Question

An automobile engine develops 133 hp(=99.18 kW) when rotating at 1820 rev/min. How much torque does it deliver?

Answer 1

In SI Units:
\[kW=\frac{2 pi \tau (RPM)}{60,000}\]\[99.13kW=\frac{2 pi \tau(1820RPM)}{60,000}\]\[\tau=5201~N·m\]
In British units:
\[HP=\frac{2 pi \tau(RPM)}{33,000 ft-lb/min}\]\[133hp=\frac{2 pi \tau(1820RPM)}{33,000 ft-lb/min}\]\[\tau=383.8~lbf·ft\]

Answer 2

err...that should have been 520.1Nm

The decimal disappeared I swear...

Answer 3

Could you please show me the formula where torque is expressed in other terms?

Answer 4

@stormfire1

Answer 5

Oh, I sorted it out.

Answer 6

ok :)

Answer 7

Thanks a lot, buddy!
One more question, how to derive the equation you gave me?
@stormfire1

Answer 8

I mean this equation
\[kW=\frac{2 \pi \tau (RPM)}{60,000}\]

Answer 9

It's essentially this equation:\[W=\tau 2piRPS\] converted to RPM

See: http://en.wikipedia.org/wiki/Torque

Answer 10

There's a more complete derivation on that site

Answer 11

Oh, next time I'll check wikipedia first :)))
Nevertheless, thanks a lot!)))

Answer 12

gotta love wiki :P