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Starting Torque of 3-Phase Induction Motor; Torque Equation of 3-Phase Induction Motor
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 Ο2)
Therefore,
$$\mathrm{\tau \propto πΈ_2πΌ_2 cos \varphi_2}$$
$$\mathrm{⇒ \tau = πΎπΈ_2πΌ_2 cos \varphi_2 … (1)}$$
Where, K is the constant of proportionality.
Starting Torque of 3-Phase Induction Motor
Let,
- Rotor resistance/Phase = π 2
- Rotor reactance/Phase at standstill = π2
- Rotor EMF/Phase at standstill = E2
∴ Rotor impedance/Phase at standstill,
$$\mathrm{π_2 = \sqrt{π _2^2 + π2^2}}$$
Rotor current/Phase at standstill,
$$\mathrm{πΌ_2 =\frac{πΈ_2}{π_2}=\frac{πΈ_2}{\sqrt{π _2^2 + π2^2}}}$$
And, Rotor power factor at standstill,
$$\mathrm{cos\varphi_2 =\frac{π _2}{π_2}=\frac{π _2}{\sqrt{π _2^2 + π2^2}}}$$
∴ Starting torque,
$$\mathrm{\tau_π = πΎπΈ_2πΌ_2 cos\varphi_2 = πΎπΈ_2 × (\frac{πΈ_2}{\sqrt{π _2^2 + π2^2}}) × (\frac{π _2}{\sqrt{π _2^2 + π2^2}})}$$
$$\mathrm{⇒ \tau_π =\frac{πΎ πΈ_2^2 π _2}{(π _2^2 + π_2^2)}… (2)}$$
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.
$$\mathrm{\tau_π =\frac{πΎ_1 π _2}{(π _2^2 + π2^2)}… (3)}$$
Where, K1 = K E22 is another constant.
Condition for Maximum Starting Torque of 3-Phase Induction Motor
As the starting torque is given by,
$$\mathrm{\tau_π =\frac{πΎ_1 π _2}{(π _2^2 + π2^2)}=\frac{πΎ_1}{(π _2 +\frac{π_2^2}{π _2})}… (4)}$$
To be the maximum of the stating torque, the denominator of the eqn. (4) should be minimum, i.e.,
$$\mathrm{\frac{π}{ππ‘}(π _2 +\frac{π_2^2}{π _2}) = 0}$$
$$\mathrm{⇒ 1 −\frac{π_2^2}{π _2^2} = 0}$$
$$\mathrm{⇒ π _2 = π2 … (5)}$$
Hence, the stating will be maximum when,
$$\mathrm{ππ¨ππ¨π«\:πππ¬π’π¬πππ§ππ/ππ‘ππ¬π = ππ¨ππ¨π«\:πππππππ§ππ/ππ‘ππ¬π\:ππ\:π¬πππ§ππ¬ππ’π₯π₯}$$
Under the condition of maximum starting torque, the rotor power factor angle Ο2 = 45° and the rotor power factor is 0.707 lagging.
Effect of Change of Supply Voltage on Starting Torque of 3-Phase Induction Motor
$$\mathrm{\because \tau_π =\frac{πΎ πΈ_2^2 π _2}{(π _2^2 + π_2^2)}}$$
$$\mathrm{\because πΈ2 \propto Supply\:voltage (π)}$$
$$\mathrm{\therefore \tau_π =\frac{πΎ π_2 π _2}{(π _2^2 + π_2^2)}}$$
$$\mathrm{⇒ \tau_π \propto π^2 … (6)}$$
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,
$$\mathrm{πΎ =\frac{Rotor\:turns/phase}{Stator\:turns/phase} =\frac{1}{2.5}= 0.4}$$
$$\mathrm{Rotor \:resistance/phase\: referred \:to \:stator, π ′_2 =\frac{π _2}{πΎ_2} =\frac{0.2}{0.4^2}= 1.25 \Omega}$$
The reactance of the rotor circuit is,
$$\mathrm{π_2 = 2πππΏ = 2π × 50 × (4 × 10^{−3}) = 1.256 \Omega;}$$
Rotor reactance/phase referred to stator,
$$\mathrm{π′_2 =\frac{π_2}{πΎ^2} =\frac{1.256}{0.4^2} = 7.85 \Omega}$$
Now, supply voltage⁄phase,
$$\mathrm{πΈ_1 =\frac{3000}{\sqrt{3}}= 1732 V}$$
Therefore, the starting torque of the motor is,
$$\mathrm{\tau_π =\frac{πΎ πΈ_2^2 π _2}{(π _2^2 + π_2^2)}=\frac{3}{2ππ_π }×\frac{πΈ_1^2π '_2}{(π ′_2)^2 + (π′2)^2}}$$
Where,
$$\mathrm{π_π =\frac{120π}{π}=\frac{120 × 50}{8}= 750 RPM = 12.5 rps}$$
And,
$$\mathrm{πΎ =\frac{3}{2ππ_π }; and πΈ2 \propto πΈ1}$$
$$\mathrm{\therefore \tau_π =\frac{3}{2ππ_π }×\frac{πΈ_1^2π '_2}{(π ′_2)^2 + (π′2)^2}}$$
$$\mathrm{= (\frac{3}{2π × 12.5}) × (\frac{(1732)^2 × 1.25}{1.252 + 7.85^2})}$$
$$\mathrm{= 2267 Nm}$$
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