482 



ELECTRICAL EQUIPMENT 



leg of the Y. The voltage drop in each part of the circuit is in 

 phase with the voltage of the short-circuited leg a-b, and the 



FIG. 299. Single-phase Short-circuit Currents 



total voltage drop is equal to c-d or 0.866#. The following equa- 

 tions may be written from the figure : 



2(e-iz)=i(y 2 +z); 

 from which we find 



or expressed in per cent reactance based on normal three-phase 

 line current I 



. 1007 



Based on these fundamental equations it is possible to solve 

 problems in cases involving a number of generating stations, a 

 network of lines, etc. As the number of generating stations 

 increases, however, the equations increase in complexity and the 

 solution becomes quite laborious. The labor is lessened some- 

 what by representing the network by an equivalent circuit with 

 the component parts expressed in per cent reactance and solving 

 either by the slide rule or by the calculating table. 



An example will iUustrate this. In Fig. 300 let GI and G 2 

 represent generators, TI and T 2 transformers with isolated neutrals 

 and TZ a transformer with grounded neutral. The percentages 

 of reactance based on 10,000 Kv.A. 100,000 volts and three-phase 

 are indicated. 



