Question

a 3250 aircraft takes 12.5 min to achieve its cruising altitude of 10 km and cruising speed of 850 km/h if the planes engines deliver, on average, 1500 hp during this time, what is the efficiency of the engines

Answer 1

The efficiency of the engines can be calculated using the formula Efficiency = (Useful Output Energy / Input Energy) x 100%. By converting the power delivered by the engines from horsepower to watts, and calculating the work done by the engines, we can determine the efficiency to be approximately 12.5%.

Step-by-step explanation:

The efficiency of the engines can be determined using the formula:

Efficiency = (Useful Output Energy / Input Energy) x 100%

In this case, the useful output energy is the work done by the engines to achieve the cruising altitude and speed.

The input energy is the power delivered by the engines.

First, we need to convert the power from horsepower (hp) to watts (W).

1 horsepower is equal to 745.7 watts.

So, the power delivered by the engines is 1500 hp x 745.7 W/hp = 1118550 W.

Now, we need to calculate the useful output energy.

The work done by the engines is equal to the force applied multiplied by the distance traveled.

The force applied can be calculated using the equation:

Force = mass x acceleration

The mass of the aircraft is not given, so we cannot directly calculate the force. However, we can use the formula for work:

Work = Force x Distance

Since the aircraft is traveling at a constant speed and altitude, the work done is equal to the power delivered multiplied by the time taken:

Work = Power x Time

So, the work done by the engines is 1118550 W x 12.5 min = 13981875 J.

Now, we can calculate the efficiency:

Efficiency = (13981875 J / 1118550 W) x 100%

≈ 12.5%

Therefore, the efficiency of the engines is approximately 12.5%.