Zadatak Za elektroenergetski sistem prikazan na slici:
a) Izvršiti klasifikaciju čvorova i napisati jednačine za proračun modula i faznih uglova napona u čvorovima usvajajući čvor 1 kao referentno-balansni. b) Napisati jednačine za proračun tokova aktivnih i reaktivnih snaga u granama 1–2, 1–3 i 2–3. c) Primenom DC modela proračunati fazne uglove napona u čvorovima, tokove aktivnih snaga u granama i aktivnu snagu injektiranja referentnog čvora. (radi se kasnije na vežbama) . NAPOMENA: Za bazne vrednosti pri svođenju veličina usvojeno U B = 220 U_B = 220 U B = 220 S B = 100 S_B = 100 S B = 100 1 2 Y g 12 = (0,25 − j2,5) r.j. S G1 V 1 = 1 r.j. θ 1 = 0° S P2 = (2 + j1) r.j. S G2 = (3 + jQ 2 ) r.j. V 2 = 1 r.j. Y g 13 = (0,5 − j4) r.j. Y g 23 = (0 − j5) r.j. 3 S P3 = (4 + j2) r.j. Poznato G 12 g = 0,25 G^g_{12} = 0{,}25 G 12 g = 0 , 25 r.j. − B 12 g = 2,5 -B^g_{12} = 2{,}5 − B 12 g = 2 , 5 r.j. G 13 g = 0,5 G^g_{13} = 0{,}5 G 13 g = 0 , 5 r.j. − B 13 g = 4 -B^g_{13} = 4 − B 13 g = 4 r.j. − B 23 g = 5 -B^g_{23} = 5 − B 23 g = 5 r.j. P P 2 = 2 P_{P2} = 2 P P 2 = 2 r.j. Q P 2 = 1 Q_{P2} = 1 Q P 2 = 1 r.j. P G 2 = 3 P_{G2} = 3 P G 2 = 3 r.j. P P 3 = 4 P_{P3} = 4 P P 3 = 4 r.j. Q P 3 = 2 Q_{P3} = 2 Q P 3 = 2 r.j. V 1 = 1 V_1 = 1 V 1 = 1 r.j. V 2 = 1 V_2 = 1 V 2 = 1 r.j. Rešenje a. Klasifikacija čvorova Klasifikacija čvora Poznato Nepoznato 1 SL (Ref.-Balansni) θ 1 \theta_1 θ 1 V 1 V_1 V 1 P 1 P_1 P 1 Q 1 Q_1 Q 1 2 PV P 2 P_2 P 2 V 2 V_2 V 2 θ 2 \theta_2 θ 2 Q 2 Q_2 Q 2 3 PQ P 3 P_3 P 3 Q 3 Q_3 Q 3 θ 3 \theta_3 θ 3 V 3 V_3 V 3
Vektor specificiranih veličina: X = [ θ 1 V 1 P 2 V 2 P 3 Q 3 ] T \boldsymbol{X} = \begin{bmatrix} \theta_1 & V_1 & P_2 & V_2 & P_3 & Q_3 \end{bmatrix}^{\mathrm{T}} X = [ θ 1 V 1 P 2 V 2 P 3 Q 3 ] T
Vektor nepoznatih veličina: Y = [ P 1 Q 1 θ 2 Q 2 θ 3 V 3 ] T \boldsymbol{Y} = \begin{bmatrix} P_1 & Q_1 & \theta_2 & Q_2 & \theta_3 & V_3 \end{bmatrix}^{\mathrm{T}} Y = [ P 1 Q 1 θ 2 Q 2 θ 3 V 3 ] T
Aktivne i reaktivne snage injektiranja:
P 1 = P G 1 ; P 2 = P G 2 − P P 2 = 3 − 2 = 1 r.j. ; P 3 = P G 3 − P P 3 = − 4 r.j. ; Q 1 = Q G 1 ; Q 2 = Q G 2 − Q P 2 = ( Q G 2 − 1 ) r.j. ; Q 3 = Q G 3 − Q P 3 = − 2 r.j. \begin{aligned} P_1 &= P_{G1}; \\ P_2 &= P_{G2} - P_{P2} = 3 - 2 = 1\ \rj; \\ P_3 &= P_{G3} - P_{P3} = -4\ \rj; \\ Q_1 &= Q_{G1}; \\ Q_2 &= Q_{G2} - Q_{P2} = (Q_{G2} - 1)\ \rj; \\ Q_3 &= Q_{G3} - Q_{P3} = -2\ \rj \end{aligned} P 1 P 2 P 3 Q 1 Q 2 Q 3 = P G 1 ; = P G 2 − P P 2 = 3 − 2 = 1 r.j. ; = P G 3 − P P 3 = − 4 r.j. ; = Q G 1 ; = Q G 2 − Q P 2 = ( Q G 2 − 1 ) r.j. ; = Q G 3 − Q P 3 = − 2 r.j. Matrica admitansi nezavisnih čvorova
Y ‾ B U S = [ Y ‾ 12 g + Y ‾ 13 g − Y ‾ 12 g − Y ‾ 13 g − Y ‾ 12 g Y ‾ 12 g + Y ‾ 23 g − Y ‾ 23 g − Y ‾ 13 g − Y ‾ 23 g Y ‾ 13 g + Y ‾ 23 g ] \ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} \ph{Y}_{12}^g + \ph{Y}_{13}^g & -\ph{Y}_{12}^g & -\ph{Y}_{13}^g \\ -\ph{Y}_{12}^g & \ph{Y}_{12}^g + \ph{Y}_{23}^g & -\ph{Y}_{23}^g \\ -\ph{Y}_{13}^g & -\ph{Y}_{23}^g & \ph{Y}_{13}^g + \ph{Y}_{23}^g \end{bmatrix} Y B U S = Y 12 g + Y 13 g − Y 12 g − Y 13 g − Y 12 g Y 12 g + Y 23 g − Y 23 g − Y 13 g − Y 23 g Y 13 g + Y 23 g Posle zamene brojčanih vrednosti ova matrica je
Y ‾ B U S = [ 0,75 − j 6,5 − 0,25 + j 2,5 − 0,5 + j 4 − 0,25 + j 2,5 0,25 − j 7,5 j 5 − 0,5 + j 4 j 5 0,5 − j 9 ] \ph{\boldsymbol{Y}}_{BUS} = \begin{bmatrix} 0{,}75 - \mathrm{j}6{,}5 & -0{,}25 + \mathrm{j}2{,}5 & -0{,}5 + \mathrm{j}4 \\ -0{,}25 + \mathrm{j}2{,}5 & 0{,}25 - \mathrm{j}7{,}5 & \mathrm{j}5 \\ -0{,}5 + \mathrm{j}4 & \mathrm{j}5 & 0{,}5 - \mathrm{j}9 \end{bmatrix} Y B U S = 0 , 75 − j 6 , 5 − 0 , 25 + j 2 , 5 − 0 , 5 + j 4 − 0 , 25 + j 2 , 5 0 , 25 − j 7 , 5 j 5 − 0 , 5 + j 4 j 5 0 , 5 − j 9 Jednačine aktivnih i reaktivnih injektiranja u čvoru 1 (referentno-balansni) su:
P 1 = G 11 V 1 2 + V 1 V 2 [ G 12 cos ( θ 1 − θ 2 ) + B 12 sin ( θ 1 − θ 2 ) ] + V 1 V 3 [ G 13 cos ( θ 1 − θ 3 ) + B 13 sin ( θ 1 − θ 3 ) ] Q 1 = − B 11 V 1 2 + V 1 V 2 [ G 12 sin ( θ 1 − θ 2 ) − B 12 cos ( θ 1 − θ 2 ) ] + V 1 V 3 [ G 13 sin ( θ 1 − θ 3 ) − B 13 cos ( θ 1 − θ 3 ) ] \begin{aligned} P_1 = G_{11}V_1^2 &+ V_1V_2\left[G_{12}\cos(\theta_1 - \theta_2) + B_{12}\sin(\theta_1 - \theta_2)\right] \\ &+ V_1V_3\left[G_{13}\cos(\theta_1 - \theta_3) + B_{13}\sin(\theta_1 - \theta_3)\right] \\ Q_1 = -B_{11}V_1^2 &+ V_1V_2\left[G_{12}\sin(\theta_1 - \theta_2) - B_{12}\cos(\theta_1 - \theta_2)\right] \\ &+ V_1V_3\left[G_{13}\sin(\theta_1 - \theta_3) - B_{13}\cos(\theta_1 - \theta_3)\right] \end{aligned} P 1 = G 11 V 1 2 Q 1 = − B 11 V 1 2 + V 1 V 2 [ G 12 cos ( θ 1 − θ 2 ) + B 12 sin ( θ 1 − θ 2 ) ] + V 1 V 3 [ G 13 cos ( θ 1 − θ 3 ) + B 13 sin ( θ 1 − θ 3 ) ] + V 1 V 2 [ G 12 sin ( θ 1 − θ 2 ) − B 12 cos ( θ 1 − θ 2 ) ] + V 1 V 3 [ G 13 sin ( θ 1 − θ 3 ) − B 13 cos ( θ 1 − θ 3 ) ] Posle zamene brojčanih vrednosti (za θ 1 = 0 \theta_1 = 0 θ 1 = 0
P 1 = 0,75 ⋅ 1 2 + 1 ⋅ 1 ( − 0,25 ⋅ cos θ 2 − 2,5 ⋅ sin θ 2 ) + 1 ⋅ V 3 ( − 0,5 cos θ 3 − 4 sin θ 3 ) Q 1 = 6,5 ⋅ 1 2 + 1 ⋅ 1 ( 0,25 ⋅ sin θ 2 − 2,5 ⋅ cos θ 2 ) + 1 ⋅ V 3 ( 0,5 sin θ 3 − 4 cos θ 3 ) \begin{aligned} P_1 = 0{,}75 \cdot 1^2 &+ 1 \cdot 1\left(-0{,}25 \cdot \cos\theta_2 - 2{,}5 \cdot \sin\theta_2\right) \\ &+ 1 \cdot V_3\left(-0{,}5\cos\theta_3 - 4\sin\theta_3\right) \\ Q_1 = 6{,}5 \cdot 1^2 &+ 1 \cdot 1\left(0{,}25 \cdot \sin\theta_2 - 2{,}5 \cdot \cos\theta_2\right) \\ &+ 1 \cdot V_3\left(0{,}5\sin\theta_3 - 4\cos\theta_3\right) \end{aligned} P 1 = 0 , 75 ⋅ 1 2 Q 1 = 6 , 5 ⋅ 1 2 + 1 ⋅ 1 ( − 0 , 25 ⋅ cos θ 2 − 2 , 5 ⋅ sin θ 2 ) + 1 ⋅ V 3 ( − 0 , 5 cos θ 3 − 4 sin θ 3 ) + 1 ⋅ 1 ( 0 , 25 ⋅ sin θ 2 − 2 , 5 ⋅ cos θ 2 ) + 1 ⋅ V 3 ( 0 , 5 sin θ 3 − 4 cos θ 3 ) Na sličan način se formiraju jednačine specificiranih injektiranja:
P 2 = 1 = 0,25 ⋅ 1 2 − 0,25 ⋅ cos θ 2 + 2,5 ⋅ sin θ 2 + 5 ⋅ V 3 ⋅ sin ( θ 2 − θ 3 ) P 3 = − 4 = 0,5 ⋅ V 3 2 − 0,5 ⋅ V 3 ⋅ cos θ 3 + 4 ⋅ V 3 ⋅ sin θ 3 + 5 ⋅ V 3 ⋅ sin ( θ 3 − θ 2 ) Q 3 = − 2 = 9 ⋅ V 3 2 − 0,5 ⋅ V 3 ⋅ sin θ 3 − 4 ⋅ V 3 ⋅ cos θ 3 − 5 ⋅ V 3 ⋅ cos ( θ 3 − θ 2 ) \begin{aligned} P_2 = 1 &= 0{,}25 \cdot 1^2 - 0{,}25 \cdot \cos\theta_2 + 2{,}5 \cdot \sin\theta_2 \\ &\quad + 5 \cdot V_3 \cdot \sin(\theta_2 - \theta_3) \\ P_3 = -4 &= 0{,}5 \cdot V_3^2 - 0{,}5 \cdot V_3 \cdot \cos\theta_3 + 4 \cdot V_3 \cdot \sin\theta_3 \\ &\quad + 5 \cdot V_3 \cdot \sin(\theta_3 - \theta_2) \\ Q_3 = -2 &= 9 \cdot V_3^2 - 0{,}5 \cdot V_3 \cdot \sin\theta_3 - 4 \cdot V_3 \cdot \cos\theta_3 \\ &\quad - 5 \cdot V_3 \cdot \cos(\theta_3 - \theta_2) \end{aligned} P 2 = 1 P 3 = − 4 Q 3 = − 2 = 0 , 25 ⋅ 1 2 − 0 , 25 ⋅ cos θ 2 + 2 , 5 ⋅ sin θ 2 + 5 ⋅ V 3 ⋅ sin ( θ 2 − θ 3 ) = 0 , 5 ⋅ V 3 2 − 0 , 5 ⋅ V 3 ⋅ cos θ 3 + 4 ⋅ V 3 ⋅ sin θ 3 + 5 ⋅ V 3 ⋅ sin ( θ 3 − θ 2 ) = 9 ⋅ V 3 2 − 0 , 5 ⋅ V 3 ⋅ sin θ 3 − 4 ⋅ V 3 ⋅ cos θ 3 − 5 ⋅ V 3 ⋅ cos ( θ 3 − θ 2 ) U prethodne tri jednačine postoje tačno tri nepoznate: V 3 V_3 V 3 θ 2 \theta_2 θ 2 θ 3 \theta_3 θ 3
b. Tokovi snaga po granama Tokovi snaga po traženim granama dobijaju se iz opštih izraza koji važe u slučaju zanemarenih otočnih elemenata:
S ‾ i j = 3 V ‾ i I ‾ i j ∗ = 3 V ‾ i Y ‾ i j g ∗ ( V ‾ i ∗ − V ‾ j ∗ ) = Y ‾ i j g ∗ U i 2 − Y ‾ i j g ∗ U ‾ i U ‾ j ∗ = Y ‾ i j g ∗ U i 2 − Y ‾ i j g ∗ ( U i / θ i ‾ ⋅ U j / − θ j ‾ ) = Y ‾ i j g ∗ U i 2 − Y ‾ i j g ∗ U i U j / θ i j ‾ , \begin{aligned} \ph{S}_{ij} &= 3\ph{V}_i\cj{\ph{I}}_{ij} = 3\ph{V}_i\ph{Y}_{ij}^{g*}(\cj{\ph{V}}_i - \cj{\ph{V}}_j) \\ &= \ph{Y}_{ij}^{g*}U_i^2 - \ph{Y}_{ij}^{g*}\ph{U}_i\cj{\ph{U}}_j = \ph{Y}_{ij}^{g*}U_i^2 - \ph{Y}_{ij}^{g*}(U_i \underline{/\theta_i} \cdot U_j \underline{/{-\theta_j}}) \\ &= \ph{Y}_{ij}^{g*}U_i^2 - \ph{Y}_{ij}^{g*}U_iU_j \underline{/\theta_{ij}}, \end{aligned} S ij = 3 V i I ij ∗ = 3 V i Y ij g ∗ ( V i ∗ − V j ∗ ) = Y ij g ∗ U i 2 − Y ij g ∗ U i U j ∗ = Y ij g ∗ U i 2 − Y ij g ∗ ( U i / θ i ⋅ U j / − θ j ) = Y ij g ∗ U i 2 − Y ij g ∗ U i U j / θ ij , Iz prethodne jednačine kada su sve veličine u relativnim jedinicama (V ‾ i = U ‾ i \ph{V}_i = \ph{U}_i V i = U i
P i j = V i 2 G i j g − V i V j ( G i j g cos θ i j + B i j g sin θ i j ) Q i j = − V i 2 B i j g − V i V j ( G i j g sin θ i j − B i j g cos θ i j ) \begin{aligned} P_{ij} &= V_i^2G_{ij}^g - V_iV_j(G_{ij}^g\cos\theta_{ij} + B_{ij}^g\sin\theta_{ij}) \\ Q_{ij} &= -V_i^2B_{ij}^g - V_iV_j(G_{ij}^g\sin\theta_{ij} - B_{ij}^g\cos\theta_{ij}) \end{aligned} P ij Q ij = V i 2 G ij g − V i V j ( G ij g cos θ ij + B ij g sin θ ij ) = − V i 2 B ij g − V i V j ( G ij g sin θ ij − B ij g cos θ ij ) odnosno
P 12 = 0,25 − 0,25 ⋅ cos ( θ 1 − θ 2 ) + 2,5 ⋅ sin ( θ 1 − θ 2 ) = 0,25 − 0,25 ⋅ cos θ 2 − 2,5 ⋅ sin θ 2 Q 12 = 2,5 + 0,25 ⋅ sin θ 2 − 2,5 ⋅ cos θ 2 P 13 = 0,5 − 0,5 ⋅ V 3 ⋅ cos θ 3 − 4 ⋅ V 3 ⋅ sin θ 3 Q 13 = 4 + 0,5 ⋅ V 3 ⋅ sin θ 3 − 4 ⋅ V 3 ⋅ cos θ 3 P 32 = 5 ⋅ V 3 ⋅ sin ( θ 3 − θ 2 ) Q 32 = 5 V 3 2 − 5 V 3 ⋅ cos ( θ 3 − θ 2 ) \begin{aligned} P_{12} &= 0{,}25 - 0{,}25 \cdot \cos(\theta_1 - \theta_2) + 2{,}5 \cdot \sin(\theta_1 - \theta_2) \\ &= 0{,}25 - 0{,}25 \cdot \cos\theta_2 - 2{,}5 \cdot \sin\theta_2 \\ Q_{12} &= 2{,}5 + 0{,}25 \cdot \sin\theta_2 - 2{,}5 \cdot \cos\theta_2 \\ P_{13} &= 0{,}5 - 0{,}5 \cdot V_3 \cdot \cos\theta_3 - 4 \cdot V_3 \cdot \sin\theta_3 \\ Q_{13} &= 4 + 0{,}5 \cdot V_3 \cdot \sin\theta_3 - 4 \cdot V_3 \cdot \cos\theta_3 \\ P_{32} &= 5 \cdot V_3 \cdot \sin(\theta_3 - \theta_2) \\ Q_{32} &= 5V_3^2 - 5V_3 \cdot \cos(\theta_3 - \theta_2) \end{aligned} P 12 Q 12 P 13 Q 13 P 32 Q 32 = 0 , 25 − 0 , 25 ⋅ cos ( θ 1 − θ 2 ) + 2 , 5 ⋅ sin ( θ 1 − θ 2 ) = 0 , 25 − 0 , 25 ⋅ cos θ 2 − 2 , 5 ⋅ sin θ 2 = 2 , 5 + 0 , 25 ⋅ sin θ 2 − 2 , 5 ⋅ cos θ 2 = 0 , 5 − 0 , 5 ⋅ V 3 ⋅ cos θ 3 − 4 ⋅ V 3 ⋅ sin θ 3 = 4 + 0 , 5 ⋅ V 3 ⋅ sin θ 3 − 4 ⋅ V 3 ⋅ cos θ 3 = 5 ⋅ V 3 ⋅ sin ( θ 3 − θ 2 ) = 5 V 3 2 − 5 V 3 ⋅ cos ( θ 3 − θ 2 ) c. DC model Ovo će se rešiti kod zadataka iz DC modela tokova snaga:
[ P 2 P 3 ] = [ P G 2 − P p 2 − P p 3 ] = − [ B 22 B 23 B 32 B 33 ] [ θ 2 θ 3 ] \begin{bmatrix} P_2 \\ P_3 \end{bmatrix} = \begin{bmatrix} P_{G2} - P_{p2} \\ -P_{p3} \end{bmatrix} = -\begin{bmatrix} B_{22} & B_{23} \\ B_{32} & B_{33} \end{bmatrix}\begin{bmatrix} \theta_2 \\ \theta_3 \end{bmatrix} [ P 2 P 3 ] = [ P G 2 − P p 2 − P p 3 ] = − [ B 22 B 32 B 23 B 33 ] [ θ 2 θ 3 ] odnosno
[ 3 − 2 − 4 ] = − [ − 7,5 5 5 − 9 ] [ θ 2 θ 3 ] , ili [ θ 2 θ 3 ] = − [ − 7,5 5 5 − 9 ] − 1 [ 1 − 4 ] = [ − 0,2584 − 0,5882 ] = [ − 14,8 ∘ − 33,7 ∘ ] \begin{gathered} \begin{bmatrix} 3 - 2 \\ -4 \end{bmatrix} = -\begin{bmatrix} -7{,}5 & 5 \\ 5 & -9 \end{bmatrix}\begin{bmatrix} \theta_2 \\ \theta_3 \end{bmatrix}, \text{ ili} \\ \begin{bmatrix} \theta_2 \\ \theta_3 \end{bmatrix} = -\begin{bmatrix} -7{,}5 & 5 \\ 5 & -9 \end{bmatrix}^{-1}\begin{bmatrix} 1 \\ -4 \end{bmatrix} = \begin{bmatrix} -0{,}2584 \\ -0{,}5882 \end{bmatrix} = \begin{bmatrix} -14{,}8^\circ \\ -33{,}7^\circ \end{bmatrix} \end{gathered} [ 3 − 2 − 4 ] = − [ − 7 , 5 5 5 − 9 ] [ θ 2 θ 3 ] , ili [ θ 2 θ 3 ] = − [ − 7 , 5 5 5 − 9 ] − 1 [ 1 − 4 ] = [ − 0 , 2584 − 0 , 5882 ] = [ − 14 , 8 ∘ − 33 , 7 ∘ ] Tokovi snaga po granama su
P 12 = − B 12 ( θ 1 − θ 2 ) = − 2,5 ⋅ ( − 0,2584 ) = 0,646 = 64,6 MW P 13 = − B 13 ( θ 1 − θ 3 ) = − 4 ⋅ ( − 0,5882 ) = 2,352 = 235,2 MW P 23 = − B 23 ( θ 2 − θ 3 ) = − 5 ⋅ ( 0,2584 − 0,5882 ) = 1,649 = 164,9 MW P 1 = P 12 + P 13 = 299,8 MW ≈ P G 1 \begin{aligned} P_{12} &= -B_{12}(\theta_1 - \theta_2) = -2{,}5 \cdot (-0{,}2584) = 0{,}646 = 64{,}6\ \text{MW} \\ P_{13} &= -B_{13}(\theta_1 - \theta_3) = -4 \cdot (-0{,}5882) = 2{,}352 = 235{,}2\ \text{MW} \\ P_{23} &= -B_{23}(\theta_2 - \theta_3) = -5 \cdot (0{,}2584 - 0{,}5882) = 1{,}649 = 164{,}9\ \text{MW} \\ P_1 &= P_{12} + P_{13} = 299{,}8\ \text{MW} \approx P_{G1} \end{aligned} P 12 P 13 P 23 P 1 = − B 12 ( θ 1 − θ 2 ) = − 2 , 5 ⋅ ( − 0 , 2584 ) = 0 , 646 = 64 , 6 MW = − B 13 ( θ 1 − θ 3 ) = − 4 ⋅ ( − 0 , 5882 ) = 2 , 352 = 235 , 2 MW = − B 23 ( θ 2 − θ 3 ) = − 5 ⋅ ( 0 , 2584 − 0 , 5882 ) = 1 , 649 = 164 , 9 MW = P 12 + P 13 = 299 , 8 MW ≈ P G 1 Rezultat Čvor 1 SL, čvor 2 PV, čvor 3 PQ; nepoznate V 3 V_3 V 3 θ 2 \theta_2 θ 2 θ 3 \theta_3 θ 3
DC model: θ 2 = − 0,2584 \theta_2 = -0{,}2584 θ 2 = − 0 , 2584 − 14,8 ∘ -14{,}8^\circ − 14 , 8 ∘ θ 3 = − 0,5882 \theta_3 = -0{,}5882 θ 3 = − 0 , 5882 − 33,7 ∘ -33{,}7^\circ − 33 , 7 ∘ P 12 = 64,6 P_{12} = 64{,}6 P 12 = 64 , 6 P 13 = 235,2 P_{13} = 235{,}2 P 13 = 235 , 2 P 23 = 164,9 P_{23} = 164{,}9 P 23 = 164 , 9 P 1 = 299,8 P_1 = 299{,}8 P 1 = 299 , 8
← Prethodni zadatak Zadatak 5: Matrica admitansi i prividne snage čvorova Sledeći zadatak → Zadatak 7: Gauss-Seidel u tri iteracije, isključen prekidač