T-S characteristics; why linear portion operation

Torque-slip characteristics

The torque produced by a three-phase induction motor at slip $s$ is

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

The curve starts at $T_{st}$ for $s=1$, rises to $T_{max}$ at $s_m = R_2/X_2$, then drops to zero at $s=0$.

 T
 |          Tmax
 |        _.-*-._
 |  Tst _-'      `-._
 |   *-'              `-.
 |                        `-.
 +--+--------+--------------*--> s
  s=1       s_m   linear   s=0
          (unstable)(stable)

• Between $s=1$ and $s_m$ the torque falls as slip rises ($T \propto 1/s$): unstable.
• Between $s_m$ and $0$ the torque is nearly proportional to slip ($T \propto s$): stable and almost linear.
• The maximum torque is $T_{max} = \frac{3E_2^2}{2\omega_s X_2}$.

Why the motor operates only in the linear portion

1. For small slip, $sX_2 \ll R_2$, so $T \approx \frac{3E_2^2}{\omega_s R_2}s \propto s$. The curve is a straight line.
2. In this region, if the load rises, the speed falls slightly, the slip rises and the torque increases in proportion. A new equilibrium is reached quickly, so the operation is stable.
3. Beyond $s_m$, a rise in load makes the speed fall, which reduces the torque, so the speed falls further and the motor stalls. The operation is unstable.
4. In the linear region the rotor resistance dominates the rotor reactance ($R_2 \gg sX_2$), so the rotor power factor is nearly unity. The slip is small (2 to 5 %), so the rotor copper loss ($= sP_g$) is small and the efficiency is high.
5. The speed remains almost constant (close to $N_s$) from no load to full load, which is the behaviour most drives need.

So the motor is run in the low-slip region of the curve, with full-load slip well below $s_m$.