BOILER ACCESSORIES: HORSE-POWER OF BOILERS The unit which we call the horse-power is arbitrary. Assuming that 30 pounds of steam are required per horse-power per hour for an average engine, this unit for boilers has been adopted.
One (1) horse-power is the evaporation of 30 pounds of water per hour, from a temperature of 100° F. into steam at 70 pounds gauge pressure. This is considered equivalent to the evaporation of 341 pounds per hour from and at 212° F. A boiler horse-power is equivalent to 33,327 B. T. U. per hour.
As all boilers do not generate steam at the same pressure and from the same temperature of feed-water, it is necessary to reduce the actual evaporation to an equivalent evaporation. Unless this is done, the relative performances of boilers cannot be compared.
For this comparison, the actual evaporation is reduced to the equivalent evaporation from and at 212° F. That is, we suppose the water to be fed at 212° and evaporated into steam at 212°.
Let W = Water actually evaporated in pounds; H = Total heat of steam above 32° F., at actual absolute pressure; T = Temperature of feed water; w = Equivalent evaporation from and at 212° F.
Since 966 B. T. U. are necessary to evaporate one pound of water from and at 212° F., the equivalent evaporation may be found from the formula, W (H + 32 — T) = 966w, or w = (W(H + 32 - T))/966.

Then the horse-power of the boiler is: H. P. = w/34.5 The above method is considerably shortened by substituting for the quantity (H + 32 - T)/966the number found in the accompanying table (page 80) which corresponds to the actual feed-water temperature and steam pressure.
For example, a boiler is required to furnish 2,100 pounds of steam per hour. If the gauge pressure is 85 pounds, and the feedwater enters at 50° F., what is the equivalent evaporation, and what is the horse-power? From the table, the factor for 85 pounds pressure and 50° F. is 1.204. Then the equivalent evaporation would be 1.204 X 2,100 = 2,528.4 pounds; and 2,528.4/34.5 = 73 (approx.) = the H. P.