### Torque Ratios

The performance of the motor is sometimes expressed in terms of comparison of various torques such as full load torque, starting torque and maximum torque. The comparison is obtained by finding out ratios of these torques.
1.1 Full load and Maximum Torque Ratio
In general,        Tα (s E22 R2)/(R22 +(s X2)2)
...                       TF.L. α (s E22 R2)/(R22 +(s X2)2)
and                    sm = Slip for maximum torque T
...                       Tα (s E22 R2)/(R22 +(s X2)2)

Dividing both numerator and denominator by X22 we get,
But                       R2/X2 = s
TF.L./T = (s x 2 sm2)/(sm x (sm2+ sf2))
TF.L./Tm   = (2 s sm)/(sm2 + sf2)
1.1 Starting Torque and Maximum Torque Ratio
Against starting with torque equation as,
T α  (s E22 R2)/(R22 +(s X2)2)
Now for Tst,        s =1
Tst  α   (E22 R2)/(R22 +( X2)2)
While for Tm,       s = sm

Dividing both numerator and denominator by X22 we get,
Substituting             R2/X2 = sm

Infact using the same method, ratio of any two torques at two different slip values can be obtained.
Sometimes using the relation, R2 = a X2 the torque ratios are expressed interms of constant a as,
TF.L./T = (a sf )/(a2+ sf2)
and                       Tst/T= 2 a/ (1 + a2)
where                    a = R2/X2 = sm
Example 1 : A 24 pole, 50 Hz, star connected induction motor has rotor resistance of 0.016 Ω per phase and rotor reactance of 0.265 Ω per phase at standstill. It is achieving its full load torque at a speed of 247 r.p.m. Calculate the ratio of
i) Full load torque to maximum torque   ii) starting torque to maximum torque
Solution : Given values are,
P = 24,   f = 50 Hz,       R2 = 0.016 Ω,   X2 = 0.265 Ω,     N = 247 r.p.m.
N= 120f / P = (120x50)/24 = 250 r.p.m.
s= (N- N)/N = (250-247)/250 = 0.012 = Full load slip
sm = R2/X2 = 0.016/0.265 = 0.06037
i)              TF.L./T= (2 sm sf )/(sm2+ sf2) = (2 x 0.06037 x 0.012)/(0.060372 + 0.0122)
ii)              Tst/T = (2 sm )/(1 + sm2) = (2 x 0.06037)/(1 + 0.060372) = 0.1203