The efficiency of an indirect heater depends upon its form, the difference in temperature between the steam and the surrounding air, and the velocity with which the air passes over the heater. Under ordinary conditions in dwelling-house work, a good form of indirect radiator will give off about 2 B. T. U. per square foot per hour for each degree difference in temperature. Assuming a steam pressure of 2 pounds and an outside temperature of zero we should have a difference in ternperature of about 220 degrees, which under the conditions stated would give an efficiency of 220 X 2 = 440 B. T. U. per hour for each square foot of radiation. By making a similar computation for 10 degrees below zero we find the efficiency to be 460. In the same manner we may calculate the efficiency for varying conditions of steam pressure and outside temperature. In the case of schoolhouses and similar buildings where large volumes of air are warmed to a moderate temperature, a somewhat higher efficiency is obtained due to the increased velocity of the air over the heaters. Where efficiencies of 440 and 460 are used for dwellings, we may substitute 600 and 620 for school-houses. This corresponds approximately to 2.7 B. T. U. per square foot per hour for a difference of 1 degree between the air and steam. Fig. 12.

The principles involved in indirect steam heating are similar to those already described in furnace heating. Part of the heat given off by the radiator must be used in warming up the air supply to the temperature of the room, and part for offsetting the loss by conduction through walls and windows. The method of computing the heating surface required, depends upon the volume of air to be supplied to the room. In the case of a schoolroom or hall, where the air quantity is large as compared with the exposed wall and window surface we should proceed as follows:

First compute the B. T. U. required for loss by conduction through walls and windows, and to this add the B. T. U. required for the necessary ventilation, and divide the sum by the efficiency of the radiators. An example will make this clear.

How many square feet of indirect radiation will be required to warm and ventilate a schoolroom in zero weather, where the heat loss by conduction through walls and windows is 36000 B. T. U. and the air supply is 100,000 cubic feet per hour? By the methods given under "Heat for Ventilation" we have

100,000 X 70 = 127,272 = 55 B. T. U. required for ventilation.

36,000 + 127,272 + 163,272 B. T. U. = the total heat required, and this in turn divided by 600 (the efficiency of indirect radiators under these conditions) gives 272 square feet of surface required.

In the case of a dwelling-house the conditions are somewhat changed, for a room having a comparatively large exposure will perhaps have only 2 or 3 occupants, so that if the small air quantity necessary in this case was used to convey the required amount of heat to the room, it would have to be raised to an excessively high temperature. It has been found by experience that the radiating surface necessary for indirect heating is about 50 per cent greater than that required for direct heating. So for this work we may compute the surface required for direct radiation and multiply the result by 1.5.

Buildings like hospitals are in a class between dwellings and school houses. The air supply is based on the number of occupants, as in schools, but other conditions conform more nearly to dwelling houses.

To obtain the radiating surface for buildings of this class, we compute the total heat required for warming and ventilation as in the case of schoolhouses, and divide this sum by the efficiencies given for dwellings, that is 440 for zero weather and 460 for 10 degrees below.

Example. A hospital ward requires 50,000 cubic feet of air per hour for ventilation, and the heat loss by conduction through walls, etc. is 100,000 B. T. U. per hour. How many square feet of indirect radiation will be required to warm the ward in zero weather.

(50,000 X 70)/ 55 = 63,636 B. T. U. for ventilation: then,

63,686 + 100,000 = 372 + square feet.

Example - A school room having 40 pupils is to be warmed and ventilated when it is 10 degrees below zero. If the heat loss by conduction is 30,000 B. T. U. per hour and the air supply is to be 40 cubic feet per minute per pupil, how many square feet of indirect radiation will be required?

B. T. U, required for ventilation per hour

= (40 X 40 X 60 X 80)/ 55 = 139,636.

Total B. T. U. required = 139,636 + 30,000 = 169,636.

Surface required = 169,636/ 620 = 273 square feet.

Example. - The heat loss from a sitting room is 11,250 B. T. U. per hour in zero weather. How many square feet of indirect radiation will be required to warm it?

From Part I, page 56, we have 11250/ 225 = 50 square feet; here, then, we need 50 X 1.5 = 75 square feet.