polyphase circuits learning goals three phase circuits

POLYPHASE CIRCUITSLEARNING GOALS

Three Phase CircuitsAdvantages of polyphase circuits

Three Phase ConnectionsBasic configurations for three phase circuits

Source/Load ConnectionsDelta-Y connections

Power RelationshipsStudy power delivered by three phase circuits

Power Factor CorrectionImproving power factor for three phase circuits

THREE PHASE CIRCUITS

))(240cos()())(120cos()(

))(cos()(

VtVtvVtVtv

VtVtv

mc

mbn

man

VoltagesPhase ousInstantane

ai

bi

ci

)240cos()()120cos()(

)cos()(

tItitItitIti

mc

mb

ma

Currents Phase Balanced

2120mV

)()()()()()()( titvtitvtitvtp ccnbbnaan power ousInstantane

TheoremFor a balanced three phase circuit the instantaneous power is constant

)(cos2

3)( WIVtp mm

Proof of TheoremFor a balanced three phase circuit the instantaneous power is constant

)(cos2

3)( WIVtp mm

)240cos()240cos()120cos()120cos(

)cos(cos)(

)()()()()()()(

tttt

ttIVtp

titvtitvtitvtp

mm

ccnbbnaan

power ousInstantane

)cos()cos(21coscos

)4802cos()2402cos()2cos(cos3

)(

tt

tIVtp mm

)120sin(sin)120cos(cos)120cos()120sin(sin)120cos(cos)120cos(

coscos

0)120cos()120cos(cos

Proof

Lemma5.0)120cos(

0)120cos()120cos(cos

)120cos()480cos()120cos()240cos(

t

THREE-PHASE CONNECTIONS

Positive sequencea-b-c

Y-connectedloads Delta connected loads

SOURCE/LOAD CONNECTIONS

BALANCED Y-Y CONNECTION

120||

120||

0||

pcn

pbn

pan

VV

VV

VV

Positive sequencephase voltages

abV

bcV

caV

Line voltages

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 VVY

cnc

Y

bnb

Y

ana Z

VIZVI

ZVI ;;

120||;120||;|| LcLbLa IIIIII

0 ncba IIII For this balanced circuit it is enough to analyze one phase

Relationship betweenphase and line voltages

LEARNING EXAMPLEFor an abc sequence, balanced Y - Y three phase circuit 30208abV

Determine the phase voltages

Balanced Y - Y


30||3 pab VV

120||

120||

0||

pcn

pbn

pan

VV

VV

VV


)3030(3

|| ab

anVV

30by lags aban VV

30208abV

6012018012060120

an

bn

an

VVV

The phasor diagram could be rotated by any angle

LEARNING EXAMPLEFor an abc sequence, balanced Y - Y three phase circuit 1020,11,)(120|| jZjZVV phaselinermsphase source

Determine line currents and load voltagesBecause circuit is balanceddata on any one phase aresufficient

0120 Chosenas reference

120120120120

0120

an

bn

an

VVV

Abc sequence

rmsAj

VI anaA

)(65.2706.565.2771.23

01201121

rmsAIrmsAI

cC

bB

)(65.2712006.5)(65.2712006.5

57.2636.22)1020( aAaAAN IjIV

rmsVVAN )(08.115.113

rmsVVrmsVV

CN

BN

)(92.11815.113)(08.12115.113

LEARNING EXTENSIONFor an abc sequence, balanced Y - Y three phase circuit

voltagesline the Find .)(90120 rmsVVan


30||3 pab VV

30by lags aban VV

30by leads anab VV

rmsVVab )(1201203

rmsVV

rmsVV

ca

bc

)(2401203

)(01203

voltagesphase the Find.)(0208 rmsVVab

30by lags aban VV

rmsVVan )(303

208

rmsVV

rmsVV

cn

bn

)(903

208

)(1503

208

LEARNING EXTENSIONFor an abc sequence, balanced Y - Y three phase circuit 38,11,)(6.2602.104|| jZjZVV phaselinermsphase load,

Determine source phase voltages

rmsV )(6.2602.104

8

3j

Currents are not required. Use inversevoltage divider

ANan Vj

jjV38

)11()38(

41.315.1

73584

3838

3849 j

jj

jj

30120anV


15012090120

cn

bn

VV

DELTA CONNECTED SOURCES Convert to an equivalent Y connection


30||3 pab VV

30by lags aban VV

120120

0

Lca

Lbc

Lab

VVVVVV

903

1503

303

Lcn

Lbn

Lan

VV

VV

VV

15012090120

30120

cn

bn

an

VVV

18020860208

60208

ca

bc

ab

VVVExample

LEARNING EXAMPLEDetermine line currents and line voltages at the loads

Source is Delta connected.Convert to equivalent Y

120120

0

Lca

Lbc

Lab

VVVVVV

903

1503

303

Lcn

Lbn

Lan

VV

VV

VV

Analyze one phase

rmsAj

IaA )(14.4938.92.41.1230)3/208(

rmsVjVAN )(71.3065.11819.4938.9)412(

Determine the other phases using the balance

rmsAIrmsAI

cC

bB

)(86.7138.9)(14.16938.9

30||3 pab VV 71.065.1183ABV

71.12065.1183

29.11965.1183

CA

BC

V

V

LEARNING EXTENSIONCompute the magnitude of the line voltage at the load


120120

0

Lca

Lbc

Lab

VVVVVV

903

1503

303

Lcn

Lbn

Lan

VV

VV

VV

Analyze one phase

301201.41.10

410jjVAN

10

1.0j

Only interested in magnitudes!

rmsVVAN )(57.11890.1077.10120||

30||3 pab VV rmsVVAB )(4.205||

DELTA-CONNECTED LOAD

Method 1: Solve directly

120||

120||

0||

pcn

pbn

pan

VV

VV

VV


210||3

90||3

30||3

pca

pbc

pab

VV

VV

VV

120||

120||

||

IZVI

IZVI

IZVI

CACA

BCBC

ABAB

currents phase Load

BCCAcC

ABBCbB

CAABaA

IIIIIIIII

currents Line

Method 2: We can also convert the delta connected load into a Y connected one. The same formulas derived for resistive circuits are applicable to impedances

3

ZZY case Balanced

ZL

ABaA

LaAY

anaA Z

VIIZVI

3/||

3/||||||

ZLZZ ||

Z 30

30||3||

lineline II

Line-phase currentrelationship

Y

Y baab RRR

)(|| 312 RRRRab

321

312 )(RRRRRRRR ba

321

213 )(RRRRRRRR cb

321

321 )(RRRRRRRR ac

SUBTRACT THE FIRST TWO THEN ADDTO THE THIRD TO GET Ra

a

b

b

a

RRRR

RR

RR 1

33

1 c

b

c

b

RRRR

RR

RR 1

21

2

REPLACE IN THE THIRD AND SOLVE FOR R1

YRRR

RRR

RRRRRR

RRRRRR

c

b

a

321

13

321

32

321

21

YR

RRRRRRR

RRRRRRRR

RRRRRRRR

a

accbba

c

accbba

b

accbba

3

2

1

tionsTransformaY

OF REVIEW

3321

RRRRRR Y

30||3||

lineline II


iprelationsh voltagephase-Line

30

||3||

phase

phaseVV

LEARNING EXTENSION

currents phase the Find .4012 aAI

19093.65093.6

7093.6

CA

BC

AB

III

LEARNING EXAMPLE

02.3752.1254.710 jZ

Delta-connected load consists of 10-Ohm resistance in serieswith 20-mH inductance. Source is Y-connected, abc sequence,120-V rms, 60Hz. Determine all line and phase currents

rmsVVan )(30120

54.7020.0602inductanceZ

rmsAjZ

VI ABAB )(98.2260.16

54.710603120

30||3||

lineline II



30

||3||

phase

phaseVV

rmsAIaA )(02.775.28

rmsAIrmsAI

CA

BC

)(98.14260.16)(02.9760.16

rmsAIrmsAI

cC

bB

)(98.11275.28)(02.12775.28

02.3717.4YZ

Alternatively, determine first the line currentsand then the delta currents

POWER RELATIONSHIPS


30

||3||

phase

phaseVV

30||3||

lineline II


*3 phasephaseTotal IVS

*3 linelineTotal IVS

*3 IVS linetotal

*3 linelineTotal IVS

lineV

lineI

Power factor anglef

flinelinetotal

flinelinetotal

IVQ

IVP

sin||||3

cos||||3

- Impedance angle

LEARNING EXAMPLE

lagging 20 anglefactor power

WPrmsVV

total

line

1200)(208||

Determine the magnitude of the line currents and the value of load impedanceper phase in the delta

flinelinetotal

flinelinetotal

IVQ

IVP

sin||||3

cos||||3

flinelinetotal IVP cos

3||||

3 rmsAI line )(54.3||

30||3||

lineline II


rmsAI )(05.2||

46.101

||||||

IVZ line

2046.101Z

lineV

lineI

Power factor anglef

- Impedance angle

lineV

LEARNING EXAMPLEFor an abc sequence, balanced Y - Y three phase circuit 1020,11,)(120|| jZjZVV phaselinermsphase source

Because circuit is balanceddata on any one phase aresufficient

0120 Chosenas reference

120120120120

0120

an

bn

an

VVV

Abc sequence

rmsAj

VI anaA

)(65.2706.565.2771.23

01201121

Determine real and reactive power per phase at the load and total real, reactive andcomplex power at the source

57.2636.22)1020( aAaAAN IjIV

rmsVVAN )(08.115.113

65.2706.508.115.113*aAANphase IVS

rmsVAjS phase )(09.25651257.2654.572

phase per P phase per Q 65.2706.50120*

aAan IVS phase source

VAj

S

78.28186.537

65.272.607

phase source

)(78.2813)(86.5373

VAQWP

source total

source total

)(6.1821||)(2.8456.1613

VASVAj

QPS

source total

source totalsource totalsource total

LEARNING EXAMPLE

rmsVVline )(208 leading 0.8pfat 12kVA :3 Load

1 pfat 10kW :2 Loadlagging 0.6pfat 24kW :1 Load

balanced isCircuit

Determine the line currents and the combined power factor

kVASlaggingpf

kWP40||

6.024

11

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

kVAPSQ 32|||||| 21

211

kVAjSinductivelagging 32241

kVAjSpf

kWP010

110

22

2 Load

kVAQkWP

pfkVAS

2.7||6.9

8.012||

3

33

3 Load

kVAjScapacitivepfleading 2.76.93

kVAkVAjSSSSTOTAL 63.29160.508.246.43321

flinelinetotal

flinelinetotal

IVQ

IVP

sin||||3

cos||||3

63.29

||||3||

f

linelinetotal IVS

laggingpf 869.0

inductive

capacitive

f

rmsAI line )(23.139||

321 SSSStotal

Continued ...

source theat factor power and voltagesline determine are impedances line the If 02.005.0 jZ lineLEARNING EXAMPLE

continued ….

rmsAI line )(23.139||

2* ||3)(3 linelinelinelinelineline IZIIZS

)(11632908 VAjSline

kVAkVAjS 63.29160.508.246.43total load

kVAjS 17.29264.53963.25508.46total source

17.29||||3||

f

linelinetotal IVS

inductive

capacitive

f

rmsVVline )(87.22013.1393

264,53

laggingpf f 873.0)17.29cos(cos

LEARNING EXTENSIONA Y -Y balanced three-phase circuit has a line voltage of208-Vrms. The total real power absorbed by the load is 12kWat pf=0.8 lagging. Determine the per-phase impedanceof the load


30

||3||

phase

phaseVV

*3 phasephaseTotal IVS

rmsVVphase )(1203

208||

*

2*||

33phase

phase

phase

phasephasetotal Z

VZV

VS

lineV

lineI

Power factor anglef

Impedance angle

87.36cos8.0 ffpf

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

kVApfPS total

total 15||

88.2||

||3||

2

total

phasephase S

VZ

87.3688.2pahseZ

301201.41.10

410jjVAN


Analyze one phase

10

1.0j

rmsVVAN )(57.11890.1077.10120||

LEARNING EXTENSIONDetermine real, reactive and complex power at both loadand source

*

2* ||33

phase

ANaAANload Z

VIVS

*

2* ||33

phasetotal

anaAansource Z

VIVS

22

22

410)410(|57.118|3

410|57.118|3

j

j

22

22

)1.4()1.10()1.41.10(|120|3

1.41.10|120|3

jj

)4961.1224(3)8.4840.212,1(3

jSjS

source

load

LEARNING EXTENSIONA 480-V rms line feeds two balanced 3-phase loads. The loads are ratedLoad 1: 5kVA at 0.8 pf laggingLoad 2: 10kVA at 0.9 pf lagging.

Determine the magnitude of the line current from the 408-V rms source

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

21 SSStotal

kVAPSQ

kWPPkVAS

0.3||

48.0

5||

21

211

11

kVAjSlaggingpf 341

kVAjSkVAPSQ

kWPPkVAS

36.4936.4||

99.0

10||

2

22

222

2

kVAjStotal 36.713 ||||3||

sin||||3

cos||||3

linelinetotal

flinelinetotal

flinelinetotal

IVS

IVQ

IVP

rmsAV

SIline

totallineq )(14.21

68.706939,14

||3||||

POWER FACTOR CORRECTION

Balanced loadLow pflagging

Similar to single phase case.Use capacitors to increase thepower factor

Keep clear about total/phasepower, line/phase voltages

oldold

old QpfS

newnew

old QpfP

added be toPower Reactive

oldnew QQQ

To use capacitors this valueshould be negative

connected are capacitors thehow on depends voltageThe

capacitorper 2CVQ

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

21sincos pfpf ff

21tan

pfpf

f

0 QlaggingfPQ tan

LEARNING EXAMPLE leading pf rms kV V Hz fline94 . 0 . 5. 34 | | , 60 : Required

626.01sincos 2 pfpf ff

MWPMVAQ

old

old

72.1802.15||

21tan

pfpf

f

0 oldQlagging

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

fPQ tan

MVAQleadingpfMWP

newnew

old 8.694.0

72.18

MVAQMVAQ

273.782.2102.158.6

capacitorper

rmskVVY35.34

capacitorconnection23

6

3105.3460210273.7

C

FC 6.48

LEARNING EXAMPLE lagging pf rms kV V Hz fline90 . 0 . 5. 34 | | , 60 : Required

626.01sincos 2 pfpf ff

MWPMVAQ

old

old

72.1802.15||

21tan

pfpf

f

0 oldQlagging

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

fPQ tan

MVAQlaggingpfMWP

newnew

old 067.990.0

72.18

MVAQMVAQ

984.1953.502.15067.9

capacitorper

rmskVVY35.34

capacitorconnection23

6

3105.3460210984.1

C

FC 26.13

Determine the currentflowing. Convert linevoltages to phase voltages

LEARNING EXAMPLEMEASURING POWER FLOW

rmskV 012 rmskV 512

Which circuit is the source and what is the average power supplied?

Equivalent 1-phase circuit

21

253

12000303

12000

21 jjVVI ANan

aA

rmsAIaA )(93.18030.270

*

*

)(3

3

aAanX

aAANY

IVS

IVS

MVASY )93.18025(2703.0123

MWPMWP

X

Y

91.413.5

fSPjQPScos||

System Y is the source

MVASX )93.18030(2703.0123

)( YXloss PPP

Phase differencesdetermine directionof power flow!

CAPACITOR SPECIFICATIONS

Capacitors for power factor correction are normally specified in VARs

given. is eother valu unless Assumeecapacitanc the know toorder in given bemust frequency and voltageThe

capacitorper

Hz

CVQ

60

|| 2

FC

53.26)1050(602

102523

6

2

FC

7.49)1020(602

105.723

6

3

FC

1.106)1010(602

10423

6

1

kVVkVV phaseline 9.195.34

LEARNING EXAMPLE

FC 6.48 needs one leading 0.94pfFor Capacity Rated Voltage (kV) Rated Q (Mvar)

1 10 42 50 253 20 7.5

Choices available

Capacitor 1 is not rated at high enoughvoltage!

Capacitor 3 is the best alternative

LEARNING BY DESIGNProposed new storerms A 170at rated wire #4ACSR

1. Is the wire suitable?

2. What capacitancewould be required tohave a composite pf =0.92 laggingCapacitors are to beY - connected

f

f

f

pf

SQ

SPjQPS

cos

sin||

cos||

*

*

3

3

lineline

phasephaseTotal

IV

IVS

321 SSSStotal

kVAjS

4205609.367001

kVAjkVAS

8665006010002

kVAjkVAS

3497208.258003

kVAkVAjStotal 57.42241716351780

rmsAV

SIline

totalline 1.101

108.13310417.2

3|||| 3

6

Wire is OK

newnew

old QpfP

oldnew QQQ

)(tan newfP kVA28.758

kVA72.8762|| CVQ capacitorper

38.13|| kVVV phase

3/108.13602

3/1072.87623

3

C F2.12

Polyphase

polyphase circuits learning goals three phase circuits

Documents