Zadatak 4. Kolika je granična "električna dužina" λ \lambda λ
G T 1 Z c = 330 Ω 2 jaka mreža P nat U 2 = Const S Gn = S Tn = 500 MVA U Gn = 15,75 kV m T = 15,75/400 kV/kV x dG = 200 % x T = 14 % Poznato S G n = S T n = 500 S_{Gn} = S_{Tn} = 500 S G n = S T n = 500 MVA U G n = 15,75 U_{Gn} = 15{,}75 U G n = 15 , 75 kV U T n = 400 U_{Tn} = 400 U T n = 400 kV x d G = 200 x_{dG} = 200 x d G = 200 % x T = 14 x_T = 14 x T = 14 % Z c = 330 Z_c = 330 Z c = 330 Ω Rešenje Jednopolna šema sistema je:
Četvorokrajnik između tačaka 1’ i 1 je:
V ‾ 1 ′ = A ‾ V ‾ 1 + B ‾ I ‾ 1 I ‾ 1 ′ = C ‾ V ‾ 1 + D ‾ I ‾ 1 ⇒ V ‾ 1 ′ = 1 ⋅ V ‾ 1 + j X d ( G + T ) I ‾ 1 I ‾ 1 ′ = 0 ⋅ V ‾ 1 + 1 ⋅ I ‾ 1 \begin{aligned} \ph{V}'_1 &= \ph{A}\ph{V}_1 + \ph{B}\ph{I}_1 \\ \ph{I}'_1 &= \ph{C}\ph{V}_1 + \ph{D}\ph{I}_1 \end{aligned} \quad\Rightarrow\quad \begin{aligned} \ph{V}'_1 &= 1 \cdot \ph{V}_1 + \mathrm{j}X_{d(G+T)}\ph{I}_1 \\ \ph{I}'_1 &= 0 \cdot \ph{V}_1 + 1 \cdot \ph{I}_1 \end{aligned} V 1 ′ I 1 ′ = A V 1 + B I 1 = C V 1 + D I 1 ⇒ V 1 ′ I 1 ′ = 1 ⋅ V 1 + j X d ( G + T ) I 1 = 0 ⋅ V 1 + 1 ⋅ I 1 Četvorokrajnik između tačaka 1 i 2 je (jednačine idealizovanog voda):
V ‾ 1 = A ‾ v V ‾ 2 + B ‾ v I ‾ 2 I ‾ 1 = C ‾ v V ‾ 2 + D ‾ v I ‾ 2 ⇒ V ‾ 1 = cos λ ⋅ V ‾ 2 + j Z c ⋅ sin λ ⋅ I ‾ 2 I ‾ 1 = j 1 Z c sin λ ⋅ V ‾ 2 + cos λ ⋅ I ‾ 2 \begin{aligned} \ph{V}_1 &= \ph{A}_v\ph{V}_2 + \ph{B}_v\ph{I}_2 \\ \ph{I}_1 &= \ph{C}_v\ph{V}_2 + \ph{D}_v\ph{I}_2 \end{aligned} \quad\Rightarrow\quad \begin{aligned} \ph{V}_1 &= \cos\lambda \cdot \ph{V}_2 + \mathrm{j}Z_c \cdot \sin\lambda \cdot \ph{I}_2 \\ \ph{I}_1 &= \mathrm{j}\frac{1}{Z_c}\sin\lambda \cdot \ph{V}_2 + \cos\lambda \cdot \ph{I}_2 \end{aligned} V 1 I 1 = A v V 2 + B v I 2 = C v V 2 + D v I 2 ⇒ V 1 I 1 = cos λ ⋅ V 2 + j Z c ⋅ sin λ ⋅ I 2 = j Z c 1 sin λ ⋅ V 2 + cos λ ⋅ I 2 To znači da je:
[ A ‾ e B ‾ e C ‾ e D ‾ e ] = [ 1 j X d ( G + T ) 0 1 ] [ cos λ j Z c ⋅ sin λ j 1 Z c sin λ cos λ ] . \begin{bmatrix} \ph{A}_e & \ph{B}_e \\ \ph{C}_e & \ph{D}_e \end{bmatrix} = \begin{bmatrix} 1 & \mathrm{j}X_{d(G+T)} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\lambda & \mathrm{j}Z_c \cdot \sin\lambda \\ \mathrm{j}\dfrac{1}{Z_c}\sin\lambda & \cos\lambda \end{bmatrix}. [ A e C e B e D e ] = [ 1 0 j X d ( G + T ) 1 ] cos λ j Z c 1 sin λ j Z c ⋅ sin λ cos λ . Posle ovih množenja dobija se
A ‾ e = cos λ − X d ( G + T ) Z c sin λ ; B ‾ e = j [ X d ( G + T ) ⋅ cos λ + Z c ⋅ sin λ ] ; C ‾ e = j 1 Z c sin λ ; D ‾ e = cos λ . \begin{aligned} \ph{A}_e &= \cos\lambda - \frac{X_{d(G+T)}}{Z_c}\sin\lambda; \\ \ph{B}_e &= \mathrm{j}\left[X_{d(G+T)} \cdot \cos\lambda + Z_c \cdot \sin\lambda\right]; \\ \ph{C}_e &= \mathrm{j}\frac{1}{Z_c}\sin\lambda; \\ \ph{D}_e &= \cos\lambda. \end{aligned} A e B e C e D e = cos λ − Z c X d ( G + T ) sin λ ; = j [ X d ( G + T ) ⋅ cos λ + Z c ⋅ sin λ ] ; = j Z c 1 sin λ ; = cos λ . To znači da je ulazna impedansa
Z ‾ 12 = V ‾ 1 I ‾ 1 = B ‾ e = X d ( G + T ) ⋅ cos λ + Z c ⋅ sin λ . \ph{Z}_{12} = \frac{\ph{V}_1}{\ph{I}_1} = \ph{B}_e = X_{d(G+T)} \cdot \cos\lambda + Z_c \cdot \sin\lambda. Z 12 = I 1 V 1 = B e = X d ( G + T ) ⋅ cos λ + Z c ⋅ sin λ . To znači da je aktivna snaga koja se može preneti ovim sistemom:
P = E G U 2 X d ( G + T ) ⋅ cos λ + Z c ⋅ sin λ sin θ = E G U 2 cos λ ( X d ( G + T ) + Z c ⋅ t g λ ) sin θ , \begin{aligned} P &= \frac{E_GU_2}{X_{d(G+T)} \cdot \cos\lambda + Z_c \cdot \sin\lambda}\sin\theta \\ &= \frac{E_GU_2}{\cos\lambda\left(X_{d(G+T)} + Z_c \cdot \mathrm{tg}\lambda\right)}\sin\theta, \end{aligned} P = X d ( G + T ) ⋅ cos λ + Z c ⋅ sin λ E G U 2 sin θ = cos λ ( X d ( G + T ) + Z c ⋅ tg λ ) E G U 2 sin θ , gde je θ \theta θ E G E_G E G V 2 V_2 V 2
U režimu prenosa prirodne snage voda, naponi i struje duž voda (odnosno na njegovom početku i kraju) se ne menjaju, pa je V 1 = V 2 = V V_1 = V_2 = V V 1 = V 2 = V P n a t = U 2 Z c P_{nat} = \frac{U^2}{Z_c} P na t = Z c U 2 V ‾ 2 = V 2 \ph{V}_2 = V_2 V 2 = V 2
Sa zamenske šeme je
E ‾ G = U + j P n a t X d ( G + T ) U = U + j U ⋅ X d ( G + T ) Z c , \ph{E}_G = U + \mathrm{j}\frac{P_{nat}X_{d(G+T)}}{U} = U + \mathrm{j}\frac{U \cdot X_{d(G+T)}}{Z_c}, E G = U + j U P na t X d ( G + T ) = U + j Z c U ⋅ X d ( G + T ) , pa je efektivna vrednost unutrašnje elektromotorne sile generatora
E G = U Z c Z c 2 + X d ( G + T ) 2 . E_G = \frac{U}{Z_c}\sqrt{Z_c^2 + X_{d(G+T)}^2}. E G = Z c U Z c 2 + X d ( G + T ) 2 . Iz prethodnog izraza za snagu prenosa, granična snaga prenosa ima se za θ = π / 2 \theta = \pi/2 θ = π /2
P = P n a t = U 2 Z c = E G U cos λ ( X d ( G + T ) + Z c ⋅ t g λ ) . P = P_{nat} = \frac{U^2}{Z_c} = \frac{E_GU}{\cos\lambda\left(X_{d(G+T)} + Z_c \cdot \mathrm{tg}\lambda\right)}. P = P na t = Z c U 2 = cos λ ( X d ( G + T ) + Z c ⋅ tg λ ) E G U . Ako se u ovu jednačinu zameni izvedeno E G E_G E G
P = U 2 Z c = U 2 Z c Z c 2 + X d ( G + T ) 2 cos λ ( X d ( G + T ) + Z c ⋅ t g λ ) . P = \frac{U^2}{Z_c} = \frac{\dfrac{U^2}{Z_c}\sqrt{Z_c^2 + X_{d(G+T)}^2}}{\cos\lambda\left(X_{d(G+T)} + Z_c \cdot \mathrm{tg}\lambda\right)}. P = Z c U 2 = cos λ ( X d ( G + T ) + Z c ⋅ tg λ ) Z c U 2 Z c 2 + X d ( G + T ) 2 . Iz ove jednačine je
cos λ ( X d ( G + T ) + Z c ⋅ t g λ ) = Z c 2 + X d ( G + T ) 2 . \cos\lambda\left(X_{d(G+T)} + Z_c \cdot \mathrm{tg}\lambda\right) = \sqrt{Z_c^2 + X_{d(G+T)}^2}. cos λ ( X d ( G + T ) + Z c ⋅ tg λ ) = Z c 2 + X d ( G + T ) 2 . Sređivanjem se dobija
( X d ( G + T ) ⋅ sin λ − Z c ⋅ cos λ ) 2 = 0 , odnosno t g λ = Z c X d ( G + T ) \left(X_{d(G+T)} \cdot \sin\lambda - Z_c \cdot \cos\lambda\right)^2 = 0, \text{ odnosno } \mathrm{tg}\lambda = \frac{Z_c}{X_{d(G+T)}} ( X d ( G + T ) ⋅ sin λ − Z c ⋅ cos λ ) 2 = 0 , odnosno tg λ = X d ( G + T ) Z c Konkretna vrednost reaktanse bloka generator-transformator je
X d ( G + T ) = ( x d G + x T ) ( % ) 100 ⋅ U G n 2 S G n = 214 100 ⋅ 400 2 500 = 684,8 Ω , \begin{aligned} X_{d(G+T)} &= \frac{(x_{dG} + x_T)(\%)}{100} \cdot \frac{U_{Gn}^2}{S_{Gn}} \\ &= \frac{214}{100} \cdot \frac{400^2}{500} = 684{,}8\ \Omega, \end{aligned} X d ( G + T ) = 100 ( x d G + x T ) ( % ) ⋅ S G n U G n 2 = 100 214 ⋅ 500 40 0 2 = 684 , 8 Ω , pa je
t g λ = Z c X d ( G + T ) = 330 684,8 = 0,482 , \mathrm{tg}\lambda = \frac{Z_c}{X_{d(G+T)}} = \frac{330}{684{,}8} = 0{,}482, tg λ = X d ( G + T ) Z c = 684 , 8 330 = 0 , 482 , odnosno
λ = 25,734 ∘ = 0,06 ∘ ⋅ L , \lambda = 25{,}734^\circ = 0{,}06^\circ \cdot L, λ = 25 , 73 4 ∘ = 0 , 0 6 ∘ ⋅ L , pa je dozvoljena dužina voda
L = 25,734 ∘ 0,06 ∘ = 428 km . L = \frac{25{,}734^\circ}{0{,}06^\circ} = 428\ \text{km}. L = 0 , 0 6 ∘ 25 , 73 4 ∘ = 428 km . Rezultat t g λ = Z c / X d ( G + T ) = 0,482 \mathrm{tg}\lambda = Z_c/X_{d(G+T)} = 0{,}482 tg λ = Z c / X d ( G + T ) = 0 , 482 λ = 25,734 ∘ \lambda = 25{,}734^\circ λ = 25 , 73 4 ∘ L = 428 L = 428 L = 428
← Prethodni zadatak Zadatak 3: Statička stabilnost sa potrošačem kao impedansom Sledeći zadatak → Zadatak 5: Statička stabilnost i metod sabiranja uglova 
