334 PRINCIPLES OF ELECTRICAL DESIGN 



Items (39) and (40). We are now in a position to draw a 

 vector diagram similar to Fig. 115 for any power-factor angle 8, 

 the calculated numerical values of the component vectors being: 



OE, = = 3,810 



E t P = 8.4 

 PE g = 116 



EgE'g = 190- 



For a given terminal voltage of 6,600 (or 3,810 volts as meas- 

 ured between terminal and neutral point) the required developed 

 voltage will be a maximum when the external power-factor angle 

 is equal to the angle E g E t P because the additional voltage to be 

 generated will then be E t E g , which, in this particular example, 



FIG. 137. Vector diagram for 8000 k.v.a. turbo-alternator. 



amounts to V(8.4) 2 + (116) 2 = 116.4 volts, the effect of the 

 armature resistance being negligible as compared with the react- 

 ance of the end connections. 



For 80 per cent, power factor, as mentioned in the specification, 

 the angle will be 36 52', and the developed voltage per phase 

 winding (see Fig. 137) is 



OE g = 



where OB = OA + AB = OE t cos + 

 and BEg = BP + PE g = OE t sin 6 + PE . 



Similarly, for calculating the full-load flux per pole (see Art. 99) 

 we have: 



"Apparent" developed voltage = OE' = \/(OJ5) 2 + (BE' g ) z 



