Question

An aeroplane is flying with a velocity U${}_p$. The plane takes in volume V of air of mass $\mu$ per second to burn mass m of fuel every second. The exhaust relative to plane has a speed of U${}_E$. Power of engine of aeroplane is:

• A. $\frac{U_E^2(\mu +m)-U_P^2(\mu -m)}{776}$
• B. $\frac{U_E(\mu +m)-U_P(\mu -m)}{746}$
• C. $\frac{U_E{U}_P(\mu +m)-U_P^2\left(\mu \right)}{746}$
• D. $\frac{U_P^2(\mu +m)-U_P{U}_E(\mu -m)}{746}$

Answer 1

The correct option is C $\frac{U_E{U}_P(\mu +m)-U_P^2\left(\mu \right)}{746}$

Velocity of exhaust relative to aeroplane is $U_E$

Velocity of aeroplane relative to ground = $U_P$

So, velocity of exhaust relative to ground = $U_E-U_P$

Mass of fuel expelled per second = m

Momentum change of fuel per second = $m{U}_P-(-m(U_E-U_P\left)\right)=m{U}_P+m(U_E-U_P)$

Momentum change of air per second = $0-(-\mu (U_E-U_P\left)\right)=\mu (U_E-U_P)$

Total chane of the momentum of the expelled mass per second = $(\mu +m)(U_E-U_P)+m{U}_P$

So, from the $newto{n}^{\prime }s3rdlawofmotion$

Momentum change of the aeroplane per second = $(\mu +m)(U_E-U_P)+m{U}_P$ = force imparted by the engine

So, the total power generated by the engine = $U_P\left(\right(\mu +m\left)\right(U_E-U_P)+m{U}_P)$

In horse power unit total power generated = $\frac{\left(U_E{U}_P\right(\mu +m)-\mu ({U_P)}^2}{746}$