NODAL ADMITTANCE MATRIX

VIR vUe

4.3 FORMULATION OF THE PROBLEM

4.3.1 NODAL ADMITTANCE MATRIX

nected to all other nodes is 1\

I.

= 2,.V

^anEn

,.,=i

^m

= 1,2,--- N

;-

'I ^,

"

,

-

^\

_,

,

t

4.6

or in matrix fonn,

I, Y" Y12 --- ^YIN

I. Y" Y•• --- ^Y••

1

I

^I

I

I

I

I I

I I'

I

^I

I I I

I I

I I ^I

I I ^I

I.

Y" Y••

---

Y••

H,

..~

.•...

^'"

- .••...

,

~

..

^,,

[

••

l

•

"t

, _~"

. ^- _{- ,.1}

"

,

, I

. ^,

-

^,

-1

'~i

,

^,_•

, -

. _.

I

'

.

" '

t

( I

"

,

I

I I

, .

,

, .

The nodal admittance matrix is a sparse matrix (a few nunber of elements

"

are non-zero for an actual power system. This is why the oomputer memol7 requirements for storing the nodal admittance matrix is very low. It need to store only a very few non-zero elements,it need not to store the zeros of the \ matrix. Again the nodal admittance being a synrnetric matrix along the leading ", diagonal,the computer need t.o store

matrix only.

•

the upper triangular nodal ~~~,' "'~

.', •.

•. '~:rt,

< "

If the intercomlection between the various nodes for a given system, and':,-:' .-\ ,'~

the admittance value for each interconnecting circuit are known,the admittance'."

.' f

...

-,

4.7

can be seen that the ^SUIII of the elements in each colunn of the the'di.agonal element of each node is the ^SUlll of the

However,it

between the nodes.

admitt.ances cormecled to it.

ii.) t,he off-<liagonal element is the negative admittance

a singular matrix and hence the rows of the matrix are linearly dependent.

,

(

\ I

I

I

'~I

"

d \

I

admittance matrix slDIlIIatesto zero which means that nodal admittance matrix is-_{• - ••}

1

_I

\

:!

::

^~.

,'I

matrix maybe assembled as follows ; i)

As discussed earlier, il. is slack bus. Based on admittance

. _,,'f -" ',1'

~ "',1

" \ .l

"""J

" I:••. ~, I

necessary to select one of the buses as

the,':'.

.•.~...•'

matrix approach,it can be said mathematicallyl.

,

..

that one of the buses should be taken as slack bus,otheI'Wise the nodal matrix","',

" ,4

is singular and carilloI. be haniled. By taking one of the buses as the'.

reference , corresponding row and =lumn are deleted from the nodal nUll.l"ix I1l1d1""I<~lthe rcdu,-".! nUltrix Ix":lomesnurt-sutgulllr,which can very eu.'iily .

SHUNTBRANCHES

"'. "'4'1~.~••.

be handled' .'

"'-,'

^-

..•...

"',

"

Shunt admittances are added 'to the diagonal elements corresponding 1;0 the' , "

',I

~<. '.,~

nodes at which they are =rmected. The off-<liagonal elements are unaffected. I' 'I

.1" -;, .,~.,., . ⁰ The addition of shunt element in this way may either strengthen or weaken th~. '

. _,.

_, ~.

Y matrix diagonal depending Ortwhether the additional elements add to or sub-.'.

,.'

stract from the diagonal terms, ""I-

representirtg series admittance s\Jllllllltion.' , • Shunt induclanees strcrtgthen U,e diagonal while shunt capacitances weaken it.,

• '

.

With this alsol.he singularity of the admittance mal.rix can be avoided.

.

^,

"

. ..

'f '

TAPPED TRANSFORMERS

~.

The tapp<.."lltransfonners operaLi,ng at off-nominal tap positions provide,!

means of exchnagirtg reactive power between networks operatirtg at different'

voltages, and between generators and the network system to whi~, they. are;

~:li'.

=rmected, it is required to reflected this irlto power balance equat10nB. .j.'

-,"

A trwlsi'onner with off-nominal turns ratios can be represented by its im- "',

.•~. r ~ •.

p<.."llencesor admittances, =nnected irl series with an ideal autotransformer as~'

'..

^'

('.

shown in fig.4.1a. An equivalent circuit can be obtairled from this rep~

-" I' sentation 1.0 be used in load flow studies. The elements of the equivalent '~

circuit, then, can be treated in the Sllmemanner as line elements.

4.8

. ^.

,

^"

'-,,

. ^.

@ 0 ^@

a: 1 Ypq

~

--

-

^Ip

^-,----

^{"0 10}

(a)

-

I'm" _(£) @ (0

\;J

A ^YpO

-

^"Ip

--

^Iq

-

^I

^p

^a

^..-

¹⁰

B

~(~-l ^{)ypq (1-1.)y}

a pq

(b)

- -

^(e)

- -

F^i.g.<1.^t Tr<.l..n~Conner representaLions.

a)Fqu.i.valenLcircui ^{I: ;} b) EquivalcenL (~ircu^{lt ;} c) Equivalen ^t

n

cireui ^t with rnramet"rs "xpr"ssed in terms of adm5.U:anccund off-nominal turns

ratios.

The parameLers of t.he equaivalent circuit, showTIin fig.4.1b, can be derived by eqw~ting the terminal currents of the transformer with correspond~

ing currents of the equivalent circuit. At bus- m the terminal current ^1m of the transformer, shown in fig.4.1a, ^1S

1111 = itn/a

where, a ¹⁸ the tUITISratio of the ideal autotransformer and ^1<n, the current flowing from t to n, is

r,

, ,

_"

_,

^,~'

\,

\.

C'.'.

"

.r r

_1.\J'

'11,e cor-r""pondlng lennin ••!' cUl',-enLs for U1e equivalent

1.

= ^{(E. E.}

^{)A +}

^{E.B ---}

^(4.7)

1.

=

^(E.

E.

^)A +

E.C

~---(4.8)

Letting

E. = ^o

^and

^E. ₌

¹ in equation (4.4) ,

.fi

.",

.

^~~

.

^[

•

~I ~

, '.

" d

• 1.1 ' I

.~:>•.r

'.

. r "

";. ,,' "r

• 4\.. "r

~-',,-,

,-

"

,.

--

,

(4.5)

--- (4.6)

--- (4.4)

) Y••

1.

=

(E. - Et

1.

= ( ^{Et - E. )}

^y••/a ---

~4.3)

Et

=

^E./a

1.

=

(aE. - J~ ) y••/a

it. = (Et - E. ) Y••

Therefore,

4.1b are

eqwltion (4.3) beL~s.

Similarly, the terminal current 1. at bus n is

Substituting for Et, equation (4.5) becomes since

Let;ting E.

=

^{0 and E.}

=

¹ in equation (4.7),

1

4.10

Im=-A

Since th,-, terminal ,-,ur.-rents ror" the tra.l1sformer and its equivalent must be equal.

,

A

=

^Y••n/a

circuit

fUm.i.JarJy,8ubHt.i.tut.ing Em = 0 and En = in boLh equations (4.6) and (4.8).

In = ^{YlIln and} In=A+C

Again,si"ce the Lerminal currents for the tr"dnsfonner and its equivalent must

Ymn

=

^A -i-

c

Substituting for A from equation (4.9) mid solving for C

C

=

Ymn - Ymn/a

= (1 - I/a) Y••n

Equating the curre"t f,"ollJeqllations (4.4 ) arid (4.7) and suootituting for A from equation (4.9),

(Em -aEn solving for ll,

Ymn /a'

=

^{(Em - En ) Y••n /a} + E ••B

(Em - aEn ) Y••n/a' - (E•• -En ) Y••n/a B

=

Em

=Yaan /a^{2 -- Yaln} /a

4. 11 !

= lla (1/a -1 ) Y.n

111eequivalent circuit with its par~"'K,LersexpresHed in tenns of the

!~Urr)B rtltio a llIld ,the tJ1u)sf()r111(~r admitt.Julce 11r"e stJown in fig.4.1c.

WlI('~Jl the orr-rJ()f1J iJII1.l tUI"rIS Y";lLi () i,H "'~pn~Henl.Jlted at ^bUH m for u trallS- former connecting m and n. - th,-, self admiLtance at bus m is

YmD

=

^{Y., + --- +Ymn la + ~.--}

y..

^{1- lla (lla - 1)YmD}

= Ym) + Ym2 +--- + Yran/a2 + --- + YIIIN

111e muLual admittallce from m to n is

Ymn = -Ymn fa

The ~..l(~I [-..ad11l.i U";UICC at b1J~ n I.S

Ynn = YDI + --- + YDm/a 1 ---- + YD • +(1-1/a )YDm

=Ynl +--- + Ynm + --- + Yn N and is unchanged. 'Ille mutual admittance frOllln to m is

Ynm = - Ynm/a

4.12

b11s sys L(~m :

.N

Ira ^=~Ymn Fn

'Y\=- i

III = J,2 --- N

OJ:

.N

I .• = Y^mm Em ^+~Yllln En

7\=1

7\*n1

•

or Em

=

1m

Ymm Ymm

En --- (4.10)

'J11C real and. reactive Ix)werat any bu~ mis-

Pm - jQm = E~ 1m

Pm - jQm or 1m = ---~

wher(~" current 1m is pOHitive when J'Jowi_n.g into lhe system.

Now,substi.tuting for 1m in "quaUon (4.10),

Em

=

1 Ymm

[Pm - j'~ . ---

1, ••

"!.m •

m

=

^l,2,---N

--- (4.11)

Equations (4.11) an, the load fJow "qua,Uon!'!w1J,.,reth"bus voltages are Ute variables. FornnJla.Ling t.h,., load l".Iowpr'oblem in t,his manner results in a set

of f1oll-l,ilH~ar (.~qlJati.()n~l.haL (:;1.11t)(~ SOIVf:d hy an jt(~r'uLive nK~Lh(xJ.

4.13

"

,,

A .,3.3 CmIPUTATIONAL FEATU]lE

A significant l'eductioJl ill U1l'~computing time for a solution will be ob- tained by lx,rforming as manyari.LhemaUc operations as possible. before in- it:iaLi.ng Lhe iLerative calculaLion. Since I',Q at load buses and Y at all buses do' not clHulge H:i..thiterations, UH.~Lf~nnYmn/Y.,. and (P. - jQ. )/Y •• can ~

I/Y"," = I••

f~qlJaLi()n (r1. 11 ) C;UI l)(~ wr'j LLcTl

..,..,",1, 2, - - -' -1'1"

---(4.12}

x

22mnlAAEn

"" -=.

i

-n.-4=rn

--- (4.13)

(Pm - jQm )l •••

Em = ---

E!

and Y••ⁿ

r.... =

^YL..n --- (4.14)

then, Lhe bus voll.:,ge "'luaLion (1.12) t.""""es

--- (4.15b)

Kl••• X

EIIl

=

~YL""n

r'"'

em ^-n=.j .•• -4=Tn

=

^{El ••- E2Ra (Let)}

En m

=

1,2, --- N (4.15a)

Now,sepenlting the real and im,,ginary parts of KL,.from equation (4.13),

KI"" = (Pm

= (1'••

jQm )I.••

jQ", )/Ymm

= ---

Gmm + jBnua

4.14

P",C•••• - QmB".. -P",B", •• -

Qmc".

= ---

+j ---

U.?m ⁺ n.:-.. U.2•

⁺

n:.

= KI1~ + jl(Ll",

KL 1m = lmag inary pal't of J<Lm

~epernting the real and innginar-y par.t of .YLmn from equation (4.14),

YLmn = Y",n I.•••