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T-S characteristics and condition for maximum torque

Short answer

Torque-slip curve is linear at low slip, peaks at sm, then decreases. Maximum torque condition is dT/ds=0, giving R2 = sm X2, so sm = R2/X2. Maximum torque equals 3E2^2/(2 omega_s X2) and is independent of rotor resistance.

Explain the torque-slip characteristics of an induction motor. Show the condition for which the maximum torque develops in the induction motor.

Torque-slip characteristics

The torque-slip curve shows the variation of the torque of a 3-phase induction motor with slip, at constant supply voltage and frequency.

Torque of a 3-phase induction motor at slip ss (per-phase equivalent):

T=3ωs⋅sE22R2R22+(sX2)2 N⋅mT = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2 + (sX_2)^2}\ \text{N·m}

where E2E_2 and X2X_2 are the rotor emf and reactance per phase at standstill and ωs\omega_s is the synchronous speed in rad/s.

 Torque T
   |          Tmax
   |         .--*--.
   |  Tst   /   |   \          Operating
   |    *--'    |    \          region
   |            |     \___ ___ (stable)
   |            |     
   +------------+-----------+----> slip s
   1 (standstill) sm       0 (sync)
  • At standstill (s=1s = 1): starting torque TstT_{st}, with high rotor current but a poor power factor.
  • At small slip: sX2≪R2sX_2\ll R_2, so T∝sT\propto s. The curve is nearly a straight line, and this is the stable operating region (full-load slip 2–5%). Load increases give a small fall in speed.
  • At s=sms = s_m: the torque reaches its maximum value TmaxT_{max}.
  • At large slip: T∝1/sT\propto 1/s, so torque decreases as ss rises. This part is unstable for normal loads.
  • At synchronous speed (s=0s = 0): torque is zero.

Condition for maximum torque

Let K=3E22ωsK = \dfrac{3E_2^2}{\omega_s}. Then

T=K sR2R22+s2X22T = K\,\frac{sR_2}{R_2^2 + s^2X_2^2}

For the maximum value, dTds=0\dfrac{dT}{ds} = 0:

dTds=K (R22+s2X22)R2−sR2(2sX22)(R22+s2X22)2=0\frac{dT}{ds} = K\,\frac{(R_2^2 + s^2X_2^2)R_2 - sR_2(2sX_2^2)}{(R_2^2 + s^2X_2^2)^2} = 0 R22+s2X22−2s2X22=0  ⇒  R22=s2X22R_2^2 + s^2X_2^2 - 2s^2X_2^2 = 0 \;\Rightarrow\; R_2^2 = s^2X_2^2 sm=R2X2s_m = \frac{R_2}{X_2}

So the torque is maximum when the rotor resistance equals the rotor reactance at slip sms_m, that is R2=smX2R_2 = s_mX_2 (the rotor power factor is then 0.7070.707 lagging).

Substituting s=sms = s_m:

Tmax=K smR22R22=3E222ωsX2T_{max} = K\,\frac{s_mR_2}{2R_2^2} = \frac{3E_2^2}{2\omega_sX_2}

TmaxT_{max} is independent of the rotor resistance, is proportional to V2V^2 and inversely proportional to X2X_2.

Also: TTmax=2ssms2+sm2\dfrac{T}{T_{max}} = \dfrac{2ss_m}{s^2 + s_m^2}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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