Induction Motors: Three-Phase and Single-Phase

What to remember

• Synchronous speed Ns = 120 f / P rpm. Slip s = (Ns − N)/Ns. Rotor frequency = s f. The air-gap power splits as Pg : rotor copper loss : mechanical power = 1 : s : (1 − s).
• Torque is proportional to s E2² R2/(R2² + (sX2)²). Maximum torque occurs at s = R2/X2. Its value does not depend on rotor resistance, but is proportional to V².
• A single-phase induction motor has no starting torque. It needs an auxiliary starting arrangement: split-phase, capacitor-start, capacitor-run or shaded-pole.

1. Construction and working

The stator has a three-phase winding in slots on a laminated core. The rotor is of two types:

• Squirrel-cage rotor: bars of aluminium or copper short-circuited by end rings. It is rugged, cheap and needs little maintenance. Slots are often skewed to reduce noise and cogging.
• Slip-ring (wound) rotor: a three-phase winding brought out to slip rings and brushes. External resistance can be added for high starting torque and low starting current, and for speed control.

A three-phase supply gives the stator a rotating magnetic field of constant magnitude (1.5 times the peak flux of one phase) rotating at synchronous speed:

Ns = 120 f / P rpm (P = number of poles).

This field cuts the rotor conductors and induces EMF. Rotor currents flow and the rotor turns in the direction of the field (Lenz's law). The rotor can never reach Ns, otherwise there would be no relative motion and no induced EMF.

Reversing the rotation: interchange any two stator supply leads.

2. Slip and rotor quantities

• Slip s = (Ns − N)/Ns. At standstill s = 1; at synchronous speed s = 0. For a normal motor, full-load slip is a few percent (small motors higher). For an induction generator (rotor above Ns), s is negative. For plugging (rotor running against the field), s lies between 1 and 2.
• Rotor speed: N = Ns (1 − s).
• Rotor EMF at slip s: E2s = s E2. Rotor reactance: X2s = s X2. Rotor frequency: f2 = s f.
• Rotor current I2 = s E2 / √(R2² + (s X2)²) = E2 / √((R2/s)² + X2²). The rotor power factor = R2/√(R2² + (sX2)²). At small slip the rotor reactance sX2 is small, so the rotor power factor is near unity. It falls as slip rises.

3. Power flow and efficiency

Let Pg be the air-gap power (rotor input).

• Rotor copper loss = s Pg
• Mechanical power developed Pm = (1 − s) Pg
• So Pg : Pcu2 : Pm = 1 : s : (1 − s)
• Rotor efficiency = Pm/Pg = 1 − s = N/Ns.
• Torque developed T = Pg/ωs = Pm/ωm, where ωs = 2πNs/60.
• Output shaft power = Pm − friction and windage loss.
• Input power − stator copper loss − stator core loss = Pg.

4. Torque-slip characteristic

T = (3/ωs) × s E2² R2 / (R2² + (s X2)²) (E2 is per-phase standstill EMF).

Stable operation lies on the part of the curve where slip is below sm (T rises with slip). Adding rotor resistance in a slip-ring motor shifts the peak torque toward higher slip (towards standstill) but does not change its value. Near full load, T ∝ s.

Ratio form: T/Tmax = 2 s sm /(s² + sm²).

Crawling: a cage motor sometimes runs steadily at about one-seventh of synchronous speed because of the seventh space harmonic. Cogging (locking): the rotor refuses to start because the numbers of stator and rotor slots are unsuitably related.

5. Equivalent circuit and tests

The per-phase equivalent circuit is like a transformer's: R1, X1 in the stator, a shunt branch (R0 and Xm) and the rotor branch with X2′ and R2′/s. The rotor branch resistance is split into R2′ (copper loss) and R2′(1 − s)/s (the load resistance representing mechanical power).

The circle diagram is drawn from these two tests and gives the full-load current, power factor, torque, maximum output and efficiency graphically.

6. Starting methods

A cage motor draws high starting current, which causes a voltage dip in the supply. Therefore reduced-voltage starting is used.

7. Speed control

Speed N = Ns (1 − s) = (120 f / P)(1 − s). Control by:

• 1. Changing the supply frequency (V/f control): the best method; the V/f ratio is held constant so flux stays constant. It uses a variable-frequency drive (inverter).
• 2. Changing the number of poles: pole-changing (consequent-pole) windings give discrete speeds (cage motors).
• 3. Varying the supply voltage: a small range; torque falls as V².
• 4. Rotor resistance control: wound-rotor motors only; speed falls, loss is high.
• 5. Slip power recovery (Kramer, Scherbius): recovers energy from the rotor circuit.

8. Single-phase induction motors

A single-phase winding produces a pulsating field. This can be resolved into two fields of equal magnitude rotating in opposite directions (double-revolving-field theory). At standstill they give equal and opposite torques, so the starting torque is zero. The motor will run in either direction if started by hand.

Cross-field theory also explains the running torque. All types need an auxiliary winding to start:

In a capacitor-start motor the auxiliary-winding current leads the main-winding current by a large angle (close to 90°) at start, which gives a higher starting torque than the resistance split-phase type. The centrifugal switch opens at about 70 to 80 percent of synchronous speed.

9. Induction generator

If the rotor is driven above synchronous speed (negative slip), the machine returns active power to the supply. It needs reactive power from the grid or a capacitor bank. It is used in wind turbines because it needs no synchronisation.

Worked example: A 6-pole, 50 Hz motor runs at 960 rpm. Ns = 120 × 50/6 = 1000 rpm; s = (1000 − 960)/1000 = 0.04; rotor frequency = 0.04 × 50 = 2 Hz. If air-gap power is 10 kW, rotor copper loss = 0.4 kW and Pm = 9.6 kW.

Exam traps

• 1. The rotor can never run at synchronous speed; the synchronous motor can.
• 2. Maximum torque is independent of rotor resistance, but the slip at maximum torque is proportional to it.
• 3. Torque is proportional to V², not V. Starting torque falls by 36 percent when voltage falls by 20 percent.
• 4. In star-delta starting, both current and torque are one-third of DOL values. Torque is not retained.
• 5. Rotor frequency is s f, not f. At standstill it equals supply frequency.
• 6. Single-phase induction motors have no starting torque; they are not self-starting.
• 7. Slip is negative in generator mode and between 1 and 2 in plugging.
• 8. Crawling is due to the seventh harmonic; cogging is due to the slot combination.

One-liners

• 1. 4-pole, 50 Hz motor: Ns = 1500 rpm.
• 2. Rotor copper loss = s × air-gap power.
• 3. Mechanical power = (1 − s) × air-gap power.
• 4. Slip-ring motor gives high starting torque with low current.
• 5. Skewed rotor slots reduce cogging and noise.
• 6. No-load power factor of an induction motor is low.
• 7. Star-delta starter reduces starting current to one-third.
• 8. V/f control is the best speed control method for cage motors.
• 9. Capacitor-start motor has a high starting torque for a single-phase motor.
• 10. Shaded-pole motor has the lowest starting torque and the lowest cost.
• 11. Induction generator needs reactive power from outside.
• 12. Reversal of rotation: swap two supply leads.

Practice questions

1. The synchronous speed of a 4-pole, 50 Hz induction motor is

   1. 1000 rpm
   2. 750 rpm
   3. 1500 rpm
   4. 3000 rpm
   Answer

   C. 1500 rpm

   Ns = 120f/P = 120×50/4 = 1500 rpm.

2. A 6-pole, 50 Hz motor runs at 960 rpm. Its slip is

   1. 4%
   2. 6%
   3. 40%
   4. 2%
   Answer

   A. 4%

   Ns = 1000; s = 40/1000 = 0.04.

3. The rotor current frequency of a motor with slip 0.04 on a 50 Hz supply is

   1. 50 Hz
   2. 2 Hz
   3. 1.2 Hz
   4. 4 Hz
   Answer

   B. 2 Hz

   f2 = s f = 0.04 × 50 = 2 Hz.

4. The slip of an induction motor at standstill is

   1. 0
   2. 0.5
   3. 1
   4. infinity
   Answer

   C. 1

   N = 0 gives s = (Ns − 0)/Ns = 1.

5. An induction motor can never run at synchronous speed because

   1. the stator would burn
   2. the frequency would double
   3. the rotor would be magnetised permanently
   4. no EMF would be induced in the rotor
   Answer

   D. no EMF would be induced in the rotor

   No relative motion means no rotor EMF and no torque.

6. Rotor copper loss of an induction motor equals

   1. air-gap power / slip
   2. (1 − slip) × air-gap power
   3. slip × shaft output
   4. slip × air-gap power
   Answer

   D. slip × air-gap power

   Pg : Pcu : Pm = 1 : s : (1 − s).

7. An induction motor has air-gap power 10 kW at slip 0.04. The mechanical power developed is

   1. 9.6 kW
   2. 10.4 kW
   3. 0.4 kW
   4. 9.0 kW
   Answer

   A. 9.6 kW

   Pm = (1 − s)Pg = 0.96 × 10 = 9.6 kW.

8. The rotor efficiency of an induction motor at slip 0.05 is

   1. 90%
   2. 95%
   3. 5%
   4. 99.5%
   Answer

   B. 95%

   Rotor efficiency = 1 − s = 0.95.

9. An induction motor has air-gap power 5 kW and synchronous angular speed 100 rad/s. The torque developed is

   1. 5 N·m
   2. 500 N·m
   3. 50 N·m
   4. 0.02 N·m
   Answer

   C. 50 N·m

   T = Pg/ωs = 5000/100 = 50 N·m.

10. Rotor copper loss is 150 W at slip 0.03. The rotor input (air-gap power) is

    1. 4.5 W
    2. 150 kW
    3. 0.5 kW
    4. 5 kW
    Answer

    D. 5 kW

    Pg = Pcu/s = 150/0.03 = 5000 W.

11. The slip at which maximum torque occurs is

    1. R2/X2
    2. X2/R2
    3. 1/(R2 X2)
    4. R2 × X2
    Answer

    A. R2/X2

    sm = R2/X2, where X2 is standstill rotor reactance.

12. A rotor has R2 = 0.2 Ω and standstill X2 = 1 Ω per phase. The slip at maximum torque is

    1. 0.8
    2. 0.2
    3. 1.2
    4. 5
    Answer

    B. 0.2

    sm = R2/X2 = 0.2.

13. Adding resistance in the rotor circuit of a slip-ring motor

    1. has no effect on starting torque
    2. increases the slip at maximum torque without changing its value
    3. decreases the maximum torque
    4. increases the maximum torque
    Answer

    B. increases the slip at maximum torque without changing its value

    Tmax is independent of R2; sm moves toward standstill.

14. If the supply voltage falls to 90% of rated, the maximum torque becomes

    1. 81%
    2. 95%
    3. 73%
    4. 90%
    Answer

    A. 81%

    T ∝ V², so 0.9² = 0.81.

15. If the supply voltage falls to 80%, the starting torque falls to

    1. 90%
    2. 40%
    3. 80%
    4. 64%
    Answer

    D. 64%

    T ∝ V²; 0.8² = 0.64.

16. Statements about torque in an induction motor: 1. Maximum torque is independent of rotor resistance. 2. Torque is proportional to the supply voltage.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    Torque is proportional to V², so only statement 1 is correct.

17. A cage motor draws 600 A on direct starting. With a star-delta starter, the line current at start is

    1. 346 A
    2. 300 A
    3. 200 A
    4. 600 A
    Answer

    C. 200 A

    Star-delta reduces line current to one-third.

18. In star-delta starting, the starting torque compared with DOL is

    1. the same
    2. one-third
    3. one-ninth
    4. three times
    Answer

    B. one-third

    Phase voltage falls by √3, torque by 3.

19. An auto-transformer starter with a 50% tap gives starting current from the supply and starting torque, each as a fraction of DOL, of

    1. one-half
    2. three-quarters
    3. one-third
    4. one-quarter
    Answer

    D. one-quarter

    Both scale as x² = 0.25.

20. The star-delta starter can be used for motors

    1. with slip rings
    2. normally connected in star
    3. normally connected in delta
    4. of single-phase type
    Answer

    C. normally connected in delta

    The motor is started in star and changed to delta for running.

21. The slip-ring induction motor is chosen over the squirrel-cage when

    1. speed must be constant
    2. maintenance must be minimal
    3. high starting torque with low starting current is needed
    4. lowest cost is needed
    Answer

    C. high starting torque with low starting current is needed

    External rotor resistance gives this.

22. The best method of speed control of a cage induction motor, with good efficiency over a wide range, is

    1. stator resistance control
    2. variable voltage, variable frequency (V/f) control
    3. rotor resistance control
    4. varying the number of slots
    Answer

    B. variable voltage, variable frequency (V/f) control

    Frequency changes Ns while V/f keeps flux constant.

23. A 2-pole motor needs a supply of what frequency for a synchronous speed of 3000 rpm?

    1. 50 Hz
    2. 60 Hz
    3. 100 Hz
    4. 25 Hz
    Answer

    A. 50 Hz

    f = Ns P/120 = 3000 × 2/120 = 50 Hz.

24. A motor has Ns = 600 rpm at 50 Hz. The number of poles is

    1. 8
    2. 12
    3. 6
    4. 10
    Answer

    D. 10

    P = 120 f / Ns = 6000/600 = 10.

25. When the poles of a 50 Hz motor are changed from 4 to 8, the synchronous speed

    1. stays the same
    2. doubles
    3. halves
    4. becomes one-fourth
    Answer

    C. halves

    Ns ∝ 1/P.

26. Crawling in a squirrel-cage motor is caused by

    1. shortage of rotor resistance
    2. third harmonic current
    3. low supply voltage
    4. the seventh space harmonic of the air-gap flux
    Answer

    D. the seventh space harmonic of the air-gap flux

    The harmonic produces a second stable speed at about one-seventh of Ns.

27. Cogging in cage motors is reduced by

    1. increasing air gap
    2. skewing the rotor slots
    3. using aluminium bars
    4. using wound rotors
    Answer

    B. skewing the rotor slots

    Skew also reduces noise.

28. The no-load test on an induction motor gives

    1. core loss and friction and windage loss
    2. full-load copper loss
    3. starting torque
    4. maximum torque
    Answer

    A. core loss and friction and windage loss

    Rotor copper loss is negligible at no load.

29. The blocked-rotor test is equivalent to the following transformer test:

    1. Sumpner's test
    2. open-circuit test
    3. polarity test
    4. short-circuit test
    Answer

    D. short-circuit test

    The rotor is short-circuited electrically and locked; s = 1.

30. Match test with parameter: (a) No-load test (b) Blocked-rotor test 1. Equivalent resistance and leakage reactance 2. Magnetising branch and rotational loss

    1. a-2, b-1
    2. a-1, b-1
    3. a-2, b-2
    4. a-1, b-2
    Answer

    A. a-2, b-1

    No-load gives the shunt branch and rotational loss; blocked-rotor gives series impedance.

31. The slip of an induction generator is

    1. zero
    2. negative
    3. between 0 and 1
    4. greater than 1
    Answer

    B. negative

    The rotor runs above synchronous speed.

32. An induction generator requires

    1. a field winding on the rotor
    2. a governor to stay in synchronism
    3. reactive power from the grid or a capacitor bank
    4. a DC exciter
    Answer

    C. reactive power from the grid or a capacitor bank

    It has no independent excitation.

33. A single-phase induction motor has zero starting torque because

    1. its supply has no frequency
    2. a single-phase winding gives a pulsating field, equal to two opposite rotating fields
    3. slip is zero at start
    4. its rotor is of the wrong material
    Answer

    B. a single-phase winding gives a pulsating field, equal to two opposite rotating fields

    Equal and opposite torques cancel at standstill.

34. Which single-phase motor has the highest starting torque?

    1. Capacitor-start motor
    2. Shaded-pole motor
    3. Resistance split-phase motor
    4. Permanent split-capacitor motor
    Answer

    A. Capacitor-start motor

    A series capacitor gives a large angle between the winding currents.

35. Which motor is typically used in a ceiling fan?

    1. Shaded-pole motor on a lathe
    2. Universal motor of a drill
    3. Three-phase slip-ring motor
    4. Permanent split-capacitor motor
    Answer

    D. Permanent split-capacitor motor

    A ceiling fan needs low starting torque and quiet running.

36. The centrifugal switch in a capacitor-start motor

    1. reverses the rotation
    2. connects the capacitor permanently
    3. disconnects the starting winding near rated speed
    4. protects against overvoltage
    Answer

    C. disconnects the starting winding near rated speed

    It opens at about 70 to 80 percent of synchronous speed.

37. Statements about single-phase motors: 1. A shaded-pole motor has a very low starting torque. 2. A capacitor-start motor has a better starting torque than a resistance split-phase motor.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both are correct standard properties.

38. A shaded-pole motor is suited to

    1. heavy hoists
    2. small fans and low-torque devices
    3. traction
    4. compressors
    Answer

    B. small fans and low-torque devices

    It is cheap and simple but has low starting torque and low efficiency.

39. The direction of rotation of a three-phase induction motor is reversed by

    1. adding a capacitor
    2. changing the voltage
    3. interchanging all three leads
    4. interchanging any two supply leads
    Answer

    D. interchanging any two supply leads

    This reverses the phase sequence and field direction.

40. The rotating field of a three-phase stator has a magnitude of

    1. 1.5 times the maximum flux of one phase
    2. equal to the flux of one phase
    3. three times the flux of one phase
    4. 0.5 times the flux of one phase
    Answer

    A. 1.5 times the maximum flux of one phase

    Resultant of three pulsating fluxes is 3/2 Φm.

41. The no-load power factor of an induction motor is low because

    1. the supply is DC
    2. friction is zero
    3. the rotor is open
    4. the magnetising current is large because of the air gap
    Answer

    D. the magnetising current is large because of the air gap

    Magnetising current is a large share of no-load current; power factor about 0.1.

42. At small slip, the rotor power factor is

    1. near 0.5 lagging
    2. near zero
    3. near unity
    4. leading
    Answer

    C. near unity

    The rotor reactance sX2 is small compared with R2.

43. The torque of an induction motor at full load, with small slip, is approximately

    1. independent of slip
    2. proportional to slip
    3. proportional to the square of slip
    4. inversely proportional to slip
    Answer

    B. proportional to slip

    For s much less than sm, T ≈ 3 E2² s/(ωs R2).

44. In the equivalent circuit, the load resistance representing mechanical power is

    1. R2′(1 − s)/s
    2. R2′ s
    3. R2′(1 + s)
    4. R2′/s²
    Answer

    A. R2′(1 − s)/s

    R2′/s = R2′ + R2′(1 − s)/s.

45. The speed of a squirrel-cage motor can be changed in steps by

    1. adding rotor resistance
    2. pole changing
    3. moving the commutator
    4. changing the brushes
    Answer

    B. pole changing

    Consequent-pole windings give discrete speeds.