474 ELECTRICAL EQUIPMENT 



The number of sections into which a bus should be divided 

 depends largely upon the individual sytem, and the conditions 

 under which it is expected to operate. In the above example, 

 with four generators on the section, the total power which an oil 

 switch may be required to rupture would be equivalent to the 

 short-circuit current of five generators or forty times the full load 

 of one generator. If the generators were rated, say, 20,000 Kv.A., 

 and the switch equipped with instantaneous relays, the switch 

 would have to rupture 40 X 20,000 X. 6 = 480,000 Kv.A. 



When dealing with the subject of bus reactors, it may be of 

 interest to consider their action a little more fully, and, in order to 

 obtain some idea of the angular relations of the currents and 

 voltages the following case will be considered. 



Assume an arrangement as illustrated in Fig. 294. The equip- 

 ment consists of four 20,000 Kv.A. generators, having a short- 

 circuit ratio of eight times normal full-load current. The bus is 

 divided in two sections by means of a reactor which will permit a 

 power transfer equivalent to one-half the capacity of one generator, 

 as shown. The power-factor of the load is 0.8 and it is assumed 

 that the generators are to carry equal loads and that the voltages 

 of the two bus sections A and B are kept the same. 



It is at once apparent that the generators on section A must 

 supply 10,000 Kv.A. through the reactor to section B, and in 

 order to limit the amount to this value, a 25 per cent reactor is 

 required, this figure being based on the rating of one generator. 

 Based on the actual transfer energy (one-half the capacity of one 

 generator), it would be 12 J per cent; thus, a total of 1250 Kv.A., 

 three-phase, or 416 Kv.A. per single reactor. 



The diagram illustrating the current and voltage relations may 

 be constructed as follows: Draw OA and OB, representing the 

 equal voltages of the two sections, in such a manner that AB, 

 which represents the voltage across the reactor, is 12 J per cent of 

 OA. Since this voltage differs in phase from the current prac- 

 tically 90 (neglecting the reactor losses), it follows that the 

 angular position of the circulating current is midway between the 

 voltages OA and OB. OC represents the current on section A 

 lagging approximately 37 (cos < = .8) behind its voltage OA, 

 while OD represents the current on section B, this, in turn, lagging 

 37 behind the voltage OB. OC and OD should be drawn to scale 

 so that their lengths represent the actual proportions between the 



