Per - phase equivalents

:Y . source + Y - load

:Ia

-

>1

Van ~

E

•

~ n ~

Ig z•n E

Ten Tbn > n

¥

Its per.

phase equivalent is given by

±¥a=zThe single phase descriptionVan ~ Ahas all information to

-

•b-

computeall 34 quantities

.

For example ,Ia= In ,

andyou can calculate

Z

Is = IIL - i2o°, IT = TIL #209 etc .

° Y . source + 1 had :

Ia a -

-

> '

I,

Van ~

Isa

-1

Ea.

Et•

~ n ~ Iz EaIb e <

-

Ten Tbn > nb

¥

Let us transform the A- had to

an equivalent Y. lad .

Tosimplify ,

Assume that Tan= vo Lo

,-

V

⇒ tab = B vo L3o°,

itIs.= But -90 : j¥€TaTea

= B vo Ll 50 :.

€t Ab

Vb

¥ =

Mfg =

beLsi,

Ea

⇐ t£÷=hzt÷tao

;Is =

EE = began:o : Ia = Ii - Is =

B¥÷ Lsi -

✓3z÷ List

Ia

Is'

>

¥,

=¥[b%L3o°- is %Lsog

= ¥, . B Vo .fs .

Lo °

=

±Eats

i. Tan = Vo Lo° drives a current

ofEa - fag,

thoughthe a- phase line

,

thatingests a

per- phase equivalent

circuit given by

+.

.¥T¥13 .

Jan~ A

•b-

Then,

the tone is theper

-

phase equivalent

of a Y . connected load with Ey = Eats i

i.e.,

q•

HintsIs EL

*¥o¥.

±III. .

•.

ZIoo . Y - A transformation .

Example # t

Consider a A- connected load with 4=345 kV,

drawing 750 mva at a pf of ons lagging .

Compute :

I Line and phase currents .

I Real § reactive powerdrawn

per phase .

I Perphase impedance .

Solution : ± |§§| = 750 MVA

⇒ BVLIL = 750 MVA

T = 750 MVA750MVA⇒ - l BT=

as uj' 256A '

T T⇒ -4

=

Is =725A.± P¢=tRe{5s¢ } c-Note the Ys in

front of Re{ Esq } .

Q¢=fIm{ 53°1 } .

NA'

t50 ° : Pg =tsRe{ 750237°} MW

¥ = } 750×0.8 MW

cost (0-8)=370. = Zoo MW .

QQ= fIm{ 7502370 } MVAR

= } 750×0.6 MVAR

= 150 MVAR .

E=fghLo=V¥Lo -451¥)↳7orin

why ?.

= 47667° R .

Iii prsiiigemainssfisonejiotineTwo 3¢ loads are connected in parallel

to

a Y . connected

lido:

°

'Funneled load 1 : 24kW at 0 's pf lagging. A - connected load 2 : 30 KVA at 0.8 pf leading

.

Find the line current phasorsII.a andIaaforthe two loads

,total

complex power I,

drawnby the loads

,and the total line

current drawn at the source . Let [Van=o.

Solution:§¢, ,=

o?4gL37° .ua. Aztywwstost

-

= 30/370 KVA .

£4,2 = 30 £370 KVA .

o : I,= §¢

, ,+ 53%2=48 kW .

Y = 480 V for both loads .

⇒

Eat= Iffy = js÷fo± A = 36 . l A .

⇒

LIT,

a=

- 37: why ? Current lags voltage .

⇒ III.al = lr5gy÷= 30×1 A = 36.1 A .

✓3 . 4 so

⇒ L II.a

= +37 ?

° : IIa= 36.1 L - 37

°

A,

Iz, a

= 36 . l L +370 A.

Total line current = IT, a

+ II a

= 36£1 L -37° + 36.1 437DA= 57.7 A .

Example#3_

An industrialplant

consists of several

3$ induction Motors and draws 300kW

at 0.6 pf lagging .How much

per.

phasereactive

powershould be injected from a 3$ capacitor

bank in parallel to the plant toimprove

the netpower factor to og lagging .

Solution : - .^

. i- age |g§p

a = cos. '

(0-6)=53

go.no?IolQtws'

(0-9)=258,

300kW

Here, sjy and 5dg are the

complex powerswith and without the

capacitorbanks .

o : 99ft (injagftuhmnby )=3oow#ttmEjima so)

= 254.7 KVAR