. Electric railway review . Figure 3.—Showing Method of Changing Cost Per Horsepower Per Year at Various Load-Factors to Equivalent Cost Per Kilowatt-Hour and Vice-Versa. are 43,200. then average demand for power is 43,200divided by 720 (the number of hours in the month), equal to60 kilowatts or 80 horsepower. Assuming further that hismaximum demand each day was just 400 horsepower, then,of course, his average maximum demand for the month will be the same amount, and the load-factor is 80 divided by400 — .2, or, as commonly expressed, 20 per cent. It the rate i er horsepower-year varies b
. Electric railway review . Figure 3.—Showing Method of Changing Cost Per Horsepower Per Year at Various Load-Factors to Equivalent Cost Per Kilowatt-Hour and Vice-Versa. are 43,200. then average demand for power is 43,200divided by 720 (the number of hours in the month), equal to60 kilowatts or 80 horsepower. Assuming further that hismaximum demand each day was just 400 horsepower, then,of course, his average maximum demand for the month will be the same amount, and the load-factor is 80 divided by400 — .2, or, as commonly expressed, 20 per cent. It the rate i er horsepower-year varies between $1H and$43, it will be evident that the variable quantity is the differ-ence between $16 and $43, or $27. The rale is therefore equalto the minimum rate ($16) plus the load-factor (.2) times thevariable ($27). Two-tenths of $27 is $, so the rate per J 7^ r-. Figure 4.—Chart Illustrating Suggested Plan ot Charging ForElectric Power. horsepower per year for the mouth will be $16 plus $ $ The total charge for the month would there-fore be 400 times $ divided by 12, or $ This isequal to cents per kilowatt-hour. If his use of the powerhad been such as to give a load-factor of 30 per cent, the rateper horsepower per year would have increased to $, butthe equivalent cost per kilowatt-hour would have decreasedto cents—a reduction of almost 25 per cent in cost perkilowatt-hour due to increasing the load-factor to 30 per cent This may readily be put in the form of an equationwhich, if the desired rate per horsepower per year is Rthe minimum rate limit is A the maximum rate limit is B and the load-factor is L is expressed by R = A -f L (B-A). This method is much more equitable than that sometimesused, of selling all the power on a kilowatt-hour basis with aguarantee from the consumer of a specified load
Size: 1934px × 1292px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1900, booksubjectstreetr, bookyear1906
