Question 3 of 6: Net force of the pipe on the fluid in a choked adiabatic duct

Question 3: Net force of the pipe on the fluid in a choked adiabatic duct (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 net axial force the pipe exerts on the fluid — that is, the resultant of the wall shear and any pressure the wall carries, taken as a single reaction on the gas.

[Figure not reproduced: Figure 3 (redrawn) as a control volume. Because the area is constant and the duct is insulated, the flow follows the Fanno line; the integrated wall shear is the single unknown in the axial momentum balance, so neither the friction factor nor the duct length is needed. See the official exam paper.]

Approach. Get the inlet Mach number, use the Fanno relations to find the sonic (exit) state, then close an axial momentum balance on the whole duct. The force is what makes the balance balance, so it comes out without ever computing the friction factor or the length.

1. Inlet Mach number. The speed of sound at the inlet is $$a_{1}=\sqrt{\gamma R T_{1}}=\sqrt{1.4(287)(333.15)}=365.9\ \text{m/s},$$ so $$M_{1}=\frac{V_{1}}{a_{1}}=\frac{110}{365.9}=0.301 .$$ The value is essentially exactly 0.30, which is the usual sign that this paper was built backwards from a table entry — a helpful check that nothing has been misread.
2. Fanno ratios at the inlet. For adiabatic constant-area flow the sonic reference state satisfies $$\frac{T}{T^{*}}=\frac{\gamma+1}{2+(\gamma-1)M^{2}},\qquad \frac{P}{P^{*}}=\frac{1}{M}\sqrt{\frac{T}{T^{*}}} .$$ At M₁ = 0.301, $$\frac{T_{1}}{T^{*}}=\frac{2.4}{2+0.4(0.301)^{2}}=1.1787,\qquad \frac{P_{1}}{P^{*}}=\frac{1}{0.301}\sqrt{1.1787}=3.611 .$$
3. The exit (sonic) state. Because the duct is choked at the exit, station 2 is the sonic reference state: $$T^{*}=\frac{333.15}{1.1787}=282.6\ \text{K},\qquad P^{*}=\frac{680}{3.611}=188.3\ \text{kPa},$$ $$V^{*}=a^{*}=\sqrt{1.4(287)(282.6)}=337.0\ \text{m/s}.$$ Friction has cut the static pressure to 28 per cent of its inlet value while tripling the velocity — the characteristic behaviour of a subsonic Fanno flow.
4. Mass flow, and a free continuity check. The inlet density is $$\rho_{1}=\frac{P_{1}}{RT_{1}}=\frac{680\,000}{287(333.15)}=7.112\ \text{kg/m}^{3},$$ so $$\dot m=\rho_{1}AV_{1}=7.112(0.013273)(110)=10.38\ \text{kg/s}.$$ Recomputing at the exit, ρ^* = 188 300/[287(282.6)] = 2.321 kg/m³ and ρ^*AV^* = 10.38 kg/s — identical, which confirms both end states at once. The stagnation temperature is likewise conserved, T₁(1 + 0.2M₁²) = 1.2T^* = 339.2 K, as an insulated duct requires.
5. Axial momentum balance on the whole duct. Taking x along the flow and calling F the net force the pipe applies to the fluid, $$P_{1}A-P^{*}A+F=\dot m\left(V^{*}-V_{1}\right).$$ Substituting, $$\dot m\left(V^{*}-V_{1}\right)=10.38(337.0-110)=2356\ \text{N},$$ $$\left(P_{1}-P^{*}\right)A=(680-188.3)\times 10^{3}\times 0.013273=6527\ \text{N},$$ $$F=2356-6527=-4169\ \text{N}.$$ The pressure drop alone would accelerate the gas far more than it actually accelerates, and the deficit is exactly what the wall absorbs: $$\boxed{\;\left|F\right|=4.17\ \text{kN, directed opposite to the flow}\;}$$ Equivalently, the fluid pushes the pipe downstream with 4.17 kN — the load the duct supports and anchors must carry.
6. Cross-check with the impulse function. Defining I = PA + ṁV, the same balance reads F = I₂ − I₁: $$I_{1}=680\,000(0.013273)+10.38(110)=10\,168\ \text{N},$$ $$I_{2}=188\,300(0.013273)+10.38(337.0)=5999\ \text{N},$$ $$I_{2}-I_{1}=-4169\ \text{N}\ \checkmark$$ This second form is worth writing down because it makes the one classic sign error — adding rather than subtracting the momentum flux — impossible to hide.
7. What the answer does not depend on. Neither the friction factor nor the duct length appears anywhere above, because the integrated wall shear is the single unknown that the momentum balance solves for. For interest, the Fanno function at the inlet is $$\frac{fL^{*}}{D}=\frac{1-M_{1}^{2}}{\gamma M_{1}^{2}}+\frac{\gamma+1}{2\gamma} \ln\!\frac{(\gamma+1)M_{1}^{2}}{2+(\gamma-1)M_{1}^{2}}=5.27,$$ so at a plausible f = 0.018 the implied length is L^* = 38 m. That number is a plausibility note only; changing f changes the length, never the force.