Induction motors: torque-speed characteristics and starting

Why it matters

The three-phase squirrel-cage induction motor drives most pumps, fans, conveyors and machine tools in Indian industry because it is cheap, rugged and needs almost no maintenance. To pick one for a mechatronic system you must know whether it can start the load, how much its speed drops under load, and how much current it pulls from the supply at switch-on.

Key ideas

• Rotating field. A balanced three-phase current in a three-phase stator winding produces a magnetic field of constant magnitude rotating at synchronous speed N_s = 120 f / P. For 50 Hz this gives 3000, 1500, 1000 and 750 rpm for 2, 4, 6 and 8 poles.
• Induction and slip. The rotor conductors (cage bars or a wound rotor) cut this field only if the rotor turns slower than the field. The relative speed is described by slip s = (N_s − N)/N_s. At standstill s = 1; at synchronous speed s = 0 and no torque is produced. Typical full-load slip is 2–6 % (lower for large motors).
• Rotor quantities scale with slip. The rotor emf and frequency are s·E₂ and s·f, so at full load the rotor currents are only a few hertz. The rotor reactance at running is s·X₂.
• Power flow. Of the air-gap power P_ag crossing to the rotor, a fraction s is lost as rotor copper loss and (1 − s) becomes gross mechanical power. This is why running at high slip (e.g. by rotor-resistance speed control) is inefficient.
• Torque–slip (torque–speed) curve. Using the approximate equivalent circuit, torque rises almost linearly with slip at small slip (the normal operating region, where the motor behaves like a stiff, nearly constant-speed drive), reaches a maximum called the breakdown (pull-out) torque at slip s_m, and then falls as slip increases towards standstill. The torque at s = 1 is the starting torque. Operation is stable only on the low-slip side of the peak, where torque falls as speed rises.
• Effect of rotor resistance. Maximum torque does not depend on rotor resistance; only the slip at which it occurs does (s_m ∝ R₂). Adding resistance to a slip-ring rotor moves the peak towards standstill and raises starting torque while reducing starting current. Double-cage and deep-bar rotors achieve a similar effect automatically.
• Effect of voltage. Torque at any slip is proportional to V², so a 10 % voltage dip cuts torque by about 19 %.
• Starting. At s = 1 the motor is like a short-circuited transformer and draws 5–8 times rated current direct-on-line (DOL). Methods:
  • DOL – full voltage; highest current and full starting torque. Used for small motors.
  • Star–delta – starts in star (phase voltage 1/√3 of delta), so both line current and torque are 1/3 of DOL values. Needs a delta-rated motor with six terminals.
  • Autotransformer – tapping ratio x gives motor current x·I_st(DOL), line current and torque x² times the DOL values.
  • Rotor resistance (slip-ring only) – reduces current and increases starting torque.
  • Soft starter / VFD – thyristor voltage ramp or variable frequency (see the next topic); a VFD can give rated torque at rated current from standstill.

Formulas

N_s = 120 f / P

• N_s synchronous speed (rpm), f supply frequency (Hz), P number of poles. ω_s = 2π N_s / 60 (rad/s).

s = (N_s − N) / N_s, f_r = s·f

• N rotor speed (rpm), f_r rotor current frequency (Hz).

P_cu2 = s·P_ag, P_mech = (1 − s)·P_ag, T = P_ag / ω_s

• P_ag air-gap power (W), P_cu2 rotor copper loss (W), P_mech gross mechanical power (W), T electromagnetic torque (N·m).

T = (3 / ω_s) · V² (R₂/s) / [ (R₁ + R₂/s)² + X² ]

• V stator phase voltage (V), R₁ stator resistance (Ω), R₂ rotor resistance referred to stator (Ω), X = X₁ + X₂ total leakage reactance referred to stator at supply frequency (Ω). Approximate equivalent circuit (magnetising branch moved to the terminals). When R₁ is neglected the bracket becomes (R₂/s)² + X².

s_m = R₂ / X, T_max = 3 V² / (2 ω_s X) (R₁ neglected)

T / T_max = 2 / (s/s_m + s_m/s) (R₁ neglected; handy ratio form)

Star–delta: I_line,start(Y) = I_line,start(Δ, DOL) / 3, T_start(Y) = T_start(DOL) / 3

Worked examples

Example 1 (standard). A 4-pole, 50 Hz induction motor runs at 1440 rpm with an air-gap power of 10 kW. Find slip, rotor frequency, rotor copper loss, gross mechanical power and torque.

1. N_s = 120 f / P = 120 × 50 / 4 = 1500 rpm.
2. s = (1500 − 1440)/1500 = 0.04.
3. f_r = s·f = 0.04 × 50 = 2 Hz.
4. P_cu2 = s·P_ag = 0.04 × 10 000 = 400 W; P_mech = 10 000 − 400 = 9600 W.
5. ω_s = 2π × 1500/60 = 157.08 rad/s; T = P_ag/ω_s = 10 000/157.08 = 63.66 N·m (same as P_mech/ω = 9600/150.80).

Answer: s = 0.04, f_r = 2 Hz, P_cu2 = 400 W, P_mech = 9.6 kW, T ≈ 63.7 N·m.

Example 2 (GATE level). A 400 V (line), 50 Hz, 4-pole, star-connected motor has R₂ = 0.2 Ω and total leakage reactance X = 1.0 Ω referred to the stator; neglect R₁ and the magnetising branch. Find (a) the slip at maximum torque, (b) the maximum torque, (c) the DOL starting torque, (d) the DOL starting line current.

1. Phase voltage V = 400/√3 = 230.94 V, V² = 53 333 V², ω_s = 157.08 rad/s.
2. (a) s_m = R₂/X = 0.2/1.0 = 0.2.
3. (b) T_max = 3V²/(2ω_s X) = 3 × 53 333/(2 × 157.08 × 1.0) = 509.3 N·m.
4. (c) At s = 1: T_st = (3/ω_s)·V²R₂/(R₂² + X²) = (3/157.08) × 53 333 × 0.2/(0.04 + 1) = 195.9 N·m. Check with the ratio form: T/T_max = 2/(1/0.2 + 0.2/1) = 2/5.2 = 0.3846, and 0.3846 × 509.3 = 195.9 N·m.
5. (d) I_st = V/√(R₂² + X²) = 230.94/1.0198 = 226.5 A (line current = phase current in star). For comparison, a delta-wound motor started through a star–delta starter draws one third of its DOL line current and develops one third of its DOL torque.
6. Note: with the ratio form, T at s = 0.04 is 2/(0.2 + 5) × 509.3 = 195.9 N·m too — the same torque as at standstill, because s·1 = s_m². Two slips give equal torque whenever their product is s_m².

Answer: s_m = 0.2, T_max ≈ 509 N·m, T_st ≈ 196 N·m, I_st ≈ 226 A (DOL).

Common mistakes

• Using the line voltage in the per-phase torque formula for a star motor (gives three times the torque). Always use phase voltage and multiply by 3 phases.
• Writing V²·R₂ instead of V²·R₂/s in the numerator; the slip must appear in both numerator and denominator.
• Using N_s in rpm instead of ω_s in rad/s in T = P_ag/ω_s.
• Thinking added rotor resistance raises maximum torque; it only shifts where it occurs.
• Saying starting torque is the maximum torque. For normal cage motors T_st is roughly 1.5–2.5 × rated but below T_max.
• Mixing up star–delta factors: line current and torque both drop to 1/3, not 1/√3.
• Assuming slip-ring starting applies to cage motors.

For GATE ME

Expect numericals on synchronous speed and slip, rotor frequency, the air-gap power split 1 : s : (1 − s), torque from the approximate equivalent circuit, slip and value of maximum torque, ratio of starting to maximum torque, and the current/torque reduction of star–delta and autotransformer starters. Conceptual MCQs test the shape of the torque–speed curve, the stable region, and the effect of voltage and rotor resistance. Practise doing the ratio form T/T_max = 2/(s/s_m + s_m/s) quickly.

Quick check

1. A 6-pole, 50 Hz motor runs at 960 rpm. What is the slip?
2. The air-gap power is 20 kW at 3 % slip. What is the rotor copper loss?
3. Does doubling rotor resistance change the maximum torque?
4. By what factor does a star–delta starter reduce starting torque?
5. Supply voltage falls by 10 %. Roughly how much does torque at a given slip fall?

Answers: 1. 0.04 (4 %). 2. 600 W. 3. No, only the slip at which it occurs (s_m doubles). 4. To one third of the DOL value. 5. To 0.81 of its value, about a 19 % drop.