Question

The speed of an aeroplane is 1200 ms^-1
1 −
The engines take in 80 kg of air per second and
mix it with 40 kg of fuel. This mixture is expelled after it ignites and it moves at a
velocity of
3000 ,ms^-1 relative to the aeroplane. Calculate the thrust of the engine.

Answer 1

Answer on Question#39826, Physics, Mechanics | Kinematics | Dynamics

The speed of an airplane is $1200\,\mathrm{ms}^{-1}$. The engines take in $80\,\mathrm{kg}$ of air per second and mix it with $40\,\mathrm{kg}$ of fuel. This mixture is expelled after it ignites and it moves at a velocity of 3000, $\mathrm{ms}^{-1}$ relative to the airplane. Calculate the thrust of the engine.

Solution

Thrust is a reaction force described quantitatively by Newton's second and third laws. When a system expels or accelerates mass in one direction, the accelerated mass will cause a force of equal magnitude but opposite direction on that system.

From Newton's second law of motion a force $F$ on an object is equal to the rate of change of its momentum

$$F = \frac{dp}{dt} = \frac{d(mv)}{dt}.$$

In our case a force $F$ can be expressed as

$$F = \frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{\Delta t}.$$

$m_1$ – mass of air per second, $m_2$ – mass of fuel per second.

$v_1 = 1200\,\frac{\mathrm{m}}{\mathrm{s}}$ – initial velocity of an air relative to the airplane, $v_2 = 3000\,\frac{\mathrm{m}}{\mathrm{s}}$ – final velocity of a fuel and an air relative to the airplane, initial velocity of a fuel relative to the airplane is 0.

Change of momentum in 1 second:

$$\Delta p = m_1(v_2 - v_1) + m_2(v_2 - 0) = (m_1 + m_2)v_2 - m_1v_1.$$

The thrust of the engine:

$$F = \frac{\Delta p}{\Delta t} = \frac{(m_1 + m_2)v_2 - m_1v_1}{\Delta t} = \frac{(80 + 40) \cdot 3000 - 80 \cdot 1200}{1} = 264\,\mathrm{kN}.$$

Answer: 264 kN.