Max torque condition and torque-slip curve

Condition for maximum torque

The torque of a three-phase induction motor is

$$T=\frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2+(sX_2)^2}$$

where $E_2$, $R_2$, $X_2$ are the rotor emf, resistance and reactance per phase at standstill, and $s$ is the slip. For a constant supply, $E_2$ is constant. Differentiate with respect to $s$ and equate to zero (equivalently, maximise $\frac{sR_2}{R_2^2+s^2X_2^2}$):

$$\frac{d}{ds}\left[\frac{sR_2}{R_2^2+s^2X_2^2}\right]=0 \Rightarrow R_2^2+s^2X_2^2-2s^2X_2^2=0$$ $$R_2^2=s^2X_2^2 \Rightarrow s_m=\frac{R_2}{X_2}$$

Result: torque is maximum when the rotor resistance equals the rotor reactance at that slip ($R_2=s_mX_2$). Substituting,

$$T_{max}=\frac{3}{\omega_s}\cdot\frac{E_2^2}{2X_2}$$

$T_{max}$ is independent of the rotor resistance, but the slip at which it occurs ($s_m=R_2/X_2$) depends on it.

Torque-slip characteristic

 Torque
   |                      Tmax
   |                    .--*--.
   |               _.-''       \
   |          _.-''             \
   |     _.-''                   \
   *--'''                         \
  Tst                              \
   +--------------------------------*---> slip
   s=1                     sm      s=0
  (start)                         (sync)

• At $s=1$ (standstill) the torque is the starting torque $T_{st}$, which is lower than $T_{max}$.
• From $s=1$ to $s_m$, torque rises to $T_{max}$ (unstable region; here $R_2\ll sX_2$, so $T\propto 1/s$).
• Between $s_m$ and zero (the normal working region), $T\propto s$ and the curve is nearly straight and steep; a small slip (2-5%) gives full-load torque, so the speed is nearly constant.
• At $s=0$ (synchronous speed) the torque is zero.
• Increasing the rotor resistance shifts $s_m$ towards 1 and improves the starting torque; when $R_2=X_2$, $T_{st}=T_{max}$.