Balanced poly phase circuits

Two and four phase systems

A two phase system is an electrical system in which the voltages of the phases are 90 degree out of time phase.

The emf’s Eac and Ebd are 90 degree out of time phase. A two phase system is the equivalent of two separate single-phase systems

that are separated 90 degree in time phase.

If the connection is made between the two windings at n and n`, the system would be called a four-phase system.

The voltages Eda, Eab, Ebc and Ecd are called the line voltages, while voltages E0a, E0b, E0c and E0d are called the phase voltages or voltages to neutral.

Eda=Ed0+E0a. Thus, the line voltages= times the phase voltages in the four phase star.

2

Two and four phase systems

The line voltages are 45 degree or 135 degree out of time phase from the phase voltages.

For four-phase mesh, the aa` line current is Iaa`=Ida+Iba. Thus, the line currents= times the phase currents degree out of time

phase in the four phase mesh2

Three phase four wire systems

The line voltages are 45 degree or 135 degree out of time phase from the phase voltages.

For four-phase mesh, the aa` line current is Iaa`=Ida+Iba. Thus, the line currents= times the phase currents degree out of time

phase in the four phase mesh2

Three phase four wire system

V an

V bn

V cn - V bn

V ab

V a b = V an - V b n

V ab = 3 V an 3 0 o

3 0 o

N is the neutral wire.Lighting loads are placed from line to neutral. Motors and other three phase power loads are connected between the three lines.


30||3

23

21||||

)120sin120(cos1||

120||0||

p

pp

p

pp

bnanab

V

jVV

jV

VVVVV

210||3

90||3

pca

pbc

VV

VV

VoltageLine ||3 pL VV


The current in any line is the same as the current in the corresponding phase.

Current in the neutral wire is obtained through the application of KCL.

These currents are equal in magnitude and displaced from one another in time phase by 120 degree.Thus the current in the neutral wire is 0 since

IIL

0 ncnbnann IIII

ncnbnann IIII

Three phase three wire system

VVL

0 0

0

1 30 2402 2

3 3 3 132 2 2 2

3 30

ab caa I I I I I II

I

j I

I j I I j

Balanced Wye loadsGiven the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power.Draw the vector diagram

ivviPI

I

ZVI

voltsjVVVvoltsjV

voltsjjV

voltsjVwattspowertotal

wattsRIphaseperpower

amperesII

voltsVV

na

nc

nb

na

nana

nabnba

nc

nb

na

PP

pL

LP

87.667.12

13.1737.12

13.537.12

1105.1901105.63120127

1105.63120sin120cos127120127

012729049683

968

7.1286

127

1273

2

22

Balanced Delta loadsGiven the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power.Draw the vector diagram

self

Power calculation in balanced systems

PLLPL

LPPpt

PLLPLL

PPpt

PPpPt

PPpp

IVIVIVP

IVIVIVP

InVnPP

IVP

cos3cos3

3cos3

cos3cos3

3cos3

cos

cos

Wye connection

Dealta connection

Volt-Amperes

LLL

L

LLLL

Pppt

IVIV

IVIV

IVvava

33

3

33

3

33

Wye connection

Dealta connection

Reactive Volt-Amperes

PLLPL

LX

PLLPLL

X

PPpX

IVIVP

IVIVP

IVP

sin3sin3

3

sin3sin3

3

sin3

Wye connection

Dealta connection

Single phase and balanced three phase power

cos5.1240sin240sin120sin120sinsinsin

240sin240sin

120sin120sin

sinsin

3

mm

mmmmmm

aba

mma

mmb

mma

IVttIVttIVttIV

PPPPttIVP

ttIVP

ttIVP

For any phase a

tIVIVttIVP

mmmm

mm

2cos2

cos2

sinsin1

Three wattmeter method

Two wattmeter method

Copper required to transmit power under fixed conditions

43

13

13

21

23

32

3

cos3cos

cos3

cos

3

1

21

23

3

1

32

312

1

31

31

31

33

11

RR

wiresofnumberwiresofnumber

coppercopper

II

RR

RIRI

II

VIVI

PPVIP

VIP

Harmonics in balanced three phase wye loads

240sin120sinsin240sin

120sin240sinsin120sin

740sin600sin360sin120sin

sinsinsinsin

7755331

7755331

7755331

7755331

tetetetee

tetetete

tetetetee

tetetetee

mmmmnb

mmmm

mmmmnb

mmmmna

harmonics 1 3 5 7 9 11 13

Phase A 0 0 0 0 0 0 0

Phase B 120 0 240 0 0 240 120

Phase C 240 0 120 0 0 120 240

Harmonics in balanced three phase wye loadsThe fundamental and all harmonics that obtained by adding a multiple of 6 to the fundamental will have the same sequence. These are first, seventh, thirteenth, nineteenth, twenty-fifth and so on.The fifth, elevenths, seventeenths, twenty-thirds have the same sequences but opposite to that of fundamentals.The third, ninth and all multiples of the third will be found to be in phase.

Eba is 30 degree ahead of ena. For the third harmonic, eba3=0. fifth, eba5 lags ena5 by 30 degree.

nabnba eee

907sin3

905sin390sin3

1507sin3

1505sin3150sin3

307sin3

305sin330sin3

77

551

77

551

77

551

tE

tEtEe

tE

tEtEe

tE

tEtEe

m

mmba

m

mmac

m

mmba

Harmonics in balanced three phase wye loads

23

22

72

52

1

27

25

23

21

mmmba

mmmmna

EEEE

EEEEE

Unbalanced system

An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero.

Unbalanced three phase Y connected load.

Line currents DO NOT add up to zero.

In= -(Ia+ Ib+ Ic) ≠ 0

4.2326.9

1205

9.15620

1.5310

6.8650120100

6.8650120100

0100

acabaa

ca

caca

bc

bcbc

ab

abab

bc

bc

ab

III

ZVI

ZVI

ZVI

jV

jV

jVExam-1

50292182

2492182

18192182

2401000866500

1201000866500

0100001000

jZjZ

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jZohmsjZjZ

jZohmsjZ

voltsjE

jE

jE

nc

cc

nc

nb

bb

nb

na

aa

na

nc

nb

na

nbnabnnaba

naaana

bb

ncnb

cnnbbbc

nbna

bnnabba

c

b

a

VVVVVZIV

IIIEEEEIZIZZ

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Z

Z

j

jjjZ

21

12

21

9.610.68

8.86.52

8.516.352822

18192182

Exam-3

The Wye-Wye system with neutral connection

cnlg

nnncncc

bnlg

nnnbnbb

anlg

nnnanaa

ZZZZIEI

ZZZZIEI

ZZZZIEI

nnc

nnncn

b

nnnbn

a

nnnan

nnccbbaa

IZ

ZIEZ

ZIEZ

ZIEIIII

Methods of checking voltage phase sequence

bcbncnbnancn

bccncnbnbn

abbnbnanan

cnbnan

VIZZIZVIZIZVIZIZ

III

0

The voltage across a lamp

cnbncnbnan

bnbccnbnabananan ZZZZZ

ZVZZVZIZ

|

The voltage across “c” lamp

anancacncn IZVIZ

Example:5

volts

IZ

volts

IZ

V

V

V

cncn

anan

ca

bc

ab

55.712.23

45.483.86240100

45.484.86

45.632238090100120100454.14101000100

240100

120100

0100

Two wattmeter method

W1 reads cosaab IV

W2 reads coscbc IV

Three Phase Power Measurement Three-wattmeter method for measuring three-phase power

dtivivivT

P

dtiiivT

dtivivivT

P

vvvvvvvvv

dtivivivT

P

dtivivivT

P

T

cccnbbbnaaanmeter

T

ccbbaan

T

cccnbbbnaaanmeter

ncnc

nbnb

nana

T

cccbbbaaameter

T

cccnbbbnaaanabc

0

0 0

0

00

00

00

0 000

0

1

1

1

1

1

Power factor in unbalanced three phase system

coscos

coscossin

tancos

cossin

..........sinsinsinsin

..........coscoscoscos

cossin

cos

1

22

22

VIofmagnitudeVI

factorpowerVector

VIVI

factorpowerVector

IVIVIVVI

VIVIVI

IVIVIVVI

IVIVIVVI

VIVI

VIfactorpowerVector

cccbbbaaa

cccbbbaaa

cccbbbaaa 283vars

283 watts

Phase a

a

-433vars

250 watts

Phase c

c

200 watts

Phase b0b

cosVI

sinVIVI

Example:11

ampII

ohmsZZ

ohmsZ

ohmsZ

EEEIZIZ

EEEIZIZ

E

E

Egiven

an

bnnc

annb

cn

bn

an

45.48864.0

454.14190100

90100454.141

454.14190100

1201000100

90100

454.141010090100

454.141901000100

120100

0100

907.57

1507.57

307.57

1

2112

22

11

2222121

1212111

Balanced poly phase circuits

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