There are two kinds of load factor: the annual, and the daily.
The annual load factor is obtained as a percentage by multiplying the number of units sold (per year) by 100, and dividing by the product of the maximum load and the number of hours in the year. The daily load factor is obtained by taking the figures for 24 hours instead of a year.
Fig. 2,719.—Load curve of plant supplying power for the operation of motors in a manufacturing district. The horizontal dotted lines show suitable power ratings. A properly designed steam plant has a large overload capacity, a hydraulic plant has a small overload capacity, and a gasoline engine plant has no overload capacity. Accordingly, the peak of the load (maximum load) may be 25 or 30 per cent. in excess of the rated capacity of a steam plant, not more than 5 or 10 per cent. in excess of the rated capacity of a hydraulic plant, not at all in excess of the rated capacity of a gas engine plant.
Ques. What must be provided in addition to the machinery required to supply the peak load?
Ans. Additional units must be installed for use in case of repairs or break down of some of the other units.
EXAMPLE.—What would be the boiler horse power required to generate 5,000 kw. under the following conditions: Efficiency of generators 85%; efficiency of engines 90%; feed water of engines and auxiliaries 15 lbs. per I. H. P.; boiler pressure 175 lbs.; temperature of feed water 150° Fahr? With a rate of combustion of 15 lbs. of coal per sq. foot of grate per hour and an evaporation (from and at 212°) of 8 lbs. of water per lb. of coal, what area of grate would be required and how much heating surface?
5,000 kw. = 5,000 ÷ .746 = 6,702 electrical horse power
To obtain this electrical horse power with alternators whose efficiency is 85% requires