Nozzles and Thrust Generation Questions and Answers

Answer: C

Multiplying gives \( 0.45 \times 50 = 22.5 \) kilonewtons. Reversers cannot turn the flow through a full half turn, so only part of the forward thrust is recovered as retarding force, which is typical of practical designs.

Answer: D

Correct answer: 27 kN

Net thrust is what the airframe actually receives:

Ram drag is the momentum of the incoming air, \( \dot{m}V_0 \), which the engine must first arrest; it is zero at rest and grows with flight speed. This is why an engine’s static rating always exceeds what it delivers in cruise.

Answer: A

The area ratio relation gives a supersonic solution near Mach 2.2 for that specific heat ratio. The same area ratio also has a subsonic solution, and which one occurs depends entirely on the pressure ratio actually applied.

Answer: B

Correct answer: Minus 16.4 kN

Applying the same expression with an exit pressure below ambient,

The negative sign is the point: the nozzle is over-expanded, ambient pressure acts inward on the exit plane, and the pressure term subtracts from the momentum thrust. Severe over-expansion also risks shocks and separation inside the divergent section, which is why fixed nozzles are sized towards the under-expanded side.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. On a test bed the flight speed is zero, so the ram drag term \( \dot{m}V_0 \) vanishes and the measured figure is the gross thrust. In flight that momentum must be arrested before the engine can add to it, and the deduction grows in proportion to speed, so net thrust falls steadily. The advertised static rating is therefore an upper bound never seen in flight.

Answer: D

Both hold and the second is the reason. That term partly compensates for the velocity the nozzle failed to produce, which is why a simple convergent nozzle remains acceptable on many engines despite being under-expanded.

Answer: A

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. A divergent passage accelerates supersonic flow but decelerates subsonic flow. Below the critical pressure ratio the throat never reaches sonic conditions, so the flow entering the divergent section is subsonic and that section acts as a diffuser, converting velocity back into pressure and reducing the thrust. The convergent-divergent form pays only where the pressure ratio is well above critical.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is why the first is the right choice. Under-expansion simply leaves some expansion to occur outside the nozzle, a modest loss of thrust. Severe over-expansion is far worse: the flow is compressed by ambient pressure, oblique or normal shocks form, and in the severe case the boundary layer separates asymmetrically inside the divergent section, producing side loads as well as a large thrust loss. Sizing towards under-expansion errs on the forgiving side.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. Thrust depends on the increment \( V_e - V_0 \). A turbojet with an exhaust at 800 m/s loses a modest fraction of that increment when the aircraft flies at 250 m/s; a high bypass engine whose exhaust is at 350 m/s loses a far larger fraction. The lapse with speed is therefore much steeper - which is exactly why high bypass engines suit subsonic transports and not fast aircraft.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the magnitude of the first. Because acoustic power varies as \( V_j^8 \), cutting the jet velocity from 800 to 400 m/s divides the noise power by \( 2^8 = 256 \), some 24 decibels. No other single measure has come close, and it is why the transition from turbojet to high bypass turbofan transformed airport noise while simultaneously improving fuel consumption.