3 Phase Circuits 1

8/12/2019 3 Phase Circuits 1

1/25

BALANCED THREEBALANCED THREE --PHASEPHASEAC CIRCUITAC CIRCUIT

Balanced Three-Phase Voltage Sources

Delta ConnectionStar Connection Balanced 3-phase Load

Delta ConnectionStar Connection

Power in a Balanced Phase Circuit


2/25

IntroductionThree Phase System


3/25

Balanced Three Phase VoltagesThree-phase voltage sources

a) wye-connecte

d source b) delta-connected source

If the voltage source have the same amplitude andfrequency and are out of phase with each other by120 o, the voltage are said to be balanced.

0VVV cnbnan =++cnbnan VVV ==

Balanced phase voltages are equal in magnitude and

out of phase with each other by 120 o


4/25

Balanced Three Phase Voltages

0p

0pcn

0pbn

0pan

120V240VV

120VV

0VV

+==

=

=abc sequence or positive sequence:

acb sequence or negative sequence:

0p

0pbn

0pcn

0pan

120V240VV

120VV0VV

+==

==

pV is the effective or rms value

Va Vb Vc

t

Vm


5/25

Balanced Three Phase LoadsTwo possible three-phase load configurations:

a) a Star or Y-connected load b) a delta-connected load

For a balanced wye connected load:

Y321 ZZZZ === === ZZZZ cba

For a balanced delta connected load:

YZ3Z == Z31ZY


6/25

Example 1

V V an = 10200 V V bn = 230200

Determine the phase sequence of the set of voltages

van = 2 200 cos ( t + 10 )

vbn = 2 200 cos ( t 230 ), v cn = 2 200 cos ( t 110 )

Solution:

The voltages can be expressed in phasor form as

We notice that Van leads Vcn by 120 and Vcn in turn leads Vbn by120 .

Hence, we have an acb sequence.V V cn = 110200

V V an = 150110

V V bn = 30110Given that , find Van and Vcn , assuming a

positive ( abc ) sequence.

Answer: V V cn = 90110


7/25

Balanced Y-Y ConnectionA balanced Y-Y system is a three phase system with a balanced Y connectedsource and balanced Y connected load.

LsY

L

s

Z Z Z Z

Z

Z Z

++=

=

=

=

l

l

Source impedanceLine impedance

Load impedance

LY ZZ =


8/25

Balanced Wye-Wye Connection

LsY

Y

L

s

ZZZZZ

Z

Z

Z

++==

=

=

=

l

l

Source impedance

Line impedance

Load impedance

Total impedance per phase

LY ZZ =


9/25

Balanced Y-Y Connection

0pca

0pbc

0pab

210V3V

90V3V

30V3V

=

=

=

Line to line voltages or line voltagesgiven that phase a voltage is referencecan be shown to be:

p L

p L

I I

V V =

= 3cabcabL VVVV ===

cnbnanp VVVV ===

120 o

Vcn

30 o

Vbn

Vab

Van


10/25


11/25

A balanced positive-sequence Y-connected 60 Hz three-phase source has

phase voltage Va=1000V. Each phase of the load consists of a 0.1-Hinductance in series with a 50- resistance.

Find the line currents, the line voltages, the power and the reactive powerdelivered to the load. Draw a phasor diagram showing line voltages, phasevoltages and the line currents. Assuming that the phase angle of V an is zero.

0

0

37

3762.627.3750=

=+=+=

j L j R Z 0

0 0

15.97 37

15.97 157 , 15.97 83

anaA

bB cC

V I Z

I I

= =

= =

&&

& &

Y-Y configuration Example:1

0 0 0 03 30 1732 30 1732 90 , 1732 150ab an bc caV V V V = = = = & & & &


12/25

Example 21- Calculate the linecurrents in the three wireY-Y system of figurebelow.

2- A Y-connected balanced three-phase generator with an impedance of

0.4+ j0.3 per phase is connected to a Y-connected balanced load withan impedance of 24 + j19 per phase. The line joining the generatorand the load has an impedance of 0 .6 + j0.7 per phase.

Assuming a positive sequence for the source voltages and that

Find: (a) the line voltages (b) the line currents

V V an030120 =


13/25

Balanced Y-Delta Connection

CA0

pca

BC0

pbc

AB0

pab

V210V3V

V90V3V

V30V3V

==

==

==

Line voltages:

==

==

==

903

1503303

AB BC CAc

AB AB BC b

ABCA ABa

I I I I

I I I I I I I I

Line currents:

=== Z V

I Z V

I Z V

I CACA BC

BC AB

AB ;;Phase currents:

A balanced Y- system consists of balanced Y connected source feeding abalanced connected load.


14/25

Balanced Y-Delta Connection

3Z

Z Y

=

)24011(IIII

240II0

ABCAABa

0ABCA

==

=

0

ABa303II =

3II pL =

cbaL IIII ===

CABCABp IIII ===

Magnitude line currents:

A single phase equivalent circuit

3 / ==

Z

V

Z

V I an

Y

ana


15/25

Y-Delta configuration: Example 31- A balanced abc sequence Y-connected source with

is connected to a -connected balanced load (8+j4) per phase.Calculate the phase and line currents.

V10100V 0an =

V V AB020180 =2-One line voltage of a balanced Y-connected source is

If the source is connected to a -connected load of , findthe phase and line currents.

Assume the abc sequence.

04020


16/25

Balanced Delta-DeltaConnection

A balanced - system is one in which both balancedsource and balanced load are connected.


17/25

Balanced Delta-Delta Connection

=

=

=

Z

V I

Z

V I

Z

V I

CACA

BC BC

AB AB

CAca

BCbc

ABab

VV

VVVV

=

=

=Line voltages:

Line currents:

3II pL =

Magnitude line currents:

3Z

Z Y=

Total impedance:

Phase currents:

==

==

==

903

1503

303

AB BC CAc

AB AB BC b

ABCA ABa

I I I I

I I I I

I I I I

A balanced - system is the one in which both balanced source and balancedload are connected.


18/25

A delta-connected sourcesupplies a delta-connectedload through wires havingimpedances of Z line=0.3+j0.4 ,the load impedance areZ=30+j6 , the balancedsource ab voltage is

Vab=1000


19/25

1- A balanced connected load having an impedance of 20-j15

is connected to a connected, positive sequence generator having

Calculate the phase currents of the load and the line currents.

V0330V 0ab =

A I a0355.22 =

2- A positive-sequence, balanced -connected source supplies abalanced -connected load. If the impedance per phase of theload is 18+ j12 and , find I AB and V AB.


20/25

Balanced Delta-Y ConnectionReplace connected source to equivalent Y connected source.

Phase voltages:

0pcn

0pbn

0pan

903

VV

1503

VV

303

VV

+=

=

=

A single phase equivalent circuit

Y

p

Y

ana

Z

V

Z

V I

0303

==


21/25

Y-Delta configuration: Example 51-A balanced Y connected load with a phase resistance of 40 and a reactance of 25 is supplies by a balanced, positivesequence connected source with a line voltage of 210 V.

Calculate the phase currents. Use V ab as reference.

V V ab015240=2-In a balanced -Y circuit,

and Z Y = (12 + j15) . Calculate the line currents.


22/25

POWER IN A BALANCED SYSTEM

)120cos(2),120cos(2,cos2 00 +=== t V vt V vt V v pCN p BN p AN For Y connected load, the phase voltage:

=ZZY

If Phase current lag phase voltage by .

)120tcos(I2i

)120tcos(I2i

)tcos(I2i

0pc

0pb

pa

+=

=

=

The phase current:


23/25

POWER IN A BALANCED SYSTEM

Total instantaneous power:

cCNbBNaANcba ivivivpppp ++=++=

= cosIV3p pp

Average power per phase:

Apparent power per phase:

Reactive power per phase:

Complex power per phase:

= sinIVQ ppp= cosIVP ppp

*ppppp IV jQPS =+=ppp IVS =


24/25

POWER IN A BALANCED SYSTEMTotal average power:

=== cosIV3cosIV3P3P LLppp

Total reactive power:

=== sinIV3sinIV3Q3Q LLpppTotal complex power:

*p

2p

p2p

*ppp Z

V3ZI3IV3S3S ====

=+=LL IV3 jQPS


25/25

Power: Example 6

1-A three-phase motor can be regarded as a balanced Y-load. Athree-phase motor draws 5.6 kW when the line voltage is 220 Vand the line current is 18.2 A. Determine the power factor of themotor.

2- Calculate the line current required for a 30-kW three-phasemotor having a power factor of 0.85 lagging if it is connected to abalanced source with a line voltage of 440 V.

3 Phase Circuits 1

Documents

Transcript of 3 Phase Circuits 1