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Electrical Machines

Induction Motors_Note(1)

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Induction Motor

• Comparing with synchronous motor No dc field current is required to run the machine.

• Instead, amortisseur windings are installed in the rotor.

• The machine is called “Induction” because the rotor voltage is induced in the rotor windings rather than physically connected by wires.

• It is possible to use an induction machine as motor or generator, but there are many disadvantages to use it as generator.

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Induction Motor Components 1) Stator: Consisting of a steel frame

that supports a hollow, cylindrical core of stacked laminations. Slots on the internal circumference of the stator house the stator winding

2) Rotor: There are two different types of induction motor rotor:Squirrel cage rotor (Cage rotor):Consists of series of conducting bars laid into slots carved in the face of rotor and shorted at either end by large shorting rings.

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Wound rotor IM

•A wound rotor has a 3-phase winding, similar to the stator winding.

•The rotor winding terminals are connected to three slip rings which turn with the rotor. The slip rings/brushes allow external resistors to be connected in series with the winding.

•The external resistors are mainly used during start-up under normal running conditions the windings short circuited externally.

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Induction Motors Operation Principle

1) The three-phase stator is supplied by balanced three-phase voltage that drives an AC magnetizing current through each phase winding.

2) The magnetizing current in each phase generates a pulsating AC flux.

3) The total flux in the machine is the sum of the three fluxes.

4) The summation of the three AC fluxes results in a rotating flux, which turns with constant speed and has constant amplitude.

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5. The rotating flux induces a voltage in the short-circuited bars of the rotor. This voltage drives current through the bars.

6. The induced voltage is proportional with the difference of motor and synchronous speed. Consequently the motor speed is less than the synchronous speed

7. The interaction of the rotating flux and the rotor current generates a force that drives the motor.

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IM Rotating FieldConsider a simple stator with 6 salient poles - windings

AN, BN, CN. The windings are mechanically spaced at 120° from each other and connected to a 3-phase source.

AC currents Ia, Ib and Ic will flow in the windings, but will be displaced in time by 120°. Each winding produces its own MMF, which creates a flux across the hollow interior of the stator. The 3 fluxes combine to produce a magnetic field that rotates at the same frequency as the supply.

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The phase current waveforms follow each other in the sequence A-B-C.This produces a clockwise rotating magnetic field.

If we interchange any two of the lines connected to the stator, the new phase sequence will be A-C-B.This will produce a counterclockwise rotating field, reversing the motor direction.

Rotating Field - Direction of rotation

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Number of Poles – Synchronous Speed

1) The rotating speed of the revolving stator flux can be reduced by increasing the number of poles (in multiples of two). In a four-pole stator, the phase groups span an angle of 90°. In a six-pole stator, the phase groups span an angle of 60°.

2) This leads to the definition of synchronous speed (rotating stator flux speed):

Ns = 120 f/pwhere Ns = synchronous speed (rpm), f = frequency of the supply (Hz), p = number of poles

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Induction Motors SlipThe difference between the synchronous

speed and rotor speed can be expressed as a percentage of synchronous speed, known as the slip:

where s = slip, Ns = synchronous speed (rpm), N

= rotor speed (rpm)

At no-load, the slip is nearly zero (<0.1%). At full load, the slip for large motors rarely exceeds 0.5%. For small motors at full load, it rarely exceeds 5%. The slip is 100% for locked rotor.

X 100 %

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Frequency Induced In the Rotor

The frequency induced in the rotor depends on the slip:

where fR = frequency of voltage and current in the rotor, f = frequency of the supply and stator field, s = slip

fNs

NNsfR

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Examples

• Example 7-1

Equivalent circuit of induction motor

• Transformer Model of an Induction Motor

• Per-phase equivalent circuit of transformer:

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Induction Motors Equivalent circuit

Stator Rotor

Xsta Xrot_t Rrot_t

Irot_tVsta

Rsta

Vsup

IstaXmRc

Air gap

Rrot_t(1-s)/s

An induction motor can be described as rotating transformer, it is input is three phase voltage and current, the output of IM is shorted out so no electrical output exist, instead the output is mechanical. The Per phase equivalent circuit of an induction motor:

Stator: Resistance and self inductance , magnetization reactance XM and core resistance RC

Rotor: resistance and reactance

Transformer model of induction motor

Magnetization curve of induction motor

Transformer Model of Induction Motor

• Note: slope of induction motor’s magneto-motive force-flux curve is much shallower than curve of a good transformer

• because there is an air gap in an induction motor which greatly increase reluctance of flux path & therefore reduces coupling between primary & secondary windings

• Higher reluctance caused by air gap means a higher magnetizing reactance XM in equivalent circuit will have a much smaller value (larger susceptance BM) than its value in an ordinary transformer.

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Power flow diagram

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Power and torque in Induction Motors

• The supply power is:

• The power transferred through the air gap by the magnetic coupling is the input power (Pin) minus the stator copper loss and the magnetizing (stator iron) loss.

cos3P VIin

s

RIP

PPPP

AG

coreSCLinAG

2223

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• The stator copper and core losses are determined using the following:

• The electrically developed power (Pdv) is the difference between the air gap power (Pag) and rotor copper loss.

mindAGconvdv

AGRCLRCL

mindRCLAGconvdv

ss

RIsP

sPPRIP

PP

)1(3)1(P

3

P

22)(

222

)(

ccore

SCL

R

E

RI2

1

12

1

3P

3P

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Induction Motors Output Power

• The subtraction of the mechanical ventilation and friction losses (Pmloss) from the developed power gives the mechanical output power (in Hp):

mload mlossdvout PPP

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Induction Motors Efficiency and Torque

• The motor efficiency:

• Motor torque:

in

out

P

P

m

outP

T

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Example(1)

A 460 V, 60 Hz, 25 hp, 4 pole, Y connected induction motor has following impedances in Ω /phase referred to stator circuit:

R1 = 0.641 Ω R2=0.332 Ω X1 = 1.106 Ω X2 = 0.464 Ω XM=26.3 Ω The total rotational losses are 1100 W, & assumed to be

constant core loss is lumped in with rotational losses. For rotor slip of 2.2 % at rated voltage & rated frequency, find:

(a) Speed (b) stator current (c) P.F. (d) Pconv & Pout (e) Tind & Tload (f) Efficiency

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IM Torque-Speed Characteristic

• How does the torque of IM change as the load changes?

• At light loads: The rotor slip is very small and so the relative motion between the rotor and magnetic field is very small and the rotor frequency is also very small. Current and ER is very small and in phase so BR is relatively small, as the rotor magnetic field is very small then the induced torque is small:

90

sink

R

netRind BB

Rotor power factor

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IM Torque-Speed Characteristic

• At heavy loads: – As load increase, the slip increase, rotor speed

falls down,– thus, more relative motion appears and

produce stronger ER, – larger rotor current IR and so rotor magnetic

field BR will be seen. – The angle of the rotor current will be also

changed. – The increase in BR tend to increase in the

torque.

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• Starting torque: is 200-250% of the full load torque (rated torque).

• Pullout torque: Occurs at the point where for an incremental increase in load the increase in the rotor current is exactly balanced by the decrease in the rotor power factor. It is 200-250 % of the full load torque.

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A typical induction motor torque-speed characteristic curve

Torque-Speed Characteristic Curve Regions

Low-slip region:

In this region the motor slip increases approximately linearly with increase load & rotor mechanical speed decreases approximately linearly with load.

• In this region rotor reactance is negligible, so rotor PF is approximately unity, while rotor current increases linearly with slip.

• The entire normal steady-state operating range of an induction motor is included in this linear low-slip region.

• Moderate-slip region

In moderate-slip region rotor frequency is higher than before, & rotor reactance is on the same order of magnitude as rotor resistance.

- In this region rotor current, no longer increases as rapidly as before and the P.F. starts to drop

- peak torque (pullout torque) of motor occurs at point where, for an incremental increase in load, increase in rotor current is exactly balanced by decrease in rotor P.F.

High-slip region: • In high- slip region, induced torque actually

decreases with increased load, since the increase in rotor current is completely overshadowed by decrease in rotor P.F.

• For a typical induction motor, pullout torque is 200 to 250 % of rated full-load torque

• And starting torque (at zero speed) is about 150% of full-load torque

• Unlike synchronous motor, induction motor can start with a full-load attached to its shaft

Comments on IM torque speed curve

1- Induced torque of motor is zero at Syn. Speed.2- Torque-speed curve is nearly linear between no

load and full load. In this range rotor resistance is much larger than its reactance so rotor current, rotor magnetic field & induced torque increase linearly with increasing slip

3- There is a maximum possible torque that cannot be exceeded (pullout torque) is 2 to 3 times rated full-load torque of motor.

Comments on IM torque speed curve (con.)

4- Starting torque on motor is slightly larger than its full-load torque, so this motor will start carrying any load that it can supply at full power

5- Note: that torque on motor for a given slip varies as square of applied voltage. This is useful in one form of induction motor speed control that will be described.

6- If rotor of induction motor driven faster than sync. Speed, direction of Tind reverses & machine become Gen. converting Pmech to Pelec.

Induction Motor/Generator Mode

• Thevenin equivalent voltage of induction motor:

Induction Motor Maximum Torque Using Thevenin Equivalent

ZTH=RTH+jXTH = jXM(R1+jX1)/[R1+j(X1+XM)]

• Since XM>>X1 and XM+X1>>R1 , Thevenin

resistance & reactance can be approximated as:

RTH ≈ R1 ( XM/ [X1+XM] ) ^2

XTH ≈ X1

• 2

222

2

)()(

I

XXs

RR

V

THTH

TH

Induction Motor Maximum Torque

In IM the maximum power transfer occurs when: R2/s=√RTH^2 + (XTH+X2)^2

So maximum slip Smax is:

Smax=R2 / √RTH^2 + (XTH+X2)^2

By applying this value of slip to torque equation

in slide (8):

2

max22

2

3

2

TH

sync TH TH TH

V

R R X X

)1(3 22 ss

RImind

Note: smax ~ R2 , Maximum torque is independent of R2

Effect of Varying Rotor Resistancein wound rotor IM

Effect of varying rotor resistance on T-ω of wound rotor

Example (2)

• A 2 pole, 50 Hz induction motor supplies 15kW to a load at a speed of 2950 r/min.

Determine:- The motor’s slip?

- The induced torque in the motor in Nm under these conditions?

- The operating speed of the motor be if its torque is doubled?

- The power will be supplied by the motor when the torque is doubled?

Solution:

(a) nsync= 120fe/p= 120x50/2=3000 r/min

s= 3000-2950/3000=0.0167 or 1.67%

(b) Tind=Pconv/ωm=15 / (2950)(2πx1/60)=48.6 N.m.

(c) In the low slip region, the torque-speed is

linear & induced torque ~ s doubling Tind slip would be 3.33 %

nm=(1-s)nsync =(1-0.0333)(3000)=2900 r/min

(d) Pconv=Tind ωm=97.2 x 2900 x 2πx1/60=29.5 kW

Example (3)

• A 460V, 25hp, 60Hz, 4-pole, Y-connected wound rotor induction motor has the following impedances in ohms per-phase referred to the stator circuit:

• R1 = 0.641 Ω R2 = 0.332 Ω• X1 = 1.106 Ω X2 = 0.464 Ω Xm = 26.3 Ω

a)What is the max torque of this motor? At what speed and slip does it occur?

b) What is the starting torque?

c) When the rotor resistance is doubled, what is the speed at which the max torque now occurs?

d) What is the new starting torque?

Solution : using Thevenin equivalent :

= 266/ √(0.641)^2+(1.106+26.3)^2= 255.2 V

=(0.641)(26.3/[1.106+26.3])^2=0.59 Ω

XTH≈X1=1.106 Ω

(a) Smax = R2 / √RTH^2 + (XTH+X2)^2

=0.332/√(0.59)^2+(1.106+0.464)^2=0.198

221 1

mTH

m

XV V

R X X

- This corresponds to a mechanical speed of :

Nm=(1-s)Nsync=(1-0.198)(1800)=1444 r/min

- The maximum torque at this speed :

= 3(255.2)^2 / 2x188.5x[0.59+√0.59^2+(1.106+0.464)^2] =229 N.m.

2

max22

2

3

2

TH

sync TH TH TH

V

R R X X

(b) Starting torque of motor found by s=1

= 3x255.2^2 x 0.332 / 188.5x[(0.59+0.332)^2+(1.106+0.464)^2]=104 N.m.

c) Rotor resistance is doubled, s at Tmax doubles

smax=0.396 , and the speed at Tmax is:

nm=(1-s)Nsync=(1-0.396)(1800)=1087 r/min

Maximum torque is still:

Tmax=229 N.m. and starting torque is :

Tstart=3(255.2)(0.664) / (188.5)[(0.59+0.664)^2+(1.106+0.464)^2] =170 N.m.

])()[(

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22

2

22

XXRR

RV

THTHsync

THstart