Physics Help Forum Power of a rotor - derivation

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 Apr 17th 2018, 09:25 AM #1 Junior Member   Join Date: Apr 2018 Posts: 1 Power of a rotor - derivation I'm trying to understand the equation P = 2(pi)TN/60, where T=torque, N=rpm... So far I understand: E = F.d = F.(2.pi.r)*(no. of revolutions) = 2T.pi.n And P = E/t = 2T.pi.n/t - but n/t is revolutions per second, so: P = 2T.pi.f f = rev/second This is where I'm having trouble, because 1 rev/sec = 60 rpm, so I'd say: P = 2T.pi.60.N but in the textbook and online it says P = 2(pi)TN/60 so I'm not sure what's going on... Anyone know what's going on?
 Apr 17th 2018, 12:14 PM #2 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 132 Power of a rotor - derivation 1 rev/sec = 60 rpm P = 2T.pi.f f = rev/second f = N/60 P = 2(pi)TN/60
Apr 17th 2018, 12:30 PM   #3
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 This is where I'm having trouble, because 1 rev/sec = 60 rpm, so I'd say: P = 2T.pi.60.N but in the textbook and online it says P = 2(pi)TN/60 so I'm not sure what's going on...

However your intial data says the shaft is running at N rpm

By definition this is N/60 rps.

So your f = N/60 and you must substitute this for f in the equation you have developed, not the other way round.

Last edited by studiot; Apr 17th 2018 at 01:33 PM.

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