Maximum Torque of Induction Motor – Derivation & Condition

We know that the expression for torque of an induction motor under running condition is given by,

\[T=\frac{KsE_{2}^{2}{{R}_{2}}}{R_{2}^{2}+{{(s{{X}_{2}})}^{2}}}\]

Let,

\[Y=\frac{1}{T}\] \[Y=\frac{R_{2}^{2}+{{(s{{X}_{2}})}^{2}}}{KsE_{2}^{2}{{R}_{2}}}\]

\[=\frac{R_{2}^{2}}{KsE_{2}^{2}{{R}_{2}}}+\frac{{{s}^{2}}X_{2}^{2}}{KsE_{2}^{2}{{R}_{2}}}\]

\[=\frac{{{R}_{2}}}{KsE_{2}^{2}}+\frac{sX_{2}^{2}}{KE_{2}^{2}{{R}_{2}}}\]

To obtain the condition for maximum torque, differentiating the above equation with respect to slip and then equating it to zero,

\[\frac{dY}{ds}=0\]

\[\frac{d}{ds}\left[ \frac{{{R}_{2}}}{KsE_{2}^{2}}+\frac{sX_{2}^{2}}{KE_{2}^{2}{{R}_{2}}} \right]=0\]

\[\frac{d}{ds}\left[ \frac{{{R}_{2}}}{KsE_{2}^{2}} \right]+\frac{d}{ds}\left[ \frac{sX_{2}^{2}}{KE_{2}^{2}{{R}_{2}}} \right]=0\]

\[\frac{{{R}_{2}}}{KsE_{2}^{2}}.\frac{d}{ds}\left[ \frac{1}{s} \right]+\frac{X_{2}^{2}}{KE_{2}^{2}{{R}_{2}}}\frac{d}{ds}\left[ s \right]=0\]

\[\frac{{{R}_{2}}}{KsE_{2}^{2}}\left[ \frac{-1}{{{s}^{2}}} \right]+\frac{X_{2}^{2}}{KE_{2}^{2}{{R}_{2}}}\left[ 1 \right]=0\]

\[\frac{-{{R}_{2}}}{KsE_{2}^{2}}+\frac{X_{2}^{2}}{KE_{2}^{2}{{R}_{2}}}=0\]

\[\frac{{{R}_{2}}}{KsE_{2}^{2}}=\frac{X_{2}^{2}}{KE_{2}^{2}{{R}_{2}}}\]

\[\frac{{{R}_{2}}}{{{s}^{2}}}=\frac{X_{2}^{2}}{{{R}_{2}}}\]

\[R_{2}^{2}={{s}^{2}}X_{2}^{2}\]

\[{{R}_{2}}=s{{X}_{2}}\]

Therefore, the condition for maximum torque in a 3-ϕ induction motor under running condition is that the rotor resistance per phase must be equal to the product of slip and the rotor reactance per phase.

i.e., R2 = sX2 Also, condition for maximum torque under running condition is slip,

\[s=\frac{{{R}_{2}}}{{{X}_{2}}}\]

Condition for Maximum Torque in 3-phase induction motor at Standstill Condition.

The starting torque (Tst) will be maximum (Tmax) when rotor resistance (R2) is equal to standstill reactance (X2).

i.e, R2 = X2

Since,

\[{{T}_{FL}}\propto \frac{{{R}_{2}}}{R_{2}^{2}+X_{2}^{2}}\]

\[{{T}_{FL}}=\frac{K{{R}_{2}}}{R_{2}^{2}+X_{2}^{2}}\]

Differentiating with respect to R2 we get,

\[\frac{d{{T}_{st}}}{d{{R}_{2}}}=K\left[ \frac{\left( R_{2}^{2}+X_{2}^{2} \right)-{{R}_{2}}2{{R}_{2}}}{{{\left( R_{2}^{2}+X_{2}^{2} \right)}^{2}}} \right]\]

\[\frac{d{{T}_{st}}}{d{{R}_{2}}}=K\left[ \frac{1}{\left( R_{2}^{2}+X_{2}^{2} \right)}-\frac{2R_{2}^{2}}{{{\left( R_{2}^{2}+X_{2}^{2} \right)}^{2}}} \right]\]

For torque to be maximum,

\[\frac{d{{T}_{st}}}{d{{R}_{2}}}=0\].

\[\frac{1}{\left( R_{2}^{2}+X_{2}^{2} \right)}-\frac{2R_{2}^{2}}{{{\left( R_{2}^{2}+X_{2}^{2} \right)}^{2}}}=0\]

\[\frac{1}{\left( R_{2}^{2}+X_{2}^{2} \right)}=\frac{2R_{2}^{2}}{{{\left( R_{2}^{2}+X_{2}^{2} \right)}^{2}}}\]

\[R_{2}^{2}+X_{2}^{2}=2R_{2}^{2}\]

\[R_{2}^{2}=X_{2}^{2}\]

\[{{R}_{2}}={{X}_{2}}\]

Now,

\[{{T}_{st}}\propto \frac{{{R}_{2}}}{R_{2}^{2}+X_{2}^{2}}\text{       }\!\![\!\!\text{ Neglecting E}_{2}^{2}\text{ }\!\!]\!\!\text{ }\]

Considering maximum condition i.e.,

When, X2 = R2 

\[{{T}_{\max }}\propto \frac{{{R}_{2}}}{R_{2}^{2}+X_{2}^{2}}\]

\[{{T}_{\max }}\propto \frac{{{R}_{2}}}{2R_{2}^{2}}\]

\[{{T}_{\max }}\propto \frac{1}{2{{R}_{2}}}\]

When, R2 = X2.

\[{{T}_{\max }}\propto \frac{{{X}_{2}}}{X_{2}^{2}+X_{2}^{2}}\]

\[{{T}_{\max }}\propto \frac{{{X}_{2}}}{2X_{2}^{2}}\]

\[{{T}_{\max }}\propto \frac{1}{2{{X}_{2}}}\]

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