{2}+\dfrac{2}{\gamma-1}}{\dfrac{2\gamma}{\gamma-1}M_1^{2}-1},$$ the stagnation-pressure loss from the Rayleigh–Pitot form of $P_{02}/P_{01}$, and a new sonic area $A^{*}_2=A_s/(A/A^{*})_{M_2}$ carries the subsonic flow to the exit. Matching the exit static pressure to 178.4 kPa gives $$\frac{A_s}{A_T}=2.01,\qquad M_1=2.202,\qquad M_2=0.5467,\qquad M_E=0.2759 .$$ The shock therefore stands where the nozzle has opened to about twice the throat area, roughly midway along the divergent section, with a stagnation-pressure loss of about 6 per cent. This is more than part (c) demands but it is the quantitative statement of the answer.
8. Part (d) — re-read the manometer for ideal expansion. At design conditions the shock disappears, the exit Mach number is 2.800 throughout and the exit static pressure equals the back-reservoir pressure, $P_b=11.06$ kPa. The throat is still choked, so $P_T=158.5$ kPa is unchanged and the manometer now spans a far larger difference: $$h=\frac{P_T-P_b}{\rho_{\text{Hg}}\,g}=\frac{(158\,485-11\,055)\ \text{Pa}}{(13\,550)(9.81)} =1.109\ \text{m}$$ $$\boxed{\;h=1.109\ \text{m}=110.9\ \text{cm of mercury}\;}$$ and it now deflects the other way, with mercury standing higher on the reservoir-(b) leg, because the back reservoir has become the low-pressure side. The mass flow is unaffected by any of this — the choked throat passes $\dot m=\rho^{*}A_T a^{*}=0.565$ kg/s in every case considered.