A British Thermal Unit, or B. T. U., has been defined as the amount of heat required to raise the temperature of one pound of water one degree F. This measure of heat enters into many of the calculations involved in the solving of problems in heating and ventilation, and one should familiarize himself with the exact meaning of the term.

## Causes Of Heat Loss

The heat loss from a building is due to the following causes; first, radiation and conduction of heat through walls and windows; second, leakage of warm air around doors and windows and through the walls themselves; and third, heat required to warm the air for ventilation.

## Loss Through Walls And Windows

The loss of heat through the walls of a building depends upon the material used, the thickness, the number of layers and the difference between the inside and outside temperatures. The exact amount of heat lost in this way is very difficult to determine theoretically, hence we depend principally on the results of experiments.

## Loss By Air Leakage

The leakage of air from a room varies from one to two or more changes of the entire contents per hour, depending upon the construction, opening of doors, etc. It is common practice to allow for one change per hour in well-constructed buildings where two walls of the room have an outside exposure. As the amount of leakage depends upon the extent of exposed wall and window surface it seems best to allow for this loss by increasing that due to conduction and radiation. The following table gives the heat losses through different thickness of walls, doors, windows, etc., in B. T. U., per square foot of surface per hour for varying differences in inside and outside temperatures.

Authorities differ considerably in the factors given for heat losses, and there are various methods for computing the same. The following figures and methods have been used extensively in actual practice and have been found to give good results when used with judgment.

 Material. Difference between inside and outside temperatures. 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 8" Brick Wall 5 9 13 18 22 27 31 36 40 45 12" Brick Wall 4 7 10 13 16 20 23 26 30 35 16" Brick Wall 3 5 8 10 13 16 19 22 24 27 20" Brick Wall 2.8 4.5 7 9 11 14 16 18 20 23 24" Brick Wall 2.5 4 6 8 10 12 14 16 18 20 28" Brick Wall 2 3.5 4.5 7 9 11 13 14 16 18 32" Brick Wall 1.5 3 5 6 8 10 11 13 15 16 Single Window 12 24 36 49 60 73 85 93 105 Double Window 8 16 24 32 40 48 56 62 70 Single Skylight 11 21 31 42 52 63 73 84 94 Double Skylight 7 14 20 28 35 42 48 56 62 1" Wooden Door 4 8 12 16 20 24 28 32 36 40 2" Wooden Door 3 5 8 11 14 17 20 23 25 28 Concrete Floor on Brick Arch 2 4 6.5 9 11 13 15 18 20 22 Wood Floor on Brick Arch 1.5 3 4.5 6 7 9 10 12 13 15 Double Wood Floor 1 2 3 4 5 6 7 8 9 10 Walls of Ordinary Wooden Dwellings 3 5 8 10 13 16 19 22 24 27

For solid stone walls multiply the figures for brick of the same thickness by 1.7. Where rooms have a cold attic above or cellar beneath, multiply the heat loss through walls and windows by 1.1. The figures given in table III. are for a southern exposure; for other exposures multiply the heat loss given in table III. by the factors given in table IV.

 TABLE IV. Exposure. Factor. N. 1.32 E. 1.12 S. 1.0 W. 1.20 N. E. 1.22 N. W. 1.26 S. E. 1.06 S. W. 1.10 N. E. S. W. or total exposure. 1.16

In order to make the use of the table clear we will give a number of examples illustrating its use.

Assuming an inside temperature of 70°, what will be the heat loss from a room having an exposed wall surface of 200 square feet and a glass surface of 50 square feet, when the outside temperature is zero. The wall is of brick, 16 inches in thickness and has a southern exposure; the windows are single.

We find from table III, that the factor for a 16" brick wall with a difference in temperature of 70° is 19, and that for glass (single window) under the same condition is 85; therefore Loss through walls = 200 X 19 = 3800 Loss through windows = 50 X 85= 4250

Total loss per hour = 8050 B. T. U.

In computing the heat loss through walls, only those exposed to the outside air are considered.

A room 15 ft. square and 10 ft. high has two exposed walls; one toward the north and the other toward the east. There are 4 windows, each 3' X 6' in size The two in the north wall are double while the other two are single. The walls are of brick, 20 inches in thickness; with an inside temperature of 70° what will be the heat loss per hour when it is 10° below zero.

Total surface = 15 X 10 X 2 = 300 Glass surface = 3 X 6 X 4 = 72

Net wall surface = 228

Difference between inside and outside temperature 80°.

Factor for 20" brick wall is 18.

Factor for single window is 93.

Factor for double window is 62.

The heat losses are as follows: Wall, 228 X 18 = 4104

Single windows, 36 X 93 = 3348 Double windows, 36 X 62 = 2232

9684 B. T. U. As one side is toward the north and the other toward the west the actual exposure is N. W. Looking in table IV. we find the correction factor for this exposure to be 1.26; therefore the total heat loss is

9684 X 1.26 = 12,201.84 B. T. U.

A dwelling house of wooden construction measures 160 ft. around the outside; it has 2 stories, each 8 ft. in height; the windows are single and the glass surface amounts to 1/5 the total exposure; the attic and cellar are unwarmed. If 8000 B. T. U. are utilized from each pound of coal burned in the furnace, how many pounds will be required per hour to maintain a temperature of 70° when it is 20° above zero outside. Total exposure = 160 X 16 = 2560 Glass surface = 2560 / 5 = 512

Net wall = 2048

Temperature difference =70 - 20 = 50° Wall 2048 X 13 = 26624 Glass 512 X 60 = 30720 As the building is exposed on all sides the factor for exposure will be the average of those for N. E. S. and W. or (1.32 + 1.12 + 1.0 + 1.20/ 4 = 1.16

The house has a cold cellar and attic so we must increase the heat loss 10% for each or 20% for both. Making these corrections we shall have 57344 X 1.16 X 1.20 = 79822 B. T. U.

One pound of coal furnishes 8000 B. T. U. then 79822/ 8000 = 9.97; or about 10 pounds of coal per hour will be required to warm the building to 70° under the conditions stated.

## Approximate Method

For dwelling houses of usual construction the following simple method may be used. Multiply the total exposed surface by 38, which will give the heat loss in B. T. U. per hour for an inside temperature of 70° in zero weather.

This factor is obtained in the following manner. Assume the glass surface to be 1/6 the total exposure, which is an average proportion.

Then each square foot of exposed surface consists of 1/6 glass and 5/6 wall and the heat loss for 70° difference in temperature would be as follows: