Exercício Diagrama Unifilar Resolvido-2021 2

I_{a2} = \frac{\bar{V}_A - \bar{V}_{m1}}{Z_y + 0,12 + j0,6} \quad \quad I_{g1} = I_{ra2} + I_{pa2} \bar{S}_{m1\varphi} = 448,667.10^3 \angle 36,86889^\circ \enspace VA \bar{S}_{g2 1\varphi} = 615.10^3 \angle 25,8949^\circ \enspace VA \bar{I}_{m1} = \frac{\bar{S}_{m1\varphi}^*}{\bar{V}_{m1}^*} \quad \quad \bar{I}_{g2} = \frac{\bar{S}_{g2 1\varphi}^*}{\bar{V}_{m1}^*} \quad \therefore \quad \bar{I}_{a2} = \frac{1}{\bar{V}_{m1}^*} \left(\bar{S}_{m1\varphi}^* - \bar{S}_{g2 1\varphi}^*\right) \quad (i) e \epsilon \bar{I}_{a2} = \frac{\bar{V}_A - \bar{V}_{m1}}{\bar{Z}_y + 0,12 + j0,6} \quad \quad (ii) \frac{1}{\bar{V}_{m1}^*} \left(\bar{S}_{m1\varphi}^* - \bar{S}_{g2 1\varphi}^*\right) = \frac{\bar{V}_A - \bar{V}_{m1}}{\bar{Z}_y + 0,12 + j0,6} \left(\bar{S}_{m1\varphi}^* - \bar{S}_{g2 1\varphi}^*\right)\left(Z_y + 0,12 + j0,6 \right) = \bar{V}_{m1}^*\left(\bar{V}_A - \bar{V}_{m1}\right) = \bar{V}_{m1}^* \bar{V}_A - |\bar{V}_{m1}|^2 (448,667.10^3 \angle 36,86889^\circ - 615.10^3 \angle 25,8949^\circ) = \left(a - jb \right)1397,69 \angle 2,96854^\circ - \left(a^2 + b^2 \right) +2556,88+j 38 957,8=−a139561 −ja 72383 t-jb 1395,81 -b, 72383 +a^2 +b^2 38 957,8 = b1395,81 -a 72,383 +2556,88 = −a 1395,81 -b, 72,383 +a^2 +b^2 a_1 = -6,274 194 \quad b_1 = 27,5853 a_2 = 1399,13 \quad b_2 = 100,469 \therefore \quad Tensao Simples: \bar{V}_{m1 an} = 140 \angle 9 \angle 4,10708^\circ \bar{V}_{m1 bn} = 140 \angle 9 \angle 12,4,10708^\circ \bar{V}_{m1 cn} = 140 \angle 9 \angle 12,9,10708^\circ \bar{V}_{m1} = 140 \angle 79 \angle 41,10708^\circ \enspace \overset{\wedge}{\mathrm{p}} \enspace fase - neutro\box_a\text{A corrente nas três fases da linha 2-3,} I_{a2} = \frac{\bar{V}_A - \bar{V}_{m1}}{Z_y + 0,12 + j0,6} \bar{I}_{r a2} = \frac{1397,69 \angle 2,96854^\circ - 1602,79 \angle 41,10708^\circ}{0,04 + j0,2 + 0,12 + j0,6} \therefore \quad \bar{I}_{a2,b} = 38,3965 \angle 163,775^\circ \enspace A \therefore \quad \bar{I}_{a2,b} = 38,3965 \angle 43,775^\circ \enspace A \bar{I}_{a2,b} = 38,3965 \angle 283,775^\circ \enspace A 4. A potência complexa fornecida pelo gerador 1 e seu fator de potência. \bar{I}_{g1} = \bar{I}_{a2} + \bar{I}_{g a2} = 38,3965 \angle 163,775^\circ + 486,593 \angle -31,86^\circ \bar{I}_{g1} = 449,164 \angle -37,2008^\circ \enspace A \bar{V}_{gL} = \bar{I}_{g1} \bar{Z}_y + \bar{V}_a = 449,164 \angle -37,2008^\circ (0,04 + j0,12) + 1397,69 \angle 2,96854^\circ \therefore \quad \bar{V}_{g1} = 1466,52 \angle 5,368^\circ \enspace V \bar{S}_{g 1 3\varphi} = 3. \bar{I}_{g1}^* \bar{V}_{g 1} = 1,976511 \angle 38,589^\circ \enspace MVA