Torque exerted by cranksahft on load

[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?

Please help!

Thanks,

Josh

 

[supersix2]

Ok let's 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.

 

[wais]

Hi Josh,

Thank you for reaching out for help with your problem. Let's break down the steps to find the torque exerted by the crankshaft on the load.

Step 1: Find the mechanical power output of the engine
As you correctly calculated, the mechanical power output of the engine is 185 horsepower. However, we need to convert this to the standard unit of power, which is watts. 1 horsepower is equivalent to 550 ft lb/s, so 185 horsepower would be 185*550 = 101,750 ft lb/s. Now, we need to convert this to watts by multiplying by 0.73756 (1 ft lb/s = 0.73756 watts), which gives us a mechanical power output of 75,004 watts.

Step 2: Calculate the torque
Torque is defined as the force applied perpendicular to a lever arm, multiplied by the length of the lever arm. In this case, the lever arm is the crankshaft and the force is the power output of the engine. So, we can calculate the torque by dividing the power output by the angular velocity (in radians per second) of the crankshaft.

To find the angular velocity, we need to convert the given engine speed of 2500 rev/min to radians per second. There are 2π radians in one revolution, so 2500 rev/min would be (2500 rev/min * 2π radians/rev) / (60 s/min) = 261.8 radians/s.

Now, we can calculate the torque by dividing the power output (75,004 watts) by the angular velocity (261.8 radians/s), which gives us a torque of 286.6 Nm.

Step 3: Removing the seconds
In your calculation, you converted the power output to ft lb/s and then multiplied by 550 to get ft lb/s^2. However, we only need to convert the power output to watts and divide by the angular velocity to get the torque in Nm. There is no need to multiply by 550 or convert to ft lb/s^2.

I hope this helps clarify the steps to find the torque exerted by the crankshaft on the load. Let me know if you have any further questions. Good luck with your problem!