Un

it 2

Th

ree

Ph

as

e C

irc

uit

s

Balanced Y-Y Connections

Phase & Line Voltages and Line Currents

Power in a Balanced System

Lecture Outline

Recaps…Three Phase Circuit

Balanced three phase system

Three generators produce

voltages

same magnitude

120 deg phase shift

()

()

() °

−−

=

°−

−=

−=

240

cos

)(

120

cos

)(

cos

)(

θω

θω

θω

tI

ti

tI

ti

tI

ti

Mb

Ma

120 deg phase shift

() °

−−

=240

cos

)(

θωt

It

iM

c

()

()

()

() °

+=

°−

=

°−

==

120

cos

240

cos

)(

120

cos

)(

cos

)(

tV

tV

tv

tV

tv

tV

tv

MM

cn

Mbn

Man

ωωωω

Y-Y

Co

nn

ecti

on

s

Cont…

Th

ree

-ph

as

e s

ys

tem

wit

h

ab

ala

nc

ed

y-c

on

ne

cte

d

so

urc

e a

nd

ab

ala

nc

ed

y-

co

nn

ec

ted

lo

ad

* Unbalanced y-connected

load

Ba

lan

ce

dU

nb

ala

nc

ed

Recaps…Approximating Impedance

in a 3-Phase System

LS

YZ

ZZ

Z+

+=

l

LS

ZZ

<<

LZ

Z<<

l

Normally

LY

ZZ

≈thus:

The Phase Voltages, V

p

a

Vab

Van

I aL

ine-t

o-n

eu

tral vo

ltag

es

(Ph

ase v

olt

ag

es)

∠=

δPV

an

V

n

b c

Vab

Vbc

Vca

Vbn

Vcn

I b

I c

°−

∠=

°−

∠=

240

120

δδ

PP

VV

cnbn

VV

0=

++

cnbn

an

VV

V

Ba

lan

ce

d s

ys

tem

The Line Voltages, V

L

a

Vab

Van

I aL

ine-t

o-l

ine v

olt

ag

es, fr

om

the n

→a→

b→

n l

oo

p

°−

∠−

°∠

=

−=

+=

120

0p

pV

V

VV

VV

Vbn

an

nb

an

ab

n

b c

Vab

Vbc

Vca

Vbn

Vcn

I b

I c

°∠

=

++

=

30

3

23

211

120

0

pp

pp

j

VV

Sim

ila

rly

°−

∠=

−=

°−

∠=

−=

210

3

90

3

pp

VV

VV

VV

VV

an

cnca

cnbn

bc

Three Phase Circuit

Ve

cto

r d

iag

ram

Vca

Vab

Vcn

30°

120°

Vbc

Vbn

Van-V

bn

The Lines Currents

( =the phase currents )

°−

∠=

°−

∠=

==

120

120

,

an

bnYan

a

IV

VI

ZVI

Applying KVL to each phase:

°−

∠=

°−

∠=

=

°−

∠=

°−

∠=

=

240

240

120

120

ca

Y

an

Ycn

a

Y

an

Ybn

b

IZ

V

ZVI

IZ

V

ZVI

0=

++

cb

aI

II

Th

us

0

0)

==

=+

+=

nn

nN

cb

an

IZ

V

II

-(I

I Thus

Three Phase Circuit

(For balanced connection)

Fo

ur

wir

e s

yste

m

Can

be

red

uced

Th

ree w

ire s

yste

m

red

uced

to

Alternative way of analyzing a

balanced Y-Y system

an

VI=

“p

er

ph

as

e” b

as

is

Sin

gle

-ph

as

e a

na

lys

is y

ield

s;

Yan

aZ

I=

°−

∠=

120

ab

II

°−

∠=

240

ac

II

Th

en

, u

se

ph

as

e s

eq

ue

nc

e t

o o

bta

in:

Example

Calculate the line currents in

the three-wire Y-Y system

Example

Ω°∠

=

++

−=

8.21

155

.16

)8

10

()2

5(Y

jj

ZCalculate the line currents in

the three-wire Y-Y system

Replace with its single equivalent

circuit:

A2.

98

81

.6

8.261

81

.6

240

°∠

=°

−∠

=°

−∠

=a

cI

I

A8.

21

81

.6

8.21

155

.16

0110

°−

∠=

°∠

°∠

==

Yan

aZV

I

A8.

141

81

.6

120

°−

∠=°

−∠

=a

bI

I

Th

ree p

hase p

ow

er

measu

rem

en

t fo

r a b

ala

nced

m

easu

rem

en

t fo

r a b

ala

nced

syste

m

Three Phase Circuit

Bala

nced

syste

m

Insta

nta

neo

us P

ow

er

0)(

)(

)(

=+

+t

it

it

ic

ba

0)(

)(

)(

=+

+t

vt

vt

vcn

bn

an

)cos(

3)(

)(

)(

)(

θ θθθp

pc

ba

IV

tp

tp

tp

tp

=+

+=

θθ

cos

3cos

33

LL

pp

pc

ba

IV

IV

PP

PP

P=

==

++

=

)cos(

3)(

)(

)(

)(

θ θθθp

pc

ba

IV

tp

tp

tp

tp

=+

+=

To

tal co

mp

lex p

ow

er

pp

pp

pC

BA

TZ

II

VS

SS

SS

2*3

33

==

=+

+=

To

tal avera

ge p

ow

er

(real p

ow

er)

Sim

ilarl

y, to

tal re

acti

ve p

ow

er

θ θθθθ θθθ

sin

3sin

33

LL

pp

pI

VI

VQ

Q=

==

Example

Determ

ine the total

average power, reactive

power and complex power

at the source and load

Example

V0

110

°∠

=p

V

A8.

21

81

.6

°−

∠=

I

Replace with its single equivalent

circuit:

Determ

ine the total

average power, reactive

power and complex power

at the source and load A

8.21

81

.6

°−

∠=

pI

)21.8

)(6.81

0110

(3

3*

°∠

°∠

−=

−=

ppI

VS

s

Thus, at the source (following the passive sign convention);

VA

)6.

834

2087

(8.

21

2247

j+

−=

°∠

−=

Real power supplied = –

2087 W

Reactive power = 834.6 VAR

Example

V0

110

°∠

=p

V

A8.

21

81

.6

°−

∠=

I

Replace with its single equivalent

circuit:

Determ

ine the total

average power, reactive

power and complex power

at the source and load A

8.21

81

.6

°−

∠=

pI

)66

.38

(12.81

)81

.6(

33

32

2*

°∠

==

=p

pp

ZI

IV

Sp

L

And at the load;

VA

)1113

1392

(66

.38

1782

j+

=°

∠=

Real power absorbed = 1392 W

Reactive power = 1113 VAR

Example

VA

S)

1113

1392

(j

L+

=

VA

S)6.

834

2087

(j

s+

−=

Replace with its single equivalent

circuit:

At the load;

pI

At the source;

pI

The difference between S

sand S

Lis absorbed by the line

impedance (5 –

j2)Ω

= (695.6

–j2

78.4

) VA

(Pow

er loss

)

Prove:

)8.

21

21

.749

()8.

21

385

.5(

)81

.6(

33

22

°−

∠=

−∠

==

lp

lZ

IS

VA

)4.

278

6.695

(j

−=

Conservation of power: S

s+

SL

+ S

l=

0

Video presentation

Thre

e-P

hase D

isconnect Sw

itch

•th

ree p

hase v

ertic

al bre

ak d

isconnect sw

itch a

ttem

pting to d

e-e

nerg

ize

an u

nlo

aded s

ection o

f transm

issio

n lin

e

•the s

witch w

as p

art o

f an e

xperim

enta

l desig

n a

nd is n

o longer in

serv

ice

•Air b

reak d

isconnect sw

itches a

re n

ot in

tended to a

ctively

sw

itch load

curr

ent

•the a

rcin

gis

due to the a

ttem

pte

d inte

rruption o

f com

para

tively

low

500kV_Switch.mpeg

345kV_Switch.mpeg

http://205.243.100.155/frames/longarc.htm

•the a

rcin

gis

due to the a

ttem

pte

d inte

rruption o

f com

para

tively

low

reactive (capacitiv

e o

r "c

harg

ing")

curr

ents

dra

wn b

y the o

pen

transm

issio

n lin

e.

•even w

ith the reduced c

urr

ent, the d

isconnect sw

itch w

as n

ot alw

ays

capable

of openin

g the c

ircuit.

•at th

e v

ery

end o

f th

e c

lip, a b

rief phase-to-p

hase p

ow

er arc

causes the

upstream

lin

e inte

rrupte

rs to o

pen, finally

extinguis

hin

g the a

rcs.