Three-phase induction motor: principle and equivalent circuit

Why it matters

The three-phase squirrel-cage induction motor drives most of the pumps, fans, compressors, conveyors and agitators in any plant — it is cheap, rugged and needs almost no maintenance. To size a motor, read its nameplate, predict its current and torque, or set up a variable-frequency drive, you need slip, the per-phase equivalent circuit and the power-flow split that follows from it.

Key ideas

Construction. A laminated stator carries a distributed three-phase winding. The rotor is either a squirrel cage (aluminium or copper bars short-circuited by end rings; no brushes, very robust) or a wound (slip-ring) rotor with a three-phase winding brought out through slip rings so that external resistance can be added for starting or speed control. The air gap is kept as small as possible (fractions of a millimetre) to keep the magnetising current and hence the no-load current low.

Rotating magnetic field. Three windings displaced 120° in space and carrying currents displaced 120° in time produce a field of constant magnitude (1.5 times the peak of one phase) rotating at synchronous speed Ns = 120·f/P. Swapping any two supply leads reverses the direction of rotation.

Slip. The field cuts the rotor conductors, induces EMFs and currents in them, and the force on these currents drags the rotor along (Lenz's law: the rotor tries to reduce the relative motion). The rotor can never reach Ns, because then there would be no relative motion, no induced current and no torque. Slip s = (Ns − N)/Ns, typically 2–6 % at full load. At standstill s = 1.

Rotor quantities depend on slip

• Rotor frequency f2 = s·f (a few hertz at full load).
• Rotor EMF per phase E2s = s·E2, where E2 is the standstill value.
• Rotor reactance X2s = s·X2.
• Rotor current I2 = s·E2 / √(R2² + (s·X2)²) = E2 / √((R2/s)² + X2²). The second form is the key step: dividing by s turns the moving rotor into a stationary circuit at supply frequency with load-dependent resistance R2/s.

Equivalent circuit (per phase, referred to stator). Exactly like a transformer: stator R1, X1; shunt magnetising branch Rc ∥ Xm; rotor X2' and a resistance R2'/s. Split R2'/s = R2' + R2'·(1 − s)/s: the first part is the real rotor copper loss, the second is a fictitious "load resistance" whose power is the mechanical power developed. Compared with a transformer, the magnetising current is much larger (25–40 % of rated, because of the air gap), so the no-load power factor is poor, and leakage reactance is larger.

Power flow. Input P_in → minus stator copper and core losses → air-gap power Pag (crosses to the rotor) → splits into rotor copper loss s·Pag and gross mechanical power (1 − s)·Pag → minus friction and windage → shaft output. So Pag : Pcu2 : Pm = 1 : s : (1 − s). Electromagnetic torque T = Pag/ωs = Pm/ωr. Large slip therefore means a large fraction of the air-gap power is burnt in the rotor — operating at high slip is inherently inefficient.

Formulas

• Ns = 120·f / P — synchronous speed (rpm); f in Hz, P poles.
• s = (Ns − N) / Ns — slip (dimensionless); N rotor speed (rpm).
• f2 = s·f ; E2s = s·E2 ; X2s = s·X2 — rotor frequency (Hz), EMF (V) and reactance (Ω) at slip s.
• I2 = E2 / √((R2/s)² + X2²) — rotor current per phase (A).
• Pag = 3·I2'²·R2'/s — air-gap power (W), three-phase.
• Pcu2 = s·Pag ; Pm = (1 − s)·Pag — rotor copper loss and gross mechanical power (W).
• T = Pag / ωs = Pm / ωr ; ωs = 2π·Ns/60 — electromagnetic torque (N·m).
• T = 3·V1²·(R2'/s) / {ωs·[(R1 + R2'/s)² + (X1 + X2')²]} — torque from the approximate circuit, V1 the per-phase voltage.

Worked examples

Example 1 (standard): slip and rotor quantities. Given: 4-pole, 50 Hz induction motor running at 1440 rpm; standstill rotor EMF 100 V per phase, rotor resistance 0.1 Ω and standstill reactance 0.5 Ω per phase.

1. Ns = 120 × 50/4 = 1500 rpm.
2. s = (1500 − 1440)/1500 = 0.04 (4 %).
3. Rotor frequency f2 = 0.04 × 50 = 2 Hz; rotor EMF E2s = 0.04 × 100 = 4 V per phase.
4. Rotor current: I2 = 100/√((0.1/0.04)² + 0.5²) = 100/√(6.25 + 0.25) = 39.2 A.

Example 2 (GATE level): power flow. Given: 4-pole, 50 Hz motor; input 40 kW; stator copper + core loss 1.5 kW; slip 4 %; friction and windage 0.96 kW.

1. Air-gap power: Pag = 40 − 1.5 = 38.5 kW.
2. Rotor copper loss: Pcu2 = s·Pag = 0.04 × 38.5 = 1.54 kW.
3. Gross mechanical power: Pm = 38.5 − 1.54 = 36.96 kW; shaft output = 36.96 − 0.96 = 36.0 kW; efficiency = 36/40 = 90 %.
4. Electromagnetic torque: ωs = 2π × 1500/60 = 157.1 rad/s; T = 38500/157.1 = 245 N·m.
5. Shaft torque: ωr = 0.96 × 157.1 = 150.8 rad/s; T_shaft = 36000/150.8 = 238.7 N·m.

Common mistakes

• Writing slip in rpm (that is slip speed Ns − N) when a fraction is asked.
• Forgetting that rotor frequency, EMF and reactance all scale with s.
• Using R2' instead of R2'/s in the torque equation numerator.
• Using line voltage in a per-phase circuit; for a star connection V_phase = V_line/√3.
• Computing torque as P_out/ωs; electromagnetic torque is Pag/ωs (or Pm/ωr), shaft torque is P_out/ωr.
• Thinking the rotor can reach synchronous speed on no load.

For GATE IN

Very regular: Ns and slip from nameplate data, rotor frequency (also asked as "frequency of rotor current" for a tachometer-type sensor), Pag : Pcu2 : Pm ratios, torque from the approximate equivalent circuit, and why the no-load power factor is low. Practise per-phase calculations with star/delta conversions.

Quick check

1. A 6-pole, 50 Hz motor runs at 960 rpm. Find s and f2.
2. Air-gap power 10 kW at s = 0.05. What is the rotor copper loss?
3. Why must there be slip?
4. Why is the magnetising current of an induction motor larger than that of a transformer?

Answers: 1. Ns = 1000 rpm, s = 0.04, f2 = 2 Hz. 2. 500 W. 3. Without relative motion there is no rotor EMF, current or torque. 4. The air gap adds a large reluctance.