Document DDV8VeyjOZj77LykYgRm41nrB

HEATING VENTILATING AIR CONDITIONING CUIDE 1943 Solution. The coefficient of transmission (U) of a plain 8 in. brick wall is 0.50 (Chapter 4, Table 3). The area (/4) is 150 sq ft. Substituting in Equation 1: Ht = 150 X 0.50 X [70 - ( - 10)] = 6000 Btu per hour. INFILTRATION HEAT LOSSES The infiltration heat losses include (1) the sensible heat loss or the heat required to warm the outside air entering by infiltration and (2) the latent heat loss or the heat equivalent of any moisture which must be added. Sensible Heat Loss The formula for the heat required to warm the outside air which enters a room by infiltration;;to the temperature of the room, is given in Equation 2. Hs = 0.24 Qd (/ - to) (2) where _. Hs -- heat required to raise temperature of a'ir leaking into building from t0 to t, Btu per hour. 0.24 = specific heat of air. ' Q -- volume of outside air entering building, cubic feet per hour (see Chapter 5). d = density of air at temperature to, pounds per cubic foot. It is sufficiently accurate "to use d -- 0.075 in which case Equation 2 reduces to L lls -- 0.018 Q (/ - t0) . (2a) ' The volume of outside air entering per hour (0 depends on the wind velocity and direction, the width of crack or size of openings, the type of openings and other factors, as explained in Chapter 5. Where the crack method is used for estimating the amount of air leakage, it is more con venient to express the heat loss due to air leakage in terms of the crack length, as follows:" , where Hs '= 6.018 QL(t - ti) = B L (t - t0) .] ; . (2b): B = air leakage per foot of crack (Chapter 5) for the'wind velocity and type of windows or door crack involved multiplied by 0.018. L = length of window or door crack to be taken'into consideration; feet.' Example 8. What is the infiltration heat loss per hour through the crack of a 3 x 5 ft double-hung wood window, based on an average non-weatherstripped window and a wind velocity of 15 mph? Assume inside and outside temperatures to be 70 F and zero respectively. ._ . Solution. According to Table 2, Chapter 5, the air leakage through a window of this type (based on hs in. crack and % in. clearance) is 39.3 cu ft per foot of crack per hour. Therefore, B -- 39.3 X 0.018 = 0.71. The length of crack (L) is (2X5) + "(3 X 3), or 19 ft; l -- 70 and tQ = 0.' Substituting in Equation 2b, H3 = 0.71 X 19 X (70 -- 0) = 944 Btu per hour, Crack Length to be Used for Computations ... The amount .of. crack used for computing the infiltration heat loss should not be less than half of the total crack in the outside walls of the 128 CHAPTER .6. HEATING-LOAD room. For a building having no partitions, whatever wind enters through the cracks on the windward, side must leave through the cracks on the leeward side. Therefore, take one-half the total crack for computing each side and end of the building. In a room with one exposed wall, take all the crack; with two exposed walls, take the wall haying the most crack; and with three of four exposed walls, take the wall having the most crack; but in no case take less than half the total crack. The total infiltration loss of a building having partitions will not be equal to the sum of the infiltration losses of the various rooms since at any given time infiltration will take place Only on the windward side or sides and not on the leeward side. Therefore, if a building-has more than one room which is divided by interior walls or partitions, it is sufficiently accurate to use half of the total infiltration losses for determining the total heat requirements. Latent Heat Loss When it is intended to add. moisture to air leaking into a room for the maintenance of proper winter comfort conditions, it is necessary to determine the heat equivalent to evaporate the required amount of water vapor, which may be calculated by the equation: ' (3) where Hi = heat required to increase moisture content of air leaking into building from mo to mi, Btu per hour. Q' = volume of outside air entering building, cubic feet per hour; d = density of air at temperature t{, pounds per cubic foot, mi = vapor density of inside air, grains per pound of dry air. m0 = vapor density of outside air, grains per pound of dry air. hfg = latent heat of vapor at ti, Btu^per pound. If the latent heat of vapor (hig) is assumed to be 1060 Btu per pound, Equation 3 reduces to . Hi - 0.0114 Q (mi - mo) (3a) Equations 2a, 2b and 3a may also be used for determining the sensible and latent heat gains due to infiltration in. cooling load computations. INSIDE TEMPERATURES The inside air temperature which must be maintained within a building is understood to be the dry-bulb temperature at the breathing line, 5 ft above the floor, or the 30-in. line, and not less than 3 ft from the outside walls. Inside air temperatures, usually specified, vary in accordance with the use to which the building is to be put and Table 1 presents values which conform with good practice. The proper dry-bulb temperature to be maintained, depends upon the . relative humidity and air motion, as explained in Chapter 2. In other words, a.person may feel warm or cool at the same dry-bulb temperature, depending on the relative humidity and air motion; The optimum.winter 129