Engine Instrumentation Questions and Answers

Answer: A

Correct answer: 600 kW

Shaft power is the product of torque and angular velocity:

\[ P = T\omega = 3000 \times 200 = 600\,000\ \mathrm{W} = 600\ \mathrm{kW} \]

The relation holds only with \( \omega \) in radians per second; a speed given in rpm must first be converted by \( \omega = 2\pi N/60 \). This is precisely why a turboprop needs both a torquemeter and a tachometer - torque alone is not power. In practice the propeller governs at constant rpm, so torque becomes a direct proxy for power at that setting.

Answer: B

Correct answer: 1.40

Engine pressure ratio is turbine discharge pressure over compressor inlet pressure:

\[ \mathrm{EPR} = \frac{P_{t7}}{P_{t2}} = \frac{42}{30} = 1.40 \]

The distractor 0.71 inverts the ratio, which is worth guarding against: EPR must exceed one, since the engine adds energy to the gas. It is used as a thrust-setting parameter on many turbofans because thrust itself cannot be measured in flight, and the pressure rise across the engine is closely related to it.

Answer: C

Correct answer: 70 per cent

Rotor speed is displayed as a percentage of a datum rather than in rpm:

\[ \frac{10\,500}{15\,000} \times 100 = 70\% \]

The datum is chosen by the manufacturer as a convenient reference near maximum continuous speed, and it differs between spools and between engine types. Displaying a percentage lets the same instrument and the same limits serve engines of quite different absolute speeds, and it means a crew reads one familiar scale whatever the installation.

Answer: D

Correct answer: 2000 kg

Fuel required is the flow rate multiplied by the time, with the time expressed in the same units as the rate:

\[ 2\ \mathrm{h}\ 30\ \mathrm{min} = 2.5\ \mathrm{h} \]

\[ 800 \times 2.5 = 2000\ \mathrm{kg} \]

The common error is to treat 30 minutes as 0.30 of an hour, giving 1840 kg - an underestimate, and in the wrong direction for safety. Note also that this is fuel to destination only: reserves, alternate and contingency fuel are additional, so the figure is never the whole requirement.

Answer: A

Correct answer: 700 degrees

A thermocouple's output depends on the difference between its hot and cold junctions:

\[ 720 - 20 = 700\ \text{degrees} \]

The instrument nonetheless displays 720, because cold junction compensation adds the reference temperature back in. Without that compensation the indication would change with cockpit temperature alone, which is why the compensation circuit is part of every thermocouple installation and why a fault in it produces an error that varies through the flight.

Answer: B

Correct answer: 945 degrees for 12 seconds

An exceedance is recorded as the peak value actually reached together with its duration, because that is what maintenance needs in order to assess the damage:

\[ 900 + 45 = 945\ \text{degrees, for } 12\ \mathrm{s} \]

The margin of 45 degrees alone does not say what temperature the turbine saw, and the limit of 900 alone says nothing about the event. Turbine blade life falls very steeply with temperature, so both the peak and the time at it govern whether a borescope inspection or a component change is required.

Answer: C

Correct answer: 28.8 kg/h

Mass flow is volumetric flow multiplied by density:

\[ \dot{m} = 40 \times 0.72 = 28.8\ \mathrm{kg/h} \]

Piston engine fuel flow is often indicated volumetrically, in litres or gallons per hour, so a conversion of this kind is needed to work in mass - which is what mass and balance and range calculations require. Because density falls as fuel warms, the same volume carries less mass on a hot day, which is exactly why turbine installations indicate mass flow directly and avoid the problem.

Answer: D

Correct answer: 11500 rpm

The percentage indication is referred to the datum speed:

\[ N = 0.92 \times 12\,500 = 11\,500\ \mathrm{rpm} \]

The distractor 13587 divides by 0.92 instead of multiplying, giving a speed above the datum for an indication below 100 per cent - an impossibility worth spotting before any arithmetic is done. Percentage display is used because it lets one scale and one set of limits serve engines whose absolute speeds differ widely, low-pressure and high-pressure spools included.

Answer: A

Correct answer: 75 per cent

The drop is measured against the original value:

\[ \frac{60 - 15}{60} \times 100 = 75\% \]

The distractor 25 per cent is the remaining fraction rather than the loss, and 300 per cent inverts the ratio. In operational terms a fall of this size is far more than a trend: at 15 psi the oil film supporting the bearings is unlikely to be maintained, so this is an immediate-action indication rather than something to monitor.

Answer: B

Correct answer: 54 kg

Thrust specific fuel consumption multiplies by the thrust being produced:

\[ \dot{m}_f = 0.6 \times 90 = 54\ \mathrm{kg/h} \]

Expressing consumption per kilonewton of thrust makes engines of different sizes comparable, in the same way that specific fuel consumption per unit power does for shaft engines. Note the figure is a rate, so it must still be multiplied by flight time to give fuel required, and TSFC itself varies with altitude, speed and thrust setting rather than being a constant for the engine.