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Zadatak 4

Matrica admitansi sa regulacionim transformatorima: rešen zadatak

Zadatak i rešenje: prof. dr Andrija T. Sarić

Analiza elektroenergetskih sistema, TF Čačak

Zadatak

Formiranje matrice admitansi nezavisnih čvorova

Na slici je prikazan uprošćeni EES. Formirati matricu admitansi čvorova Y‾BUS\ph{\boldsymbol{Y}}_{BUS} posmatranog sistema u kome su zanemarene aktivne otpornosti (r≈0r \approx 0, g≈0g \approx 0).

12Vod 1: L12 = 100 kmSG1SG2 = (150 + jQG2) MVAU2 = 220 kVxv = 0,4 Ω/kmVod 2: L13 = 150 km4Vod 3: L23 = 80 km5T1n1 = −7T2n2 = −33Sp3 = (270 + j90) MVASnT1 = SnT2 = 150 MVAm = 220 kV ± 12×1,25 %/115 kVxT(n = 0) = 10 % (rT ≈ 0)

Poznato

  • L12=100L_{12} = 100km
  • L13=150L_{13} = 150km
  • L23=80L_{23} = 80km
  • xv=0,4x_v = 0{,}4Ω/km
  • SnT1=SnT2=150S_{nT1} = S_{nT2} = 150MVA
  • Un1T=220U_{n1}^{T} = 220kV
  • Un2T=115U_{n2}^{T} = 115kV
  • Δa=1,25\Delta a = 1{,}25% po stepenu (±12)
  • xT(n=0)=10x_T(n = 0) = 10%
  • n1=−7n_1 = -7
  • n2=−3n_2 = -3
  • Pp3=270P_{p3} = 270MW
  • Qp3=90Q_{p3} = 90MVAr
  • PG2=150P_{G2} = 150MW
  • U2=220U_2 = 220kV
  • UB=220U_B = 220kV
  • SB=100S_B = 100MVA

Rešenje

Proračun parametara sistema:

XV1=xVL12=0,4⋅100=40 Ω;Y‾V1=1jXV1=1j40=−j0,025 S\begin{gathered} X_{V1} = x_VL_{12} = 0{,}4 \cdot 100 = 40\ \Omega; \\ \ph{Y}_{V1} = \frac{1}{\mathrm{j}X_{V1}} = \frac{1}{\mathrm{j}40} = -\mathrm{j}0{,}025\ \text{S} \end{gathered}
XV2=xVL13=0,4⋅150=60 Ω;Y‾V2=1jXV2=1j60=−j0,0167 S\begin{gathered} X_{V2} = x_VL_{13} = 0{,}4 \cdot 150 = 60\ \Omega; \\ \ph{Y}_{V2} = \frac{1}{\mathrm{j}X_{V2}} = \frac{1}{\mathrm{j}60} = -\mathrm{j}0{,}0167\ \text{S} \end{gathered}
XV3=xVL23=0,4⋅80=32 Ω;Y‾V3=1jXV3=1j32=−j0,03125 S\begin{gathered} X_{V3} = x_VL_{23} = 0{,}4 \cdot 80 = 32\ \Omega; \\ \ph{Y}_{V3} = \frac{1}{\mathrm{j}X_{V3}} = \frac{1}{\mathrm{j}32} = -\mathrm{j}0{,}03125\ \text{S} \end{gathered}
XT1=XT2=XT=xTUnT2SnT=0,1⋅2202150=32,26 Ω;Y‾T1=1jXT1=1j32,26=−j0,031 S\begin{gathered} X_{T1} = X_{T2} = X_T = x_T\frac{U_{nT}^2}{S_{nT}} = 0{,}1 \cdot \frac{220^2}{150} = 32{,}26\ \Omega; \\ \ph{Y}_{T1} = \frac{1}{\mathrm{j}X_{T1}} = \frac{1}{\mathrm{j}32{,}26} = -\mathrm{j}0{,}031\ \text{S} \end{gathered}

Bazne vrednosti impedanse i admitanse su:

ZB=UB2SB=2202100=484 Ω;YB=1ZB=1484=2,0661⋅10−3 SZ_B = \frac{U_B^2}{S_B} = \frac{220^2}{100} = 484\ \Omega; \quad Y_B = \frac{1}{Z_B} = \frac{1}{484} = 2{,}0661 \cdot 10^{-3}\ \text{S}

Relativne vrednosti izračunatih admitansi su:

Y‾V1[r.j.]=Y‾V1[SI]YB=−j0,0252,0661⋅10−3=−j12,10 r.j.Y‾V2[r.j.]=−j0,01672,0661⋅10−3=−j8,07 r.j.Y‾V3[r.j.]=−j0,031252,0661⋅10−3=−j15,12 r.j.Y‾T[r.j.]=−j0,0312,0661⋅10−3=−j15 r.j.\begin{aligned} \ph{Y}_{V1}[\rj] &= \frac{\ph{Y}_{V1}[\text{SI}]}{Y_B} = \frac{-\mathrm{j}0{,}025}{2{,}0661 \cdot 10^{-3}} = -\mathrm{j}12{,}10\ \rj \\ \ph{Y}_{V2}[\rj] &= \frac{-\mathrm{j}0{,}0167}{2{,}0661 \cdot 10^{-3}} = -\mathrm{j}8{,}07\ \rj \\ \ph{Y}_{V3}[\rj] &= \frac{-\mathrm{j}0{,}03125}{2{,}0661 \cdot 10^{-3}} = -\mathrm{j}15{,}12\ \rj \\ \ph{Y}_T[\rj] &= \frac{-\mathrm{j}0{,}031}{2{,}0661 \cdot 10^{-3}} = -\mathrm{j}15\ \rj \end{aligned}

Zamenska šema regulacionog transformatora je:

12Y12Y01Y02
Y‾12=Y‾Ta;Y‾10=1−aa2Y‾T;Y‾20=a−1aY‾T\ph{Y}_{12} = \frac{\ph{Y}_T}{a}; \quad \ph{Y}_1^0 = \frac{1 - a}{a^2}\ph{Y}_T; \quad \ph{Y}_2^0 = \frac{a - 1}{a}\ph{Y}_T

Za prethodnu šemu nenominalni odnosi transformacije su:

T1: a1=1+n1Δa=1−71,25100=0,9125T2: a2=1+n2Δa=1−31,25100=0,9625\begin{aligned} \text{T1: } a_1 &= 1 + n_1\Delta a = 1 - 7\frac{1{,}25}{100} = 0{,}9125 \\ \text{T2: } a_2 &= 1 + n_2\Delta a = 1 - 3\frac{1{,}25}{100} = 0{,}9625 \end{aligned}

Sada se mogu izračunati parametri ekvivalenata regulacionih transformatora:

Y‾T1, 12=Y‾T1a1=−j150,9125=−j16,44 r.j.Y‾T1, 10=1−a1a12Y‾T1=−j1−0,91250,91252⋅15=−j1,576 r.j.Y‾T1, 20=a1−1a1Y‾T1=−j0,9125−10,9125⋅15=j1,44 r.j.\begin{aligned} \ph{Y}_{T1,\,12} &= \frac{\ph{Y}_{T1}}{a_1} = -\mathrm{j}\frac{15}{0{,}9125} = -\mathrm{j}16{,}44\ \rj \\ \ph{Y}_{T1,\,1}^0 &= \frac{1 - a_1}{a_1^2}\ph{Y}_{T1} = -\mathrm{j}\frac{1 - 0{,}9125}{0{,}9125^2} \cdot 15 = -\mathrm{j}1{,}576\ \rj \\ \ph{Y}_{T1,\,2}^0 &= \frac{a_1 - 1}{a_1}\ph{Y}_{T1} = -\mathrm{j}\frac{0{,}9125 - 1}{0{,}9125} \cdot 15 = \mathrm{j}1{,}44\ \rj \end{aligned}
Y‾T2, 12=Y‾T2a2=−j150,9625=−j15,58 r.j.Y‾T2, 10=1−a2a22Y‾T2=−j1−0,96250,96252⋅15=−j0,607 r.j.Y‾T2, 20=a2−1a2Y‾T2=−j0,9625−10,9625⋅15=j0,58 r.j.\begin{aligned} \ph{Y}_{T2,\,12} &= \frac{\ph{Y}_{T2}}{a_2} = -\mathrm{j}\frac{15}{0{,}9625} = -\mathrm{j}15{,}58\ \rj \\ \ph{Y}_{T2,\,1}^0 &= \frac{1 - a_2}{a_2^2}\ph{Y}_{T2} = -\mathrm{j}\frac{1 - 0{,}9625}{0{,}9625^2} \cdot 15 = -\mathrm{j}0{,}607\ \rj \\ \ph{Y}_{T2,\,2}^0 &= \frac{a_2 - 1}{a_2}\ph{Y}_{T2} = -\mathrm{j}\frac{0{,}9625 - 1}{0{,}9625} \cdot 15 = \mathrm{j}0{,}58\ \rj \end{aligned}

Sada je zamenska šema sistema

Zamenska šema sistema u r.j.: vod 1–2 −j12,1, vod 1–4 −j8,07, vod 2–5 −j15,125; T1 između čvorova 4 i 3 redno −j16,44, otočno −j1,576 kod čvora 4 i j1,44 kod čvora 3; T2 između 5 i 3 redno −j15,58, otočno −j0,607 kod čvora 5 i j0,58 kod čvora 3; potrošnja −2,7 − j0,9 u čvoru 3, S_G1 u čvoru 1 i S_G2 = (1,5 + jQ_G2) r.j. u čvoru 2

Matrica admitansi nezavisnih čvorova za ovu jednopolnu šemu je

Y‾BUS=[−j20,17j12,10j8,070j12,1−j27,22500j15,12500−j30j16,44j15,58j8,070j16,44−j26,0900j15,125j15,580−j31,315]\ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} -\mathrm{j}20{,}17 & \mathrm{j}12{,}1 & 0 & \mathrm{j}8{,}07 & 0 \\ \mathrm{j}12{,}1 & -\mathrm{j}27{,}225 & 0 & 0 & \mathrm{j}15{,}125 \\ 0 & 0 & -\mathrm{j}30 & \mathrm{j}16{,}44 & \mathrm{j}15{,}58 \\ \mathrm{j}8{,}07 & 0 & \mathrm{j}16{,}44 & -\mathrm{j}26{,}09 & 0 \\ 0 & \mathrm{j}15{,}125 & \mathrm{j}15{,}58 & 0 & -\mathrm{j}31{,}315 \end{bmatrix}

Rezultat

Dijagonalni elementi: Y‾11=−j20,17\ph{Y}_{11} = -\mathrm{j}20{,}17, Y‾22=−j27,225\ph{Y}_{22} = -\mathrm{j}27{,}225, Y‾33=−j30\ph{Y}_{33} = -\mathrm{j}30, Y‾44=−j26,09\ph{Y}_{44} = -\mathrm{j}26{,}09, Y‾55=−j31,315\ph{Y}_{55} = -\mathrm{j}31{,}315 r.j.

Van dijagonale: Y‾12=j12,1\ph{Y}_{12} = \mathrm{j}12{,}1, Y‾14=j8,07\ph{Y}_{14} = \mathrm{j}8{,}07, Y‾25=j15,125\ph{Y}_{25} = \mathrm{j}15{,}125, Y‾34=j16,44\ph{Y}_{34} = \mathrm{j}16{,}44, Y‾35=j15,58\ph{Y}_{35} = \mathrm{j}15{,}58 r.j.

Metod u predavanju

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Teme

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