第 二 題

利用 $I_S = V_R Y (1 + \frac{ZY}{4}) + I_R (1 + \frac{ZY}{2})$ 或 KCL：$I_S = I_R + V_R \frac{Y}{2} + V_S \frac{Y}{2}$ $I_{C1} = V_S \frac{Y}{2} = (156.72 + j46.44) \times 10^3 \times j4.18 \times 10^{-4} = -19.41 + j65.51 , \text{A}$ $I_{line} = I_R + V_R \frac{Y}{2} = (627.6 - j303.9) + j55.51 = 627.6 - j248.39 , \text{A}$

1. 總線路常數： $R = r \times l = 0.0364 \times 210 = 7.644 , \Omega$ $L_{total} = L \times l = 0.973 \times 10^{-3} \times 210 = 0.20433 , \text{H}$

$S_S = 3 V_S I_S^* = 3 (156.72 + j46.44) \times 10^3 \times (608.19 + j182.88) = 3 (95316 + j28660 + j28244 - 8493) = 3 (86823 + j56904) = 260.47 , \text{MW} + j170.71 , \text{MVAR}$ 送電端三相輸入功率 $P_S = 260.47 , \text{MW}$

輸電效率 $\eta = \frac{P_R}{P_S} \times 100% = \frac{250}{260.47} \times 100% = 95.98%$