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1 3 PHASE INDUCTION MOTOR 1 PHASE OF 3 PHASE STATOR WINDING CAGE ROTOR AIR GAP STATOR MAGNETIC FIELD ROTOTATING AT SYNCHRONOUS SPEED ROTOR FOLLOWING STAOR FIELD S R S S R S N N N s n n n s
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Induction Motors 2015

Apr 26, 2023

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Page 1: Induction Motors 2015

1

3 PHASE INDUCTION MOTOR

1 PHASE OF 3 PHASE STATOR WINDING

CAGE ROTOR

AIR GAPSTATOR

MAGNETIC FIELD ROTOTATING AT SYNCHRONOUS

SPEED

ROTOR FOLLOWING STAOR FIELD

S

RS

S

RS

NNNs

nnns

Page 2: Induction Motors 2015

2

STEEL LAMINATIONS INSERTED BETWEEN ROTOR

BARS

COPPER ROTOR BARS

BARS BRAZED ONTO END RINGS

SQUIRREL CAGE ROTOR OF A 3 PHASE INDUCTION MOTOR.

INDUCTION MOTORS

Page 3: Induction Motors 2015

3

3 PHASE STATOR WINDINGS

3 PHASE STATOR CONNECTIONS

CAGE ROTOR

OUTPUT SHAFT

COOLING FINS

AIR GAP

ROTOR BEARINGS

Page 4: Induction Motors 2015

44

3 PHASE STATOR WINDINGS

3 PHASE STATOR CONNECTIONS

CAGE ROTOR

OUTPUT SHAFT

COOLING FINS

AIR GAP

ROTOR BEARINGS

INDUCTION MOTORS

Page 5: Induction Motors 2015

5

3 PHASE INDUCTION MOTOR

•THE 3 PHASE STATOR WINDINGS PRODUCE A MAGNETIC FIELD THAT ROTATES AT SYNCHRONOUS SPEED NS .

• f = SUPPLY FREQUENCY p = NUMBER OF MAGNETIC POLE PAIRS.

• THE ROTATING MAGNETIC FIELD INDUCES A CURRENT FLOW IN THE COPPER ROTOR BARS. THIS CURRENT CREATES A ROTOR MAGNETIC FIELD• ROTOR TORQUE IS PRODUCED BY THE INTERACTION OF THE STATOR AND ROTOR MAGNETIC FIELDS•THE ROTOR ACCELERATES UP TO A STEADY ROTOR SPEED NR .•THE ROTOR SPEED IS LESS THAN SYNCHRONOUS SPEED. •THE SPEED DIFFERENCE IS CALLED THE SLIP SPEED.•THE SLIP s IS DEFINED BY:

pfNS

S

RS

NNNs

Page 6: Induction Motors 2015

66

V1

I2I1

PER PHASE EQUIVALENT CIRCUIT OF THE INDUCTION MOTORTHE

R1 R2X1 sX2

E1sE2

R1 = STATOR RESISTANCEX1 = STATOR LEAKAGE REACTANCER2 = ROTOR RESISTANCEX2 = ROTOR LEAKAGE REACTANCEE1 = INDUCED VOLTAGE ACROSS THE STATOR WINDINGE2 = INDUCED VOLTAGE ACROSS THE ROTOR WINDINGV1= STATOR SUPPLY VOLTAGEs = SLIP =

STATOR ROTOR

S

RS

NNN

Page 7: Induction Motors 2015

777

V1

I2I1

R1 X1 X2

E1E2

PER PHASE EQUIVALENT CIRCUIT OF THE INDUCTION MOTORTHE

sR2

STATOR ROTOR

22

2

22

22

jXsR

EjsXR

sEI

Page 8: Induction Motors 2015

8888

V1

I2I1

R1 X1 X2

E1E2

PER PHASE EQUIVALENT CIRCUIT OF THE INDUCTION MOTORTHE

2R

)1(2 ssR

22

2

22

22

2

)1(

)1(

jXssRR

EI

ssRR

sR

STATOR ROTOR

Page 9: Induction Motors 2015

99999

V1

I2I1

R1 X1 X2

E1E2

PER PHASE EQUIVALENT CIRCUIT OF THE INDUCTION MOTORTHE

2R

)1(2 ssR

POWERMECHANICALssRI

HEATASLOSSPOWERRI

ssRIRIPAG

)1()(

)(

)1()()(

222

22

2

2222

22

STATOR ROTOR

THE POWER TRANSFERRED FROM THE STATOR TO THE ROTOR = AIR GAP POWER PAG

AGP

Page 10: Induction Motors 2015

101010101010

V1

I2I1

R1 X1 X2

E1E2

PER PHASE EQUIVALENT CIRCUIT OF THE INDUCTION MOTORTHE

2R

)1(2 ssR

AGM

AGLOSS

AG

PsssRIPOWERMECHANICALP

sPRIPHEATASLOSSPOWER

POWERGAPAIRsRIP

)1(1)(

)(

)(

222

22

2

222

STATOR ROTOR

AGP

Page 11: Induction Motors 2015

11111111111111

V1

I2I1

R1 X1 X2

E1E2

PER PHASE EQUIVALENT CIRCUIT OF THE INDUCTION MOTORTHE

2R

)1(2 ssR

AG

AG

AG

PsPOWERMECHANICALTOTAL

sPHEATASLOSSPOWERLOSSCOPPERROTOR

PPOWERGAPAIR

)1(

STATOR ROTOR

AGP

IMPORTANT EQUATIONS :

Page 12: Induction Motors 2015

12

TUTORIAL EXAMPLE 1

A 4 POLE 50 Hz 3 PHASE INDUCTION MOTOR RUNS WITH A SLIP s = 0.08THE TOTAL AIR GAP POWER = 18 kW.THE TOTAL WINDAGE AND FRICTION LOSS = 828 WTHE PER PHASE ROTOR CURRENT I2 = 69.3 AMPS.CALCULATE: i)THE SYNCHRONOUS SPEEDii)THE ROTOR SPEEDiii)THE COPPER POWER LOSS PER PHASEiv)THE MECHANICAL POWER PER PHASEv)THE TOTAL MECHANICAL POWER vi)THE TOTAL MECHANICAL OUTPUT POWER

Page 13: Induction Motors 2015

13

i) THE SYNCHRONOUS SPEED

ii) THE ROTOR SPEED

iii) THE COPPER POWER LOSS PER PHASE

RPMsNNNNNs SRS

RS 1380)08.01(1500)1(

WsPRIPHASEPERLOSSPOWERCOPPER AG 480600008.0)( 22

2

TUTORIAL EXAMPLE 1: SOLUTION

3POWERGAPAIRTOTAL

PHASEPERPOWERGAPAIRPAG

RPMN

pfN

S

S

150060250

Page 14: Induction Motors 2015

1414

iv) THE MECHANICAL POWER PER PHASE

v) THE TOTAL MECHANICAL POWER

vi) THE TOTAL MECHANICAL OUTPUT = THE TOTAL MECHANICAL POWER – TOTAL WINDAGE AND FRICTION POWER LOSS

TUTORIAL EXAMPLE 1: SOLUTION

kWWPsPHASEPERPOWERMECHANICAL AG

52.555206000)08.01()1(

kWPOWERMECHANICALTOTAL 56.1652.53

kW732.15828.056.16