An aircraft weighing 70,000 kg is climbing out after take-off. During this climb the aircraft flies with an angle of attack of 2 degrees , while the pilot observes an angle of between the view from the cockpit and the horizon. Furthermore, it is known that during this climb out the aircraft has a drag of 20 kN.

Question

What is the thrust (in kN) needed for this climb-out?:

unanswered

Answer 1

To determine the thrust needed for the aircraft's climb-out, we need to account for both the weight of the aircraft and the drag acting on it.

Given Data:

• Weight of the aircraft (\( W \)) = 70,000 kg
• Drag (\( D \)) = 20 kN

Force Components:

The forces acting on the aircraft during the climb-out are:

1. Weight (W): Acts downward with a force equivalent to the weight of the aircraft.
2. Drag (D): Acts opposite to the direction of thrust.
3. Thrust (T): Acts forward in the direction of motion.
4. Lift (L): Acts upward, which is required to counteract the weight of the aircraft.

Calculating Weight in Newtons:

Convert the weight of the aircraft into Newtons: \[ W = m \cdot g = 70,000 , \text{kg} \times 9.81 , \text{m/s}^2 = 686,700 , \text{N} \text{ (or 686.7 kN)} \]

Lift during Climb-out:

In a steady climb, the lift must equal the weight of the aircraft. Therefore, we have: \[ L = W = 686.7 , \text{kN} \]

Thrust Equation:

During the climb-out, the net force acting on the aircraft can be expressed as: \[ T - D = W \cdot \sin(\alpha) \] Where \( \alpha \) is the angle of climb (which can typically be interpreted as the angle of attack in this context, though it may not directly equal the angle of attack mentioned in the problem). However, if we assume straight and level climb out with only overcoming drag, we simplify as follows:

When climbing out, the thrust must overcome both the drag and provide the necessary vertical component to counteract the weight of the aircraft. In this case of simple ascent:

For a steady climb: \[ T = D + (W \cdot \sin(\alpha)) \] The sine of the angle of attack (2 degrees) can be calculated: \[ \sin(2^{\circ}) \approx 0.0349 \] So: \[ W \cdot \sin(2^\circ) = 686.7 \cdot 0.0349 = 23.99 , \text{kN} \]

Final Thrust Calculation:

Putting it all together, the thrust needed is: \[ T = D + (W \cdot \sin(2^\circ)) \] \[ T = 20 , \text{kN} + 23.99 , \text{kN} \approx 43.99 , \text{kN} \]

Conclusion:

The thrust needed for the climb-out is approximately 44 kN when rounded to the nearest whole number.