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Zadatak 3

Relativne jedinice, zamenska šema za tokove snaga: rešen zadatak

Zadatak i rešenje: prof. dr Andrija T. Sarić

Analiza elektroenergetskih sistema, TF Čačak

Zadatak

Normalizovana zamenska šema sistema

Za elektroenergetski sistem na slici formirati zamensku šemu za proračun tokova snaga sa svim veličinama u relativnim jedinicama.

Podaci o transformatorima i vodovima:

  • T1: mn=400 kV/231 kVm_n = 400\ \text{kV}/231\ \text{kV}; Sn=400S_n = 400 MVA; uk=12u_k = 12 %
  • T2: mn=220 kV/115 kVm_n = 220\ \text{kV}/115\ \text{kV}; Sn=150S_n = 150 MVA; uk=12u_k = 12 %
  • T3: mn=400 kV/115 kV/31,5 kVm_n = 400\ \text{kV}/115\ \text{kV}/31{,}5\ \text{kV}; Sn=300 MVA/300 MVA/100 MVAS_n = 300\ \text{MVA}/300\ \text{MVA}/100\ \text{MVA}; uk=12 %/15 %/10 %u_k = 12\ \%/15\ \%/10\ \%
  • V1: xv=0,4x_v = 0{,}4 Ω/km; Lv=100L_v = 100 km
  • V2: xv=0,4x_v = 0{,}4 Ω/km; Lv=50L_v = 50 km
G+T1V12T25110 kVT1220 kV400 kVG+T3V24T3630 kV

Poznato

  • Un1T1=400U_{n1}^{T1} = 400kV
  • Un2T1=231U_{n2}^{T1} = 231kV
  • SnT1=400S_n^{T1} = 400MVA
  • ukT1=12u_k^{T1} = 12%
  • Un1T2=220U_{n1}^{T2} = 220kV
  • Un2T2=115U_{n2}^{T2} = 115kV
  • SnT2=150S_n^{T2} = 150MVA
  • ukT2=12u_k^{T2} = 12%
  • Unp,T3=400U_{n}^{p,T3} = 400kV
  • Uns,T3=115U_{n}^{s,T3} = 115kV
  • Unt,T3=31,5U_{n}^{t,T3} = 31{,}5kV
  • Snp,T3=300S_{n}^{p,T3} = 300MVA
  • Sns,T3=300S_{n}^{s,T3} = 300MVA
  • Snt,T3=100S_{n}^{t,T3} = 100MVA
  • ukps,T3=12u_k^{ps,T3} = 12%
  • ukpt,T3=15u_k^{pt,T3} = 15%
  • ukst,T3=10u_k^{st,T3} = 10%
  • xv=0,4x_v = 0{,}4Ω/km
  • LV1=100L_{V1} = 100km
  • LV2=50L_{V2} = 50km

Rešenje

Proračun parametara transformatora i vodova:

XT1=ukT1%100⋅UnT12SnT1=12100⋅4002400=48 Ω\begin{aligned} X_{T1} &= \frac{u_{kT1}\%}{100} \cdot \frac{U_{nT1}^2}{S_{nT1}} \\ &= \frac{12}{100} \cdot \frac{400^2}{400} = 48\ \Omega \end{aligned}
XT2=ukT2%100⋅UnT22SnT2=12100⋅2202150=38,72 Ω\begin{aligned} X_{T2} &= \frac{u_{kT2}\%}{100} \cdot \frac{U_{nT2}^2}{S_{nT2}} \\ &= \frac{12}{100} \cdot \frac{220^2}{150} = 38{,}72\ \Omega \end{aligned}
XT3ps=ukT3ps%100⋅(UnT3p)2SnT3p=12100⋅4002300=64 Ω\begin{aligned} X_{T3}^{ps} &= \frac{u_{kT3}^{ps}\%}{100} \cdot \frac{\left(U_{nT3}^{p}\right)^2}{S_{nT3}^{p}} \\ &= \frac{12}{100} \cdot \frac{400^2}{300} = 64\ \Omega \end{aligned}
XT3pt=ukT3pt%100⋅(UnT3p)2SnT3t=15100⋅4002100=240 Ω\begin{aligned} X_{T3}^{pt} &= \frac{u_{kT3}^{pt}\%}{100} \cdot \frac{\left(U_{nT3}^{p}\right)^2}{S_{nT3}^{t}} \\ &= \frac{15}{100} \cdot \frac{400^2}{100} = 240\ \Omega \end{aligned}
XT3st=ukT3st%100⋅(UnT3p)2SnT3t=10100⋅4002100=160 Ω\begin{aligned} X_{T3}^{st} &= \frac{u_{kT3}^{st}\%}{100} \cdot \frac{\left(U_{nT3}^{p}\right)^2}{S_{nT3}^{t}} \\ &= \frac{10}{100} \cdot \frac{400^2}{100} = 160\ \Omega \end{aligned}
XT3p=12(XT3ps+XT3pt−XT3st)=12(64+240−160)=72 ΩXT3s=12(XT3ps+XT3st−XT3pt)=12(64+160−240)=−8 ΩXT3t=12(XT3pt+XT3st−XT3ps)=12(160+240−64)=168 Ω\begin{aligned} X_{T3}^{p} &= \tfrac{1}{2}\left( X_{T3}^{ps} + X_{T3}^{pt} - X_{T3}^{st} \right) \\ &= \tfrac{1}{2}\left( 64 + 240 - 160 \right) = 72\ \Omega \\[1ex] X_{T3}^{s} &= \tfrac{1}{2}\left( X_{T3}^{ps} + X_{T3}^{st} - X_{T3}^{pt} \right) \\ &= \tfrac{1}{2}\left( 64 + 160 - 240 \right) = -8\ \Omega \\[1ex] X_{T3}^{t} &= \tfrac{1}{2}\left( X_{T3}^{pt} + X_{T3}^{st} - X_{T3}^{ps} \right) \\ &= \tfrac{1}{2}\left( 160 + 240 - 64 \right) = 168\ \Omega \end{aligned}
XV1=xvLV1=0,4⋅100=40 ΩXV2=xvLV2=0,4⋅50=20 Ω\begin{aligned} X_{V1} &= x_v L_{V1} = 0{,}4 \cdot 100 = 40\ \Omega \\ X_{V2} &= x_v L_{V2} = 0{,}4 \cdot 50 = 20\ \Omega \end{aligned}

Zatim se proverava da li se petlja zatvara:

220 kV115 kV⋅115 kV400 kV⋅400 kV231 kV=220 kV231 kV,\begin{aligned} &\frac{220\ \text{kV}}{115\ \text{kV}} \cdot \frac{115\ \text{kV}}{400\ \text{kV}} \cdot \frac{400\ \text{kV}}{231\ \text{kV}} \\ &\quad = \frac{220\ \text{kV}}{231\ \text{kV}}, \end{aligned}

što znači da se petlja ne zatvara. Dakle, sigurno se negde mora dodati fiktivni transformator.

Izbor baznih vrednosti (bazne vrednosti jednake nominalnim naponima):

SB=100 MVA;UB400=400 kV;UB220=220 kV;UB110=110 kV.\begin{gathered} S_B = 100\ \text{MVA}; \quad U_B^{400} = 400\ \text{kV}; \\ U_B^{220} = 220\ \text{kV}; \quad U_B^{110} = 110\ \text{kV}. \end{gathered}

Na osnovu ovih vrednosti izračunavaju se:

ZB400=(UB400)2SB=4002100=1600 Ω;YB400=1ZB400=11600=0,625⋅10−3 S\begin{aligned} Z_B^{400} &= \frac{\left(U_B^{400}\right)^2}{S_B} = \frac{400^2}{100} = 1600\ \Omega; \\ Y_B^{400} &= \frac{1}{Z_B^{400}} = \frac{1}{1600} = 0{,}625 \cdot 10^{-3}\ \text{S} \end{aligned}
ZB220=(UB220)2SB=2202100=484 Ω;YB220=1ZB220=1484=2,066⋅10−3 S\begin{aligned} Z_B^{220} &= \frac{\left(U_B^{220}\right)^2}{S_B} = \frac{220^2}{100} = 484\ \Omega; \\ Y_B^{220} &= \frac{1}{Z_B^{220}} = \frac{1}{484} = 2{,}066 \cdot 10^{-3}\ \text{S} \end{aligned}
ZB110=(UB110)2SB=1102100=121 Ω;YB110=1ZB110=1121=8,26⋅10−3 S.\begin{aligned} Z_B^{110} &= \frac{\left(U_B^{110}\right)^2}{S_B} = \frac{110^2}{100} = 121\ \Omega; \\ Y_B^{110} &= \frac{1}{Z_B^{110}} = \frac{1}{121} = 8{,}26 \cdot 10^{-3}\ \text{S}. \end{aligned}

Na osnovu ovih baznih vrednosti mogu se izračunati relativne vrednosti susceptansi grana

y‾T1=Y‾T1YB400=1jXT1⋅1YB400=1j48⋅10,625⋅10−3=−j33,33 r.j.\begin{aligned} \ph{y}_{T1} &= \frac{\ph{Y}_{T1}}{Y_B^{400}} = \frac{1}{\mathrm{j}X_{T1}} \cdot \frac{1}{Y_B^{400}} \\ &= \frac{1}{\mathrm{j}48} \cdot \frac{1}{0{,}625 \cdot 10^{-3}} = -\mathrm{j}33{,}33\ \rj \end{aligned}
y‾T2=Y‾T2YB220=1jXT2⋅1YB220=1j38,72⋅12,066⋅10−3=−j12,50 r.j.\begin{aligned} \ph{y}_{T2} &= \frac{\ph{Y}_{T2}}{Y_B^{220}} = \frac{1}{\mathrm{j}X_{T2}} \cdot \frac{1}{Y_B^{220}} \\ &= \frac{1}{\mathrm{j}38{,}72} \cdot \frac{1}{2{,}066 \cdot 10^{-3}} = -\mathrm{j}12{,}50\ \rj \end{aligned}
y‾T3p=Y‾T3pYB400=1jXT3p⋅1YB400=1j72⋅10,625⋅10−3=−j22,22 r.j.\begin{aligned} \ph{y}_{T3}^{p} &= \frac{\ph{Y}_{T3}^{p}}{Y_B^{400}} = \frac{1}{\mathrm{j}X_{T3}^{p}} \cdot \frac{1}{Y_B^{400}} \\ &= \frac{1}{\mathrm{j}72} \cdot \frac{1}{0{,}625 \cdot 10^{-3}} = -\mathrm{j}22{,}22\ \rj \end{aligned}
y‾T3s=Y‾T3sYB400=1jXT3s⋅1YB400=1−j8⋅10,625⋅10−3=j200 r.j.\begin{aligned} \ph{y}_{T3}^{s} &= \frac{\ph{Y}_{T3}^{s}}{Y_B^{400}} = \frac{1}{\mathrm{j}X_{T3}^{s}} \cdot \frac{1}{Y_B^{400}} \\ &= \frac{1}{-\mathrm{j}8} \cdot \frac{1}{0{,}625 \cdot 10^{-3}} = \mathrm{j}200\ \rj \end{aligned}
y‾T3t=Y‾T3tYB400=1jXT3t⋅1YB400=1j168⋅10,625⋅10−3=−j9,524 r.j.\begin{aligned} \ph{y}_{T3}^{t} &= \frac{\ph{Y}_{T3}^{t}}{Y_B^{400}} = \frac{1}{\mathrm{j}X_{T3}^{t}} \cdot \frac{1}{Y_B^{400}} \\ &= \frac{1}{\mathrm{j}168} \cdot \frac{1}{0{,}625 \cdot 10^{-3}} = -\mathrm{j}9{,}524\ \rj \end{aligned}
y‾V1=Y‾V1YB220=1jXV1⋅1YB220=1j40⋅12,066⋅10−3=−j12,10 r.j.\begin{aligned} \ph{y}_{V1} &= \frac{\ph{Y}_{V1}}{Y_B^{220}} = \frac{1}{\mathrm{j}X_{V1}} \cdot \frac{1}{Y_B^{220}} \\ &= \frac{1}{\mathrm{j}40} \cdot \frac{1}{2{,}066 \cdot 10^{-3}} = -\mathrm{j}12{,}10\ \rj \end{aligned}
y‾V2=Y‾V2YB400=1jXV2⋅1YB400=1j20⋅10,625⋅10−3=−j80 r.j.\begin{aligned} \ph{y}_{V2} &= \frac{\ph{Y}_{V2}}{Y_B^{400}} = \frac{1}{\mathrm{j}X_{V2}} \cdot \frac{1}{Y_B^{400}} \\ &= \frac{1}{\mathrm{j}20} \cdot \frac{1}{0{,}625 \cdot 10^{-3}} = -\mathrm{j}80\ \rj \end{aligned}

Sada se mora videti gde treba umetnuti fiktivne transformatore zbog svođenja veličina u relativne jedinice

Sistem iz zadatka sa nominalnim odnosima transformatora: T1 m_n = 400 kV/231 kV između sabirnice 2 (220 kV) i sabirnice 4 (400 kV), T2 m_n = 220 kV/115 kV između sabirnica 2 i 5 (110 kV), i tronamotajni T3 m_n = 400 kV/115 kV/31,5 kV od sabirnice 4 do sabirnica 5 i 6 (30 kV).

Na osnovu ove slike može se formirati sledeća jednopolna šema sa uvedenim fiktivnim transformatorima

Jednopolna šema sa admitansama u r.j. i fiktivnim transformatorima: y_V1 između sabirnica 1 i 2; y_T2 i fiktivni transformator n_T2 = 115/110 = 1,045 do sabirnice 5; od sabirnice 2 fiktivni transformator n_T1 = 231/220 = 1,05 i y_T1 do sabirnice 4 (400 kV); y_V2 između sabirnica 3 i 4; od sabirnice 4 grana y_T3^p do zvezdišta T3, odakle y_T3^s i fiktivni transformator n_T3^s = 115/110 = 1,045 vode do sabirnice 5 (110 kV), a y_T3^t i fiktivni transformator n_T3^t = 31,5/31 = 1,016 do sabirnice 6 (31 kV).

Sa slike se vidi da u jednopolnoj šemi nema idealnih transformatora usled različitih baznih napona pojedinih mreža, ali su ostali idealni transformatori koji su posledica razlike baznih napona mreže i sekundarnih napona pojedinih transformatora.

Ovi transformatori se mogu eliminisati na sledeći način:

Grana od sabirnice 2 do sabirnice 5: admitansa y_T2, pa idealni transformator n_T2 : 1.
Ista grana kao nesimetrični π-ekvivalent: redna admitansa y′_T2 = n_T2·y_T2 između sabirnica 2 i 5, otočna y_T2^01 = (1 − n_T2)·y_T2 na strani sabirnice 2 i otočna y_T2^02 = n_T2(n_T2 − 1)·y_T2 na strani sabirnice 5.

Modifikovani parametri nesimetričnog π-ekvivalenta transformatora T2 sa prethodne slike su:

y‾T2′=nT2y‾T2=−j1,045⋅12,50 r.j.=−j13,0625 r.j.y‾T202=nT2(nT2−1)y‾T2=−j1,045(1,045−1)⋅12,50 r.j.=−j0,5878 r.j.y‾T201=(1−nT2)y‾T2=−j(1−1,045)⋅12,50 r.j.=j0,5625 r.j.\begin{aligned} \ph{y}'_{T2} &= n_{T2}\ph{y}_{T2} \\ &= -\mathrm{j}1{,}045 \cdot 12{,}50\ \rj \\ &=-\mathrm{j}13{,}0625\ \rj \\ \ph{y}_{T2}^{02} &= n_{T2}(n_{T2} - 1)\ph{y}_{T2} \\ &= -\mathrm{j}1{,}045(1{,}045 - 1) \cdot 12{,}50\ \rj \\ &=-\mathrm{j}0{,}5878\ \rj \\ \ph{y}_{T2}^{01} &= (1 - n_{T2})\ph{y}_{T2} \\ &= -\mathrm{j}(1 - 1{,}045) \cdot 12{,}50\ \rj \\ &=\mathrm{j}0{,}5625\ \rj \end{aligned}

Na sličan način se proračunavaju i parametri nesimetričnih π-ekvivalenata ostalih transformatora

y‾T1′=nT1y‾T1=−j1,05⋅33,33 r.j.=−j34,9965 r.j.y‾T101=(1−nT1)y‾T1=−j(1−1,05)⋅33,33 r.j.=j1,6665 r.j.y‾T102=nT1(nT1−1)y‾T1=−j1,05(1,05−1)⋅33,33 r.j.=−j1,7498 r.j.\begin{aligned} \ph{y}'_{T1} &= n_{T1}\ph{y}_{T1} \\ &= -\mathrm{j}1{,}05 \cdot 33{,}33\ \rj \\ &=-\mathrm{j}34{,}9965\ \rj \\ \ph{y}_{T1}^{01} &= (1 - n_{T1})\ph{y}_{T1} \\ &= -\mathrm{j}(1 - 1{,}05) \cdot 33{,}33\ \rj \\ &=\mathrm{j}1{,}6665\ \rj \\ \ph{y}_{T1}^{02} &= n_{T1}(n_{T1} - 1)\ph{y}_{T1} \\ &= -\mathrm{j}1{,}05(1{,}05 - 1) \cdot 33{,}33\ \rj \\ &=-\mathrm{j}1{,}7498\ \rj \end{aligned}
y‾T3′s=nT3sy‾T3s=j1,045⋅200 r.j.=j209 r.j.y‾T3s01=(1−nT3s)y‾T3s=j(1−1,045)⋅200 r.j.=−j9 r.j.y‾T3s02=nT3s(nT3s−1)y‾T3s=j1,045(1,045−1)⋅200 r.j.=j9,405 r.j.\begin{aligned} \ph{y}'^{s}_{T3} &= n_{T3}^{s}\ph{y}_{T3}^{s} \\ &= \mathrm{j}1{,}045 \cdot 200\ \rj \\ &=\mathrm{j}209\ \rj \\ \ph{y}_{T3}^{s01} &= (1 - n_{T3}^{s})\ph{y}_{T3}^{s} \\ &= \mathrm{j}(1 - 1{,}045) \cdot 200\ \rj \\ &=-\mathrm{j}9\ \rj \\ \ph{y}_{T3}^{s02} &= n_{T3}^{s}(n_{T3}^{s} - 1)\ph{y}_{T3}^{s} \\ &= \mathrm{j}1{,}045(1{,}045 - 1) \cdot 200\ \rj \\ &=\mathrm{j}9{,}405\ \rj \end{aligned}
y‾T3′t=nT3ty‾T3t=−j1,016⋅9,524 r.j.=−j9,6764 r.j.y‾T3t01=(1−nT3t)y‾T3t=−j(1−1,016)⋅9,524 r.j.=j0,1524 r.j.y‾T3t02=nT3t(nT3t−1)y‾T3t=−j1,016(1,016−1)⋅9,524 r.j.=−j0,1548 r.j.\begin{aligned} \ph{y}'^{t}_{T3} &= n_{T3}^{t}\ph{y}_{T3}^{t} \\ &= -\mathrm{j}1{,}016 \cdot 9{,}524\ \rj \\ &=-\mathrm{j}9{,}6764\ \rj \\ \ph{y}_{T3}^{t01} &= (1 - n_{T3}^{t})\ph{y}_{T3}^{t} \\ &= -\mathrm{j}(1 - 1{,}016) \cdot 9{,}524\ \rj \\ &=\mathrm{j}0{,}1524\ \rj \\ \ph{y}_{T3}^{t02} &= n_{T3}^{t}(n_{T3}^{t} - 1)\ph{y}_{T3}^{t} \\ &= -\mathrm{j}1{,}016(1{,}016 - 1) \cdot 9{,}524\ \rj \\ &=-\mathrm{j}0{,}1548\ \rj \end{aligned}

Uvažavajući prethodno izračunate modifikovane parametre, nova jednopolna zamenska šema sistema je

Normalizovana zamenska šema: y_V1 između sabirnica 1 i 2; od sabirnice 2 ka sabirnici 5 redna y′_T2 sa otočnim y_T2^01 i y_T2^02; od sabirnice 2 ka sabirnici 4 redna y′_T1 sa otočnim y_T1^02 na strani sabirnice 2 i y_T1^01 na strani sabirnice 4; y_V2 između sabirnica 3 i 4; od sabirnice 4 y_T3^p do zvezdišta T3, odakle redna y′^s_T3 sa otočnim y_T3^s01 i y_T3^s02 vodi do sabirnice 5, a redna y′^t_T3 sa otočnim y_T3^t01 i y_T3^t02 do sabirnice 6.

Rezultat

Normalizovana zamenska šema, sve u r.j. na SB=100S_B = 100 MVA: y‾V1=−j12,10\ph{y}_{V1} = -\mathrm{j}12{,}10, y‾V2=−j80\ph{y}_{V2} = -\mathrm{j}80, y‾T3p=−j22,22\ph{y}_{T3}^{p} = -\mathrm{j}22{,}22;

T1: y‾T1′=−j34,9965\ph{y}'_{T1} = -\mathrm{j}34{,}9965, y‾T101=j1,6665\ph{y}_{T1}^{01} = \mathrm{j}1{,}6665, y‾T102=−j1,7498\ph{y}_{T1}^{02} = -\mathrm{j}1{,}7498; T2: y‾T2′=−j13,0625\ph{y}'_{T2} = -\mathrm{j}13{,}0625, y‾T201=j0,5625\ph{y}_{T2}^{01} = \mathrm{j}0{,}5625, y‾T202=−j0,5878\ph{y}_{T2}^{02} = -\mathrm{j}0{,}5878;

T3: y‾T3′s=j209\ph{y}'^{s}_{T3} = \mathrm{j}209, y‾T3s01=−j9\ph{y}_{T3}^{s01} = -\mathrm{j}9, y‾T3s02=j9,405\ph{y}_{T3}^{s02} = \mathrm{j}9{,}405, y‾T3′t=−j9,6764\ph{y}'^{t}_{T3} = -\mathrm{j}9{,}6764, y‾T3t01=j0,1524\ph{y}_{T3}^{t01} = \mathrm{j}0{,}1524, y‾T3t02=−j0,1548\ph{y}_{T3}^{t02} = -\mathrm{j}0{,}1548.

Metod u predavanju

Slični zadaci

Teme

Relativne jedinice i svođenje veličina · Tokovi snaga · Transformatori

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