Zadatak 11. Za EES sistem prikazan na slici proveriti dinamičku stabilnost generatora za slučaj trofaznog kratkog spoja na početku jednog od dva voda i vremenom isključenja voda u kvaru t i t_i t i
G T 1 A l v = 200 km Z ce B T 2 C jaka mreža P C = 0,8·ΣP nat cos φ C = 0,9 ind S Gn = S T1n = 1,2·ΣP nat m T1 = U Gn /1,05·U vn kV/kV x dG = 30 % x T1 = 12 % H = 8 s S T2n = 1,08·ΣP nat m T2 = U vn /U Mn x T2 = 12 % U C = U Mn Poznato S G n / ∑ P n a t = S T 1 n / ∑ P n a t = 1,2 S_{Gn}/\sum P_{nat} = S_{T1n}/\sum P_{nat} = 1{,}2 S G n / ∑ P na t = S T 1 n / ∑ P na t = 1 , 2 U T 1 n / U v n = 1,05 U_{T1n}/U_{vn} = 1{,}05 U T 1 n / U v n = 1 , 05 x d G = 30 x_{dG} = 30 x d G = 30 % x T 1 = 12 x_{T1} = 12 x T 1 = 12 % H = 8 H = 8 H = 8 s l v = 200 l_v = 200 l v = 200 km S T 2 n / ∑ P n a t = 1,08 S_{T2n}/\sum P_{nat} = 1{,}08 S T 2 n / ∑ P na t = 1 , 08 x T 2 = 12 x_{T2} = 12 x T 2 = 12 % P C / ∑ P n a t = 0,8 P_C/\sum P_{nat} = 0{,}8 P C / ∑ P na t = 0 , 8 cos φ C = 0,9 \cos\varphi_C = 0{,}9 cos φ C = 0 , 9 ind. t i = 0,2 t_i = 0{,}2 t i = 0 , 2 s Rešenje Jednopolna šema sistema je:
U zamenskoj šemi generator je predstavljen tranzijentnom reaktansom X d G ′ X'_{dG} X d G ′ E ‾ G ′ \ph{E}'_G E G ′ Z c e = Z c / 2 Z_{ce} = Z_c/2 Z ce = Z c /2 Z c Z_c Z c U c U_c U c
Ako se elementi predstave preko odgovarajućih četvorokrajnika dobija se:
Kada se srede ovi četvorokrajnici dobija se
[ A ‾ e B ‾ e C ‾ e D ‾ e ] = [ 1 j X ( d G + T 1 ) ′ 0 1 ] [ cos λ j Z c e ⋅ sin λ j 1 Z c e sin λ cos λ ] [ 1 j X T 2 0 1 ] . \begin{bmatrix} \ph{A}_e & \ph{B}_e \\ \ph{C}_e & \ph{D}_e \end{bmatrix} = \begin{bmatrix} 1 & \mathrm{j}X'_{(dG+T1)} \\ 0 & 1 \end{bmatrix}\begin{bmatrix} \cos\lambda & \mathrm{j}Z_{ce} \cdot \sin\lambda \\ \mathrm{j}\dfrac{1}{Z_{ce}}\sin\lambda & \cos\lambda \end{bmatrix}\begin{bmatrix} 1 & \mathrm{j}X_{T2} \\ 0 & 1 \end{bmatrix}. [ A e C e B e D e ] = [ 1 0 j X ( d G + T 1 ) ′ 1 ] cos λ j Z ce 1 sin λ j Z ce ⋅ sin λ cos λ [ 1 0 j X T 2 1 ] . Posle ovih množenja dobija se:
A ‾ e = cos λ − X ( d G + T 1 ) ′ Z c e sin λ ; B ‾ e = j [ X T 2 ⋅ cos λ − X T 2 Z c e X ( d G + T 1 ) ′ ⋅ sin λ + Z c e ⋅ sin λ + X ( d G + T ) ′ ⋅ cos λ ] ; C ‾ e = j 1 Z c e sin λ ; D ‾ e = cos λ − X T 2 Z c e sin λ . \begin{aligned} \ph{A}_e &= \cos\lambda - \frac{X'_{(dG+T1)}}{Z_{ce}}\sin\lambda; \\ \ph{B}_e &= \mathrm{j}\left[X_{T2} \cdot \cos\lambda - \frac{X_{T2}}{Z_{ce}}X'_{(dG+T1)} \cdot \sin\lambda + Z_{ce} \cdot \sin\lambda + X'_{(dG+T)} \cdot \cos\lambda\right]; \\ \ph{C}_e &= \mathrm{j}\frac{1}{Z_{ce}}\sin\lambda; \\ \ph{D}_e &= \cos\lambda - \frac{X_{T2}}{Z_{ce}}\sin\lambda. \end{aligned} A e B e C e D e = cos λ − Z ce X ( d G + T 1 ) ′ sin λ ; = j [ X T 2 ⋅ cos λ − Z ce X T 2 X ( d G + T 1 ) ′ ⋅ sin λ + Z ce ⋅ sin λ + X ( d G + T ) ′ ⋅ cos λ ] ; = j Z ce 1 sin λ ; = cos λ − Z ce X T 2 sin λ . Parametri zamenske šeme su:
X ( d G + T 1 ) ′ = x d G ′ + x T 1 ( % ) 100 ⋅ U G n 2 S G n ⋅ ( 1,05 U v n ) 2 U G n 2 = 0,42 1,2 ∑ P n a t ⋅ ( 1,05 U v n ) 2 = 0,3859 U v n 2 ∑ P n a t ; X T 2 = x T 2 ( % ) 100 ⋅ U T 2 n 2 S T 2 n = 0,111 U v n 2 ∑ P n a t . \begin{aligned} X'_{(dG+T1)} &= \frac{x'_{dG} + x_{T1}(\%)}{100} \cdot \frac{U_{Gn}^2}{S_{Gn}} \cdot \frac{\left(1{,}05U_{vn}\right)^2}{U_{Gn}^2} \\ &= \frac{0{,}42}{1{,}2\sum P_{nat}} \cdot \left(1{,}05U_{vn}\right)^2 = 0{,}3859\dfrac{U_{vn}^2}{\sum P_{nat}}; \\ X_{T2} &= \frac{x_{T2}(\%)}{100} \cdot \frac{U_{T2n}^2}{S_{T2n}} = 0{,}111\dfrac{U_{vn}^2}{\sum P_{nat}}. \end{aligned} X ( d G + T 1 ) ′ X T 2 = 100 x d G ′ + x T 1 ( % ) ⋅ S G n U G n 2 ⋅ U G n 2 ( 1 , 05 U v n ) 2 = 1 , 2 ∑ P na t 0 , 42 ⋅ ( 1 , 05 U v n ) 2 = 0 , 3859 ∑ P na t U v n 2 ; = 100 x T 2 ( % ) ⋅ S T 2 n U T 2 n 2 = 0 , 111 ∑ P na t U v n 2 . Radne veličine koje odgovaraju zadatom radnom stanju sistema na sabirnicama C su:
P = 0,8 ∑ P n a t ; Q = P t g φ = 0,8 ∑ P n a t 1 − cos 2 φ cos φ = 0,8 ∑ P n a t ⋅ 0,484 = 0,3872 ∑ P n a t . \begin{gathered} P = 0{,}8\sum P_{nat}; \\ Q = P\,\mathrm{tg}\varphi = 0{,}8\sum P_{nat}\frac{\sqrt{1 - \cos^2\varphi}}{\cos\varphi} = 0{,}8\sum P_{nat} \cdot 0{,}484 = 0{,}3872\sum P_{nat}. \end{gathered} P = 0 , 8 ∑ P na t ; Q = P tg φ = 0 , 8 ∑ P na t cos φ 1 − cos 2 φ = 0 , 8 ∑ P na t ⋅ 0 , 484 = 0 , 3872 ∑ P na t . U prethodnim izrazima ∑ P n a t \sum P_{nat} ∑ P na t
Ugao između elektromotorne sile E ‾ G ′ \ph{E}'_G E G ′ U ‾ C \ph{U}_C U C θ 0 \theta_0 θ 0 U ‾ C \ph{U}_C U C U ‾ C = U C / 0 ∘ ‾ = U M n U v n U M n / 0 ∘ ‾ = U v n / 0 ∘ ‾ \ph{U}_C = U_C\underline{/0^\circ} = U_{Mn}\frac{U_{vn}}{U_{Mn}}\underline{/0^\circ} = U_{vn}\underline{/0^\circ} U C = U C / 0 ∘ = U M n U M n U v n / 0 ∘ = U v n / 0 ∘ E ‾ G ′ = E G ′ / θ 0 ‾ \ph{E}'_G = E'_G\underline{/\theta_0} E G ′ = E G ′ / θ 0
Ova promenljiva može se naći iz prve jednačine koja važi za četvorokrajnik sa prethodne slike:
E ‾ G f ′ = A ‾ e V ‾ C + B ‾ e I ‾ , \ph{E}'_{Gf} = \ph{A}_e\ph{V}_C + \ph{B}_e\ph{I}, E G f ′ = A e V C + B e I , gde su:
A ‾ e = cos λ − X ( d G + T ) ′ Z c e sin λ = cos ( 12 ∘ ) − 0,3859 U v n 2 ∑ P n a t ∑ P n a t ⋅ U v n 2 sin ( 12 ∘ ) = 0,89815 ; λ = 0,06 ⋅ 200 = 12 ∘ ; Z c e = U v n 2 ∑ P n a t ; \begin{aligned} \ph{A}_e &= \cos\lambda - \frac{X'_{(dG+T)}}{Z_{ce}}\sin\lambda \\ &= \cos(12^\circ) - \frac{0{,}3859U_{vn}^2\sum P_{nat}}{\sum P_{nat} \cdot U_{vn}^2}\sin(12^\circ) = 0{,}89815; \\ \lambda &= 0{,}06 \cdot 200 = 12^\circ; \\ Z_{ce} &= \dfrac{U_{vn}^2}{\sum P_{nat}}; \end{aligned} A e λ Z ce = cos λ − Z ce X ( d G + T ) ′ sin λ = cos ( 1 2 ∘ ) − ∑ P na t ⋅ U v n 2 0 , 3859 U v n 2 ∑ P na t sin ( 1 2 ∘ ) = 0 , 89815 ; = 0 , 06 ⋅ 200 = 1 2 ∘ ; = ∑ P na t U v n 2 ; B ‾ e = j [ X T 2 ⋅ cos λ − X T 2 Z c e X ( d G + T ) ′ ⋅ sin λ + Z c e ⋅ sin λ + X ( d G + T ) ′ ⋅ cos λ ] = j ⋅ cos λ [ X T 2 + X ( d G + T ) ′ ( 1 − X T 2 Z c e t g λ ) + Z c e ⋅ t g λ ] = j ⋅ cos ( 12 ∘ ) [ 0,3859 U v n 2 ∑ P n a t ( 1 − 0,111 U v n 2 ∑ P n a t U v n 2 ∑ P n a t t g ( 12 ∘ ) ) + U v n 2 ∑ P n a t t g ( 12 ∘ ) + 0,111 U v n 2 ∑ P n a t ] = j ⋅ 0,684 U v n 2 ∑ P n a t \begin{aligned} \ph{B}_e &= \mathrm{j}\left[X_{T2} \cdot \cos\lambda - \frac{X_{T2}}{Z_{ce}}X'_{(dG+T)} \cdot \sin\lambda + Z_{ce} \cdot \sin\lambda + X'_{(dG+T)} \cdot \cos\lambda\right] \\ &= \mathrm{j} \cdot \cos\lambda\left[X_{T2} + X'_{(dG+T)}\left(1 - \frac{X_{T2}}{Z_{ce}}\mathrm{tg}\lambda\right) + Z_{ce} \cdot \mathrm{tg}\lambda\right] \\ &= \mathrm{j} \cdot \cos(12^\circ)\left[0{,}3859\dfrac{U_{vn}^2}{\sum P_{nat}}\left(1 - \frac{0{,}111\dfrac{U_{vn}^2}{\sum P_{nat}}}{\dfrac{U_{vn}^2}{\sum P_{nat}}}\mathrm{tg}(12^\circ)\right) + \dfrac{U_{vn}^2}{\sum P_{nat}}\mathrm{tg}(12^\circ) + 0{,}111\dfrac{U_{vn}^2}{\sum P_{nat}}\right] \\ &= \mathrm{j} \cdot 0{,}684\dfrac{U_{vn}^2}{\sum P_{nat}} \end{aligned} B e = j [ X T 2 ⋅ cos λ − Z ce X T 2 X ( d G + T ) ′ ⋅ sin λ + Z ce ⋅ sin λ + X ( d G + T ) ′ ⋅ cos λ ] = j ⋅ cos λ [ X T 2 + X ( d G + T ) ′ ( 1 − Z ce X T 2 tg λ ) + Z ce ⋅ tg λ ] = j ⋅ cos ( 1 2 ∘ ) 0 , 3859 ∑ P na t U v n 2 1 − ∑ P na t U v n 2 0 , 111 ∑ P na t U v n 2 tg ( 1 2 ∘ ) + ∑ P na t U v n 2 tg ( 1 2 ∘ ) + 0 , 111 ∑ P na t U v n 2 = j ⋅ 0 , 684 ∑ P na t U v n 2 Struja I ‾ \ph{I} I
I ‾ = S ‾ ∗ V ‾ C ∗ = S e − j φ V v n = P cos φ V v n e − j φ = 0,8 ∑ P n a t cos φ ⋅ V v n e − j φ = 0,889 ∑ P n a t V v n e − j 25,84 ∘ , \begin{aligned} \ph{I} &= \frac{\ph{S}^*}{\ph{V}_C^*} = \frac{Se^{-\mathrm{j}\varphi}}{V_{vn}} = \frac{\dfrac{P}{\cos\varphi}}{V_{vn}}e^{-\mathrm{j}\varphi} = \frac{0{,}8\sum P_{nat}}{\cos\varphi \cdot V_{vn}}e^{-\mathrm{j}\varphi} \\ &= 0{,}889\frac{\sum P_{nat}}{V_{vn}}e^{-\mathrm{j}25{,}84^\circ}, \end{aligned} I = V C ∗ S ∗ = V v n S e − j φ = V v n cos φ P e − j φ = cos φ ⋅ V v n 0 , 8 ∑ P na t e − j φ = 0 , 889 V v n ∑ P na t e − j 25 , 8 4 ∘ , Kada se ova struja zameni u izraz E ‾ G ′ \ph{E}'_G E G ′
E ‾ G ′ = 0,89815 ⋅ U v n + j 0,684 U v n 2 ∑ P n a t ⋅ 0,889 ∑ P n a t U v n e − j 25,84 ∘ = 1,2853 ⋅ U v n ⋅ e j 25 ∘ 12 ′ \begin{aligned} \ph{E}'_G &= 0{,}89815 \cdot U_{vn} + \mathrm{j}0{,}684\dfrac{U_{vn}^2}{\sum P_{nat}} \cdot 0{,}889\frac{\sum P_{nat}}{U_{vn}}e^{-\mathrm{j}25{,}84^\circ} \\ &= 1{,}2853 \cdot U_{vn} \cdot e^{\mathrm{j}25^\circ12'} \end{aligned} E G ′ = 0 , 89815 ⋅ U v n + j 0 , 684 ∑ P na t U v n 2 ⋅ 0 , 889 U v n ∑ P na t e − j 25 , 8 4 ∘ = 1 , 2853 ⋅ U v n ⋅ e j 2 5 ∘ 1 2 ′ To znači da je δ 1 = 25 ∘ 12 ′ \delta_1 = 25^\circ12' δ 1 = 2 5 ∘ 1 2 ′
Kontrola ovog rezultata vrši se preko dinamičke karakteristike ustaljenog stanja sa koje se ugao δ 1 \delta_1 δ 1
E G ′ U C B ‾ e sin δ 1 = P m = 0,8 ∑ P n a t , \frac{E'_GU_C}{\ph{B}_e}\sin\delta_1 = P_m = 0{,}8\sum P_{nat}, B e E G ′ U C sin δ 1 = P m = 0 , 8 ∑ P na t , odakle je
sin δ 1 = 0,8 ∑ P n a t ⋅ B ‾ e E G ′ U C . \sin\delta_1 = \frac{0{,}8\sum P_{nat} \cdot \ph{B}_e}{E'_GU_C}. sin δ 1 = E G ′ U C 0 , 8 ∑ P na t ⋅ B e . Dinamička karakteristika za vreme kvara je P ( k ) = 0 P^{(k)} = 0 P ( k ) = 0
Dinamička karakteristika posle isključenja voda u kvaru nalazi se preko koeficijenta ekvivalentnog četvorokrajnika B ‾ e i \ph{B}_{ei} B e i i i i Z c e → Z c Z_{ce} \to Z_c Z ce → Z c Z c = 2 Z c e = 2 U v n 2 ∑ P n a t Z_c = 2Z_{ce} = 2\dfrac{U_{vn}^2}{\sum P_{nat}} Z c = 2 Z ce = 2 ∑ P na t U v n 2
To znači da je:
B ‾ e i = j ⋅ cos λ [ X T 2 + X ( d G + T ) ′ ( 1 − X T 2 Z c t g λ ) + Z c ⋅ t g λ ] = j ⋅ cos ( 12 ∘ ) [ 0,3859 U v n 2 ∑ P n a t ( 1 − 0,111 U v n 2 ∑ P n a t 2 U v n 2 ∑ P n a t t g ( 12 ∘ ) ) + 2 U v n 2 ∑ P n a t t g ( 12 ∘ ) + 0,111 U v n 2 ∑ P n a t ] = j ⋅ 0,896 U v n 2 ∑ P n a t \begin{aligned} \ph{B}_{ei} &= \mathrm{j} \cdot \cos\lambda\left[X_{T2} + X'_{(dG+T)}\left(1 - \frac{X_{T2}}{Z_c}\mathrm{tg}\lambda\right) + Z_c \cdot \mathrm{tg}\lambda\right] \\ &= \mathrm{j} \cdot \cos(12^\circ)\left[0{,}3859\dfrac{U_{vn}^2}{\sum P_{nat}}\left(1 - \frac{0{,}111\dfrac{U_{vn}^2}{\sum P_{nat}}}{2\dfrac{U_{vn}^2}{\sum P_{nat}}}\mathrm{tg}(12^\circ)\right) + 2\dfrac{U_{vn}^2}{\sum P_{nat}}\mathrm{tg}(12^\circ) + 0{,}111\dfrac{U_{vn}^2}{\sum P_{nat}}\right] \\ &= \mathrm{j} \cdot 0{,}896\dfrac{U_{vn}^2}{\sum P_{nat}} \end{aligned} B e i = j ⋅ cos λ [ X T 2 + X ( d G + T ) ′ ( 1 − Z c X T 2 tg λ ) + Z c ⋅ tg λ ] = j ⋅ cos ( 1 2 ∘ ) 0 , 3859 ∑ P na t U v n 2 1 − 2 ∑ P na t U v n 2 0 , 111 ∑ P na t U v n 2 tg ( 1 2 ∘ ) + 2 ∑ P na t U v n 2 tg ( 1 2 ∘ ) + 0 , 111 ∑ P na t U v n 2 = j ⋅ 0 , 896 ∑ P na t U v n 2 Dinamička karakteristika snaga-ugao u normalnom radnom režimu (pre kvara) je:
P 1 = E G ′ U C B ‾ e sin δ = 1,2853 ⋅ U v n 2 0,684 U v n 2 ∑ P n a t sin δ = 1,9 ⋅ ∑ P n a t ⋅ sin δ . \begin{aligned} P_1 &= \frac{E'_GU_C}{\ph{B}_e}\sin\delta = \frac{1{,}2853 \cdot U_{vn}^2}{0{,}684\dfrac{U_{vn}^2}{\sum P_{nat}}}\sin\delta \\ &= 1{,}9 \cdot \sum P_{nat} \cdot \sin\delta. \end{aligned} P 1 = B e E G ′ U C sin δ = 0 , 684 ∑ P na t U v n 2 1 , 2853 ⋅ U v n 2 sin δ = 1 , 9 ⋅ ∑ P na t ⋅ sin δ . Dinamička karakteristika za vreme kvara je:
P ( k ) = 0. P^{(k)} = 0. P ( k ) = 0. Dinamička karakteristika posle kvara je:
P 2 = E G ′ U C B ‾ e i sin δ = 1,2853 ⋅ U v n 2 0,896 U v n 2 ∑ P n a t sin δ = 1,434 ⋅ ∑ P n a t ⋅ sin δ . \begin{aligned} P_2 &= \frac{E'_GU_C}{\ph{B}_{ei}}\sin\delta = \frac{1{,}2853 \cdot U_{vn}^2}{0{,}896\dfrac{U_{vn}^2}{\sum P_{nat}}}\sin\delta \\ &= 1{,}434 \cdot \sum P_{nat} \cdot \sin\delta. \end{aligned} P 2 = B e i E G ′ U C sin δ = 0 , 896 ∑ P na t U v n 2 1 , 2853 ⋅ U v n 2 sin δ = 1 , 434 ⋅ ∑ P na t ⋅ sin δ . Ilustracija ovih krivih snaga-ugao je:
Sa slike je:
sin δ 2 = 0,8 ⋅ ∑ P n a t 1,434 ⋅ ∑ P n a t = 0,5577 ( δ 2 = 33 ∘ 54 ′ i δ g r = 180 ∘ − δ 2 = 146 ∘ 06 ′ ) . \begin{gathered} \sin\delta_2 = \frac{0{,}8 \cdot \sum P_{nat}}{1{,}434 \cdot \sum P_{nat}} = 0{,}5577 \\ \left(\delta_2 = 33^\circ54' \text{ i } \delta^{gr} = 180^\circ - \delta_2 = 146^\circ06'\right). \end{gathered} sin δ 2 = 1 , 434 ⋅ ∑ P na t 0 , 8 ⋅ ∑ P na t = 0 , 5577 ( δ 2 = 3 3 ∘ 5 4 ′ i δ g r = 18 0 ∘ − δ 2 = 14 6 ∘ 0 6 ′ ) . Iz jednakosti površina je:
cos δ i s = P m ( δ g r − δ 1 ) + P 2 M cos δ g r P 2 M = 0,8 ∑ P n a t ( δ g r − δ 1 ) + 1,434 ∑ P n a t ⋅ cos δ g r 1,434 ∑ P n a t = 0,8 ( 146,10 − 25,2 ) 3,14 180 + 1,434 ⋅ cos ( 146,10 ) 1,434 = 0,347 , \begin{aligned} \cos\delta_{is} &= \frac{P_m(\delta^{gr} - \delta_1) + P_{2M}\cos\delta^{gr}}{P_{2M}} \\ &= \frac{0{,}8\sum P_{nat}(\delta^{gr} - \delta_1) + 1{,}434\sum P_{nat} \cdot \cos\delta^{gr}}{1{,}434\sum P_{nat}} \\ &= \frac{0{,}8(146{,}10 - 25{,}2)\dfrac{3{,}14}{180} + 1{,}434 \cdot \cos(146{,}10)}{1{,}434} \\ &= 0{,}347, \end{aligned} cos δ i s = P 2 M P m ( δ g r − δ 1 ) + P 2 M cos δ g r = 1 , 434 ∑ P na t 0 , 8 ∑ P na t ( δ g r − δ 1 ) + 1 , 434 ∑ P na t ⋅ cos δ g r = 1 , 434 0 , 8 ( 146 , 10 − 25 , 2 ) 180 3 , 14 + 1 , 434 ⋅ cos ( 146 , 10 ) = 0 , 347 , pa je δ i s = 69,72 ∘ \delta_{is} = 69{,}72^\circ δ i s = 69 , 7 2 ∘
Kritično vreme isključenja kvara je
t i s = 2 H ( δ i s − δ 1 ) π f 0 P m S G n = 4 H ( δ i s − δ 1 ) ω s P m S G n = 0,345 s . t_{is} = \sqrt{\frac{2H(\delta_{is} - \delta_1)}{\pi f_0\dfrac{P_m}{S_{Gn}}}} = \sqrt{\frac{4H(\delta_{is} - \delta_1)}{\omega_s\dfrac{P_m}{S_{Gn}}}} = 0{,}345\ \text{s}. t i s = π f 0 S G n P m 2 H ( δ i s − δ 1 ) = ω s S G n P m 4 H ( δ i s − δ 1 ) = 0 , 345 s . Pošto se tropolni kratak spoj isključuje posle 0,20 s, generator je stabilan.
Rezultat δ i s = 69,72 ∘ \delta_{is} = 69{,}72^\circ δ i s = 69 , 7 2 ∘ t i s = 0,345 t_{is} = 0{,}345 t i s = 0 , 345 > t i = 0,20 > t_i = 0{,}20 > t i = 0 , 20
← Prethodni zadatak Zadatak 10: Dinamička stabilnost, kratak spoj na ogranku Sledeći zadatak → Zadatak 12: Kritično vreme obostranog isključenja voda 
