Chevrons on a nozzle trailing edge reduce noise by
Answer: B
The serrations generate streamwise vortices that break up the strong shear layer where most noise is produced. The penalty is a small thrust loss, which is accepted because noise limits often bind more tightly than performance does.
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Specific thrust is related to the nozzle by
Answer: C
Thrust per unit mass flow is essentially the velocity increment the engine imparts. A turbojet produces a fast jet and high specific thrust, while a high bypass turbofan produces a slower jet and correspondingly low specific thrust.
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Installed thrust differs from bare engine test bed thrust because
Answer: D
A test bed measures the engine alone under controlled conditions, while an installed engine must also pay for spillage, nacelle drag and any air or power taken for the airframe. The difference is accounted for carefully in performance guarantees.
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An engine passes \( 80 \ \mathrm{kg/s} \) with an exit velocity of \( 500 \ \mathrm{m/s} \) while flying at \( 200 \ \mathrm{m/s} \). The net thrust, ignoring pressure terms and fuel flow, is
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.
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For the same engine at rest on the ground, the thrust is
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.
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A nozzle of exit area \( 0.5 \ \mathrm{m^2} \) has an exit pressure of \( 120 \ \mathrm{kPa} \) against an ambient of \( 101 \ \mathrm{kPa} \). The pressure thrust is
Answer: C
Correct answer: 9.5 kN
The pressure thrust is the excess of exit over ambient pressure acting on the exit area:
\[ F_p = (p_e - p_a)A_e = (120 - 101)\times10^3 \times 0.5 = 9500 \text{ N} = 9.5 \text{ kN} \]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.
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The critical pressure ratio is \( ((\gamma+1)/2)^{\gamma/(\gamma-1)} \). For \( \gamma = 1.33 \) it is approximately
Answer: D
Correct answer: 1.85
Substituting \( \gamma = 1.33 \), which is representative of hot combustion products,
\[ \left(\frac{\gamma+1}{2}\right)^{\frac{\gamma}{\gamma-1}} = (1.165)^{4.03} \approx 1.85 \]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.
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An engine has a nozzle pressure ratio of 3.2 against a critical ratio of 1.85. The nozzle is
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.
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Exhaust gas at 900 K expands isentropically to an exit velocity given by \( V = \sqrt{2 c_p \Delta T} \). With \( c_p = 1150 \) and a drop of 150 K, the velocity is approximately
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.
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A nozzle has a coefficient of velocity of 0.98. Its ideal exit velocity of \( 600 \ \mathrm{m/s} \) becomes
Answer: C
Correct answer: 588 m/s
The coefficient of velocity is the ratio of actual to ideal exit velocity:
\[ V_e = C_v V_{ideal} = 0.98 \times 600 = 588 \text{ m/s} \]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.
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