1 chapter 1. three-phase system. 1.1: review of single-phase system the sinusoidal voltage v 1 (t) =...

1

Chapter 1.

Three-Phase System

1.1: Review of Single-Phase System

The Sinusoidal voltagev1(t) = Vm sin t

i

v1

Load

AC generator

v2

2

3

1.1: Review of Single-Phase System

The Sinusoidal voltagev(t) = Vm sin t

whereVm = the amplitude of the sinusoid = the angular frequency in radian/s t = time

v(t)

Vm

-Vm

t

4

v(t)

Vm

-Vm

t

2

TT

1f

f2The angular frequency in radians per second

5

A more general expression for the sinusoid (as shown in the figure):

v2(t) = Vm sin (t + )

where is the phase

v(t)

Vm

-Vm

t

V1 = Vm sin t

V2 = Vm sin t + )

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 (ωt ± 180o) = - sin ωt

cos (ωt ± 180o) = - cos ωt

sin (ωt ± 90o) = ± cos ωt

cos (ωt ± 90o) = + sin ωt

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 (ωt + θ)

Time domain Phasor domain

Time domain

mVV Phasor domain

)cos( tVm mV

)sin( tVm

om 90V

)cos( tmI mI

)sin( tmIo

m 90I

8

Time domain

Phasor domain

v(t)

Vm

-Vm

t

V1 = Vm sin t

V2 = Vm sin t + )

θ

V1

V2

v2(t) = Vm sin (t + )

v1(t) = Vm sint

mVV2

01 mVV

or

or 01 rmsVV

rmsVV2

9

1.1.1: Instantaneous and Average Power

The instantaneous power is the power at any instant of time.

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

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

Using the trigonometric identity, gives

)cos()cos()( ivmmivmm t2IV2

1IV

2

1t p

10

The average power is the average of the instantaneous power over one period.

T

dttpT

P0

)(1

)cos( ivmmIV2

1P

p(t)

t

)cos( ivmmIV2

1

mmIV2

1

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( ivP rmsrmsIV

2

VV m

rms

2

II m

rms

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.

rmsrmsIVS

)sin( ivQ rmsrmsIV

13

The power factor is the cosine of the phase difference between voltage and current.

The complex power:

)cos( ivfactor Power S

P

iv

jQP

rmsrms IV

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.

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 apart around the stator.

Y’

BY

B’

Stator

Rotor

Y

R

B

R

R’

N

S

16

V or v is generally represented a voltage, but to differentiate the emf voltage of generator from voltage drop in a circuit, it is convenient to use e or E for induced (emf) voltage.

1717

v(t)

t

vR

vY vB

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

eR = EmR sin t

eY = EmY sin (t -120o)

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

ER

Three-phase Load

Three-phase AC generator

VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN

18


eR = EmR sin ωt

eY = EmY sin (ωt -120o)

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

In phasor domain:

ER = ERrms

0o

EY = EYrms

-120o

EB = EBrms

120o

Phase voltagePhase voltage

120o

-120o

0o

ERrms = EYrms = EBrms = Ep ERrms = EYrms = EBrms = Ep Magnitude of phase voltage19

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN

Line voltageLine voltage

ERY

ERY = ER - EYERY = ER - EY

21


ERY = ER - EYERY = ER - EY

120o

-120o

0o

-EY

ERY= Ep 0o - Ep -120o

= 1.732Ep

ERY

30o= √3 Ep

= EL 30o

30o

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN


EYB

EYB = EY - EBEYB = EY - EB


EYB = EY - EBEYB = EY - EB

120o

-120o

0o

-EB

EYB

= Ep -120o - Ep120o

= 1.732Ep

EYB

-90o

-90o= √3 Ep

= EL -90o

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

IN


EBR

EBR = EB - EREBR = EB - ER

2525


EBR = EB - EREBR = EB - ER

120o

-120o

0o

-ER

EBR

= Ep 120o- Ep 0o

= 1.732Ep

EBR

150o

150o= √3 Ep

= EL 150o

For star connected supply, EL= √3 Ep

26

120o

-120o

0o

Phase voltagesPhase voltages

ER = Ep

0o

EY = Ep

-120o

EB = Ep

120o

Line voltagesLine voltages

ERY = EL 30o

EYB = EL -90o

EBR = EL 150o

It can be seen that the phase voltage ER is reference.

2727

Phase voltagesPhase voltages

ER = Ep

-30o

EY = Ep

-150o

EB = Ep

90o

Line voltagesLine voltages

ERY = EL 0o

EYB = EL -120o

EBR = EL 120o

Or we can take the line voltage ERY as reference.

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

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

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 Loads a) 4-wire system b) 3-wire system

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

Delta-Connected Balanced Loads a) 3-wire system

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

INZN

VN


VN = INZNVN = INZNVoltage drop across neutral impedance:

1.1

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

INZN

VN


IR + IY + IB= IN

Applying KCL at star pointApplying KCL at star point

1.2

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

INZN

VN


Applying KVL on R-phase loopApplying KVL on R-phase loop

33

ER

Three-phase Load


IR

VR ZR

INZN

VN

Applying KVL on R-phase loopApplying KVL on R-phase loop

ER – VR – VN = 0

ER – IRZR – VN = 0

IR = Thus ER – VN

ZR

1.3


34

35

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

INZN

VN


Applying KVL on Y-phase loopApplying KVL on Y-phase loop

36Three-phase

Load


VYEY

IY

ZY

INZN

VN

Applying KVL on Y-phase loopApplying KVL on Y-phase loop


EY – VY – VN = 0

EY – IYZY – VN = 0IY =

Thus EY – VN

ZY

1.4

Three-phase Load


EB

IB

ZB

VB

INZN

VN


Applying KVL on B-phase loopApplying KVL on B-phase loop

EB – VB – VN = 0

EB – IBZB – VN = 0IB =

Thus EB – VN

ZB

1.5

37

38


IR + IY + IB= IN

Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into Eq. 1.1:

=EB – VN

ZB

+ EY – VN

ZY

ER – VN

ZR

+ VN

ZN

ER – VN EY – VN + + EB – VN = VN

ZNZR ZR ZY ZY ZB ZB

ER

ZR

+ EY

ZY

+ EB

ZB

=1

ZN

+ 1

ZR

+ 1

ZY

VN + 1

ZB

39


ER

ZR

+ EY

ZY

+ EB

ZB

=1

ZN

+ 1

ZR

+ 1

ZY

VN + 1

ZB

VN =

ER

ZR

+ EY

ZY

+ EB

ZB

1

ZN

+ 1

ZR

+ 1

ZY

+ 1

ZB

1.6


VN =

ER

ZR

+ EY

ZY

+ EB

ZB

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 and supply star point of three-phase system.

We have to determine the value of VN in order to find the values of currents and voltages of star connected loads of three-phase system. 40

ExampleExample

ER

Three-phase Load

ZY= 2 Ω

IR

VR

EY

EB

ZR = 5 Ω

IY

IB

ZB = 10 Ω

VB

INZN =10 Ω

VN

Find the line currents IR ,IY and IB. Also find the neutral current IN.

EL = 415 volt

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

INZN

VN


ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB

VN


No neutral line = open circuit , ZN = ∞No neutral line = open circuit , ZN = ∞ 43

44

VN =

ER

ZR

+ EY

ZY

+ EB

ZB

1

ZN

+ 1

ZR

+ 1

ZY

+ 1

ZB

1.6


∞=ZN ∞

1

∞= 0

45

VN =

ER

ZR

+ EY

ZY

+ EB

ZB

1

ZR

+ 1

ZY

+ 1

ZB

1.7


ExampleExample

ER

Three-phase Load

ZY= 2 Ω

IR

VR

EY

EB

ZR = 5 Ω

IY

IB

ZB = 10 Ω

VBVN

EL = 415 volt

Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB.

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

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IY

IB

ZB ZY

VB


ER

Three-phase Load


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy


ER

Three-phase Load


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy

ERY=VRY

EYB =VYB

EBR =VBR

50


Phase currentsPhase currents

30o

Ir = VRY

ZRY

=ERY

ZRY

=EL

ZRY

-90o

Iy = VYB

ZYB

=EYB

ZYB

=EL

ZYB

150o

Ib = VBR

ZBR

=EBR

ZBR

=EL

ZBR


ER

Three-phase Load


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy

ERY=VRY

EYB =VYB

EBR =VBR

Line currentsLine currents

IR = Ir Ib -

= EL

ZRY

30o- 150oEL

ZBR

IY = Iy Ir -

= EL

ZYB

-90o- 30oEL

ZRY


ER

Three-phase Load


IR

EY

EB

IY

IB

VRY

ZRYZBR

ZYB

VYB

VBR

Ir

IbIy

ERY=VRY

EYB =VYB

EBR =VBR

Line currentsLine currents

IB = Ib Iy -

= EL

ZBR

150o- -90oEL

ZYB

ZRYZBR

ZYB

ZR

ZB

ZY

Star to delta conversion

ZRY = ZRZY + ZYZB + ZBZR

ZB

ZYB = ZRZY + ZYZB + ZBZR

ZR

ZBR = ZRZY + ZYZB + ZBZR

ZY

ExampleExample

ER

Three-phase Load

ZY= 2 Ω

IR

VR

EY

EB

ZR = 5 Ω

IY

IB

ZB = 10 Ω

VBVN

Find the line currents IR ,IY and IB .

EL = 415 volt

Use star-delta conversion.

ER

Three-phase Load


VY

IR

VR

EY

EB

ZR

IR

IB

ZB ZY

VB

INZN

VN


= 0 Ω

VN = INZN = IN(0) = 0 voltVN = INZN = IN(0) = 0 volt 55

56


For 4-wire three-phase system, VN is equal to 0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5 become,

IB = EB

ZB

1.5EB – VN

IY = EY

ZY

1.4EY – VN

IR = ER

ZR

1.3ER – VN

ExampleExample

ER

Three-phase Load

ZY= 2 Ω

IR

VR

EY

EB

ZR = 5 Ω

IY

IB

ZB = 10 Ω

VB

IN

VN

Find the line currents IR ,IY and IB . Also find the neutral current IN.

EL = 415 volt

58

v(t)

t

vR

vY vB


eR = EmR sin t

eY = EmY sin (t -120o)

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

59

1.4: Phase sequencesRYB and RBY

120o

-120o

120oVR

VY

VB

o

)rms(RR 0VV

o

)rms(YY 120VV

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

(b) RBY or negative sequence

120o

-120o

120oVR

VB

VY

o

)rms(RR 0VV

o

)rms(BB 120VV

ormsY

ormsYY

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

1.5: Connection in Three Phase System

R

Y

B

ZR

1.5.1: Star Connection a) Three wire system

62

Star Connection b) Four wire system

VRN

VBN VYN

ZR

Y B

R

BN

Y

63

Wye connection of Load

Z1

Z3

Z2

R

B

Y

NLoad

Z3

12

R

Y

B

Load

N

64

1.5.2: Delta Connection

R

Y

B

Y

B

R

65

Delta connection of load

Zc

Za

Zb

R

B

Y

Load

c b

Za

R

Y

B

Load

66

1.6: Balanced Load Connection in 3-Phase System

67

ExampleExample

ER

Three-phase Load

ZY= 20 Ω

IR

VR

EY

EB

ZR = 20 Ω

IY

IB

ZB = 20 Ω

VBVN

EL = 415 volt

Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB.

Wye-Connected Balanced Loads b) Three wire system

68


VN = = 0 voltVN = = 0 volt

VR = ER

VY = EY

VB = EB

69

ExampleExample

ER

Three-phase Load

ZY= 20 Ω

IR

VR

EY

EB

ZR = 20 Ω

IY

IB

ZB = 20 Ω

VB

IN

VN

Find the line currents IR ,IY and IB . Also find the neutral current IN.

EL = 415 volt

1.6.1: Wye-Connected Balanced Loadsa) Four wire system

70

VRN

VBN

Z1

Z 2Z

3

R

B

N

Y

VYN

IR

IY

IB

IN

BYRN IIII

For balanced load system, IN = 0 and Z1 = Z2 = Z3

3

o

BNB

2

o

YNY

1

o

RNR

Z

120VI

Z

120VI

Z

0VI

BNYNRNphasa

phasaBN

phasaYN

phasaRN

VVVVwhere

120VV

120VV

0VV

1.6.1: Wye-Connected Balanced Loadsa) Four wire system

71


R

Y

B

Z1

Z 2 Z3

IR

IY

IB

VRY

VYB

VBR S

0III BYR

3

o

BSB

2

o

YSY

1

o

RSR

Z

120VI

Z

120VI

Z

0VI

BSYSRSphasa

phasaBS

phasaYS

phasaRS

VVVVwhere

240VV

120VV

0VV

72

1.6.2: Delta-Connected Balanced Loads

Z

R

Y

B

VRY

VYB

VBR

IR

IRYIBR

IYB

IB

IY

Phase currents:

3

o

BRBR

2

o

YBYB

1

o

RYRY

Z

120VI

Z

120VI

Z

0VI

Line currents:

YBBRB

RYYBY

BRRYR

III

III

III

lineBYR

phasaBRYBRY

IIIIand

IIIIwhere

73

1.7: Unbalanced Loads

74

1.7.1: Wye-Connected Unbalanced LoadsFour wire system

VRN

VBN

Z1

2 3

R

B

N

Y

VYN

IR

IY

IB

IN

BYRN IIII

For unbalanced load system, IN 0 and Z1 Z2 Z3

3

o

BNB

2

o

YNY

1

o

RNR

Z

120VI

Z

120VI

Z

0VI

120VV

120VV

0VV

phasaBN

phasaYN

phasaRN

75

1.7.2: Delta-Connected Unbalanced Loads

Z

R

Y

B

VRY

VYB

VBR

IR

IRYIBR

IYB

IB

IY

Phase currents:

3

o

BRBR

2

o

YBYB

1

o

RYRY

Z

120VI

Z

120VI

Z

0VI

Line currents:

YBBRB

RYYBY

BRRYR

III

III

III

120VV

120VV

0VV

phasaBN

phasaYN

phasaRN

76

1.8 Power in a Three Phase System

77

Power Calculation

The three phase power is equal the sum of the phase powers

P = PR + PY + PB

If the load is balanced: P = 3 Pphase = 3 Vphase Iphase cos θ

78

1.8.1: Wye connection system:

I phase = I L and

Real Power, P = 3 Vphase Iphase cos θ

Reactive power, Q = 3 Vphase Iphase sin θ

Apparent power, S = 3 Vphase Iphase

or S = P + jQ

phaseLL VV 3

WattIV LLL cos3

VARIV3 LLL sin

VAIV3 LLL

79

1.8.2: Delta connection system

VLL= Vphase

P = 3 Vphase Iphase cos θ

phaseL I3I

WattIV LLL cos3

80

1.9: Three phase power measurement

81

Power measurement

In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters.

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

8383

watt-meter connection

Current coil (low impedance)

voltage coil (high impedance)

W

ExampleExample

ER

Three-phase Load

Ω

IR

VR

EY

EB

ZR = 5

IYIB

ZB = 20

VB

IN

VN

Find the three-phase total power, PT.

EL = 415 volt

a) Four wire system

30o

ZY = 10 90o

45o Ω

Ω

WR

WB

WY

ExampleExample

ER

Three-phase Load

Ω

IR

VR

EY

EB

ZR = 5

IYIB

ZB = 20

VN


EL = 415 volt

b) Three wire system

30o

ZY = 10 90o

45o Ω

Ω

WR

WB

WY

ExampleExample

ER

Three-phase Load

Ω

IR

VR

EY

EB

ZR = 5

IYIB

ZB = 20

VN


EL = 415 volt

b) Three wire system

30o

ZY = 10 90o

45o Ω

Ω

WR

WB

WY

VB

87

Three Phase Circuit Three wire system,

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

A

A

V

V

W

W

Phase A

Phase B

Phase C

VAB = VA - VB

VCB = VC - VB

IA

IC

PAB

PCBCBABT PPP

8888

The three phase power (3-wire system) is the sum of the two watt-meters reading

CBABT PPP Proving:

Instantaneous power:

pA = vA iA

pB = vB iB

pC = vC iC

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 + vB iB +vC iC

= vA iA + vB iB +vC iC = vA iA + vB (-iA -iC) +vCiC

8989

The three phase power (3-wire system) is the sum of the two watt-meters reading

CBABT PPP

Proving:

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 (-iA –iC) +vCiC

= (vA – vB )iA + (vC – vB )iC

= vAB iA + vCBiC

A

A

V

V

W

W

Phase A

Phase B

Phase C

VAB = VA - VB

VCB = VC - VB

IA

IC

PAB

PCB

9090

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.

1 chapter 1. three-phase system. 1.1: review of single-phase system the sinusoidal voltage v 1 (t) =...

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