Question 40: Minimum Power for Helicopter Take-off
Problem Statement
We need to find the power $P$ required to lift a helicopter of mass $m$ using a rotor of diameter $d$, assuming air density $\rho$.
1. Lifting Force (Momentum Transfer)
To hover or take off, the rotor must push air downwards, creating an upward reaction force (Thrust) equal to the helicopter’s weight.
$$F_{thrust} = mg$$Let $v$ be the velocity of the air pushed downwards. The mass of air moved per unit time ($\dot{m}$) through the rotor area $A$ is:
$$\dot{m} = \text{Density} \times \text{Volume Flow Rate} = \rho A v$$The rate of momentum change imparted to the air is:
$$F_{thrust} = \dot{m} v = (\rho A v) v = \rho A v^2$$Equating Thrust to Weight:
$$\rho A v^2 = mg \implies v = \sqrt{\frac{mg}{\rho A}}$$2. Power Requirement
The following analysis is as per book, but I think the Power requirement should be double the answer as only half the energy delivered to blades should reach the air as rest half would be dissipated due to inelastic collision between blades and air molecules
The power $P$ is the kinetic energy imparted to the air per unit time:
$$P = \frac{1}{2} \dot{m} v^2 = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} \rho A v^3$$Substituting the expression for $v$ derived above:
$$P = \frac{1}{2} \rho A \left( \sqrt{\frac{mg}{\rho A}} \right)^3 = \frac{1}{2} \rho A \frac{(mg)^{3/2}}{(\rho A)^{3/2}}$$ $$P = \frac{1}{2} \frac{(mg)^{3/2}}{\sqrt{\rho A}}$$3. Geometry Substitution
The area of the rotor with diameter $d$ is $A = \frac{\pi d^2}{4}$. Substituting this into the power equation:
$$P = \frac{1}{2} \frac{(mg)^{3/2}}{\sqrt{\rho \frac{\pi d^2}{4}}}$$ $$P = \frac{1}{2} \frac{(mg)^{3/2}}{\frac{d}{2} \sqrt{\pi \rho}}$$The factor of $1/2$ cancels out: