Induction Motor Torque-Speed Characteristics

• Locked-rotor torque (starting torque): This is the torque generated at zero speed when the rotor is stationary (locked). It is critical for starting the motor, especially in machines with heavy initial loads.
• Pull-up torque: The minimum torque produced by the motor during the acceleration phase, before it reaches synchronous speed.
• Break-down torque: Maximum torque available during acceleration period before the motor reaches its full-load (nominal) torque
• Full-load torque: The torque required to produce the rated power at full-load speed.
• Service factor (S.F.) The percentage multiplier of rated power (or load) that the motor can operate for a short time without overheating.

Examples

Induction Motor Spec Example Question

For the motor shown below:

• What would the pole number likely be?
• What would be the approximate rated slip value?
• What would be the approximate rated torque?

We can get the pole number from the synchronous speed of 1800.

$$\frac{120f}{P}=1800,f=60⟹P=4$$

We can get the slip value by noting that $\omega ={\omega }_r=1720$:

$${\omega }_s=1800⟹s=\frac{{\omega }_s-\omega }{{\omega }_s}=\frac{1800-1720}{1800}=0.044=4.4\%$$

And for torque:

$$T_r=\frac{P_r}{{\omega }_r}=\frac{0.5\text{ A}\times 550\text{ V}}{1720\times \frac{2\pi }{60}}=1.53\text{ lbf}\cdot \text{ft}$$

Induction Motor Example Question

Induction motor draws 25 A for a slip of 1%, and the corresponding torque is 2 Nm.

• Estimate the torque when the load is increased so that the motor slip is 3%.
• If the synchronous speed is 3600 rpm, how much does the speed decrease by changing the slip from 1% to 3%?

We can use the fact that torque-speed characteristics are approximately linear around rated speed and synchronous speed.

$$T_{3\%}=3\times T_{1\%}=3\times 2=6\text{ Nm}$$

The speed will be reduced by the corresponding slip change (i.e. 2%).

$$\begin{array}{rl}\Delta \omega & =\omega -{\omega }^{\prime }\\ & =(1-s){\omega }_s-(1-s^{\prime }){\omega }_s\\ & =(s^{\prime }-s){\omega }_s\\ & =(0.03-0.01)\times 3600\\ & =0.02\times 3600\\ & =72\text{ rpm}\end{array}$$