Question

An airplane flies horizontally at 80 m/s. Its propeller delivers 1300 N of thrust (forward force) to overcome aerodynamic drag (backward force). Using dimensional reasoning and unity conversion ratios, calculate the useful power delivered by the propeller in units of kW and horsepower.

Answer 1

Power in kW is 104 kW

Power in horsepower is 139.41 hp

Solution:

As per the question:

Velocity of the airplane,
`v_(a) = 80 m/s`

Force exerted by the propeller,
`F_(p) = 1300 N`

Now,

The useful power that the propeller delivered,
`P_(p)`:

`P_(p) = (Energy)/(time, t)`

Here, work done provides the useful energy

Also, Work done is the product of the displacement, 'x' of an object when acted upon by some external force.

Thus

`P_(p) = (F_(p)* x)/(time, t)`

`P_(p) = (F_(p)* x)/(time, t)`

`P_(p) = F_(p)* v_(a)`

Now, putting given values in it:

`P_(p) = 1300* 80 = 104000 W = 104 kW`

In horsepower:

1 hp = 746 W

Thus

`P_(p) = (104000)/(746) = 139.41 hp`