A gear box has the input power 10kw at 2400 rev/min. The gear ratio is a reduction of 2 to 1.

A gear box has the input power 10kw at 2400 rev/min. The gear ratio is a reduction of 2 to 1. The efficiency is 98%. What is the holding torque required?

The Power to gear 1 is 10Kw. The power received by gear 2 is a little bit lower, it means, 0.98*10 Kw ( 9.8 Kw).

You should remember that Power is = Torque (T) * rotational speed (w). In such a way,

T2 * W2 = 9.8 Kw

But W2 = 2 W1. In this way W2 = 2 * 2400 rev./min. ( 251 rad/seg )

Therefore:

T2 = 9800 (J/s)/W2 = 9800 J/s / 251 rad / seg =
= 39.04 J or 39.04 N * m

The holding torque ( or load torque ) is 39.04 N*m