Document J3OOXwRdLdB3Vw9ZMmJpMQLG6

American Society of Heating and Ventilating Engineers Guide, 1934 conditions maintained in the room must be provided by the tempering coils, preheater coils and reheater coils. If a positive pressure is not main tained in the room or space to be conditioned, the normal infiltration of outside cold air will take place in this room, and the outlet temperature, together with the required air volume at this temperature, must be suf ficient to provide for both the infiltration and transmission losses. Volume of Outside Air . The volume of outside air required for ventilation or air conditioning purposes may be determined from data in Chapter 2. In no case shall less than 10 cfm per person be introduced. The heat required to warm the outside air introduced for ventilation purposes (H0) may be determined by means of the following formula: where H0 = 0.24 (t - to) M0 (1) 0.24 = specific heat of air at constant pressure. / = room temperature, degrees Fahrenheit. to -- outside temperature, degrees Fahrenheit. M0 = weight of outside air to be introduced per hour in pounds = doQo- Qo = volume of outside air to be introduced, cubic feet per hour. do = density of air at to, pounds per cubic foot. \ Example 1. A building in which the temperature to be maintained is 70 F requires 10,000 cfm. If the outside temperature is 20 F, how much heat will be required to warm the air introduced for ventilation purposes to the room temperature? Solution. Qo = 10,000 X 60 = 600,000 cfh; dQ = 0.08276 (Table 1, Chapter 41); Mo = 0.08276 X 600,000 = 49,656 lb; t = 70 F; to = 20 F; Ha = 0.24 X (70 - 20) X 49,656 = 595,872 Btu per hour. Temperature of Air Leaving Registers If the system is to function only as a heating^ystem, that is, entirely as a recirculating one, the temperature of the air leaving the register outlets must be assumed. For public buildings, these temperatures may range from 100 to 120 F, whereas for factories and industrial buildings the out let or register temperature may be as high as 140 F. In no case should the . outlet temperature exceed these values. For ventilating or conditioning systems, the temperature of the air leaving the supply outlets may be estimated by means of the following formula: _ 55.2H h------- 7i------h t (2) where ty = outlet temperature, degrees Fahrenheit. H. = heat loss of room or space to be conditioned, Btu per hour. Q = total volume of air to be introduced at the temperature, t, cubic feet per hour. . If the outlet temperature (ty) as determined from Equation 2 exceeds 120 F for public buildings, or 140 F for factories or industrial buildings, this temperature should be assumed using the maximum permissible 296 Chapter 22--Central Fan Heating Systems temperature and the volume of air to be introduced into the room or space determined by means of the following equation: 55.2H Q = (ty -- t) (3) Example 2. The heat loss of a certain auditorium to be conditioned is 100,000 Btu per hour. The ventilating requirements are 90,000 cu ft per hour and the room temperature 70 F. Determine the outlet temperature. Solution. Substituting in Formula 2, Inasmuch as this temperature is excessive, it will be necessary to assume the outlet temperature, which will be taken as 120 F, and to calculate the amount of air to be introduced into the room at this temperature to provide for the heat loss. Substituting in Equation 3, q = 55'^2Q = n.400 cfh (at temperature t) Weight of Air to be Circulated The total weight of air to be introduced into the room or space to be heated or conditioned (M) is given by the following formulae: 'M 0.24(Iy -- t) (4) M = Mo + Mr '(5) (6) II f9- where d = density of air at temperature t, pounds per cubic foot. d0 = density of air at temperature to, pounds per cubic foot. Qo = volume of outside air at temperature to. Mo -- weight of outside air, pounds. Mi = weight of recirculated air, pounds. Example 3. Using the data of Example 2 and an outside temperature of 20 F, what will be the values of M, Ma and Mil Solution, d = 0.07495, do = 0.08276, Q = 110,400, Q,, = 90,000, H = 100,000. M 100,000 = 8,333 lb 0.24 X (120 - 70) Mo = 0.08276 X 90,000 = 7,448 lb Mi = M -- M0 = 8,333 - 7,448 = 885 lb Temperature Loss in Ducts The allowances to be made for loss in transit through the duct system (Q are as follows: 1. When the duct system is located in the enclosure to which the air is being delivered, as in a factory, it may be assumed that there is no loss between the reheater coil and the point or points of discharge into the enclosure. 297.