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)  

Let                     s_f  = Full load slip
.^..                       T_F.L. α (s_f  E₂² R₂)/(R₂² +(s_f  X₂)²)  

and                    s_m = Slip for maximum torque T_m 

.^..                       T_m  α (s_m  E₂² R₂)/(R₂² +(s_f  X₂)²)

       Dividing both numerator and denominator by X₂² we get,

But                       R₂/X₂ = s_m 

                            T_F.L./T_m  = (s_f  x 2 s_m²)/(s_m x (s_m²+ s_f²))

                             T_F.L./T_m   = (2 s_f  s_m)/(s_m² + s_f²)

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 T_st,        s =1
                           T_st  α   (E₂² R₂)/(R₂² +( X₂)²)

While for T_m,       s = s_m

 

       Dividing both numerator and denominator by X₂² we get,

      Substituting             R₂/X₂ = s_m

 

       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,
                            T_F.L./T_m  = (a s_f )/(a²+ s_f²)

and                       T_st/T_m  = 2 a/ (1 + a²)

where                    a = R₂/X₂ = s_m

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_s  = 120f / P = (120x50)/24 = 250 r.p.m.
                s_f  = (N_s  - N)/N_s  = (250-247)/250 = 0.012 = Full load slip
                 s_m = R₂/X₂ = 0.016/0.265 = 0.06037
i)              T_F.L./T_m  = (2 s_m s_f )/(s_m²+ s_f²) = (2 x 0.06037 x 0.012)/(0.06037² + 0.012²)
ii)              T_st/T_m  = (2 s_m )/(1 + s_m²) = (2 x 0.06037)/(1 + 0.06037²) = 0.1203

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