A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters: R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm=30.74 Ω Rotational losses are 2450W. Core loss is neglected. The motor drives a mechanical load at a speed of 1170 rpm.
Ns = 1200rpm
slip = 0.025
Line current = 26.87 angle -0.38
Pin = 20.745KW
Airgap power = 19.71KW
Converted power = 19.21KW
Torque = 156.85Nm
Pout = 16.76KW
efficiency = 80.79%
Find : 1. Pullout slip
2. Pull out torque
3. Start torque
I need the solution with full calculations and formulas.
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