Three-Phase and Single-Phase Induction Motors
Rotating magnetic field, synchronous speed and slip, equivalent circuit and power flow, torque-slip characteristic, maximum and starting torque, starting methods, speed control and single-phase motors.
The induction motor is the workhorse of industry: rugged, cheap and simple. Its rotor current is induced by the stator's rotating field, and it always runs slightly below synchronous speed.
Rotating field and slip
- Three-phase currents in a distributed winding create a field rotating at N_s = 120 f/P rpm.
- Slip s = (N_s − N)/N_s. Full-load slip is typically 2–5%.
- Rotor frequency f_r = s f; rotor EMF and reactance at slip s are s × their standstill values.
Power flow
From the equivalent circuit (rotor resistance R₂/s):
- Air-gap power P_ag = stator input − stator losses.
- Rotor copper loss = s P_ag.
- Mechanical power developed = (1 − s) P_ag.
- So P_ag : rotor Cu loss : P_mech = 1 : s : (1 − s).
- Torque T = P_ag/ω_s.
Torque–slip characteristic
T ∝ s R₂ E₂²/(R₂² + (s X₂)²)
- For small slip, T ∝ s (linear region where the motor runs). For large slip, T ∝ 1/s.
- Maximum (breakdown) torque at s_max = R₂/X₂; its value is independent of R₂.
- Starting torque is maximum when R₂ = X₂ (slip-ring motors add external rotor resistance for this).
- T ∝ V²: a 10% voltage drop reduces torque by about 19%.
Starting methods
| Method | Starting current (vs DOL) | Starting torque (vs DOL) |
|---|---|---|
| Direct-on-line (DOL) | 1 | 1 |
| Star–delta | 1/3 | 1/3 |
| Autotransformer (tap x) | x² | x² |
| Rotor resistance (slip ring) | Reduced | Increased |
Speed control
Stator voltage control (small range), V/f control with inverters (constant flux, wide range), pole changing, rotor resistance control, and slip power recovery.
Single-phase induction motors
A single-phase winding produces a pulsating field (two counter-rotating fields: double revolving field theory), so there is no starting torque. Starting methods: split-phase, capacitor-start, capacitor-run, shaded pole.
Worked example
A 4-pole, 50 Hz motor runs at a slip of 4% with air-gap power 10 kW.
- N_s = 1500 rpm; N = 1440 rpm; rotor frequency = 2 Hz.
- Rotor copper loss = 400 W; mechanical power = 9.6 kW.
Practice
A 6-pole, 50 Hz induction motor runs at 960 rpm. What is the slip?
Show answer and explanation
Answer: 0.04N_s = 1000 rpm; s = 40/1000 = 0.04.
An induction motor has R₂ = 0.2 Ω and X₂ = 1 Ω (rotor values at standstill). At what slip does maximum torque occur?
Show answer and explanation
Answer: 0.2s_max = R₂/X₂ = 0.2.
With star-delta starting, the starting torque compared with DOL starting is:
Show answer and explanation
Answer: CPhase voltage falls by √3 in star, so torque (∝ V²) falls to 1/3, as does the line current.
The supply voltage of an induction motor falls by 10%. By roughly what percentage does its torque at a given slip fall? (Give 1 decimal place.)
Show answer and explanation
Answer: 19T ∝ V², so the torque becomes 0.81 of its value: a 19% drop.