Induction motor torque and condition for maximum torque

The torque of an induction motor is proportional to the product of the stator flux, the rotor current and the rotor power factor:

$$T \propto \phi\, I_2 \cos\phi_2$$

Derivation of running torque

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

• rotor frequency $f_2 = s f$
• rotor emf $= sE_2$
• rotor reactance $= sX_2$

Rotor current and power factor:

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

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

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

The same result follows from power: air-gap power $P_{ag} = 3I_2^2 R_2/s$ and $T = P_{ag}/\omega_s$, giving $K = 3/\omega_s = 3/(2\pi n_s)$ with $n_s$ in rps.

Torque–slip characteristic

 T
 |        Tmax
 |       .--.
 |     .'    '.
 |    /        '.
 |   /           '-._
 |  /                '-- Tst
 | / T ~ s   |  T ~ 1/s
 |/__________|___________ s
 0          sm          1
 N=Ns                 N=0

• Small $s$: $T \propto s$ (straight line) — stable running region.
• Large $s$: $T \propto 1/s$ (hyperbola) — unstable region.
• At $s = 1$: starting torque.

Condition for maximum torque

$E_2$, $R_2$, $X_2$ are constant, so $T$ is maximum when $\dfrac{dT}{ds} = 0$. Write

$$T = \frac{K E_2^2 R_2}{\dfrac{R_2^2}{s} + sX_2^2}$$

$T$ is maximum when the denominator $y = R_2^2/s + sX_2^2$ is minimum:

$$\begin{aligned} \frac{dy}{ds} &= -\frac{R_2^2}{s^2} + X_2^2 = 0 \\ \Rightarrow\ R_2 &= s X_2 \\ \Rightarrow\ s_m &= \frac{R_2}{X_2} \end{aligned}$$

($d^2y/ds^2 = 2R_2^2/s^3 > 0$, so it is a minimum of $y$, i.e. maximum of $T$.)

Condition: torque is maximum when rotor resistance equals rotor reactance at that slip, $R_2 = s X_2$.

Substituting $s_m$:

$$T_{max} = \frac{K (R_2/X_2) E_2^2 R_2}{R_2^2 + R_2^2} = \frac{K E_2^2}{2X_2}$$

Results:

1. $T_{max}$ does not depend on $R_2$; only the slip at which it occurs does.
2. $T_{max} \propto E_2^2 \propto V^2$ and $\propto 1/X_2$.
3. For maximum starting torque, $s_m = 1$, i.e. $R_2 = X_2$ (done with external rotor resistance in slip-ring motors).