Zadatak 12. U kom vremenu treba obostrano jednovremeno isključiti vod V1, na čijem se početku dogodio tropolni kratak spoj, da bi sistem bio dinamički stabilan, ako je generator nominalno pobuđen i ako odaje u sistem aktivnu snagu od 300 MW.
Potrošački centri su jednaki i pri naponu na njihovim sabirnicama od 220 kV svaki uzima po 150 MW pri cos φ = 0,95 i n d \cos\varphi = 0{,}95_{ind} cos φ = 0 , 9 5 in d
G T Z P = Const Z P = Const V 1 V 2 V 3 V 4 l v1 = l v2 = l v3 = l v4 = 150 km x v = 0,42 Ω/km U = 220 kV jaka mreža S Gn = S Tn = 500 MVA cos φ Gn = 0,9 U Gn = 15,75 kV m T = 15,75/231 kV/kV x dG = 30 % x T = 12 % H = 8 s Poznato S G n = S T n = 500 S_{Gn} = S_{Tn} = 500 S G n = S T n = 500 MVA cos φ G n = 0,9 \cos\varphi_{Gn} = 0{,}9 cos φ G n = 0 , 9 U G n = 15,75 U_{Gn} = 15{,}75 U G n = 15 , 75 kV U T n = 231 U_{Tn} = 231 U T n = 231 kV x d G = 30 x_{dG} = 30 x d G = 30 % x T = 12 x_T = 12 x T = 12 % H = 8 H = 8 H = 8 s l v 1 = l v 2 = l v 3 = l v 4 = 150 l_{v1} = l_{v2} = l_{v3} = l_{v4} = 150 l v 1 = l v 2 = l v 3 = l v 4 = 150 km x v = 0,42 x_v = 0{,}42 x v = 0 , 42 Ω/km U = 220 U = 220 U = 220 kV P G = 300 P_G = 300 P G = 300 MW P P = 150 P_P = 150 P P = 150 MW cos φ P = 0,95 \cos\varphi_P = 0{,}95 cos φ P = 0 , 95 ind. Rešenje Parametri elemenata sistema su:
X ( d G + T 1 ) ′ = x d G ′ + x T 1 ( % ) 100 ⋅ U G n 2 S G n = 30 + 12 100 ⋅ 231 2 500 = 44,823 Ω . \begin{aligned} X'_{(dG+T1)} &= \frac{x'_{dG} + x_{T1}(\%)}{100} \cdot \frac{U_{Gn}^2}{S_{Gn}} \\ &= \frac{30 + 12}{100} \cdot \frac{231^2}{500} = 44{,}823\ \Omega. \end{aligned} X ( d G + T 1 ) ′ = 100 x d G ′ + x T 1 ( % ) ⋅ S G n U G n 2 = 100 30 + 12 ⋅ 500 23 1 2 = 44 , 823 Ω. X v = X v 1 = X v 2 = X v 3 = X v 4 = x v L = 0,42 ⋅ 150 = 63 Ω ; X_v = X_{v1} = X_{v2} = X_{v3} = X_{v4} = x_vL = 0{,}42 \cdot 150 = 63\ \Omega; X v = X v 1 = X v 2 = X v 3 = X v 4 = x v L = 0 , 42 ⋅ 150 = 63 Ω ; Z ‾ p = U 2 P / cos φ = 220 2 150 / 0,95 ( cos φ + j sin φ ) = ( 291 + j 95,6 ) Ω . \begin{aligned} \ph{Z}_p &= \frac{U^2}{P/\cos\varphi} = \frac{220^2}{150/0{,}95}(\cos\varphi + \mathrm{j}\sin\varphi) \\ &= (291 + \mathrm{j}95{,}6)\ \Omega. \end{aligned} Z p = P / cos φ U 2 = 150/0 , 95 22 0 2 ( cos φ + j sin φ ) = ( 291 + j 95 , 6 ) Ω. Jednopolna šema sistema je:
Kako su tačke A i B na istom potencijalu, onda se šema može uprostiti kao:
Sa prethodne slike je:
Z ‾ 11 ( 1 ) = j X ( d G + T 1 ) ′ + j X v 2 + j X v 2 ⋅ Z ‾ p 2 j X v 2 + Z ‾ p 2 = 105,07 / 87 ∘ 07 ′ ‾ Ω ; \begin{aligned} \ph{Z}_{11}^{(1)} &= \mathrm{j}X'_{(dG+T1)} + \frac{\mathrm{j}X_v}{2} + \frac{\mathrm{j}\dfrac{X_v}{2} \cdot \dfrac{\ph{Z}_p}{2}}{\mathrm{j}\dfrac{X_v}{2} + \dfrac{\ph{Z}_p}{2}} \\ &= 105{,}07\underline{/87^\circ07'}\ \Omega; \end{aligned} Z 11 ( 1 ) = j X ( d G + T 1 ) ′ + 2 j X v + j 2 X v + 2 Z p j 2 X v ⋅ 2 Z p = 105 , 07 /8 7 ∘ 0 7 ′ Ω ; Z ‾ 12 ( 1 ) = j X ( d G + T 1 ) ′ + j X v 2 + j X v 2 + ( j X ( d G + T 1 ) ′ + j X v 2 ) ⋅ j X v 2 Z ‾ p 2 = 113,69 / 97 ∘ 33 ′ ‾ Ω ; \begin{aligned} \ph{Z}_{12}^{(1)} &= \mathrm{j}X'_{(dG+T1)} + \mathrm{j}\frac{X_v}{2} + \mathrm{j}\frac{X_v}{2} \\ &\quad + \frac{\left(\mathrm{j}X'_{(dG+T1)} + \mathrm{j}\dfrac{X_v}{2}\right) \cdot \mathrm{j}\dfrac{X_v}{2}}{\dfrac{\ph{Z}_p}{2}} \\ &= 113{,}69\underline{/97^\circ33'}\ \Omega; \end{aligned} Z 12 ( 1 ) = j X ( d G + T 1 ) ′ + j 2 X v + j 2 X v + 2 Z p ( j X ( d G + T 1 ) ′ + j 2 X v ) ⋅ j 2 X v = 113 , 69 /9 7 ∘ 3 3 ′ Ω ; Elektromotorna sila iza sinhrone reaktanse (E ‾ G ′ \ph{E}'_G E G ′ U G n U_{Gn} U G n
E ‾ G ′ = U G n + Q X d G ′ U G n + j P X d G ′ U G n = 18,31 kV , \begin{aligned} \ph{E}'_G &= U_{Gn} + \frac{QX'_{dG}}{U_{Gn}} + \mathrm{j}\frac{PX'_{dG}}{U_{Gn}} \\ &= 18{,}31\ \text{kV}, \end{aligned} E G ′ = U G n + U G n Q X d G ′ + j U G n P X d G ′ = 18 , 31 kV , Domaći: Proveriti ovu vrednost ili svedeno na stranu mreže nominalnog napona 220 kV:
E G ′ = 18,31 ⋅ 231 15,75 = 268,5 kV . E'_G = 18{,}31 \cdot \frac{231}{15{,}75} = 268{,}5\ \text{kV}. E G ′ = 18 , 31 ⋅ 15 , 75 231 = 268 , 5 kV . Karakteristika električne odate snage generatora neposredno pre nastanka kvara je:
P G i = E G i 2 Z i i sin ψ i i + ∑ j = 1 j ≠ i n E G i E G j Z i j sin ( δ i j − ψ i j ) , (Izvedeno u drugom zadatku) P_{Gi} = \frac{E_{Gi}^2}{Z_{ii}}\sin\psi_{ii} + \sum_{\substack{j=1 \\ j \ne i}}^{n}\frac{E_{Gi}E_{Gj}}{Z_{ij}}\sin(\delta_{ij} - \psi_{ij}), \quad \text{(Izvedeno u drugom zadatku)} P G i = Z ii E G i 2 sin ψ ii + j = 1 j = i ∑ n Z ij E G i E G j sin ( δ ij − ψ ij ) , (Izvedeno u drugom zadatku) ili u ovom slučaju
P 1 = E G ′ 2 Z 11 ( 1 ) sin ψ 11 ( 1 ) + E G ′ U G n Z 12 ( 1 ) sin ( δ 12 − ψ 12 ( 1 ) ) , P_1 = \frac{E_G'^2}{Z_{11}^{(1)}}\sin\psi_{11}^{(1)} + \frac{E'_GU_{Gn}}{Z_{12}^{(1)}}\sin(\delta_{12} - \psi_{12}^{(1)}), P 1 = Z 11 ( 1 ) E G ′2 sin ψ 11 ( 1 ) + Z 12 ( 1 ) E G ′ U G n sin ( δ 12 − ψ 12 ( 1 ) ) , gde je iz prethodnog proračuna:
ψ 11 ( 1 ) = 90 ∘ − β 11 ( 1 ) = 90 ∘ − 87 ∘ 07 ′ = 2 ∘ 53 ′ ; Z ‾ 11 ( 1 ) = Z 11 ( 1 ) / β 11 ( 1 ) ‾ ; ψ 12 ( 1 ) = 90 ∘ − β 12 ( 1 ) = 90 ∘ − 97 ∘ 33 ′ = − 7 ∘ 33 ′ ; Z ‾ 12 ( 1 ) = Z 12 ( 1 ) / β 12 ( 1 ) ‾ . \begin{aligned} \psi_{11}^{(1)} &= 90^\circ - \beta_{11}^{(1)} = 90^\circ - 87^\circ07' = 2^\circ53'; && \ph{Z}_{11}^{(1)} = Z_{11}^{(1)}\underline{/\beta_{11}^{(1)}}; \\ \psi_{12}^{(1)} &= 90^\circ - \beta_{12}^{(1)} = 90^\circ - 97^\circ33' = -7^\circ33'; && \ph{Z}_{12}^{(1)} = Z_{12}^{(1)}\underline{/\beta_{12}^{(1)}}. \end{aligned} ψ 11 ( 1 ) ψ 12 ( 1 ) = 9 0 ∘ − β 11 ( 1 ) = 9 0 ∘ − 8 7 ∘ 0 7 ′ = 2 ∘ 5 3 ′ ; = 9 0 ∘ − β 12 ( 1 ) = 9 0 ∘ − 9 7 ∘ 3 3 ′ = − 7 ∘ 3 3 ′ ; Z 11 ( 1 ) = Z 11 ( 1 ) / β 11 ( 1 ) ; Z 12 ( 1 ) = Z 12 ( 1 ) / β 12 ( 1 ) . To znači da je kriva snaga-ugao
P 1 = 34,51 + 520 ⋅ sin ( δ 12 + 7 ∘ 33 ′ ) . P_1 = 34{,}51 + 520 \cdot \sin(\delta_{12} + 7^\circ33'). P 1 = 34 , 51 + 520 ⋅ sin ( δ 12 + 7 ∘ 3 3 ′ ) . Ugao δ 12 \delta_{12} δ 12 E ‾ G ′ \ph{E}'_G E G ′ U G n U_{Gn} U G n
Ugao δ 12 ( 1 ) \delta_{12}^{(1)} δ 12 ( 1 ) P 1 = 300 P_1 = 300 P 1 = 300
300 = 268,5 2 105,07 ⋅ sin ( 2 ∘ 53 ′ ) + 268,5 ⋅ 220 113,69 sin ( δ 12 ( 1 ) + 7 ∘ 33 ′ ) . \begin{aligned} 300 &= \frac{268{,}5^2}{105{,}07} \cdot \sin(2^\circ53') \\ &\quad + \frac{268{,}5 \cdot 220}{113{,}69}\sin(\delta_{12}^{(1)} + 7^\circ33'). \end{aligned} 300 = 105 , 07 268 , 5 2 ⋅ sin ( 2 ∘ 5 3 ′ ) + 113 , 69 268 , 5 ⋅ 220 sin ( δ 12 ( 1 ) + 7 ∘ 3 3 ′ ) . Rešenje prethodne jednačine je
δ 12 ( 1 ) = 23 ∘ 11 ′ . \delta_{12}^{(1)} = 23^\circ11'. δ 12 ( 1 ) = 2 3 ∘ 1 1 ′ . Za vreme tropolnog kratkog spoja odata snaga generatora je P ( k ) = 0 P^{(k)} = 0 P ( k ) = 0
Zamenska šema sistema posle isključenja voda V1 je
Sa prethodne slike je:
Z ‾ 11 ( 2 ) = j X ( d G + T 1 ) ′ + j X v + j X v ⋅ Z ‾ p j X v + Z ‾ p = 165,53 / 86 ∘ 21 ′ ‾ Ω ; \begin{aligned} \ph{Z}_{11}^{(2)} &= \mathrm{j}X'_{(dG+T1)} + \mathrm{j}X_v + \frac{\mathrm{j}X_v \cdot \ph{Z}_p}{\mathrm{j}X_v + \ph{Z}_p} \\ &= 165{,}53\underline{/86^\circ21'}\ \Omega; \end{aligned} Z 11 ( 2 ) = j X ( d G + T 1 ) ′ + j X v + j X v + Z p j X v ⋅ Z p = 165 , 53 /8 6 ∘ 2 1 ′ Ω ; Z ‾ 12 ( 2 ) = j X ( d G + T 1 ) ′ + j X v + j X v + ( j X ( d G + T 1 ) ′ + j X v ) ⋅ j X v Z ‾ p = 178,97 / 96 ∘ 46 ′ ‾ Ω ; \begin{aligned} \ph{Z}_{12}^{(2)} &= \mathrm{j}X'_{(dG+T1)} + \mathrm{j}X_v + \mathrm{j}X_v \\ &\quad + \frac{\left(\mathrm{j}X'_{(dG+T1)} + \mathrm{j}X_v\right) \cdot \mathrm{j}X_v}{\ph{Z}_p} \\ &= 178{,}97\underline{/96^\circ46'}\ \Omega; \end{aligned} Z 12 ( 2 ) = j X ( d G + T 1 ) ′ + j X v + j X v + Z p ( j X ( d G + T 1 ) ′ + j X v ) ⋅ j X v = 178 , 97 /9 6 ∘ 4 6 ′ Ω ; Karakteristika električne odate snage generatora posle kvara je:
P 2 = E G ′ 2 Z 11 ( 2 ) sin ψ 11 ( 2 ) + E G ′ U G n Z 12 ( 2 ) sin ( δ 12 − ψ 12 ( 2 ) ) , P_2 = \frac{E_G'^2}{Z_{11}^{(2)}}\sin\psi_{11}^{(2)} + \frac{E'_GU_{Gn}}{Z_{12}^{(2)}}\sin(\delta_{12} - \psi_{12}^{(2)}), P 2 = Z 11 ( 2 ) E G ′2 sin ψ 11 ( 2 ) + Z 12 ( 2 ) E G ′ U G n sin ( δ 12 − ψ 12 ( 2 ) ) , gde je iz prethodnog proračuna:
ψ 11 ( 2 ) = 90 ∘ − β 11 ( 2 ) = 90 ∘ − 86 ∘ 21 ′ = 3 ∘ 39 ′ ; Z ‾ 11 ( 2 ) = Z 11 ( 2 ) / β 11 ( 2 ) ‾ ; ψ 12 ( 2 ) = 90 ∘ − β 12 ( 2 ) = 90 ∘ − 96 ∘ 46 ′ = − 6 ∘ 46 ′ ; Z ‾ 12 ( 2 ) = Z 12 ( 2 ) / β 12 ( 2 ) ‾ . \begin{aligned} \psi_{11}^{(2)} &= 90^\circ - \beta_{11}^{(2)} = 90^\circ - 86^\circ21' = 3^\circ39'; && \ph{Z}_{11}^{(2)} = Z_{11}^{(2)}\underline{/\beta_{11}^{(2)}}; \\ \psi_{12}^{(2)} &= 90^\circ - \beta_{12}^{(2)} = 90^\circ - 96^\circ46' = -6^\circ46'; && \ph{Z}_{12}^{(2)} = Z_{12}^{(2)}\underline{/\beta_{12}^{(2)}}. \end{aligned} ψ 11 ( 2 ) ψ 12 ( 2 ) = 9 0 ∘ − β 11 ( 2 ) = 9 0 ∘ − 8 6 ∘ 2 1 ′ = 3 ∘ 3 9 ′ ; = 9 0 ∘ − β 12 ( 2 ) = 9 0 ∘ − 9 6 ∘ 4 6 ′ = − 6 ∘ 4 6 ′ ; Z 11 ( 2 ) = Z 11 ( 2 ) / β 11 ( 2 ) ; Z 12 ( 2 ) = Z 12 ( 2 ) / β 12 ( 2 ) . To znači da je
P 2 = 27,7 + 330 ⋅ sin ( δ 12 + 6 ∘ 46 ′ ) . P_2 = 27{,}7 + 330 \cdot \sin(\delta_{12} + 6^\circ46'). P 2 = 27 , 7 + 330 ⋅ sin ( δ 12 + 6 ∘ 4 6 ′ ) . Granični ugao određuje se iz jednačine:
P m = 300 = P 2 = 27,7 + 330 ⋅ sin ( δ 12 g r + 6 ∘ 46 ′ ) . P_m = 300 = P_2 = 27{,}7 + 330 \cdot \sin(\delta_{12}^{gr} + 6^\circ46'). P m = 300 = P 2 = 27 , 7 + 330 ⋅ sin ( δ 12 g r + 6 ∘ 4 6 ′ ) . Rešenje prethodne jednačine je
δ 12 g r = 117 ∘ 39 ′ . \delta_{12}^{gr} = 117^\circ39'. δ 12 g r = 11 7 ∘ 3 9 ′ . Ilustracija krivih snaga-ugao je:
Sa prethodne slike se vidi da se kritični ugao isključenja kvara dobija iz jednakosti površina
P m ( δ i s − δ 12 ( 1 ) ) = ∫ δ i s δ 12 g r ( P 2 − P 0 ) ⋅ d δ 12 , P_m\left(\delta_{is} - \delta_{12}^{(1)}\right) = \int\limits_{\delta_{is}}^{\delta_{12}^{gr}}\left(P_2 - P_0\right) \cdot d\delta_{12}, P m ( δ i s − δ 12 ( 1 ) ) = δ i s ∫ δ 12 g r ( P 2 − P 0 ) ⋅ d δ 12 , odnosno
300 ( δ i s − 23 ∘ 11 ′ ⋅ 3,14 180 ) = ∫ δ i s 117 ∘ 39 ′ ( 27,7 + 330 ⋅ sin ( δ 12 + 6 ∘ 46 ′ ⋅ 3,14 180 ) − 300 ) ⋅ d δ 12 . \begin{aligned} &300\left(\delta_{is} - 23^\circ11' \cdot \frac{3{,}14}{180}\right) \\ &\quad = \int\limits_{\delta_{is}}^{117^\circ39'}\left(27{,}7 + 330 \cdot \sin\left(\delta_{12} + 6^\circ46' \cdot \frac{3{,}14}{180}\right) - 300\right) \cdot d\delta_{12}. \end{aligned} 300 ( δ i s − 2 3 ∘ 1 1 ′ ⋅ 180 3 , 14 ) = δ i s ∫ 11 7 ∘ 3 9 ′ ( 27 , 7 + 330 ⋅ sin ( δ 12 + 6 ∘ 4 6 ′ ⋅ 180 3 , 14 ) − 300 ) ⋅ d δ 12 . Sređivanjem prethodne jednačine dobija se
330 ⋅ cos ( δ i s + 6 ∘ 46 ′ ⋅ 3,14 180 ) = 27,7 ⋅ δ i s + 251,1696 330 ⋅ cos ( δ i s + 0,1181 ) = 27,7 ⋅ δ i s + 251,1696. \begin{aligned} 330 \cdot \cos\left(\delta_{is} + 6^\circ46' \cdot \frac{3{,}14}{180}\right) &= 27{,}7 \cdot \delta_{is} + 251{,}1696 \\ 330 \cdot \cos\left(\delta_{is} + 0{,}1181\right) &= 27{,}7 \cdot \delta_{is} + 251{,}1696. \end{aligned} 330 ⋅ cos ( δ i s + 6 ∘ 4 6 ′ ⋅ 180 3 , 14 ) 330 ⋅ cos ( δ i s + 0 , 1181 ) = 27 , 7 ⋅ δ i s + 251 , 1696 = 27 , 7 ⋅ δ i s + 251 , 1696. Prethodna jednačina je transcendentna, pa se rešenje ne može eksplicitno izraziti. Metodom probanja (ili numerički) dobija se
δ i s = 29 ∘ 40 ′ . \delta_{is} = 29^\circ40'. δ i s = 2 9 ∘ 4 0 ′ . Kritično vreme jednovremenog obostranog isključenja voda V1 je
t i s = 2 H ( δ i s − δ 12 ( 1 ) ) π f 0 P m S G n = 4 H ( δ i s − δ 12 ( 1 ) ) ω s P m S G n = 0,139 s . t_{is} = \sqrt{\frac{2H\left(\delta_{is} - \delta_{12}^{(1)}\right)}{\pi f_0\dfrac{P_m}{S_{Gn}}}} = \sqrt{\frac{4H\left(\delta_{is} - \delta_{12}^{(1)}\right)}{\omega_s\dfrac{P_m}{S_{Gn}}}} = 0{,}139\ \text{s}. t i s = π f 0 S G n P m 2 H ( δ i s − δ 12 ( 1 ) ) = ω s S G n P m 4 H ( δ i s − δ 12 ( 1 ) ) = 0 , 139 s . Rezultat δ i s = 29 ∘ 40 ′ \delta_{is} = 29^\circ40' δ i s = 2 9 ∘ 4 0 ′ t i s = 0,139 t_{is} = 0{,}139 t i s = 0 , 139
← Prethodni zadatak Zadatak 11: Dinamička stabilnost, vod kao četvorokrajnik Sledeći zadatak → Zadatak 13: Najopasnija vrsta kratkog spoja za stabilnost 
