Departamento de Engenharia Elétrica
A forma mais frequente de estudo de faltas assimétricas é através de componente simétricas.
A ideia básica das componentes simétricas é a decomposição de um sistema desbalanceado em um determinado número de sistemas balanceados.
A teoria das componentes simétricas permite representar grandezas desbalanceadas.
$\scriptsize {I}_{\rm{a,1}} = | I_{\rm{a,1}} | \angle{0^\circ}$
$\scriptsize {I}_{\rm{b,1}} = | I_{\rm{b,1}} | \angle{240^\circ} { = a^2 {I}_{\rm{a,1}}}$
$\scriptsize {I}_{\rm{c,1}} = | I_{\rm{c,1}} | \angle{120^\circ} { = a {I}_{\rm{a,1}} }$
$\scriptsize {I}_{\rm{a,2}} = | I_{\rm{a,2}} | \angle{0^\circ}$
$\scriptsize {I}_{\rm{b,2}} = | I_{\rm{b,2}} | \angle{120^\circ} { = a {I}_{\rm{a,2}}}$
$\scriptsize {I}_{\rm{c,2}} = | I_{\rm{c,2}} | \angle{240^\circ} { = a^2 {I}_{\rm{a,2}} }$
$\scriptsize {I}_{\rm{a,0}} = {I}_{\rm{b,0}} = {I}_{\rm{c,0}}$
$\scriptsize {I}_{\rm{a,1}} = | I_{\rm{a,1}} | \angle{0^\circ}$
$\scriptsize {I}_{\rm{b,1}} = | I_{\rm{b,1}} | \angle{240^\circ} { = a^2 {I}_{\rm{a,1}}}$
$\scriptsize {I}_{\rm{c,1}} = | I_{\rm{c,1}} | \angle{120^\circ} { = a {I}_{\rm{a,1}} }$
$\scriptsize {I}_{\rm{a,2}} = | I_{\rm{a,2}} | \angle{0^\circ}$
$\scriptsize {I}_{\rm{b,2}} = | I_{\rm{b,2}} | \angle{120^\circ} { = a {I}_{\rm{a,2}}}$
$\scriptsize {I}_{\rm{c,2}} = | I_{\rm{c,2}} | \angle{240^\circ} { = a^2 {I}_{\rm{a,2}} }$
$\scriptsize {I}_{\rm{a,0}} = {I}_{\rm{b,0}} = {I}_{\rm{c,0}}$
Pelo princípio de superposição, tem-se que:
$\scriptsize {I}_{\rm{a}} = {I}_{\rm{a,0}} + {I}_{\rm{a,1}} +{I}_{\rm{a,2}} \\ \scriptsize {I}_{\rm{b}} = {I}_{\rm{b,0}} + {I}_{\rm{b,1}} +{I}_{\rm{b,2}} \\ \scriptsize {I}_{\rm{c}} = {I}_{\rm{c,0}} + {I}_{\rm{c,1}} +{I}_{\rm{c,2}} $
$\huge \rightarrow $
$\scriptsize {I}_{\rm{a}} = {I}_{\rm{a,0}} + {I}_{\rm{a,1}} +{I}_{\rm{a,2}} \\ \scriptsize {I}_{\rm{b}} = {I}_{\rm{a,0}} + a^2 {I}_{\rm{a,1}} + a {I}_{\rm{a,2}} \\ \scriptsize {I}_{\rm{c}} = {I}_{\rm{a,0}} + a {I}_{\rm{a,1}} + a^2 {I}_{\rm{a,2}} $
$\scriptsize {I}_{\rm{a}} = {I}_{\rm{a,0}} + {I}_{\rm{a,1}} +{I}_{\rm{a,2}} \\ \scriptsize {I}_{\rm{b}} = {I}_{\rm{a,0}} + a^2 {I}_{\rm{a,1}} + a {I}_{\rm{a,2}} \\ \scriptsize {I}_{\rm{c}} = {I}_{\rm{a,0}} + a {I}_{\rm{a,1}} + a^2 {I}_{\rm{a,2}} $
$\huge \rightarrow $
$\scriptsize \left[ {\begin{array}{*{20}c} {I}_{\rm{a}} \\ {I}_{\rm{b}} \\ {I}_{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 1 & a^2 & a \\ 1 & a & a^2 \end{array}} \right] \left[ {\begin{array}{*{20}c} {I}_{\rm{a,0}} \\ {I}_{\rm{a,1}} \\ {I}_{\rm{a,2}} \end{array}} \right] $
$\scriptsize {I}_{\rm{a,b,c}} = A {I}_{\rm{0,1,2}}$
Expressando $\small {I}_{\rm{0,1,2}}$ e $\small {V}_{\rm{0,1,2}}$ em função de $\small {I}_{\rm{a,b,c}}$ e $\small {V}_{\rm{a,b,c}}$, respectivamente, tem-se que:
$\scriptsize {I}_{\rm{0,1,2}} = A^{-1} {I}_{\rm{a,b,c}}$
$\scriptsize \left[ {\begin{array}{*{20}c} {I}_{\rm{0}} \\ {I}_{\rm{1}} \\ {I}_{\rm{2}} \end{array}} \right] = \frac{1}{3} \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{array}} \right] \left[ {\begin{array}{*{20}c} {I}_{\rm{a}} \\ {I}_{\rm{b}} \\ {I}_{\rm{c}} \end{array}} \right] $
$\scriptsize {V}_{\rm{0,1,2}} = A^{-1} {V}_{\rm{a,b,c}}$
$\scriptsize \left[ {\begin{array}{*{20}c} {V}_{\rm{0}} \\ {V}_{\rm{1}} \\ {V}_{\rm{2}} \end{array}} \right] = \frac{1}{3} \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{array}} \right] \left[ {\begin{array}{*{20}c} {V}_{\rm{a}} \\ {V}_{\rm{b}} \\ {V}_{\rm{c}} \end{array}} \right] $
Rede completa:
Rede simplificada:
$\scriptsize \left[ {\begin{array}{*{20}c} {\hspace{-0.1cm} {E}_{\rm{a}} \hspace{-0.1cm} } \\ {\hspace{-0.1cm} {E}_{\rm{b}} \hspace{-0.1cm} } \\ {\hspace{-0.1cm} {E}_{\rm{c}} \hspace{-0.1cm} } \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} {\hspace{-0.1cm} {\rm{E}}_a \angle{0^\circ} } \\ {\hspace{-0.1cm} {\rm{E}}_b \angle{240^\circ}} \\ {\hspace{-0.1cm} {\rm{E}}_c \angle{120^\circ}} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} \hspace{-0.1cm} 1 \hspace{-0.1cm} \\ \hspace{-0.1cm} a^2 \hspace{-0.1cm} \\ \hspace{-0.1cm} a \hspace{-0.1cm} \\ \end{array}} \right] {E}_{\rm{a}}$
$\scriptsize {E}_{\rm{a}} - {Z}_{\rm{S}} {I}_{\rm{a}} - {V}_{\rm{a}} - {Z}_{\rm{n}} {I}_{\rm{n}} = 0 \\ \scriptsize {E}_{\rm{b}} - {Z}_{\rm{S}} {I}_{\rm{b}} - {V}_{\rm{b}} - {Z}_{\rm{n}} {I}_{\rm{n}} = 0 \\ \scriptsize {E}_{\rm{c}} - {Z}_{\rm{S}} {I}_{\rm{c}} - {V}_{\rm{c}} - {Z}_{\rm{n}} {I}_{\rm{n}} = 0 \\ \scriptsize {I}_{\rm{n}} = {I}_{\rm{a}} + {I}_{\rm{b}} + {I}_{\rm{c}}$
$\scriptsize {E}_{\rm{a}} - {Z}_{\rm{S}} {I}_{\rm{a}} - {V}_{\rm{a}} - {Z}_{\rm{n}} {I}_{\rm{n}} = 0 \\ \scriptsize {E}_{\rm{b}} - {Z}_{\rm{S}} {I}_{\rm{b}} - {V}_{\rm{b}} - {Z}_{\rm{n}} {I}_{\rm{n}} = 0 \\ \scriptsize {E}_{\rm{c}} - {Z}_{\rm{S}} {I}_{\rm{c}} - {V}_{\rm{c}} - {Z}_{\rm{n}} {I}_{\rm{n}} = 0 \\ \scriptsize {I}_{\rm{n}} = {I}_{\rm{a}} + {I}_{\rm{b}} + {I}_{\rm{c}}$
$\scriptsize \left[ {\begin{array}{*{20}c} { {V}_{\rm{a}}} \\ {{{V}_{\rm{b}} }} \\ {{V}_{\rm{c}} } \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} { {E}_{\rm{a}}} \\ {{{E}_{\rm{b}} }} \\ {{E}_{\rm{c}} } \\ \end{array}} \right] - \left[ {\begin{array}{*{20}c} {{Z}_{\rm{S}} + {Z}_{\rm{n}} } & {{Z}_{\rm{n}} } & {{Z}_{\rm{n}} } \\ {{Z}_{\rm{n}} } & {{Z}_{\rm{S}} + {Z}_{\rm{n}} } & {{Z}_{\rm{n}} } \\ {{Z}_{\rm{n}} } & {{Z}_{\rm{n}} } & {{Z}_{\rm{S}} + {Z}_{\rm{n}} } \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} {I}_{\rm{a}} \\ {I}_{\rm{b}} \\ {I}_{\rm{c}} \\ \end{array}} \right]$
$\scriptsize {V}_{\rm{a,b,c}} = {E}_{\rm{a,b,c}} - {Z}_{\rm{a,b,c}} {I}_{\rm{a,b,c}}$
$\scriptsize {V}_{\rm{a,b,c}} = {E}_{\rm{a,b,c}} - {Z}_{\rm{a,b,c}} {I}_{\rm{a,b,c}}$
$\scriptsize {\rm{A}} {V}_{\rm{0,1,2}} = {\rm{A}} {E}_{\rm{0,1,2}} - {Z}_{\rm{a,b,c}} {\rm{A}} {I}_{\rm{0,1,2}}$
$\scriptsize {V}_{\rm{0,1,2}} = {E}_{\rm{0,1,2}} - {\rm{A}}^{-1} {Z}_{\rm{a,b,c}} {\rm{A}} {I}_{\rm{0,1,2}}$
$\scriptsize {Z}_{\rm{0,1,2}} = {\rm{A}}^{-1} {Z}_{\rm{a,b,c}} {\rm{A}}$
$\scriptsize \boxed{ {Z}_{\rm{0}} = {Z}_{\rm{S}} + 3 {Z}_{\rm{n}} } $
$\scriptsize \boxed{ {Z}_{\rm{1}} = {Z}_{\rm{S}} } $
$\scriptsize \boxed{ {Z}_{\rm{2}} = {Z}_{\rm{S}} } $
$\scriptsize \boxed{ {V}_{\rm{0}} = - {Z}_{\rm{0}} {I}_{\rm{0}} }$
$\scriptsize \boxed{ {V}_{\rm{1}} = {E}_{\rm{a}} - {Z}_{\rm{1}} {I}_{\rm{1}} }$
$\scriptsize \boxed{ {V}_{\rm{2}} = - {Z}_{\rm{2}} {I}_{\rm{2}} }$
Força eletromotriz do gerador balanceada $\scriptsize \rightarrow$ $\small E_0 = E_2 = 0$ e $\small E_1 = E_a$
$\scriptsize \boxed{ {Z}_{\rm{0}} = {Z}_{\rm{S}} + 3 {Z}_{\rm{n}} }$
$\scriptsize \boxed{ {Z}_{\rm{1}} = {Z}_{\rm{S}} }$
$\scriptsize \boxed{ {Z}_{\rm{2}} = {Z}_{\rm{S}} }$
$\scriptsize \boxed{ {V}_{\rm{0}} = - {Z}_{\rm{0}} {I}_{\rm{0}} }$
$\scriptsize \boxed{ {V}_{\rm{1}} = {E}_{\rm{a}} - {Z}_{\rm{1}} {I}_{\rm{1}} }$
$\scriptsize \boxed{ {V}_{\rm{2}} = - {Z}_{\rm{2}} {I}_{\rm{2}} }$
$\scriptsize \left. {\begin{array}{*{20}c} { {I}_{\rm{a}}}, \ \ {{{I}_{\rm{b}} }}, \ \ {{I}_{\rm{c}} } \\ { {V}_{\rm{a}}}, \ \ {{{V}_{\rm{b}} }}, \ \ {{V}_{\rm{c}} } \\ \end{array}} \right\rbrace {\large ?} $
$\scriptsize {\large \downarrow} $
$\scriptsize \left. {\begin{array}{*{20}c} { {I}_{\rm{0}}}, \ \ {{{I}_{\rm{1}} }}, \ \ {{I}_{\rm{2}} } \\ { {V}_{\rm{0}}}, \ \ {{{V}_{\rm{1}} }}, \ \ {{V}_{\rm{2}} } \\ \end{array}} \right\rbrace {\large ?} $
$\scriptsize {{I}_{\rm{a}} = {I}_{\rm{0}} + {I}_{\rm{1}} +{I}_{\rm{2}} } \\ \scriptsize {I}_{\rm{b}} = {I}_{\rm{0}} + a^2 {I}_{\rm{1}} + a {I}_{\rm{2}} \\ \scriptsize {I}_{\rm{c}} = {I}_{\rm{0}} + a {I}_{\rm{1}} + a^2 {I}_{\rm{2}} $
$\scriptsize {{V}_{\rm{a}} = {V}_{\rm{0}} + {V}_{\rm{1}} +{V}_{\rm{2}} } \\ \scriptsize {V}_{\rm{b}} = {V}_{\rm{0}} + a^2 {V}_{\rm{1}} + a {V}_{\rm{2}} \\ \scriptsize {V}_{\rm{c}} = {V}_{\rm{0}} + a {V}_{\rm{1}} + a^2 {V}_{\rm{2}} $
$\scriptsize {V}_{\rm{a}} = {Z}_{F} {I}_{\rm{a}} $
$\scriptsize { {I}_{\rm{b}} = {I}_{\rm{c}} = 0 } $
$\scriptsize {I}_{\rm{0,1,2}} = {\rm{A~}}^{-1} {I}_{\rm{a,b,c}} $
$\tiny \left[ {\begin{array}{*{20}{c}} {{I}_0}\\ {{I}_1}\\ {{I}_2} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a}&{a^2}\\ {1}&{a^2}&{a} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I}_{\rm{a}}}\\ {0}\\ {0} \end{array}} \right] $
$\tiny \boxed{ {{I}_0} = {{I}_1} = {{I}_2} = \frac{1}{3} {{I}_a} } $
$\scriptsize {V}_{\rm{a}} = {V}_{\rm{0}} + {V}_{\rm{1}} + {V}_{\rm{2}} $
$\scriptsize \boxed{ {V}_{\rm{0}} = - {Z}_{\rm{0}} {I}_{\rm{0}} }$ $\scriptsize \boxed{ {V}_{\rm{1}} = {E}_{\rm{a}} - {Z}_{\rm{1}} {I}_{\rm{1}} }$ $\scriptsize \boxed{ {V}_{\rm{2}} = - {Z}_{\rm{2}} {I}_{\rm{2}} }$
$\scriptsize {V}_{\rm{a}} = {E}_{\rm{a}} - ( {Z}_{\rm{0}} + {Z}_{\rm{1}} + {Z}_{\rm{2}} ) {I}_{\rm{0}} $
$\scriptsize {V}_{\rm{a}} = {Z}_{F} {I}_{\rm{a}} $
$\scriptsize \boxed{ {{I}_0} = {{I}_1} = {{I}_2} = \frac{1}{3} {{I}_a} } $
$\scriptsize {V}_{\rm{a}} = 3{Z}_{F} {I}_{\rm{0}} $
$\scriptsize 3{Z}_{F} {I}_{\rm{0}} = {E}_{\rm{a}} - ( {Z}_{\rm{0}} + {Z}_{\rm{1}} + {Z}_{\rm{2}} ) {I}_{\rm{0}} $
$\tiny \boxed{ {I}_{\rm{0}} = \frac{{E}_{\rm{a}}}{{Z}_{\rm{0}} + {Z}_{\rm{1}} + {Z}_{\rm{2}} + 3{Z}_{F}} }$
$\scriptsize \boxed{ {{I}_0} = {{I}_1} = {{I}_2} = \frac{1}{3} {{I}_a} } $
$\scriptsize \boxed{ {I}_{\rm{0}} = \frac{{E}_{\rm{a}}}{{Z}_{\rm{0}} + {Z}_{\rm{1}} + {Z}_{\rm{2}} + 3{Z}_{F}} }$
$\scriptsize \left. {\begin{array}{*{20}c} { {I}_{\rm{a}}}, \ \ {{{I}_{\rm{b}} }}, \ \ {{I}_{\rm{c}} } \\ { {V}_{\rm{a}}}, \ \ {{{V}_{\rm{b}} }}, \ \ {{V}_{\rm{c}} } \\ \end{array}} \right\rbrace {\large ?} $
$\scriptsize {\large \downarrow} $
$\scriptsize \left. {\begin{array}{*{20}c} { {I}_{\rm{0}}}, \ \ {{{I}_{\rm{1}} }}, \ \ {{I}_{\rm{2}} } \\ { {V}_{\rm{0}}}, \ \ {{{V}_{\rm{1}} }}, \ \ {{V}_{\rm{2}} } \\ \end{array}} \right\rbrace {\large ?} $
$\scriptsize {{I}_{\rm{a}} = {I}_{\rm{0}} + {I}_{\rm{1}} +{I}_{\rm{2}} } \\ \scriptsize {I}_{\rm{b}} = {I}_{\rm{0}} + a^2 {I}_{\rm{1}} + a {I}_{\rm{2}} \\ \scriptsize {I}_{\rm{c}} = {I}_{\rm{0}} + a {I}_{\rm{1}} + a^2 {I}_{\rm{2}} $
$\scriptsize {{V}_{\rm{a}} = {V}_{\rm{0}} + {V}_{\rm{1}} +{V}_{\rm{2}} } \\ \scriptsize {V}_{\rm{b}} = {V}_{\rm{0}} + a^2 {V}_{\rm{1}} + a {V}_{\rm{2}} \\ \scriptsize {V}_{\rm{c}} = {V}_{\rm{0}} + a {V}_{\rm{1}} + a^2 {V}_{\rm{2}} $
$\scriptsize {V}_{\rm{b}} - {V}_{\rm{c}} = {Z}_{\rm{F}} {I}_{\rm{b}} $
$\scriptsize { {I}_{\rm{b}} + {I}_{\rm{c}} = 0 } $
$\scriptsize { {I}_{\rm{a}} = 0 } $
$\tiny \left[ {\begin{array}{*{20}{c}} {{I}_0}\\ {{I}_1}\\ {{I}_2} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a}&{a^2}\\ {1}&{a^2}&{a} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {0}\\ {{I}_{\rm{b}}}\\ {-{I}_{\rm{b}}} \end{array}} \right] $
$\scriptsize { {I}_{\rm{0}} = 0 } $
$\scriptsize {I}_{\rm{1}} = (1/3) (a - a^2) {I}_{\rm{b}} $
$\scriptsize {I}_{\rm{2}} = (1/3) (a^2 - a) {I}_{\rm{b}} $
$\scriptsize \boxed{ {I}_{\rm{1}} = - {I}_{\rm{2}} } $
$\tiny \left[ {\begin{array}{*{20}{c}} {{V}_a}\\ {{V}_b}\\ {{V}_c} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V}_0}\\ {{V}_1}\\ {{V}_2} \end{array}} \right] $
$\scriptsize {V}_{\rm{b}} - {V}_{\rm{c}} = {Z}_{\rm{F}} {I}_{\rm{b}} \\ \scriptsize \ \ \ \ \ \ \ \ \ \ \ \ \ = (a^2 - a)( {V}_{\rm{1}} - {V}_{\rm{2}} ) $
$\scriptsize \boxed{ {V}_{\rm{1}} = {E}_{\rm{a}} - {Z}_{\rm{1}} {I}_{\rm{1}} } $
$\scriptsize \boxed{ {V}_{\rm{2}} = - {Z}_{\rm{2}} {I}_{\rm{2}} } $
$\scriptsize \boxed{ {I}_{\rm{2}} = - {I}_{\rm{1}} } $
$\scriptsize ( a^2 - a ) [ {E}_{\rm{a}} - ( {Z}_{\rm{1}} + {Z}_{\rm{2}} ) {I}_{\rm{1}} ] = {Z}_{\rm{F}} {I}_{\rm{b}} $
$\scriptsize \boxed{ {I}_{\rm{1}} = (1/3) (a - a^2) {I}_{\rm{b}}}; \ \tiny (a - a^2)(a^2 - a) = 3 $
$\tiny \boxed{ {I}_{\rm{1}} = \frac{E_{\rm{a}}}{{Z}_{\rm{1}} + {Z}_{\rm{2}} + {Z}_{\rm{F}} } } $
$\scriptsize \boxed{ {I}_{\rm{1}} = - {I}_{\rm{2}} } $
$\scriptsize \boxed{ {I}_{\rm{1}} = \frac{E_{\rm{a}}}{{Z}_{\rm{1}} + {Z}_{\rm{2}} + {Z}_{\rm{F}} } } $
$\scriptsize \left. {\begin{array}{*{20}c} { {I}_{\rm{a}}}, \ \ {{{I}_{\rm{b}} }}, \ \ {{I}_{\rm{c}} } \\ { {V}_{\rm{a}}}, \ \ {{{V}_{\rm{b}} }}, \ \ {{V}_{\rm{c}} } \\ \end{array}} \right\rbrace {\large ?} $
$\scriptsize {\large \downarrow} $
$\scriptsize \left. {\begin{array}{*{20}c} { {I}_{\rm{0}}}, \ \ {{{I}_{\rm{1}} }}, \ \ {{I}_{\rm{2}} } \\ { {V}_{\rm{0}}}, \ \ {{{V}_{\rm{1}} }}, \ \ {{V}_{\rm{2}} } \\ \end{array}} \right\rbrace {\large ?} $
$\scriptsize {{I}_{\rm{a}} = {I}_{\rm{0}} + {I}_{\rm{1}} +{I}_{\rm{2}} } \\ \scriptsize {I}_{\rm{b}} = {I}_{\rm{0}} + a^2 {I}_{\rm{1}} + a {I}_{\rm{2}} \\ \scriptsize {I}_{\rm{c}} = {I}_{\rm{0}} + a {I}_{\rm{1}} + a^2 {I}_{\rm{2}} $
$\scriptsize {{V}_{\rm{a}} = {V}_{\rm{0}} + {V}_{\rm{1}} +{V}_{\rm{2}} } \\ \scriptsize {V}_{\rm{b}} = {V}_{\rm{0}} + a^2 {V}_{\rm{1}} + a {V}_{\rm{2}} \\ \scriptsize {V}_{\rm{c}} = {V}_{\rm{0}} + a {V}_{\rm{1}} + a^2 {V}_{\rm{2}} $
$\scriptsize {V}_{\rm{b}} = {V}_{\rm{c}} = {Z}_{\rm{F}} ( {I}_{\rm{b}} + {I}_{\rm{c}} ) $
$\scriptsize {I}_{\rm{b}} = {I}_{\rm{0}} + a^2 {I}_{\rm{1}} + a {I}_{\rm{2}}$ $\scriptsize {I}_{\rm{c}} = {I}_{\rm{0}} + a {I}_{\rm{1}} + a^2 {I}_{\rm{2}} $
$\scriptsize {V}_{\rm{b}} = 3 {Z}_{\rm{F}} {I}_{\rm{0}} $
$\scriptsize {V}_{\rm{b}} = {V}_{\rm{0}} + a^2 {V}_{\rm{1}} + a {V}_{\rm{2}}$ $\scriptsize {V}_{\rm{c}} = {V}_{\rm{0}} + a {V}_{\rm{1}} + a^2 {V}_{\rm{2}} $ $\scriptsize {V}_{\rm{1}} = {V}_{\rm{2}} $ $\scriptsize 3 {Z}_{\rm{F}} {I}_{\rm{0}} = {V}_{\rm{0}} + ( a^2 + a ) {V}_{\rm{1}} $ $\scriptsize 3 {Z}_{\rm{F}} {I}_{\rm{0}} = {V}_{\rm{0}} - {V}_{\rm{1}} $
$\scriptsize \boxed{ {V}_{\rm{0}} = - {Z}_{\rm{0}} {I}_{\rm{0}} } $
$\scriptsize \boxed{ {V}_{\rm{1}} = {E}_{\rm{a}} - {Z}_{\rm{1}} {I}_{\rm{1}} } $
$\tiny \boxed{ {I}_{\rm{0}} = - \frac{{E}_{\rm{a}} - {Z}_{\rm{1}}{I}_{\rm{1}}}{ {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } $
$\scriptsize \boxed{ {V}_{\rm{1}} = {V}_{\rm{2}} } $
$\scriptsize \boxed{ {V}_{\rm{1}} = {E}_{\rm{a}} - {Z}_{\rm{1}} {I}_{\rm{1}} } $
$\scriptsize \boxed{ {V}_{\rm{2}} = - {Z}_{\rm{2}} {I}_{\rm{2}} } $
$\scriptsize { {I}_{\rm{2}} = - \frac{{E}_{\rm{a}} - {Z}_{\rm{1}}{I}_{\rm{1}}}{ {Z}_{\rm{2}} } } $
$\scriptsize \boxed{ {I}_{\rm{0}} = - \frac{{E}_{\rm{a}} - {Z}_{\rm{1}}{I}_{\rm{1}}}{ {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } $
$\scriptsize {I}_{\rm{a}} = {I}_{\rm{0}} +{I}_{\rm{1}} + {I}_{\rm{2}} = 0$
$\tiny \boxed{ {I}_{\rm{1}} = \frac{{E}_{\rm{a}}}{ {Z}_{\rm{1}} + \frac{ {Z}_{\rm{2}} ({Z}_{\rm{0}} + 3 {Z}_{\rm{F}} ) }{ {Z}_{\rm{2}} + {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } } $
$\scriptsize \boxed{ {I}_{\rm{0}} = - \frac{{E}_{\rm{a}} - {Z}_{\rm{1}}{I}_{\rm{1}}}{ {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } $
$\scriptsize \boxed{ {I}_{\rm{1}} = \frac{{E}_{\rm{a}}}{ {Z}_{\rm{1}} + \frac{ {Z}_{\rm{2}} ({Z}_{\rm{0}} + 3 {Z}_{\rm{F}} ) }{ {Z}_{\rm{2}} + {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } } $
Considere uma falta na Barra 3 do circuito da figura acima. Sabendo que as impedâncias de sequência zero, positiva e negativa ($\small {Z}_{\rm{3}}^{\rm{0}}$, $\small {Z}_{\rm{3}}^{\rm{1}}$ e $\small {Z}_{\rm{3}}^{\rm{2}}$), vistas desde a Barra 3, são iguais a $\small j$0,35, $\small j$0,22 e $\small j$0,22, respectivamente, calcular as correntes de falta na própria Barra 3 ($\small {I}_{\rm{3}}^{\rm{a}}$, $\small {I}_{\rm{3}}^{\rm{b}}$ e $\small {I}_{\rm{3}}^{\rm{c}}$) para os três tipos de faltas assimétricas.
Falta Linha-Terra:
As compoentes de sequência da corrente de falta são:
$\small {I}_{\rm{3}}^{\rm{0}} = {I}_{\rm{3}}^{\rm{1}} = {I}_{\rm{3}}^{\rm{2}} = \frac{{{V}_{\rm{3}}^{\rm{a}}(0) }}{{Z}_{\rm{33}}^{\rm{1}} + {Z}_{\rm{33}}^{\rm{2}} + {Z}_{\rm{33}}^{\rm{0}} + 3{Z}_{F}}$
$\small {I}_{\rm{3}}^{\rm{0}} = {I}_{\rm{3}}^{\rm{1}} = {I}_{\rm{3}}^{\rm{2}} = \frac{{1,0 }}{j0,22 + j0,22 + j0,35 + 3(j0,1)}$
$\small {I}_{\rm{3}}^{\rm{0}} = {I}_{\rm{3}}^{\rm{1}} = {I}_{\rm{3}}^{\rm{2}} = -j0,9174$ pu
Logo as correntes de falta são:
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{a}}\\ {I}_{\rm{3}}^{\rm{b}}\\ {I}_{\rm{3}}^{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -j2,7523\\ 0\\ 0 \end{array}} \right] $ pu
Falta Linha-Linha:
As compoentes de sequência da corrente de falta são:
$\small {I}_{\rm{3}}^{\rm{0}} = 0 $
$\small {I}_{\rm{3}}^{\rm{1}} = - {I}_{\rm{3}}^{\rm{2}} = \frac{{{V}_{\rm{3}}^{\rm{a}}(0) }}{{Z}_{\rm{33}}^{\rm{1}} + {Z}_{\rm{33}}^{\rm{2}} + {Z}_{F}}$
$\small {I}_{\rm{3}}^{\rm{1}} = - {I}_{\rm{3}}^{\rm{2}} = \frac{{1,0 }}{j0,22 + j0,22 + j0,1}$
$\small {I}_{\rm{3}}^{\rm{1}} = - {I}_{\rm{3}}^{\rm{2}} = -j1,8519$ pu
Logo as correntes de falta são:
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{a}}\\ {I}_{\rm{3}}^{\rm{b}}\\ {I}_{\rm{3}}^{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ -3,2075\\ 3,2075 \end{array}} \right] $ pu
Falta Linha-Linha-Terra:
As compoentes de sequência da corrente de falta são:
$\small {I}_{\rm{3}}^{\rm{1}} = \frac{{V}_{\rm{3}}^{\rm{a}}(0)}{ {Z}_{\rm{33}}^{\rm{1}} + \frac{ {Z}_{\rm{33}}^{\rm{2}} \left( { {Z}_{\rm{33}}^{\rm{0}} + 3 {Z}_{\rm{F}} } \right) }{ {Z}_{\rm{33}}^{\rm{2}} + {Z}_{\rm{33}}^{\rm{0}} + 3{Z}_{\rm{F}} } } = \frac{1,0}{ j0,22 + \frac{ j0,22 \left( { j0,35 + 3 (j0,1) } \right) }{ j0,22 + j0,35 + 3(j0,1) } } = -j2,6017 $ pu
$\small {I}_{\rm{3}}^{\rm{2}} = - \frac{{V}_{\rm{3}}^{\rm{a}}(0) - {Z}_{\rm{33}}^{\rm{1}}{I}_{\rm{3}}^{\rm{1}}}{ {Z}_{\rm{33}}^{\rm{2}}} = - \frac{1,0 - (j0,22)(-j2,6017)}{ j0,22 } = j1,9438 $ pu
$\small {I}_{\rm{3}}^{\rm{0}} = - \frac{{V}_{\rm{3}}^{\rm{a}}(0) - {Z}_{\rm{33}}^{\rm{1}}{I}_{\rm{3}}^{\rm{1}}}{ {Z}_{\rm{33}}^{\rm{0}} + 3{Z}_{\rm{F}} } = - \frac{1,0 - (j0,22)(-j2,6017)}{ j0,35 + 3(j0,1) } = j0,6579 $ pu
Logo as correntes de falta são:
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{a}}\\ {I}_{\rm{3}}^{\rm{b}}\\ {I}_{\rm{3}}^{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 4,058 \angle{165,93^\circ}\\ 4,058 \angle{14,07^\circ} \end{array}} \right] $ pu
Quando a rede é balanceada, é possível calcular a matriz $\small Z_{\rm{Barra}}$ separadamente para as redes de sequência zero, positiva e negativa.
Para uma falta na barra $\small k$, o elemento $\small kk$ da matriz $\small Z_{\rm{Barra}}$ é equivalente à impedância de Thèvenin ao ponto em falta.
Para obter a solução de faltas assimétricas, são obtidos os elementos das matrizes impedância de barra para cada sequência ($\small Z_{\rm{kk}}^0$, $\small Z_{\rm{kk}}^1$ e $\small Z_{\rm{kk}}^2).$
A seguir, serão apresentadas as expressões usadas em faltas assimétricas através da matriz $\small Z_{\rm{Barra}}$.
Equações do gerador em falta:
$\scriptsize \boxed{ {{I}_0} = {{I}_1} = {{I}_2} = \frac{1}{3} {{I}_a} } $
$\scriptsize \boxed{ {I}_{\rm{0}} = \frac{{E}_{\rm{a}}}{{Z}_{\rm{0}} + {Z}_{\rm{1}} + {Z}_{\rm{2}} + 3{Z}_{F}} }$
$\scriptsize \boxed{ I_{\rm{k}}^0 = I_{\rm{k}}^1 = I_{\rm{k}}^2 } $
$\scriptsize \boxed{ I_{\rm{k}}^0 = \frac{V_{\rm{k}}(0)}{Z_{\rm{kk}}^0 + Z_{\rm{kk}}^1 +Z_{\rm{kk}}^2 + 3Z_{\rm{F}} } } $
Equações do gerador em falta:
$\scriptsize \boxed{ {I}_{\rm{1}} = - {I}_{\rm{2}} } $
$\scriptsize \boxed{ {I}_{\rm{1}} = \frac{E_{\rm{a}}}{{Z}_{\rm{1}} + {Z}_{\rm{2}} + {Z}_{\rm{F}} } } $
$\scriptsize \boxed{ I_{\rm{k}}^1 = - I_{\rm{k}}^2 } $
$\scriptsize \boxed{ I_{\rm{k}}^1 = \frac{V_{\rm{k}}(0)}{Z_{\rm{kk}}^1 +Z_{\rm{kk}}^2 + Z_{\rm{F}} } } $
Equações do gerador em falta:
$\scriptsize \boxed{ {I}_{\rm{0}} = - \frac{{E}_{\rm{a}} - {Z}_{\rm{1}}{I}_{\rm{1}}}{ {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } $
$\scriptsize \boxed{ {I}_{\rm{1}} = \frac{{E}_{\rm{a}}}{ {Z}_{\rm{1}} + \frac{ {Z}_{\rm{2}} ({Z}_{\rm{0}} + 3 {Z}_{\rm{F}} ) }{ {Z}_{\rm{2}} + {Z}_{\rm{0}} + 3{Z}_{\rm{F}} } } } $
$\scriptsize \boxed{ I_{\rm{k}}^0 = - \frac{ V_{\rm{k}}(0) - Z_{\rm{kk}}^1 I_{\rm{k}}^1 }{Z_{\rm{kk}}^0 + 3Z_{\rm{F}}} } $
$\scriptsize \boxed{ I_{\rm{k}}^1 = \frac{ V_{\rm{k}}(0) }{ Z_{\rm{kk}}^1 + \frac{Z_{\rm{kk}}^2 (Z_{\rm{kk}}^0 + 3 Z_{\rm{F}})}{Z_{\rm{kk}}^2 + Z_{\rm{kk}}^0 + 3Z_{\rm{F}}} } } $
$\scriptsize Z_{\rm{Barra}}^{\rm{0}} = \left[ {\begin{array}{*{20}c} j0,1820 & j0,0545 & j0,1400 \\ j0,0545 & j0,0864 & j0,0650 \\ j0,1400 & j0,0650 & j0,3500 \end{array}} \right] $
$\scriptsize Z_{\rm{Barra}}^{\rm{1}} = Z_{\rm{Barra}}^{\rm{2}} = \left[ {\begin{array}{*{20}c} j0,1450 & j0,1050 & j0,1300 \\ j0,1050 & j0,1450 & j0,1200 \\ j0,1300 & j0,1200 & j0,2200 \end{array}} \right] $
Considere uma falta na Barra 3 do circuito da figura acima. Sabendo que as componentes de sequência da matriz $\small Z_{\rm{Barra}}$ são as dadas acima, calcular as correntes de falta na Barra 3 ($\small {I}_{\rm{3}}^{\rm{a}}$, $\small {I}_{\rm{3}}^{\rm{b}}$ e $\small {I}_{\rm{3}}^{\rm{c}}$), assim como as tensões de falta nas três barras ($\small {V}_{\rm{1}}^{\rm{abc}}$, $\small {V}_{\rm{2}}^{\rm{abc}}$ e $\small {V}_{\rm{3}}^{\rm{abc}}$), para os três tipos de faltas assimétricas.
Falta Linha-Terra:
As compoentes de sequência da corrente de falta são:
$\small {I}_{\rm{3}}^{\rm{0}} = {I}_{\rm{3}}^{\rm{1}} = {I}_{\rm{3}}^{\rm{2}} = \frac{{{V}_{\rm{3}}^{\rm{a}}(0) }}{{Z}_{\rm{33}}^{\rm{1}} + {Z}_{\rm{33}}^{\rm{2}} + {Z}_{\rm{33}}^{\rm{0}} + 3{Z}_{F}}$
$\small {I}_{\rm{3}}^{\rm{0}} = {I}_{\rm{3}}^{\rm{1}} = {I}_{\rm{3}}^{\rm{2}} = \frac{{1,0 }}{j0,22 + j0,22 + j0,35 + 3(j0,1)}$
$\small {I}_{\rm{3}}^{\rm{0}} = {I}_{\rm{3}}^{\rm{1}} = {I}_{\rm{3}}^{\rm{2}} = -j0,9174$ pu
Logo as correntes de falta são:
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{a}}\\ {I}_{\rm{3}}^{\rm{b}}\\ {I}_{\rm{3}}^{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -j2,7523\\ 0\\ 0 \end{array}} \right] $ pu
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -j0,9174\\ -j0,9174\\ -j0,9174 \end{array}} \right] $ pu
As componentes de sequência das tensões durante a falta são:
$\scriptsize {V}_{\rm{1}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} - {Z}_{\rm{13}}^{\rm{0}} {I}_{\rm{3}}^{\rm{0}} \\ {V}_{\rm{1}}^{\rm{1}}(0) - {Z}_{\rm{13}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{13}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -j0,140(-j0,9174)\\ 1,0 - j0,130(-j0,9174)\\ -j0,130(-j0,9174) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -0,1284\\ 0,8807\\ -0,1193 \end{array}} \right] $ pu
$\scriptsize {V}_{\rm{2}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} - {Z}_{\rm{23}}^{\rm{0}} {I}_{\rm{3}}^{\rm{0}} \\ {V}_{\rm{2}}^{\rm{1}}(0) - {Z}_{\rm{23}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{23}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -j0,065(-j0,9174)\\ 1,0 - j0,120(-j0,9174)\\ -j0,120(-j0,9174) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -0,0596\\ 0,8899\\ -0,1101 \end{array}} \right] $ pu
$\scriptsize {V}_{\rm{3}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} - {Z}_{\rm{33}}^{\rm{0}} {I}_{\rm{3}}^{\rm{0}} \\ {V}_{\rm{3}}^{\rm{1}}(0) - {Z}_{\rm{33}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{33}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -j0,350(-j0,9174)\\ 1,0 - j0,220(-j0,9174)\\ -j0,220(-j0,9174) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} -0,3211\\ 0,7982\\ -0,2018 \end{array}} \right] $ pu
Logo as tensões durante a falta são:
$\small {V}_{\rm{1}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} -0,1284\\ 0,8807\\ -0,1193 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,633\angle{0^\circ}\\ 1,0046\angle{-120,45^\circ}\\ 1,0046\angle{120,45^\circ} \end{array}} \right] $ pu
$\small {V}_{\rm{2}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} -0,0596\\ 0,8899\\ -0,1101 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,7207\angle{0^\circ}\\ 0,9757\angle{-117,43^\circ}\\ 0,9757\angle{117,43^\circ} \end{array}} \right] $ pu
$\small {V}_{\rm{3}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} -0,3211\\ 0,7982\\ -0,2018 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,2752\angle{0^\circ}\\ 1,0647\angle{-125,56^\circ}\\ 1,0647\angle{125,56^\circ} \end{array}} \right] $ pu
Falta Linha-Linha:
As compoentes de sequência da corrente de falta são:
$\small {I}_{\rm{3}}^{\rm{0}} = 0 $
$\small {I}_{\rm{3}}^{\rm{1}} = - {I}_{\rm{3}}^{\rm{2}} = \frac{{{V}_{\rm{3}}^{\rm{a}}(0) }}{{Z}_{\rm{33}}^{\rm{1}} + {Z}_{\rm{33}}^{\rm{2}} + {Z}_{F}}$
$\small {I}_{\rm{3}}^{\rm{1}} = - {I}_{\rm{3}}^{\rm{2}} = \frac{{1,0 }}{j0,22 + j0,22 + j0,1}$
$\small {I}_{\rm{3}}^{\rm{1}} = - {I}_{\rm{3}}^{\rm{2}} = -j1,8519$ pu
Logo as correntes de falta são:
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{a}}\\ {I}_{\rm{3}}^{\rm{b}}\\ {I}_{\rm{3}}^{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ -3,2075\\ 3,2075 \end{array}} \right] $ pu
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ -j1,8519\\ j1,8519 \end{array}} \right] $ pu
As componentes de sequência das tensões durante a falta são:
$\scriptsize {V}_{\rm{1}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} 0 \\ {V}_{\rm{1}}^{\rm{1}}(0) - {Z}_{\rm{13}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{13}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 1,0 - j0,130(-j1,8519)\\ -j0,130(j1,8519) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 0,7593\\ 0,2407 \end{array}} \right] $ pu
$\scriptsize {V}_{\rm{2}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} 0 \\ {V}_{\rm{2}}^{\rm{1}}(0) - {Z}_{\rm{23}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{23}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 1,0 - j0,120(-j1,8519)\\ -j0,120(j1,8519) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 0,7778\\ 0,2222 \end{array}} \right] $ pu
$\scriptsize {V}_{\rm{3}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} 0 \\ {V}_{\rm{3}}^{\rm{1}}(0) - {Z}_{\rm{33}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{33}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 1,0 - j0,220(-j1,8519)\\ -j0,220(j1,8519) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 0,5926\\ 0,4074 \end{array}} \right] $ pu
Logo as tensões durante a falta são:
$\small {V}_{\rm{1}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0\\ 0,7593\\ 0,2407 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1,0\angle{0^\circ}\\ 0,672\angle{-138,07^\circ}\\ 0,672\angle{138,07^\circ} \end{array}} \right] $ pu
$\small {V}_{\rm{2}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0\\ 0,7778\\ 0,2222 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1,0\angle{0^\circ}\\ 0,6939\angle{-136,10^\circ}\\ 0,6939\angle{136,10^\circ} \end{array}} \right] $ pu
$\small {V}_{\rm{3}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0\\ 0,5923\\ 0,4074 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1,0\angle{0^\circ}\\ 0,5251\angle{-162,21^\circ}\\ 0,5251\angle{162,21^\circ} \end{array}} \right] $ pu
Falta Linha-Linha-Terra:
As compoentes de sequência da corrente de falta são:
$\small {I}_{\rm{3}}^{\rm{1}} = \frac{{V}_{\rm{3}}^{\rm{a}}(0)}{ {Z}_{\rm{33}}^{\rm{1}} + \frac{ {Z}_{\rm{33}}^{\rm{2}} \left( { {Z}_{\rm{33}}^{\rm{0}} + 3 {Z}_{\rm{F}} } \right) }{ {Z}_{\rm{33}}^{\rm{2}} + {Z}_{\rm{33}}^{\rm{0}} + 3{Z}_{\rm{F}} } } = \frac{1,0}{ j0,22 + \frac{ j0,22 \left( { j0,35 + 3 (j0,1) } \right) }{ j0,22 + j0,35 + 3(j0,1) } } = -j2,6017 $ pu
$\small {I}_{\rm{3}}^{\rm{2}} = - \frac{{V}_{\rm{3}}^{\rm{a}}(0) - {Z}_{\rm{33}}^{\rm{1}}{I}_{\rm{3}}^{\rm{1}}}{ {Z}_{\rm{33}}^{\rm{2}}} = - \frac{1,0 - (j0,22)(-j2,6017)}{ j0,22 } = j1,9438 $ pu
$\small {I}_{\rm{3}}^{\rm{0}} = - \frac{{V}_{\rm{3}}^{\rm{a}}(0) - {Z}_{\rm{33}}^{\rm{1}}{I}_{\rm{3}}^{\rm{1}}}{ {Z}_{\rm{33}}^{\rm{0}} + 3{Z}_{\rm{F}} } = - \frac{1,0 - (j0,22)(-j2,6017)}{ j0,35 + 3(j0,1) } = j0,6579 $ pu
Logo as correntes de falta são:
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{a}}\\ {I}_{\rm{3}}^{\rm{b}}\\ {I}_{\rm{3}}^{\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 4,058 \angle{165,93^\circ}\\ 4,058 \angle{14,07^\circ} \end{array}} \right] $ pu
$\small \left[ {\begin{array}{*{20}{c}} {I}_{\rm{3}}^{\rm{0}}\\ {I}_{\rm{3}}^{\rm{1}}\\ {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,6579\\ -j2,6017\\ j1,9438 \end{array}} \right] $ pu
As componentes de sequência das tensões durante a falta são:
$\scriptsize {V}_{\rm{1}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} - {Z}_{\rm{13}}^{\rm{0}} {I}_{\rm{3}}^{\rm{0}} \\ {V}_{\rm{1}}^{\rm{1}}(0) - {Z}_{\rm{13}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{13}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} - j0,140(-j0,6579)\\ 1,0 - j0,130(-j2,6017)\\ -j0,130(j1,9438) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,0921\\ 0,6618\\ 0,2527 \end{array}} \right] $ pu
$\scriptsize {V}_{\rm{2}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} - {Z}_{\rm{23}}^{\rm{0}} {I}_{\rm{3}}^{\rm{0}} \\ {V}_{\rm{2}}^{\rm{1}}(0) - {Z}_{\rm{23}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{23}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} - j0,065(-j0,6579)\\ 1,0 - j0,120(-j2,6017)\\ -j0,120(j1,9438) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,0428\\ 0,6878\\ 0,2333 \end{array}} \right] $ pu
$\scriptsize {V}_{\rm{3}}^{\rm{012}} = \left[ {\begin{array}{*{20}{c}} - {Z}_{\rm{33}}^{\rm{0}} {I}_{\rm{3}}^{\rm{0}} \\ {V}_{\rm{3}}^{\rm{1}}(0) - {Z}_{\rm{33}}^{\rm{1}} {I}_{\rm{3}}^{\rm{1}} \\ - {Z}_{\rm{33}}^{\rm{2}} {I}_{\rm{3}}^{\rm{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} - j0,350(-j0,6579)\\ 1,0 - j0,220(-j2,6017)\\ -j0,220(j1,9438) \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,2303\\ 0,4276\\ 0,4276 \end{array}} \right] $ pu
Logo as tensões durante a falta são:
$\small {V}_{\rm{1}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0,0921\\ 0,6618\\ 0,2527 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1,0066\angle{0^\circ}\\ 0,5088\angle{-135,89^\circ}\\ 0,5088\angle{135,89^\circ} \end{array}} \right] $ pu
$\small {V}_{\rm{2}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0,0428\\ 0,6878\\ 0,2333 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0,9638\angle{0^\circ}\\ 0,5740\angle{-136,70^\circ}\\ 0,5740\angle{136,70^\circ} \end{array}} \right] $ pu
$\small {V}_{\rm{3}}^{\rm{abc}} = \left[ {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {1}&{a^2}&{a}\\ {1}&{a}&{a^2} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0,2303\\ 0,4276\\ 0,4276 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1,0855\angle{0^\circ}\\ 0,1974\angle{-180^\circ}\\ 0,1974\angle{180^\circ} \end{array}} \right] $ pu
Departamento de Engenharia Elétrica