An Alpha Jet (see figure) is approaching an airport with a speed of 100 knots. At this altitude and speed, each of the two engines (assuming there is no bypass flow) provides 10 kN of thrust while expelling the gas out of the engine at a speed of 291.5 m/s. To provide this amount of thrust the mass flow through a single engine is 27.21 kg/s. The following questions are all for 1 engine!

Question

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You are asked to answer the following questions.

What is the flight speed in m/s?:

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Calculate the power available (of 1 engine) in MW:

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Calculate the jet power in MW:

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What is the jet efficiency () in %?:

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If the thermal efficiency of this engine is given to be 65.3%, what is the total efficiency of this engine in %?:

Answer 1

Let's go through the calculations step by step.

1. Flight Speed Conversion

First, convert the flight speed from knots to meters per second (m/s).

1 knot = 0.514444 m/s

\[ \text{Flight speed} = 100 \text{ knots} \times 0.514444 \frac{m}{s} = 51.4444 \text{ m/s} \]

2. Power Available from 1 Engine

The thrust produced by one engine is \(10 \text{ kN} = 10,000 \text{ N}\) and the flight speed is \(51.4444 \text{ m/s}\).

The available power (P) from the engine can be calculated using the formula:

\[ P = \text{Thrust} \times \text{Flight speed} \]

Substituting the values in:

\[ P = 10,000 \text{ N} \times 51.4444 \text{ m/s} = 514444.4 \text{ W} \]

To convert to megawatts (MW), we divide by \(10^6\):

\[ P = \frac{514444.4}{10^6} \text{ MW} = 0.5144 \text{ MW} \]

3. Jet Power

Jet power can be calculated using the mass flow rate and jet velocity:

\[ \text{Jet Power} (P_j) = \frac{1}{2} \times \dot{m} \times v_e^2 \]

where:

• \(\dot{m} = 27.21 \text{ kg/s}\) (mass flow rate)
• \(v_e = 291.5 \text{ m/s}\) (exhaust velocity)

Calculating:

\[ P_j = \frac{1}{2} \times 27.21 \text{ kg/s} \times (291.5 \text{ m/s})^2 \]

Calculating \((291.5)^2\):

\[ (291.5)^2 = 85061.25 \text{ m}^2/\text{s}^2 \]

Now substituting back into jet power formula:

\[ P_j = \frac{1}{2} \times 27.21 \times 85061.25 = 1156003.40 \text{ W} \]

Convert to MW:

\[ P_j = \frac{1156003.40}{10^6} \text{ MW} = 1.1560 \text{ MW} \]

4. Jet Efficiency

Jet efficiency (\(\eta_j\)) can be calculated using:

\[ \eta_j = \frac{\text{Power available}}{\text{Jet Power}} \times 100 \]

Substituting values:

\[ \eta_j = \frac{0.5144 \text{ MW}}{1.1560 \text{ MW}} \times 100 = 44.49% \]

5. Total Efficiency

Total efficiency (\(\eta_t\)) is the product of jet efficiency (\(\eta_j\)) and thermal efficiency (\(\eta_{th}\)):

\[ \eta_t = \eta_j \times \eta_{th} \]

Given \(\eta_{th} = 65.3%\):

First, convert both efficiencies into their decimal form:

\[ \eta_j = 0.4449 \quad \text{and} \quad \eta_{th} = 0.653 \]

Now calculate total efficiency:

\[ \eta_t = 0.4449 \times 0.653 = 0.2901 \text{ or } 29.01% \]

Summary of Results

• Flight speed: 51.44 m/s
• Power available (1 engine): 0.5144 MW
• Jet power: 1.1560 MW
• Jet efficiency: 44.49%
• Total efficiency: 29.01%

These results should answer all the questions presented.