Direct Current Transmission Links

Abstract

In explaining the mechanism of power transmission by direct current it will be convenient to begin by comparing it with the mechanism of an a.c. transmission link. We shall use the index 1 for the sending end and the index 2 for the receiving end. Let the angle between the voltages U₁ and U₂ be δ and the reactance of the transmission link X. Disregarding the ohmic resistances we obtain for the power transfer by the a.c. link

$$ {P_1} = {P_2} = \frac{{{U_1}{U_2}}}{X}\sin \delta $$

(9.1)

$$ {Q_1} = \frac{{{U_1}({U_1} - {U_2}\cos \delta )}}{X};\;\;\;\;\;{Q_2} = \frac{{{U_2}({U_2} - {U_1}\cos \delta )}}{X}. $$

(9.2)