Three phase induction motors

The three-phase induction motor is the workhorse of industry — about two-thirds of industrial electrical load is induction motors driving pumps, fans, compressors, conveyors and mills. It is simple, rugged, cheap and needs little maintenance. This topic covers construction (squirrel cage and slip-ring/phase-wound), the rotating-field principle and slip, the equivalent circuit, torque–slip characteristics and the effect of rotor resistance, starting methods, the no-load and blocked-rotor tests, and speed control.

Key concepts

• Construction: stator — laminated core with a 3-phase distributed winding; rotor — laminated core with either:
  • Squirrel cage: bars of aluminium/copper short-circuited by end rings; slots skewed to reduce noise and cogging. Cheap, robust, no slip rings; constant rotor resistance.
  • Phase-wound (slip-ring): 3-phase star winding brought out to slip rings so external resistance can be added. Higher starting torque, lower starting current, speed control possible; costlier, more maintenance.
• Principle: balanced 3-phase currents in the stator create a rotating magnetic field (RMF) of constant magnitude ($1.5\Phi_m$) at synchronous speed $N_s = 120f/P$. It cuts the rotor conductors, induces emf and current; the rotor follows the field (Lenz) but always slower — it is an "asynchronous" machine. Reversal: swap any two supply phases.
• Slip: $s = (N_s - N)/N_s$; typically 2–5% at full load. Rotor frequency $f_r = sf$ (a few hertz at full load).
• Equivalent circuit (per phase, referred to stator): $R_1, X_1$ — stator; shunt $R_0 \parallel X_m$ — core loss and magnetising (large magnetising current, 25–40%, because of the air gap → poor no-load pf); rotor $R_2', X_2'$; load represented by $R_2'(1-s)/s$.
• Power flow: air-gap power $P_g$ → rotor copper loss $sP_g$ → mechanical power $(1-s)P_g$. So $P_g : P_{cu2} : P_m = 1 : s : (1-s)$.
• Torque: $T = \dfrac{3}{\omega_s}\cdot\dfrac{sE_2^2R_2}{R_2^2 + (sX_2)^2}$.
  • Low slip region: $T \propto s$ (linear, stable operating region).
  • High slip region: $T \propto 1/s$.
  • Maximum (breakdown/pull-out) torque at $s_m = R_2/X_2$; its value $T_m \propto E_2^2/(2X_2)$ is independent of rotor resistance.
  • $T \propto V^2$ — a 10% voltage drop reduces torque by about 19%.
• Effect of rotor resistance: increasing $R_2$ shifts the peak to higher slip (lower speed) without changing $T_{max}$; starting torque increases until $R_2 = X_2$, when starting torque = maximum torque. Hence slip-ring motors use added rotor resistance at start; double-cage and deep-bar rotors achieve the same automatically.
• Starting methods (starting current with DOL is 5–7× rated):

Method	Starting current	Starting torque
DOL	$I_{sc}$	$T_{st}$ (full) — small motors (≲5 kW)
Star–delta	1/3 of DOL (line)	1/3 of DOL
Auto-transformer (tap x)	$x^2$ of DOL	$x^2$ of DOL
Stator resistance/reactance	reduced by x	$x^2$
Rotor resistance (slip ring)	low	high
Soft starter / VFD	controlled	controlled

• No-load test: rated voltage, no load → no-load input = core loss + friction/windage (+ small Cu); gives $R_0, X_m$.
• Blocked-rotor test: rotor locked, reduced voltage to circulate rated current → full-load copper loss and equivalent $R, X$ (like the transformer SC test).
• Speed control: from stator side — V/f control (VFD; keeps flux constant; most common today), pole changing (two-speed motors, Dahlander), stator voltage control (fans). From rotor side (slip ring) — rotor resistance, slip-power recovery (Kramer, Scherbius), cascade.
• Crawling (running stable at about 1/7 of $N_s$ due to 7th harmonic) and cogging (refusal to start when stator and rotor slots equal/aligned) — reduced by skewing and proper slot combination.

High-yield points

• $N_s = 120f/P$; $s = (N_s - N)/N_s$; rotor frequency $f_r = sf$.
• $P_g : P_{cu2} : P_m = 1 : s : (1 - s)$.
• Maximum torque is independent of $R_2$; it occurs at $s_m = R_2/X_2$.
• Starting torque equals maximum torque when $R_2 = X_2$.
• $T \propto V^2$; star–delta starting gives 1/3 of DOL current and torque.
• Swapping any two supply phases reverses rotation.

Formulas

$N_s = \frac{120 f}{P},\quad s = \frac{N_s - N}{N_s},\quad P_{cu2} = s\,P_g,\quad P_m = (1-s)P_g$

$\frac{T_{st}}{T_{fl}} = \left(\frac{I_{st}}{I_{fl}}\right)^2 s_{fl}\ \text{(DOL)}$

Worked example

A 4-pole, 50 Hz motor runs at 1440 rpm with rotor input 10 kW. Find slip, rotor Cu loss, mechanical power and rotor frequency.

1. $N_s = 1500$ rpm; $s = 60/1500 = 0.04$.
2. $P_{cu2} = 0.04\times10 = 0.4$ kW.
3. $P_m = 9.6$ kW.
4. $f_r = 0.04\times50 = 2$ Hz.

Exam tips

• Maximum torque is independent of $R_2$; slip at maximum torque is proportional to $R_2$.
• Star–delta: current and torque both 1/3 of DOL.
• $T \propto V^2$; rotor frequency = sf.
• Swapping two phases reverses rotation.
• Blocked-rotor test ≈ transformer short-circuit test; no-load test ≈ open-circuit test.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. In the equivalent circuit of an induction motor referred to the stator, the resistance $R_2'(1-s)/s$ represents the:

• A. Core loss
• B. Stator copper loss
• C. Rotor copper loss
• D. Mechanical power developed (load)

Answer: D. The power absorbed by $R_2'(1-s)/s$ is the gross mechanical power developed, $P_{mech} = (1-s)P_{ag}$.

2. A squirrel-cage motor draws 6 times its full-load current on direct-on-line starting and has a full-load slip of 4 %. With a star–delta starter, the starting torque as a fraction of full-load torque is:

• A. 0.48
• B. 0.24
• C. 2.0
• D. 1.44

Answer: A. $T_{st}/T_{fl} = (I_{st}/I_{fl})^2 s_{fl} = 36\times0.04 = 1.44$ on DOL; star–delta reduces it to one third: $1.44/3 = 0.48$.

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