balanced three-phase circuits osman parlaktuna osmangazi university eskisehir, turkey

BALANCED THREE-PHASE CIRCUITS

Osman ParlaktunaOsmangazi University

Eskisehir, TURKEY

Circuit Analysis II Spring 2005 Osman Parlaktuna

BALANCED THREE-PHASE VOLTAGES

A set of balanced three-phase voltages consists of three sinusoidalvoltages that have identical amplitudes and frequencies but are out ofphase with each other by exactly 1200. The phases are referred to as a, b, and c, and usually the a-phase is taken as the reference.

abc (positive) phase sequence: b-phase lags a-phase by 1200,and c-phase leads a-phase by 1200.

Va

Vc

Vb

V

V

V

a m

b m

c m

V

V

V

-120

+120

00

0

0


acb (negative) phase sequence: b-phase leads a-phase by 1200,and c-phase lags a-phase by 1200.

Va

Vc

Vb

V

V

V

a m

b m

c m

V

V

V

+120

-120

00

0

0

Another important characteristic of a set of balanced three-phase voltages is that the sum of the voltages is zero.

V V Va b c 0


THREE-PHASE VOLTAGE SOURCES

A three-phase voltage source is a generator with three separatewindings distributed around the periphery of the stator.

The rotor of the generator is an electro-magnet driven at synchronous speed bya prime mover. Rotation of the electro-magnet induces a sinusoidal voltagein each winding.The phase windings are designed so thatthe sinusoidal voltages induced in them are equal in amplitude and out of phasewith each other by 1200. The phase windings are stationary with respect tothe rotating electromagnet.


There are two ways of interconnecting the separate phase windingsto form a 3-phase source: in either a wye (Y) or a delta ().


ANALYSIS OF THE Y-Y CIRCUIT

V V V V V V VN N a n

A a ga

N b n

B b gb

N c n

C c gcZ Z Z Z Z Z Z Z Z Z0 1 1 1

0


BALANCED THREE-PHASE CIRCUIT

The voltage sources form a set of balanced three-phase voltages, this means Va

,n, Vb

,n, Vc

,n are a set of

balanced three-phase voltages. The impedance of each phase of the voltage source

is the same: Zga=Zgb=Zgc. The impedance of each line conductor is the same:

Z1a=Z1b=Z1c. The impedance of each phase of the load is the

same: ZA=ZB=ZC

There is no restriction on the impedance of the neutral line, its value has no effect on the system.


If the circuit is balanced, the equation can be written as

VV V V

Na n b n c n

A a ga B b gb C c gc

Z Z Z

Z Z Z Z Z Z Z Z Z Z

( )1 1

0

1 1 1

Since the system is balanced,

V V V V a n b n c n N0 0 and

This is a very important result. If VN=0, there is no potential differencebetween n and N, consequently the current in the neutral line is zero.Hence we may either remove the neutral line from a balanced Y-Y circuit (I0=0) or replace it with a perfect short circuit between n and N(VN=0). Both are convenient to use when modeling balanced 3-phase circuits.


When the system is balanced, the three line currents are

I I IaAa n

bBb n

cCc n

Z Z Z V V V

, ,

The three line currents from a balanced set of three-phase currents,that is, the current in each line is equal in amplitude and frequencyand is 1200 out of phase with the other two line currents. Thus, ifwe calculate IaA and know the phase sequence, we can easily writeIbB and IcC.


SINGLE-PHASE EQUIVALENT CIRCUITWe can use line current equations to construct an equivalent circuitfor the a-phase of the balanced Y-Y circuit. Once we solve this circuit,we can easily write the voltages and currents of the other two phases.Caution: The current in the neutral conductor in this figure is IaA, which is not the same as the current in the neutral conductor of thebalanced three-phase circuit, which is I0=IaA+IbB+IcC


LINE-TO-LINE AND LINE-TO-NEUTRAL VOLTAGES

The line-to-line voltages are VAB, VBC, andVCA. The line-to-neutral voltages are VAN,VBN, and VCN. Using KVL

V V V

V V V

V V V

AB AN BN

BC BN CN

CA CN AN

For a positive (abc) phase sequence wherea-phase taken as reference

V V

V

AN BN

CN

V V

V

-120

+120

00 0

0


V

V

V

AB

BC

AB

V V V

V V V

V V V

0 -120 30

-120 120 - 90

120 0 150

0 0 0

0 0 0

0 0 0

3

3

3

• The magnitude of the line-to-line voltage is3 times the magnitude of the line-to-neutralvoltage.• The line-to-line voltages form a balanced three-phase set of voltages.•The set of line-to-line voltages leads the setof line-to-neutral voltages by 300

•For a negative sequence, the set of line-to-linevoltages lags the set of line-to-neutral voltagesby 300


EXAMPLEA balanced three-phase Y-connected generator with positive sequencehas an impedance of 0.2+j0.5 Ω/ and an internal voltage of 120V/ .The generator feeds a balanced three-phase Y-connected load having an impedance of 39+j28 Ω/. The line connecting the generator to theload is 0.8+j1.5 Ω/. The a-phase internal voltage of the generator is specified as the reference phasor.

Calculate the line currents IaA, IbB, IcC

I

I

I

aA

bB

cC

A

A

A

1202 4 36 87

2 4 156 87

2 4

0

0

0

40 + j30

83.13

0

0

. .

. .

.


Calculate the three phase voltages at the load VAN, VBN, VCN

V

V

Vj

CN

BN

AN

0

0

00

81.118 22.115

19.121 22.115

19.1 22.115)87.36 4.2)(2839(

V

V

V

Calculate the line voltages VAB, VBC, VCA

V V

V

V

AB AN

BC

CA

V

V

V

( ) .

.

.

3 30 199 58

199 58

199 58

0 28.81

- 91.19

148.81

0

0

0


Calculate the phase voltages at the terminals of the generatorVan, Vbn, Vcn

V

V

Vj

cn

bn

an

0

0

00

68.119 9.118

32.120 9.118

32.0 9.118)87.36 4.2)(5.02.0(120

V

V

V

Calculate the line voltages at the terminals of the generatorVab, Vbc, Vca

V V

V

V

ab an

bc

ca

V

V

V

( ) . .

. .

. .

3 30 20594 29 68

20594 90 32

20594 149 68

0 0

0

0

Backfill: Resistive Delta/Wye Conversion


Delta/Wye continued...


Other way,Wye to Delta...



ANALYSIS OF THE Y- CIRCUIT

If the load in a three-phase circuit is connected in a delta, it can betransformed into a Y by using the to Y transformation. When the loadis balanced, the impedance of each leg of the Y is ZY=Z/3. After thistransformation, the a-phase can be modeled by the previous method.When a load is connected in a delta, the current in each leg of the deltais the phase current, and the voltage across each leg is the phase voltage.


To demonstrate the relationship between the phase currents and linecurrents, assume a positive phase sequence and

I I IAB BC CAI I I -120 120 00 0 0, ,

I I I

I I I

I I I

aA AB CA

bB BC AB

cC CA BC

I I I

I I I

I I I

120 - 30

-120 -150

120 -120 90

0 3

0 3

3

0 0 0

0 0 0

0 0 0

The magnitude of the line currents is times the magnitude of the phase currents and the set of line currents lags the set of phase currents by 300. For the negative sequence, line currents lead thephase currents by 300.

3


Positive sequence Negative sequence


EXAMPLEThe Y-connected load of the previous example feeds a -connected loadthrough a distribution line having an impedance of 0.3+j0.9/. The loadimpedance is 118.5+j85.8 /.

Transforming load into Y and drawing a-phase of the circuit gives

118 5 858

339 5 28 6

. .. . /

jj


Calculate the line currents IaA, IbB, and IcC

I

I I

aA

bB cC

jA

A A

120 0

40 302 4 36 87

2 4 156 87 2 4 8313

00

0 0

. .

. . . .

Calculate the phase voltages at the load terminals:Because the load is connected, the phase voltages are the same asthe line voltages.

V

V V

V V

AN

AB AN

BC CA

j V

V

V V

( . . )( . . ) . .

. .

. . . .

39 5 28 6 2 4 36 87 117 04 0 96

3 30 202 72 29 04

202 72 90 96 202 72 149 04

0 0

0 0

0 0


Calculate the phase currents of the load.

I I

I I

AB aA

BC CA

A

A A

1

330 139 6 87

139 126 87 139 11313

0 0

0 0

=

. .

. . . .

Calculate the line voltages at the source terminals

V

V V

V V

an

ab an

bc ca

j V

V

V V

( . . . . ) .

. .

. . . .

39 8 29 5)(2 4 36 87 118 9

3 30 20594 29 68

20594 90 32 20594 149 68

0

0 0

0 0

- 0.32

0


AVERAGE POWER IN A BALANCED Y LOAD

P

P

P

A AN aA vA iA

B BN bB vB iB

C CN cC vC iC

V I

V I

V I

cos( )

cos( )

cos( )

In a balanced three-phase system, themagnitude of each line-to-neutral voltageis the same as is the magnitude of each phasecurrent. The argument of the cosine functionsis also the same for all three phases.

V

I

AN BN CN

aA bB cC

vA iA vB iB vC iC

V V V

I I I


For a balanced system, the power delivered to each phase of theload is the same

P P P P V I

P P V I

PV

I V I

A B C

T

TL

L L L

cos

cos

( ) cos cos

3 3

33

3

Where PT is the total power delivered to the load.


COMPLEX POWER IN A BALANCED Y LOAD

Q V I

Q Q V IT

sin

sin3 3

S

S P jQ

S S V I

AN aA BN bB CN cC

T L L

V I V I V I V I

V I

* * * *

*

3 3

Reactive power in a balanced system

Complex power in a balanced system


POWER CALCULATIONS IN A BALANCED DELTA LOAD

P

P

P

AB AB AB vAB ivAB

BC BC BC vBC ivBC

CA CA CA vCA ivCA

V I

V I

V I

cos( )

cos( )

cos( )

For a balanced system

V V V

I I I

AB BC CA

AB BC CA

vAB iAB vBC iBC vCA iCA

V

I

cosIVPPPP CABCAB


P P V I

PIV V I

T

TL

L L L

3 3

33

3

cos

( ) cos cos

Q V I

Q Q V IT

sin

sin3 3

S P jQ

S S V IT L L

V I*

3 3


INSTANTANEOUS POWER IN THREE-PHASE CIRCUITS

In a balanced three-phase circuit, the total instantaneous power isinvariant with time. Thus the torque developed at the shaft of a three-phase motor is constant, which means less vibration in machinery powered by three-phase motors.

Considering a-phase as the reference, for a positive phasesequence, the instantaneous power in each phase becomes

p v i V I t t

p v i V I t t

p v i V I t t

p p p p V I

A AN aA m m

B BN bB m m

C CN cC m m

T A B C m m

cos cos( )

cos( )cos( )

cos( )cos( )

. cos

120 120

120 120

15

0 0

0 0


EXAMPLEA balanced three-phase load requires 480 kW at a lagging power factorof 0.8. The line has an impedance of 0.005+j0.025/. The line voltageat the terminals of the load is 600V

Single phase equivalent

Calculate the magnitude of theline current.

600

3160 120 10

577 35 36 87

577 35 36 87

577 35

3

0

0

I

I

I

aA

aA

aA

L

j

A

A

I A

*

*

( )

. .

. .

.

-


Calculate the magnitude of the line voltage at the sending end of the line.

V V I

V

an AN L aA

L an

Z j

V

V V

600

30 005 0 025)(577 35

357 51 157

3 619 23

0

( . . . )

. .

.

- 36.87

0

Calculate the power factor at the sending end of the line

pf

= Lagging

cos[ . ( . )]

cos . .

157 36 87

38 44 0 783

0 0

0


EXAMPLE

Determine all the load currents:

Transforming to Y as 60/3=20, and drawing a-phase of the circuit gives:

Vrms

Arms

AN

aA

00

00

0 90)12)(0 5.7(

0 5.730//204

0 120

V

I


In the original Y connected load

ArmsArms

Arms

CNBN

AN

00

00

012 3 012- 3

0 330

0 90

II

I

For the original delta-connected load

ArmsArms

Arms

Vrms

CABC

AB

AB

00

00

000

150 6.2 90 6.2

30 6.260

30 88.155

30 88.155300 390

II

I

V

35Eeng 224

Unbalanced Three Phase Systems 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

36Eeng 224

37Eeng 224

Three Phase Power Measurement Two-meter method for measuring three-phase power

38Eeng 224

Residential Wiring

Single phase three-wire residential wiring

balanced three-phase circuits osman parlaktuna osmangazi university eskisehir, turkey

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