Alternate Current Transformer. 



19 



factors. This would not be a satisfactory method for determining 

 the transformer numerics, as their values so obtained would 

 depend on the correct reading of small diti'erences. 



23. As an illustration of how closely this theory agrees 

 with practice I will discuss the following manufacturer's speci- 

 fication of the performance of a 10 K.W. Westinghouse O.D. 

 ti'ansformer. 



Primary volts, 2100 : — Frequency, 60 periods per sec: — 



Output, 10 K.W.:— Iron loss, 138 watts:— 



Copper loss at full load, 159 watts : — 



Efficiency (per cent.) 



Full load - - 97.1 

 97.05 

 96.55 



I load 



load 



Regulation (per cent.) 



Power factor - 1.0 - 1.65 



- .9 - 2.45 

 „ - .8 - 2.65 



- .6 - 2.80 



I load - - - 94.4 



In the first place we will find whether any definite values of 

 X and T will simultaneously satisfy the four equations in these 

 two quantities obtained from the four observations of drop for 

 different power factors. 



If we substitute x forj'X and /forjvT in the general expi-es- 

 sion in § 20 for the drop per cent. (R say) we get 



Cos'4> ' 



R = 100- jcSin(/)+ /Cos</)4-;v:'^(Siu'^<^+ —-^ \ + x t H'm cfiCos, (fi 



/''(Cos^</,+ 



Sin^(^ 



'^)) 



By means of this equation x and / were determined from the 

 first two observations of the regulation and found to be 



x = .0218, /=.016. 



Using these values and calculating R for the power factors .8 

 and .6, we found that when 



Cos<^ = .8 R = 2.66 



Cos</> = .6 R = 2.78 

 which agree very closely with the observed values of R, namely 

 2.65 and 2.80. 



As in each case the full load is lOCosqi K.W. (see § 21) and 

 the above 



Full load 



y= 



PoCos^ 



