Discussion on “The interconnection of alternating-current power stations”

Abstract

Mr. J. S. Peck : There are several problems which are of special interest and must occur in almost every case. Suppose it is wished to transmit a certain amount of energy, at a certain power factor, from one station to another. The question is : What voltage difference is required between the two stations, and also what would be the power factor of the load on the sending station, assuming that a certain power factor is required on the receiving station ? Again, if the boost is fixed, or the difference in voltages which is allowed between the sending and receiving station has been settled, what is the maximum amount of power in kilowatts that can be delivered at a certain power factor ? The third question is : Supposing that a voltage difference is not allowed between the two stations, what will be the power factor, assuming that it is desired to deliver a certain amount of energy at one station or the other ? These are the most usual problems in connection with this subject, and the authors have given certain formulae for determining, in some cases with considerable exactitude, in others approximately, the solutions of these problems. I think it would have added considerably to the value of the paper if they had worked out definite examples, and I suggest that they should do so in their reply. It seems to me that the problem can be looked at in this way. There are only four voltages which come into question, i.e. the voltage at the receiving end, the voltage at the sending end, the ohmic drop and the reactive drop in the line. If it be assumed that a certain current or a certain load has to be transmitted, and the power factor at which that must be received be fixed, then the voltage at the receiving end and the power factor can be determined at once. From the kilowatts and power factor the current can be found and, knowing the constants of the interconnector, the ohmic drop and the reactive drop can be calculated. By applying the Mershon chart it would be possible, in a very few moments, to solve almost any problem that might arise. Fig. A shows the Mershon chart. The numerals 10, 20, 30, etc., represent power factors and voltage. Suppose it is desired to deliver a certain amount of energy at 80 per cent power factor and assume that under this load the ohmic drop on the line (IB) is 6 per cent and the reactive drop (IX) is 10 per cent. From the intersection of the 80 per cent power-factor line with the 100 per cent voltage circle, draw the horizontal line IB-6 p;r cent. From the end of IB erect the vertical line IX = 1 0 per cent. Then IZ= the impedancedrop in per cent and E-± = the value and phase angle of the voltage at the sending station. It will be seen at once that the value of E± is 111 per cent and the power factor at the sending end 77-5 per cent, approximately. This means that the voltage at the sending end must be 11 per cent higher than that at the receiving end. This is the same construction as the authors show in Fig. 1, the advantage of the chart being that the results may be read off without dra...