chapter 1-three phase

7/31/2019 Chapter 1-Three Phase

1/90

1

Chapter 1.

Three-Phase System


2/90

1.1: Review of Single-Phase System

The Sinusoidal voltage

v1(t) = Vm sin wt

i

v1

Load

ACgenerator

v2

2


3/90

3

1.1: Review of Single-Phase System

The Sinusoidal voltage

v(t) = Vm sin wtwhere

Vm = the amplitude of the sinusoid

w = the angular frequency in radian/s

t = time

v(t)

Vm

-Vm

wt


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4

v(t)

Vm

-Vm

t

w 2T

T

1f

f2wThe angular frequency in radians per second


5/90

5

A more general expression for the sinusoid (as

shown in the figure):

v2(t) = Vm sin (wt + )where is the phase

v(t)

Vm

-Vm

wt

V1= V

msin wt

V2= V

msin wt + )


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6

A sinusoid can be expressed in either sine or cosine form.

When comparing two sinusoids, it is expedient to express

both as either sine or cosine with positive amplitudes.

We can transform a sinusoid from sine to cosine form or

vice versa using this relationship:

sin (t 180o) = - sin t

cos (t 180o) = - cos t

sin (t 90o

) = cos tcos (t 90o) = + sin t


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7

Sinusoids are easily expressed in terms of phasors.

A phasor is a complex number that represents the

amplitude and phase of a sinusoid.

v(t) = Vm cos (t + )

Time domain Phasor domain

Time domain

mVV Phasor domain

)cos( wtVm mV

)sin( wtVmo

m 90V

)cos( wtmI mI

)sin( wtmIo

m 90I


8/90

8

Time domain

Phasor domain

v(t)

Vm

-Vm

wt

V1= V

msin wt

V2

= Vm

sin wt + )

V1

V2

v2(t) = Vm sin (wt + )

v1(t) = Vm sinwt

m

VV2

01 mVV

or

or 01 rmsVV

rms

VV2


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9

1.1.1: Instantaneous and Average Power

The instantaneous power is the power at any instantof time.

p(t) = v(t) i(t)

Where v(t) = Vm cos (wt + v)i(t) = Im cos (wt + i)

Using the trigonometric identity, gives

)cos()cos()( ivmmivmm t2IV2

1IV

2

1t wp


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10

The average power is the average of the

instantaneous power over one period.

T

dttpT

P0

)(1

)cos( ivmmIV21 P

p(t)

t

)cos( ivmmIV2

1

mmIV2

1


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11

The effective value is the root mean square (rms) of

the periodic signal.The average power in terms of the rms values is

Where

)cos(iv

P rm srms

IV

2

VV mrm s

2

II

m

rm s


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12

1.1.2: Apparent Power, Reactive Power and

Power Factor

The apparent power is the product of the rms values

of voltage and current.

The reactive power is a measure of the energy

exchange between the source and the load reactive

part.

rm srm sIVS

)sin( ivQ rm srm sIV


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13

The power factor is the cosine of the phase difference

between voltage and current.

The complex power:

)cos( ivfactorPower S

P

iv

jQP

rm srm s IV


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14

1.2: Three-Phase System

In a three phase system the source consists of three

sinusoidal voltages. For a balanced source, the three

sources have equal magnitudes and are phase

displaced from one another by 120 electrical degrees.

A three-phase system is superior economically and

advantage, and for an operating of view, to a single-

phase system. In a balanced three phase system the

power delivered to the load is constant at all times,whereas in a single-phase system the power pulsates

with time.


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15

1.3: Generation of Three-Phase

Three separate windings or coils with terminals R-R,Y-Y and B-B are physically placed 120o apartaround the stator.

Y

BY

B

Stator

Rotor

Y

R

B

R

R

N

S


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16

Vor vis generally represented a voltage, butto differentiate the emf voltage of generatorfrom voltage drop in a circuit, it is convenient touse eor Efor induced (emf) voltage.


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1717

v(t)

wt

vR

vY

vB

The instantaneous e.m.f. generated in phase R, Y and B:

eR= EmR sin wteY= EmY sin (wt -120

o)

eB= EmB sin (wt -240o)= EmBsin (wt +120

o)


18/90

ER

Three-phase

Load

Three-phaseAC generator

VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN

18


19/90


eR= EmR sin t

eY= EmY sin (t -120o)

eB= EmB sin (t -240o) = EmBsin (t +120

o)

In phasor domain:

ER= ERrms 0o

EY= EYrms -120o

EB= EBrms 120o

Phase voltage

120o

-120o

0o

ERrms= EYrms= EBrms= Ep Magnitude of phase voltage19


20/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN

Line voltage

ERY

ERY

= ER

- EY


21/90

21

Line voltage

ERY

= ER

- EY

120o

-120o

0o

-EYERY

= Ep 0o - Ep -120

o

= 1.732Ep

ERY

30o= 3 Ep

= EL

30o

30o


22/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN

Line voltage

EYB

EYB

= EY

- EB


23/90

Line voltage

EYB

= EY

- EB

120o

-120o

0o

-EB

EYB

= Ep -120o - Ep

120o

= 1.732Ep

EYB

-90o

-90o= 3 Ep

= EL

-90o


24/90

ER

Three-phaseLoad


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN

Line voltage

EBR

EBR = EB- ER


25/90

2525

Line voltage

EBR

= EB

- ER

120o

-120o

0o

-ER

EBR= Ep 120

o - Ep 0o

= 1.732Ep

EBR

150o

150o= 3 Ep

= EL

150o

For star connected supply, EL= 3 Ep


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26

120o

-120o

0o

Phase voltages

ER= Ep 0o

EY= Ep -120o

EB= Ep 120o

Line voltages

ERY= EL 30o

EYB= EL -90o

EBR= EL 150o

It can be seen that the phasevoltage ER is reference.


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2727

Phase voltages

ER= Ep -30o

EY= Ep -150o

EB= Ep 90o

Line voltages

ERY= EL 0o

EYB= EL -120o

EBR= EL 120o

Or we can take the line voltageERYas reference.


28/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

ERY

Delta connected Three-Phase supply

ERY= ER = Ep 0o


29/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

EYB

EBR

For delta connected supply, EL= Ep

Delta connected Three-Phase supply


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30

Connection in Three Phase System

4-wire system (neutral line with impedance)

3-wire system (no neutral line )

4-wire system (neutral line without impedance)

Star-Connected Balanced Loadsa)4-wire system b) 3-wire system

3-wire system (no neutral line ), delta connected load

Delta-Connected Balanced Loads

a) 3-wire system


31/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN ZN

VN


VN= INZNVoltage drop across neutralimpedance: 1.1


32/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN ZN

VN


IR + IY+ IB= IN

Applying KCL at star point

1.2


33/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN ZN

VN


Applying KVL on R-phase loop

33


34/90

ER

Three-phase

Load


IR

VR ZR

IN ZN

VN

Applying KVL on R-phase loop

ERVRVN= 0

ERIRZRVN= 0

IR=

ThusERVN

ZR

1.3


34


35/90

4 i ( l li i h i d )


36/90

36

Three-phase

Load


VYEY

IY

ZY

IN ZN

VN

Applying KVL on Y-phase loop


EYVYVN= 0

EYIYZYVN= 0IY=

ThusEYVN

ZY

1.4

4 i t ( t l li ith i d )


37/90

Three-phase

Load


EB

IB

ZB

VB

IN ZN

VN


Applying KVL on B-phase loop

EBVBVN= 0

EBIBZBVN= 0IB=

ThusEBVN

ZB

1.5

37



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38


IR + IY+ IB= IN

Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into

Eq. 1.1:

=EBVN

ZB+

EYVN

ZY

ERVN

ZR +

VN

ZN

ERVN EYVN+ + EBVN =VN

ZNZR ZR ZY ZY ZB ZB

ER

ZR+

EY

ZY+

EB

ZB=

1

ZN+

1

ZR+

1

ZY

VN +1

ZB



39/90

39


ER

ZR+

EY

ZY+

EB

ZB=

1

ZN+

1

ZR+

1

ZY

VN +1

ZB

VN =

ERZR

+ EYZY

+ EBZB

1

ZN

+1

ZR

+1

ZY

+1

ZB

1.6



40/90


VN =

ERZR

+ EYZY

+ EBZB

1

ZN

+1

ZR

+1

ZY

+1

ZB

1.6

VN is the voltage drop across neutral line impedance

or the potential different between load star point andsupply star point of three-phase system.

We have to determine the value of VN in order to find thevalues of currents and voltages of star connected loads of

three-phase system. 40


41/90

Example

ER

Three-phase

Load

ZY= 2

IR

VR

EY

EB

ZR= 5

IY

IB

ZB= 10

VB

IN ZN=10

VN

Find the line currents IR,IYand IB. Also find

the neutral current IN.

EL = 415 volt

Th h 3 wire system (no neutral line )


42/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN ZN

VN


Th h 3 wire system (no neutral line )


43/90

ER

Three-phase

Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

VN


No neutral line = open circuit , ZN=

43

3 wire system (no neutral line )


44/90

44

VN =

ER

ZR+ E

Y

ZY+ E

B

ZB

1

ZN +

1

ZR +

1

ZY +

1

ZB

1.6


=ZN

1

= 0

3 wire system (no neutral line )


45/90

45

VN =

ER

ZR+ E

Y

ZY+ E

B

ZB

1

ZR +

1

ZY +

1

ZB

1.7



46/90

Example

ER

Three-phase

Load

ZY= 2

IR

VR

EY

EB

ZR= 5

IY

IB

ZB= 10

VBVN

EL = 415 volt


the voltages VR, VY and VB.

3 wire system (no neutral line ) delta connected load


47/90

3-wire system (no neutral line ),delta connected load

ER

Three-phaseLoad


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

3 wire system (no neutral line ) delta connected load


48/90


ER

Three-phaseLoad


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy

3-wire system (no neutral line ) delta connected load


49/90


ER

Three-phaseLoad


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy

ERY=VRY

EYB =VYB

EBR =VBR



50/90

50


Phase currents

30oIr=

VRY

ZRY=

ERY

ZRY=

EL

ZRY

-90oIy=

VYB

ZYB=

EYB

ZYB=

EL

ZYB

150oIb

=VBR

ZBR=

EBR

ZBR=

EL

ZBR



51/90


ER

Three-phaseLoad


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy

ERY=VRY

EYB =VYB

EBR =VBR

Line currents

IR= Ir Ib-

=EL

ZRY

30o-

150oEL

ZBR

IY= Iy Ir-

=EL

ZYB

-90o-

30oEL

ZRY


52/90


53/90


54/90

Example

ER

Three-phase

Load

ZY= 2

IR

VR

EY

EB

ZR= 5

IY

IB

ZB= 10

VBVN

Find the line currents IR,IYand IB .

EL = 415 volt

Use star-delta conversion.

Three-phase 4-wire system (neutral line without impedance)


55/90

ER

Three-phase

Load

Three phaseAC generator

VY

IR

VR

EY

EB

ZR

IR

IB

ZB ZY

VB

INZ

N

VN

4 wire system (neutral line without impedance)

=0

VN = INZN= IN(0)= 0 volt55

4-wire system (neutral line without impedance)


56/90

56

4 wire system (neutral line without impedance)

For 4-wire three-phase system, VN

is equal to0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5become,

IB=EB

ZB

1.5EBVN

IY=EY

ZY

1.4EYVN

IR=ER

ZR1.3

ERVN


57/90

Example

ER

Three-phase

Load

ZY= 2

IR

VR

EY

EB

ZR= 5

IY

IB

ZB= 10

VB

IN

VN



EL = 415 volt

(t)


58/90

58

v(t)

wt

vR

vY

vB


eR= EmR sinw

teY= EmY sin (wt -120

o)

eB= EmB sin (wt -240o)= EmBsin (wt +120

o)


59/90

59

1.4: Phase sequencesRYB and RBY

120o

-120o

120o VR

VY

VB

wo

)rms(RR0VV

o

)rms(YY120VV

o

)rms(B

o

)rms(BB

120V

240VV

VR leads VY, which in turn leads VB.This sequence is produced when the rotor rotates in

the counterclockwise direction.

(a) RYB or positive sequence


60/90

60

(b) RBY or negative sequence

120o

-120o

120

o

VR

VB

VY

w

o

)rms(RR0VV

o

)rms(BB120VV

o

rmsY

o

rmsYY

V

V

120

240

)(

)(

V

VR leads VB, which in turn leads VY.This sequence is produced when the rotor rotates in

the clockwise direction.


61/90


62/90

62

Star Connectionb) Four wire system

VRN

VBN

VYN

ZR

ZY ZB

R

B

NY


63/90

63

Wye connection of Load

Z1

Z3

Z2

R

B

Y

NLoad

Z3

Z1

Z2

R

Y

B

Load

N


64/90

64

1.5.2: Delta Connection

R

Y

B

Y

B

R


65/90

65

Delta connection of load

Zc

Za

Zb

R

B

Y

Load

Zc

Zb

Za

R

Y

B

Load


66/90

Wye-Connected Balanced Loads


67/90

67

Example

ER

Three-phase

Load

ZY= 20

IR

VR

EY

EB

ZR= 20

IY

IB

ZB= 20

VBVN

EL = 415 volt


the voltages VR, VY and VB.

Wye Connected Balanced Loadsb) Three wire system


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68

Wye-Connected Balanced Loadsb) Three wire system

VN = = 0 volt

VR= ER

VY= EY

VB= EB

1.6.1: Wye-Connected Balanced Loads

) i


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69

Example

ER

Three-phase

Load

ZY= 20

IR

VR

EY

EB

ZR= 20

IY

IB

ZB= 20

VB

IN

VN



EL = 415 volt

a)Four wire system

1.6.1: Wye-Connected Balanced Loads


70/90

70

VRN

VBN

Z1

Z2 Z3

R

B

N

Y

VYN

IR

IY

IB

IN

BYRNIIII

For balanced load system,

IN = 0 and Z1 = Z2 = Z3

3

o

BN

B

2

o

YN

Y

1

o

RN

R

Z

120VI

Z

120V

I

Z

0VI

BNYNRNphasa

phasaBN

phasaYN

phasaRN

VVVVwhere

120VV

120VV

0VV

a)Four wire system


71/90

71

Wye-Connected Balanced Loadsb) Three wire system

R

Y

B

Z1

Z 2 Z3

IR

IY

IB

VRY

VYB

VBR S

0IIIBYR

3

o

BS

B

2

o

YS

Y

1

o

RS

R

Z

120VI

Z

120VI

Z

0VI

BSYSRSphasa

phasaBS

phasaYS

phasaRS

VVVVwhere

240VV

120VV

0VV


72/90

72

1.6.2: Delta-Connected Balanced Loads

ZZ

Z

R

Y

B

VRY

VYB

VBR

IR

IRY

IBR

IYB

IB

IY

Phase currents:

3

o

BR

BR

2

o

YB

YB

1

o

RY

RY

Z

120VI

Z

120VI

Z

0VI

Line currents:

YBBRB

RYYBY

BRRYR

III

III

III

lineBYR

phasaBRYBRY

IIIIand

IIIIwhere


73/90


74/90

74

1.7.1: Wye-Connected Unbalanced LoadsFour wire system

VRN

VBN

Z1

Z2 Z3

R

B

N

Y

VYN

IR

IY

IB

IN

BYRNIIII

For unbalanced load system,

IN

0 and Z1

Z2

Z3

3

o

BN

B

2

o

YN

Y

1

o

RN

R

Z

120VI

Z

120V

I

Z

0VI

120VV

120VV

0VV

phasaBN

phasaYN

phasaRN


75/90

75

1.7.2: Delta-Connected Unbalanced Loads

Z

Z

Z

R

Y

B

VRY

VYB

VBR

IR

IRY

IBR

IYB

IB

IY

Phase currents:

3

o

BR

BR

2

o

YB

YB

1

o

RY

RY

Z

120VI

Z

120VI

Z

0VI

Line currents:

YBBRB

RYYBY

BRRYR

III

III

III

120VV

120VV

0VV

phasaBN

phasaYN

phasaRN


76/90

76

1.8 Power in a Three Phase

System


77/90

77

Power Calculation

The three phase power is equal the sum ofthe phase powers

P = PR+ PY+ PB

If the load is balanced:

P = 3 Pphase = 3 Vphase Iphase cos

1.8.1: Wye connection system:


78/90

78

I phase= I L and

Real Power, P = 3 Vphase Iphase cos

Reactive power,

Q = 3 Vphase

Iphase

sin

Apparent power,

S = 3 Vphase Iphase

or S = P + jQ

phaseLL VV 3

WattIV LLL cos3

VARIV3 LLL sin

VAIV3 LLL


79/90

79

1.8.2: Delta connection system

VLL= Vphase

P = 3 Vphase Iphase cos

phaseL I3I

WattIV LLL cos3


80/90

80

1.9: Three phase power

measurement


81/90

81

Power measurement

In a four-wire system (3 phases and a

neutral) the real power is measured using

three single-phase watt-meters.

Three Phase Circuit


82/90

82

Three Phase Circuit

Four wire system,

Each phase measured separately

A

A

V

W

W

Phase A

Phase B

Phase C

VAN

IA

IC

V

A

V

W

IB

VBN

VCN

Neutral (N)

PA

PB

PC

watt-meter connection


83/90

8383

Current coil (low impedance)

voltage coil (high impedance)

W

Examplea)Four wire system


84/90

Example

ER

Three-phase

Load

IR

VR

EY

EB

ZR= 5

IYIB

ZB= 20

VB

IN

VN

Find the three-phase total power, PT.

EL = 415 volt30o

ZY= 10 90o

45o

WR

WB

WY


85/90

Exampleb)Three wire system


86/90

Example

ER

Three-phase

Load

IR

VR

EY

EB

ZR= 5

IYIB

ZB= 20

VN

Find the three-phase total power, PT.

EL = 415 volt30o

ZY= 10 90o

45o

WR

WB

WY

VB

Th Ph Ci it


87/90

87

Three Phase Circuit

Three wire system,

The three phase power is the sum of the two watt-

meters readingA

A

V

V

W

W

Phase A

Phase B

Phase C

VAB

= VA

- VB

VCB

= VC

- VB

IA

IC

PAB

PCBCBABT PPP

CBABT PPP Proving:


88/90

8888

The three phase power (3-wire system) is the

sum of the two watt-meters reading

Instantaneous power:

pA = vA iA

pB= vBiB

pC= vCiC

A

A

V

W

W

Phase A

Phase B

Phase C

VAN

IA

IC

V

A

V

W

IB

VBN

VCN

Neutral (N)

PA

PB

PC

pT= pA + pB+ pC= vA iA + vBiB+vCiC

= vA iA + vBiB+vCiC= vA iA + vB(-iA -iC)+vCiC

Proving:


89/90

8989

The three phase power (3-wire system) is the

sum of the two watt-meters reading

CBABT PPP

Instantaneous power:A

A

V

W

W

Phase A

Phase B

Phase C

VAN

IA

IC

V

A

V

W

IB

VBN

VCN

Neutral (N)

PA

PB

PC

pT= pAB+ pCB

pT= vA iA + vB(-iAiC) +vCiC

= (vA vB)iA + (vC vB)iC

= vABiA + vCBiC

A

A

V

V

W

W

Phase A

Phase B

Phase C

VAB

= VA

- VB

VCB

= VC

- VB

IA

IC

PAB

PCB


90/90

Power measurement

In a four-wire system (3 phases and a

neutral) the real power is measured using

three single-phase watt-meters.

In a three-wire system (three phases

without neutral) the power is measured

using only two single phase watt-meters.

The watt-meters are supplied by the line

current and the line-to-line voltage.

chapter 1-three phase

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