# Torque exerted by cranksahft on load

1. Feb 1, 2005

### sisigsarap

I am in need of help regarding a problem which asks for the torque exerted by the crankshaft on the load.

Initialy I am given that a gasoline engine operating at 2500 rev/min takes in energy 7.89 X 10^3 J and exhausts 4.58 X 10^3 J for each revolution of the crankshaft.

I found the mechanical power output of the engine to be 185 horsepower.

This was found by:

((7.89X10^3J - 4.58X10^3J) * 2500 rev * .738 ft lb/s) / (60s * 550 ft lb/s)

I converted the 185 hp to ft lb/s by multiplying it by 550. Then I converted the lb to newtons and the ft to meters so I have Newton Meter/s. I am not sure if I am doing this correctly, or if I am, how to get rid of the seconds?

Thanks,

Josh

2. Feb 2, 2005

### supersix2

Ok lets look at the units here. I'm going to use the FLT system (Force, Length, Time).

A Joule is FL
Your unit of power is FL/T
And of course your unit of time is T

So the units for you equation are

((FL - FL) * (1) * (FL/T)) / (T * (FL/T)) = FL/T

The equation basically equals FL/T because horsepower is just another way of expressing ft lb/s.

Because you only multiplied by revolutions, which are unitless (thats why its just a 1) I think you'll find that your units cancel out properly and you are only left with FL/T like you are supposed to.

I checked your math myself and I got 185 hp as well.

Another way of doing it would be to divided your rpm (units of 1/T because revolutions are unitless) by 60s to begin with and that would get your extra seconds out of the denominator so your equation would look like this

((FL - FL) * (1/T)(T/T) * (FL/T)) / FL/T

And you will find it equals FL/T like it should.

I'm pretty sure you used the right method but got confused by the units. If I'm wrong, I apologize, I'm sure someone else will correct me.