Zadatak 8
Gauss-Seidel metoda, dve iteracije, napon 230 kV: rešen zadatak Zadatak prof. dr Andrije T. Sarića. Rešenje: Miloš Sarić
Analiza elektroenergetskih sistema , TF Čačak
Zadatak U sistemu na slici elektrana A je regulaciona i održava napon na sabirnicama 1 od U 1 = 230 U_1 = 230 U 1 = 230 Gauss (-Seidel) -ove metode za proračun tokova snaga u dve iteracije.
Pitanje za razmišljanje: Kako bi se računalo kada reaktivna snaga Q 3 Q_3 Q 3 A 1 L 13 = 100 km 3 B S G3 = (130 + j60) MVA L 12 = 80 km 2 L 23 = 90 km S P2 = (120 + j60) MVA z v = (0,1 + j0,4) Ω/km y 0 v = (0 + j2,8)·10 −6 S/km Poznato U 1 = 230 U_1 = 230 U 1 = 230 kV L 12 = 80 L_{12} = 80 L 12 = 80 km L 13 = 100 L_{13} = 100 L 13 = 100 km L 23 = 90 L_{23} = 90 L 23 = 90 km r v = 0,1 r_v = 0{,}1 r v = 0 , 1 Ω/km x v = 0,4 x_v = 0{,}4 x v = 0 , 4 Ω/km b v 0 = 2,8 b_v^0 = 2{,}8 b v 0 = 2 , 8 ·10⁻⁶ S/km P P 2 = 120 P_{P2} = 120 P P 2 = 120 MW Q P 2 = 60 Q_{P2} = 60 Q P 2 = 60 MVAr P G 3 = 130 P_{G3} = 130 P G 3 = 130 MW Q G 3 = 60 Q_{G3} = 60 Q G 3 = 60 MVAr Rešenje Šta se traži i kojom metodom Traže se naponi sabirnica 2 i 3 posle dve iteracije Gauss-Seidel -ove metode. Sabirnica 1 je referentno-balansna: elektrana A drži njen napon na 230 kV, a njen fazni ugao je referentni, 0°. Sabirnica 2 je potrošačka, a sabirnica 3 generatorska sa zadatim i aktivnim i reaktivnim snagama, pa su obe PQ-čvorovi i za obe se računa napon.
Računa se kao u prethodnom zadatku, bez svođenja na relativne jedinice: naponi u kV (linijski), admitanse u S, snage u MVA. Tada su i ( P i − j Q i ) / U ‾ i ∗ (P_i - \mathrm{j}Q_i)/\ph{U}_i^* ( P i − j Q i ) / U i ∗ Y ‾ i j U ‾ j \ph{Y}_{ij}\ph{U}_j Y ij U j 3 \sqrt{3} 3 Gauss-Seidel -ova formula važi sa linijskim naponima. Za početne vrednosti uzima se ravan profil, U ‾ 2 ( 0 ) = U ‾ 3 ( 0 ) = 230 \ph{U}_2^{(0)} = \ph{U}_3^{(0)} = 230 U 2 ( 0 ) = U 3 ( 0 ) = 230 / 0 ∘ ‾ \underline{/0^\circ} / 0 ∘
Parametri vodova Y ‾ 12 g = 1 z ‾ v L 12 = 1 8 + j 32 = ( 0,007353 − j 0,02941 ) S Y ‾ 13 g = 1 z ‾ v L 13 = 1 10 + j 40 = ( 0,005882 − j 0,02353 ) S Y ‾ 23 g = 1 z ‾ v L 23 = 1 9 + j 36 = ( 0,006536 − j 0,02614 ) S \begin{aligned} \ph{Y}_{12}^g &= \frac{1}{\ph{z}_vL_{12}} = \frac{1}{8 + \mathrm{j}32} = (0{,}007353 - \mathrm{j}0{,}02941)\ \text{S} \\ \ph{Y}_{13}^g &= \frac{1}{\ph{z}_vL_{13}} = \frac{1}{10 + \mathrm{j}40} = (0{,}005882 - \mathrm{j}0{,}02353)\ \text{S} \\ \ph{Y}_{23}^g &= \frac{1}{\ph{z}_vL_{23}} = \frac{1}{9 + \mathrm{j}36} = (0{,}006536 - \mathrm{j}0{,}02614)\ \text{S} \end{aligned} Y 12 g Y 13 g Y 23 g = z v L 12 1 = 8 + j 32 1 = ( 0 , 007353 − j 0 , 02941 ) S = z v L 13 1 = 10 + j 40 1 = ( 0 , 005882 − j 0 , 02353 ) S = z v L 23 1 = 9 + j 36 1 = ( 0 , 006536 − j 0 , 02614 ) S Otočne admitanse, po polovina na svakom kraju voda:
Y ‾ 12 g 0 2 = y ‾ v 0 2 L 12 = j 2,8 ⋅ 10 − 6 2 ⋅ 80 = j 1,12 ⋅ 10 − 4 S Y ‾ 13 g 0 2 = j 2,8 ⋅ 10 − 6 2 ⋅ 100 = j 1,4 ⋅ 10 − 4 S Y ‾ 23 g 0 2 = j 2,8 ⋅ 10 − 6 2 ⋅ 90 = j 1,26 ⋅ 10 − 4 S \begin{aligned} \frac{\ph{Y}_{12}^{g0}}{2} &= \frac{\ph{y}_v^0}{2}L_{12} = \mathrm{j}\frac{2{,}8 \cdot 10^{-6}}{2} \cdot 80 = \mathrm{j}1{,}12 \cdot 10^{-4}\ \text{S} \\ \frac{\ph{Y}_{13}^{g0}}{2} &= \mathrm{j}\frac{2{,}8 \cdot 10^{-6}}{2} \cdot 100 = \mathrm{j}1{,}4 \cdot 10^{-4}\ \text{S} \\ \frac{\ph{Y}_{23}^{g0}}{2} &= \mathrm{j}\frac{2{,}8 \cdot 10^{-6}}{2} \cdot 90 = \mathrm{j}1{,}26 \cdot 10^{-4}\ \text{S} \end{aligned} 2 Y 12 g 0 2 Y 13 g 0 2 Y 23 g 0 = 2 y v 0 L 12 = j 2 2 , 8 ⋅ 1 0 − 6 ⋅ 80 = j 1 , 12 ⋅ 1 0 − 4 S = j 2 2 , 8 ⋅ 1 0 − 6 ⋅ 100 = j 1 , 4 ⋅ 1 0 − 4 S = j 2 2 , 8 ⋅ 1 0 − 6 ⋅ 90 = j 1 , 26 ⋅ 1 0 − 4 S Matrica admitansi nezavisnih čvorova Dijagonalni element je zbir rednih i otočnih admitansi grana u čvoru, a vandijagonalni negativna redna admitansa grane između dva čvora:
Y ‾ B U S = [ Y ‾ 12 g + Y ‾ 13 g + Y ‾ 12 g 0 2 + Y ‾ 13 g 0 2 − Y ‾ 12 g − Y ‾ 13 g − Y ‾ 12 g Y ‾ 12 g + Y ‾ 23 g + Y ‾ 12 g 0 2 + Y ‾ 23 g 0 2 − Y ‾ 23 g − Y ‾ 13 g − Y ‾ 23 g Y ‾ 13 g + Y ‾ 23 g + Y ‾ 13 g 0 2 + Y ‾ 23 g 0 2 ] \ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} \ph{Y}_{12}^g + \ph{Y}_{13}^g + \dfrac{\ph{Y}_{12}^{g0}}{2} + \dfrac{\ph{Y}_{13}^{g0}}{2} & -\ph{Y}_{12}^g & -\ph{Y}_{13}^g \\[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] -\ph{Y}_{13}^g & -\ph{Y}_{23}^g & \ph{Y}_{13}^g + \ph{Y}_{23}^g + \dfrac{\ph{Y}_{13}^{g0}}{2} + \dfrac{\ph{Y}_{23}^{g0}}{2} \end{bmatrix} Y B U S = Y 12 g + Y 13 g + 2 Y 12 g 0 + 2 Y 13 g 0 − Y 12 g − Y 13 g − Y 12 g Y 12 g + Y 23 g + 2 Y 12 g 0 + 2 Y 23 g 0 − Y 23 g − Y 13 g − Y 23 g Y 13 g + Y 23 g + 2 Y 13 g 0 + 2 Y 23 g 0 Brojno, u S:
Y ‾ 11 = 0,01324 − j 0,05269 , Y ‾ 12 = − 0,007353 + j 0,02941 , Y ‾ 13 = − 0,005882 + j 0,02353 , Y ‾ 22 = 0,01389 − j 0,05532 , Y ‾ 23 = − 0,006536 + j 0,02614 , Y ‾ 33 = 0,01242 − j 0,04941 \begin{aligned} \ph{Y}_{11} &= 0{,}01324 - \mathrm{j}0{,}05269, \quad \ph{Y}_{12} = -0{,}007353 + \mathrm{j}0{,}02941, \\ \ph{Y}_{13} &= -0{,}005882 + \mathrm{j}0{,}02353, \quad \ph{Y}_{22} = 0{,}01389 - \mathrm{j}0{,}05532, \\ \ph{Y}_{23} &= -0{,}006536 + \mathrm{j}0{,}02614, \quad \ph{Y}_{33} = 0{,}01242 - \mathrm{j}0{,}04941 \end{aligned} Y 11 Y 13 Y 23 = 0 , 01324 − j 0 , 05269 , Y 12 = − 0 , 007353 + j 0 , 02941 , = − 0 , 005882 + j 0 , 02353 , Y 22 = 0 , 01389 − j 0 , 05532 , = − 0 , 006536 + j 0 , 02614 , Y 33 = 0 , 01242 − j 0 , 04941 Gauss-Seidel-ov izraz Napon svakog PQ-čvora računa se iz njegove jednačine injektiranja, sa najnovijim naponima ostalih čvorova; za čvor 2 se snaga uzima sa minusom, jer je čvor potrošački, a za čvor 3 sa plusom, jer elektrana B daje snagu u mrežu:
U ‾ 2 ( k + 1 ) = 1 Y ‾ 22 [ − P P 2 − j Q P 2 U ‾ 2 ( k ) ∗ − Y ‾ 21 U ‾ 1 − Y ‾ 23 U ‾ 3 ( k ) ] U ‾ 3 ( k + 1 ) = 1 Y ‾ 33 [ P G 3 − j Q G 3 U ‾ 3 ( k ) ∗ − Y ‾ 31 U ‾ 1 − Y ‾ 32 U ‾ 2 ( k + 1 ) ] \begin{aligned} \ph{U}_2^{(k+1)} &= \frac{1}{\ph{Y}_{22}}\left[-\frac{P_{P2} - \mathrm{j}Q_{P2}}{\ph{U}_2^{(k)*}} - \ph{Y}_{21}\ph{U}_1 - \ph{Y}_{23}\ph{U}_3^{(k)}\right] \\ \ph{U}_3^{(k+1)} &= \frac{1}{\ph{Y}_{33}}\left[\frac{P_{G3} - \mathrm{j}Q_{G3}}{\ph{U}_3^{(k)*}} - \ph{Y}_{31}\ph{U}_1 - \ph{Y}_{32}\ph{U}_2^{(k+1)}\right] \end{aligned} U 2 ( k + 1 ) U 3 ( k + 1 ) = Y 22 1 [ − U 2 ( k ) ∗ P P 2 − j Q P 2 − Y 21 U 1 − Y 23 U 3 ( k ) ] = Y 33 1 [ U 3 ( k ) ∗ P G 3 − j Q G 3 − Y 31 U 1 − Y 32 U 2 ( k + 1 ) ] U drugom izrazu stoji već izračunati U ‾ 2 ( k + 1 ) \ph{U}_2^{(k+1)} U 2 ( k + 1 ) Gauss-Seidel -ove metode u odnosu na Gauss -ovu, koja bi uzela U ‾ 2 ( k ) \ph{U}_2^{(k)} U 2 ( k )
1. iteracija Članovi u zagradi za U ‾ 2 ( 1 ) \ph{U}_2^{(1)} U 2 ( 1 )
− 120 − j 60 230 = − 0,5217 + j 0,2609 Y ‾ 21 U ‾ 1 = − 1,691 + j 6,765 Y ‾ 23 U ‾ 3 ( 0 ) = − 1,503 + j 6,013 \begin{aligned} -\frac{120 - \mathrm{j}60}{230} &= -0{,}5217 + \mathrm{j}0{,}2609 \\ \ph{Y}_{21}\ph{U}_1 &= -1{,}691 + \mathrm{j}6{,}765 \\ \ph{Y}_{23}\ph{U}_3^{(0)} &= -1{,}503 + \mathrm{j}6{,}013 \end{aligned} − 230 120 − j 60 Y 21 U 1 Y 23 U 3 ( 0 ) = − 0 , 5217 + j 0 , 2609 = − 1 , 691 + j 6 , 765 = − 1 , 503 + j 6 , 013 U ‾ 2 ( 1 ) = ( − 0,5217 + j 0,2609 ) − ( − 1,691 + j 6,765 ) − ( − 1,503 + j 6,013 ) 0,01389 − j 0,05532 = ( 224,267 − j 7,992 ) kV = 224,409 ∠ − 2,041 ∘ kV \begin{aligned} \ph{U}_2^{(1)} &= \frac{(-0{,}5217 + \mathrm{j}0{,}2609) - (-1{,}691 + \mathrm{j}6{,}765) - (-1{,}503 + \mathrm{j}6{,}013)}{0{,}01389 - \mathrm{j}0{,}05532} \\ &= (224{,}267 - \mathrm{j}7{,}992)\ \text{kV} = 224{,}409\angle{-2{,}041^\circ}\ \text{kV} \end{aligned} U 2 ( 1 ) = 0 , 01389 − j 0 , 05532 ( − 0 , 5217 + j 0 , 2609 ) − ( − 1 , 691 + j 6 , 765 ) − ( − 1 , 503 + j 6 , 013 ) = ( 224 , 267 − j 7 , 992 ) kV = 224 , 409∠ − 2 , 04 1 ∘ kV Za U ‾ 3 ( 1 ) \ph{U}_3^{(1)} U 3 ( 1 ) U ‾ 2 ( 1 ) \ph{U}_2^{(1)} U 2 ( 1 )
130 − j 60 230 = 0,5652 − j 0,2609 Y ‾ 31 U ‾ 1 = − 1,353 + j 5,412 Y ‾ 32 U ‾ 2 ( 1 ) = − 1,257 + j 5,915 \begin{aligned} \frac{130 - \mathrm{j}60}{230} &= 0{,}5652 - \mathrm{j}0{,}2609 \\ \ph{Y}_{31}\ph{U}_1 &= -1{,}353 + \mathrm{j}5{,}412 \\ \ph{Y}_{32}\ph{U}_2^{(1)} &= -1{,}257 + \mathrm{j}5{,}915 \end{aligned} 230 130 − j 60 Y 31 U 1 Y 32 U 2 ( 1 ) = 0 , 5652 − j 0 , 2609 = − 1 , 353 + j 5 , 412 = − 1 , 257 + j 5 , 915 U ‾ 3 ( 1 ) = ( 0,5652 − j 0,2609 ) − ( − 1,353 + j 5,412 ) − ( − 1,257 + j 5,915 ) 0,01242 − j 0,04941 = ( 235,797 + j 4,995 ) kV = 235,850 ∠ 1,214 ∘ kV \begin{aligned} \ph{U}_3^{(1)} &= \frac{(0{,}5652 - \mathrm{j}0{,}2609) - (-1{,}353 + \mathrm{j}5{,}412) - (-1{,}257 + \mathrm{j}5{,}915)}{0{,}01242 - \mathrm{j}0{,}04941} \\ &= (235{,}797 + \mathrm{j}4{,}995)\ \text{kV} = 235{,}850\angle{1{,}214^\circ}\ \text{kV} \end{aligned} U 3 ( 1 ) = 0 , 01242 − j 0 , 04941 ( 0 , 5652 − j 0 , 2609 ) − ( − 1 , 353 + j 5 , 412 ) − ( − 1 , 257 + j 5 , 915 ) = ( 235 , 797 + j 4 , 995 ) kV = 235 , 850∠ 1 , 21 4 ∘ kV 2. iteracija Isti izrazi, sa U ‾ 2 ( 1 ) \ph{U}_2^{(1)} U 2 ( 1 ) U ‾ 3 ( 1 ) \ph{U}_3^{(1)} U 3 ( 1 ) Y ‾ 21 U ‾ 1 \ph{Y}_{21}\ph{U}_1 Y 21 U 1
− 120 − j 60 U ‾ 2 ( 1 ) ∗ = − 0,5249 + j 0,2862 Y ‾ 23 U ‾ 3 ( 1 ) = − 1,672 + j 6,132 U ‾ 2 ( 2 ) = ( 226,564 − j 5,580 ) kV = 226,633 ∠ − 1,411 ∘ kV \begin{aligned} -\frac{120 - \mathrm{j}60}{\ph{U}_2^{(1)*}} &= -0{,}5249 + \mathrm{j}0{,}2862 \\ \ph{Y}_{23}\ph{U}_3^{(1)} &= -1{,}672 + \mathrm{j}6{,}132 \\ \ph{U}_2^{(2)} &= (226{,}564 - \mathrm{j}5{,}580)\ \text{kV} = 226{,}633\angle{-1{,}411^\circ}\ \text{kV} \end{aligned} − U 2 ( 1 ) ∗ 120 − j 60 Y 23 U 3 ( 1 ) U 2 ( 2 ) = − 0 , 5249 + j 0 , 2862 = − 1 , 672 + j 6 , 132 = ( 226 , 564 − j 5 , 580 ) kV = 226 , 633∠ − 1 , 41 1 ∘ kV 130 − j 60 U ‾ 3 ( 1 ) ∗ = 0,5565 − j 0,2427 Y ‾ 32 U ‾ 2 ( 2 ) = − 1,335 + j 5,96 U ‾ 3 ( 2 ) = ( 236,626 + j 6,190 ) kV = 236,707 ∠ 1,499 ∘ kV \begin{aligned} \frac{130 - \mathrm{j}60}{\ph{U}_3^{(1)*}} &= 0{,}5565 - \mathrm{j}0{,}2427 \\ \ph{Y}_{32}\ph{U}_2^{(2)} &= -1{,}335 + \mathrm{j}5{,}96 \\ \ph{U}_3^{(2)} &= (236{,}626 + \mathrm{j}6{,}190)\ \text{kV} = 236{,}707\angle{1{,}499^\circ}\ \text{kV} \end{aligned} U 3 ( 1 ) ∗ 130 − j 60 Y 32 U 2 ( 2 ) U 3 ( 2 ) = 0 , 5565 − j 0 , 2427 = − 1 , 335 + j 5 , 96 = ( 236 , 626 + j 6 , 190 ) kV = 236 , 707∠ 1 , 49 9 ∘ kV Kad se iteracije nastave, naponi se ustale na U ‾ 2 = 227,346 ∠ − 1,217 ∘ \ph{U}_2 = 227{,}346\angle{-1{,}217^\circ} U 2 = 227 , 346∠ − 1 , 21 7 ∘ U ‾ 3 = 237,006 ∠ 1,593 ∘ \ph{U}_3 = 237{,}006\angle{1{,}593^\circ} U 3 = 237 , 006∠ 1 , 59 3 ∘ Newton-Raphson -ov metod. Napon čvora 3 je viši od 230 kV jer elektrana B, osim aktivne, daje i 60 MVAr reaktivne snage u mrežu.
Pitanje za razmišljanje Ako Q 3 Q_3 Q 3 P 3 P_3 P 3 U 3 U_3 U 3
Q 3 ( k ) = − I m { U ‾ 3 ( k ) ∗ ( Y ‾ 31 U ‾ 1 + Y ‾ 32 U ‾ 2 ( k + 1 ) + Y ‾ 33 U ‾ 3 ( k ) ) } , \begin{aligned} Q_3^{(k)} = -\mathrm{Im}\big\{\ph{U}_3^{(k)*}\big(&\ph{Y}_{31}\ph{U}_1 + \ph{Y}_{32}\ph{U}_2^{(k+1)} \\ &+ \ph{Y}_{33}\ph{U}_3^{(k)}\big)\big\}, \end{aligned} Q 3 ( k ) = − Im { U 3 ( k ) ∗ ( Y 31 U 1 + Y 32 U 2 ( k + 1 ) + Y 33 U 3 ( k ) ) } , sa njom se izračuna novi napon, a od njega se zadrži samo fazni ugao, dok se modul vrati na zadatu vrednost U 3 U_3 U 3
Rezultat Posle 1. iteracije: U ‾ 2 ( 1 ) = 224,409 ∠ − 2,041 ∘ \ph{U}_2^{(1)} = 224{,}409\angle{-2{,}041^\circ} U 2 ( 1 ) = 224 , 409∠ − 2 , 04 1 ∘ U ‾ 3 ( 1 ) = 235,850 ∠ 1,214 ∘ \ph{U}_3^{(1)} = 235{,}850\angle{1{,}214^\circ} U 3 ( 1 ) = 235 , 850∠ 1 , 21 4 ∘
Posle 2. iteracije: U ‾ 2 ( 2 ) = 226,633 ∠ − 1,411 ∘ \ph{U}_2^{(2)} = 226{,}633\angle{-1{,}411^\circ} U 2 ( 2 ) = 226 , 633∠ − 1 , 41 1 ∘ U ‾ 3 ( 2 ) = 236,707 ∠ 1,499 ∘ \ph{U}_3^{(2)} = 236{,}707\angle{1{,}499^\circ} U 3 ( 2 ) = 236 , 707∠ 1 , 49 9 ∘
← Prethodni zadatak Zadatak 7: Gauss-Seidel u tri iteracije, isključen prekidač Sledeći zadatak → Zadatak 9: Jakobijan Newton-Raphson-a, mreža od četiri čvora