EMLAB

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11. Polyphase circuits

EMLAB

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1. Three Phase CircuitsAdvantages of polyphase circuits

2. Three Phase ConnectionsBasic configurations for three phase circuits

3. Source/Load ConnectionsDelta-Wye connections

4. Power RelationshipsStudy power delivered by three phase circuits

5. Power Factor CorrectionImproving power factor for three phase circuits

Contents

EMLAB

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11.1 Three phase circuits

))(240cos()(

))(120cos()(

))(cos()(

VtVtv

VtVtv

VtVtv

mcn

mbn

man

Instantaneous voltage

EMLAB

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이행선

R1

1k

+-

R1

1k

1kWW102 RIP rmsLoss 100V(rms)

05.0R

A10rmsI

05.0R

Disadvantage of single phase two wire system

1. If the load increases to 2kW, Ploss becomes 40W. 2. If each wire can sustain heat only to 5W power loss, the cross-

section area should increase 4 times, whereby cost for copper wires increase 4 times.

100V(rms)

R1

1k

+- 2kWW402 RIP rmsLoss 100V(rms)

05.0R

A20rmsI

05.0R

100V(rms)

R1

1k

EMLAB

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이행선

5

R1

1k+-

1kW100V(rms)

05.0R

05.0R+- 100V(rms) 1kW

R1

1k

R1

1k

A10rmsI

100V(rms)

100V(rms)

1. Although the power load increases to 2kW, Ploss remains 10W.2. With 2 times power load, only one copper wire is added.

A0rmsI

Single-phase 3 wire circuits

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Advantage of three phase-four wire circuit

R1

1k

+-

1kW

05.0RrmsaI A010

05.0R

+-

1kW

R1

1k

R1

1k

1. If power load increases to 3kW, the Ploss for a wire is still 5W. 2. With three times larger power load, only one wires need to be added.

W52 RIP rmsLoss

+-

05.0R

R1

1k

05.0R

1kW

rmsbI A12010

rmscI A24010

W52 RIP rmsLoss

W52 RIP rmsLoss

A0rmsIW02 RIP rmsLoss

rmsanV V0100

rmsbnV V120100

rmscnV V240100

EMLAB

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Three phase generator

EMLAB

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Three phase electric motor

1. Power output of single phase circuit

2. Power output of three phase circuit

EMLAB

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Three phase circuits

))(240cos()(

))(120cos()(

))(cos()(

VtVtv

VtVtv

VtVtv

mc

mbn

man

VoltagesPhase ousInstantane

ai

bi

ci

)240cos()(

)120cos()(

)cos()(

tIti

tIti

tIti

mc

mb

ma

Currents Phase Balanced

2120mV

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

TheoremFor a balanced three phase circuit the instantaneous power is constant

cos3)(cos2

3)( rmsrmsmm IVW

IVtp

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

)(cos2

3)( WIV

tp mm

)240cos()240cos(

)120cos()120cos(

)cos(cos

)(

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

power ousInstantane

tt

tt

tt

IVtp

titvtitvtitvtp

mm

ccnbbnaan

)cos()cos(2

1coscos

)4802cos(

)2402cos(

)2cos(cos3

)(

t

t

t

IVtp mm

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

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

coscos

Proof

0)120cos()120cos(cos

Lemma

0)120cos()120cos(cos

)120cos()480cos(

)120cos()240cos(

t

EMLAB

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Three-phase connections

Positive sequencea-b-c

Y-connected loadsDelta connected loads

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voltageLine;LV

current Line;LI

voltagePhase;PV

current Phase;PI

voltagePhase;PV

current Phase;PI

Line voltage, phase voltage

Phase current : current that flows from a voltage source.Phase voltage : voltage drop across a voltage source.

(current that flows in a transmission line.)

(voltage difference between two lines.)

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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

2

3

2

1||||

)120sin120(cos1||

120||0||

p

pp

p

pp

bnanab

V

jVV

jV

VV

VVV

210||3

90||3

pca

pbc

VV

VV

VoltageLine ||3 pL VVY

cnc

Y

bnb

Y

ana Z

VI

Z

VI

Z

VI ;;

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

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

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120

120

anV

bnV

cnV

Y-connected source

Phase voltage

120

120anV

bnV

cnV

Line voltage

30

In a balanced Y-Y connection, phase currents are equal to line currents.

phaseline VV 3

phaseline II

EMLAB

1515Δ- connected source → Y-connected source

120abV

bcV

caV

Line voltage

30

With Δ-connected sources, we can arbitrarily set the reference point and transform to Y-connected sources.

phaseline VV

phaseline II 3

3032

3,

2

3

)90sin,90cos()30sin,30(cos

phasephase

phasephase

bcabline

II

II

III

abI

bcI

caI

120

120anV

bnV

cnV

30

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Example 11.2 For an abc sequence, balanced Y - Y three phase circuit

1020,11,)(120|| jZjZVV phaselinermsphase source

Determine line currents and load voltages

Because circuit is balanceddata on any one phase issufficient

0120Chosenas reference

120120

120120

0120

an

bn

an

V

V

V

Abc sequence

rmsA

j

VI anaA

)(65.2706.5

65.2771.23

0120

1121

rmsAI

rmsAI

cC

bB

)(65.2712006.5

)(65.2712006.5

57.2636.22)1020( aAaAAN IjIV

rmsVVAN )(08.115.113

rmsVV

rmsVV

CN

BN

)(92.11815.113

)(08.12115.113

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Example 11.4 Determine line currents and line voltages at the loads

Source is Delta connected.Convert to equivalent Y

120

120

0

Lca

Lbc

Lab

VV

VV

VV

903

1503

303

Lcn

Lbn

Lan

VV

VV

VV

Analyze one phase

rmsAj

IaA )(14.4938.92.41.12

30)3/208(

rmsVjVAN )(71.3065.11819.4938.9)412(

Determine the other phases using the balance

rmsAI

rmsAI

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

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Delta-connected load

Method 1: Solve directly

120||

120||

0||

pcn

pbn

pan

VV

VV

VV

Positive sequencephase voltages

210||3

90||3

30||3

pca

pbc

pab

VV

VV

VV

120||

120||

||

currents phase Load

IZ

VI

IZ

VI

IZ

VI

CACA

BCBC

ABAB

BCCAcC

ABBCbB

CAABaA

III

III

III

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

VI

IZ

VI

3/||

3/||||

||

ZLZZ ||

Z 30

30

||3||

lineline II

Line-phase currentrelationship

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Y baab RRR )(|| 312 RRRRab

321

312 )(

RRR

RRRRR ba

321

213 )(

RRR

RRRRR cb

321

321 )(

RRR

RRRRR ac

Subtract the first two then add to the third to get Ra

a

b

b

a

R

RRR

R

R

R

R 13

3

1 c

b

c

b

R

RRR

R

R

R

R 12

1

2

Replace in the third and solve for R1

Y

RRR

RRR

RRR

RRR

RRR

RRR

c

b

a

321

13

321

32

321

21

Y

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

a

accbba

c

accbba

b

accbba

3

2

1

321 RRRR

Δ- connected load → Y-connected load

3

RRY

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Δ- connected load → Y-connected load

Z

Z

Z

3

ZZY

3

ZZY

3

ZZY

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Example 11.4

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

54.710

603120

30

||3||

lineline II

Line-phase current relationship

iprelationsh

voltagephase-Line

30

||3||

phase

phaseVV

rmsAIaA )(02.775.28

rmsAI

rmsAI

CA

BC

)(98.14260.16

)(02.9760.16

rmsAI

rmsAI

cC

bB

)(98.11275.28

)(02.12775.28

02.3717.4YZ

Alternatively, determine first the line currentsand then the delta currents

EMLAB

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Power relationships

iprelationsh

voltagephase-Line

30

||3||

phase

phaseVV

30

||3||

lineline II

Line-phase currentrelationship

*3 phasephaseTotal IVS

*3 linelineTotal IVS

*3 IVS linetotal

*3 linelineTotal IVS

flinelinetotal

flinelinetotal

IVQ

IVP

sin||||3

cos||||3

lineV

lineI

Power factor anglef

- Impedance angle

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lagging 20 anglefactor power

WP

rmsVV

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

Line-phase currentrelationship

rmsAI )(05.2||

46.101

||

||||

I

VZ line

2046.101Z

lineV

Example 11.5

lineV

lineI

Power factor anglef

- Impedance angle

EMLAB

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For an abc sequence, balanced Y - Y three phase circuit

1020,11,)(120|| jZjZVV phaselinermsphase source

Because circuit is balanceddata on any one phase issufficient

0120Chosenas reference

120120

120120

0120

an

bn

an

V

V

V

Abc sequence

rmsA

j

VI anaA

)(65.2706.5

65.2771.23

0120

1121

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

65.2706.50120*aAan IVS phase source

VAj

S

78.28186.537

65.272.607

phase source

)(78.2813

)(86.5373

VAQ

WP

source total

source total

)(6.1821||

)(2.8456.1613

VAS

VAj

QPS

source total

source totalsource totalsource total

Pper_phase Qper_phase

Example 11.6

EMLAB

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Power factor correction

Balanced loadLow pf lagging

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

Keep clear about total/phasepower, line/phase voltages

oldold

old Qpf

S

newnew

old Qpf

P

added be Power to Reactiveoldnew QQQ

To use capacitors this valueshould be negative

connected are capacitors the

howon depends voltageThe

2capacitorper CVQ

f

f

f

pf

SQ

SP

jQPS

cos

sin||

cos||

21sincos pfpf ff

21tan

pf

pff

0 QlaggingfPQ tan

EMLAB

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MWP

MVAQ

old

old

72.18

02.15

0 oldQlagging

MVAQleadingpf

MWPnew

new

old 8.694.0

72.18

Example 11.9 leadingpfrmskVVHzf line 94.0 :Required.5.34||,60

MjMjMjQPold 157.18)78.0178.0(24 2 S

3L

C

VV

MjMjMjP pfnew 8.67.1894.0

94.0117.18)tan1(

2

S

2

1 3382.2102.158.6 CCC VCIVMVAQ

Fk

M

VC

MC

C

5.48)5.34(602

82.21

3

82.2122

EMLAB

*VIS

cos

}Re{cos

S

S

VIP

sin

}Im{sin

S

S

VIQ

rmslinean kVVV 20||3

1

78.0cos

*IVS

94.0cos

CVjII

IVVIQQ

sinsin

sinIVQ

2VCQ

2

)(

VVI

IVIVS*

**

Cj

CVj