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Zadatak 5

Matrica admitansi i prividne snage čvorova: rešen zadatak

Zadatak i rešenje: prof. dr Andrija T. Sarić

Analiza elektroenergetskih sistema, TF Čačak

Zadatak

Za deo elektroenergetskog sistema na slici poznati su naponi na sabirnicama i impedanse vodova. Odrediti:

  • a) Matricu admitansi čvorova Y‾BUS\ph{\boldsymbol{Y}}_{BUS}
  • b) Prividne snage na sabirnicama 1, 2 i 3.
1S1L12 = 60 km2S2L23 = 80 km3S3U1 = 115 kVθ1 = 0°U2 = 110 kVθ2 = −4°U3 = 118 kVθ3 = 1°zv = (0,13 + j0,41) Ω/kmy0v = (0 + j2,7)·10−6 S/km

Poznato

  • rv=0,13r_v = 0{,}13Ω/km
  • xv=0,41x_v = 0{,}41Ω/km
  • bv0=2,7b_v^0 = 2{,}7·10⁻⁶ S/km
  • L12=60L_{12} = 60km
  • L23=80L_{23} = 80km
  • U1=115U_1 = 115kV
  • θ1=0\theta_1 = 0°
  • U2=110U_2 = 110kV
  • θ2=−4\theta_2 = -4°
  • U3=118U_3 = 118kV
  • θ3=1\theta_3 = 1°

Rešenje

a. Proračun parametara sistema:

Z‾12g=z‾vL12=(0,13+j0,41)⋅60=(7,8+j24,6) Ω;Y‾12g=1Z‾12g=(0,011711−j0,036936) S\begin{gathered} \ph{Z}_{12}^g = \ph{z}_vL_{12} = (0{,}13 + \mathrm{j}0{,}41) \cdot 60 = (7{,}8 + \mathrm{j}24{,}6)\ \Omega; \\ \ph{Y}_{12}^g = \frac{1}{\ph{Z}_{12}^g} = (0{,}011711 - \mathrm{j}0{,}036936)\ \text{S} \end{gathered}
Z‾23g=z‾vL23=(0,13+j0,41)⋅80=(10,4+j32,8) Ω;Y‾23g=1Z‾23g=(0,0087837−j0,0277027) S\begin{gathered} \ph{Z}_{23}^g = \ph{z}_vL_{23} = (0{,}13 + \mathrm{j}0{,}41) \cdot 80 = (10{,}4 + \mathrm{j}32{,}8)\ \Omega; \\ \ph{Y}_{23}^g = \frac{1}{\ph{Z}_{23}^g} = (0{,}0087837 - \mathrm{j}0{,}0277027)\ \text{S} \end{gathered}
Y‾12g0=y‾v0L12=j2,7⋅10−6⋅60=j1,62⋅10−4 SY‾23g0=y‾v0L23=j2,7⋅10−6⋅80=j2,16⋅10−4 S\begin{aligned} \ph{Y}_{12}^{g0} &= \ph{y}_v^0L_{12} = \mathrm{j}2{,}7 \cdot 10^{-6} \cdot 60 = \mathrm{j}1{,}62 \cdot 10^{-4}\ \text{S} \\ \ph{Y}_{23}^{g0} &= \ph{y}_v^0L_{23} = \mathrm{j}2{,}7 \cdot 10^{-6} \cdot 80 = \mathrm{j}2{,}16 \cdot 10^{-4}\ \text{S} \end{aligned}

Ekvivalentna šema sistema je

Ekvivalentna šema: π-šeme vodova 1–2 i 2–3, redne admitanse Y_12^g i Y_23^g i otočne Y_12^g0/2 i Y_23^g0/2 na oba kraja, snage S1, S2 i S3

Matrica admitansi nezavisnih čvorova za ovu jednopolnu šemu je

Y‾BUS=[Y‾12g+Y‾12g02−Y‾12g0−Y‾12gY‾12g+Y‾23g+Y‾12g02+Y‾23g02−Y‾23g0−Y‾23gY‾23g+Y‾23g02]\ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} \ph{Y}_{12}^g + \dfrac{\ph{Y}_{12}^{g0}}{2} & -\ph{Y}_{12}^g & 0 \\[2ex] -\ph{Y}_{12}^g & \ph{Y}_{12}^g + \ph{Y}_{23}^g + \dfrac{\ph{Y}_{12}^{g0}}{2} + \dfrac{\ph{Y}_{23}^{g0}}{2} & -\ph{Y}_{23}^g \\[2ex] 0 & -\ph{Y}_{23}^g & \ph{Y}_{23}^g + \dfrac{\ph{Y}_{23}^{g0}}{2} \end{bmatrix}

Posle zamene brojčanih vrednosti ova matrica je

Y‾BUS=[0,011711−j0,036856−0,011711+j0,0369360−0,011711+j0,0369360,0204955−j0,0644506−0,0087837+j0,02770270−0,0087837+j0,02770270,0087837−j0,0275947][S]\ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} 0{,}011711 - \mathrm{j}0{,}036856 & -0{,}011711 + \mathrm{j}0{,}036936 & 0 \\ -0{,}011711 + \mathrm{j}0{,}036936 & 0{,}0204955 - \mathrm{j}0{,}0644506 & -0{,}0087837 + \mathrm{j}0{,}0277027 \\ 0 & -0{,}0087837 + \mathrm{j}0{,}0277027 & 0{,}0087837 - \mathrm{j}0{,}0275947 \end{bmatrix} [\text{S}]

b. Proračun prividne snage na (objasniti proračun preko linijskih i faznih veličina)

Sabirnica 1

S‾1=U‾1[Y‾11U‾1+Y‾12U‾2+Y‾13U‾3]∗=U12Y‾11∗+U‾1U‾2∗Y‾12∗=1152⋅(0,011711+j0,036856)+115⋅110ej4∘(−0,011711−j0,036936)=(39,7+j10,97) MVA\begin{aligned} \ph{S}_1 &= \ph{U}_1\left[\ph{Y}_{11}\ph{U}_1 + \ph{Y}_{12}\ph{U}_2 + \ph{Y}_{13}\ph{U}_3\right]^* = U_1^2\cj{\ph{Y}}_{11} + \ph{U}_1\cj{\ph{U}}_2\cj{\ph{Y}}_{12} \\ &= 115^2 \cdot (0{,}011711 + \mathrm{j}0{,}036856) \\ &\quad + 115 \cdot 110e^{\mathrm{j}4^\circ}(-0{,}011711 - \mathrm{j}0{,}036936) \\ &= (39{,}7 + \mathrm{j}10{,}97)\ \text{MVA} \end{aligned}

Sabirnica 2

S‾2=U‾2[Y‾21U‾1+Y‾22U‾2+Y‾23U‾3]∗=U‾2U‾1∗Y‾21∗+U22Y‾22∗+U‾2U‾3∗Y‾23∗=−(77,31+j24,20) MVA\begin{aligned} \ph{S}_2 &= \ph{U}_2\left[\ph{Y}_{21}\ph{U}_1 + \ph{Y}_{22}\ph{U}_2 + \ph{Y}_{23}\ph{U}_3\right]^* \\ &= \ph{U}_2\cj{\ph{U}}_1\cj{\ph{Y}}_{21} + U_2^2\cj{\ph{Y}}_{22} + \ph{U}_2\cj{\ph{U}}_3\cj{\ph{Y}}_{23} \\ &= -(77{,}31 + \mathrm{j}24{,}20)\ \text{MVA} \end{aligned}

Sabirnica 3

S‾3=U‾3[Y‾31U‾1+Y‾32U‾2+Y‾33U‾3]∗=U‾3U‾2∗Y‾32∗+U32Y‾33∗=(40,06+j16,08) MVA\begin{aligned} \ph{S}_3 &= \ph{U}_3\left[\ph{Y}_{31}\ph{U}_1 + \ph{Y}_{32}\ph{U}_2 + \ph{Y}_{33}\ph{U}_3\right]^* \\ &= \ph{U}_3\cj{\ph{U}}_2\cj{\ph{Y}}_{32} + U_3^2\cj{\ph{Y}}_{33} = (40{,}06 + \mathrm{j}16{,}08)\ \text{MVA} \end{aligned}

Gubici u sistemu su:

ΔS‾=S‾1+S‾2+S‾3=(2,45+j2,85) MVA\Delta\ph{S} = \ph{S}_1 + \ph{S}_2 + \ph{S}_3 = (2{,}45 + \mathrm{j}2{,}85)\ \text{MVA}

Rezultat

S‾1=(39,7+j10,97)\ph{S}_1 = (39{,}7 + \mathrm{j}10{,}97) MVA, S‾2=−(77,31+j24,20)\ph{S}_2 = -(77{,}31 + \mathrm{j}24{,}20) MVA, S‾3=(40,06+j16,08)\ph{S}_3 = (40{,}06 + \mathrm{j}16{,}08) MVA

Gubici u sistemu: ΔS‾=(2,45+j2,85)\Delta\ph{S} = (2{,}45 + \mathrm{j}2{,}85) MVA

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