Propulsion System Classification Questions and Answers

Answer: A

Thrust is the mass flow times the velocity increment, \( 100 \times (400-250) = 15\,000 \) newtons. It is the change in velocity that matters rather than the exhaust velocity itself, which is why a fast jet produces no thrust at all once the aircraft matches its speed.

Answer: B

Dividing gives \( 18\,000/60 = 300 \) newtons per kilogram per second. A high bypass turbofan sits far below this figure and needs a much larger engine for the same total thrust, which is exactly why nacelle diameters grew as bypass ratios rose.

Answer: C

Dividing gives \( 0.9/50 = 0.018 \) kilograms per kilonewton second. This is the figure that dominates a transport aircraft's operating cost, which is why decades of development went into reducing it through higher bypass ratios and pressure ratios.

Answer: D

Bypass ratio is bypass flow divided by core flow, \( 360/40 = 9 \). Adding the two flows gives the total intake of 400 kilograms per second, and dividing that by the core flow would give 10, which is the commonest error in this calculation.

Answer: A

Substituting gives \( 2 \times 250/(250+450) = 500/700 \). Reducing the jet velocity towards the flight speed raises this figure towards unity, but the thrust falls to zero as it does so, which is the trade every engine cycle has to settle.

Answer: B

Overall efficiency is the product, \( 0.45 \times 0.70 = 0.315 \). Because the two multiply, a weakness in either one caps the result, which is why the turbofan pursued propulsive efficiency without giving up the thermal efficiency of its core.

Answer: C

Thrust power is \( 1500 \times 0.85 = 1275 \) kilowatts, and dividing by the flight speed gives \( 1275\,000/120 \). The relation shows why a propeller gives enormous thrust at low speed and why that advantage disappears as the aircraft accelerates.

Answer: D

Correct answer: 36 kN

\[ 120 \times 0.30 = 36\ \mathrm{kN}. \] Thrust is proportional to the mass flow the engine can swallow, and that is proportional to the density, so a density ratio of 0.30 leaves 30 per cent of the sea level thrust. It is why an engine that is comfortably powerful at sea level becomes marginal at altitude, and why cruise altitude is limited by thrust as much as by anything else.

Answer: A

Thrust power is thrust times flight speed, \( 30\,000 \times 240 = 7.2 \times 10^6 \) watts. Thrust alone says nothing about the work being done, which is why a static thrust figure is meaningless as a measure of an engine's useful output in cruise.

Answer: B

The rotational speed is \( 1500 \times 2\pi/60 = 157 \) radians per second, and multiplying by the radius gives 393 metres per second. Adding the forward speed vectorially takes the tip past the speed of sound, which is what limits propeller aircraft.