Propulsion Integration Questions and Answers

Answer: A

The moment is thrust times arm, \( 110 \times 7.5 = 825 \) kN m. Adding the drag of the failed engine typically increases this by ten to twenty per cent, and the fin and rudder must be able to balance the total at the minimum control speed.

Answer: B

Correct answer: 0.96

Pressure recovery is the ratio of total pressure delivered to the fan face to the free stream total pressure:

\[ \eta = \frac{96}{100} = 0.96 \]

A subsonic pitot inlet would normally do better than this, above 0.98. The figure matters because thrust is roughly proportional to it and rather more than proportionally sensitive: a loss of a few per cent in recovery costs a larger percentage of thrust, which is why inlet design receives attention out of proportion to its apparent simplicity.

Answer: C

Correct answer: 0.50 m

Subtract the nacelle radius from the height of its centreline:

\[ c = 2.1 - \frac{3.2}{2} = 2.1 - 1.6 = 0.5\ \mathrm{m} \]

Half a metre is tight, and it is the static figure - the working clearance is less once the oleos compress, the tyres deflect and the aircraft banks or the wing flexes. This is the constraint that grows harder with every increase in bypass ratio, and the reason nacelle bottoms are often flattened.

Answer: D

Correct answer: 840 kN m

The worst case is the loss of an outboard engine, since the moment is thrust times arm and the outboard arm is the larger:

\[ N = 70 \times 12.0 = 840\ \mathrm{kN\,m} \]

Losing an inboard engine would give only \( 70 \times 6.0 = 420 \) kN m. The fin and rudder must be sized for the worst case, so it is the outboard failure that determines them, and this is one reason four-engine aircraft carry proportionally large fins.

Answer: A

Correct answer: 1.33

Keep the total thrust the same. The twin has two engines of \( 2T \) each at 8 m, so losing one leaves a moment of \( 2T \times 8 = 16T \). The four-engine aircraft has engines of \( T \) each, and losing the outboard gives \( T \times 12 = 12T \).

\[ \frac{16T}{12T} = 1.33 \]

So the twin suffers the larger moment despite its shorter arm, because each of its engines carries twice the thrust. Losing one of two is a far greater proportional loss than losing one of four.

Answer: B

Correct answer: 530 kN m

Convert the mass to weight and multiply by the arm:

\[ W = 6000 \times 9.81 = 58.9\ \mathrm{kN} \]

\[ M = 58.9 \times 9.0 = 530\ \mathrm{kN\,m} \]

This moment acts downward at 1 g, opposing the upward bending the lift produces, so it relieves the wing root and allows a lighter structure. The relief grows with load factor, since both the lift and the engine's inertia scale together, which is why the benefit is greatest exactly where the structure is most highly loaded.

Answer: C

Correct answer: 108 kN

Take the recovered fraction per engine, then double it:

\[ 120 \times 0.45 = 54\ \mathrm{kN}, \qquad 54 \times 2 = 108\ \mathrm{kN} \]

Forty-five per cent is typical, because a reverser turns the fan flow through rather less than 180 degrees and does nothing with the core stream on most installations. Note that this force is not credited in the certified landing distance, which assumes brakes alone - the reversers are a margin, not a requirement.

Answer: D

Correct answer: 23 per cent

Frontal area varies with the square of the diameter:

\[ \frac{A_2}{A_1} = \left(\frac{3.1}{2.8}\right)^2 = 1.107^2 = 1.226 \]

an increase of about 23 per cent. The diameter has risen only 11 per cent, which is the distractor offered, but drag follows the area rather than the diameter. This square-law growth is why nacelle drag and ground clearance both become pressing as bypass ratio rises.

Answer: A

Correct answer: 2.00 m\(^2\)

From the continuity relation \( \dot{m} = \rho A V \):

\[ A = \frac{\dot{m}}{\rho V} = \frac{180}{0.60 \times 150} = \frac{180}{90} = 2.00\ \mathrm{m}^2 \]

The capture area must suit the whole flight envelope, and the mass flow the engine demands varies enormously between takeoff and cruise. An inlet sized for one condition spills air at another, and that spillage carries a drag penalty - which is one of the compromises inlet designers are paid to manage.

Answer: B

Correct answer: 3.25 kN

Apply the percentage to the engine's thrust:

\[ 130 \times 0.025 = 3.25\ \mathrm{kN} \]

The loss arises because bleed air is taken from the compressor after work has already been done on it; that work is then thrown away rather than expanded through the turbine. On a twin the total loss is 6.5 kN, which is why bleedless architectures - electrically driven cabin compressors and anti-icing - have become attractive despite their own weight.