3phase Circuit

8/10/2019 3phase Circuit

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Three-phase power circuits

Lecture Notes: EEE

Instructor: Dr. U D Dwivedi


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Introduction

Almost all electric power generation and most of the power

transmission in the world is in the form of three-phase ACcircuits. A three-phase AC system consists of three-phase

generators, transmission lines, and loads.

There are two major advantages of three-phase systems over

a single-phase system:1) More power per kilogram of metal from a three-phase

machine;

2) Power delivered to a three-phase load is constant at all

time, instead of pulsing as it does in a single-phase system.

The first three-phase electrical system was patented in 1882

by John Hopkinson -British physicist, electrical engineer,Fellow of the Royal Society.
http://en.wikipedia.org/wiki/United_Kingdomhttp://en.wikipedia.org/wiki/United_Kingdom


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Generation of three-phase voltages

and currents

A three-phase generatorconsists of three single-

phase generators with

voltages of equal

amplitudes and phasedifferences of 1200.


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and currents

Each of three-phasegenerators can be connected

to one of three identical loads.

This way the system would

consist of three single-phasecircuits differing in phase

angle by 1200.

The current flowing to each

load can be found as

VI

Z=


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and currents

Therefore, the currents flowing in each phase are

0

0

0

0

120120

240240

A

B

A

V

I IZ

VI I

Z

VI I

Z

= =

= =

= =

(1)

(2)

(3)


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and currents

We can connect the negative (ground) ends of the threesingle-phase generators and loads together, so they share

the common return line (neutral).


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_

__

++

+

n

a

bc

c

b

a

V 0

V -120V -240

Wye Connected

Source

Wye Connected Source


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a

bc

a

b

c

+

+

+

_

_

_Delta

Source

Delta Source

Vab = | Vab| 0

Vbc

= Vab

-120

Vca= Vab-240


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Zl

Zl

Zl

ZL

ZL ZL

a

n

bc

A

B C

N

Wye Wye System


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Zl

Zl

Zl

a

bc

A

B C

ZL

ZLZ L

+

+

+

_

_

_

Delta Delta System


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Zl

Zl

Zl

a

bc

A

B C

+

+

+

_

_

_

ZL

ZL

ZL

Delta Wye System


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_

__

++

+

n

a

b

c

c

b

a

V 0

V -120V -240

A

B C

ZZ

Z

IaA ICA

IBC

IAB

Wye Delta System


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Generation of three-phase

voltages and currents

As long as the three loads are equal, the return current in

the neutral is zero!

Such three-phase power systems (equal magnitude, phase differences

of 1200, identical loads) are called balanced.

In a balanced system, the neutral is unnecessary!

Phase Sequenceis the order in which the voltages in the individual

phases peak.

abc acb


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Voltages and currents

There are two types of connections in three-phase circuits: Yand .

Each generator and each load can be either Y- or -connected. Anynumber of Y- and -connected elements may be mixed in a power system.

Phase quantities- voltages and currents in a given phase.

Line quantities voltages between the lines and currents in the lines

connected to the generators.


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1. Y-connection

Assuming

a resistiveload


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1. Y-connection (cont)

0

0

0

0

120

240

an

bn

cn

V V

V V

V V

=

=

=

0

0

0

0120

240

a

b

c

I II I

I I

=

=

=

Since we assume a resistive load:

(6)

(7)


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1. Y-connection (cont 2)

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

L

I I

=

Voltages are:

0

0

0 1 3 3 30 120

2 2 2 23 1

32 2

303

aab bV V V V V V jV V j V

j V

V

V

= = = = +

= + =

(8)

(9)


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1. Y-connection (cont 3)

3LL

V V

=

Magnitudes of the line-to-line voltages and the line-to-neutral voltages are related as:

In addition, the line voltages are

shifted by 300with respect to the

phase voltages.

(10)

In a connection with abc

sequence, the voltage of a

line leadsthe phase voltage.


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Van

Vbn

Vcn- Vbn

Vab

Vab = Van - Vbn

Vab = 3 Van 30o

30

o


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( )

=

=

=

=

=

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 VV


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2. -connection

0

0

0

0

120

240

ab

bc

ca

V V

V V

V V

=

=

=

0

0

0

0

120

240

ab

bc

ca

I I

I I

I I

=

=

=

assuming a

resistive load:

(3.14.1) (11)


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2. -connection (cont)

LLV V=

The currents are:

0 0

0

1 30 240

2 2

3 3 3 132 2 2 2

3 30

ab caa I I I I I I I

I

j I

I j I I j

= = +

= = =

=

3LI I=

(12)

(13)

(14)The magnitudes:


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For the connections with the abcphase sequences, the current of a

line lagsthe corresponding phase current by 300(see Figure below).

For the connections

with the acbphase

sequences, the line

current leadsthe

corresponding phase

current by 300.


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Power relationships

For a balanced Y-connected load with the impedance Z= Z0:

0

0

( ) 2 sin

( ) 2 sin( 120 )

( ) 2 sin( 240 )

an

bn

cn

v t V t

v t V t

v t V t

=

=

=

and voltages:

The currents can be found:

0

0

( ) 2 sin( )

( ) 2 sin( 120 )

( ) 2 sin( 240 )

a

b

c

i t I t

i t I t

i t I t

=

=

=

(3.17.1)

(3.17.2)


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Power relationships

The instantaneous power is:

( ) ( ) ( )p t v t i t=

Therefore, the instantaneous power supplied to each phase is:

0 0

0 0

( ) ( ) ( ) 2 sin( ) sin( )

( ) ( ) ( ) 2 sin( 120 ) sin( 120 )

( ) ( ) ( ) 2 sin( 240 ) sin( 240 )

a an a

b bn b

c cn c

p t v t i t VI t t

p t v t i t VI t t

p t v t i t VI t t

= =

= =

= =

(15)

(16)

Since

[ ]1

sin sin cos( ) cos( )2

= + (17)


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Power relationships

Therefore

[ ]0

0

( ) cos cos(2 )

( ) cos cos(2 240 )

( ) cos cos(2 480 )

a

b

c

p t VI t

p t VI t

p t VI t

=

=

=

The total power on the load

( ) ( )( ) 3 cos( )tot a b cp t p tp t Vp t I + + ==

The pulsing components cancel each other because of

1200phase shifts.

(18)

(19)


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Power relationships

The instantaneous

power in phases.

The total power

supplied to the load isconstant.


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Power relationships

Phase quantities in each phase of a Y- or -connection.

23 cos 3 cosP V I I Z = =

23 sin 3 sinQ V I I Z = =

23 3S V I I Z = =

Real

Reactive

Apparent

Note: these equations are valid for balanced loads only.

(20)

(21)

(22)


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Power relationships

Line quantities: Y-connection.

Power consumed by a load: 3 cosP V I =

Since for this load 3L LLI I and V V = =

Therefore: 3 cos3

LLL

VP I =

Finally: 3 cosLL LP V I =

(23)

(24)

(25)

(26)

Note: these equations are valid for balanced loads only.


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Power relationships

Line quantities: -connection.

Power consumed by a load: 3 cosP V I = (27)

Since for this load 3L LLI I and V V = = (28)

Note: these equations were derived for a balanced load.

Therefore: 3 cos3

LLL

IP V =

Finally: 3 cosLL LP V I =

(29)

(30)

Same as for a Y-connected load!


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Power relationships

Line quantities: Y- and -connection.

3 sinLL LQ V I =

3 LL LS V I=

Reactive power

Apparent power

(31)

(32)

Note: is the angle between the phase voltage and the phase

current the impedance angle.


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Analysis of balanced systems

We can determine voltages, currents, and

powers at various points in a balanced circuit.

Consider a Y-connected generator and load via

three-phase transmission line.

For a balanced Y-connected system,

insertion of a neutral does not change the

system.

All three phases are identical except of 1200

shift. Therefore, we can analyze a single

phase (per-phase circuit).

Limitation: not valid for -connections


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Analysis of balanced systems

A -connected circuit can be analyzed via the transform of impedances by theY-transform. For a balanced load, it states that a -connected load consistingof three equal impedances Zis equivalent to a Y-connected load with the

impedances Z/3. This equivalence implies that the voltages, currents, and

powers supplied to both loads would be the same.


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Analysis of balanced systems: Ex

Example 3-1:

for a 208-V three-

phase ideally balanced

system, find:

a) the magnitude of

the line current IL;b) The magnitude of

the loads line and

phase voltages VLL

and VL;

c) The real, reactive,

and the apparentpowers consumed

by the load;

d) The power factor of

the load.


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Both, the generator and the load are Y-

connected, therefore, its easy to construct

a per-phase equivalent circuit

a) The line current:0 0 0

0

0

120 0 120 0 120 07.94 37.1

(0.06 0.12) (12 9) 12.06 9.12 15.12 37.1L

L load

VI A

Z Z j j j

= = = = =

+ + + + +

b) The phase voltage on the load:

0 0 0 0(7.94 37.1 )(12 9) (7.94 37.1 )(15 36.9 ) 119.1 0.2L L LV I Z j V = = + = =

The magnitude of the line voltage on the load:

3 206.3LL LV V V= =

Btw: is the load inductive or capacitive??


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c) The real power consumed by the load:

03 cos 3 119.1 7.94 cos 36.9 2270load

P V I W

= = =

The reactive power consumed by the load:

03 sin 3 119.1 7.94 sin 36.9 1702 var loadQ V I = = =

The apparent power consumed by the load:

3 3 119.1 7.94 2839loadS V I VA = = =

d) The load power factor:

0cos cos 36.9 0.8loadPF lagging= = =


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One-line diagrams

Since, in a balanced system, three

phases are similar except of the

1200phase shift, power systems

are frequently represented by a

single line showing all threephases of the real system.

This is a one-line diagram.

Such diagrams usually include allthe major components of a power

system: generators, transformers,

transmission lines, loads.


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Using the power triangle

If we can neglect the impedance of the transmission line, an important

simplification in the power calculation is possible

If the generator voltage in the system is known,

then we can find the current and power factor at

any point in the system as follows:

1. The line voltages at the generator and the loads

will be identical since the line is lossless.

2. Real and reactive powers on each load.

3. The total real and reactive powers supplied to

all loads from the point examined.

4. The system power factor at that point using thepower triangle relationship.

5. Line and phase currents at that point.

We can treat the line voltage as constant and use the power triangle method to

quickly calculate the effect of adding a load on the overall system and power factor.

3phase Circuit

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