Compressible Flow Basics Questions and Answers

Answer: C

The exit temperature is about 484 K.

For an adiabatic nozzle with no work, the stagnation enthalpy is constant:

\[ \begin{aligned} T_{2} &= T_{1}+\frac{V_{1}^{2}-V_{2}^{2}}{2c_{p}} \\ &= 500+\frac{10^{2}-180^{2}}{2 \times 1008} \\ &= 500-16.0 = 484\ \text{K} \end{aligned} \]

The air speeds up, so its temperature must fall: 516 K and 532 K have the kinetic energy term with the wrong sign. 468 K doubles the drop.

Answer: D

The exit area is about 12.9 cm².

Take R = 287 J/kg.K for air. The mass flow rate at inlet is:

\[ \begin{aligned} \rho_{1} &= \frac{300\,000}{287 \times 500} = 2.091\ \text{kg/m}^{3} \\ \dot{m} &= 2.091 \times 0.008 \times 10 = 0.1673\ \text{kg/s} \end{aligned} \]

At exit, with T₂ = 484 K from the previous question:

\[ \begin{aligned} \rho_{2} &= \frac{100\,000}{287 \times 484} = 0.720\ \text{kg/m}^{3} \\ A_{2} &= \frac{0.1673}{0.720 \times 180} = 12.9\ \text{cm}^{2} \end{aligned} \]

4.4 cm² is what results from ignoring the density change (80 × 10/180). 90.1 and 56.3 cm² would make the exit larger than or close to the inlet, which an accelerating subsonic nozzle does not do.

Answer: B

The gas exerts a force of \(750\pi\) N on the pipe.

Apply the momentum equation to the gas between the two sections. The pipe area is \(A = \pi/4\) m², and the mass flow is \(\dot{m} = \rho V A = 1 \times 100 \times \pi/4 = 25\pi\) kg/s, the same at both sections. With \(F\) the force of the pipe on the gas in the flow direction,

\[ \begin{aligned} (p_{1} - p_{2})A + F &= \dot{m}(V_{2} - V_{1}) \\ 10^{4} \times \frac{\pi}{4} + F &= 25\pi \times 70 \\ F &= 1750\pi - 2500\pi = -750\pi\ \text{N} \end{aligned} \]

The pipe pushes the gas backwards with \(750\pi\) N, so the gas pushes the pipe forwards with \(750\pi\) N.

The other values do not satisfy this balance; leaving out the momentum change would give \(2500\pi\), and leaving out the pressure force \(1750\pi\), neither of which is offered.

Answer: B

The reason is a normal shock.

For supersonic flow the area-velocity relation reverses the familiar subsonic behaviour: in a divergent channel a supersonic stream accelerates and its pressure falls. So in smooth, isentropic flow the pressure at B would be lower than at A. A higher pressure at B means the flow has been compressed abruptly between A and B, and the mechanism the standard theory gives for that is a normal shock: across it the flow drops to subsonic speed and the static pressure jumps.

The first option uses the incompressible rule that a wider channel slows the flow; that is true for subsonic flow and wrong for supersonic flow, where density changes dominate. Viscous effects and boundary layer separation are dissipative effects the textbook treatment of this channel leaves out; the official key accepts only the normal shock.

Answer: C

At Location 2, \(u_{2} \gt u_{1}\) and \(P_{2} \lt P_{1}\).

Find the Mach number at Location 1:

\[ \begin{aligned} a_{1} &= \sqrt{\gamma P_{1}/\rho_{1}} \\ &= \sqrt{1.4 \times 10^{5}/1.2} \\ &\approx 341.6\ \text{m/s} \\ M_{1} &= 400/341.6 \approx 1.17 \end{aligned} \]

The flow is supersonic. For isentropic flow in a duct, area and velocity are linked by

\[ \frac{dA}{A} = (M^{2} - 1)\frac{du}{u} \]

With \(M \gt 1\), an increasing area (\(dA \gt 0\)) gives \(du \gt 0\): a supersonic stream speeds up in a diverging duct. In isentropic flow pressure falls as velocity rises, so \(P_{2} \lt P_{1}\).

\(u_{2} \lt u_{1}\) with \(P_{2} \gt P_{1}\) is the subsonic diffuser behaviour, the trap if the Mach number is not checked. The two mixed options break the isentropic link between velocity and pressure.

Answer: A

The average velocity increases along the flow.

In steady flow through a uniform pipe the mass flow rate \(\rho VA\) is constant, so \(\rho V\) is constant. The flow is pressure driven, so pressure falls along the pipe, and at constant temperature an ideal gas's density \(\rho = p/RT\) falls with it. Hence \(V\) must rise.

Velocity would stay constant only for an incompressible fluid, which the question rules out by saying density changes. A decrease would need density to rise downstream. The entrance-region option describes the shape of the velocity profile, not the average velocity.