Zadatak Na slici je prikazan EES u kome su admitanse svih vodova jednake i iznose Y ‾ g = ( 1 − j 10 ) \ph{Y}^g = (1 - \mathrm{j}10) Y g = ( 1 − j 10 )
Izračunati elemente matrice Y ‾ B U S \ph{\boldsymbol{Y}}_{BUS} Y B U S Usvajajući čvor 1 za referentno-balansni izvršiti klasifikaciju preostalih čvorova na PQ i PV čvorove. Odrediti dimenzije Jakobijan matrice za proračun tokova snaga Newton-Raphson -ovom metodom. Izračunati vrednosti elemenata ∂ ( Δ P 2 ) ∂ θ 2 \frac{\partial(\Delta P_2)}{\partial \theta_2} ∂ θ 2 ∂ ( Δ P 2 ) ∂ ( Δ Q 3 ) ∂ V 3 \frac{\partial(\Delta Q_3)}{\partial V_3} ∂ V 3 ∂ ( Δ Q 3 ) ∂ ( Δ P 4 ) ∂ V 4 \frac{\partial(\Delta P_4)}{\partial V_4} ∂ V 4 ∂ ( Δ P 4 ) Newton-Raphson -ovom metodom. NAPOMENA: Za početne vrednosti nepoznatih u prvoj iteraciji usvojiti V ( 0 ) = 1 V^{(0)} = 1 V ( 0 ) = 1 θ ( 0 ) = 0 \theta^{(0)} = 0 θ ( 0 ) = 0 Flat start ). 1 V 1 = 1,05 r.j. θ 1 = 0° 3 S P3 = (0,4 + j0,05) r.j. 2 S P2 = (0,6 + j0,2) r.j. S G2 = (1,1 + jQ G2 ) r.j. V 2 = 1,05 r.j. 4 S P4 = (0,6 + j0,2) r.j. Y g Y g Y g Y g Y g = (1 − j10) r.j. Poznato G g = 1 G^g = 1 G g = 1 r.j. − B g = 10 -B^g = 10 − B g = 10 r.j. V 1 = 1,05 V_1 = 1{,}05 V 1 = 1 , 05 r.j. V 2 = 1,05 V_2 = 1{,}05 V 2 = 1 , 05 r.j. P G 2 = 1,1 P_{G2} = 1{,}1 P G 2 = 1 , 1 r.j. P P 2 = 0,6 P_{P2} = 0{,}6 P P 2 = 0 , 6 r.j. Q P 2 = 0,2 Q_{P2} = 0{,}2 Q P 2 = 0 , 2 r.j. P P 3 = 0,4 P_{P3} = 0{,}4 P P 3 = 0 , 4 r.j. Q P 3 = 0,05 Q_{P3} = 0{,}05 Q P 3 = 0 , 05 r.j. P P 4 = 0,6 P_{P4} = 0{,}6 P P 4 = 0 , 6 r.j. Q P 4 = 0,2 Q_{P4} = 0{,}2 Q P 4 = 0 , 2 r.j. Rešenje a.
Y ‾ B U S = G B U S + j B B U S = [ 2 − 1 − 1 0 − 1 3 − 1 − 1 − 1 − 1 3 − 1 0 − 1 − 1 2 ] + j [ − 20 10 10 0 10 − 30 10 10 10 10 − 30 10 0 10 10 − 20 ] \begin{gathered} \ph{\boldsymbol{Y}}_{BUS} = \boldsymbol{G}_{BUS} + \mathrm{j} \boldsymbol{B}_{BUS} \\ = \begin{bmatrix} 2 & -1 & -1 & 0 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ 0 & -1 & -1 & 2 \end{bmatrix} \\ + \mathrm{j} \begin{bmatrix} -20 & 10 & 10 & 0 \\ 10 & -30 & 10 & 10 \\ 10 & 10 & -30 & 10 \\ 0 & 10 & 10 & -20 \end{bmatrix} \end{gathered} Y B U S = G B U S + j B B U S = 2 − 1 − 1 0 − 1 3 − 1 − 1 − 1 − 1 3 − 1 0 − 1 − 1 2 + j − 20 10 10 0 10 − 30 10 10 10 10 − 30 10 0 10 10 − 20 b.
Čvor 1: V 1 = 1,05 V_1 = 1{,}05 V 1 = 1 , 05 θ 1 = 0 \theta_1 = 0 θ 1 = 0 P 1 = ? P_1 = ? P 1 = ? Q 1 = ? Q_1 = ? Q 1 = ? SL čvor Čvor 2: V 2 = 1,05 V_2 = 1{,}05 V 2 = 1 , 05 P 2 = P G 2 − P p 2 = 0,5 P_2 = P_{G2} - P_{p2} = 0{,}5 P 2 = P G 2 − P p 2 = 0 , 5 Q 2 = ? Q_2 = ? Q 2 = ? θ 2 = ? \theta_2 = ? θ 2 = ? PV čvor Čvor 3: P 3 = − P p 3 = − 0,4 P_3 = -P_{p3} = -0{,}4 P 3 = − P p 3 = − 0 , 4 Q 3 = − 0,05 Q_3 = -0{,}05 Q 3 = − 0 , 05 V 3 = ? V_3 = ? V 3 = ? θ 3 = ? \theta_3 = ? θ 3 = ? PQ čvor Čvor 4: P 4 = − P p 4 = − 0,6 P_4 = -P_{p4} = -0{,}6 P 4 = − P p 4 = − 0 , 6 Q 4 = − 0,2 Q_4 = -0{,}2 Q 4 = − 0 , 2 V 4 = ? V_4 = ? V 4 = ? θ 4 = ? \theta_4 = ? θ 4 = ? PQ čvor
c.
n = 4 , 2 N P Q + N P V = 2 ⋅ 2 + 1 = 5 n = 4 , \quad 2 N_{PQ} + N_{PV} = 2 \cdot 2 + 1 = 5 n = 4 , 2 N P Q + N P V = 2 ⋅ 2 + 1 = 5 Znači, Jacobian matrica je petog reda.
J = [ ∂ ( Δ P 2 ) ∂ θ 2 ∂ ( Δ P 2 ) ∂ θ 3 ∂ ( Δ P 2 ) ∂ θ 4 ∂ ( Δ P 2 ) ∂ V 3 ∂ ( Δ P 2 ) ∂ V 4 ∂ ( Δ P 3 ) ∂ θ 2 ∂ ( Δ P 3 ) ∂ θ 3 ∂ ( Δ P 3 ) ∂ θ 4 ∂ ( Δ P 3 ) ∂ V 3 ∂ ( Δ P 3 ) ∂ V 4 ∂ ( Δ P 4 ) ∂ θ 2 ∂ ( Δ P 4 ) ∂ θ 3 ∂ ( Δ P 4 ) ∂ θ 4 ∂ ( Δ P 4 ) ∂ V 3 ∂ ( Δ P 4 ) ∂ V 4 ∂ ( Δ Q 3 ) ∂ θ 2 ∂ ( Δ Q 3 ) ∂ θ 3 ∂ ( Δ Q 3 ) ∂ θ 4 ∂ ( Δ Q 3 ) ∂ V 3 ∂ ( Δ Q 3 ) ∂ V 4 ∂ ( Δ Q 4 ) ∂ θ 2 ∂ ( Δ Q 4 ) ∂ θ 3 ∂ ( Δ Q 4 ) ∂ θ 4 ∂ ( Δ Q 4 ) ∂ V 3 ∂ ( Δ Q 4 ) ∂ V 4 ] K L ′ H F ′ J = \overset{\boldsymbol{H} \hspace{12em} \boldsymbol{F}'}{\underset{\boldsymbol{K} \hspace{12em} \boldsymbol{L}'}{\left[ \begin{array}{ccc:cc} \frac{\partial(\Delta P_2)}{\partial \theta_2} & \frac{\partial(\Delta P_2)}{\partial \theta_3} & \frac{\partial(\Delta P_2)}{\partial \theta_4} & \frac{\partial(\Delta P_2)}{\partial V_3} & \frac{\partial(\Delta P_2)}{\partial V_4} \\[1ex] \frac{\partial(\Delta P_3)}{\partial \theta_2} & \frac{\partial(\Delta P_3)}{\partial \theta_3} & \frac{\partial(\Delta P_3)}{\partial \theta_4} & \frac{\partial(\Delta P_3)}{\partial V_3} & \frac{\partial(\Delta P_3)}{\partial V_4} \\[1ex] \frac{\partial(\Delta P_4)}{\partial \theta_2} & \frac{\partial(\Delta P_4)}{\partial \theta_3} & \frac{\partial(\Delta P_4)}{\partial \theta_4} & \frac{\partial(\Delta P_4)}{\partial V_3} & \frac{\partial(\Delta P_4)}{\partial V_4} \\[1ex] \hdashline \frac{\partial(\Delta Q_3)}{\partial \theta_2} & \frac{\partial(\Delta Q_3)}{\partial \theta_3} & \frac{\partial(\Delta Q_3)}{\partial \theta_4} & \frac{\partial(\Delta Q_3)}{\partial V_3} & \frac{\partial(\Delta Q_3)}{\partial V_4} \\[1ex] \frac{\partial(\Delta Q_4)}{\partial \theta_2} & \frac{\partial(\Delta Q_4)}{\partial \theta_3} & \frac{\partial(\Delta Q_4)}{\partial \theta_4} & \frac{\partial(\Delta Q_4)}{\partial V_3} & \frac{\partial(\Delta Q_4)}{\partial V_4} \end{array} \right]}} J = K L ′ ∂ θ 2 ∂ ( Δ P 2 ) ∂ θ 2 ∂ ( Δ P 3 ) ∂ θ 2 ∂ ( Δ P 4 ) ∂ θ 2 ∂ ( Δ Q 3 ) ∂ θ 2 ∂ ( Δ Q 4 ) ∂ θ 3 ∂ ( Δ P 2 ) ∂ θ 3 ∂ ( Δ P 3 ) ∂ θ 3 ∂ ( Δ P 4 ) ∂ θ 3 ∂ ( Δ Q 3 ) ∂ θ 3 ∂ ( Δ Q 4 ) ∂ θ 4 ∂ ( Δ P 2 ) ∂ θ 4 ∂ ( Δ P 3 ) ∂ θ 4 ∂ ( Δ P 4 ) ∂ θ 4 ∂ ( Δ Q 3 ) ∂ θ 4 ∂ ( Δ Q 4 ) ∂ V 3 ∂ ( Δ P 2 ) ∂ V 3 ∂ ( Δ P 3 ) ∂ V 3 ∂ ( Δ P 4 ) ∂ V 3 ∂ ( Δ Q 3 ) ∂ V 3 ∂ ( Δ Q 4 ) ∂ V 4 ∂ ( Δ P 2 ) ∂ V 4 ∂ ( Δ P 3 ) ∂ V 4 ∂ ( Δ P 4 ) ∂ V 4 ∂ ( Δ Q 3 ) ∂ V 4 ∂ ( Δ Q 4 ) H F ′ d.
P i = V i 2 G i i + V i ∑ j ∈ α i [ V j ( G i j cos ( θ i − θ j ) + B i j sin ( θ i − θ j ) ) ] \begin{aligned} P_i = V_i^2 G_{ii} &+ V_i \sum_{j \in \alpha_i} \Big[ V_j \big( G_{ij} \cos(\theta_i - \theta_j) \\ &+ B_{ij} \sin(\theta_i - \theta_j) \big) \Big] \end{aligned} P i = V i 2 G ii + V i j ∈ α i ∑ [ V j ( G ij cos ( θ i − θ j ) + B ij sin ( θ i − θ j ) ) ] J 11 = ∂ ( Δ P 2 ) ∂ θ 2 = V 2 ∑ j = 1 j ≠ 2 4 V j [ − G 2 j sin ( θ 2 − θ j ) + B 2 j cos ( θ 2 − θ j ) ] \begin{aligned} J_{11} &= \frac{\partial(\Delta P_2)}{\partial \theta_2} \\ &= V_2 \sum_{\substack{j = 1 \\ j \neq 2}}^{4} V_j \Big[ -G_{2j} \sin(\theta_2 - \theta_j) \\ &\qquad + B_{2j} \cos(\theta_2 - \theta_j) \Big] \end{aligned} J 11 = ∂ θ 2 ∂ ( Δ P 2 ) = V 2 j = 1 j = 2 ∑ 4 V j [ − G 2 j sin ( θ 2 − θ j ) + B 2 j cos ( θ 2 − θ j ) ] J 11 ( 0 ) = V 2 V 1 [ − G 21 sin ( θ 2 ( 0 ) − θ 1 0 ) + B 21 cos ( θ 2 ( 0 ) − θ 1 ) ] + V 2 V 3 ( 0 ) [ − G 23 sin ( θ 2 ( 0 ) − θ 3 ( 0 ) ) + B 23 cos ( θ 2 ( 0 ) − θ 3 ( 0 ) ) ] + V 2 V 4 ( 0 ) [ − G 24 sin ( θ 2 ( 0 ) − θ 4 ( 0 ) ) + B 24 cos ( θ 2 ( 0 ) − θ 4 ( 0 ) ) ] \begin{aligned} J_{11}^{(0)} = {} &V_2 V_1 \Big[ -G_{21} \sin(\overset{0}{\cancel{\theta_2^{(0)} - \theta_1}}) \\ &\qquad + B_{21} \cos(\cancel{\theta_2^{(0)} - \theta_1}) \Big] \\ &+ V_2 V_3^{(0)} \Big[ -G_{23} \sin(\cancel{\theta_2^{(0)} - \theta_3^{(0)}}) \\ &\qquad + B_{23} \cos(\cancel{\theta_2^{(0)} - \theta_3^{(0)}}) \Big] \\ &+ V_2 V_4^{(0)} \Big[ -G_{24} \sin(\cancel{\theta_2^{(0)} - \theta_4^{(0)}}) \\ &\qquad + B_{24} \cos(\cancel{\theta_2^{(0)} - \theta_4^{(0)}}) \Big] \end{aligned} J 11 ( 0 ) = V 2 V 1 [ − G 21 sin ( θ 2 ( 0 ) − θ 1 0 ) + B 21 cos ( θ 2 ( 0 ) − θ 1 ) ] + V 2 V 3 ( 0 ) [ − G 23 sin ( θ 2 ( 0 ) − θ 3 ( 0 ) ) + B 23 cos ( θ 2 ( 0 ) − θ 3 ( 0 ) ) ] + V 2 V 4 ( 0 ) [ − G 24 sin ( θ 2 ( 0 ) − θ 4 ( 0 ) ) + B 24 cos ( θ 2 ( 0 ) − θ 4 ( 0 ) ) ] J 11 ( 0 ) = V 2 V 1 B 21 + V 2 V 3 ( 0 ) B 23 + V 2 V 4 ( 0 ) B 24 = 1,05 ⋅ 1,05 ⋅ 10 + 1,05 ⋅ 1 ⋅ 10 + 1,05 ⋅ 1 ⋅ 10 = 32,025 \begin{aligned} J_{11}^{(0)} &= V_2 V_1 B_{21} + V_2 V_3^{(0)} B_{23} \\ &\quad + V_2 V_4^{(0)} B_{24} \\ &= 1{,}05 \cdot 1{,}05 \cdot 10 + 1{,}05 \cdot 1 \cdot 10 \\ &\quad + 1{,}05 \cdot 1 \cdot 10 = 32{,}025 \end{aligned} J 11 ( 0 ) = V 2 V 1 B 21 + V 2 V 3 ( 0 ) B 23 + V 2 V 4 ( 0 ) B 24 = 1 , 05 ⋅ 1 , 05 ⋅ 10 + 1 , 05 ⋅ 1 ⋅ 10 + 1 , 05 ⋅ 1 ⋅ 10 = 32 , 025 Q i = − V i 2 B i i + V i ∑ j ∈ α i [ V j ( G i j sin ( θ i − θ j ) − B i j cos ( θ i − θ j ) ) ] \begin{aligned} Q_i = -V_i^2 B_{ii} &+ V_i \sum_{j \in \alpha_i} \Big[ V_j \big( G_{ij} \sin(\theta_i - \theta_j) \\ &- B_{ij} \cos(\theta_i - \theta_j) \big) \Big] \end{aligned} Q i = − V i 2 B ii + V i j ∈ α i ∑ [ V j ( G ij sin ( θ i − θ j ) − B ij cos ( θ i − θ j ) ) ] J 44 = ∂ ( Δ Q 3 ) ∂ V 3 = − 2 B 33 V 3 + ∑ j = 1 j ≠ 3 4 V j [ G 3 j sin ( θ 3 − θ j ) − B 3 j cos ( θ 3 − θ j ) ] \begin{aligned} J_{44} &= \frac{\partial(\Delta Q_3)}{\partial V_3} \\ &= -2 B_{33} V_3 + \sum_{\substack{j = 1 \\ j \neq 3}}^{4} V_j \Big[ G_{3j} \sin(\theta_3 - \theta_j) \\ &\qquad - B_{3j} \cos(\theta_3 - \theta_j) \Big] \end{aligned} J 44 = ∂ V 3 ∂ ( Δ Q 3 ) = − 2 B 33 V 3 + j = 1 j = 3 ∑ 4 V j [ G 3 j sin ( θ 3 − θ j ) − B 3 j cos ( θ 3 − θ j ) ] J 44 ( 0 ) = − 2 B 33 V 3 ( 0 ) + V 1 [ G 31 sin ( θ 3 ( 0 ) − θ 1 0 ) − B 31 cos ( θ 3 ( 0 ) − θ 1 ) ] + V 2 [ G 32 sin ( θ 3 ( 0 ) − θ 2 ( 0 ) ) − B 32 cos ( θ 3 ( 0 ) − θ 2 ( 0 ) ) ] + V 4 ( 0 ) [ G 34 sin ( θ 3 ( 0 ) − θ 4 ( 0 ) ) − B 34 cos ( θ 3 ( 0 ) − θ 4 ( 0 ) ) ] \begin{aligned} J_{44}^{(0)} = {} &-2 B_{33} V_3^{(0)} \\ &+ V_1 \Big[ G_{31} \sin(\overset{0}{\cancel{\theta_3^{(0)} - \theta_1}}) \\ &\qquad - B_{31} \cos(\cancel{\theta_3^{(0)} - \theta_1}) \Big] \\ &+ V_2 \Big[ G_{32} \sin(\cancel{\theta_3^{(0)} - \theta_2^{(0)}}) \\ &\qquad - B_{32} \cos(\cancel{\theta_3^{(0)} - \theta_2^{(0)}}) \Big] \\ &+ V_4^{(0)} \Big[ G_{34} \sin(\cancel{\theta_3^{(0)} - \theta_4^{(0)}}) \\ &\qquad - B_{34} \cos(\cancel{\theta_3^{(0)} - \theta_4^{(0)}}) \Big] \end{aligned} J 44 ( 0 ) = − 2 B 33 V 3 ( 0 ) + V 1 [ G 31 sin ( θ 3 ( 0 ) − θ 1 0 ) − B 31 cos ( θ 3 ( 0 ) − θ 1 ) ] + V 2 [ G 32 sin ( θ 3 ( 0 ) − θ 2 ( 0 ) ) − B 32 cos ( θ 3 ( 0 ) − θ 2 ( 0 ) ) ] + V 4 ( 0 ) [ G 34 sin ( θ 3 ( 0 ) − θ 4 ( 0 ) ) − B 34 cos ( θ 3 ( 0 ) − θ 4 ( 0 ) ) ] J 44 ( 0 ) = − 2 B 33 V 3 ( 0 ) − V 1 B 31 − V 2 B 32 − V 4 ( 0 ) B 34 = 29 \begin{aligned} J_{44}^{(0)} &= -2 B_{33} V_3^{(0)} - V_1 B_{31} \\ &\quad - V_2 B_{32} - V_4^{(0)} B_{34} \\ &= 29 \end{aligned} J 44 ( 0 ) = − 2 B 33 V 3 ( 0 ) − V 1 B 31 − V 2 B 32 − V 4 ( 0 ) B 34 = 29 Za domaći: Uraditi J 35 = ∂ ( Δ P 4 ) ∂ V 4 J_{35} = \frac{\partial(\Delta P_4)}{\partial V_4} J 35 = ∂ V 4 ∂ ( Δ P 4 ) Iterativni postupak proračuna:
1. Inicijalne vrednosti promenljivih stanja X ( 0 ) \boldsymbol{X}^{(0)} X ( 0 )
2. Vrši se popravka rešenja:
Δ X ( k + 1 ) = − [ J ( k ) ] − 1 F ( X ( k ) ) , F ( X ( k ) ) = P i , Q i ( X ( k ) ) − P i s p , Q i s p \begin{gathered} \Delta \boldsymbol{X}^{(k+1)} = -\left[ \boldsymbol{J}^{(k)} \right]^{-1} \boldsymbol{F}\left( \boldsymbol{X}^{(k)} \right) , \\ \boldsymbol{F}\left( \boldsymbol{X}^{(k)} \right) = P_i, Q_i\left( \boldsymbol{X}^{(k)} \right) - P_i^{sp}, Q_i^{sp} \end{gathered} Δ X ( k + 1 ) = − [ J ( k ) ] − 1 F ( X ( k ) ) , F ( X ( k ) ) = P i , Q i ( X ( k ) ) − P i s p , Q i s p 3. koja koriguje rešenje:
X ( k + 1 ) = X ( k ) + Δ X ( k + 1 ) , \boldsymbol{X}^{(k+1)} = \boldsymbol{X}^{(k)} + \Delta \boldsymbol{X}^{(k+1)} , X ( k + 1 ) = X ( k ) + Δ X ( k + 1 ) , 4. posle čega se vrši provera kriterijuma konvergencije:
∣ X ( k + 1 ) − X ( k ) ∣ ≤ ε . \left| \boldsymbol{X}^{(k+1)} - \boldsymbol{X}^{(k)} \right| \le \varepsilon . X ( k + 1 ) − X ( k ) ≤ ε . Rezultat Čvor 2 je PV čvor (P 2 = 0,5 P_2 = 0{,}5 P 2 = 0 , 5
U prvoj iteraciji iz ravnog profila: ∂ ( Δ P 2 ) ∂ θ 2 = 32,025 \frac{\partial(\Delta P_2)}{\partial \theta_2} = 32{,}025 ∂ θ 2 ∂ ( Δ P 2 ) = 32 , 025 ∂ ( Δ Q 3 ) ∂ V 3 = 29 \frac{\partial(\Delta Q_3)}{\partial V_3} = 29 ∂ V 3 ∂ ( Δ Q 3 ) = 29
← Prethodni zadatak Zadatak 8: Gauss-Seidel metoda, dve iteracije, napon 230 kV Sledeći zadatak → Zadatak 10: Stott-ov raspregnuti postupak, jedna iteracija