## 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 E_{1 }and E_{2}. Hence E_{2 }is also consatnt. Similarly R_{2}, X_{2 }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 E

_{2}^{2}R_{2})/(R_{2}^{2}+(s X_{2})^{2}) While carrying out differential remember that E

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.

_{2}, R_{2}, X_{2 }andk 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 E

_{2}^{2}R_{2})/(R_{2}^{2}+ s^{2}X_{2}^{2}) …….Writing (s X_{2})^{2}= s^{2}X_{2}^{2}As both numerator and denominator contains s terms, differential T with

respect to s using the rule of differentiation for u/v.

**.**k s E

^{.}._{2}

^{2}R

_{2}(2s X

_{2}

^{2}) – (R

_{2}

^{2}+ s

^{2}X

_{2}

^{2})(k E

_{2}

^{2}R

_{2}) = 0

**.**2 s

^{.}.^{2}k X

_{2}

^{2}E

_{2}

^{2}R

_{2}– R

_{2}

^{2}k E

_{2}

^{2}R

_{2}– k s

^{2}X

_{2}

^{2}E

_{2}

^{2}R

_{2}= 0

**.**k s

^{.}.^{2}X

_{2}

^{2}E

_{2}

^{2}R

_{2}– R

_{2}

^{2}k X

_{2}

^{2}R

_{2 }= 0

**.**s

^{.}.^{2}X

_{2}

^{2 }– R

_{2}

^{2}= 0 Taking k E

_{2}

^{2}R

_{2}common.

**.**s

^{.}.^{2}= R

_{2}

^{2}/X

_{2}

^{2}

**.**s = R

^{.}._{2}/X

_{2}Neglecting negative slip

This is the slip at which the torque is maximum and is denoted as s

_{m}.**.**s

^{.}._{m}= R

_{2}/X

_{2}

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 }= R_{2}/X_{2}in the torque equation. It is denoted by T_{m}. T

_{m }= (k s_{m }E_{2}^{2}R_{2})/(R_{2}^{2}+(s_{m }X_{2})^{2}) From the expression of T

_{m}, it can be observed that1. 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 R

_{2}. But the slip at which it occurs i.e. speed at which it occurs depends on the value of rotor resistance R_{2}.**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 ? and 0.1 ?

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, R

_{2 }= 0.01 ?, X_{2 }= 0.1 ? 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**.**E

^{.}._{2 }= (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 E

_{2}

^{2}R

_{2})/(R

_{2}

^{2}+( X

_{2})

^{2}) where k = 3/(2 ? n

_{s})

n

_{s }= N_{s}/60 = 1500/60 = 25 r.p.s.**.**k = 3/(2? x 25) = 0.01909

^{.}.**.**T

^{.}._{st }= ( 0.01909 x 57.735

^{2}x 0.01 )/( 0.01

^{2}+ 0.1

^{2}) = 63.031 N-m

ii) Slip at which maximum torque occurs is,

s

_{m }= R_{2}/X_{2}= 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 E_{2}^{2})/(2 X_{2}) = (0.01909 x 57.735^{2})/(2 x 0.1) = 318.16 N-mv) Full load slip, s

_{f }= 0.04 as % s_{f }= 4 %**.**T

^{.}._{f.l. }= (k s

_{f }E

_{2}

^{2}R

_{2})/(R

_{2}

^{2}+(s

_{f }X

_{2})

^{2}) = (0.01909 x 0.04 x 57.735

^{2}x 0.01)/( 0.01

^{2}+ (0.04 x 0.1)

^{2})

= 219.52 N-m

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