74 Scientific Proceedings^ Royal Dublin Society. 



x^ x^ x^ 

 cosa.= 1 -2". + 4|- g^ +... 



x^ x^ x7 

 ,,na.r = . _ _ + - - ^-, + 



7. The constants for the single-phase or synunetiical three-phase line are 

 determined by the following forniuhe. Tliey refer to one conductor : — 



,, -043 , ., „ -070 , .,,,.. 



11 = — — ohms per mile for copper ; or — j- ohms per mile tor aluminium. 



-£ = I 8 + 74'1 logio - I X lO"*^ henrys per mile. 



^=5x10"" mhos per mile, approxin\ately. 



^, 3-88 X 10-» „ , ., 



o = = — farads per mile. 



log:o- 



8. We commence with the determination of the best size of conductor. 



First we seek to determine the amount of power lost in the year and its value. 

 The average loss of power in the line for the year depends, not on the average 

 current, but on the square root of mean square of the current. 



We require to know this square root of mean square value of the current for 

 the year. 



Let it be G amperes per line. If the demand for power is steady and con- 

 tinuous night and day throughout the year, then G is the value of the current 

 transmitted. 



But in general the power demand is not steady and continuous, and it is 

 necessary to estimate the square root of mean square value of the current 

 transmitted. 



Thus, for example, if the various values of the current during the year, placed 

 in order of magnitude for their appropriate times, give a straight line, having an 

 arithmetical mean value h and extreme variations of Id from the mean, the effective 

 value of the current may be found thus : — 



J jet y be the value of the current at any time t. Then 



y = (^' - F) + 2^'^,, 

 where 7' is the whole period of time or year. The root mean square value of y is 



^ -'■ Jo 



As another example : if the variations of the current transmitted throughout 

 the year follow a sine law with a complete period of a year, or if the various 

 values of the current during the year, placed in order of magnitude for their 

 appropriate times, are equivalent to a sine load, we obtain Jk^ + hji'' for the 

 effective value of the current. Here k is the arithmetical mean value of the 

 current for the year, and k' is the amplitude of the sine curve. Then 



y = k + k sm-^, 



