Ratios of Full-Load, Starting and Maximum Torques of 3-Phase Induction Motor

Digital ElectronicsElectronElectronics & Electrical

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

$$\mathrm{\tau_{𝐹.𝐿} \propto\frac{𝑠𝑅_2}{𝑅_2^2} + (𝑠𝑋_2)^2… (1)}$$

Where, s is slip corresponds to full-load.

The starting torque is given by,

$$\mathrm{\tau_𝑠 \propto \frac{𝑅2}{𝑅_2^2 + 𝑋_2^2} … (2)}$$

And the maximum torque is given by,

$$\mathrm{\tau_𝑚 \propto\frac{1}{2 𝑋_2}… (3)}$$

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,

$$\mathrm{\frac{\tau_𝑚}{\tau_{𝐹.𝐿}}=\frac{(𝑅_{2}⁄𝑋_{2})^2 + 𝑠^2}{2 𝑠 (𝑅_{2}⁄𝑋_{2})}}$$

$$\mathrm{⇒\frac{\tau_𝑚}{\tau_{𝐹.𝐿}}=}$$

$$\mathrm{\frac{𝑠_𝑚^2 + 𝑠^2}{2 𝑠 𝑠_𝑚}… (4)}$$

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,

$$\mathrm{\frac{\tau_𝑚}{\tau_𝑠}=\frac{(𝑅_{2}⁄𝑋_{2})^2 + 1}{2 (𝑅_{2}⁄𝑋_{2})}}$$

$$\mathrm{⇒\frac{\tau_𝑚}{\tau_𝑠}=\frac{𝑠_𝑚^2 + 1}{2 𝑠_𝑚}… (5)}$$

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,

$$\mathrm{Slip\:at\:maximum\:torque, 𝑠𝑚 =\frac{𝑅_2}{𝑋_2}=\frac{0.02}{0.2}= 0.1 \Omega}$$

$$\mathrm{\therefore\frac{\tau_𝑚}{\tau_{𝐹.𝐿}}=\frac{𝑠_𝑚^2 + 𝑠^2}{2 𝑠 𝑠_𝑚}=\frac{0.1^2 + 0.03^2}{2 × 0.03 × 0.1}= 1.82}$$

(ii) Ratio of maximum torque to starting torque −

$$\mathrm{\frac{\tau_𝑚}{\tau_𝑠}=\frac{𝑠_𝑚^2 + 1}{2 𝑠_𝑚}=\frac{0.1^2 + 1}{2 × 0.1}= 5.05}$$

raja
Published on 26-Aug-2021 08:15:38
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