Propulsion System Classification Questions and Answers

Answer: C

Correct answer: 800 kg/s

\[ \dot{m} = \frac{200{,}000\ \mathrm{N}}{250\ \mathrm{N\,per\,kg/s}} = 800\ \mathrm{kg/s}. \] Converting kN to N first is the step to be careful with. A specific thrust of 250 is characteristic of a high bypass turbofan, and 800 kg/s of air is why such an engine has a fan well over two metres across - the low specific thrust has to be paid for in frontal area.

Answer: D

Thrust is mass flow times exhaust velocity, \( 200 \times 3000 = 600\,000 \) newtons. There is no flight speed term because the rocket takes in nothing, which is precisely why its thrust is independent of speed and altitude alike.

Answer: A

Thrust power is \( 40\,000 \times 250 = 10 \) megawatts, and dividing by 0.35 gives about 28.6. Nearly two thirds of the fuel energy leaves as heat and residual kinetic energy, which is a fair summary of where a jet engine's energy actually goes.

Answer: B

With a bypass ratio of 5 the total is six times the core flow, so \( 300/6 = 50 \) kilograms per second. Dividing by the bypass ratio instead gives 60 and is the error that follows from forgetting the core flow is part of the total.

Answer: C

Both statements are true and the second is the reason for the first. Every air breathing type stops producing thrust where there is no oxygen, and carrying the oxidiser is precisely the mass penalty that limits a rocket's endurance so severely.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are correct, and the second is the reason for the first. Propulsive efficiency is \[ \eta_p = \frac{2}{1 + V_j/V}, \] so it improves as the jet velocity approaches the flight velocity. Producing a given thrust by moving a large mass of air slowly rather than a small mass quickly does exactly that, and less energy is left behind in the wake - which appears at the aircraft as lower fuel consumption.

Answer: A

The first is correct but the second is not: thrust depends on the velocity increment, which would be zero. That tension is exactly why a practical cycle settles on a jet velocity somewhat above flight speed rather than chasing efficiency alone.

Answer: B

Both are true and the second is the reason. Having no compressor makes the engine mechanically simple and capable of very high speeds, but it must be accelerated to operating speed by some other means before it will work at all.

Answer: C

The first is correct and the second is not, because the two multiply rather than add. A thermally excellent core matched to a badly chosen jet velocity gives a disappointing result, which is why both must be pursued together.

Answer: D

Both hold and the second is the dominant reason. Falling temperature partly offsets it by improving the cycle, so the real lapse is gentler than a simple density scaling would predict, though the trend is unmistakable.