A Method to Teach the Design and Operation Qf a Distribution System

8/13/2019 A Method to Teach the Design and Operation Qf a Distribution System

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IEEE T r a n s a c t i o n o n Power A p p a r a t u s a n d S y s t e m s , V o l . P A S - 10 3 , N o . 7 , J u l y 1 9 8 4A METHOD T O T EAC H T HE DESIGN AN DOPERATION Q F A DISTRIBUTION SYSTEM

W . H . KerstingSenior MemberNe w Mexico State UniversityLas C r u c e s , Ne w Mexico 8 8 0 0 3

INTRODUCTIONTh e undergraduate po w e r option program a t Ne wMexico State University consists o f five courses. O fthese f i v e , one i s required o f all electrical engineer-i n g majors, one i s a block elective a n d three ares e n i o r e l e c t i v e s . T h e block elective i s a method b ywhich an electrical engineering major m u s t select t h r e e

courses from a l i s t o f s i x . Th e l i s t i n c l u d e s onecourse i n each o f t h e s i x option areas o f f e r e d b y t h edepartment. T h e f i v e power courses a r e :

1. Electric Machinery required of allmajors2 . P o w e r T r an s mi s si o n L i ne s block elective3 . Power Systems Analysis senior elective4. Distribution Systems senior elective5 . Power System Protection senior electiveThe f i r s t three courses are offered every semes-t e r and are taken sequentially b y t h e s t u d e n t s . Th el a s t t w o courses are offered i n alternate s e m e s t e r s and

may b e taken along with t h e P o w e r Systems Analysiscourse.

Th e purpose o f this paper i s t o discuss t h e Dis-tribution Systems course. I n particular a c o m p u t e rprogram will be described t h a t has b e e n developed t oh e l p demonstrate some d es ig n a nd operating character-i s t i c s o f a distribution s y s t e m .

T H E DISTRIBUTION SYSTEMS COURSEOn e o f the positive t h i n g s about the power programa t Ne w Mexico State i s t h e industry advisory committeethat i s an i n t e g r a l p a r t o f t h e industry sponsoredgraduate program. O ne i mp o rt an t function o f t h e advi-

sory committee i s to a nn ua ll y r ev ie w the power c u r r i c u -lum ( u n d e r g r a d u a t e a n d g r a d u a t e ) and make recommend-ations f o r c h a n g e s , additions and/or deletions. It- w ast h i s a dv i so r y c om m it t ee t h a t m a de t h e recommendationnearly fifteen years ago f o r t h e creation o f a coursei n Distribution Systems. I t was further r e c o m m e n d e dthat the course s h o u l d b e of f ered as a senior electivef o r terminating students or as a f i r s t s e m e s t e r grad-u a t e course f o r students starting t h e graduate program.The reason f o r t h e need o f a Distribution Systemscourse was simply t h a t t h e majority o f new electricalengineers g o i n g to work f o r a utility company are firstassigned t o w o r k i n t h e distribution department. Ont h e surface i t may appear t ha t t h i s i s a m uc h t o ospecialized topic to appear as a senior elective letalone as a f i r s t s e m e s t e r graduate course. H o w e v e r , i t

8 4 WM 179-8 A paper r e c o m m e n d e d a nd approvedb y the IEEE P o w e r Engineering E d u c a t i o n C o m m i t t e eo f the IEEE P o w er Engineering Society f o r p r e s e n -tation at the IEEE/PES 1 9 8 4 Winter M e e t i n g , Dallas,T e x a s , January 2 9 February 3 , 1 9 84. Manuscriptsubmitted August 1 2 , 1 . 9 8 3 ; m a d e available f o rprinting N o v e m b e r 1 1 , 1983.

h a s been found t h a t t h i s course provides a g r e a t m e c h a -nism f o r review o f some basics of power s y s t e m s t h e o r yand a marvelous w a y i n which t o develop and expand u p o nt h e b a s i c s .Th e goals o f t h e course are straight f o r w a r d :1 . Define t h e basic c o m p o n e n t s starting with t h ebulk power substation a n d terminating at t h ec u s t o m e r ' s mete r .2 . Define and discuss t h i s t hi ng c al le d l o a d .3 . Develop typical s u b s t a t i o n l a y o u t s .4 . D e v e l o p t y p i c a l l a y o u t s o f substransmissiona n d distribution feeder systems.5 . Define a c c e p t a b l e s t e a d y - s t a t e o p e r a t i n gc o n d i t i o n s .6 . D ev e lo p m e th o ds o f analyzing the steady-stateoperating conditions o f unbalanced t h r e e - p h a s esy stem s.I n t h e process of achieving the goals i t i s neces

sary f o r t h e students to review material fr o m previouspower courses and-then to expand upon these basics i no r d e r to understand the design and operation of ani n h e r e n t l y u n ba l an c ed t h re e -p ha s e distribution s y s t e m .The review process i s what m a k e s this a good terminalc o u i r s e f o r graduating s e n i o r s or a good f i r s t s e m e s t e rg - - r a d u a t e course f or students returning to s c h o o l a f t e rworking i n i n d u i s t r y or f o r new graduate students who didtheir undergraduate w o r k a t some o t h e r u n i v e r s i t y .

On e of the difficulties o f teaching the Distribu-tion Systems course, a nd m o s t power s y s t e m s analysisc o u r s e ' s , i s n o t being able t o have a real s y s t e m toexperiment w i t h in t h e l a b o r a t o r y . Without such a reals y s t e m to e xp er im en t w i th some simulation means i s nec-essary. As i s t r u e i n m o s t cases like t h i s , the digi-tal c o m p u t e r ha s b ec om e a very p o p ul a r d ev ic e u se d tosimulate real s y s t e m s .

During the last t e n years a Radial Three-PhaseP o w e r Flow Program ( R A D F L O W ) has been developed t osimulate t h e design a n d operation of a distributionsy stem . This program ha s b ec om e a very effective toolused b y the students to solve regular h o m e w o r k problemsand assist i n t h e s e m e s t e r long project of the designand analysis of a distribution sy stem .THE RADIAL THREE-PHASE POWER F L O W PROGRAM

The first ve r s i o n o f RADFLOW was developed t e nyears ago and was a b as ic G a us s- S ei d el program expandedto model all three phases o f a d i s t r i b u t i o n system.1] Unfortunately t h i s first ve r s i o n was handicappedb y the v e r y poor convergence c h a r a c t e r i s t i c s o f G a u s s -Seidel on rad ial systems. The students sp ent more timefighting t h e convergence problems t h a n they d i d l e a r n -i n g a b o u t d i s t r i b u t i o n s s y , s t e m s . The f ir st v er si on w a sreplaced by the-present RADFLOW program w h i c h u t i l i z e sa modified l ad de r n e tw o rk theory in the i t e r a t i v e rou-t i n e . This method displays outstanding c o n v e r g e n c echaracteristics for r a d i a l systems. [ 2 ] B e f o r edescribing t h e m o di f i e d l a d d e r t h e o r y , i t i s necessaryto develop t h e line s e c t i o n m o d e l a n d t h e l oa d modelused by the program.

0 0 1 8 - 9 5 1 0 / 8 4 / 0 7 0 0 - 19 4 5 0 1 . 0 0 1 9 8 4 IEEE

1 9 4 5


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1 9 4 6Line S ec ti on M od e l

I n t h e prerequisite course ( P o w e r TransmissionL i n e s ) t h e s t u d e n t s have been introduced t o t h e calcu-lation o f transmission line i m p e d a n c e . I n t h a t courset h e t e x t E l e m e n t s o f P o w e r S y s t e m s Analysis i s u s e d .3 ] Assumptions a r e made i n t h a t t e x t i n o r d e r t oreduce t h e complexity of t h e calculation o f linei n d u c t a n c e . T h e t w o f u n d a m e n t a l a s s u m p t i o n s a r e :

1 . Balanced t h r e e - p h a s e currents2 . Complete transposition o f c o n d u c t o r sWith these t w o a s s u m p t i o n s a s i m p l e equation i s derivedf o r p h a s e i n d u c t a n c e which i n c l u d e s t h e s e l f i n d u c t a n c ea n d t h e e f f e c t o f t h e m u t u a l i n d u c t a n c e with t h e othert w o p h a s e s . F or balanced three-phase system analysist h i s i s a viable method a n d t h e one most commonly u s e di n i n d u s t r y f o r t h e determination o f i m pe da n ce v a lu e st o u s e i n po wer f lo w s t u d i e s o f transmission s y s t e m s .H o w e v e r , a distribution system d o e s not l e n d itself t oeither o f t h e two assumptions. Because o f t h e domi-nance o f single-phase l o a d s t h e assumption o f balancedt h r e e - p h a s e currents i s not j u s t i f i e d . Distributionl i n e s a r e never t r a n s p o s e d nor c a n i t b e assumed t h a tt h e c o n d u c t o r c o n f i g u r a t i o n i s a n e q u i l a t e r a l t r i a n g l e .When t h e t w o assumptions a re thrown o u t i t i s necessaryt o i n t r o d u c e t h e students t o a more s o p h i s t i c a t e d andaccurate method o f calculating line i m p e d a n c e s . Th em e t h o d presently being t a u g h t i s t h e application o fC a r s o n ' s equations t o a t h r e e - p h a s e g r o u n d e d f o u r wirewye s y s t e m . [ 4 ] C a r s o n ' s equations a r e n ot d er iv ed i nt o t a l , b u t i t i s possible t o d e v e l o p them i n form basedupon t h e derivation o f inductance a s developed i n t h ep r e v i o u s c o u r s e . T h i s development i s an example o fwhat was previously described a s a n ic e f ea tu re o f t h ec o u r s e . T h e s t u d e n t s a r e required t o r e v i e w t h e deri-va t i o n of i n d u c t a n c e a n d then t o s e e t h a t methodexpanded t o a very a c c u r a t e , though c o m p u t a t i o n a l l yd i f f i c u l t , method o f calculating l i n e i m p e d a n c e s .

C a r s o n ' s equations a l l o w f o r t h e computation o fc o n d u c t o r s e l f impedance a n d t h e m u t u a l i m p e d a n c e sbetween conductors f o r any number o f c o nd u ct o rs a b ov eg r o u n d . For a t h r e e - p h a s e , f o u r w ir e g ro un de d wye sys-tem a s i m p l e model i s shown i n F i g u r e 1 .

a

b~~~~~~~~a ab

C . c_ I ~p I n .

g\ \\ \\\sF i g . 1 . T h r e e - P h a s e , Four Wi re L ine Model

F or t h i s f o u r wire s y s t e m , C a r s o n ' s e q u a t i o n s willl e a d t o t h e d e v e l o p m e n t o f a 4 x 4 i m p e d a n c e m a t r i x .T h i s matrix i s u s e d t o c a l c u l a t e conductor voltaged r o p s as s h o w n i n Equation 1 .

IV a g - Va gI Za aV b g V b , g = Z b aIv v I ZIcg cV caV V I Z n a

L

z I z t a b a c a nZb Z b ' c Z b ' nc b c c c n

z t zt z tn b n c nn

x

I 1aI b II Ii c iL n

I n a g r o u n d e d neutral system:V -V, = 0ng n g

( 1 )

( 2 )B y substituting Equation 2 i n t o Equation 1 , t h e 4 x 4matrix ca n be reduced t o a 3 x 3 m a t r ix as shown i nE q u a t i o n 3 .

aa' ag aaab ac aV b b ? = Vb g - Vbg9 = Z b a Zb b Z b c x I b ( 3 )L c c i L ' c g C T Lca c b cc c j

Th e relationship between t h e Carson equation i m p e d a n c e so f Equation 1 a n d t h e f i n a l impedances o f Equation 3 i sg i v e n b y :

t IZ . Z, Z i n n jZ i j = z i j nn n ( 4 )Figure 2 represents t h e final model of a t h r e e - p h a s e ,f o u r wire g r o u n d e d wye l i n e as defined b y E q u a t i o n 3 .

F i g . 2 . Final Three-Phase Line ModelT h e s a m e m e t h o d o l o g y i s used t o model d o u b l e - p h a s eand s i n g l e - p h a s e l i n e s e c t i o n s . In t h e case of ad o u b l e - p h a s e line i n v o l v i n g p h a s e s a a n d c , for e x a m p l e ,C a r s o n ' s e q u a t i o n s lead t o a 3 x 3 matrix ( a , c , n ) whichi s t h e n reduced t o a 2 x 2 m a t r i x ' . i s i n g Equation 4 .Th e 2 x 2 matrix i s then e x p a n d e d t o a 3 x 3 b y placingz e r o s i n t h e r o w and column of the m i s s i n g p h a s e ( b i nt h i s e x a m p l e ) . T h e r e s u l t i n g m a t r ix e q u a t i o n i s showna s Equation 5 .a a ag a g aa ac

V b b = = 0 x ( 5 )c c V c g - c g c a cc I

F or a s i n g l e - p h a s e l i n e , C a r s o n ' s equations result i n a2 x 2 m a t r i x . At t h i s point t h e 2 x 2 could b e r e d u c e dt o a s in gl e t er m using Equation 4 . H o w e v e r , research a tthe P u b l i c Service Company of Ne w Mexico h a s shown t h a ti n a s i n g l e - p h a s e line nearly a l l o f t h e phase c u r r e n treturns through t h e neutral c o n d u c t o r . [ 5 ] With t h ereturn current through t h e n e u t r a l , a d if f er en t p ro ce -dur e i s f o l l o w e d i n reducing t h e 2 x 2 - m a t r i x .

9 Ix \ % N \ k \ \ \ , -,


3/8

1 9 4 7A single-phase line consisting o f a p h a s e a andg r o u n d e d neutral n i s shown i n Figure 3 .

Iz I1a

an

J--z IInfg \\ \ N N N \ \\ N N N ,N \N \\\\\

F i g . 3 . S i n g l e Phase Line SectionF r o m Figure 3 t h e equation f o r voltage d r o p on phase acan b e written a s :

V =V -Vaa an a i n=Z ' I + Zt I - Z ' I - Z' I ( 6 )aa a a n n nn n na a

Because a l l o f t h e p h a s e current returns i n t h e n e u t r a lc o n d u c t o r :I = - I ( 7 )n a

Substituting Equation 7 into Equation 6 leads t o :V , = V - V aa a n a ' n

= ( Z + Z ' - Z ' - Z ' ) Iaa nn an na aS i n g l e - p h a s e i m p e d a n c e i s n ow d e f i n e d :

z = z + z t - z s _ z tl 4 a a n n an na

( 8 )

( 9 )Because t h e program t o b e d e s c r i b e d assumes t h a t a l lline sections are m ode le d b y a 3 x 3 m a t r i x , t h es i n g l e - p h a s e l i n e ( p h a s e a f o r e x a m p l e ) i s m o d e l e d b yt h e matrix s h o w n i n Equation 1 0 where t h e missingp h a s e s ( b and c ) a re r e p r e s e n t e d b y z e r o s i n t h e appro-priate r o w s a nd c o l u m n s .

v a a ' v a -ai z 1 I ab b | = = x 0 ( 1 0 )

T h e e n d r e s u l t i s t h a t a l l l i n e s e c t i o n s , b e t h e yt h r e e - p h a s e , d o u b l e - p h a s e or s i n g l e - p h a s e can useFigure 2 a s t h e m o d e l . Equation 3 r e p r e s e n t s t h e gen-e r a l e q u a t i o n u s e d t o c a l c u l a t e voltage d r o p on a lines e g m e n t . I n cases o f d o u b l e - p h a s e a n d single-phasel i n e s , o n l y t h e p h y s i c a l l y present p h a s e s w il l hav ecurrent f lo w a nd voltage d r o p s . This i s accomplishedb y s e t t i n g t o zero t h e appropriate t e r m s f o r t h emissing p h a s e s .Load Model

Al l l o a d s are a s s u m e d t o be f i x e d complex power( S = P + j Q ) b y p ha s e . I t i s f u r t h e r assumed t h a t a l lt h r e e - p h a s e l o a d s a re wye connected and a l l d o u b l e -p h a s e a n d s i n g l e - p h a s e l oa ds a re connected line t og r o u n d e d n e u t r a l . T h e t h r e e - p h a s e load model i s showni n F i g u r e 4 .

F i g . 4 . Load ModelI n Figure 4 t h e three-phase l o a d does not have t o b eb a l a n c e d . T h a t i s S a , S b a n d S c can b e different valuesor even z e r o . In f a c t , double-phase and s i n g l e - p h a s eloads a r e modeled b y setting t h e values o f t h e c o m p l e xpower t o z e r o f o r t h e non-exist p h a s e s . Figure 4 t h e nr e p r e s e n t s t h e l o a d model f o r a l l l o a d s .

I n Figure 4 i t i s a s s u m e d t h a t t h e c o m p l e x powerso f each p h a s e are known a n d t h a t t h e l i n e - t o - n e u t r a lvoltages have been s p e c i f i e d . ( M o r e on t h i s later i nt h e p a p e r ) . With t h i s k n o w n , t h e l o a d currents a r edetermined b y :I = ( S a i a)a a a n

( 1 1 )I b = ( S b / V b n ) *I = ( S / V ) *c c c mTH E LADDER THEORY ITERATIVE METHOD

Time a nd sp ace d o not p e r m i t a d e t a i l e d e x p l a n a t i o no f t h e iterative t e c h n i q u e u s e d i n t h e p r o g r a m . A com-p l e t e development and d e s c r i p t i o n can be f o u n d i nReference 2 A g e n e r a l o u t l i n e will have t o s u f f i c ef o r t h i s p a p e r .The iterative t e c h n i q u e u s e d i s an a d a p t a t i o n o f t h eL a d d e r Network Theory o f l i n e a r s y s t e m s as t a u g h t i nbasic s o p h o m o r e l e v e l circuit courses. In t h a tmethod a linear l a d d e r n e t w o r k , as shown in F i g u r e 5 , i sanalyzed by first a s s u m i n g a v o l t a g e - a t node n . C u r r e n t ,I n , i s c a l c u l a t e d using E q u a t i o n 1 2 .

I = Y V ( 1 2 )n n n

F i g . 5 . Linear Ladder NetworkWith I n k n o w n , t h e v o l t a g e d r o p i n t h e l i n e section( n - l t o n ) can b e c a l c u l a t e d and i n tu rn V n . l i s calcu-a t e d . No w t h e l o a d c u r r e n t a t n - l ca n b e d e t e r m i n e d a n dt h i s i s a d d e d t o t h e p r e v i o u s line current s o t h e n e x tline section voltage d r o p c a n b e c a l c u l a t e d . T h i s pro-cess i s r e p e a t e d until t h e v o l t a g e at t h e s o u r c e ( V 1 )

, a


4/8

1 9 4 8h a s been c a l c u l a t e d . The calculated V 1 i s c o m p a r e d t ot h e known i n p u t v o l t a g e ( E s ) a n d a correction f a c t o r( C F ) derived b y :

CF = E s / V ( 1 3 )Because the n e t w o r k is l i n e a r all c a l c u l a t e d voltagesa n d c u r r e n t s are multiplied b y this c o r r e c t i o n factora n d a f i n a l solution i s a c h i e v e d .

A distribution s y s t e m i s n o n - l i n e a r because loadsare assumed to be c o n s t a n t complex p o w e r . The e x a c tl a d d e r t h e o r y method can n o t b e a p p l i e d . H o w e v e r , amodification to t h e procedure j u s t d e s c r i b e d can bea p p l i e d . Figure 6 shows a simple n o n - l i n e a r l a d d e r n e t -w o r k . Note the only difference b e t w e e n this a nd thelinear network i s t h a t loads are c o n s t a n t complex p o w e rrather t h a n c o n s t a n t a d m i t t a n c e as before.

F i g . 6 . Non-Linear L a d d e r N e t w o r kTo g e t started a voltage Vn i s assumed a t n o d e n . On asingle-phase basis the l oad c u r r e n t I n i s calcu lated:

n= ( S n / V n ) ( 1 4 )As b e f o r e , t h e s y s t e m can be analyzed working b a c ktowards t h e source using Equation 14 to c a l c u l a t e a llload c u r r e n t s . I n time V1 will be calculated an d com-p a r e d to t h e known source voltage E s . Because t hes y s t e m i s non-linear a simple c o r r e c t i o n factor can n o tbe determined and then applied to all o th er c al cu la t edc u r r e n t s and v o l t a g e s . Instead a delta voltage isdetermined b y :

AV = E V1 ( 1 5 )The calculated delta voltage is applied t o the a s s u m e dvoltage a t Vn u sin g Equation 1 6 .

v = V + AV ( 1 6 ). n n e w n o l dWith the new Vn t h e s y s t e m i s analyzed by working f r o mnod e nback to the source node . This p r o c e dur e is con-tinu ed u n t i l the c a l c u l a t e d voltage V1 is w i t h i n aspecified tolerance of the k n o w n source voltage E s .

The ba si c p r o c e dur e i s expanded to a three-phasea n a l y s i s b y u si ng E qu at io n 11 t o c a l c u l a t e phase l o a dc u r r e n t s and u si ng E qu at io n 3 t o c a l c u l a t e three-phaseline voltage d r o p s . Recall that in the development o fEquations 3 a nd 11 that t he re were no r e s t r i c t i o n splaced on t h e lines and/or loads being balanced . R e m e m -ber also t h a t Equations 3 and 1 1 can b e u s e d to m o d e ldouble-phase a nd single-phase l in es a nd l oa d s.

V4V 1

S 2

F i g . 7 . Main Feeder With A T a pI n s u c h a s y s t e m V 4 would f i r s t b e a ss um ed a nd V2 c a l c u -l a t e d . Because b u s 2 i s a t a p b u s , t h e calculated V 2now becomes a k n o w n v o l t a g e . The p r o c e dur e a t t h i sp o i n t i s t o assume a voltage V 3 a t b u s 3 . As before avoltage V ' i s calculated. N o w the same p r o c e d u r e i su s e d as i s u s e d a t t h e source b u s . T h a t i s , V2 calcu-l a t e d i s compared to the previously calculated V 2 . I ft h e t w o are n o t within tolerance, a delta voltage i sc a l c u l a t e d a n d a p p l i e d t o V 3 a n d a new value o f V i sc a l c u l a t e d . T h i s continues until V2 and V2 are withint o l e r a n c e . When V 2 and V 2 are within t o l e r a n c e t h esolution c o n t i n u e s back t o the source bu s as b e f o r e .Once again t h e p r o c e dur e o f comparing t h e calculatedvoltage to t h e known source voltage i s repeated. I ft h e s e t w o voltages are n o t w i t h i n tolerance, t h e d e l t avoltage i s applied to t h e m o s t r e m o t e b u s voltage a n dt h e t o t a l procedure i s repeated. Each calculation o f asource voltage i s t h e completion o f a m a j o r iterationwhile t h e calculation o f a t a p voltage i s k n o w n as a s u b - i t e r a t i o n . As s u c h , one major iteration may haveseveral hundred sub-iterations; depending upon t h e com-plexity o f t h e s ys t e m.

Th e p r e s e n t version of RADFLOW i s dimensioned f o r40 0 b u s e s . Input consists of line da ta cards w h i c h willcontain t h e f r o m a n d t o bus n u m b e r s , t h e d i s t a n c e o ft h e l i n e and a configuration code number t h a t i s used b yt h e c o m p u t e r t o assign t h e proper impedance m at r i x tot h e l i n e s e c t i o n . Bu s data cards contain t h e c o m p l e xpowers b y phase a n d t h e value o f s h u n t capacitors, i fp r e s e n t , a t each b u s .

The program ha s been modified many t i m e s over t h eyears to make i t as user friendly as possible and to a d dadditional s y s t e m m o d e l s . T h e m o s t r e c e n t additionshave been t h e inclusion o f s t e p voltage regulators a n dline t r a n s f o r m e r s . T h e line transformer addition wasinstalled because some s y s t e m s being studied b y t h e s t u -d e n t s will change voltage levels on l on g r ur al f e e d e r s .Each year brings more modifications t h a t add t o t h e com-plexity of th e program but also a d d s to t h e usefulnessin helping t h e s tu de nt s b et te r understand the design a n do p e r a t i o n of a distribution s y s t e m .

RADFLOW APPLICATIONSRADFLOW can be and is u s e d in a v a r i e t y o f ways i nthe Distribution S y s t e m s course F i g u r e 8 shows a oneline diagram o f a s m a l l 4 . 16 k V d i s t r i b u t i o n systemThis t e s t system w i l l b e u s e d to demonstrate some of theapplications o f RADFLOW.

A typical d i s t r i b u t i o n s y s t e m is r a r e l y a p u r eladder network. That i s , typical d i s t r i b u t i o n systemsw i l l h av e l at er al t a p s o ff the primary m a i n f e e d e rand the l a t e r a l t a p s t h e m s e l v e s w i l l a l s o b e t a p p e d .This complicates the solution t ec hn iq u e u s in g the laddert h e o r y , b ut d o e s n o t m a k e it i m p o s s i b l e . Figure 7 s h o w sa s e c t i o n o f a m a i n f ee de r w it h a l a t e r a l t a p .

S 3V 3 -


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F i g . 8 . One-Line D i a g r a m of T e s t S y s t e mI n Figure 8 t h e distances between buses i s given inf e e t . The va l ue s o f th e loads are given as eitherthree-phase or single-phase k V A .Ba se Ca se

Th e first study performed will be referred t o asth e b a s e case . I f t h i s was a new system i t w o u l d b enecessary t o select conductor sizes and configurationsand to determine t h e t p h a s i n g of l o a d s . F o r the pur-poses of this paper i t will be assumed that this i s anexisting s y s t e m with all o f those decisions alreadymade. H o w e v e r , i n t h e Distribution Systems courseleaving these d e c i s i o n s up t o t h e students i s an excel-lent exercise to help t h e students d e v e l o p an u n d e r -standing and appreciation o f what i s involved in layingo u t a new s y s t e m .

F o r t h e t e s t s y s t e m i t will b e assumed that th eprimary main conductors are 336,400 2 6 / 7 ACSR a n d alllaterals ( t h r e e - p h a s e , double-phase a nd single-phase)are 1 / 0 ACSR. T h e configuration o f t h e conductors onthe poles are s h o w n i n Figure 9 .

4 . 5 ' :-2.14 'A~b~1

n T2 4 '

Three-Phase

t. - 7 1b f ~ L

n n _ +2 4 ' 2 4 '

D o uble - P h ase S ingle - P h ase

F i g . 9 . Pole Configuration

1 9 4 9The 3 x 3 i m p e d a n c e matrices a r e first determined a soutlined p r e v i o u s l y . RADFLOW r e q u i r e s t h a t each conduc-t o r size a n d c o n f i g u r a t i o n be a s s i g n e d a s p e c i f i c c o d e .Table I g i v e s t h i s b r e a k d o w n .

Table IConductor C o n f i g u r a t i o n CodesConductor Size Phases

3 3 6 , 4 0 01 / 01 / 01 / 01 / 01 / 01 / 01 / 0

a , b , c , na , b , c , na ,nb , nc,na,c,na , b , nb , c , n

Code12345678

U s i n g Table I e x p l a i n s t h e numbers underlined o n t h eone-line diagram o f Figure 8 . F o r e x a m p l e , the l i n esection from bus 1 7 0 to bus 1 5 0 i s a code 8 which m e a n si t i s a d o u b l e - p h a s e line c o n s i s t i n g of p h a s e s b and ca n d n e u t r a l .After t h e data has be e n entered and the program

r u n , attention i s turned to t h e output. The output f r o mRADFLOW i s broken into four p a r t s .1 . L i s t i n g o f line and load d a t a . The line dataoutput g i v e s the p e r - u n i t 3 x 3 i m p e d a n c ematrix f o r each line section.2 . V o l t a g e Profile -- T h i s i s a quick summary oft h e p e r - u n i t v o l t a g e magnitude and angle i nd e g r e e s b y p h a s e f o r each b u s and t h e va l ueof the v o l t a g e unbalance at the b u s . V o l t a g eunbalance i s defined a s the ratio o f n e g a t i v esequence voltage t o positive sequence v o l t a g e .3 . Radial Load Fl o w Solution - - This i s a t y p i c a loutput g i v i n g all essential bus information andt h e line flows i n t o and o u t o f t h e b u s . Th ep r o g r a m has t h e option t o either o ut put lineflow i n kW a n d kVAR or current magnitude i n

amps and a n g l e i n d e g r e e s .4 . S u m m a r y - - This i s t h e final ouput and i n c l u d e s :a . R e q u i r e d number o f iterations f o r solutionb . V o l t a g e tolerances usedc . Total system i n p u t b y p h a s ed . Total system load b y p h a s ee . Total system losses b y p h a s ef . List o f capacitors b y bus and p h a s eg . List o f overloaded linesF i g u r e 1 0 shows the Voltage Profile f o r t h e first t e s tsystem r u n , F i g u r e 1 1 g i v e s a s a m p l e o f t h e Load Fl o wSolution output and Figure 1 2 shows t h e Summary p a g e .

Armed w i t h all of t h i s output, what does t h e stu -dent d o ? Th e first c o n c e r n i s the voltage p r o f i l e .Prior t o making t he f i r s t run some general g u i d e l i n e shave been established i n class a s t o what will c o n s t i t u t ea n acceptable voltage profile. Th e t w o basic r e q u i r e -m e n t s ar e:

1 . N o m o r e than a 3 % voltage d r o p f r o m the sub-station t o the last b u s . With a 1 . 0 p e r - u n i tvoltage at t h e s u b s t a t i o n , - - t h i s means t h a tthe minimum acceptable per-unit voltage i s 0 . 9 7 .2 . Voltage unbalance can not exceed 2 % .As can b e seen f r o m Figure 1 0 , the v o l t a g e d r o p criteriaha s not b e e n satisfied but the voltage unbalance i sa c c e p t a b l e .

I n order t o solve the voltage drop p r o b l e m , shuntcapacitors n e e d t o b e installed. Th e voltage d r o p sbelow 0 . 9 7 per-unit s t a r t i n g at bus 2 0 , t h e r e f o r e , t h a t


6/8

1 9 5 0O L r A G E P R OF L

B U S NAM E P HA S E AM AGNIT U D E AM NL E

500 SUBSTA350 T AP110 T AP1 4 0 BUS 1 4 04 0 BUS 4080 BUS 809 0 BUS 9 03 0 BUS 3 0

2 2 0 rAP1 70 BUS 1 7 0150 B U S I SO1 2 0 BUS 1 2 01 00 END1 3 0 B U S 1 3 01 60 BUS 1 602 0 BUS 20400 TIEPT70 BUS 706 0 BUS 601 0 BUS 1 0

200 BUS 200I 80 BUS 180210 BUS 2 1 01 9 0 BUS 1 9 050 BUS 50

1.0000000 . 9 8 6 4 6 10.9842460.982054

0.980679

0 . 9 8 2 8 7 80 . 9 8 1 3 3 60.981336

0 9816890 . 9 7 6 1 7 00.9749720 . 9 6 7 0 7 80.9670780.963414

0.9609330.9616550.9581800.956170

0.000000-0.553486-0.647968-0.675814

-0.688453

-0.697955-0.708954-0.708954

- 0 . 7 1 1 4 5 5-0.644149-0.657873-0.722773-0.722773-0.725213

-0.725367-0.800105- 0 . 8 3 9 7 4 t-0.872335

PHASE BM A G N I T U D E A N G L E1.0000000 . 9 9 0 1520 . 9 8 7 8 7 80 985395

-120.000000- 1 2 0 . 4 5 1 4 1 6-120.560425- 1 2 0 . 5 6 4 8 9 6

0 . 9 8 4 3 4 6 - 1 2 0 . 5 8 9 2 6 40.9825S90 . 9 8 6 5 4 20 . 9 8 4 6 6 10 9825560. 9 7 8 8 8 70 . 9 8 5 3 6 60 . 9 8 2 9 8 40.9818060 . 9 7 6 9 9 4

-120.608994-120.634399-120.6612851 2 0 . 6 9 2 5 9 6- 1 2 0 . 7 3 2 3 1 5-120.644135- 1 2 0 . 4 6 2 8 3 0-120.472549-120.464584

0.976994 -120.4645840 . 9 7 2 6 5 4 - 1 2 0 . 4 1 4 7 4 90 . 9 7 0 2 4 6 - 1 2 0 . 4 0 7 3 6 40.975347 - 1 2 0 . S 1 5 9 9 10.973353 - 1 2 0 . 56 6 2 6 90 . 9 7 2 8 6 5 - 1 2 0 . 5 8 2 3 8 20 . 9 7 1 9 4 1 -120.592499

PHASE CMAGNIT UD E ANGL E1.0000000 . 9 8 8 0 1 10.9857030.9822750 . 9 7 8 6 0 10 . 9 76 76 20.982477

120 0000001 19 .5 0 6 4 3 91 1 9 . 3 8 5 8 9 51 1 9 . 3 3 2 1 8 41 1 9 . 2924651 1 9 . 2 7 2 4 6 11 1 9 . 341202

0.984897 1 1 9 . 3 4 2 4 0 70.984238 1 1 9 . 3 3 3 7 7 10.983517 1 1 9 . 3 1 3 2 7 80 . 9 8 3 8 2 80 . 9 8 0 1 9 90.9791250.9734630.9734630 . 9 6 6 4 8 10.9650860.9642340.9737120 . 9 7 4 0 8 30.974083

1 1 9 . 327545119.4613041 1 9 . 4463351 1 9 . 4 3 05 4 21 1 9 . 4 3 0 5 4 21 19 . 3 6 0 1 6 8119.3448031 1 9 . 3 54 752119.469498119.4932711 1 9 . 4 9 3 2 7 1

0.954760 0 888023

F i g . 1 0 . Test S y s t e m Voltage Profile_ _ RADIAL LO D PLOw SOUT O

T Es r r s r T M F O R T B H 1983 SU M M ER POWER M E E r I N G P A P E RB U S NAME P H ASE A PLEASE B PHASE C (UNITS)500 S U B S TA P . U . VOLTA G E: 1.000007 0.000019 1.000004 - 1 1 9 - 9 9 9 6 1.000000 120-0000 (MAG ANG)

LOCAL LOAD: 0.000 0.000 0.000 0.000 0.000 0.000 (E M X V A R )STATIC CAP: 0 . 0 0 0 0.000 0.000 1 E V A R )TO BUS 350: 2 75. 133 - 2 9 . 4 7 6 240.752 - 1 4 9 . 6 60 252.409 9 0 . 3 6 8 I A M P M G )

350 T A P P.U. VOLTAGE: 0 . 9 8 6 4 6 1 -0.553486 0.990152 -120.4514 0.988011 119.5064 AG A N G )LOCAL LOAD: 0.000 0.000 0.000 0.000 0.000 0.000 (E M KVAR)S H U N T C A P .: 0.000 0.000 0.000 1 E V A R )

F R O M B U S 500: 2 75. 133 - 2 9 . 4 7 6 240.752 -1 4 9 .660 252.409 90 . 36 8 (AMP DE GTO B us 110: 95.453 - 2 6 .5 40 - 1 0 5 . 9 7 6 - 146 . 499 106.107 9 3 . 4 6 6 (AMP D E G CTo H U S 130: 1 7 9 . 8 7 1 - 3 1 . 0 3 4 135.064 - 1 5 2 . 1 4 0 146.569 8 8 . 1 2 6 (AMP D E G )11 0 T AP P.U. VOLTAGE: 0 . 9 8 4 2 4 6 -0 . 6 4796 8 0.987878 -120.5604 0.985703 119 . 3859 (NAG A 1 G )

LOCAL L O A D : 0.000 0.000 0.000 0.000 0.000 0.000 (O f E V A R )SUNT CAP.: 0 - 0 0 0 0.000 0.000 E V A R )F R O M B U S 350: 95.453 - 2 6 .5 40 105.976 - 1 4 6 . 4 9 9 106.107 9 3 . 4 6 6 (AYP D E G )T O B U S 140: 4 2 . 4 2 6 - 2 6 .5 24 42 .3 15 - 146. 42 9 6 3 . 7 8 0 93.457 (AMP D E G )TO B U S 220: 53.027 -26.552 6 3 . 6 6 1 - 1 4 6 . 545 4 2 . 3 2 7 9 3 .4 7 8 (AMP D E G )

14 0 B U S 1 40 P . U . VOLTAGE: 0.982054 -0 .6 75814 0.98S395 -120.5649 0 . 98 2 2 75 1 1 9 . 3 3 2 2 { ( A G T V )LOCAL LOAD:SHUT C A P . :

FROM BUS 1 1 0:TOBUS 4 0:To BUS 90:

45.00042 . 4 2 62 1.228

2 1 *7940-000-26.524-26.530

45*00042.31521.188

2 1 o 7 9 40 . 0 0 0-146.429-146.451

45.000 21.794 KVAR)0.000 1 K V A R )63.780 93.457 AMP D E G )42.586 93.441 AMP DE G )0.000 0.000 t ( A P DE G )

Fig. 1 1 . Test S y s t e m Samp l e L o a d F l o w S o l u t i o n

_ _ __ __SUMMARY__4 I T E R A T I O N S REQUZRED FOR C ONVE R G E N C E ( 26 7 S U B - I TE R A TI ON S R E Q U I R E D )VOLTAGE TOLERANCE:R E A L : 0.000010Z M A G : 0 . 0 0 0 0 1 0

S c W KVARTOTAL SYSTEM INPUT: PHASE A: 575.279 325 157PHASE B: 502.471 286.142PHASE C: 526.945 299.739r o r A L SYSTEM LOAD: PHASE A: 561.099 307 248PHASE B: 493.599 274.556PHASE C: 516.099 285.453TOTAL SiUNT CAP.: PHASE A: 0 000 0 000PHASE B: 0.000 0.000PHASE C: 0.000 0 000

TOTAL SYSTEM LOSSES: PHASE A: 14 179 17 910PHASE B: 8.872 11586PHASE C: 10.846 14.286

K VA PP660.812 0.870563578.234 0.868975606.229 0.869217639.713 0.877110564.819 0.873906589.781 0.875069

0.000 0.0000000.000 0.0000000.000 0.00000022.843 0.62072914.593 0.60796317.936 0.604690

F i g . 12. T e s t System S u m m a r y

8 UNA L A A C E0.00010 . 1 t 1 0 80 . 1 1 670 . 1 2 6 0

0.129 10 . 1 0 9 40 . 1 08 90 . 0 6 7 50. 1 1 3 80 . 2 2 2 90 - 2 2 8 20 . 3 3 2 90 . 3 3 2 90.33310 . 3 4 4 20.46360 . 5 4 1 90.6018


7/8

i s a logical location to install a three-phase shuntcapacitor b a n k . The next question i s what s i z e ? F r o mp r e vi o us courses the s tu d en ts h av e b ee n t au gh t t h a t agood s ta r ti ng p o in t i s to install capacitor b a n k s sucht h a t the total reactive'power f l o w i n t o the bus i s s u p -plied b y t h e capacitor bank. The ou tp u t at bus 20 ( n o tshown h e r e ) shows that t h e complex p o w e r -fl o w i n t o bus2 0 from bus 130 on p h h s e a i s 315.9 k W and 189.0 kVAR.Since capacitors are normally rated in increments of2 5 k V A R , a 2 0 0 kVAR per p ha se c ap ac it or b an k w ill beinstalled a t bu s 2 0 .

When t h e program i s r u n w i t h the 200 kVAR perphase capacitor b a n k a l l voltages a r e a b o v e 0.97 e x c e p tf o r t h e phase a voltage from b u s 1 8 0 down to bus 50.When a 5 0 kVAR capacitor bank i s i n s t a l l e d a t b u s 5 0 onphase a, t h e voltage a t bus 5 0 b ec om es 0. 97 1 p e r - u n i t .Th e s y s t e m w ith t h e capacitor b an ks i ns ta ll ed a tbuses 20 and 50 constitutes what i s now r e f e r r e d t o asthe f ina l b as e c a s e . I t i s interesting to comparethe summary o f losses f o r the initial run and thisf i n a l b a s e c a s e . Table I I g i v e s t h i s breakdown:

Table I IPower Loss ComparisonInitial Final

Phase aPhase bPhase cTotal

14.18 k W8. 87 k W1 0 . 8 5 k w33.90 k W11.01 k W7 . 3 0 k W8.96 kW2 7 . 2 7 k W

Obviously there i s a lesson to be le ar ne d i no bs er vi ng h ow t he lo ss es by phase vary and how t h etotal losses are r ed uce d b y n ea rl y 2 0 % w i t h t h e instal-lation of t h e capacitor b a n k s . A n ic e a ss ig nm en t a tthis time i s t o get t h e students involved w i t h theeconomics a n d determine t h e p ay - ba ck p er io d for t h ecapacitor c o s t s b a s e d u p o n the loss s a v i n g s .Contingency Study

On ce the f i n a l b a s e case i s established i t i s inter-esting to determine how t h e s y s t e m will'perform shouldt h e substation be o u t of service. I n such an i n s t a n c e ,i t would b e n e c e s s a r y to f e e d t h e s y s t e m through a t i epoint t o an adjacent f e e d e r . In t h e t e s t s y s t e m such at i e point exists a t B u s 4 0 0 .R AD F LO W h as t h e option to re-define t h e sourcepoint once t h e base case h a s b e e n established. I n t h i scase th e s y s t e m i s run a ga in o nl y with b u s 4 00 d ef in eda s the s o u r c e point. An analysis of the o u t p u t showsthe lowest voltage to be 0.9615 per-unit a t b u s 1 2 0 ,maximun voltage unbalance i s 0.423 a t b u s 2 1 0 a n dtotal s y s t e m losses to be 33.61 kW This would b e anacceptable operating condition for a c o n t i n g e n c y opera-t i o n . I t i s interesting t h a t t h e m ax i mu m v ol ta g e

u n b a l a n c e i s reduced from the 0.552 value a t b u s 2 1 0in t h e final base c a s e .L oa d G ro wt h

Another feature of the program i s t h e ability t oautomatically increase the s y s t e m l o a d s . At input t i m ee a c h bus i s d ef in ed t o. be i n an area. Th e programallows f o r a maximum of t e n areas. Once the base caseha s b e e n r u n , the user has t h e option to specify a per-c e n t a g e growth r a t e for each area a n d t h e number o fyears such a growth r a t e i s a n t i c i p a t e d . T h e programwill increase the-loads and t h e run t h e s t u d y .

Fo r t h e t e s t s y s t e m three areas were defined a n d ,for s i m p l i c i t y , i t was assumed e a c h area would grow a ta r a t e o f 5 % for t e n years. Output f o r t h e study

1 9 5 1s h o w s t ha t t he minimum voltage i s 0 . 9 4 0 3 a t bus 5 0 ,v o l t a g e u n b a l a n c e i s 0 . 9 4 2 a t b u s 210 and total systeml o s s e s are 7 7 . 9 0 k W . T he s um ma ry shows t h a t no linesare o v e r l o a d e d . T h i s study indicates t h a t t h e systemcan experience s u c h a growth rate and the o n l y s e r i o u sproblem w i l l b e t h e lo w voltages which can b e correctedby t h e additionof s h u n t capacitors o r , p e r h a p s , t h einstallation o f a voltage r e g u l a t o r .Vo l ta g e R eg u la t or s

S t e p voltage r e g u l a t o r s are work horses o n m o s tdistribution f e e d e r s . Regulators a re found in t h e sub-stations in t h e form o f load t a p changing ( L T C ) trans-f o r m e r s or separate three-phase o r s i n g l e - p h a s e u n i t s .S t e p r e g u l a t o r s often a re f o u n d out i n t h e s y s t e m . Thisi s particularly t r u e on l on g r ur al f e e d e r s .A s t e p v o l t a g e regulator i s a dynamic device i nt h a t t h e amount o f b u c k i n g o r b o o s t i n g o f v o l t a g e i sa function o f t h e l o a d . Th e intelligence f o r t h e s t e pr e g u l a t o r i s p ro vi de d b y t h e line d r o p compensator a ndt h e associated c o n t r o l c i r c u i t . I t i s f a r b e y o n d t h escope o f t h i s paper t o develop t h e theory a n d o p e r a t i o no f step voltage r e g u l a t o r s . However, i n t h e Distribu-tion Systems course q u i t e a few lectures ar e p r e s e n t e d

s t a r t i n g with a review o f a u t o - t r d n s f o r m e r s a n d c o n c l u d -i n g with t h e theory o f t h e determination o f R and Xs e t t i n g s f o r t h e l i n e d r o p compensator.R ADF LOW a ll ow s f or t h e m o d e l i n g o f u p t o s e v e ns t e p - r e g u l a t o r s . These r e g u l a t o r s m ay b e in a seriesa l o n g t h e m a in f e e d e r a n d / o r p l a c e d o n l a t e r a l s .Required i n p u t data for each r e g u l a t o r c o n s i s t s o f :1 . Regulator location ( b u s n u m b e r )2 . P o t e n t i a l a nd c ur re nt t r a n s f o r m e r ratios3 . Location o f t h e l o a d c e n t e r ( b u s n u m b e r )4 . Voltage level desired a t t h e l o a d center5 . B a n d w i d t h s e t t i n g6 . Option d e s i r e dOption 1 : L oa d c en te r voltage i s held con-s t a n t . RADFLOW a d j u s t s taps and

c a l c u l a t e s t h e r e q u i r e d R a n d Xsettings f o r t h e c o m p e n s a t o r .Option 2 : R a n d X s e t t i n g s ar e s p e c i f i e d .RADFLOW adjusts t a p s a n d c a l c u l a t e s .t h e r e s u l t i n g load center v o l t a g e .I n t h e previous s t u d y o f t h e test system i t w a sf o u n d , i n t h e l o a d growth s t u d y , lo w voltages occurreds t a r t i n g at b u s 1 3 0 and continuing t o t h e e n d o f t h ev a r i o u s l a t e r a l s . F o r d e m o n s t r a t i o n purposes t h r e es i n g l e - p h a s e step r e g u l a t o r s are installed at b u s 1 3 0 .B u s 7 0 i s d e f i n e d as t h e l o a d c e n t e r . O p t i o n 1 i ss p e c i f i e d a n d t h e v o l t a g e l e v e l set t o 1 2 2 volts( 1 . 0 1 6 7 per u n i t ) w i t h a 2 v o l t b a n d w i d t h .

I D u T P u rrEsr SYSTEM PO R r T E 1 9 8 3 S U E M M E R P R W E R M E E r T I N PAPEa

C o L P E sARo E URrUiMMAsUS _ Y M P E D A N C E II N V O L T S ) VOLTAGE IVOLTS r R E A N S P O N M E R LOADP R O M TO E S i S T A S N C E REACTACE LEVEL EANGWIO7 POrENrIAL CUARENr OPTZON C E . N T R1 3 0 4130 0.000 0.000 1 22. 2 . 2400. 100. 10C O M P E N S A r t O R OPE'RAr Nc rA P&U S IMPEDANCE S I N VOLTS) P O S I T I O NvA_c RArJxG-- -jrA L oJADimGC-PROM tO A E C A 0 C R E S I S T A N C E REACTANCE A 0 C

13 0 4130 2 5 0 . 2 5 0 . 2 5 0 . 0 0 1 . 4 4 8 . 4 8 9 . 1.053 0.561 1 0 S 1 0

F i g . 1 3 . Vo l t ag e R eg u la t or Summary ReportTh e V o l t a g e R e g u l a t o r S u m m a r y report s h o w n i nFigure 1 3 g i v e s a summary o f t h e i n p u t data and t h e


8/8

1 9 5 2operating conditions o f t h e r e g u l a t o r s consisting o ft h e kVA l o a d f o r each p h a s e and the t a p position ofe a c h phase r e g u l a t o r . Included a re the R and X s e t t i n g si n volts f o r t h e l i n e drop compensator. The voltageprofile summary f o r this study shows t h e regulatoroutput voltage on phase a i s 1 . 0 3 1 5 per-unit and t h evoltage at b u s 7 0 o n p h a s e a i s 1 .0155 p e r -unit o r1 2 1 . 8 7 volts o n t h e regulator b a s e .Extreme U n ba l an c ed L o ad i ng

A distribution system with extreme unbalanced load-ing will display some u n u s u a l operating characteristics.T o d e m o n s t r a t e t w o seemingly strange p h e n o m e n a , a 5 0 0 0f t . 1 / 0 t h r e e - p h a s e line i s c o nn e ct e d b e tw e en b u s 100a n d a n e w b u s 1 1 1 i n t h e test s y s t e m . A t b u s 1 1 1 aseverely unbalanced load o f 5 0 0 kVA on phase a , a n d 5 0k V A on phases b and c i s connected. P o w e r factor i sassumed t o be 0 . 9 l a gg i n g . Table I I I summarizes t h et w o e n d b u s voltages.Table I I IB u s Voltages for Unbalanced Lo a di n g

Not o n l y i s the p o w e r loss b y p h a s e wrong, but the t o t a lthree-phase p o w e r loss i s wrong. This i s anotherexample o f ho w t h e students ca n be reviewed t o t h e f a c tt h a t I 2 R only works i n ve r y special cases.A d m i t t e d l y , the severe unbalance placed o n t h e t e s tsystem i s not ve r y r e a l i s t i c . However, many times int h e past t h e voltage rise and the negative p o w e r lossha ve bee n observed when studying actual d i s t r i b u t i o n

s y s t e m s .CONCLUSIONS

RADFLOW has p r o v e n t o be an invaluable tool inteaching the Distribution Systems c o u r s e . In using theprogram the students are f orced t o r e vi e w m a t e r i a l f r o mp r e vi o us courses in o r d e r t o develop the required d atafor the program. This i s one of t h e goals of thecourse as m e n ti o ne d p r ev i ou s ly . Normal a nd a b n o r m a lo p e r a t i n g conditions can be demonstrated w i t h o u t requir-i n g t h e students t o spend hours doing hand c a l c u l a t i o n sor requiring t h e instructor t o de r i ve underivableequations.

Phase-a Phase-bB u s 1 0 0 0 . 9 4 0 4 / - 2 . 5 0 . 9 9 8 1 / - 1 2 0 . 9B u s 1 1 1 0 . 7 7 9 7 / - 5 . 6 1 . 0 1 2 7 / - 1 2 1 . 2

1 7 . 7 % drop 1 . 5 % rise

Phase-c0 . 9 8 1 9 / 1 1 9 . 60 . 9 6 3 8 / 1 2 0 . 61 . 8 % d r o p

As i s e x p e c t e d , phase a ha s experienced a severe v o l t a g ed r o p . The unexpected i s t h e voltage rise o n phase b .Phase c experiences a normal voltage d r o p . The voltagerise phenomena i s interesting a n d i s c o r r e c t . Th ereason f o r t h e rise i s t h e manner i n which t h e mutualcoupling between phases occurs. A ve r y good exercisef o r t h e students i s t o require them t o sketch t h e phasord i a g r a m t o h e l p explain ho w t h e voltage rise o c c u r s .T h e s ec on d s tr an ge condition displayed by t h esevere unbalanced loading i s t h e power flow over t h eline from b u s 1 0 0 t o bus 1 1 1 . A summary o f t h e real

power flows i s given i n Table I V .T a b l e I V

L e a v i n g B u s 1 0 0 fArriving a t B u s 1 1 1 dP ow er L os sTotal Loss = 7 8 . 2 3 1 k IAs m i g h t b e expected,l o s s ( 1 4 . 7 % ) , phase cb u t p h a s e b h a s a n e g ab r i g h t young engineerpower l o s s has t o b e iaccurate m o d e l i n g o f tan unusual actual c o n cf e r r e d from.phase a t cs o l i d connection i n m ttransferred across a t

One l a s t observatistudents f i r m l y b e l i e sd e t e r m i n e d b y t a k i n g Idone f o r t h e unbalancea r e d e t e r m i n e d :

Various versions of t h e p r o g r a m ar e p r e s e n t l y b e i n gused b y a number o f utility companies and a n e n g i n e e r i n gconsulting f i r m . I t has b e e n gratifying t o know t h a tm an y times t h e p r o g r a m h a s p r e d i c t e d strange o p e r a t i n gconditions t h a t h a v e been confirmed b y actual f i e l dt e s t s . [ 7 ] A g r e a t l y modified version o f RADFLOW hasrecently been developed under a contract t o a computercompany t h a t provides c o m p u t e r services t o electriccooperatives i n Ne w M e x i c o , Arizona a n d C o l o r a d o . [ 6 ]T h a t program allows a cooperative t o run studies basedup o n actual billing d a t a . Again r e s u l t s o f s o m e o fthese s t u d i e s have been verified b y field tests.

As time goes o n , RADFLOW will continue t o c h a n g e i nan effort t o m o r e closely model a ct ua l e q ui p me n t ando p e r a t i n g conditions of a distribution f e e d e r . WithoutR A D F L O W , t h e teaching o f t h e Distribution Systemscourse would b e boring t o t h e students and i m p o s s i b l ef o r t h e instructor.REFERENCES

U l r o w e r F l o w s [ 1 ] W . H . Kersting and S . A . S e e k e r , A P r o g r a m t oPhase-a Phase-b Phase-c S t u d y t h e E f f e c t s o f Mutual C o u p l i n g and Unbal-anced Loading o n a Distribution S y s t e m , IEEE5 2 7 . 7 6 0 kW 4 4 . 2 3 6 k W 4 6 . 2 3 4 k W Conference P a p e r C.75 0 4 7 - 6 . Presented a t the5 5 0 . 0 0 0 kW 4 5 . 0 0 0 k W 4 5 . 0 0 0 kW Wi nt er P ow er M ee ti ng , Ne w York, J a n ua r y 1 9 7 5 .1 2 ] W . H . Kersting and D . L . M e n d i v e , A n Application7 7 . 7 6 0 k W - 0 . 7 6 4 k W 1 . 2 3 4 kW of L ad de r N et w or k Theory t o t h e Solution of Three-Phase Radial Lo ad- F lo w P r o b l e m s , IEEE ConferenceP a p e r . Presented at t h e Winter P o w e r M e e t i n g ,' N e w Y o r k , J a n u a r y 1 9 7 6 .p h a s e a h a s a r a t h e r l a r gower l o s s ( [ 3 ] W . D . Stev enson. Elements o f P o w e r S y s t e mhasanormalpower l os s s ( - 1 . 7 ) (2.7Analysis, 4 t h e d . Ne w York: M c G r a w - H i l l , 1 9 8 2 .O r t i v e n p o e l n o s s ( - 1 . 7 % ) . g a n y [ 4 ] J . R . C a r s o n , W a v e P r o p a g a t i o n i n O v e r h e a d Wireso n e r r o r

tu t

k n o w s h e r o n c e a g a t i v h e and Ground R e t u r n , Bell S y s t e m T e c h n i c a l J o u r n a l ,L n e r r o r . H o w e v e r , o n c e a g a i n theVo 5196t h e mutual c o u p l i n g has s u r f a c e d Io.5 96[ 5 ] R . E . M c C o t t e r , T h e Accurate Determination o fl i t i o n . P owe r has b ee n t ra ns - Distribution L i n e I m p e d a n c e s , M a s t e r o f Sciencep h a s e b without t h e benefit o f a T h e s i s , Ne w Mexico State U n i v e r s i t y , 1 9 8 2 .i c h t h e s a m e m a n n e r that power i s [ 6 ] T . A . C a r l s o n , D e v e l o p m e n t o f a R a d i a l D i s t r i b u -t r a n s f o r m e r . tion P o w e r F lo w P r o g r a m , M a s t e r o f ScienceL o n o n t h e p o w e r l o s s . Most Technical R e p o r t , N ew M ex ic o State U n i v e r s i t y ,t e t h a t p o w e r l o s s c a n a l w a y s b e 1 9 8 2 .[ 7 ] R . W . F i s c h e r , V o l t a g e Unbalance on Three-PhaseE2Rfora c h p h a s e . W h e n t h i s i s Distribution S y s t e m s , Southeastern Electrica d condition t h e following losses E- ~ ~ ~ ~ ~ ~ ~ E x c h a n g e , 1 9 7 7 Annual Conference.

Phase a : 7 5 . 6 1 2 kWPhase 0 . 4 4 8 kWP h a s e c : 0 . 4 9 5 kWT o t a l 7 6 . 5 5 5 kW

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A Method to Teach the Design and Operation Qf a Distribution System

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