HP to wheel size conversion 1

[820currie]

I am trying to turn a generator which requires 30hp to turn at 500rpm. What amount of Force or Torque (I am not sure what the best measurement would be) would I need to apply to a 2 inch shaft to achieve this?

The shaft will be attached to a 2 ft diamter steel wheel pulley.

Thanks for any help-
820currie

 

[desertfox]

Hi 820currie

The answer to your question is:-

power= 2*pi*N*T/60

were power = 30hp*746watts

pi=3.142

N=rpm

T=torque (Newton metres)


rearranging the formula we get:-


T= power*60/(2*pi*500) T= 427.426 Nm = 3773.822lbin

if your applying your torque to the 2FT pulley

then the force you need to exert
is 3773.822lbin/12in" = 314.5lbf

regards desertfox

 

[820currie]

desertfox:

Thanks a bunch for the help.

Sanity question. It sounds like 315 lbf is pretty extreme. Should I get a bigger pully or something to help reduce the amount of force required? Would that help? Is there a formula to use to help reduce the amount of force required to turn this object?


Thanks for your help.
820currie

 

[desertfox]

Hi 820currie

You can reduce the force req by increasing the pulley radius.

torque= force * radius.

However if you change the pulley size you will have to also change the pulley size with which you drive it with.

regards desertfox