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Torque-slip curve: starting, maximum and operating region

Short answer

Torque-slip curve: stable region s < sm (T ∝ s, sX2 ≪ R2), unstable region s > sm (T ∝ 1/s). Start torque Tst = K E2² R2/(R2²+X2²). Maximum torque at sm = R2/X2, Tmax = K E2²/(2X2), independent of R2. Normal operation is on linear part with slip 2–5%.

Draw and explain the torque-slip (speed) characteristics of 3-phase induction motor, showing clearly the starting torque, maximum torque and normal operating region.

The torque–slip (speed) characteristic of a 3-phase induction motor is the graph of developed torque against slip (or speed) at constant supply 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
 |              B  Tmax
 |            .---.
 |          .'     '.
 |    Tfl  /         '.
 |   ...../ A          '._
 |       /|               '-.  C
 |      / |                  '-- Tst
 |     /  |    unstable          |
 |    /   | <-- region -->       |
 |   /    |                      |
 |  /     |                      |
 |_/______|______________________|__ s
 0  sfl   sm                     1
 N=Ns                           N=0
 |<-normal->|
   operating

Starting torque (point C, s=1s = 1, N=0N = 0)

Tst=KE22R2R22+X22T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2}

At standstill the rotor reactance X2X_2 is large compared with R2R_2, so rotor power factor is low and starting torque is moderate (about 1.5–2 times full-load torque for cage motors), even though starting current is 5–7 times full-load. Increasing R2R_2 (slip-ring motors) raises TstT_{st}, up to TmaxT_{max} when R2=X2R_2 = X_2.

Maximum torque (point B)

Putting dT/ds=0dT/ds = 0 gives R2=sX2R_2 = sX_2:

sm=R2X2,Tmax=KE222X2s_m = \frac{R_2}{X_2}, \qquad T_{max} = \frac{K E_2^2}{2X_2}

TmaxT_{max} is the pull-out (breakdown) torque, usually 2–3 times full-load torque. It is independent of R2R_2 and proportional to V2V^2. If load torque exceeds TmaxT_{max}, the motor stalls.

Regions of the curve

  1. Normal operating (stable) region, 0<s<sm0 < s < s_m: sX2≪R2sX_2 \ll R_2, so T≈(KE22/R2)sT \approx (KE_2^2/R_2)s, i.e. T∝sT \propto s — almost a straight line. The full-load point A lies here, at slip 2–5%. If load increases, speed falls a little and torque rises to match: stable, nearly constant speed.
  2. Unstable region, sm<s<1s_m < s < 1: sX2≫R2sX_2 \gg R_2, so T≈KE22R2/(sX22)T \approx KE_2^2R_2/(sX_2^2), i.e. T∝1/sT \propto 1/s — a rectangular hyperbola. A rise in load lowers speed, which lowers torque further, so the motor slows to a stop. The motor only passes through this region while accelerating.
  3. Beyond the motoring range: s<0s < 0 (N>NsN > N_s) gives generating action; s>1s > 1 (rotor driven against the field) gives braking (plugging).
PointSlipTorque
Synchronous speed00
Full load0.02–0.05TflT_{fl}
MaximumR2/X2R_2/X_2KE22/(2X2)KE_2^2/(2X_2)
Starting1KE22R2/(R22+X22)KE_2^2R_2/(R_2^2 + X_2^2)

✓ Answer checked by Claude Opus 5.5 (high effort)

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