Three-phase Induction Motor - Starting Torque (Torque Equation)

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The torque (τ) developed by the rotor of a 3-phase induction motor is directly proportional to −

  • Rotor current (I2)
  • Rotor EMF (E2)
  • Rotor circuit power factor (cos 2)

Therefore,

$$\mathrm{\tau \: \propto \: E_{2}I_{2} \: \cos \varphi_{2}}$$

$$\mathrm{\Rightarrow \: \tau \: = \: KE_{2}I_{2} \: \cos \varphi_{2} \:\quad\: \dotso \: (1)}$$

Where, K is the constant of proportionality.

Starting Torque of 3-Phase Induction Motor

Let,

  • Rotor resistance/Phase = R2
  • Rotor reactance/Phase at standstill = X2
  • Rotor EMF/Phase at standstill = E2

∴ Rotor impedance/Phase at standstill,

$$\mathrm{Z_{2} \: = \: \sqrt{R_{2}^{2} \: + \: X_{2}^{2}}}$$

Rotor current/Phase at standstill,

$$\mathrm{I_{2} \: = \: \frac{E_{2}}{Z_{2}} \: = \: \frac{E_{2}}{\sqrt{R_{2}^{2} \: + \: X_{2}^{2}}}}$$

And, Rotor power factor at standstill,

$$\mathrm{\cos \varphi_{2} \: = \: \frac{R_{2}}{Z_{2}} \: = \: \frac{R_{2}}{\sqrt{R_{2}^{2} \: + \: X_{2}^{2}}}}$$

∴ Starting torque,

$$\mathrm{\tau_{s} \: = \: KE_{2}I_{2} \: \cos \varphi_{2} \: = \: KE_{2} \: \times \: \left(\frac{E_{2}}{\sqrt{R_{2}^{2} \: + \: X_{2}^{2}}} \right) \: \times \: \left(\frac{R_{2}}{\sqrt{R_{2}^{2} \: + \: X_{2}^{2}}} \right)}$$

$$\mathrm{\Rightarrow \: \tau_{s} \: = \: \frac{KE_{2}^{2}R_{2}^{2}}{(R_{2}^{2} \: + \: X_{2}^{2})} \:\quad\: \dotso \: (2)}$$

In general, the stator supply voltage V is constant so that the magnetic flux per pole set up by the stator is also constant. Therefore, the EMF E2 in the rotor will be constant.

$$\mathrm{\tau_{s} \: = \: \frac{K_{1}R_{2}}{(R_{2}^{2} \: + \: X_{2}^{2})} \:\quad\: \dotso \: (3)}$$

Where, K1 = K E22 is another constant.

Condition for Maximum Starting Torque of 3-Phase Induction Motor

As the starting torque is given by,

$$\mathrm{\tau_{s} \: = \: \frac{K_{1}R_{2}}{(R_{2}^{2} \: + \: X_{2}^{2})} \: = \: \frac{K_{1}}{\left(R_{2} \: + \: \frac{X_{2}^{2}}{R_{2}}\right)} \:\quad\: \dotso \: (4)}$$

To be the maximum of the stating torque, the denominator of the eqn. (4) should be minimum, i.e.,

$$\mathrm{\frac{d}{dt}(R_{2} \: + \: \frac{X_{2}^{2}}{R_{2}}) \: = \: 0}$$

$$\mathrm{\Rightarrow \: 1 \: - \: \frac{X_{2}^{2}}{R_{2}^{2}} \: = \: 0}$$

$$\mathrm{\Rightarrow \: R_{2} \: = \: X_{2} \:\quad\: \dotso \: (5)}$$

Hence, the stating will be maximum when,

$$\mathrm{\text{Rotor Resistance / Phase } \: = \: \text{Rotor Reactance / Phase at standstill}}$$

Under the condition of maximum starting torque, the rotor power factor angle 2 = 45° and the rotor power factor is 0.707 lagging.

Effect of Change of Supply Voltage on Starting Torque of 3-Phase Induction Motor

$$\mathrm{\because \: \tau_{s} \: = \: \frac{KE_{2}^{2}R_{2}}{(R_{2}^{2} \: + \: X_{2}^{2})}}$$

$$\mathrm{\because \: E2 \: \propto \: \text{Supply Voltage (V)}}$$

$$\mathrm{\therefore \: \tau_{s} \: = \: \frac{KV_{2}R_{2}}{(R_{2}^{2} \: + \: X_{2}^{2})}}$$

$$\mathrm{\Rightarrow \: \tau_{s} \: \propto \: V^{2} \:\quad\: \dotso \: (6)}$$

Hence, the starting torque is directly proportional to the square of the supply voltage. Therefore, the starting torque is very sensitive to changes in the value of the supply voltage.

Important −

  • Starting Torque of Squirrel Cage Motors − For the squirrel cage motors, the starting torque is very low about 1.5 to 2 times of the full-load value.
  • Starting Torque of Wound Rotor Motors − In case of slip ring induction motors, the resistance of the rotor circuit can be increased by inserting external resistance. By adding the proper value of the external resistance (i.e., R2 = X2), maximum starting torque can be obtained.

Numerical Example

A 100 kW, 3 kV, 50 Hz, 8-pole, star connected induction motor has a star connected slip ring rotor with a turn ratio of 2.5 (stator/rotor). The rotor resistance is 0.2 Ω/phase and its per phase leakage inductance is 4 mH. The stator impedance may be neglected. Find the starting torque on rated voltage with short circuited slip rings.

Solution

Trasformation ratio,

$$\mathrm{K \: = \: \frac{Rotor \: turns / phase}{Stator \: turns / phase} \: = \:\frac{1}{2.5} \: = \: 0.4}$$

$$\mathrm{\text{Rotor Resistance / Phase referred to stator, } \: R'_{2} \: = \: \frac{R_{2}}{K_{2}} \: = \: \frac{0.2}{(0.4)^{2}} \: = \: 1.25 \Omega}$$

The reactance of the rotor circuit is,

$$\mathrm{X_{2} \: = \: 2\pi fL \: = \: 2\pi \: \times \: 50 \: \times \: (4 \: \times \: 10^{-3}) \: = \: 1.256 \: \Omega}$$

Rotor reactance/phase referred to stator,

$$\mathrm{X'_{2} \: = \: \frac{X_{2}}{K^{2}} \: = \: \frac{1.256}{0.4^{2}} \: = \: 7.85 \: \Omega}$$

Now, supply voltage/phase,

$$\mathrm{X_{1} \: = \: \frac{3000}{\sqrt{3}} \: = \: 1732 \: V}$$

Therefore, the starting torque of the motor is,

$$\mathrm{\tau_{s} \: = \: \frac{K E^{2}_{2} R_{2}}{R_{2}^{2} \: + \: X_{2}^{2}} \: = \: \frac{3}{2\pi N_{s}} \: \times \: \frac{E_{1}^{2} R'_{2}}{(R'_{2})^{2} \: + \: (X'2)^{2}}}$$

Where,

$$\mathrm{N_{s} \: = \: \frac{120f}{P} \: = \: \frac{120 \: \times \: 50}{8} \: = \: 750 \: RPM \: = \: 12.5 \:rps}$$

And,

$$\mathrm{K \: = \: \frac{3}{2\pi N_{s}} \: ; \: \text{ and } \: E2 \: \propto \: E1}$$

$$\mathrm{\therefore \: \tau_{s} \: = \: \frac{3}{2\pi N_{s}} \: \times \: \frac{E^{2}_{1} R'_{2}}{(R'_{2})^{2} \: + \: (X'_{2})^{2}}}$$

$$\mathrm{= \: \left(\frac{3}{2 \pi \: \times \: 12.5}\right) \: \times \: \left(\frac{(1732)^{2} \: \times \: 1.25}{1.252 \: + \: (7.85)^{2}}\right)}$$

$$\mathrm{= \: 2267 \: Nm}$$

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