A twin-engined propeller aircraft is equipped with two engines, each rated at 98 kW shaft power. During cruise flight at 72 m/s at an altitude of 3400 metres, the aircraft experiences a lift coefficient of 0.555. Its wing surface area is 28 square metres, with an average chord length of 2 metres. Its zero-lift drag coefficient is 0.01 and its span efficiency (or Oswald) factor is 0.7.

Determine the propulsive efficiency (in percent) in this flight condition. Do not include the % sign in your answer.

1 answer

To determine the propulsive efficiency of the aircraft, we first need to calculate the induced drag, the total drag, and then find the thrust required at cruise flight to overcome this drag. We can then use this to evaluate the propulsive efficiency.

Step 1: Calculate the Lift

The lift \( L \) can be calculated using the following formula: \[ L = \frac{1}{2} \rho V^2 S C_L \] where:

  • \( \rho \) is the air density at the altitude (3400 m),
  • \( V \) is the speed (72 m/s),
  • \( S \) is the wing surface area (28 m²),
  • \( C_L \) is the lift coefficient (0.555).

The air density at 3400 m can be approximated (ISA model) to be about \( \rho = 0.908 , \text{kg/m}^3 \).

Thus, calculating the lift:

\[ L = \frac{1}{2} (0.908 , \text{kg/m}^3) (72 , \text{m/s})^2 (28 , \text{m}^2) (0.555) \]

Calculating \( L \):

\[ L = 0.5 \times 0.908 \times 5184 \times 28 \times 0.555 \approx 0.5 \times 0.908 \times 5184 \times 15.54 \approx 0.5 \times 0.908 \times 80397.12 \approx 0.5 \times 72931.29 \approx 36465.645 , \text{N} \]

Step 2: Calculate the Induced Drag Coefficient \( C_Di \)

The induced drag can be estimated using the following equation: \[ C_{Di} = \frac{C_L^2}{\pi A e} \] where:

  • \( A \) is the aspect ratio given by \( \frac{S}{b} \). Assuming a width \( b \) (wingspan), we need not explicitly calculate it since \( S = b \times \text{chord} \) gives us an idea of \( b \).

Calculating \( A \): Assume the wingspan \( b = \frac{S}{\text{average chord}} = \frac{28}{2} = 14 , \text{m} \) \[ A = \frac{b^2}{S} = \frac{14^2}{28} = \frac{196}{28} = 7 \]

Now substituting values to compute \( C_{Di} \):

\[ C_{Di} = \frac{(0.555)^2}{\pi \times 7 \times 0.7} \approx \frac{0.308025}{15.4006} \approx 0.02001 \]

Step 3: Calculate Total Drag Coefficient

The total drag coefficient \( C_D \) can be calculated as: \[ C_D = C_{D0} + C_{Di} \] So: \[ C_D = 0.01 + 0.02001 \approx 0.03001 \]

Step 4: Calculate Total Drag Force

The drag force \( D \) can be calculated with: \[ D = \frac{1}{2} \rho V^2 S C_D \] Calculating \( D \): \[ D = \frac{1}{2} \times 0.908 \times (72)^2 \times 28 \times 0.03001 \]

Solving this: \[ D = 0.5 \times 0.908 \times 5184 \times 28 \times 0.03001 \approx 0.5 \times 0.908 \times 5184 \times 0.84028 \approx 0.5 \times 0.908 \times 4355.918 \approx 1978.5763 , \text{N} \]

Step 5: Calculate Thrust and Propulsive Efficiency

The total shaft power available from the engines is: \[ P_{total} = 2 \cdot 98 , \text{kW} = 196 , \text{kW} \]

Now, convert to watts: \[ P_{total} = 196,000 , \text{W} \]

The required thrust \( T \) relates to power and velocity: \[ P = T \cdot V \] Thus converting drag to thrust gives: \[ T = D \approx 1978.5763 , \text{N} \]

Calculating propulsive efficiency \( \eta_p \): \[ \eta_p = \frac{T \cdot V}{P_{total}} = \frac{1978.5763 \times 72}{196000} \]

Calculating: \[ \eta_p = \frac{142656.7016}{196000} \approx 0.726 \]

Finally, converting to percentage: \[ \eta_p \approx 0.726 \times 100 \approx 72.6 \] Thus, the propulsive efficiency is \( 72.6 \) (without the percent sign).

\[ \text{Answer: } 72.6 \]