Induction motor torque equation and torque-slip characteristic

The torque developed by a 3-phase induction motor is proportional to the stator flux, the rotor current and the rotor power factor: $T \propto \phi I_2\cos\phi_2$.

Derivation

At standstill let rotor emf $= E_2$, rotor resistance $= R_2$, rotor reactance $= X_2$ (per phase). When running at slip $s$:

• rotor emf $E_{2r} = sE_2$
• rotor reactance $X_{2r} = sX_2$ (since $f_2 = sf$)
• rotor resistance remains $R_2$

     R2       sX2
  +-[===]----[~~~]--+
  |                 |
 (~) sE2     I2 ->  |
  +-----------------+

$$I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}}$$

Since $E_2 \propto \phi$, $\phi = E_2/k_1$:

$$\begin{aligned} T_R &= k\,\phi\, I_2\cos\phi_2 \\ &= k\frac{E_2}{k_1}\cdot\frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + s^2 X_2^2} \end{aligned}$$

From the power method, $T_R = \dfrac{P_{ag}}{\omega_s} = \dfrac{3I_2^2R_2/s}{2\pi N_s/60}$, so $K = \dfrac{3}{2\pi N_s/60}$ (N·m when $E_2$ in volts, $R_2, X_2$ in ohms).

Torque–slip characteristic

 T
 |          Tmax
 |         .--.
 |       .'    '.
 |  Tfl /        '.
 |  ...*           '-._
 |    /|               '-- Tst
 |   / |  T ~ 1/s
 |  /  |  (unstable)
 | / T~s
 |/(stable)
 |_____|_______________|__ s
 0     sm              1
 N=Ns                 N=0

1. At $s = 0$ ($N = N_s$): $T = 0$.
2. Low slip ($s < s_m$): $sX_2 \ll R_2$, so $T \approx \dfrac{KE_2^2}{R_2}s$, i.e. $T \propto s$. Nearly straight line. Normal operating region; stable; speed nearly constant (full-load slip 2–5%).
3. Maximum torque: $dT/ds = 0$ gives $s_m = R_2/X_2$ and $T_{max} = \dfrac{KE_2^2}{2X_2}$, independent of $R_2$, proportional to $V^2$. Usually 2–3 times full-load torque.
4. High slip ($s > s_m$): $sX_2 \gg R_2$, so $T \approx \dfrac{KE_2^2R_2}{sX_2^2}$, i.e. $T \propto 1/s$ (rectangular hyperbola). Unstable: extra load reduces speed and torque further, and the motor stalls.
5. At $s = 1$: starting torque $T_{st} = \dfrac{KE_2^2R_2}{R_2^2 + X_2^2}$.

Increasing $R_2$ moves $s_m$ towards 1 (higher starting torque) without changing $T_{max}$; reducing supply voltage lowers the whole curve as $V^2$.