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3-Phase Circuits

Balanced 3-phase

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Balanced Complex Numbers

Consider the

complex roots of

unity:

rootstheofsumthefindThen

j

ofrootstheallFind

3

3 1

:

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...3,2,1,0;120112011

3601

31

3

!!!@

!@!

nnn

mxZzxZz

n

o

o

mm

o

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!@!

!

3

1

3

3

00

2

3

2

1

,2

3

2

1

,011

2401,1201,011

:

:

jroots

jjj

or

arerootshe

thatshowcanWe

ooo

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o01

o1201

o2401

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!@

!

!

!@

!

!!

3

1

3

00

0,

2

1

2

3,

2

1

2

3

2701,1501,301

301201

903601

9010

jroots

jjj

nj

n

jj

ooo

oo

oo

o

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It is evident from above that the sum of

any three roots of the cube-root of any

complex number is absolutely zero.

This is the essence of a balanced three

phase system and indeed the most

powerful tool in power transmission.

The results below are self-evident.

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Consider a general complex number z

!@

!

!

!

!!

3

1

3

1

3

3

1

2

3

1

1

3

1

3

00

)23

21(

)2

3

2

1(

)01(

120.....0

jroots

jZz

jZz

jZz

nZzZz oo

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Balances 3-phase Currents

Let us use the above discovery and

KCL:

Suppose we have three currents suchthat

:

int

;,,sup

240,120,0

321

321

sconclusion

erestingakewebelowshownas

loadatoiiiplytodecideweIf

IiIiIi ooo

@

!!!

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nI

1i

2i

3-phase

supply

3-phase

load

3i

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Clearly,

systemphasebalancedain

ireneutralNOal aysisThere

ireneutralhavetoneedlessI

I

iii

ButiiiI

n

n

n

@

!

!@

!

!

3

;0

0

0

,

321

321

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Advantages of balanced 3-phase

supply

1.The elimination of a neutral wire is clearly

the biggest saving.

2. If we wanted to supply the same amount of

power using single phases, we would haveused 6 wires! But we have managed with just

3 of them.

3. The fact that line voltages are higher means

that the line currents are lower hence reduced

losses.

RI2

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Line & Phase Values

In practice, when a 3-phase system is

used, we do have both the currents &

voltages as 3-phase. Their values may

be given as phase or line values.

Let us consider a balanced 3-phase

voltage system.

Just as we did in the case of currents,

the voltage is given by:

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Line & Phase Values

.&

;

)2

3

2

1(240

)2

3

2

1

(120

)01(0

3

2

1

earthirephasebet een

measuredvaluephaseVhere

jVVv

jVVv

jVVv

o

o

o

!

!!

!!

!!

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Line & Phase Values

!@

!!!

!!!

!!!

00

1503)2

3

2

3(

903)30(

303)

2

3

2

3(

1331

3223

2112

jvoltagesline

jvvv

jvvv

jvvv

o

o

o

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Line & Phase Values

It is however usual to measure voltage

between one phase & another phase. So

the value is that of one phase with

respect to the other. Since it is measured

between one phase & another it is

referred as the line voltage.

The line voltages must the phasordifferences as shown below:

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Phasors for Phase & Line Voltages

1v

2v

3v 12v

23v

31v

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Phase & Line Voltages

The following conclusions are made:

1. The line voltages also constitute a balanced 3-

phase system.

2.

351,63

000,11

3

000,11

41632403

240

:

.3

}!!@

!!

}!!@

!!

!

voltagephase

voltageline

voltageline

voltagephase

Exa ples

voltagephasevoltageline

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Line & Phase Values

Please note that:

1. In a 3-phase system; the line values

(voltage & current) are the ones specified.

2. It is very dangerous to touch any two line

wire (because of much higher voltages).

3. All values are rms.

4. Red, Yellow, Blue represent . 321 ,, vvv

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Line & Phase Values

While it is easy to identify phase & line

voltages; caution is taken on phase &

line currents.

It is easier to assume that phase & line

currents are the same.

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Star & Delta 3-ph representation

There are two distinct ways of

representing 3-phase voltages/currents;

Star or Delta. [Star is sometimes calledWye].

Let us re-draw the phasors for 3-phase

voltages.

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Star & Delta 3-ph representation

Phasors

1v

2v

3v

1v

2v

3v

Star

Delta

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Star & Delta 3-ph connections

Power connections

1v

2v

3v

1v

2v

3v

Star

Delta

R

Y

B

B

R

Y

n

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It is noted that:

A Star connection has a neutral point (n)and can be accessed when need arises.But the Delta connection does not haveone.

Hence in a Delta connection only the line

voltages can be measured. But in theStar connection both line & phasevoltages are measurable.

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Star-Star connection

Supply-load connections

1i

supply Load

n n

2i

3i

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Star-Delta connection


1i

supply

Loadn

2i

3i

1I

2I

3I

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Line & Phase currents

In the star connection; the line current=phase

current.

But in Delta connection; the line current DOES

NOT equal to phase current e.g.

But we may use KCL to find the relationshipsbetween line and phase currents.

11 Ii {

0311 ! IIi

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Power in 3-phase

We know that is 1-phase;

)(

,

cos3cos3

3

.,

;cos3

;3

cos

L

LL

LLLL

IIassu e

valueslineareIwhere

IIpowerotal

valuesphaseareIwhere

I

powerotal

phaseinut

IPower

!

!!@

!

!

UU

U

U

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3-phase exercises

Exercise 3:

A 0.6MW, 416V, 3-phase, 50Hz load

has a power factor of 0.5. Determine the necessary capacitors and

show their connections so as to improve

the power factor to 0.9.

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Solution 1

The 3-phase voltages differ by angle only:

)120sin()240sin()(

)120sin()(

)sin()(

3

2

1

oo

o

wtwttv

wttv

wttv

!!

!@

!

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Solution 2

connectionstarif

currentphaselineincurrent

I

I

ILoad

WLoadWply

L

L

LL

!!

}!@

!@

!

!!@!

@

0.168311

3200

5.0**10*11*310*6.1

cos3

6.12*8.02sup

36

U

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Solution 3

The power factor improvement is done

by connecting a capacitor between @

phase & the neutral point.

[Though it is possible to connect a

capacitor between phases but then the

capacitors are more expensive because

of the higher (line) voltages].

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Star-Star connection


1i

Supply Load

n n

2i

3i

C

C

C

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Solution 3

Since @ capacitor is connected across a

phase & neutral; phase voltages must be

used.

We are at liberty to use the phasor

diagram studied earlier.

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Solution 3

1

--

U Jc- V

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Solution 3

)tancos(sin

cos

cos

coscos

sinsin

9.0cos

5.0cos

1

1

1

JUU

J

U

JU

JU

J

U

!@

!@

!

!

!

!

II

I

I

II

III

C

C

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Solution 3

388,13416

10

5.0*416*3

10*6.0

cos3

cos3

6

6

$

!!@

!

!

U

U

L

L

LL

V

I

IIhere

IV

But

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Solution 3

fIZIV

capacitoracrossVoltage

I

T

JU

2

1*

9.865

]10

19

9

5

2

3[3416

10

]9.0

sin5.0[sin3416

10

6

6

!!!

$

!

!@

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Solution 3

.

18390@3

1839001839.0

3416*100100

100

marketonavailablevaluable

highernearestusepraticeIn

Fofcap

FF

IV

I

I

@

!$

!!@

!

Q

Q

TT

T

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Question 4

A 440V, 3-ph, Y-connected source has

two loads connected as:

One load is balanced in @ phase withZ=10+j5.

The second load is balanced in @ phase

with Z=15+j0.

(i) Find the average power to @ load.

(ii) Find the total power delivered.

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kWVI

phaseino erZ

V

Z

Vi

5.1510*5.15cos3

3

0

3

1

$!

!

!

!

U

U

U

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Total load =(10+j5)+(15+j0)=25+j5=Z

kW

Z

iV po erTotal

ZZi

ZjZ

o

phasephase

phase

45.7

3.11cos650

440cos*

3

440*3

440*3

cos**3

3

4400

3440

25

5arctan;525

2

}

!!

!@

!

!@

!!!

U

U

UU

UU

Documents

EE1101 Circ N 5