Three-Phase Induction Motor: Slip, Power Flow and Torque Calculations
The three-phase squirrel cage induction motor drives most of the pumps, compressors, fans, conveyors and machine tools in Indian industry. It has no brushes or commutator, needs very little maintenance, and starts on its own from a three-phase supply. Its operation rests on one idea: a rotating magnetic field that drags a short-circuited rotor along behind it.
This article covers synchronous speed, slip, rotor frequency, the power flow from input to shaft, the torque-slip characteristic and a full worked example for a 7.5 kW, 4-pole motor.
Rotating Magnetic Field and Slip
Three stator windings displaced by 120 electrical degrees and fed with three-phase currents produce a magnetic field of constant magnitude that rotates at synchronous speed Ns. The field cuts the rotor bars and induces currents in them. These currents, interacting with the field, produce torque that turns the rotor in the same direction as the field.
The rotor can never reach synchronous speed. If it did, there would be no relative motion, no induced current and no torque. The difference between synchronous and actual speed, as a fraction of synchronous speed, is the slip.
Ns = 120 f / P (rpm) s = (Ns − N) / Ns N = Ns (1 − s) Rotor current frequency fr = s × f
| Poles | Ns at 50 Hz | Typical full-load speed |
|---|---|---|
| 2 | 3000 rpm | about 2850 to 2950 rpm |
| 4 | 1500 rpm | about 1420 to 1480 rpm |
| 6 | 1000 rpm | about 940 to 980 rpm |
| 8 | 750 rpm | about 700 to 735 rpm |
Full-load slip is typically a few percent, smaller for large efficient motors and larger for small ones. The nameplate speed tells you the slip directly: a 4-pole motor marked 1440 rpm has s = (1500 − 1440)/1500 = 0.04, and its rotor currents alternate at only 0.04 × 50 = 2 Hz.
Power Flow from Supply to Shaft
Electrical input first supplies stator copper and iron losses. The remainder crosses the air gap to the rotor as air-gap power Pg. The rotor copper loss is exactly slip times air-gap power, and the rest is converted to mechanical power. Friction and windage then reduce it to the useful shaft output.
Input = √3 VL IL cos φ Pg = Input − stator copper loss − stator iron loss Rotor copper loss = s × Pg Mechanical power developed Pm = (1 − s) × Pg Shaft output = Pm − friction and windage Pg : rotor Cu loss : Pm = 1 : s : (1 − s) Torque (N·m) = 9.55 × Power (W) / Speed (rpm)
The ratio 1 : s : (1 − s) is very useful. It shows that a motor running at high slip wastes a large share of its air-gap power as heat in the rotor. This is why reducing voltage to control the speed of a cage motor is inefficient.
Worked Example: 7.5 kW, 415 V, 4-Pole Motor
A 415 V, 50 Hz, 4-pole induction motor delivers 7.5 kW at 1440 rpm. Friction and windage loss is 180 W and stator losses (copper plus iron) total 450 W. Power factor at full load is 0.85. Find air-gap power, rotor copper loss, input, efficiency, line current and shaft torque.
- Slip: s = (1500 − 1440)/1500 = 0.04.
- Mechanical power developed: Pm = 7500 + 180 = 7680 W.
- Air-gap power: Pg = Pm/(1 − s) = 7680/0.96 = 8000 W.
- Rotor copper loss: s × Pg = 0.04 × 8000 = 320 W.
- Input power: 8000 + 450 = 8450 W.
- Efficiency: 7500/8450 = 88.8%.
- Line current: I = 8450/(√3 × 415 × 0.85) = 8450/611 = 13.8 A.
- Shaft torque: T = 9.55 × 7500/1440 = 49.7 N·m.
| Stage | Power (W) |
|---|---|
| Electrical input | 8450 |
| Stator losses | 450 |
| Air-gap power | 8000 |
| Rotor copper loss | 320 |
| Mechanical power developed | 7680 |
| Friction and windage | 180 |
| Shaft output | 7500 |
Torque-Slip Characteristic
Using the rotor equivalent circuit with standstill rotor EMF E2, rotor resistance R2 and standstill reactance X2 per phase, the torque is:
T = k × s E2² R2 / (R2² + (s X2)²) Maximum (breakdown) torque occurs at s_max = R2 / X2 T_max = k × E2² / (2 X2) (independent of R2) Near synchronous speed (small s): T ∝ s (linear) At large s: T ∝ 1/s
Three useful conclusions follow. First, in the normal running region torque is almost proportional to slip, so speed drops only slightly with load. Second, maximum torque does not depend on rotor resistance, but the slip at which it occurs does; adding external resistance in a slip-ring rotor raises starting torque. Third, torque is proportional to E2², so a 10% supply voltage dip reduces torque by about 19%, which can stall a heavily loaded pump.
Common Mistakes
- Calculating rotor copper loss as s × input power instead of s × air-gap power.
- Using the line current from the nameplate with phase voltage in the power formula.
- Assuming a motor reverses by swapping all three phases; swapping any two reverses rotation.
- Forgetting that a lightly loaded induction motor has a poor power factor because its magnetising current is nearly constant.
Summary
Synchronous speed is 120f/P and the rotor always runs slightly slower, at slip s. Air-gap power splits into rotor copper loss and mechanical power in the ratio s : (1 − s). Torque rises almost linearly with slip in the running region up to a breakdown value fixed by rotor reactance. With these equations you can read any motor nameplate and calculate its torque, current and losses.