Three-phase Induction Motor - Ratios of Full-Load, Starting and Maximum Torques

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Three-phase Induction Motor Full-Load Torque

For a 3-phase induction motor, the full-load torque is given by,

$$\mathrm{\tau_{F.L} \:\propto\: \frac{sR_2}{R_{2}^{2}} \: + \: (sX_{2})^{2} \:\: \dotso \: (1)}$$

Where, s is slip corresponds to full-load.

The starting torque is given by,

$$\mathrm{\tau_{s} \: \propto \: \frac{R2}{R_{2}^{2} \: + \: X_{2}^{2}} \:\: \dotso \: (2)}$$

And the maximum torque is given by,

$$\mathrm{\tau_{m} \: \propto \: \frac{1}{2 X_{2}} \:\: \dotso \: (3)}$$

Therefore,

(1) Ratio of maximum torque to full-load torque −

$$\mathrm{\frac{\tau_{m}}{\tau_{F.L}} \:=\: \frac{R_{2}^{2} \: + \: (sX_{2})^{2}}{2 s R_{2} X_{2}}}$$

Dividing the numerator and denominator on RHS by $X_{2}^{2}$, we have,

$$\mathrm{\frac{\tau_{m}}{\tau_{F.L}} \:=\: \frac{(R_{2} / X_{2})^{2} \:+\: s^{2}}{2 s (R_{2} / X_{2})}}$$

$$\mathrm{\Rightarrow \: \frac{\tau_{m}}{\tau_{F.L}} \: = \: \frac{s_{m}^{2} \:+\: s^{2}}{2 s s_{m}} \:\: \dotso \: (4)}$$

Where, $\mathrm{s_{m} \:=\: \frac{R_{2}}{X_{2}}}$ (Slip Corresponding to Maximum Torque)

(2) Ratio of maximum torque to starting torque −

$$\mathrm{\frac{\tau_{m}}{\tau_{s}} \:=\: \frac{R_{2}^{2} \:+\: X_{2}^{2}}{2 R_{2} X_{2}}}$$

Dividing the numerator and denominator on RHS by $X_{2}^{2}$, we have,

$$\mathrm{\frac{\tau_{m}}{\tau_{s}} \:=\: \frac{(R_{2} / X_{2})^{2} \:+\: 1}{2 (R_{2} / X_{2})}}$$

$$\mathrm{\Rightarrow \: \frac{\tau_{m}}{\tau_{s}} \:=\: \frac{s_{m}^{2} \:+\: 1}{2 s_{m}} \:\: \dotso \: (5)}$$

Numerical Example

A 50 Hz, 6-pole induction motor has full-load slip of 3%. The rotor resistance and standstill reactance are 0.02 Ω and 0.2 Ω per phase respectively. Determine −

(i) the ratio of maximum torque to full-load torque, and

(ii) the ratio of maximum torque to starting torque.

Solution

(i) Ratio of maximum torque to full-load torque −

Here,

$$\mathrm{\text{Slip at maximum torque, } \:sm \: = \: \frac{R_{2}}{X_{2}} \:=\: \frac{0.02}{0.2} \:=\: 0.1 \: \Omega}$$

$$\mathrm{\therefore \: \frac{\tau_{m}}{\tau_{F.L}} \:=\: \frac{s_{m}^{2} \:+\: s^{2}}{2 s s_{m}} \:=\: \frac{(0.1)^{2} \:+\: (0.03)^{2}}{2 \: \times \: 0.03 \: \times \: 0.1} \:=\: 1.82}$$

(ii) Ratio of maximum torque to starting torque −

$$\mathrm{\frac{\tau_{m}}{\tau_{s}} \:=\: \frac{s_{m}^{2} \: + \: 1}{2 s_{m}} \:=\: \frac{(0.1)^{2} \: + \: 1}{2 \: \times \: 0.1} \:=\: 5.05}$$