Torque and RPM 1

[Kumar G]

Dear Members,
We request to some one clear our Clarification, A shaft rotating with 26400 NM @ 3.15 rpm, Can we attached a gear box to attain 50 to 52 rpm , while do like that what will be the Torque of the final output shaft @ 50 TO 52 RPM AND how much electric power can be drawn from that shaft at mentioned final output.

Thanks & Warm Regard,
Kumar.

 

[3DDave]

Torque * radians/second = power, constant (excepting friction losses.)

3.15 Revolutions/Minute *1 Minute/60 Seconds * 2PI Radians/revolution = 0.33 radians/sec

26400N-m*0.33 radians/sec = 8700 N-m/sec = 8700 Watts input.

Frictionless Torque_out = (RPM_in/RPM_out)*Torque_in; (3.15 RPM/52 RPM)*26400 N-m = 1600 N-m.

However since this is a step-up in speed of (52/3.14) = 16.5, this is more than typically works, so 2 stages of ~4:1 will be required with about a 10% loss of torque due to friction, so around 1440 N-m output;

I'll let you figure the output power from the first equation. The electrical power will depend on the efficiency of the generator or alternator at that RPM.