Nozzles and Thrust Generation Questions and Answers

Answer: A

Net thrust is the mass flow times the velocity increment, \( 80 \times (500-200) = 24\,000 \) newtons. The gross thrust would be 40 kilonewtons, and the 16 kilonewton difference is exactly the ram drag of the incoming air.

Answer: B

With no flight speed there is no ram drag, so gross and net thrust are equal at \( 80 \times 500 = 40\,000 \) newtons. That is precisely why a static thrust figure overstates what the engine delivers once the aircraft is moving.

Answer: C

Correct answer: 9.5 kN

The pressure thrust is the excess of exit over ambient pressure acting on the exit area:

This term appears when the nozzle is choked and the flow leaves under-expanded, the remaining expansion taking place outside in the free jet. The 19 kN answer omits the area, and 60 kN uses the exit pressure rather than the difference.

Answer: D

Correct answer: 1.85

Substituting \( \gamma = 1.33 \), which is representative of hot combustion products,

so the nozzle chokes once the total-to-ambient pressure ratio reaches about 1.85. For cold air, with \( \gamma = 1.4 \), the figure is 1.89. The reciprocal, 0.53, is the critical pressure ratio written the other way up, as static over total.

Answer: A

The available ratio well exceeds critical, so a convergent nozzle chokes and the remaining expansion occurs outside it. Fitting a divergent section would convert that residual pressure into velocity and recover part of the loss.

Answer: B

Substituting gives \( \sqrt{2 \times 1150 \times 150} = \sqrt{345\,000} \approx 587 \) metres per second. Every kelvin of temperature drop that the nozzle fails to convert appears as lost exit velocity and therefore lost thrust.

Answer: C

Correct answer: 588 m/s

The coefficient of velocity is the ratio of actual to ideal exit velocity:

The two per cent shortfall represents friction and non-uniformity in the nozzle. A coefficient of velocity above unity is impossible, since the ideal is the isentropic case, so the answers above 600 m/s can be rejected without calculation.

Answer: D

Correct answer: 21 kN

With thrust proportional to density,

Mass flow through a given engine is proportional to the density of the air it swallows, and thrust follows it directly, which is why an engine at 11 km produces roughly a third of its sea level thrust. This lapse is what sets the cruise altitude ceiling and why take-off thrust ratings are quoted at sea level.

Answer: A

Correct answer: 200 N per kg/s

Specific thrust is thrust divided by air mass flow:

Two hundred is typical of a high bypass turbofan, which makes its thrust from a large mass of air given a small velocity increment. A turbojet works near 800, passing a quarter of the air at four times the increment, which is why it is so much smaller and so much less efficient at subsonic speed.

Answer: B

Halving gives \( (1/2)^8 = 1/256 \), a reduction of about 24 decibels. That extraordinary sensitivity is why raising bypass ratio, which lowers jet velocity, transformed aircraft noise more than any acoustic treatment could.