Having determined the heat loss from a building by the methods given, it is a simple matter to compute the size of grate necessary to burn a sufficient quantity of coal to furnish the amount of heat required for warming.

As a matter of illustration we may consider the heat delivered to the rooms as made up of two parts; first, that required to warm the outside air up to 70° (the temperature of the rooms) and second, the quantity which must be added to this to offset the loss through walls and windows. Air is usually delivered at the registers at about 140 degrees under zero conditions outside; this air leaves the rooms by leakage at a temperature of 70 degrees, (the normal inside temperature) having lost one-half its heat by conduction, radiation, etc., so that the heat given to the entering air must be twice that which we have computed for loss through walls, etc.

Example. - The loss through the walls and windows of a building is found to be 80000 B. T. U. per hour in zero weather, what will be the size of furnace required to maintain an inside temperature of 70 degrees?

From the above we have the total heat required, equal to 80000 X 2 = 160000 B. T. U. per hour. If we assume that 8000 B. T. U. are utilized per pound of coal, then 160000/ 8000 = 20 pounds of coal required per hour, and if 5 pounds can be burned on each square foot of grate per hour, then 20/ 5 = 4 square feet required. A fire pot 28 inches in diameter has an area of 4.27 square feet and is the size we should use.

The following table will be found useful in determining the diameter of fire pot required:




















If the outside temperature is below zero the method of computation becomes slightly different. We have seen that in zero weather a certain quantity of heat is required to raise the temperature of the entering air from zero to 70°, the temperature of the room, and that a second quantity must then be added to raise the temperature of the air to 140°, which is the usual temperature of delivery at the registers. This last quantity is to offset that lost by radiation and conduction, and must equal the heat loss from the building as computed by the factors given in tables III. and IV. The air has been raised through 140 degrees and 70/140 of the heat supplied has been used to raise it to the temperature of the room and has been lost by leakage; while the remaining 70/140 an equal amount, has been given up by radiation and conduction. In this case we have only to compute the heat loss for radiation and conduction by the rules given and multiply this result by 2 to obtain the total amount of heat to be supplied by the furnace.

Now take a case where it is 10 degrees below zero. If the air is delivered to the rooms at 140 degrees as before, it must be warmed through 150 degrees. Of the heat supplied 80/ 150 has been used to raise the temperature of the outside air to that of the room, and only 70/ 150 for loss by radiation and conduction. As in the preceding example, this latter quantity must equal the computed heat loss through walls and windows; and as it is only 70/ 150 or .466 of the total amount of heat required we must multiply it by 1/ .466 = 2.14 instead of by 2 as in the first case where the outside temperature is zero.

In the same manner multiply by 2.28 for 20 degrees below zero and by 2.42 for 30 degrees.

Example. - A brick house is 20 feet wide, and has 4 stories, each being 10 feet in height. The house is one of a block and is exposed only at the front and rear. The walls are 16 inches thick and the block is so sheltered that no correction need be made for exposure. Single windows make up 1/8 the total exposed surface. Fig.re for cold attic but warm basement. What area of grate surface will be required for a furnace to keep the house at 70° when it is 10° below zero outside ? Exposed surface = 2 X 20 X 40 = 1600 sq. ft. Loss through windows = 200 X 93 = 18,600 B. T. U Loss through walls = 1400 X 22 = 30,800 Loss through walls and windows = 49,400 Entire heat loss, allowing for attic

= 49,400 X 1.1 = 54,340 B. T. U. B. T. U. required = 54,340 X 2.14 = 116,288

116,288 = 2.90 sq. ft. of grate. Ans.


5 X 8000