Three phase Induction Motor torque/slip HELP NEEDED

[callum76wood]

Hi, I'm really stuck on this assignment question and could do with a hand on the step by step process of how to get the answers.

A three phase induction motor as 2 poles, a standstill voltage of 121 v/phase (Er), rotor resistance of 0.15Ω/phase (Rr) and a standstill rotor reactance of 1.2Ω/phase (Xr)

1. Plot the torque against slip characteristic when the supply frequency is 50Hz for slip values of 1.0, 0.75, 0.5, 0.25, & 0.05

2. Determine the current in each phase of the rotor
A) At start
B) At maximum torque
C) At full load torque if this occurs at a slip of 4%

 

[Babadag]

Neglecting skin and proximity effect in rotor winding-or squirrel cage-
Irot=s*Er/sqrt(Rr^2+s^2*Xr^2) [A] Er[V] ,Rr,Xr [ohm]
Tqmot=3*Irot^2*Rr/s/(2*pi()*f/p) [Nm] p=no.of pole pairs=1
For s=1 [at start] Ir=100 A Tq= 14.33 Nm
For s=0.04 [rated] Ir=30.73 A Tq= 1691 Nm

 

[Babadag]

Sorry, I forgot the maximum torque slip calculation.
The maximum torque slip it is the slip calculated from dTq/ds=0.
If Tq=3*I^2*R/s/(2*PI()*f/p) and I=s*Er/sqrt(R^2+s^2*X^2) then
Tq =3*s^2*Er^2/(R^2+s^2*X^2)/s*R/(2*pi()*f/p)
dTq/ds=3*Er^2/(2*pi()*f/p)*[(R^2+s^2*X^2)-2*s^2*X^2]=0
X^2*s^2=R^2
s=R/X
sm=0.15/1.2=0.125

 

[Babadag]

correction:
s=0.04 Tq= 33.81 !
s=0.125 Tq= 58.25
Sorry!