Double-Cage Induction Motor Torque-Slip Characteristics and Cage Torques Comparison

Torque-Slip Characteristics of Double-Cage Induction Motor

In a double-cage induction motor, it is assumed that the two cages develop two separate torques. Thus, the total torque developed in the double-cage induction motor is equal to the sum of the two cage torques. The torque-slip characteristics of the two cages and the total torque of the motor is shown in the figure.

By changing the individual cage resistances and leakage reactances, the resultant torque-speed characteristics can be modified according to the requirement. The resistances can be changed by changing the cross-sectional area of the rotor bars while the leakage reactance can be changed by changing the width of the slot openings and the depth of the inner cage.

Comparison of Cage Torques of Double-Cage Induction Motor

Refer the equivalent circuit of the double-cage induction motor shown in Figure-1 −

The per phase power developed by the inner rotor cage is given by,

$$\mathrm{P_{di} \: = \: \frac{I′_{ri}^{2} \: R'_{ri}}{s} \:\: \dotso \: (1)}$$

And the per phase power developed by the outer rotor cage is,

$$\mathrm{P_{do} \:=\: \frac{I′_{ro}^2 \: R'_{ro}}{s} \:\: \dotso \: (2)}$$

Therefore, the total power developed per phase by both the cages is,

$$\mathrm{P_{d} \:=\: P_{di} \:+\: P_{do} \:=\: \frac{I′_{ri}^{2} \: R'_{ri}}{s} \: +\: \frac{I′_{ro}^2 \: R'_{ro}}{s} \:\: \dotso \: (3)}$$

Also,

$$\mathrm{I′_{ro} \:=\: \frac{E′_r}{Z′_{ro}} \:\: \dotso \: (4)}$$

$$\mathrm{I′_{ri} \:=\: \frac{E′_r}{Z′_{ri}} \:\: \dotso \: (5)}$$

$$\mathrm{Z′_{ro} \:=\: \sqrt{\left(\frac{R′_{ro}}{s}\right)^{2} \:+\: (X′_{ro})^{2}} \:\: \dotso \: (6)}$$

And,

$$\mathrm{Z′_{ri} \:=\: \sqrt{\left(\frac{R′_{ri}}{s}\right)^{2} \:+\: (X′_{ri})^{2}} \:\: \dotso \: (7)}$$

Now, let,

• τ_di = Torque developed by the inner cage
• τ_do = Torque developed by the outer cage
• τ_d = Total torque developed by the two cages

Thus, the total power developed and hence the total developed torque is given by,

$$\mathrm{P_{d} \:=\: 2\pi \: n_{s} \: \tau_{d}}$$

$$\mathrm{\therefore \: \tau_{d} \:=\: \frac{P_d}{2\pi \:n_{s}}}$$

$$\mathrm{\Rightarrow \: \tau_{d} \:=\: \frac{1}{2 \pi \: n_{s}}\left(\frac{I′_{ri}^{2}\:R'_{ri}}{s} \:+\: \frac{I′_{ro}^{2}\:R'_{ro}}{s}\right) \:\: \dotso \: (8)}$$

Hence, the ratio of the outer cage torque and the inner cage will be,

$$\mathrm{\frac{\tau_{do}}{\tau_{di}} \:=\: \frac{\left(\frac{R′_{ro}}{s}\right)^{2} \:+\: (X′_{ro})^{2}}{\left(\frac{R′_{ro}}{s}\right)^{2} \:+\: (X′_{ri})^{2}} \:\: \dotso \: (9)}$$