THE TRANSFORMER. 



2I 3 



The all-day efficiency of a transformer (or of any machine) is 

 defined as the quotient obtained by dividing the total energy de- 

 livered by the machine in a day by the total energy delivered to 

 the machine in a day. For example, consider a transformer of 

 which the iron loss is 300 watts and the copper loss at full load 

 is also 300 watts, the full-load output of the transformer being 

 10 kilowatts. The transformer is connected to the mains all day 

 and the total energy lost in the core in a day is 24 x 300 watt- 

 hours. The transformer is used at full load for I ^ hours each 

 day, during which time the copper loss is I j x 300 watt-hours, 



0.25 



0.50 0.75 1.00 



fractions of full load 



Fig. 178. 



and the copper loss during the remaining 22*/ hours is negligible. 

 The total energy delivered by the transformer in a day is 

 10,000 x I J/2 watt-hours, and the total energy delivered to the 

 transformer in a day is of course [10,000 x I J^ -f 24 x 300 

 + i j x 300] watt-hours, so that the all-day efficiency of the 

 transformer under the given conditions is 15,000 watt-hours di- 

 vided by 22,650 watt-hours, or 66.2 per cent. 



Calculation of efficiency. Given the details of design of a 

 transformer to calculate its efficiency for given value and fre- 

 quency of E 1 (or E"} and given current output /". 



