Originally Posted by sganesh

Dear Experts, If an exciter static and dynamic moment are known, how to know the maximum screen with material TPH, on which this exciter can be operated ? Please assume that motor rpm will be 900. How to know the motor KW ?

Thank you and many regards,

PLease find attached an example of calculations.

M x r value for Exciter:= 48,18 kgm

M= Out of balance mass in the mechanism (kg)

r = Radius of gyration (mm)

Operating speed:950 rpm (highest considered)

Motor speed:1490 rpm

Speed ratio:1490/950 = 1,5684

Allow 10% for friction, higher viscosity of lubrication at starting and cold weather etc and consider the requirement of the starting torque at the mechanism shaft as:

48,18 x 1,1 = 53,0 kgm and transferred to the motor shaft this will be:

53,0 / 1,5684 = 33,79 kgm

The starting torque of these motors is 250% of the FLT (Full load torque)

FLT = 33,79 / 2,5 = 13,52 kgm

Motor power required = RPM x T / 950 = 1490 x 13,52 / 950 = 21,21 kW

21,21 kW gives nearest standard motor = 30 kW

Adding 5,2% for elevation at 1650 m above sea level = 30 x 1,052 = 31,56 kW.

Nearest standard motor = 37 kW ■