Condition Of Maximum Torque In Three Phase Induction Motor

From the torque equation, it is clear that torque depends on slip at which motor is running. The supply voltage to the motor is usually rated and constant and there exists a fixed ratio between E₁ and E₂. ence E₂ is also consatnt. Similarly R₂, X₂ and n_s are constants for the induction motor.

Hence while finding the condition for maximum torque, remember that the only parameter which controls the torque is slip s.

Mathematically for the maximum torque we can write,

dT/ds = 0
where        T = (k s E₂² R₂)/(R₂² +(s X₂)²)

While carrying out differential remember that E₂, R₂, X₂ and k are constants. The only variable is slip s. As load on motor changes, its speed changes and hence slip changes. This slip decides the torque produced corresponding to the load demand.

T = (k s E₂² R₂)/(R₂² + s² X₂²)              …….Writing (s X₂)²  = s² X₂²

As both numerator and denominator contains s terms, differential T with respect to s using the rule of differentiation for u/v.

^..              k s E₂² R₂ (2s X₂²) – (R₂² + s² X₂²)(k E₂² R₂) = 0
.^..              2 s² k X₂² E₂² R₂ – R₂² k E₂² R₂ – k  s²  X₂² E₂² R₂ = 0
.^..               k s²  X₂² E₂² R₂ – R₂² k X₂² R₂ = 0
.^..               s²  X₂² – R₂² = 0                  Taking k E₂² R₂ common.
.^..               s²  = R₂²/X₂²
.^..               s = R₂/X₂                 Neglecting negative slip

This is the slip at which the torque is maximum and is denoted as s_m.

.^..                s_m= R₂/X₂

It is the ratio of standstill per values values of resistance and reactance of rotor, when the torque produced by the induction motor is at its maximum.

1.1 Magnitude of Maximum Torque

This can be obtained by substituting s_m = R₂/X₂   in the torque equation. It is denoted by T_m.

T_m = (k s_m E₂² R₂)/(R₂² +(s_m X₂)²)

From the expression of T_m, it can be observed that

1. It is inversely proportional to the rotor reactance.
2. It is directly proportional to the square of the rotor induced e.m.f. at standstill.
3. The most interesting observation is, the maximum torque is not dependent on the rotor resistance R₂.

But the slip at which it occurs i.e. speed at which it occurs depends on the value of rotor resistance R₂.

Example 1 : A 400 V, 4 pole, 3 phase, 50 Hz star connected induction motor has a rotor resistance and reactance per phase equal to 0.01 Ω and 0.1 Ω respectively. Determine i) Starting torque ii) slip at which maximum torque will occur iii) speed at which maximum torque will occur iv) maximum torque v) full load torque if full load slip is 4 %. Assume ratio of stator to rotor turns as 4.

Solution : The given values are,

P = 4,       f = 50 Hz,        stator turns/ rotor turns = 4,    R₂ = 0.01 Ω,      X₂ = 0.1 Ω
E_1line = stator line voltage = 400 V
E_1ph = E_1line/√3 = 400/√3 = 230.94 V                  …………star connection
K = E_2ph/E_1ph = Rotor turns/ Stator turns = 1/4
.^..                      E₂ = (1/4) x E_1ph = 230.94/4 = 57.735 V
N_s  = 120f/P = 120×50 / 4 = 1500 r.p.m.
i) At start,          s =1
.^..                     T_st  = (k  E₂² R₂)/(R₂² +( X₂)²)       where k = 3/(2 π n_s)
n_s = N_s/60 = 1500/60 = 25 r.p.s.
.^..                      k = 3/(2π x 25) = 0.01909
.^..                      T_st  = ( 0.01909 x 57.735² x 0.01 )/( 0.01² + 0.1² ) = 63.031 N-m

ii) Slip at which maximum torque occurs is,
s_m  = R₂/X₂ = 0.01/0.1 = 0.1
%s_m  = 0.1 x 100 = 10%

iii) Speed at which maximum torque occurs is speed corresponding to,
N = N_s (1 – s_m ) = 1500 (1 – 0.1) = 1350 r.p.m.

iv) The maximum torque is,
T_m  = (k E₂²)/(2 X₂) = (0.01909 x 57.735²)/(2 x 0.1) = 318.16 N-m

v) Full load slip,    s_f  = 0.04     as % s_f  = 4 %
.^.. T_f.l. = (k s_f  E₂² R₂)/(R₂² +(s_f  X₂)²)    = (0.01909 x 0.04 x 57.735² x 0.01)/( 0.01² + (0.04 x 0.1)²)
= 219.52 N-m