Question

Design a rigid coupling is used to connect a 45 kW,1440 rpm electric motor to a 

centrifugalpump. The starting torque of the motor is225% of the rated torque

Answer 1

power $P = 45KW = 45,000 Nm$

RPM N = 1440

$T = \frac{P \times 60}{2\pi N} = \frac{45000 \times 60}{2\pi 1440} = 298.41 Nm$

As starting torque of the motor is 225% of rated torque.

$T_{max} = 2.25 \times T = 2.25 \times 298.41 = 671.42 Nm$