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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
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
1. Stabilirea schemei monoflare şi a datelor iniiale de reea asistem!l!i electroenergetic
Schema mn!la"# de cne$iuni a sistemului TEST1
%
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
". Re#re$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
.
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
7n ca"e : Ri$ este "e8isten5a liniei elect"ice9 7n Ω Lr R oik =
% i$ ; "eactan5a inducti
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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 ]
-
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
*> 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 ]
)
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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]
+
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".".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
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
In ca8ul GS9 in
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Cnductanta c"espunde pute"ii acti
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
5.6ormarea matricii admitanelor nodale ce interine 7n model!lmatematic al regim!l!i #ermanent #rin metoda tensi!nilor nodale
1%
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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.
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
*(* 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*
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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-
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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)
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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,
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Se staFilesc
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¿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
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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 ]
%%
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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 ]
%.
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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 ))]=¿
%*
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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*
%-
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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
Test convergen ţă-2
|S1(2 )−S
1
(1)|=√ P12+Q
1
2→ ε=16,81
Test convergen ţă-3
|S1(3 )−S
1
(2 )|=√ P12+Q
1
2→ ε=21,2
%)
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
Test convergen ţă-4
|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
%+
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
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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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
" 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$Jnd
so8tKare!l NE;L%N
.'
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
TaFel "e8ultate suma"e
.1
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
TaFel "e8ultate 7n ndu"i
TaFel "e8ultate pe elemente
.%
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
1A. Inter#retarea re$!ltatelor i concl!$ii
..
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Ingineria Sistemelor Electroenergetice anul: 4 grupa:2
n cad"ul p"iectului sa studiat "e6imul pe"manent pent"u sistemul elect"ene"6etic TEST&1 9 cu metda Seidel&Gauss Ji cu aut"ul sBta"e&ului NEPLAN ca"e Blsestepent"u studiul de "e6im pe"manent metda Netn&Raphsn 9 ia" "e8ultatele F5inute!ind ap"piate ca
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