Zadatak 8. Za EES jednopolno prikazan na slici odrediti jednačine kretanja rotora generatora pri tropolnom kratkom spoju na potrošačkim sabirnicama. Koliki je ugao rotora generatora posle 0,25 s od nastanka kvara?
G 1 T 1 l v = 100 km x v = 0,42 Ω/km P 1 = 80 MW Q 1 = 25 MVAr S = (200 + j95) MVA T 2 G 2 S G1n = S T1n = 120 MVA U G1n = 10,5 kV m T1 = 10,5/231 kV/kV x dG1 = 30 % x T1 = 11 % H = 7 s U r = 220 kV P 2 = 120 MW Q 2 = 70 MVAr S G2n = S T2n = 150 MVA U G2n = 15,75 kV m T2 = 231/15,75 kV/kV x dG2 = 26 % x T2 % = 11 % T j = 6,5 s Poznato S G 1 n = S T 1 n = 120 S_{G1n} = S_{T1n} = 120 S G 1 n = S T 1 n = 120 MVA U G 1 n = 10,5 U_{G1n} = 10{,}5 U G 1 n = 10 , 5 kV U T 1 n = U T 2 n = 231 U_{T1n} = U_{T2n} = 231 U T 1 n = U T 2 n = 231 kV x d G 1 = 30 x_{dG1} = 30 x d G 1 = 30 % x T 1 = 11 x_{T1} = 11 x T 1 = 11 % H = 7 H = 7 H = 7 s S G 2 n = S T 2 n = 150 S_{G2n} = S_{T2n} = 150 S G 2 n = S T 2 n = 150 MVA U G 2 n = 15,75 U_{G2n} = 15{,}75 U G 2 n = 15 , 75 kV x d G 2 = 26 x_{dG2} = 26 x d G 2 = 26 % x T 2 = 11 x_{T2} = 11 x T 2 = 11 % T j = 6,5 T_j = 6{,}5 T j = 6 , 5 s l v = 100 l_v = 100 l v = 100 km x v = 0,42 x_v = 0{,}42 x v = 0 , 42 Ω/km P 1 = 80 P_1 = 80 P 1 = 80 MW Q 1 = 25 Q_1 = 25 Q 1 = 25 MVAr P 2 = 120 P_2 = 120 P 2 = 120 MW Q 2 = 70 Q_2 = 70 Q 2 = 70 MVAr U r = 220 U_r = 220 U r = 220 kV t = 0,25 t = 0{,}25 t = 0 , 25 s Rešenje Reaktanse pojedinih elemenata svedene na naponski nivo voda su:
X G 1 ′ = x G 1 ′ ( % ) 100 ⋅ U G 1 n 2 S G 1 n ⋅ 1 m T 1 2 = 30 100 ⋅ 10,5 2 120 ⋅ 231 2 10,5 2 = 133,402 Ω ; \begin{aligned} X'_{G1} &= \frac{x'_{G1}(\%)}{100} \cdot \frac{U_{G1n}^2}{S_{G1n}} \cdot \frac{1}{m_{T1}^2} \\ &= \frac{30}{100} \cdot \frac{10{,}5^2}{120} \cdot \frac{231^2}{10{,}5^2} = 133{,}402\ \Omega; \end{aligned} X G 1 ′ = 100 x G 1 ′ ( % ) ⋅ S G 1 n U G 1 n 2 ⋅ m T 1 2 1 = 100 30 ⋅ 120 10 , 5 2 ⋅ 10 , 5 2 23 1 2 = 133 , 402 Ω ; X G 2 ′ = x G 2 ′ ( % ) 100 ⋅ U G 2 n 2 S G 2 n ⋅ 1 m T 2 2 = 26 100 ⋅ 15,75 2 150 ⋅ 231 2 15,75 2 = 92,492 Ω ; \begin{aligned} X'_{G2} &= \frac{x'_{G2}(\%)}{100} \cdot \frac{U_{G2n}^2}{S_{G2n}} \cdot \frac{1}{m_{T2}^2} \\ &= \frac{26}{100} \cdot \frac{15{,}75^2}{150} \cdot \frac{231^2}{15{,}75^2} = 92{,}492\ \Omega; \end{aligned} X G 2 ′ = 100 x G 2 ′ ( % ) ⋅ S G 2 n U G 2 n 2 ⋅ m T 2 2 1 = 100 26 ⋅ 150 15 , 7 5 2 ⋅ 15 , 7 5 2 23 1 2 = 92 , 492 Ω ; X T 1 = u k 1 ( % ) 100 ⋅ U T 1 n 2 S T 1 n = 11 100 ⋅ 231 2 120 = 48,914 Ω ; X_{T1} = \frac{u_{k1}(\%)}{100} \cdot \frac{U_{T1n}^2}{S_{T1n}} = \frac{11}{100} \cdot \frac{231^2}{120} = 48{,}914\ \Omega; X T 1 = 100 u k 1 ( % ) ⋅ S T 1 n U T 1 n 2 = 100 11 ⋅ 120 23 1 2 = 48 , 914 Ω ; X T 2 = u k 2 ( % ) 100 ⋅ U T 2 n 2 S T 2 n = 11 100 ⋅ 231 2 150 = 39,131 Ω ; X_{T2} = \frac{u_{k2}(\%)}{100} \cdot \frac{U_{T2n}^2}{S_{T2n}} = \frac{11}{100} \cdot \frac{231^2}{150} = 39{,}131\ \Omega; X T 2 = 100 u k 2 ( % ) ⋅ S T 2 n U T 2 n 2 = 100 11 ⋅ 150 23 1 2 = 39 , 131 Ω ; X v = x v L = 0,42 ⋅ 100 = 42 Ω . X_v = x_vL = 0{,}42 \cdot 100 = 42\ \Omega. X v = x v L = 0 , 42 ⋅ 100 = 42 Ω. Ekvivalentna šema sistema u normalnom radnom režimu (pre kvara) je:
Prethodna šema može se svesti na:
j X 1 = j X G 1 ′ + j X T 1 + j X v = 224,316 Ω j X 2 = j X G 2 ′ + j X T 2 = 131,623 Ω \begin{aligned} \mathrm{j}X_1 &= \mathrm{j}X'_{G1} + \mathrm{j}X_{T1} + \mathrm{j}X_v = 224{,}316\ \Omega \\ \mathrm{j}X_2 &= \mathrm{j}X'_{G2} + \mathrm{j}X_{T2} = 131{,}623\ \Omega \end{aligned} j X 1 j X 2 = j X G 1 ′ + j X T 1 + j X v = 224 , 316 Ω = j X G 2 ′ + j X T 2 = 131 , 623 Ω Tranzijentne elektromotorne sile generatora su:
E ‾ G 1 ′ = U r + Q 1 X 1 U r + j P 1 X 1 U r = ( 245,49 + j 81,57 ) kV = 258,687 ⋅ e j 18,38 ∘ kV ; \begin{aligned} \ph{E}'_{G1} &= U_r + \frac{Q_1X_1}{U_r} + \mathrm{j}\frac{P_1X_1}{U_r} \\ &= (245{,}49 + \mathrm{j}81{,}57)\ \text{kV} = 258{,}687 \cdot e^{\mathrm{j}18{,}38^\circ}\ \text{kV}; \end{aligned} E G 1 ′ = U r + U r Q 1 X 1 + j U r P 1 X 1 = ( 245 , 49 + j 81 , 57 ) kV = 258 , 687 ⋅ e j 18 , 3 8 ∘ kV ; E ‾ G 2 ′ = U r + Q 2 X 2 U r + j P 2 X 2 U r = ( 261,88 + j 71,79 ) kV = 271,543 ⋅ e j 15,33 ∘ kV . \begin{aligned} \ph{E}'_{G2} &= U_r + \frac{Q_2X_2}{U_r} + \mathrm{j}\frac{P_2X_2}{U_r} \\ &= (261{,}88 + \mathrm{j}71{,}79)\ \text{kV} = 271{,}543 \cdot e^{\mathrm{j}15{,}33^\circ}\ \text{kV}. \end{aligned} E G 2 ′ = U r + U r Q 2 X 2 + j U r P 2 X 2 = ( 261 , 88 + j 71 , 79 ) kV = 271 , 543 ⋅ e j 15 , 3 3 ∘ kV . Ugao između elektromotorne sile E ‾ G 1 ′ \ph{E}'_{G1} E G 1 ′ U ‾ r \ph{U}_r U r
θ 1 ( 0 ) = θ E G 1 − θ U r = δ 1 ( 0 ) = 18,38 ∘ − 0 = 18,38 ∘ . \theta_1^{(0)} = \theta_{E_{G1}} - \theta_{U_r} = \delta_1^{(0)} = 18{,}38^\circ - 0 = 18{,}38^\circ. θ 1 ( 0 ) = θ E G 1 − θ U r = δ 1 ( 0 ) = 18 , 3 8 ∘ − 0 = 18 , 3 8 ∘ . Ugao između elektromotorne sile E ‾ G 2 ′ \ph{E}'_{G2} E G 2 ′ U ‾ r \ph{U}_r U r
θ 2 ( 0 ) = θ E G 2 − θ U r = δ 2 ( 0 ) = 15,33 ∘ − 0 = 15,33 ∘ . \theta_2^{(0)} = \theta_{E_{G2}} - \theta_{U_r} = \delta_2^{(0)} = 15{,}33^\circ - 0 = 15{,}33^\circ. θ 2 ( 0 ) = θ E G 2 − θ U r = δ 2 ( 0 ) = 15 , 3 3 ∘ − 0 = 15 , 3 3 ∘ . Ugao između elektromotornih sila E ‾ G 1 ′ \ph{E}'_{G1} E G 1 ′ E ‾ G 2 ′ \ph{E}'_{G2} E G 2 ′
δ ( 0 ) = δ 1 ( 0 ) − δ 2 ( 0 ) = 18,38 ∘ − 15,33 ∘ = 3,05 ∘ . \delta^{(0)} = \delta_1^{(0)} - \delta_2^{(0)} = 18{,}38^\circ - 15{,}33^\circ = 3{,}05^\circ. δ ( 0 ) = δ 1 ( 0 ) − δ 2 ( 0 ) = 18 , 3 8 ∘ − 15 , 3 3 ∘ = 3 , 0 5 ∘ . Mehaničke snage turbina T1 i T2 koje odgovaraju mehaničkim snagama generatora su
P m 1 = P 1 = 80 MW ; P m 2 = P 2 = 120 MW ; \begin{gathered} P_{m1} = P_1 = 80\ \text{MW}; \\ P_{m2} = P_2 = 120\ \text{MW}; \end{gathered} P m 1 = P 1 = 80 MW ; P m 2 = P 2 = 120 MW ; Promena ugla rotora generatora određena je njihovim jednačinama kretanja:
d 2 δ 1 d t 2 = ω 2 H 1 S G 1 n ( P m 1 − P e 1 ) ; d 2 δ 2 d t 2 = ω T j 2 S G 2 n ( P m 2 − P e 2 ) . \begin{gathered} \frac{d^2\delta_1}{dt^2} = \frac{\omega}{2H_1S_{G1n}}\left(P_{m1} - P_{e1}\right); \\ \frac{d^2\delta_2}{dt^2} = \frac{\omega}{T_{j2}S_{G2n}}\left(P_{m2} - P_{e2}\right). \end{gathered} d t 2 d 2 δ 1 = 2 H 1 S G 1 n ω ( P m 1 − P e 1 ) ; d t 2 d 2 δ 2 = T j 2 S G 2 n ω ( P m 2 − P e 2 ) . S obzirom na vrstu kvara (tropolni kratak spoj) i zanemarene aktivne otpornosti, generatori se posle kvara potpuno rasterećuju (P e 1 = 0 P_{e1} = 0 P e 1 = 0 P e 2 = 0 P_{e2} = 0 P e 2 = 0
U tom slučaju jednačine kretanja rotora generatora postaju diferencijalne jednačine drugog reda sa konstantnim koeficijentima:
d 2 δ 1 d t 2 = ω 2 H 1 S G 1 n P m 1 = k 1 ; d 2 δ 2 d t 2 = ω T j 2 S G 2 n P m 2 = k 2 . \begin{gathered} \frac{d^2\delta_1}{dt^2} = \frac{\omega}{2H_1S_{G1n}}P_{m1} = k_1; \\ \frac{d^2\delta_2}{dt^2} = \frac{\omega}{T_{j2}S_{G2n}}P_{m2} = k_2. \end{gathered} d t 2 d 2 δ 1 = 2 H 1 S G 1 n ω P m 1 = k 1 ; d t 2 d 2 δ 2 = T j 2 S G 2 n ω P m 2 = k 2 . Rešenja prethodnih diferencijalnih jednačina su:
δ 1 = k 1 t 2 2 + δ 1 ( 0 ) ; δ 2 = k 2 t 2 2 + δ 2 ( 0 ) , \begin{gathered} \delta_1 = k_1\frac{t^2}{2} + \delta_1^{(0)}; \\ \delta_2 = k_2\frac{t^2}{2} + \delta_2^{(0)}, \end{gathered} δ 1 = k 1 2 t 2 + δ 1 ( 0 ) ; δ 2 = k 2 2 t 2 + δ 2 ( 0 ) , gde su:
k 1 = ω 2 H 1 S G 1 n P m 1 = 314 2 ⋅ 7 ⋅ 120 ⋅ 80 = 14,96 rad s 2 = 857,14 ∘ s 2 ; k_1 = \frac{\omega}{2H_1S_{G1n}}P_{m1} = \frac{314}{2 \cdot 7 \cdot 120} \cdot 80 = 14{,}96\ \frac{\text{rad}}{\text{s}^2} = 857{,}14\ \frac{^\circ}{\text{s}^2}; k 1 = 2 H 1 S G 1 n ω P m 1 = 2 ⋅ 7 ⋅ 120 314 ⋅ 80 = 14 , 96 s 2 rad = 857 , 14 s 2 ∘ ; k 2 = ω T j 2 S G 2 n P m 2 = 314 6,5 ⋅ 150 ⋅ 120 = 38,66 rad s 2 = 2215,38 ∘ s 2 . k_2 = \frac{\omega}{T_{j2}S_{G2n}}P_{m2} = \frac{314}{6{,}5 \cdot 150} \cdot 120 = 38{,}66\ \frac{\text{rad}}{\text{s}^2} = 2215{,}38\ \frac{^\circ}{\text{s}^2}. k 2 = T j 2 S G 2 n ω P m 2 = 6 , 5 ⋅ 150 314 ⋅ 120 = 38 , 66 s 2 rad = 2215 , 38 s 2 ∘ . Zamenom brojčanih vrednosti je:
δ 1 = 857,14 t 2 2 + 18,38 ; δ 2 = 2215,38 t 2 2 + 15,33. \begin{gathered} \delta_1 = 857{,}14\frac{t^2}{2} + 18{,}38; \\ \delta_2 = 2215{,}38\frac{t^2}{2} + 15{,}33. \end{gathered} δ 1 = 857 , 14 2 t 2 + 18 , 38 ; δ 2 = 2215 , 38 2 t 2 + 15 , 33. Ugao između rotora dva generatora je
δ 12 = δ 2 − δ 1 = 2215,38 t 2 2 + 15,33 − 857,14 t 2 2 − 18,38 = 1358,24 t 2 2 − 3,05 \begin{aligned} \delta_{12} = \delta_2 - \delta_1 &= 2215{,}38\frac{t^2}{2} + 15{,}33 - 857{,}14\frac{t^2}{2} - 18{,}38 \\ &= 1358{,}24\frac{t^2}{2} - 3{,}05 \end{aligned} δ 12 = δ 2 − δ 1 = 2215 , 38 2 t 2 + 15 , 33 − 857 , 14 2 t 2 − 18 , 38 = 1358 , 24 2 t 2 − 3 , 05 Posle t = 0,25 t = 0{,}25 t = 0 , 25
δ 12 ( t = 0,25 s ) = 1358,24 0,25 2 2 − 3,05 = 39,40 ∘ . \delta_{12}(t = 0{,}25\ \text{s}) = 1358{,}24\frac{0{,}25^2}{2} - 3{,}05 = 39{,}40^\circ. δ 12 ( t = 0 , 25 s ) = 1358 , 24 2 0 , 2 5 2 − 3 , 05 = 39 , 4 0 ∘ . Rezultat δ 1 = 857,14 t 2 2 + 18,38 ∘ \delta_1 = 857{,}14\dfrac{t^2}{2} + 18{,}38^\circ δ 1 = 857 , 14 2 t 2 + 18 , 3 8 ∘ δ 2 = 2215,38 t 2 2 + 15,33 ∘ \delta_2 = 2215{,}38\dfrac{t^2}{2} + 15{,}33^\circ δ 2 = 2215 , 38 2 t 2 + 15 , 3 3 ∘ δ 12 = 39,40 ∘ \delta_{12} = 39{,}40^\circ δ 12 = 39 , 4 0 ∘
← Prethodni zadatak Zadatak 7: Kritično vreme isključenja i jednake površine Sledeći zadatak → Zadatak 9: Dvopolni kratak spoj sa zemljom i kriva njihanja 
