Zadatak Na slici je prikazan deo EES-a za koji je potrebno odrediti
a) Napone na sabirnicama 2 i 3 posmatranog sistema primenom Gauss-Seidel -ove metode za proračun tokova snaga u tri iteracije bez svođenja na relativne vrednosti, a za slučaj kada je isključen prekidač P, b) Prividnu snagu na sabirnicama 1. 1 S 1 L 12 = 30 km 2 L 23 = 40 km 3 S p3 = (30 + j20) MVA P 4 U 1 = 117 kV θ 1 = 0° z v = (0,13 + j0,41) Ω/km y 0 v = (0 + j2,7)·10 −6 S/km Poznato r v = 0,13 r_v = 0{,}13 r v = 0 , 13 Ω/km x v = 0,41 x_v = 0{,}41 x v = 0 , 41 Ω/km b v 0 = 2,7 b_v^0 = 2{,}7 b v 0 = 2 , 7 ·10⁻⁶ S/km L 12 = 30 L_{12} = 30 L 12 = 30 km L 23 = 40 L_{23} = 40 L 23 = 40 km U 1 = 117 U_1 = 117 U 1 = 117 kV P p 3 = 30 P_{p3} = 30 P p 3 = 30 MW Q p 3 = 20 Q_{p3} = 20 Q p 3 = 20 MVAr Rešenje Proračun parametara sistema:
Z ‾ V 1 = z ‾ v L 1 = ( 3,9 + j 12,3 ) Ω ; Y ‾ V 1 = 1 Z ‾ V 1 = ( 0,023423 − j 0,073874 ) S Z ‾ V 2 = z ‾ v L 2 = ( 5,2 + j 16,4 ) Ω ; Y ‾ V 2 = 1 Z ‾ V 2 = ( 0,017567 − j 0,055405 ) S \begin{gathered} \ph{Z}_{V1} = \ph{z}_vL_1 = (3{,}9 + \mathrm{j}12{,}3)\ \Omega; \quad \ph{Y}_{V1} = \frac{1}{\ph{Z}_{V1}} = (0{,}023423 - \mathrm{j}0{,}073874)\ \text{S} \\ \ph{Z}_{V2} = \ph{z}_vL_2 = (5{,}2 + \mathrm{j}16{,}4)\ \Omega; \quad \ph{Y}_{V2} = \frac{1}{\ph{Z}_{V2}} = (0{,}017567 - \mathrm{j}0{,}055405)\ \text{S} \end{gathered} Z V 1 = z v L 1 = ( 3 , 9 + j 12 , 3 ) Ω ; Y V 1 = Z V 1 1 = ( 0 , 023423 − j 0 , 073874 ) S Z V 2 = z v L 2 = ( 5 , 2 + j 16 , 4 ) Ω ; Y V 2 = Z V 2 1 = ( 0 , 017567 − j 0 , 055405 ) S Y ‾ V 1 0 2 = y ‾ v 0 2 L 1 = j 2,7 ⋅ 10 − 6 2 ⋅ 30 = j 0,405 ⋅ 10 − 4 S Y ‾ V 2 0 2 = y ‾ v 0 2 L 2 = j 2,7 ⋅ 10 − 6 2 ⋅ 40 = j 0,54 ⋅ 10 − 4 S \begin{aligned} \frac{\ph{Y}_{V1}^0}{2} &= \frac{\ph{y}_v^0}{2}L_1 = \mathrm{j}\frac{2{,}7 \cdot 10^{-6}}{2} \cdot 30 = \mathrm{j}0{,}405 \cdot 10^{-4}\ \text{S} \\ \frac{\ph{Y}_{V2}^0}{2} &= \frac{\ph{y}_v^0}{2}L_2 = \mathrm{j}\frac{2{,}7 \cdot 10^{-6}}{2} \cdot 40 = \mathrm{j}0{,}54 \cdot 10^{-4}\ \text{S} \end{aligned} 2 Y V 1 0 2 Y V 2 0 = 2 y v 0 L 1 = j 2 2 , 7 ⋅ 1 0 − 6 ⋅ 30 = j 0 , 405 ⋅ 1 0 − 4 S = 2 y v 0 L 2 = j 2 2 , 7 ⋅ 1 0 − 6 ⋅ 40 = j 0 , 54 ⋅ 1 0 − 4 S Ekvivalentna šema sistema je:
Matrica admitansi nezavisnih čvorova za ovu jednopolnu šemu je
Y ‾ B U S = [ Y ‾ V 1 + Y ‾ V 1 0 2 − Y ‾ V 1 0 − Y ‾ V 1 Y ‾ V 1 + Y ‾ V 2 + Y ‾ V 1 0 2 + Y ‾ V 2 0 2 − Y ‾ V 2 0 − Y ‾ V 2 Y ‾ V 2 + Y ‾ V 2 0 2 ] \ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} \ph{Y}_{V1} + \dfrac{\ph{Y}_{V1}^0}{2} & -\ph{Y}_{V1} & 0 \\[2ex] -\ph{Y}_{V1} & \ph{Y}_{V1} + \ph{Y}_{V2} + \dfrac{\ph{Y}_{V1}^0}{2} + \dfrac{\ph{Y}_{V2}^0}{2} & -\ph{Y}_{V2} \\[2ex] 0 & -\ph{Y}_{V2} & \ph{Y}_{V2} + \dfrac{\ph{Y}_{V2}^0}{2} \end{bmatrix} Y B U S = Y V 1 + 2 Y V 1 0 − Y V 1 0 − Y V 1 Y V 1 + Y V 2 + 2 Y V 1 0 + 2 Y V 2 0 − Y V 2 0 − Y V 2 Y V 2 + 2 Y V 2 0 Posle zamene brojčanih vrednosti ova matrica je
Y ‾ B U S = [ 0,023423 − j 0,073833 − 0,023423 + j 0,073874 0 − 0,023423 + j 0,073874 0,040991 − j 0,129185 − 0,017567 + j 0,055405 0 − 0,017567 + j 0,055405 0,017567 − j 0,055351 ] [ S ] \ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} 0{,}023423 - \mathrm{j}0{,}073833 & -0{,}023423 + \mathrm{j}0{,}073874 & 0 \\ -0{,}023423 + \mathrm{j}0{,}073874 & 0{,}040991 - \mathrm{j}0{,}129185 & -0{,}017567 + \mathrm{j}0{,}055405 \\ 0 & -0{,}017567 + \mathrm{j}0{,}055405 & 0{,}017567 - \mathrm{j}0{,}055351 \end{bmatrix} [\text{S}] Y B U S = 0 , 023423 − j 0 , 073833 − 0 , 023423 + j 0 , 073874 0 − 0 , 023423 + j 0 , 073874 0 , 040991 − j 0 , 129185 − 0 , 017567 + j 0 , 055405 0 − 0 , 017567 + j 0 , 055405 0 , 017567 − j 0 , 055351 [ S ] a. Naponi sabirnica 2 i 3 Pošto je čvor 1 referentno-balansni (po tekstu zadatka) i ako se usvoji V ‾ 2 ( 0 ) = 110 / 3 kV / 0 ∘ ‾ \ph{V}_2^{(0)} = 110/\sqrt{3}\ \text{kV}\, \underline{/0^\circ} V 2 ( 0 ) = 110/ 3 kV / 0 ∘ V ‾ 3 ( 0 ) = 105 / 3 kV / 0 ∘ ‾ \ph{V}_3^{(0)} = 105/\sqrt{3}\ \text{kV}\, \underline{/0^\circ} V 3 ( 0 ) = 105/ 3 kV / 0 ∘
1. iteracija V ‾ 2 ( 1 ) = 1 Y ‾ 22 [ P 2 f − j Q 2 f V ‾ 2 ( 0 ) ∗ − Y ‾ 21 V ‾ 1 − Y ‾ 23 V ‾ 3 ( 0 ) ] ili \ph{V}_2^{(1)} = \frac{1}{\ph{Y}_{22}}\left[\frac{P_{2f} - \mathrm{j}Q_{2f}}{\ph{V}_2^{(0)*}} - \ph{Y}_{21}\ph{V}_1 - \ph{Y}_{23}\ph{V}_3^{(0)}\right] \text{ ili} V 2 ( 1 ) = Y 22 1 [ V 2 ( 0 ) ∗ P 2 f − j Q 2 f − Y 21 V 1 − Y 23 V 3 ( 0 ) ] ili U ‾ 2 ( 1 ) 3 = 1 Y ‾ 22 [ P 2 / 3 − j Q 2 / 3 U ‾ 2 ( 0 ) ∗ / 3 − Y ‾ 21 U ‾ 1 3 − Y ‾ 23 U ‾ 3 ( 0 ) 3 ] \frac{\ph{U}_2^{(1)}}{\sqrt{3}} = \frac{1}{\ph{Y}_{22}}\left[\frac{P_2/3 - \mathrm{j}Q_2/3}{\ph{U}_2^{(0)*}/\sqrt{3}} - \ph{Y}_{21}\frac{\ph{U}_1}{\sqrt{3}} - \ph{Y}_{23}\frac{\ph{U}_3^{(0)}}{\sqrt{3}}\right] 3 U 2 ( 1 ) = Y 22 1 [ U 2 ( 0 ) ∗ / 3 P 2 /3 − j Q 2 /3 − Y 21 3 U 1 − Y 23 3 U 3 ( 0 ) ] ( P 2 f − j Q 2 f ) / V ‾ 2 ( 0 ) ∗ = 0 (P_{2f} - \mathrm{j}Q_{2f})/\ph{V}_2^{(0)*} = 0 ( P 2 f − j Q 2 f ) / V 2 ( 0 ) ∗ = 0 U ‾ 2 ( 1 ) = 1 Y ‾ 22 [ P 2 − j Q 2 U ‾ 2 ( 0 ) ∗ − Y ‾ 21 U ‾ 1 − Y ‾ 23 U ‾ 3 ( 0 ) ] = ( 111,93148 − j 0,023588 ) kV \begin{aligned} \ph{U}_2^{(1)} &= \frac{1}{\ph{Y}_{22}}\left[\frac{P_2 - \mathrm{j}Q_2}{\ph{U}_2^{(0)*}} - \ph{Y}_{21}\ph{U}_1 - \ph{Y}_{23}\ph{U}_3^{(0)}\right] \\ &= (111{,}93148 - \mathrm{j}0{,}023588)\ \text{kV} \end{aligned} U 2 ( 1 ) = Y 22 1 [ U 2 ( 0 ) ∗ P 2 − j Q 2 − Y 21 U 1 − Y 23 U 3 ( 0 ) ] = ( 111 , 93148 − j 0 , 023588 ) kV U ‾ 3 ( 1 ) = 1 Y ‾ 33 [ − Y ‾ 31 U ‾ 1 − Y ‾ 32 U ‾ 2 ( 1 ) − P 3 − j Q 3 U ‾ 3 ( 0 ) ∗ ] = ( 107,41603 − j 3,75231 ) kV \begin{aligned} \ph{U}_3^{(1)} &= \frac{1}{\ph{Y}_{33}}\left[-\ph{Y}_{31}\ph{U}_1 - \ph{Y}_{32}\ph{U}_2^{(1)} - \frac{P_3 - \mathrm{j}Q_3}{\ph{U}_3^{(0)*}}\right] \\ &= (107{,}41603 - \mathrm{j}3{,}75231)\ \text{kV} \end{aligned} U 3 ( 1 ) = Y 33 1 [ − Y 31 U 1 − Y 32 U 2 ( 1 ) − U 3 ( 0 ) ∗ P 3 − j Q 3 ] = ( 107 , 41603 − j 3 , 75231 ) kV Gauss: U ‾ 2 ( 0 ) \ph{U}_2^{(0)} U 2 ( 0 ) U ‾ 2 ( 1 ) \ph{U}_2^{(1)} U 2 ( 1 ) 2. iteracija U ‾ 2 ( 2 ) = 1 Y ‾ 22 [ − Y ‾ 21 U ‾ 1 − Y ‾ 23 U ‾ 3 ( 1 ) ] = ( 112,96727 − j 1,63309 ) kV U ‾ 3 ( 2 ) = 1 Y ‾ 33 [ − Y ‾ 31 U ‾ 1 − Y ‾ 32 U ‾ 2 ( 2 ) − P 3 − j Q 3 U ‾ 3 ( 1 ) ∗ ] = ( 108,43549 − j 5,11885 ) kV \begin{aligned} \ph{U}_2^{(2)} &= \frac{1}{\ph{Y}_{22}}\left[-\ph{Y}_{21}\ph{U}_1 - \ph{Y}_{23}\ph{U}_3^{(1)}\right] = (112{,}96727 - \mathrm{j}1{,}63309)\ \text{kV} \\ \ph{U}_3^{(2)} &= \frac{1}{\ph{Y}_{33}}\left[-\ph{Y}_{31}\ph{U}_1 - \ph{Y}_{32}\ph{U}_2^{(2)} - \frac{P_3 - \mathrm{j}Q_3}{\ph{U}_3^{(1)*}}\right] \\ &= (108{,}43549 - \mathrm{j}5{,}11885)\ \text{kV} \end{aligned} U 2 ( 2 ) U 3 ( 2 ) = Y 22 1 [ − Y 21 U 1 − Y 23 U 3 ( 1 ) ] = ( 112 , 96727 − j 1 , 63309 ) kV = Y 33 1 [ − Y 31 U 1 − Y 32 U 2 ( 2 ) − U 3 ( 1 ) ∗ P 3 − j Q 3 ] = ( 108 , 43549 − j 5 , 11885 ) kV 3. iteracija U ‾ 2 ( 3 ) = 1 Y ‾ 22 [ − Y ‾ 21 U ‾ 1 − Y ‾ 23 U ‾ 3 ( 2 ) ] = ( 113,404 − j 2,21948 ) kV U ‾ 3 ( 3 ) = 1 Y ‾ 33 [ − Y ‾ 31 U ‾ 1 − Y ‾ 32 U ‾ 2 ( 3 ) − P 3 − j Q 3 U ‾ 3 ( 2 ) ∗ ] = ( 108,876 − j 5,6158 ) kV \begin{aligned} \ph{U}_2^{(3)} &= \frac{1}{\ph{Y}_{22}}\left[-\ph{Y}_{21}\ph{U}_1 - \ph{Y}_{23}\ph{U}_3^{(2)}\right] = (113{,}404 - \mathrm{j}2{,}21948)\ \text{kV} \\ \ph{U}_3^{(3)} &= \frac{1}{\ph{Y}_{33}}\left[-\ph{Y}_{31}\ph{U}_1 - \ph{Y}_{32}\ph{U}_2^{(3)} - \frac{P_3 - \mathrm{j}Q_3}{\ph{U}_3^{(2)*}}\right] \\ &= (108{,}876 - \mathrm{j}5{,}6158)\ \text{kV} \end{aligned} U 2 ( 3 ) U 3 ( 3 ) = Y 22 1 [ − Y 21 U 1 − Y 23 U 3 ( 2 ) ] = ( 113 , 404 − j 2 , 21948 ) kV = Y 33 1 [ − Y 31 U 1 − Y 32 U 2 ( 3 ) − U 3 ( 2 ) ∗ P 3 − j Q 3 ] = ( 108 , 876 − j 5 , 6158 ) kV b. Prividna snaga na sabirnicama 1 S ‾ 1 = U ‾ 1 [ Y ‾ 11 U ‾ 1 + Y ‾ 12 U ‾ 2 ] ∗ = U 1 2 Y ‾ 11 ∗ + U ‾ 1 U ‾ 2 ∗ Y ‾ 12 ∗ = ( 28,9 + j 24,5 ) MVA \begin{aligned} \ph{S}_1 &= \ph{U}_1\left[\ph{Y}_{11}\ph{U}_1 + \ph{Y}_{12}\ph{U}_2\right]^* = U_1^2\cj{\ph{Y}}_{11} + \ph{U}_1\cj{\ph{U}}_2\cj{\ph{Y}}_{12} \\ &= (28{,}9 + \mathrm{j}24{,}5)\ \text{MVA} \end{aligned} S 1 = U 1 [ Y 11 U 1 + Y 12 U 2 ] ∗ = U 1 2 Y 11 ∗ + U 1 U 2 ∗ Y 12 ∗ = ( 28 , 9 + j 24 , 5 ) MVA Rezultat U ‾ 2 ( 3 ) = ( 113,404 − j 2,21948 ) \ph{U}_2^{(3)} = (113{,}404 - \mathrm{j}2{,}21948) U 2 ( 3 ) = ( 113 , 404 − j 2 , 21948 ) U ‾ 3 ( 3 ) = ( 108,876 − j 5,6158 ) \ph{U}_3^{(3)} = (108{,}876 - \mathrm{j}5{,}6158) U 3 ( 3 ) = ( 108 , 876 − j 5 , 6158 )
S ‾ 1 = ( 28,9 + j 24,5 ) \ph{S}_1 = (28{,}9 + \mathrm{j}24{,}5) S 1 = ( 28 , 9 + j 24 , 5 )
← Prethodni zadatak Zadatak 6: Klasifikacija čvorova i DC model tokova snaga Sledeći zadatak → Zadatak 8: Gauss-Seidel metoda, dve iteracije, napon 230 kV 