Nozzles and Thrust Generation Questions and Answers

Answer: D

The propulsive efficiency is 0.67.

For a jet engine with negligible fuel mass flow and a fully expanded exhaust, propulsive efficiency is thrust power divided by the rate of increase of the air’s kinetic energy:

\[ \begin{aligned} \eta_{p} &= \frac{2V_{a}}{V_{a} + V_{j}} = \frac{2}{1 + V_{j}/V_{a}} \\ &= \frac{2}{1 + 2} = 0.67 \end{aligned} \]

0.50 is the ratio \(V_{a}/V_{j}\), not the efficiency. 1.00 would require the jet to leave at flight speed, which produces no thrust. 0.33 is \(1/(1 + V_{j}/V_{a})\), the formula without its factor 2.

Answer: C

Raising the reservoir pressure leaves \(u\), \(M\) and \(T\) unchanged, while \(p\) and \(\rho\) increase.

A choked convergent nozzle has \(M = 1\) at its exit, and once choked it stays at \(M = 1\) however high the reservoir pressure goes. With adiabatic flow, the exit temperature is fixed by the reservoir temperature alone, and so is the exit velocity:

\[ \begin{aligned} T &= \frac{2T_{0}}{\gamma + 1} \\ u &= \sqrt{\gamma RT} \end{aligned} \]

Neither depends on \(P_{0}\). The exit pressure, however, is a fixed fraction of the reservoir pressure, \(p = P_{0}(2/(\gamma+1))^{\gamma/(\gamma-1)}\), so it rises with \(P_{0}\); then \(\rho = p/RT\) rises too, and that is how the mass flow of a choked nozzle increases.

Any option in which \(u\), \(M\) or \(T\) increases is therefore wrong, and the option keeping \(\rho\) constant contradicts the ideal gas law with \(p\) rising at fixed \(T\).