Document jNEO8VyowN9Z5RoRXQkkDv9B9

American Society of Heating and Ventilating Engineers Guide, 1937 sary for the ventilating requirements. The heat required to raise this air to the conditions maintained in the room must be provided by the tem pering coils, preheater coils, and reheater coils. If a positive pressure is not maintained in the room or space to be conditioned, the normal in. filtration 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 sufficient to provide for both 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 3. In no case shall less than 10 cfm per person be introduced. The heat required to warm the outside air introduced for ventilation purposes (Ha) may be determined by means of the following formula; where H0 = 0.24 (t - to) Mo (l) 0.24 = specific heat of air at constant pressure. t = room temperature, degrees Fahrenheit. to = outside temperature, degrees Fahrenheit. Mo = weight of outside air to be introduced per hour, in pounds = 60 doQoQo = volume of outside air to be introduced, cubic feet per minute. do = density of air at to, pounds per cubic foot. Example 1. A building in which the temperature to be maintained sit 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. 10,000 X 60 = 600,000 cfh; do = 0.08273 (Table 1, Chapter 1); M0 = 0.08273 X 600,000 = 49,656 lb; t = 70 F; to = 20 F; H0 = 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 system, 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: 7h H OOdQX 0.24 +. (2) where s ty = outlet temperature, degrees Fahrenheit. H = heat loss of room or space to conditioned, Btu per hour. Q = total volume of air to be introduced at the temperature t, cubic feet per minute. d = density of air, pounds per cubic foot. If the outlet temperature (ty) as determined from Equation 2 exceeds 120 F for public buildings, or 140 F for factories or industrial buildings, 174 T^^Central Systems for Heating and Humidifying these respective outlet temperatures should be used as factors in the following equation to'determine the volume of air to be introduced into the room or space 0 =6_0__d__X 0.g24 (ty - 1) (3) Example S. The heat loss of a certain auditorium to be conditioned is 100,000 Btu per hour. The ventilating requirements are 1,500 cfm and the room temperature 70 F. Determine the outlet temperature. Solution. Substituting in Formula 2, . 100,000 ___________ + 70 = 131.7 F hr = 60 X 0.07492 X 1500 X 0.24 , ,, this temperature is excessive, it will be necessary to assume an outlet Inasmuch as t P tk 12o p and to calculate the amount of air to be I'SSfthl'r^m^ thi^temperature to provide for the heat loss. Subst.tuting in Equation 3, ,, ________ 100,000_________ -- Q - 60 'x 0.07492 X 0.24 (12U - vu) 1850 cfm (at temperature t) Weight of Air to be Circulated The total weight of air (M) to be introduced into the room or space to be heated or conditioned is given by the following formulae; MTtl ___0.:2__4(-fy-------0- = 60 dQ M = Mo + Mt Mo -- 60 doQo (4) (6) (6) where d = density of air at temperature t, pounds per cubic foot, do = density of air at temperature fo, pounds per-cubic foot. Qo = volume of outside air at temperature to, cubic feet per minute. Mo = weight of outside air, pounds per hour. Mi = weight of recirculated air, pounds per hour. Example S. Using the data of Example 2 and an outside temperature of 20 F, what will be the values of M, Mo and Afr? Solution, d = 0.07492; do = 0.08273; Q = 1850; Qo = 1500; H = 100,000. M 100,000 = 0.24 X (120 - 70) = 8,333 lb Mo = 0.08273 X 60 X 1500 7,448 lb Mr = M - Mo = 8,333 - 7,448 = 885 lb Temperature Loss in Ducts _ The allowances (4) to be made for temperature drop through the duct system 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 fee assumed that there is no loss between the reheater coil and the point or points of discharge into the enclosure. 175