Zadatak Ilustracija svođenja veličina na jedinstven naponski nivo 1. Odrediti napone U 1 U_1 U 1 U 2 U_2 U 2 U 3 U_3 U 3 E g E_g E g I 1 I_1 I 1 I 2 I_2 I 2 I 3 I_3 I 3 I P I_P I P
svođenjem na naponski nivo mreže I, svođenjem na naponski nivo mreže II, svođenjem na naponski nivo mreže III. G 1 1 T 1 2 L v = 30 km x v = 0,38 Ω/km 3 T 2 S P = (6 + j2) MVA U P = 9,8 kV III II I S nG = S nT1 = 30 MVA U nG = 10,5 kV m T1 = 10/36,5 kV/kV X G = 110 % X T1 % = 10 % S nT2 = 20 MVA m T2 = 35/10 kV/kV X T2 % = 10 % Poznato P P = 6 P_P = 6 P P = 6 MW Q P = 2 Q_P = 2 Q P = 2 MVAr U P = 9,8 U_P = 9{,}8 U P = 9 , 8 kV L v = 30 L_v = 30 L v = 30 km x v = 0,38 x_v = 0{,}38 x v = 0 , 38 Ω/km S n G = S n T 1 = 30 S_{nG} = S_{nT1} = 30 S n G = S n T 1 = 30 MVA U n G = 10,5 U_{nG} = 10{,}5 U n G = 10 , 5 kV U n 1 T 1 = 10 U_{n1}^{T1} = 10 U n 1 T 1 = 10 kV U n 2 T 1 = 36,5 U_{n2}^{T1} = 36{,}5 U n 2 T 1 = 36 , 5 kV X G = 110 X_G = 110 X G = 110 % X T 1 % = 10 X_{T1}\% = 10 X T 1 % = 10 % S n T 2 = 20 S_{nT2} = 20 S n T 2 = 20 MVA U n 1 T 2 = 35 U_{n1}^{T2} = 35 U n 1 T 2 = 35 kV U n 2 T 2 = 10 U_{n2}^{T2} = 10 U n 2 T 2 = 10 kV X T 2 % = 10 X_{T2}\% = 10 X T 2 % = 10 % Rešenje Uobičajeno je da se svođenje vrši na onaj naponski nivo koji ima najviše elemenata, ili na naponski nivo koji ima najviše podataka.
Rezultati ne zavise od načina svođenja veličina!
Proračun originalnih parametara:
X G = x G U n G 2 S n G = 1,1 ⋅ 10,5 2 30 = 4,04 Ω X_G = x_G \frac{U_{nG}^2}{S_{nG}} = 1{,}1 \cdot \frac{10{,}5^2}{30} = 4{,}04\ \Omega X G = x G S n G U n G 2 = 1 , 1 ⋅ 30 10 , 5 2 = 4 , 04 Ω X T 1 = x T 1 U n T 1 2 S n T 1 = 0,1 ⋅ 36,5 2 30 = 4,44 Ω X_{T1} = x_{T1} \frac{U_{nT1}^2}{S_{nT1}} = 0{,}1 \cdot \frac{36{,}5^2}{30} = 4{,}44\ \Omega X T 1 = x T 1 S n T 1 U n T 1 2 = 0 , 1 ⋅ 30 36 , 5 2 = 4 , 44 Ω X V = x V L = 0,38 ⋅ 30 = 11,4 Ω X_V = x_V L = 0{,}38 \cdot 30 = 11{,}4\ \Omega X V = x V L = 0 , 38 ⋅ 30 = 11 , 4 Ω X T 2 = x T 2 U n T 2 2 S n T 2 = 0,1 ⋅ 35 2 20 = 6,125 Ω X_{T2} = x_{T2} \frac{U_{nT2}^2}{S_{nT2}} = 0{,}1 \cdot \frac{35^2}{20} = 6{,}125\ \Omega X T 2 = x T 2 S n T 2 U n T 2 2 = 0 , 1 ⋅ 20 3 5 2 = 6 , 125 Ω Odnosi svođenja:
U s v , I U s v , I I = 1 m T 2 = 10 35 = 0,2857 ≈ 0,286 ; U s v , I U s v , I I I = 1 m T 2 ⋅ 1 m T 1 = 10 35 ⋅ 36,5 10 = 1,043 \begin{aligned} \frac{U^{sv,I}}{U^{sv,II}} &= \frac{1}{m_{T2}} = \frac{10}{35} = 0{,}2857 \approx 0{,}286; \\ \frac{U^{sv,I}}{U^{sv,III}} &= \frac{1}{m_{T2}} \cdot \frac{1}{m_{T1}} \\ &= \frac{10}{35} \cdot \frac{36{,}5}{10} = 1{,}043 \end{aligned} U s v , I I U s v , I U s v , I I I U s v , I = m T 2 1 = 35 10 = 0 , 2857 ≈ 0 , 286 ; = m T 2 1 ⋅ m T 1 1 = 35 10 ⋅ 10 36 , 5 = 1 , 043 Opšti izrazi za svođenje napona i impedansi:
U j s v , i = U j U s v , i U s v , j ; Z ‾ j s v , i = Z ‾ j ( U s v , i U s v , j ) 2 U_j^{sv,i} = U_j \frac{U^{sv,i}}{U^{sv,j}}; \qquad \ph{Z}_j^{sv,i} = \ph{Z}_j \left( \frac{U^{sv,i}}{U^{sv,j}} \right)^2 U j s v , i = U j U s v , j U s v , i ; Z j s v , i = Z j ( U s v , j U s v , i ) 2 Primenom prethodnih izraza dobijaju se svedene impedanse (reaktanse):
X G s v , I = X G ( U s v , I U s v , I I I ) 2 = 4,04 ⋅ 1,043 2 = 4,395 Ω X T 1 s v , I = X T 1 ( U s v , I U s v , I I ) 2 = 4,44 ⋅ 0,286 2 = 0,363 Ω X V s v , I = X V ( U s v , I U s v , I I ) 2 = 11,4 ⋅ 0,286 2 = 0,932 Ω X T 2 s v , I = X T 2 ( U s v , I U s v , I I ) 2 = 6,125 ⋅ 0,286 2 = 0,5 Ω \begin{aligned} X_G^{sv,I} &= X_G \left( \frac{U^{sv,I}}{U^{sv,III}} \right)^2 \\ &= 4{,}04 \cdot 1{,}043^2 = 4{,}395\ \Omega \\[1ex] X_{T1}^{sv,I} &= X_{T1} \left( \frac{U^{sv,I}}{U^{sv,II}} \right)^2 \\ &= 4{,}44 \cdot 0{,}286^2 = 0{,}363\ \Omega \\[1ex] X_V^{sv,I} &= X_V \left( \frac{U^{sv,I}}{U^{sv,II}} \right)^2 \\ &= 11{,}4 \cdot 0{,}286^2 = 0{,}932\ \Omega \\[1ex] X_{T2}^{sv,I} &= X_{T2} \left( \frac{U^{sv,I}}{U^{sv,II}} \right)^2 \\ &= 6{,}125 \cdot 0{,}286^2 = 0{,}5\ \Omega \end{aligned} X G s v , I X T 1 s v , I X V s v , I X T 2 s v , I = X G ( U s v , I I I U s v , I ) 2 = 4 , 04 ⋅ 1 , 04 3 2 = 4 , 395 Ω = X T 1 ( U s v , I I U s v , I ) 2 = 4 , 44 ⋅ 0 , 28 6 2 = 0 , 363 Ω = X V ( U s v , I I U s v , I ) 2 = 11 , 4 ⋅ 0 , 28 6 2 = 0 , 932 Ω = X T 2 ( U s v , I I U s v , I ) 2 = 6 , 125 ⋅ 0 , 28 6 2 = 0 , 5 Ω Zamenska šema sa veličinama svedenim na naponski nivo I je:
I ‾ p s v , I = I ‾ p = S ‾ p ∗ 3 U p = 6 − j 2 3 ⋅ 9,8 = ( 0,354 − j 0,118 ) kA \begin{aligned} \ph{I}_p^{sv,I} = \ph{I}_p &= \frac{\cj{\ph{S}_p}}{\sqrt{3}\, U_p} = \frac{6 - \mathrm{j}2}{\sqrt{3} \cdot 9{,}8} \\ &= (0{,}354 - \mathrm{j}0{,}118)\ \text{kA} \end{aligned} I p s v , I = I p = 3 U p S p ∗ = 3 ⋅ 9 , 8 6 − j 2 = ( 0 , 354 − j 0 , 118 ) kA U ‾ 3 s v , I = U ‾ p + 3 ⋅ j X T 2 s v , I I ‾ p s v , I = 9,8 + 3 ⋅ j 0,5 ⋅ ( 0,354 − j 0,118 ) = ( 9,902 + j 0,306 ) kV \begin{aligned} \ph{U}_3^{sv,I} &= \ph{U}_p + \sqrt{3} \cdot \mathrm{j} X_{T2}^{sv,I} \ph{I}_p^{sv,I} \\ &= 9{,}8 + \sqrt{3} \cdot \mathrm{j}0{,}5 \cdot (0{,}354 - \mathrm{j}0{,}118) \\ &= (9{,}902 + \mathrm{j}0{,}306)\ \text{kV} \end{aligned} U 3 s v , I = U p + 3 ⋅ j X T 2 s v , I I p s v , I = 9 , 8 + 3 ⋅ j 0 , 5 ⋅ ( 0 , 354 − j 0 , 118 ) = ( 9 , 902 + j 0 , 306 ) kV U ‾ 2 s v , I = U ‾ 3 s v , I + 3 ⋅ j X V s v , I I ‾ p s v , I = ( 10,093 + j 0,879 ) kV U ‾ 1 s v , I = U ‾ 2 s v , I + 3 ⋅ j X T 1 s v , I I ‾ p s v , I = ( 10,167 + j 1,102 ) kV E ‾ G s v , I = U ‾ 1 s v , I + 3 ⋅ j X G s v , I I ‾ p s v , I = ( 11,064 + j 3,799 ) kV \begin{aligned} \ph{U}_2^{sv,I} &= \ph{U}_3^{sv,I} + \sqrt{3} \cdot \mathrm{j} X_V^{sv,I} \ph{I}_p^{sv,I} \\ &= (10{,}093 + \mathrm{j}0{,}879)\ \text{kV} \\[1ex] \ph{U}_1^{sv,I} &= \ph{U}_2^{sv,I} + \sqrt{3} \cdot \mathrm{j} X_{T1}^{sv,I} \ph{I}_p^{sv,I} \\ &= (10{,}167 + \mathrm{j}1{,}102)\ \text{kV} \\[1ex] \ph{E}_G^{sv,I} &= \ph{U}_1^{sv,I} + \sqrt{3} \cdot \mathrm{j} X_G^{sv,I} \ph{I}_p^{sv,I} \\ &= (11{,}064 + \mathrm{j}3{,}799)\ \text{kV} \end{aligned} U 2 s v , I U 1 s v , I E G s v , I = U 3 s v , I + 3 ⋅ j X V s v , I I p s v , I = ( 10 , 093 + j 0 , 879 ) kV = U 2 s v , I + 3 ⋅ j X T 1 s v , I I p s v , I = ( 10 , 167 + j 1 , 102 ) kV = U 1 s v , I + 3 ⋅ j X G s v , I I p s v , I = ( 11 , 064 + j 3 , 799 ) kV Provera prethodnih rezultata se može izvršiti primenom sledećeg izraza
I ‾ p s v , I = E ‾ G f s v , I Z ‾ e k v = E ‾ G s v , I 3 Z ‾ e k v \ph{I}_p^{sv,I} = \frac{\ph{E}_{Gf}^{sv,I}}{\ph{Z}_{ekv}} = \frac{\ph{E}_G^{sv,I}}{\sqrt{3}\, \ph{Z}_{ekv}} I p s v , I = Z e k v E G f s v , I = 3 Z e k v E G s v , I gde je Z ‾ e k v \ph{Z}_{ekv} Z e k v
Z ‾ e k v = j ( X G s v , I + X T 1 s v , I + X V s v , I + X T 2 s v , I ) + V p 2 S ‾ p ∗ = ( 14,406 + j 10,997 ) Ω \begin{aligned} \ph{Z}_{ekv} &= \mathrm{j}\left( X_G^{sv,I} + X_{T1}^{sv,I} + X_V^{sv,I} + X_{T2}^{sv,I} \right) \\ &\quad + \frac{V_p^2}{\cj{\ph{S}_p}} \\ &= (14{,}406 + \mathrm{j}10{,}997)\ \Omega \end{aligned} Z e k v = j ( X G s v , I + X T 1 s v , I + X V s v , I + X T 2 s v , I ) + S p ∗ V p 2 = ( 14 , 406 + j 10 , 997 ) Ω Zamenom Z ‾ e k v \ph{Z}_{ekv} Z e k v
I ‾ p s v , I = E ‾ G f s v , I Z ‾ e k v = ( 11,064 + j 3,799 ) 3 ⋅ ( 14,406 + j 10,997 ) = ( 0,354 − j 0,118 ) kA \begin{aligned} \ph{I}_p^{sv,I} &= \frac{\ph{E}_{Gf}^{sv,I}}{\ph{Z}_{ekv}} = \frac{(11{,}064 + \mathrm{j}3{,}799)}{\sqrt{3} \cdot (14{,}406 + \mathrm{j}10{,}997)} \\ &= (0{,}354 - \mathrm{j}0{,}118)\ \text{kA} \end{aligned} I p s v , I = Z e k v E G f s v , I = 3 ⋅ ( 14 , 406 + j 10 , 997 ) ( 11 , 064 + j 3 , 799 ) = ( 0 , 354 − j 0 , 118 ) kA Proračun stvarnih veličina iz njihovih svedenih vrednosti polazi od napred izvedenih izraza
U j s v , i = U j U s v , i U s v , j U_j^{sv,i} = U_j \frac{U^{sv,i}}{U^{sv,j}} U j s v , i = U j U s v , j U s v , i To znači da je
U ‾ 3 = U ‾ 3 s v , I U s v , I I U s v , I = ( 9,902 + j 0,306 ) ⋅ 1 0,286 = ( 34,622 + j 1,07 ) kV ; U ‾ 2 = U ‾ 2 s v , I U s v , I I U s v , I = ( 10,093 + j 0,879 ) ⋅ 1 0,286 = ( 35,29 + j 3,073 ) kV ; U ‾ 1 = U ‾ 1 s v , I U s v , I I I U s v , I = ( 10,167 + j 1,102 ) ⋅ 1 1,043 = ( 9,748 + j 1,057 ) kV ; E ‾ G = E ‾ G s v , I U s v , I I I U s v , I = ( 11,064 + j 3,799 ) ⋅ 1 1,043 = ( 10,608 + j 3,642 ) kV ; \begin{aligned} \ph{U}_3 &= \ph{U}_3^{sv,I} \frac{U^{sv,II}}{U^{sv,I}} \\ &= (9{,}902 + \mathrm{j}0{,}306) \cdot \frac{1}{0{,}286} \\ &= (34{,}622 + \mathrm{j}1{,}07)\ \text{kV}; \\[1ex] \ph{U}_2 &= \ph{U}_2^{sv,I} \frac{U^{sv,II}}{U^{sv,I}} \\ &= (10{,}093 + \mathrm{j}0{,}879) \cdot \frac{1}{0{,}286} \\ &= (35{,}29 + \mathrm{j}3{,}073)\ \text{kV}; \\[1ex] \ph{U}_1 &= \ph{U}_1^{sv,I} \frac{U^{sv,III}}{U^{sv,I}} \\ &= (10{,}167 + \mathrm{j}1{,}102) \cdot \frac{1}{1{,}043} \\ &= (9{,}748 + \mathrm{j}1{,}057)\ \text{kV}; \\[1ex] \ph{E}_G &= \ph{E}_G^{sv,I} \frac{U^{sv,III}}{U^{sv,I}} \\ &= (11{,}064 + \mathrm{j}3{,}799) \cdot \frac{1}{1{,}043} \\ &= (10{,}608 + \mathrm{j}3{,}642)\ \text{kV}; \end{aligned} U 3 U 2 U 1 E G = U 3 s v , I U s v , I U s v , I I = ( 9 , 902 + j 0 , 306 ) ⋅ 0 , 286 1 = ( 34 , 622 + j 1 , 07 ) kV ; = U 2 s v , I U s v , I U s v , I I = ( 10 , 093 + j 0 , 879 ) ⋅ 0 , 286 1 = ( 35 , 29 + j 3 , 073 ) kV ; = U 1 s v , I U s v , I U s v , I I I = ( 10 , 167 + j 1 , 102 ) ⋅ 1 , 043 1 = ( 9 , 748 + j 1 , 057 ) kV ; = E G s v , I U s v , I U s v , I I I = ( 11 , 064 + j 3 , 799 ) ⋅ 1 , 043 1 = ( 10 , 608 + j 3 , 642 ) kV ; Napomena: tačke b i c rešiti za domaći. Rezultat Svođenjem na naponski nivo mreže I: I ‾ p = ( 0,354 − j 0,118 ) \ph{I}_p = (0{,}354 - \mathrm{j}0{,}118) I p = ( 0 , 354 − j 0 , 118 ) U ‾ 3 = ( 34,622 + j 1,07 ) \ph{U}_3 = (34{,}622 + \mathrm{j}1{,}07) U 3 = ( 34 , 622 + j 1 , 07 ) U ‾ 2 = ( 35,29 + j 3,073 ) \ph{U}_2 = (35{,}29 + \mathrm{j}3{,}073) U 2 = ( 35 , 29 + j 3 , 073 ) U ‾ 1 = ( 9,748 + j 1,057 ) \ph{U}_1 = (9{,}748 + \mathrm{j}1{,}057) U 1 = ( 9 , 748 + j 1 , 057 ) E ‾ G = ( 10,608 + j 3,642 ) \ph{E}_G = (10{,}608 + \mathrm{j}3{,}642) E G = ( 10 , 608 + j 3 , 642 )
Sledeći zadatak → Zadatak 2: Petljasta mreža u praznom hodu, svođenje veličina 