Loss in Slab Construction
The heat loss for houses and small buildings constructed on a concrete slab at or near grade level is computed on the basis of heat loss per foot of exposed edge. For example, a concrete slab measuring 20 ft X 25 ft would have an exposed edge of 90 ft. This represents the measurement completely around the perimeter of the exposed edge of the slab. The heat loss will depend upon the thickness of insulation along the exposed edge and the outside design temperature range. This type of information is available from ASHRAE publications. To obtain the heat loss in Btu/h, simply multiply the total length of exposed edge by the heat loss in Btu/h per lineal foot (lin ft). For example, a 90ft exposed edge with 2inedge insulation at an outdoor design temperature of 35° would be calculated as follows:
90 lin ft X 45 Btu/h/lin ft = 4050 Btu/h Infiltration Heat Loss
During the heating season, a portion of heat loss is due to the infiltration of cooler outside air into the interior of the structure through cracks around doors and windows and other openings that are not a part of the ventilating system. The amount of air entering the structure by infiltration is important in estimating the requirements of the heating system, but the composition of this air is equally important.
A pound of air is composed of both dry air and moisture particles, which are combined (not mixed) so that each retains its individual characteristics. The distinction between these two basic components of air is important, because each is involved with a different type of heat: dry air with specific heat, and moisture content with latent heat.
The heating system must be designed with the capability of warming the cooler infiltrated dry air to the temperature of the air inside the structure. The amount of heat required to do this is referred to as the sensible heat loss and is expressed in Btu/h. The two methods used for calculating heat loss by air infiltration are: (1) the crack method, and (2) the airchange method.
The Crack Method
The crack method is the most accurate means of calculating heat loss by infiltration, because it is based on actual air leakage through cracks around windows and doors and takes into consideration the expected wind velocities in the area in which the structure is located. The airchange method (see below) does not consider wind velocities, which makes it a less accurate means of calculation.
Calculating heat loss by air infiltration with the crack method involves the following basic steps:
1. Determine the type of window or door (see Table 43).
2. Determine the wind velocity and find the air leakage from Table 43.












Industrial pivoted, V^in crack 
52 
108 
176 
244 
304 
372 
Architectural projected, Vhin crack 
15 
36 
62 
86 
112 
139 
Architectural projected, 3/64in crack 
20 
52 
88 
116 
152 
182 
Residential casement, Vfe4in crack 
6 
18 
33 
47 
60 
74 
Residential casement, V32in crack 
14 
32 
52 
76 
100 
128 
Heavy casement section, projected, Vfe4in crack 
3 
10 
18 
26 
36 
48 
Heavy casement section, projected, V32in crack 
8 
24 
38 
54 
72 
92 
30 
88 
145 
186 
221 
242 
Hollow Metal, Vertically Pivoted Window 
* Expressed in cubic feet per foot of crack per hour. The infiltration rate through cracks around closed doors is generally estimated at twice that calculated for a window.
00 SO 
Courtesy ASHRAE 1960 Guide
3. Calculate the lineal feet of crack.
4. Determine the design temperature difference.
The data obtained in these four steps are used in the following formula:
H = 0.018 X Q(ti — to) X L where
H = heat loss, or heat required to raise the temperature of air leaking into the structure to the level of the indoor temperature (t) expressed in Btu per hour.
Q = volume of air entering the structure, expressed in cubic feet per hour (Step 2 above). t = indoor temperature. to = outdoor temperature.
0.018 = specific heat of air (0.240) times density of outdoor air (approximately 0.075).
L = lineal feet of crack.
Determine the infiltration heat loss per hour through the crack of a 3 ft X 5 ft average doublehung, nonweatherstripped, wood window based on a wind velocity of 20 mph. The indoor temperature is 70°F, and the outdoor temperature 20°F.
The air leakage for a window of this type at a wind velocity of
20 mph is 59 ft3 per foot of crack per hour. This will be the value of Q in the air infiltration formula. The lineal feet of the crack is (2 X 5) plus (3 X 3), or 19 ft (the value of L in the formula). t. = 70°F, and t = 20°F. Substituting these data in the air infiltration formula gives the following results:
H = 0.018 X Q(ti — to) X L = 0.018 X 59(70 — 20) X 19 = 1.062 X 50 X 19 = 1008.9 Btu/h = 1009 Btu/h
Table 44 represents a typical wall infiltration chart for a number of different types of wall construction. Like the air infiltration rate through windows, it is also based on wind velocity (see Table 43).
Table 44 Infiltration Through Various Types of Wall Construction

Courtesy ASHRAE 1960 Guide 
AirChange Method
In the airchange method, the amount of air leakage (i. e., infiltration) is calculated on the basis of an assumed number of air changes per hour per room. The number of air changes will depend upon the type of room and the number of walls exposed to the outdoors. Table 45 is an example of a typical airchange chart used in the airchange method.
Table 45 Fresh Air Requirements
Type of Building or Room
Attic spaces (for cooling)
Boiler room Churches, auditoriums College classrooms Dining rooms (hotel)
Engine rooms
Factory buildings (ordinary
Manufacturing)
Factory buildings (extreme fumes or moisture)
Foundries
Minimum Air Cubic Feet of
Changes per Air per Minute
Hour per Occupant
1215
1520
8 2030
2530
5
46
24
1015
(continued)
Type of Building or Room
Galvanizing plants Garages (repair)
Garages (storage)
Homes (night cooling)
Hospitals (general)
Hospitals (children’s)
Hospitals (contagious diseases) Kitchens (hotel)
Kitchens (restaurant)
Libraries (public)
Laundries Mills (paper)
Mills (textile—general buildings) Mills (textile—dyehouses) Offices (public)
Offices (private)
Pickling plants Pump rooms Schools (grade)
Schools (high)
Restaurants Shops (machine)
Shops (paint)
Shops (railroad)
Shops (woodworking) Substations (electric)
Theaters
Turbine rooms (electric)
Warehouses
Waiting rooms (public)
Minimum Air Cubic Feet of
Changes per Air per Minute
Hour per Occupant
2030
2030
4 6
9 17
4050
3540
8090
10 20 1020 4
1015
1520
4
1520
3
4
1015
5
1525
3035
812
5
1520
5
5
5 10
1015
510
2
4
Posted in Audel HVAC Fundamentals Volume 1 Heating Systems, Furnaces, and Boilers