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Torque-slip characteristic and linear operating region

Short answer

The T–s curve is linear (T ∝ s) for low slip and hyperbolic (T ∝ 1/s) beyond pull‑out. The motor operates only in the linear region because there dT/ds > 0, ensuring stable operation, high efficiency, low current, and near‑constant speed.

Explain T-s characteristics of 3-phase induction motor. Why the induction motor operates only in linear portion of T-s characteristics?

The T–s characteristic shows how the torque developed by a 3-phase induction motor varies with slip (speed) at constant voltage and frequency. It follows from

T=K s E22R2R22+(sX2)2T = \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2}
 T
 |          Tmax
 |         .--.
 |       .'    '.
 |      /  B     '.
 |     /           '-._
 |    / A              '-- Tst (C)
 |   /     |
 |  /stable|  unstable
 | /       |
 |/________|______________ s
 0        sm              1
 (Ns)                    (N=0)

Regions of the curve

  1. Low-slip region (0<s<sm0 < s < s_m): sX2sX_2 is small compared with R2R_2, so
T≈KE22R2 s⇒T∝sT \approx \frac{K E_2^2}{R_2}\, s \quad\Rightarrow\quad T \propto s

The curve is almost a straight line through the origin (region OA–B). 2. Maximum torque (B): at sm=R2/X2s_m = R_2/X_2, Tmax=KE22/(2X2)T_{max} = KE_2^2/(2X_2) — pull-out torque, usually 2–3 times full-load torque. 3. High-slip region (sm<s≤1s_m < s \le 1): sX2≫R2sX_2 \gg R_2, so

T≈KE22R2sX22⇒T∝1sT \approx \frac{K E_2^2 R_2}{s X_2^2} \quad\Rightarrow\quad T \propto \frac{1}{s}

The curve is a rectangular hyperbola, ending at starting torque TstT_{st} at s=1s = 1 (C). 4. For s<0s < 0 the machine generates; for s>1s > 1 it brakes (plugging).

Why the motor operates only in the linear portion

  1. Stability: in the linear region, if load torque increases, the speed falls slightly, slip rises and the motor torque rises to meet the new load; a new steady point is reached. So dT/ds>0dT/ds > 0 gives stable operation. In the region beyond sms_m, an increase in load reduces speed, slip increases and the motor torque falls; the speed keeps falling until the motor stalls. Operation there is unstable.
  2. Near-constant speed: in the linear region slip is only 2–5%, so speed changes very little from no load to full load, which most drives need.
  3. Efficiency: rotor copper loss =s×Pag= s \times P_{ag}. At low slip this loss is small; at high slip most air-gap power is wasted as heat in the rotor.
  4. Current and heating: rotor current rises sharply with slip; at high slip the current is near starting value (5–7 × FL), which would overheat the machine.
  5. Power factor: rotor power factor R2/R22+(sX2)2R_2/\sqrt{R_2^2 + (sX_2)^2} is high at low slip and poor at high slip.

Hence the motor is designed so that full-load torque lies on the linear part, well below TmaxT_{max}, leaving a margin for temporary overloads.

✓ Answer checked by Claude Opus 5.5 (high effort)

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