356 PRINCIPLES OF ELECTRICAL DESIGN 



cient internal magnetic leakage to prevent mechanical injury 

 due to excessive magnetic forces on short-circuit. 



Items (72) and (73). Seeing that the calculation of efficiency 

 under different conditions of loading was illustrated in Art. 63 

 when working out the example in dynamo design, it will not be 

 necessary to cover the same ground a second time. We shall, 

 therefore, determine the full-load efficiency only (item (72)). 



The windage and bearing friction loss is very difficult to esti- 

 mate; but some approximate figures were given in Art. 111. We 

 shall assume 1.1 per cent, of the total k.v.a. output for the loss 

 due (mainly) to the friction of the air passing through the axial 

 ducts, and also to friction in the outside bearing. It is assumed 

 that the losses in the bearing on the side of the prime mover are 

 included in the steam-turbine efficiency. The power necessary 

 to drive the blower is not included in the above estimate of the 

 windage and friction loss. 



For the correct calculation of iron losses, the reader is referred 

 to Art. 60, Chap. IX where the method of determining the tooth 

 losses was explained ; but since the maximum tooth density under 

 full-load conditions, as indicated by curve C of Fig. 141 does not 

 differ appreciably from the maximum of the open-circuit curve 

 (A), we shall not trouble to correct the tooth losses as previously 

 calculated. 



With reference to the iron in the body of the stator, the flux 

 indicated by the area of the load flux curve C of Fig. 141 does 

 not all enter the core below the slots, because this total flux in- 

 cludes the slot-leakage flux, as explained in Art. 95. The posi- 

 tion of the conductors carrying the maximum current coincides 

 with the zero point on the armature m.m.f. curve. The point 

 where the resultant m.m.f. curve crosses the datum line is the 

 neutral zone, under the given load conditions. This is the 

 position 17 degrees in Figs. 141 and 142; and the current in the con- 

 ductor will be approximately 700 X cos 53 = 420 amp. The slot- 

 leakage flux corresponding to this particular current the total 

 not the "equivalent" flux can be calculated as explained in 

 Art. 96, Chap. XIII, when deriving formula (104) which, how- 

 ever, gives the "equivalent" and not the total slot leakage flux. 

 If $. is the calculated slot fliux, and 3> is the total flux per pole 

 in the air gap, then the flux actually carried by the section of 



<!> 

 the armature iron below the slots is -^ $ This correction 



