Proiect SEE

8/18/2019 Proiect SEE

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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2

PRO IECT

Disciplina: SISTEME ELECTROENERGETICE

Ingineria sistemelor electroenergetice

ANUL IV

TEMA DE PROIECT :

STUDIUL REGIMULUI PERMANENT DE UN!"I#NARE AL SEE

Student

1


2/34


1. Stabilirea schemei monoflare şi a datelor iniiale de reea asistem!l!i electroenergetic

Schema mn!la"# de cne$iuni a sistemului TEST1

%


3/34


". Re#re$entarea elementelor sistem!l!i

Pa"amet"ii liniil" elect"ice:

LINI%

r0

&'()m*

x0

&'()m*

b0

&+S()m*

L &)m* I max

&%*1&% '(')) '(*') %(++ 1)(% ,+-1&* '(')) '(*') %(++ %+(, ,+-%&. '(')) '(*') %(+/ 1*(* ,+'.&- '(')) '(*') %(+/ %-(% ,+-*&- '(')) '(*') %(+% 11(+ ,+'

Pa"amet"ii cncent"a0i ai liniil" elect"ice :

R=r0

∙ L 23

X = x0

∙ L 23

B=b0

∙ L 4S3

LINI%

R &'* , &'* - &+S*

1&% 1('),% )(-++% **(/+*'1&* 1(/*1* 11(.%+* ++(%/.'%&. '(,-'* -(/*)* *'('.%'

.&- 1()).% 1'(%.1% +'('-)'*&- '(++%% *(+-'% .1(/%*'

".1 Calc!l!l #arametrilor electrici şi determinarea schemelor echialenteale liniilor electrice/

Impedan5a ln6itudinala a liniei elect"ice:

ik ik ik jX R z +=

Ω3

Admitan5a ln6itudinal# :

2222

1

ik ik

ik

ik ik

ik

ik

ik X R

X j

X R

R

z y

+−

+==

S3

.


4/34


7n ca"e : Ri$ este "e8isten5a liniei elect"ice9 7n Ω Lr R oik =

% i$ ; "eactan5a inducti


5/34

Ingineria Sistemelor Electroenergetice anul: 4 grupa:2¿0.0140− j 0.0860[S ]

Y 140= j

B140

2= j

77.2830

2= j38.6415[ μS]

.> linia 2-3:

Z 23= R

23+ j X

23=0.9504+ j5.8464 [ Ω ]

Y 23=

R23

R232 + X 232 − j

X 23

R232 + X 232 =

0.9504

0.95042+5.84642− j

5.8464

0.95042+5.84642=0.0271− j0.1666 [S]

Y 230= j

B230

2= j

40.0320

2= j 20.0160[ μS ]

-


6/34


*> linia 3-5:

Z 35= R

35+ j X

35=1.6632+ j10.2312 [ Ω ]

Y 35=

R35

R35

2 + X 352 − j

X 35

R35

2 + X 352 =

1.6632

1.66322+10.23122− j

10.2312

1.66322+10.23122=¿

¿0.0155− j0.0952 [S]

Y 350= j

B350

2= j

70.0560

2= j35.0280[ μS]

-> linia 4-5:

Z 45= R

45+ j X

45=0.7722+ j 4.7502 [ Ω ]

Y 45=

R45

R45

2 + X 45

2 − j

X 45

R45

2 + X 45

2 =

0.7722

0.77222+4.75022

− j 4.7502

0.77222+4.75022

=¿

¿0.0333− j0.2051 [S]

Y 450= j

B450

2= j

31.8240

2= j15.9120 [ μS ]

)


7/34


LINIA

PARAMETRII

L1&%Z

12=1.0692+ j6.5772 [ Ω ]

Y 12=0.0241− j0.1481 [S]

Y 120= j22.4370[ μS ]

L1&*Z

14=1.8414+ j11.3274 [ Ω ]

Y 14=0.0140− j 0.0860[S ]

Y 140= j38.6415[ μS ]

L%&.Z

23=0.9504+ j5.8464 [ Ω ]

Y 23=0.0271− j0.1666[S ]

Y 230= j 20.0160[ μS ]

L.&-Z

35=1.6632+ j10.2312 [ Ω ]

Y 35=0.0155− j0.0952[S ]

Y 350= j35.0280[ μS ]

L*&-

Z 45=0.7722+ j4.7502 [ Ω ]

Y 45=0.0333− j0.2051 [S ]

Y 450= j15.9120[ μS]

+


8/34


".".Modelarea generatoarelor sincrone şi a cons!matorilor com#lecşi

".".1Generatoarele sincrone

Datele de catal6 ale 6ene"ata"el" sinc"ne9 din sistem sunt:

Pn g

&pute"ea nminal? M@3

S n g&pute"ea apa"enta nminala MVA3

U n&tensiunea nminala V3

csϕ n

&Bact"ul de pute"e nminal

X X Xd d'

d

'', ,

&"eactantele sinc"n?9 t"an8it"ie9 sup"at"an8it"ie 3

In "e6im pe"manent n"mal de Bunctina"e9 6ene"ata"ele sinc"ne ;ca"e sunt elemente acti


9/34


In ca8ul GS9 in


10/34


Cnductanta c"espunde pute"ii acti


11/34


Funcţionaea in condiţii de e!icienţ" #i andamen$ a con%uma$oi&o e%$e condiţiona$" dea%iguaea &a one&e ace%$oa a unui ni(e& de $en%iune coe%)un*"$o+a one&e con%uma$oi&o$euie ca

U min


12/34


5.6ormarea matricii admitanelor nodale ce interine 7n model!lmatematic al regim!l!i #ermanent #rin metoda tensi!nilor nodale

1%


13/34


Reg!li de scriere directa a matricii admitantelor nodale /

O"ice te"menY

ii

de pe dia6nala p"incipala este e6al cu sumaadmitantel" latu"il" incidente la ndul i∈n(

Y ii

=

Y Yi 4 0

481

n

i 4 481

n

+∑ ∑

.(%>

O"ice te"men Y

i 4

nedia6nal este e6al cu admitanta latu"ii de le6atu"adint"e ndul i si ndul k9 cnside"ata cu semnul minus(

Y ki = Y ik =

&Y i 4

.(.>

Daca int"e ndu"ile i si k nu e$ista latu"a de le6atu"a9 te"menulY

i 4

= '(

%lgoritm!l #entr! 8ormarea matricii admitantelor nodale se Fa8ea8ape "e6ulile de sc"ie"e mai sus mentinate pent"u calculul di"ect al te"menil"

dia6nali si nedia6nali si cup"inde u"mat"ii pasi:

1(Ini5iali8a"ea cu 8e" a (

1.


14/34


*(* Calculul pa"amet"il" cnsumat"il" cmplecsi daca acestia semdelea8a p"in admitanta cnstanta>(

-(Calculul te"menil" dia6naliY

ii

a mat"icii admitan5el" ln6itudinale(

)(Calculul te"menil" nedia6nali Blsind "ela5iileY

ik =

&Y

i 4

siY

ki =

&Y

4 i

(

+(Dac# nu s&au citit datele "eBe"ita"e la tate latu"ile se staFileJtel l = + 1

Ji se "epet# pe"a5iile 7ncepKnd cu pasul .(

/(Tip#"i"ea mat"icei admitan5el" ndale [Y N N ]

[Y NN ]=[Y

11Y

12Y

13Y

14Y

15

Y 21

Y 22

Y 23

Y 24

Y 25

Y 31

Y 32

Y 33

Y 34

Y 35

Y 41

Y 42

Y 43

Y 44

Y 45

Y 51

Y 52

Y 53

Y 54

Y 55

]Calc!l!l termenilor matricei admitanelor nodale/

!alculul termenilor 'iagonali:

Y

11= (Y 12+Y 14 )+( Y 120+Y 140 )=0.0381− j 0.2340[S ]

Y 22=(Y 12+Y 23 )+ (Y 120+Y 230 )=0.0512− j0.3147 [S]

Y 33=(Y 35+Y 23 )+(Y 350+Y 230)=0.0426− j 0.2617[S ]

Y 44=( Y 14+Y 45 )+(Y 140+Y 450)=0.0473− j0.2910 [S ]

Y 55=(Y 35+Y 45 )+(Y 350+Y 450 )=0.0488− j 0.3002[S ]

!alculul termenilor ne'iagonali:

1*


15/34


Y 12=−Y

12=−0.0241+ j0.1481 [S]

Y 13=0

Y 14=−Y 14=−0.0140+ j0.0860 [S]

Y 15=0

Y 21=−Y

12=−0.0241+ j0.1481 [S]

Y 23=−Y

23=−0.0271+ j 0.1666[S ]

Y 24=0

Y 25=0

Y 31=0

Y 32=−Y

23=−0.0271+ j 0.1666[S ]

Y 34=0

Y 35=−Y

35=−0.0155+ j 0.0952[S ]

Y 41=−Y

14=−0.0140+ j0.0860 [S]

Y 42=0

Y 43=0

Y 45=−Y 45=−0.0333+ j0.2051[S]

Y 51=0

Y 52=0

1-


16/34


Y 53=−Y

35=−0.0155+ j 0.0952[S ]

Y 54=−Y

45=−0.0333+ j0.2051[S]

9.6orm!larea model!l!i matematic al regim!l!i #ermanent

#rin metoda tensi!nilor nodale

Ecua5iile Filan5ului 7n ndu"i se pt sc"ie :

93:9,:PP ii θ U =

Pi = P Pgi ci−

i ( 19%9n

*(1'>

93:9,:QQ ii θ U =

Q i = Q Qgi ci−

i ( 19%9n

Ecua5iile de Filan5 *(1'> pent"u pute"ile acti Ji de da" du# ecua5ii al6eF"iceecua5iile Filan5ului de pute"i> ca"e e$p"im# le6#tu"a dint"e acestea(

Pent"u a F5ine mdelul matematic al "e6imului pe"manent este necesa"#cmpleta"ea sistemului de ecua5ii .()> cu 7nc# %n ecua5ii( Acestea se F5inspeci!cKnd pent"u !eca"e nd du# dint"e m#"imile elect"ice de sta"e(

n Bunc5ie de m#"imile speci!cate sau impuse9 nod!rile sistem!l!ielectroenergetic s!nt 7m#:rite 7n !rm:toarele categorii/

1 Nod!ri de ti#!l ;< sa! nod!ri cons!matoare

1)


17/34


Sunt ndu"ile pent"u ca"e se speci!c# pute"ile ndale Pi Ji )i Ji se dete"min# p"incalculul mdului Ui Ji a"6umentul θ i ale Ba8"ului tensiunii ndale( Dea"ecemdulul tensiunii "e8ult# p"in calcul 7n Bunc5ie de

Pg i

9& suma pute"il" acti


18/34


Ve"i!ca"ea Filan5ului de pute"e acti mdulel" tensiunil" 7n ndu"ile de tip

6ene"at" i∈ng

9 se elimin# din ecuatiile *(1'> ecuatiile de Filan5 ale pute"il""eacti


19/34


TaFelul -(1(

Ti#!l nod!l!iM:rimi

c!nosc!te

M:riminec!nosc!t

e

N!m:r

Nod generator;= P9U

Qmin9Qmax θ9 H

ng

Nod cons!mator;<

P9H

P='9H=' U9θ nC

Nod de echilibr!=

U9 θ P9H 1

Calc!l!l regim!l!i #ermanent a !n!i sistem electroenergeticconst: 7n determinarea set!l!i de m:rimi nec!nosc!te la niel!lnod!rilor sistem!l!i.

Sta*ilirea tipului +i a 'atelor ini,iale ale no'urilor sistemului electroenergeticTEST-

./-/ Sta*ilirea tipurilor 'e no'uri pe *a0a 'atelor ini,iale cunoscute(

TaFel -(%

N!m:r nod

M:rimic!nosc!teim#!se3

M:riminec!nosc!te

Ti#!l nod!l!i

1 U9 θ P9HNd de

echiliF"u

%P9U

Qmin9Qmax θ9 H Nd 6ene"at"

.P9U

Qmin9Qmax θ9 H Nd 6ene"at"

* P9H U9 θNd

cnsumat"

- P9H U9 θNd

cnsumat"

./2/ Determinarea m1rimilor electrice impuse n no'uri 3 sta*ilirea 'atelor no'ale

1,


20/34


Se staFilesc


21/34


¿U 1

∙ [(Y 11¿ ∙U 1¿ (0 ))+ (Y 12¿ ∙ U 2¿ (0) )+(Y 13¿ ∙ U 3¿ (0) )+(Y 14¿ ∙ U 4¿ (0) )+(Y 15¿ ∙ U 5¿ (0) )]=¿

S1=18,62+ j 109,5[ MVA ]

U 4=

1

Y 44 [

P4− jQ4U

4

¿ (0) −(Y 41 ∙U 1

(0))+(Y 42 ∙ U 2(0))+(Y 43 ∙ U 3

(0 ))+(Y 45 ∙U 5(0)) ]=¿

U 4 ="14BD254A> &)F*

U 5=

1

Y 55 [ P5− jQ5U

5

¿ (0 ) −(Y 51 ∙U 1

(0))+(Y 52 ∙U 2(0))+ (Y 53 ∙ U 3

(0))+(Y 54 ∙ U 4(1)) ]=¿

U 5 @"1D45"2"4>B &)F*

Q2=ℑ {U 2(0)∙ (Y 21¿ ∙U 1¿(0)+Y 22¿ ∙U 2¿(0)+Y 23¿ ∙U 3¿(0)+Y 24¿ ∙ U 4¿(1 )+Y 25¿ ∙ U 5¿(1)) }

Q2=− j 78,68 [ MVAr ]

Q3=ℑ {U 3(0)∙ (Y 31¿ ∙U 1¿(0)+Y 32¿ ∙U 2¿(0)+Y 33¿ ∙U 3¿(0)+Y 34¿ ∙ U 4¿(1 )+Y 35¿ ∙U 5¿(1)) }

Q3= j20,2 [ MVAr ]

U 2=

1

Y 22

∙( P2− j Q2U 2

¿(0) −Y 21 ∙U 1(0)−Y

23∙ U

3

(0 )−Y 24

∙ U 4

(1)−Y 25

∙ U 5

(1))U

2=220,23+ j1,40 [ kV ]

U 3=

1

Y 33

∙( P3− j Q3U 3

¿(0) −Y 31 ∙ U 1(0 )−Y

32∙ U

2

(1)−Y 34

∙ U 4

(1 )−Y 35

∙ U 5

(1 ))

%1


22/34


U 3=220,33+ j2 [ kV ]

Iteraţia-2:

S1

(1 )=U i ∙∑k =1

n

Y ik ¿

∙ U k ¿=¿

¿U 1

∙ [(Y 11¿ ∙U 1¿ (1) )+(Y 12¿ ∙ U 2¿ (1) )+(Y 13¿ ∙U 3¿ (1 ))+(Y 14¿ ∙U 4¿ (1 ))+ (Y 15¿ ∙ U 5¿ (1) )]=¿

S1=55,54+ j141,23 [ MVA ]

U 4= 1

Y 44 [

P4− jQ

4

U 4

¿ (1) −(Y 41 ∙U 1(1))+(Y 42 ∙ U 2(

1))+(Y 43 ∙U 3(

1))+(Y 45 ∙ U 5(

1))]=¿

U 4 ="1B4952949A &)F*

U 5=

1

Y 55 [ P5− jQ5U

5

¿ (1 ) −(Y 51 ∙U 1

(1))+(Y 52 ∙U 2(1))+(Y 53 ∙ U 3

(1))+ (Y 54 ∙U 4(2))]=¿

U 5

@"1B4A2545 &)F*

Q2=ℑ {U 2(1 )∙ (Y 21¿ ∙ U 1¿(1)+Y 22¿ ∙ U 2¿(1)+Y 23¿ ∙ U 3¿(1)+Y 24¿ ∙ U 4¿(2)+Y 25¿ ∙ U 5¿(2))}

Q2=− j 71,69 [ MVAr ]

Q3=ℑ {U 3(1)∙ (Y 31¿ ∙ U 1¿(

1)+Y 32¿ ∙ U 2¿(1)+Y 33¿ ∙ U 3¿(1)+Y 34¿ ∙ U 4¿(2)+Y 35¿ ∙ U 5¿(2))}

Q3= j46,62 [ MVAr ]

%%


23/34


U 2=

1

Y 22

∙( P2− j Q2U 2

¿(1) −Y 21∙U 1(1)−Y

23∙U

3

(1)−Y 24

∙ U 4

(2)−Y 25

∙ U 5

(2))U

2=220,40+ j2,45 [ kV ]

U 3=

1

Y 33

∙( P3− j Q3U 3

¿(1) −Y 31 ∙ U 1(1)−Y

32∙ U

2

(2)−Y 34

∙ U 4

(2)−Y 35

∙ U 5

(2))U

3=220,31+ j1,84 [ kV ]

Iteraţia-3:

S1

(2 )=U i ∙∑k =1

n

Y ik ¿

∙ U k ¿=¿

¿U 1

∙ [(Y 11¿ ∙U 1¿ (2) )+(Y 12¿ ∙ U 2¿ ( 2) )+(Y 13¿ ∙U 3¿ (2 ))+(Y 14¿ ∙U 4¿ (2 ))+(Y 15¿ ∙ U 5¿ (2) ) ]=¿

S1=51,72+ j 158,47[ MVA ]

U 4= 1

Y 44 [

P4− jQ

4

U 4

¿ (2) −(Y 41 ∙U 1(2)

)+(Y 42 ∙ U 2(2 )

)+(Y 43 ∙U 3(2)

)+(Y 45 ∙ U 5(2)

)]=¿

U 4 ="19452B4"9 &)F*

U 5=

1

Y 55 [ P5− jQ5U

5

¿ (2 ) −(Y 51 ∙U 1

(2))+(Y 52 ∙U 2(2))+(Y 53 ∙ U 3

(2))+(Y 54 ∙U 4(3))]=¿

U 5

@"1B4AD2949> &)F*

Q2=ℑ {U 2(2)∙ (Y 21¿ ∙ U 1¿(2)+Y 22¿ ∙ U 2¿(2)+Y 23¿ ∙ U 3¿(2)+Y 24¿ ∙ U 4¿(3)+Y 25¿ ∙ U 5¿(3)) }

Q2=− j71,06 [ MVAr ]

%.


24/34


Q3=ℑ {U 3(2)∙ (Y 31¿ ∙U 1¿(2)+Y 32¿ ∙ U 2¿(2)+Y 33¿ ∙ U 3¿(2)+Y 34¿ ∙ U 4¿(3)+Y 35¿ ∙ U 5¿(3))}

Q3= j60,25 [ MVAr ]

U 2=

1

Y 22

∙( P2− j Q2U 2

¿(2) −Y 21 ∙U 1(2)−Y

23∙ U

3

(2)−Y 24

∙ U 4

(3 )−Y 25

∙ U 5

(3 ))U

2=220,39+ j2,36 [ kV ]

U 3=

1

Y 33

∙

(

P3− j Q3U

3

¿(2) −Y 31 ∙ U 1(2)−Y

32∙ U

2

(3)−Y 34

∙ U 4

(3 )−Y 35

∙ U 5

(3 )

)U 3=220,24+ j1,36 [ kV ]

Iteraţia-4:

S1

(3 )=U i ∙∑k =1

n

Y ik ¿

∙ U k ¿=¿

¿U 1

∙ [(Y 11¿ ∙U 1¿ (3 ))+ (Y 12¿ ∙ U 2¿ (3) )+(Y 13¿ ∙ U 3¿ (3 ) )+(Y 14¿ ∙ U 4¿ (3 ) )+(Y 15¿ ∙U 5¿ (3 )) ]=¿

S1=70,93+ j167,44 [ MVA ]

U 4=

1

Y 44 [ P4− jQ4U

4

¿ (3) −(Y 41 ∙U 1

(3))+(Y 42 ∙ U 2(3 ))+(Y 43 ∙ U 3

(3))+ (Y 45 ∙U 5(3))]=¿

U 4 ="194B12B41 &)F*

U 5=

1

Y 55 [ P5− jQ5U

5

¿ (3 ) −(Y 51 ∙U 1

(3))+(Y 52 ∙U 2(3))+(Y 53 ∙ U 3

(3 ))+(Y 54 ∙U 4(4 ))]=¿

%*


25/34


U 5 @"194B>2B41 &)F*

Q2=ℑ {U 2(3 )∙ (Y 21¿ ∙U 1¿(3)+Y 22¿ ∙ U 2¿(3)+Y 23¿ ∙ U 3¿(3)+Y 24¿ ∙U 4¿(4 )+Y 25¿ ∙ U 5¿(4)) }

Q2=− j70,95 [ MVAr ]

Q3=ℑ {U 3(3)∙ (Y 31¿ ∙U 1¿(3)+Y 32¿ ∙ U 2¿(3)+Y 33¿ ∙ U 3¿(3)+Y 34¿ ∙U 4¿(4 )+Y 35¿ ∙ U 5¿(4)) }

Q3= j67,34 [ MVAr ]

U 2=

1

Y 22

∙( P2− j Q2U 2

¿(3) −Y 21 ∙U 1(3)−Y

23∙ U

3

(3)−Y 24

∙ U 4

(4)−Y 25

∙ U 5

(4 ))U

2=220,35+ j2,11 [kV ]

U 3=

1

Y 33

∙( P3− j Q3U 3

¿(3) −Y 31 ∙ U 1(3 )−Y

32∙ U

2

(4 )−Y 34

∙U 4

(4 )−Y 35

∙ U 5

(4))U

3=220,17+ j 0,94 [ kV ]

Iteraţia-5:

S1

(4 )=U i ∙∑k =1

n

Y ik ¿

∙ U k ¿=¿

¿U 1

∙ [(Y 11¿ ∙U 1¿ (4 ) )+(Y 12¿ ∙U 2¿ (4 ) )+(Y 13¿ ∙ U 3¿ ( 4) )+(Y 14¿ ∙ U 4¿ (4 ) )+(Y 15¿ ∙U 5¿ (4 ) )]=¿

S1=89,19+ j172,30 [ MVA ]

U 4=

1

Y 44 [

P4− jQ4U

4

¿ (4) −(Y 41 ∙U 1

(4 ))+(Y 42 ∙U 2(4 ))+(Y 43 ∙ U 3

(4 ))+(Y 45 ∙ U 5(4))]=¿

U 4 ="1945"24A> &)F*

%-


26/34


U 5=

1

Y 55 [ P5− jQ5U

5

¿ (4 ) −(Y 51 ∙U 1

(4))+(Y 52 ∙ U 2(4))+(Y 53 ∙ U 3

(4))+(Y 54 ∙ U 4(5 ))]=¿

U 5 @"194>2B4B5 &)F*

Q2=ℑ {U 2(4) ∙ (Y 21¿ ∙ U 1¿(4)+Y 22¿ ∙ U 2¿(4 )+Y 23¿ ∙ U 3¿(4)+Y 24¿ ∙U 4¿(5)+Y 25¿ ∙U 5¿(5)) }

Q2=− j 70,97 [ MVAr ]

Q3=ℑ {U 3(4) ∙ (Y 31¿ ∙ U 1¿(4)+Y 32¿ ∙ U 2¿(4 )+Y 33¿ ∙U 3¿(4)+Y 34¿ ∙U 4¿(5)+Y 35¿ ∙U 5¿(5)) }

Q3= j70,83 [ MVAr ]

U 2=

1

Y 22

∙( P2− j Q2U 2

¿(4) −Y 21 ∙U 1(4 )−Y

23∙ U

3

(4)−Y 24

∙ U 4

(5 )−Y 25

∙ U 5

(5))U

2=220,32+ j1,89 [ kV ]

U 3= 1

Y 33

∙( P3− j Q3U 3

¿(4 ) −Y 31 ∙ U 1(4)−Y

32∙U

2

(5)−Y 34

∙U 4

(5)−Y 35

∙ U 5

(5))U

3=220,12+ j0,62 [ kV ]

Test convergen ţă-1

|S1(1 )−S

1

(0 )|=√ P12+Q

1

2→ ε=46,68


|S1(2 )−S

1

(1)|=√ P12+Q

1

2→ ε=16,81


|S1(3 )−S

1

(2 )|=√ P12+Q

1

2→ ε=21,2

%)


27/34



|S1(4)−S

1

(3)|=√ P12+Q

1

2→ ε=18,89

. Heterminarea circ!la iilor de #!tere acti: i reacti: #rinelementele sistem!l!i

Rela0ii de calcul:

S ij=U i ∙ (U i−U j )¿

∙ ij¿ [ MVA ]

S ji=U j ∙ (U j−U i )¿

∙ ji¿ [ MVA ]

S ij= Pij+ j Qij [ MVA ]

Pij=ℜ { Sij } [ M! ]

Qij=ℑ {Sij } [ MVAr ]

Latura : 23-

S21=U 2 ∙ (U 2−U 1 )¿

∙ 21

¿ =36,04− j71,18 [ MVA ]

P21=36,04 [ M! ]

Q21=−71,18 [ MVAr ]

Latura : 23

%+


28/34


S23=U 2 ∙ (U 2−U 3 )¿

∙ 23

¿ =52,37+0,47 [ MVA ]

P23=52,37 [ M! ]

Q23=0,47 [ MVAr ]

Latura : 3.

S35=U 3 ∙ (U 3−U 5 )¿

∙ 35

¿ =175,21+ j 79,2 [ MVA ]

P35=175,21 [ M! ]

Q35=79,20 [ MVAr ]

Latura : .34

S54=U 5 ∙ (U 5−U 4 )¿

∙ 54

¿ =4,04− j7,18 [ MVA ]

P54=4,04 [ M! ]

Q54=7,18 [ MVAr ]

Latura : -34

S14=U 1 ∙ ( U 1−U 4 )¿

∙ 14

¿ =168,15+ j 111,11 [ MVA ]

P14=168,15 [ M! ]

Q14=111,11 [ MVAr ]

%/


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D. Calc!l!l #ierderilor de #!tere

Rela0ii de calcul:

" Sij=|S ij+S ji|[ MVA ]

" Pij=ℜ{ " S ij} [ M! ]

" Qij=ℑ {" S ij } [ MVAr ]

Linia : -32

" S12=|S12+S21|=0,1405+ j0,86338 [ MVA ]

" P12=ℜ { " S12}=0,140 [ M! ]

" Q12=ℑ {" S12 }= j0,863 [ MVAr ]

Linia : 23

" S23=|S23+S32|=0.053794+ j0.3307 [ MVA ]

" P23=ℜ {" S23 }=0,0537 [ M! ]

" Q23=ℑ {" S23 }= j 0,3307 [ MVAr ]

Linia : 3.

" S35=|S35+S53|=1.2726+ j7.8162 [ MVA ]

" P35=ℜ {" S35 }=1,272 [ M! ]

" Q35=ℑ {" S35 }= j7,816 [ MVAr ]

Linia : 43.

%,


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" S45=|S45+S54|=0.00113+ j0.00697 [ MVA ]

" P45=ℜ {" S45 }=0,00113 [ M! ]

" Q45=ℑ {" S45}= j 0,00697 [ MVAr ]

Linia : -34

" S14=|S14+S41|=1.5171+ j9.3195 [ MVA ]

" P14=ℜ {" S14 }=1,517 [ M! ]

" Q14=ℑ {" S14 }= j 9,319 [ MVAr ]

. Calc!l!l regim!l!i #ermanent al Reelei TEST1 !tili$Jnd

so8tKare!l NE;L%N

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TaFel "e8ultate suma"e

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TaFel "e8ultate 7n ndu"i

TaFel "e8ultate pe elemente

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1A. Inter#retarea re$!ltatelor i concl!$ii

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