Computing Total Heat Loss

The commonly accepted procedure for calculating the total heat loss from a structure is to calculate the heat loss for each room or space separately and then add the totals.

The heat loss calculation worksheet you use should contain a column in which each room or space can be identified separately. Under the identification are listed those structural sections through which heat transmission losses occur. When applicable, these will include all or most of the following:

1. Walls

2. Glass

3. Ceiling

4. Floor

5. Door(s)

TYPE OF ROOM OR SPACE TYPE OF STRUCTURAL SECTION (   BEDROOM WALLS /   N0.1 GLASS ARB A   CEILING   FLOOR   DOOR AREA   INFILTRATION

In addition to the five struc­tural sections listed above, each identified room or space should also include a line for heat loss by air infiltration. Figure 4-1 illustrates how your worksheet should appear at this point.

Now that you have identi­fied the room or space, listed Figure 4-1 Tabulation worksheet. the structural sections through

TYPE OF ROOM OR SPACE	TYPE OF STRUCTURAL SECTION	NET AREA/ AIR VOLUME	COEFFICIENT	TEMP. D IFF.	HEAT LOSS IBTU/hl
BEDROOM	WALLS	235 SO. fr.	O. Zi	90	5599
N0.1	SiASS AXE/k	45 SAFI	0.4S	90	J 823
	CEILINф	300 so. n.	0.17	40	30*40
	FLOOR.	—	——
	DOOR AREA	—	——-
	INFILTRATION	/F0ФCFH	0.018	90	2816

Figure 4-2 Tabulation worksheet.

Which heat loss occurs, and calculated the rate of air infiltration, you must determine the net surface area and the U-value for each structural section. The design temperature difference must also be determined and entered for each structural section. Except where surfaces are exposed to partially heated spaces (e. g., a garage, attic, or basement), each structural section will have the same design tem­perature difference. An example of how your worksheet should look at this point is illustrated in Figure 4-2.

Each identified room or space should also include the calculated air infiltration heat loss (see Infiltration Heat Loss following). The amount of heat loss due to air infiltration will depend upon the size, type, and number of windows and doors and other variables.

The bedroom given as an example in Figure 4-1 has two exposed walls (8 ft X 15 ft and 8 ft X 20 ft). The number of air changes suggested for a room with two exposed walls is 1V2 changes per hour.

The volume of air infiltration for the bedroom can be calculated as follows:

Vo1. No. ceiling area X ceiling height

Of air = of air X————————— ^—————

Infiltration changes

„ r 300 ft2 X 8 ft = 1.5 X 2

Knowing that the volume of air infiltration is 1800 ft3/h, the heat loss in Btu/h can be calculated as follows:

Heat loss = 0.018 X Q(ti — to) = 0.018 X 1800 X 90 = 2916 Btu

The total heat loss for bedroom No. 1 (as it would be designated on the worksheet) is 12,378 Btu/h. If there is a door in one of the outside walls, its heat loss also has to be calculated and entered on the worksheet. Furthermore, the door area (in square feet) would have to be subtracted from the surface area for the walls, because the latter is always a net figure.

After calculating the heat loss for each room or space in the structure, the results are added to obtain the total heat loss (in Btu/h) for the structure.

Posted in Audel HVAC Fundamentals Volume 1 Heating Systems, Furnaces, and Boilers