Dr Audih al-faouryPhiladelphia univers i tyElectr i cal department

2014-20151Dr.Audih alfaoury

Basics Concepts

Three-phase systems1. Introduction: Three-phase systems are commonly

used in generation, transmission and distribution of electric power. A three-phase system is a generator

and load pair, in which the generator produces three sinusoidal voltages of equal magnitude and frequency

but differing in phase by 120° from each other. And the system may be positive (ABC) or negative (ACB)

Fig 1. The three phase system may be 3 wires no neutral point or 4 wires system with neutral point .

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3Fig.1

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Fig.2

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The phase voltages va(t), vb(t) and vc(t) of positive system are

as follows

Whereas the corresponding phasors in complex are

( ) ( )( ) ( )

cos( 0 )

cos 120 cos 240

cos 240 , or cos 120

oa a

b a b a

c a c a

v V t

v V t or v V t

v V t v V t

ω

ω ω

ω ω

= ±

= − = +

= − = +

o o

o o

0 00 0

120 120

240 240

. .

for positive system for negative system

j ja a a a

j jb a b a

j jc a c a

V V e V V e

V V e V V e

V V e V V e

−

−

= =

= =

= =

o o

o o

64474486447448

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IV

ZI

V

ZI

V

Zaa

bb

cc= = =

( )I I I IZ

V V Vn a b c a b c= + + = + +1

For three-phase system if all impedances are identical, Za = Zb = Zc = Z ,then Such a load is called a

balanced load and the currents are:

.

For the figure above by using KCL, we have

In

Ia Za Va

Ic Zc Vc

Ib Zb Vb

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( )( )

[ ]

0 120 240

n

cos 0 sin 0 cos120 sin120 cos 240 sin 240

1 3 1 31 0 0 .

2 2

:

( ) ,

I 0

2

2

oj j ja b c a

o oa

a

where

Setting the result for curren

V V

t

V V e e e

KCL in the circuit before we obtai

V j

n

V j j j

j

− −+ + = + + =

= + + − + − =

= + − − − + = ÷ ÷

=

o o

o o o o

Since the current flo

wing though the fo

urth wire is zero, the neutral wire can be

removed and the system is named three wires system

n’ n

Ia Z Va

Ic Z Vc

Ib Z Vb

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To determine the line voltages Vab, Vbc, Vca and

relation with phase voltage . Using KVL, we obtain

( ) ( ) ( )1

1

30

0 120

3

2tan2 322

2 3 22 tan .2 3

cos0 sin 0 cos120 sin120

1 3 3 3 3 3(1 0) ( )

2 2 2 2 2 2

3 3.

23

2

o oj j o o o oa b a a

j

a aa

j

ab

jaa

V V V V e e V j j

V j V j V e

V e V e

−

−

−

÷ ÷ ÷ ÷

÷ ÷

= − = − = + − + =

= + − − − = + = + = ÷ ÷ ÷ ÷

= + = ÷ ÷

o

303 jab aV V e=

o30

30

3

3

jbc b

jca c

V V e

V V e

=

=

o

o

.

Similarly ;

180ab

* 180 ;

180o

oab ba

j oba ab ab

note V V reversing the subscripts of the voltage or current gives out of the original

V V e V V

≠

= = ∠ = −

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The phasor diagram of the phase and line voltage is shown as:

The line voltages Vab, Vbc, Vca form a symmetrical set of

phasors leading by 30° the set representing the phase voltages and they are 3 times greater.

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( )

( ) ( ) ( ).

3(Y) ab

an ab an

line V

phase V line I phase I

VV and I IFor = =

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( )

( ) ( ) ( )(

3) ab

an ab an

line I

phase V line V phase I

IV V andor IF = =∆

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For the side we get

( ) ( )

(2 )

Since 0

Hence 3

3

T13

herefo 3re

ab Y a b ca Y c a

ab ca Y a b c

a b c a b c

ab ca Y a

ab c

Y

Ya

Y

a

Y

V Z I I V Z I I

V V Z I I I

I I I I I I

V V Z I

V V

Z

Z

orZ

I

Z Z

Z

∆ ∆

∆

= − = −− = − −

+ + = ⇒ = − −

=

− =

=

−= =

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Example: A balance 3Φ system has Vab=173.2 0o

V and a(Y) load is connected with ZL=10 20o .Assuming ABC system find the voltage and current for all phases.Solution:

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ]

[ ]

173.2 0 V , 173.2 240 V , 173.2 120 V

173.2 173.2 173.20 30 V , 240 30 V , 120 30 V

3 3 3:

100 3010 50 A

10 20

100 21010 190 A

10 20

100

o o oab bc ca

o o o o o oan bn cn

ooan

an oL

oobn

bn oL

anan

L

V V V

V V V

And the current is

VI

Z

VI

Z

VI

Z

= ∠ = ∠ = ∠

= ∠ − = ∠ − = ∠ −

∠ −= = = ∠ −∠

∠= = = ∠∠

= = [ ]9010 70 A

10 20

oo

o

∠ = ∠∠

Vab

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Three-phase unbalanced system

c

b

a

Vn

Zn

c’

a’

b’Zp

Zp

Zp

In

n’n

Ia ZaVa

Ic ZcVc

Ib ZbVb

If the load impedance Z are not identical, an unbalanced system is produced. An unbalanced Y-connected system

is shown in Figure. Now we consider a more realistic case where the wires are represented by impedances Zp and the

neutral wire connecting n and n’ is represented by impedance Zn .

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Using the node n as referance, we express the currents Ia, Ib, Ic and In in

terms of the node voltage Vn

IV V

Z Zaa n

a p=

−+

IV V

Z Zbb n

b p=

−+

IV V

Z Zcc n

c p=

−+

IV

Znn

n=

.

The node equation (In=Ia+Ib+Ic=0), then; In-Ia-Ib-Ic=0.Thus;

V

Z

V V

Z Z

V V

Z Z

V V

Z Zn

n

a n

a p

b n

b p

c n

c p− −

+− −

+− −

+= 0

Solving this equation for Vn

V

V

Z Z

V

Z Z

V

Z Z

Z Z Z Z Z Z Z

n

a

a p

b

b p

c

c p

n a p b p c p

=+

++

++

++

++

++

1 1 1 1

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Example:For the network of Figure draw the phasor diagram showing voltages and currents and write down expressions for total line currents Ia, Ib,Ic, and the neutral current In.

Solution:Note that the power factor of the three-phase load is expressed with respect to the phase current and voltage.

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Power in three-phase circuitsIn the balanced systems, the average power consumed by

each load branch is the same and given by

cosav eff effP V I φ=%

The total average power consumed by the three phase load is the sum of those consumed by each branch, hence, we have

* where Veff is the effective value of the phase voltage, Ieff is the effective value of the phase current and φ

is the angle of the impedance

3 3 cosav av eff effP P V I φ= =%

( ) Leffeff II =( ) effLeff V3V =

In the balanced Y systems, the phase current has the same magnitude as the line current , and the line voltage has the effective value

which is:

( ) ( ) ( ) ( )3 cos 3 c s3

oeff L

av eff L eff effL L

VP V II φ φ= =

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In power term the active power is: [ ]3 . cos ,LLLLP V I Wφ=

( ) ( )3 sinav eff effL LQ V I φ=

[ ]3 . sin ,LL LLQ V I VArφ=In power term the reactive is:

.

Same for reactive power

The square root of the sum of (P) and (Q) is the apparent power (S)

[ ]2 2 2 2( . cos ) ( . sin ) ,S P jQ P Q V I V I V I VAφ φ= + = + = + =

1

2 2

cos , .sin

,and cos cos(tan )cos sin

From powe triangle we have P S Q S

P Q Q P PS

P S P Q

φ φ

φφ φ

−

= =

= = = = =+

Dr.Audih alfaoury

Advantages of 3φ Power

Can transmit more power for same amount of wire (twice as much as single phase).

Total torque produced by 3φ machines is constant, so less vibration.

Three phase machines use less material for same power rating.

Three phase machines start more easily than single phase machines.

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Complex power:If the voltage and current are expressed as:

V V and I Iα β= ∠ = ∠Then the product of the voltage and current is the apparent power and in polar form is:

*. j jS V I P jQ V e I e V Iα β α β−= = − = × = ∠ −This quantity called the complex power and in rectangular form is :

( ) ( )*. .cos .sinS V I V I V Iα β α β= = − + −

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Current and voltage at synchronous machine1- If the synchronous machine working as generator and since the internal impedance in which the resistance is very small compared with its reactance ,then the resistance may be neglected and Z=jX, the internal voltage and current become: :

.

.

, ,

a a g a

a a g a

a aa

b c b c

n a b c

E V jX I

V E jX I

E VI

jX

Same for V V I and I

I I I I

=

=

−

= +

+

+

−

=

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2- If the synchronous machine working as motor The internal voltage and current become: :

.

.

, ,

a a m a

a a m a

a aa

m

aa

m

b c b c

n a b c

E V jX I

V E jX I

E VI

jX

if the E= 0,then

VI

jX

Same for V V I and I

I I I I

==

+=

=

=

−

+ +

+

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Direction of power flowThe relation between P,Q ,bus voltage or generated

voltage with respect to P and Q signs is important because the direction inform us whether the power

flow is being generated or absorbed (with V and I are specified).

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1 2100 0 ( ) 100 30 ( )o oE V and E V= ∠ = ∠

0 5Z j= + Ω

Example consider tow machine source where the internal voltage for machines are

and

Determine whether the machine is generated or consuming real and reactive power and the amount of power absorbed by Z.

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Solution:

1 2(100 0) (86.6 j50 13.4 50

10 2.68 10.35 1955 5

oE E j jI j A

Z j j

− + − + −= = = = − − = ∠

Since the current is from 1 to 2 then for 1 is (-I ) and for 2 is (+I ),then:

* *1 1 1 1

*2 2 2 2

1 2

1 2

( ) 100(10 2.68) 1000 268 VA

( ) (86.6 50)( 10 2.68) 1000 j268

1000 1000 0 W

= 268 268 53 r

V

6 va

A

S P jQ E I j j

S P j

P P P

Q Q Q

Q E I j j

= + = − = + = −

= + = = + − + = − −= + = − =

+ = − − = −

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Machine 1: (P) is positive and (Q) is negative then the machine consumed active power and delivered reactive power (motor).Machine 2: (P)is negative and ( Q) is negative then machine generate active power and delivered reactive power ( generator).The sum of active power is zero ,which means the load connected between tow machine consumed reactive power with 536 var which required the reactance of 5 Ω.

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Operator (a): Is designate to reduce the angle calculation, such the operator is a complex number of unit magnitude and angle of 120o and define as:

120

2 240

3 360

2

1 120 1. 1(cos120 sin120 ) 0.5 0.866

1 240 1. 1(cos 240 sin 240 ) 0.5 0.866

1 360 1. 1(cos360 sin360 ) 1 0 1 0

1 0

o

o

o

o j o o

o j o o

o j o o o

a e j j

a e j j

a e j j

a a

= ∠ = = + = − +

= ∠ = = + = − −

= ∠ = = + = + = ∠+ + =

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Impedance and reactance of power system diagrams

Per-phase reactance diagram by omitting all resistance and shunt admittance

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Exercises

1 120 45 100 15

. .

oa b

ab

In a sigle circuit V and V with respect

to a referance node o Find V in polar form

− = ∠ = ∠−

2 2 2

2

1 , 1 , ,

Evaluate the following in polar form

a a a a a j ja a

−− − + + + +

3 120 210 V

10 60 A.

.

oan

oan

A voltage source E and current through the source

is given by I Find the values of P and Q also state wether

the source is delivering or receiving each

− = − ∠

= ∠

4 1 2 0 5

.

If the impedance between machines and in the figure Z j

determine :

a) whether each machine in generating or comsuming power.

b)whether each machine is receiving or supplying

positive reactor power and the amount

c)the value of

− = − Ω

.P and Q absorbed by the impedance

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5 10 15 3

208 .

.

o

ca

Three identical impedances of are Y connected to balanced

line voltages of V Specify all the line and phase voltage and the current

as phasors in polar form with V as reference for a phase sequance of abc

φ− ∠ − Ω

6 3 10 30 .

416 90 , .

o

obc cn

In abalanced system the Y connected impedances are

If V V specify I in polar form

φ− − ∠ Ω= ∠

7 15

8 6 .

2 5

110

A balanced load consisting of pure resistances of per phase is

in parallel with abalanced Y load having phase impedances of j

Identical impedances of j are in each of the three lines conneting

the combined loads to a V thr

− ∆ Ω+ Ω

+ Ωee phase supply.Find the current draw

form the supply and line voltage at the combined loads.

Dr.Audih alfaoury