# Running Torque of Three-Phase Induction Motor

## Torque of 3-Phase Induction Motor under Running Condition

Let the rotor circuit of 3-phase induction motor at standstill has per phase resistance R2, per phase reactance X2 and per phase induced EMF E2. If ‘s’ is the slip under running condition of the motor, then,

$$\mathrm{Rotor \:reactance/phase , š′_2 = š š_2}$$

$$\mathrm{Rotor \:EMF/phase , šø′_2 = š šø_2}$$

$$\mathrm{\therefore \:Rotor \:impedance/phase , š′_2 = \sqrt{š _2^2 + (š š_2)^2}}$$

$$\mathrm{Rotor\:impedance/phase ,š¼′_2 =\frac{šø'_2}{š′_2}=\frac{šø'_2}{\sqrt{š _{2}^{2} + (š š_2)^2}}… (1)}$$

$$\mathrm{Rotor\: power \:factor, cos \varphi′_2 =\frac{š _2}{š′_2}=\frac{š _2}{\sqrt{š _{2}^{2} + (š š_2)^2}}… (2)}$$

Therefore,

$$\mathrm{Running\:torque, \tau_š \propto šø′_2 š¼′_2 cos \varphi′_2 … (3) +}$$

$$\mathrm{\because šø′_2 \propto Magnetic\:flux (\varphi)}$$

$$\mathrm{\therefore \tau_š = š¾ \varphi š¼′_2 cos \varphi′_2}$$

$$\mathrm{⇒ \tau_š = š¾ \varphi ×\frac{šø'_2}{\sqrt{š _{2}^{2} + (š š_2)^2}}\times\frac{š _2}{\sqrt{š _{2}^{2} + (š š_2)^2}}}$$

$$\mathrm{⇒ \tau_š =\frac{š¾ \varphi š šø_2 š _2}{š _2^2 + (š š_2)^2}… (4)}$$

$$\mathrm{\because šø_2 \propto \varphi}$$

$$\mathrm{\therefore \tau_š =\frac{š¾ š šø_2^2 š _2}{š _2^2 + (š š_2)^2}… (5)}$$

Eqn. (4) gives the value of running torque. It can be seen that

• The running torque is directly proportional to slip, i.e., if slip increases, the torque will increase and vice-versa.
• The running torque is directly proportional to square of supply voltage since (E2 ∝ V).

As the running torque of a 3-phase induction motor is given by,

$$\mathrm{\tau_š =\frac{š¾ \varphi š šø_2 š _2}{š _2^2 + (š š_2)^2}}$$

Since the supply voltage (V) is constant, then stator flux and hence the EMF E2 will be constant.

$$\mathrm{\therefore \tau_š =\frac{š¾_1 š š _2}{š _2^2 + (š š_2)^2}… (6)}$$

Where, K1 = K Ļ E2 is a constant.

$$\mathrm{⇒ \tau_š =\frac{š¾_1 š _2}{\frac{š _2^2}{š }+ š š_2^2}… (7)}$$

To be the maximum of the running torque, the denominator of eqn. (6) should be minimum. Hence, differentiating denominator of eqn. (6) with respect to slip 's' and equating it to zero, i.e.,

$$\mathrm{\frac{š}{šš } (\frac{š _2^2}{š }+ š š_2^2) = 0}$$

$$\mathrm{⇒ −\frac{š _2^2}{š ^2}+ š_2^2 = 0}$$

$$\mathrm{⇒ š _2 = š š_2 … (8)}$$

Thus, for maximum torque under running conditions,

$$\mathrm{Rotor\:Resistance/Phase= Per\:unit\:slip \times standstill\:rotor\:reactance\:per\:phase}$$

Now,

$$\mathrm{\tau_š \propto\frac{š š _2}{š _2^2 + (š š_2)^2}}$$

Putting R2 = s X2 in the above expression, the maximum torque under running condition is given by,

$$\mathrm{⇒ \tau_š \propto\frac{1}{2 š_2}… (9)}$$

The slip corresponding to maximum torque is given by,

$$\mathrm{š _š =\frac{š _2}{š_2}… (10)}$$

## Numerical Example

A 6-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 0.03 Ω per phase and standstill reactance of 0.5 Ω per phase. Determine the slip corresponding to maximum torque and speed of the motor at which maximum torque is developed.

## Solution

(1) Slip corresponding to maximum running torque is,

$$\mathrm{š _š =\frac{š _2}{š_2}=\frac{0.03}{0.5}= 0.06 Ω}$$

(2) Speed of motor corresponding to maximum torque is,

$$\mathrm{Synchronous\:speed, š_š =\frac{120š}{š}=\frac{120 \times 50}{6}= 1000 RPM}$$

$$\mathrm{\therefore š_š = š_š (1 − š _š) = 1000 × (1 − 0.06) = 940 RPM}$$