ece 325 electric energy system components 6- three...

1

Instructor:

Kai Sun

Fall 2015

ECE 325 – Electric Energy System Components6- Three-Phase Induction Motors

2

Content

(Materials are from Chapters 13-15)

• Components and basic principles

• Selection and application

• Equivalent circuits

3

• 3-phase, 60Hz, AC induction motor

• Rated power: 30.0 hp (=22.4kW)

• Service factor: 1.15 (i.e. 115% of

rated power for short-term use)

• Rated full-load current: 35.0A

• Rated voltage: 460V

• Rated speed: 1765rpm

• Continuous duty at ambient 40oC

• Full-load nominal efficiency: 93.0%

• Frame size: shaft height of 28/4=7

inches; body’s length of 6

• Insulation class: F (155oC for 20,000

hours lifetime)

• NEMA design: B (normal starting

torque combined with a low current)

• Start inrush kVA: G (5.60-6.29kVA)

http://www.pdhonline.org/courses/e156/e156content.pdf

Understanding the Nameplate

http://www.pdhonline.org/courses/e156/e156content.pdf

4

Principle components

• Stator (stacked laminations)

• Rotor (stacked laminations)

− Wound rotor

− Squirrel-cage rotor

• Air gap (0.4mm-4mm)

5

Stator

• Hollow, cylindrical core made up of stacked laminations.

• Stator winding is placed in the evenly spaced slots punched out of

the internal circumference of the laminations.

7

Wound Rotor

• Has a 3-phase winding (usually in

Y-connection) uniformly distributed

in rotor slots, similar to the stator

winding

• Rotor winding terminals are

connected to three slip-rings which

revolve with the rotor.

• 3 stationary brushes connect the

rotor winding to external resistors

during the start-up period and are

short-circuited during normal

operations.

8

Squirrel-Cage Rotor

• More adopted in induction motors

• Instead of a winding, copper bars are

pushed into the slots in the laminations

• Welded to two copper end-rings, all

bars are short-circuited at two ends, so

as to resemble a “squirrel cage”

• In small and medium motors, the

“cage” (bars and end-rings) are made of

aluminum.

10

Principle of Operation: Faraday’s Law and Lorentz Force

• A magnet moving above a conducting ladder induces a voltage E=Blv and a current

I in the conductor underneath according to Faraday’s law (the right-hand rule).

• Cut by the moving flux, the current-carrying conductor experiences a Lorentz force

(the left-hand rule) always acting in a direction to drag it along the moving magnet

• When the system reaches a steady state, the ladder moves in the same direction as

the magnet but has a lower speed <v (why?)

• Roll up the ladder into a cylindrical squirrel-cage rotor and replace the magnet by a

stator an induction motor

Parallel conductors are short-circuited

by end-bars A and B

Conductors are hort-circuited

by end-rings A and B)

11

Rotating magnetic field

• Consider a stator having 6 salient poles with Y-connected stator windings

carrying balanced 3-phase alternating currents.

Positive currents: always flowing

into terminals A, B and C( ) 10cos 10cos( )

( ) 10cos( 120 ) 10cos( 120 )

( ) 10cos( 120 ) 10cos( 120 )

a

b

c

i t t

i t t

i t t

12

Rotating Magnetic Field: Instant 1

• =0: 6 poles together produce a magnetic field having

essentially one broad N pole and one broad S pole

13


• 60o: the magnetic field moves CW by 60o; its angular speed equals

14


• 120o: the magnetic field moves CW by 120o; its angular speed equals

15

Rotating Magnetic Field

• For the stator producing a magnetic field having one pair of N-S poles

– The field rotates 360o during one cycle of the stator current.

– The speed of the rotating field is necessarily synchronized with the frequency f

of the source, so it is called synchronous speed =60f (r/min)

– With the phase sequence A-B-C, the magnetic field rotates CW.

– If we interchange any two of the three lines connected to the stator, the new

phase sequence will be A-C-B and the magnetic field will rotate CCW at the

synchronous speed

16

Phase group• In practice, instead of using a single coil per pole, the

coil is sub-divided into 2 or more coils lodged and

staggered in adjacent slots connected in series, i.e. a

phase group

• Each phase group produces one N/S pole, so using

more phase groups allows us to increase the number

of poles (denoted by p)

• No. of groups = No. of phases No. of poles = 3 p

17

Number of N-S poles and synchronous speed

• Synchronous speed ns=120f / p [r/min]

f : frequency of the source [Hz]

p: number of poles

If f=60Hz,

ns=120f / 4 =30f =1800r/min ns=120f / 8 =15f=900r/min

18

Starting of an induction motor1. The magnetic field produced by the stator

current rotates CW at synchronous speed ns.

2. Relative to the magnetic field, the rotor rotates

CCW.

3. According to Faraday’s Law (right-hand rule),

the rotor windings (bars) have induced

voltages in the directions as indicated.

4. Large circulating currents are created in rotor

windings (bars) by the induced voltages

5. The currents have the maximum values at the

staring point because the rotor has the

maximum speed relative to the magnetic field.

6. The rotor windings rotates CW subjected to

Lorentz forces (left-hand rule) in the directions

as indicated

7. The rotor will accelerate until the total Lorentz

force equals the friction

8. Once the rotor starts rotating CW, its speed

relative to the filed will decrease, so its

winding currents and Lorentz forces will

decrease

Can the rotor reach ns?

• If it did, then its currents and Lorentz

forces would be zero and friction

would slow the rotor down, so the

answer is: No, the rotor’s speed n<ns at

a steady state.

• Since friction is very small, the rotor

speed n ns at no load conditions

19

Slip

• Slip: is the difference between synchronous speed ns and rotor

speed n, expressed as a per unit or percent of synchronous speed.

f: frequency of the source connected to the stator [Hz]

f2: frequency of the voltage and current induced in the rotor [Hz]

ns=120f / p: synchronous speed [r/min]

n: rotor speed [r/min]

r=n/30, s=ns/30 : corresponding angular speeds of n and ns [rad/s]

2 (pu)s s r

s s

n n fs

n f

20

Example 13-3

• A 0.5 hp, 6-pole induction motor is excited by a 3-phase, 60 Hz source. Calculate

the slip, and the frequency of the rotor current under the following conditions:

a. At standstill

b. Motor turning at 500r/min in the same direction as the revolving field

c. Motor turning at 500r/min in the opposite direction to the revolving field

d. Motor turning at 2000r/min in the same direction as the revolving field

Solution:

ns=120f/p=12060/6 = 1200 r/min

a. s= (ns-n)/ns=(1200-0)/1200 =1 pu

f2=sf=60Hz

b. s= (1200-500)/1200 = 0.583 pu

f2=sf=0.583 60Hz = 35 Hz

c. s= [1200- (-500)]/1200 =1.417 pu

f2=sf=1.417 60Hz = 85 Hz

d. s= [1200- 2000)]/1200 = -0.667 pu

f2=sf=-0.667 60Hz = -40 Hz

Slip Rotor speed

& freq

Operating mode

s=1n=0

f2=fTransformer (standstill)

0<s<10<n<ns

0<f2<fMotor

s>1n<0

f2>fBrake

s<0n>ns

f2<0

Generator (reversed

phase sequence A-C-B)

21

Equivalent circuits: wound-rotor induction motor1. When the motor is at standstill (i.e. n=0, s=1), it acts exactly like a

conventional transformer. Assume a Y-connection for both the stator and rotor,

and a turns ratio of 1:1. Consider the per-phase equivalent circuit:

Eg: source voltage, line to neutral,

x1, r1: stator leakage reactance and winding resistance

x2, r2: rotor leakage reactance and winding resistance

Xm, Rm: magnetizing reactance and resistance modeling losses of iron,

windage and friction

Rx: external resistance, connecting one slip-ring to the neutral of the rotor

Note: Io may reach 40% of Ip due to the air gap between the stator and rotor (much bigger

than the air gap between the P and S windings of a transformer), so we cannot eliminate

the magnetizing branch.

22

For motors exceeding 2 hp

RX

R2 =r2+RX

For motors smaller than 2 hp

RX

R2 =r2+RX

23

2. When the motor runs at a slip s, i.e. n=(1-s)ns

|E2|=|sE1|, |I2|=|I1|, jx2 jsx2 r2 and RX are the same

2 1 1 22

2 22 2 2 2 22 2

| | 0 | |= arctan

j | j | ( )

E sE sE sxI

R sx R sx RR sx

Note: the phasors on the primary side (E1 and I1) and the secondary side (E2 and I2)

cannot be drawn in one phasor diagram because they have different frequencies

RX

f sf

1 : 1

R2 =r2+RX

24

• I1 and I2 have the same effective value even

though they have different frequencies

|I1|=|I2|

• sE1 and E2 have the same effective values

|sE1|=|E2|

• E1 leads I1 and E2 leads I2 both by

Phasor diagrams on the rotor and stator

25

Simplified equivalent circuit: referred to the stator side

11 2

2 2j

sEI I

R sx

1 1

1

2 2 2/ j

E EI

R s x Z

def1

2 2 2

1

/ jE

Z R s xI

Note: The value of R2/s will vary from R2 to as the motor goes

from start-up (s=1) to ns (s=0)

jx =jx1+jx2

E1

26

Eg

Ip

Io

r1jx1 jx2

R2

RmjXm

I1

2

1 sR

s

jx

2R

s

ImIf

• Active power supplied to the rotor:

• Total I2R losses in the rotor circuit

• Mechanical power developed by the motor

• Torque developed by the motor

• Total power absorbed by the motor:

• Stator iron losses

• Stator copper losses

2 2

2 21 1

| | | || | ( )

g g

m m

E E RS j I r jx

R X s

2

2 2 21 1 1

| || | | |

g

m

E RP I r I

R s

2

1 2| | /rP I R s

2

1 1| |jsP I r

2

1 2| |jr rP I R sP

2

1 2

1(1 ) | |m r jr r

sP P P P s I R

s

9.55 9.55 (1 ) 9.55

(1 )

m r r

s s

P P s PT

n n s n

2| | /f g mP E R

Note: The torque only depends on Pr

28

Eg

Ip

Io

r1jx1 jx2

R2

RmjXm

I1

2

1 sR

s

jx

2R

sIm

If

=accos(P/|S|)

=actan(x/r1)

Z1=r1+jx

Phasor diagram of the induction motor

S=P+jQ

29

Breakdown (maximum) Torque

When |Z1|=|R2/s|, Pr and torque T both

reach their maximum values

I1

Z1I1

Eg

I1R2/s

1 2| | | |

cos2 2

gE I R

s

2

1| |b

Rs

Z

22

1 2

2

1

9.55 | |9.55 9.55 | | /

4 | | cos2

grb

s ss

EP I R sT

n nn Z

1

1

| || |

2 | | cos2

g

b

EI

Z

1 1

| || | cos

2 2

gEI Z

30

Example 13-5• A 3-phase induction motor having a synchronous speed of 1200 r/min draws

80 kW from a 3-phase feeder. The copper losses and iron losses in the stator

amount to 5 kW. If the motor runs at 1152 r/min, calculate

a. The active power transmitted to the rotor

Pr=Pe – Pjs – Pf=80 – 5 =75 kW

b. The rotor I2R losses, i.e. Pjr

s=(ns-n)/ns=(1200-1152)/1200=0.04

Pjr=sPr=0.0475=3 kW

c. The mechanical power developed

Pm=Pr – Pjr =75 – 3 =72 kW

d. The mechanical power delivered to the load, knowing that the

windage and friction losses equal to 2 kW

PL=Pm - PV=72 – 2 =70 kW

e. The efficiency of the motor

=PL/Pe=70/80=87.5%

f. The torque developed by the motor

Tm=9.55 Pr/ns= 9.5575000/1200=597 Nm

31

Two-wattmeter method to measure 3-phase active power

P1=|Vac||Ia|cos( - 30o)=ELILcos( - 30o)

P2=|Vbc||Ib|cos( + 30o)=ELILcos( + 30o)

3-phase load

in Y or

connection

3 13 1 23 2 3( )Q P PS jP P Pj

3 cos( 30 ) cos( 30 ) 3 cos( 30 ) cos( 30 )

2cos cos30 3 2sin sin 30

3 c s no si3 L L

L L L L

L L L L

L L

S E I j E I

E I j E I

EE IjI

Proof:

32

a. Power supplied to the rotor

Pe=P1+P2=70 kW

b. Rotor I2R losses Pjr

r1=0.34/2=0.17

Pjs=3I12r1=37820.17=3.1 kW

Pr=Pe – Pjs – Pf=70 – 3.1 – 2 =64.9 kW

s=(ns-n)/ns=(1800-1763)/1800=0.0205

Pjr=sPr=0.020564.9=1.33 kW

1763 r/min

Example 13-7

c. Mechanical power supplied to the load

Pm=Pr – Pjr =64.9 – 1.33 =63.5 kW

PL=Pm - PV= 63.5 – 1.2 =62.3 kW

=62.3 1.34 = 83.5 hp

d. Efficiency

=PL/Pe=62.3/70=89%

e. Torque developed at 1763 r/m

Tm=9.55 Pr/ns= 9.55649000/1800

=344 Nm

A 3-phase induction motor having a nominal rating of 100 hp (~75 kW) and a

synchronous speed of 1800 r/min is connected to a 3-phase 600 V source.

Resistance between two stator terminals =0.34. Calculate

Two-wattmeter method

33

Two practical squirrel-cage induction motors

34

no-load torque

Torque-Speed Curve 2

2

2

21

9.55 | |9.55

( )

gr

ss

s

s

E RPT

n R nZ n n

n n

Stable operating

Region (n, T)

2 2 21 1 1, | | , / ( )s

r g

s

n n R Rs P I I E Z

n s s

• The curve is nearly linear between no-load and full-load because s is small and

R2/s is big2 2

2

2 2

9.55 | | 9.55 | |( )

g g

s

s s

E ET n n s

R n R n 1 2 /gI sE R

35

Eg=440/1.73 V

R2=1.2

Z1=1.5+6j

ns=1800 r/min2

2

221

9.55 | |

| | ( )

g

ss

s

E R

R n

n

T

nZ nn

Torque-Speed Curve

21

1

g

s

s

E

RI

n

nZ

n

Current-Speed Curve

2

2

2

1 2

9.55 | |

/

g

s

E R

Z R nT

s s

Torque-Slip Curve

The 5 hp induction motor

36

Torque-Speed Curve: 5 hp motor

s I1 Pr T n

• When the motor is stalled, i.e. locked-rotor condition, the current is 5-6 times the

full-load current, making I2R losses 25-36 times higher than normal, so the rotor

must never remain locked for more than a few second

• Small motors (15 hp and less) develop their breakdown torque at about 80% of ns

37

Torque-Speed Curve: 5000 hp motor

Big motors (1500+ hp) :

• Relatively low starting

(locked-rotor) torque

• Breakdown torque at

about 98% of ns

• Rated n is close to ns

38

Effect of rotor resistance

R2

2.5R2

10R2

Rated Torque =15Nm

Rated torque =15Nm

Rated torque =15Nm

Current at the rated torque



39

• When R2 increases, starting (locked-rotor) torque TLR increases, starting current

I1,LR decreases, and breakdown torque Tb remains the same

• Pros & Cons with a high rotor resistance R2

– It produces a high starting torque TLR and a relatively low starting current I1,LR

– However, at the rated torque, it produces a rapid fall-off in speed with

increasing load because the torque-speed curve becomes flat, and the motor

has high copper losses and low efficiency and tends to overheat

• Solution

– For a squirrel-cage induction motor, design the rotor bars in a special way so

that the rotor resistance R2 is high at starting and low under normal operations

– If the rotor resistance needs to be varied over a wide range, a wound-rotor

motor needs to be used.

Effects of rotor resistance

2 2

2

2

2

2

2

1 1

9.55 | | 9.55 | |

| | | |

g g

s s

LR

E E

Z n

RT R

R Z n

2

2

1

9.55 | |

4 | | cos2

g

b

sn Z

TE

2

2

1| |jr IP R1

1,

2

g

LR

E

ZI

R

40

Effects of the terminal voltage: Problem 13-33• A 3-phase, 300kW, 2300V 60Hz, 1780 r/min induction motor is used to

drive a compressor. The motor has a full-load efficiency of 92% and power

factor of 86%. If the terminal voltage rises by 20% while the motor operates

at full load, how does each of the following change?

2

2

2

9.55 | |( )

g

FL s FL

s

ET n n

R n

2

22

1

9.55 | |

| |

g

LR

s

ET R

Z n

1,

1 2

g

LR

EI

Z R

2

2

1

9.55 | |

4 | | cos2

g

b

s

ET

n Z

Pm, T I1,LR

nFL Tb

I1,FL Average

motor

temperature

PF Flux per pole

Exciting

current

TLR Iron losses

Q=|I1|2x+|Eg|

2/Xm

Same

(Q)

About the

same

(Pf Pjs Pjr)

44%

20%

44%

About the

same (iron

windings )

20%

>20%

(considering

saturation)

41

Asynchronous generatorConnect the 5 hp, 1800 r/min, 60Hz motor

to a 440 V, 3-phase line and drive it at a

speed of 1845 r/min

s=(ns-n)/ns=(1800-1845)/1800= -0.025

R2/s=1.2/(-0.025)=-48 <0

The negative resistance indicates that power

is flowing from the rotor to the stator

(asynchronous generator mode)

Power flow from the rotor to the stator:

|E|= 440/1.73=254 V

|I1|=|E|/ |-48+1.5+ j6|=254/46.88= 5.42A

Pr=|I1|2R2/s= -1410 W <0

1410W flows from the rotor to the stator

Mechanical power and torque input to its

shaft:

Pjr=|I1|2R2=35.2 W

Pm=Pr+Pjr=1410 + 35.2=1445W

T=9.55 Pm/n=22.3 Nm

Total active power delivered to the line:

Pjs=|I1|2r1=44.1 W

Pf =|E|2/Rm=71.1 W

Pe=Pr-Pjs-Pf= 1410-44.1-71.7=1294 W

P3=3Pe=3882W

Reactive power absorbed from the line:

Q3=(|I1|2x+|E|2/Xm) 3 =(176+586) 3

=2286 var

Complex power delivered to the line:

S3=P3 - Q3=3882 - j2286 VA

cos=86.2%

Efficiency of this asynchronous generator

=Pe/Pm=1294/1445=89.5%

42

Tests to determine the equivalent circuit

• Estimate r1, r2, Xm, Rm and x (note: r2+RX=R2 where RX is the

external resistance)

1. No-load test

2. Locked-rotor test

Eg

Ip

Io

r1jx1 jx2

r2/s

RmjXm

I1

ImIf

RX/s

jx

R2/s

43

No-load test

At no-load, slip s0

R2/s is high, I1<<Io

Steps:

1. Measure stator resistance RLL

between any two terminals

(assuming a Y connection)

r1=RLL/2

2. Run the motor at no-load using

rated line-to-line voltage ENL.

Measure no-load current INL

and 3-phase active power PNL

3NL NL NLS E I

2 2

NL NL NLQ S P

2 2

1 13 3NL f NL f NL NLP P I r P P I r

2 2

2

1

( / 3)

( ) / 3 3

NL NLm

f NL NL

E ER

P P P I r

(Ignoring PV)

2 2( / 3)

( ) / 3

NL NLm

NL NL

E EX

Q Q

44

Locked-rotor test

When the rotor is locked,

slip s=1, IP >>Io r2 =r2/sR2/s, neglect the magnetizing branch

Steps:

1. Apply reduced 3-phase voltage

ELR to the stator so that the stator

current Ip the rated value

2. Measure line-to-line voltage

ELR, current ILR and 3-phase

active power PLR

3LR LR LRS E I

2 2

LR LR LRQ S P

2 2 2

1 2 2 13 3 / (3 )LR LR LR LR LRP I r I r r P I r 23

LR

LR

Qx

I

45

Example 15-1

46

Homework Assignment #5

• Read Chapters 13.0-13.16, 15.0-15.9

• Questions:

– 13-16, 13-23, 13-24, 13-28, 15-2, 15-3, 15-4, 15-5

• Due date:

– hand in your solution to Denis at MK 205 or by email

([email protected]) before the end of 11/4 (Wed)

mailto:[email protected]

ece 325 electric energy system components 6- three...

Documents