Torque equation of 3-phase induction motor

Derivation of torque

Let $E_2$ = rotor emf per phase at standstill, $R_2$, $X_2$ = rotor resistance and standstill reactance per phase, and $s$ = slip.

At slip $s$, the rotor emf is $sE_2$ and the reactance is $sX_2$, so the rotor current per phase is

$$I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}$$

Rotor power factor: $\cos\phi_2 = \frac{R_2}{\sqrt{R_2^2+(sX_2)^2}}$.

The torque equals the rotor power input (air-gap power) $P_g$ divided by the synchronous angular speed:

$$T = \frac{P_g}{\omega_s} = \frac{3I_2^2 R_2/s}{\omega_s}$$

Substituting $I_2$:

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}\ \text{N-m}, \qquad \omega_s = \frac{2\pi N_s}{60}$$

Since $E_2 \propto V_1$, $T \propto V_1^2$.

• Starting torque ($s=1$): $T_{st} = \frac{3}{\omega_s}\cdot\frac{E_2^2R_2}{R_2^2+X_2^2}$.

Torque-slip characteristic

 T
 |          Tmax
 |        _.-*-._
 |  Tst _-'      `-._
 |   *-'              `-.
 +--+--------+-----------*--> s
  s=1       s_m         s=0

Condition for maximum torque

For maximum $T$, $\frac{dT}{ds} = 0$. Writing $T \propto \frac{s R_2}{R_2^2 + s^2X_2^2}$:

$$\frac{(R_2^2 + s^2X_2^2)R_2 - sR_2(2sX_2^2)}{(R_2^2+s^2X_2^2)^2} = 0 \;\Rightarrow\; R_2^2 = s^2X_2^2$$ $$s_m = \frac{R_2}{X_2}$$

Maximum torque occurs when the rotor resistance equals the rotor reactance at that slip ($R_2 = s_mX_2$). Substituting:

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

The ratio $\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}$ is useful in numericals.