2018年電験1種　電力管理問4 - 数学・物理・電気主任技術者試験、情報処理技術者試験等お勉強の記録

(1)

近似値 $\displaystyle{=\frac{\theta_A-\theta_B}{x_1} = \frac{\pi}{12}\times 10 = \frac{5 \times 3.1416}{6}=2.618 \Doteq 2.62 \mathrm{p.u.} }$

$\displaystyle{P_1=\frac{V_A V_B}{x_1} \sin(\theta_A-\theta_B) = \frac{1}{x_1} \sin \frac{\pi}{12} = 10 \sqrt{\frac{1- \cos \frac{\pi}{6}}{2}} }$

$\displaystyle{= 10 \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} = 5 \sqrt{2-\sqrt{3}} \Doteq 2.588 \Doteq 2.59\mathrm{p.u.}}$

(2)

$\displaystyle{P_1= \frac{\theta_A - \theta_B}{x_1} = \frac{\theta_A}{x_1}[\mathrm{p.u.}]}$

$\displaystyle{P_2= \frac{\theta_A - \theta_C}{x_2} [\mathrm{p.u.}]}$

$\displaystyle{P_3= \frac{\theta_C - \theta_B}{x_3} = \frac{\theta_C}{x_3}[\mathrm{p.u.}]}$

$\displaystyle{P_A = G_A - L_A = P_1 + P_2 = \frac{\theta_A}{x_1} + \frac{\theta_A - \theta_B}{x_2} = \left( \frac{1}{x_1}+\frac{1}{x_2}\right)\theta_A - \frac{\theta_C}{x_2}[\mathrm{p.u.}]}$

$\displaystyle{P_C = G_C - L_C = -P_2 + P_3 = -\frac{\theta_A-\theta_C}{x_2} + \frac{\theta_C}{x_3} = - \frac{\theta_A}{x_2} +\left( \frac{1}{x_2}+\frac{1}{x_3}\right)\theta_C[\mathrm{p.u.}]}$

(3)

$\displaystyle{P_A = \left( \frac{1}{x_1}+\frac{1}{x_2}\right)\theta_A - \frac{\theta_C}{x_2} = \left( \frac{1}{0.05} + \frac{1}{0.02} \right) \theta_A -\frac{\theta_C}{0.02} }$

$\displaystyle{ = 70 \theta_A -50 \theta_C = G_A - L_A = 20-10=10}$

$7 \theta_A - 5 \theta_C = 1 \mathrm{p.u.}$

$\displaystyle{ P_C = - \frac{\theta_A}{x_2} +\left( \frac{1}{x_2}+\frac{1}{x_3}\right)\theta_C = - \frac{\theta_A}{0.02}+ \left(\frac{1}{0.02}+\frac{1}{0.03}\right) \theta_C }$