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 90-ft exposed edge with 2-in-edge 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 air-change 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 air-change 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 4-3).

2. Determine the wind velocity and find the air leakage from Table 4-3.

Table 4-3 Infiltration Rate Through Various Types of Windows*
Type of Window	Remarks	Wind Velocity, Miles per Hour
10	15	20 25	30
Double-Hung Wood Sash Windows (Unlocked) Around frame in masonry wall—not caulked 3 8 14 20 27 35 Around frame in masonry wall—caulked 1 2 3 4 5 6 Around frame in wood-frame construction 2 63 11 17 23 30 Total for average window, non-weather-stripped, yi6-in crack and %4-in clearance; includes wood frame leakage 7 21 39 59 80 104 Ditto, weather-stripped 4 13 24 36 49 63 Total for poorly fitted window, non-weather-stripped, 3/32-in crack and 3/32-in clearance; includes wood frame leakage 27 69 111 154 199 249 Ditto, weather-stripped 6 19 34 51 71 92 Double-Hung Non-weather-stripped, locked 20 45 70 96 125 154 Metal Windows Non-weather-stripped, unlocked 20 47 74 104 137 170 Weather-stripped, unlocked 6 19 32 46 60 76

Industrial pivoted, V^-in crack	52	108	176	244	304	372
Architectural projected, Vh-in crack	15	36	62	86	112	139
Architectural projected, 3/64-in crack	20	52	88	116	152	182
Residential casement, Vfe4-in crack	6	18	33	47	60	74
Residential casement, V32-in crack	14	32	52	76	100	128
Heavy casement section, projected, Vfe4-in crack	3	10	18	26	36	48
Heavy casement section, projected, V32-in 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 double-hung, non-weather-stripped, 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 4-4 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 4-3).

Table 4-4 Infiltration Through Various Types of Wall Construction Type of Wall Wind Velocity, Miles per Hour 5 10 15 20 25 30 Brick Wall 8V2 in Plain 2 4 8 12 19 23 Plastered 0.02 0.04 0.07 0.11 0.16 0.24 Plain 1 4 7 12 16 21 13 in Plastered 0.01 0.01 0.03 0.04 0.07 0.10 Frame Wall, Lath and Plaster 0.03 0.07 0.13 0.18 0.23 0.26

Courtesy ASHRAE 1960 Guide

Air-Change Method

In the air-change 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 4-5 is an example of a typical air-change chart used in the air­change method.

Table 4-5 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

12-15

15-20

8 20-30

25-30

5

4-6

2-4

10-15

(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

20-30

20-30

4- 6

9- 17

40-50

35-40

80-90

10- 20 10-20 4

10-15

15-20

4

15-20

3

4

10-15

5

15-25

30-35

8-12

5

15-20

5

5

5- 10

10-15

5-10

2

4

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