Document ZJ64bkkbNb7XrjbnNLJg5Gy0d

American Society of Heating ond Ventilating Engineers Guide, 1934 south street and the entrance windows and the first windows on the west street are 9 ft high. The overall heat transmission factors, V, for the different walls of the restaurant are determined by the methods outlined in Chapter 5 and are found to be the following: Exterior walls, 0.221; interior wall, 0.296; floor, 0.433; and ceiling, 0.299. Determine the dehumidification, cooling and reheating loads, considering the outside dry-bulb temperature to be 95 F and the outside design wet-bulb temperature to be 75 F. Solution. According to Table 2, Chapter 2, the inside dry-bulb temperature should be 80 F and the wet-bulb temperature should be 65 F. The heat gain to be considered may be calculated as follows; South Wall BUILDING Total area = 27 X 15 = 405 sq ft Glass (including entrance windows) 2 (9 X 9) + 2 (5 X 9) + (door) (6 X 9)= 306 sq ft Net wall area = 405 -- 306 = 99 sq ft Transmission factor for wall = 0.221 Transmission factor for glass = 1.13 Temperature difference = 15 + 25 (for sun effect) = 40 deg 99 X 0.221 X 40 = 875 Btu per hour 306 X 1.13 X 40 = 13831 Btu per hour Heat Gain, Blu per Hour 14706 West Wall Total area = 100 X 15 = 1500 sq ft Glass 15 X (2 X 5) + (9 X 5) = 195 sq ft Net wall area = 1500 -- 195 = 1305 sq ft 1305 X 0.221 X 40 = 11536 Btu per hour 195 X 1.13 X 40 = 8814 Btu per hour 20350 North Wall (no sun effect) Total wall area = 27 X 15 = 405 sq ft Glass 2 X (2 X 5) = 20 sq ft Net wall area = 385 sq ft 385 X 0.221 X 15 = 1276 Btu per hour 20 X 1.13 X 15 = 339 Btu per hour 1615 East Wall (interior, no sun effect) Total wall = 100 X 15 = 1500 sq ft 1500 X 0.296 X 15 = 6660 Fig. 1. Floor Plan of Restaurant Chapter 9--Central Fan Air Conditioning Systems Fig. 2. Diagram of Dehumidifier Method Floor 100 X 27 = 2700 sq ft 2700 X 0.433 X 15 = 17537 Ceiling 100 X 27 = 2700 sq ft 2700 X 0.299 X 15 = 12110 INFILTRATION The windows are sealed in metal frames set in concrete and the infiltration may be considered negligible, particularly as the prevailing wind is at a minimum when the temperature is high. The revolving entrance door, however, does present an infiltration proTbhleemd.oor is 8 ft high and 6 ft wide and during rush hours can be assumed to revolve approximately one turn per person entering. This figure is purely an assumption but the percentage error would not be more than 2 or 3 per cent of the total heat load. The volume of the door = --4j--L X 8 = 226 cu ft 226 X 300 W__i_th__3_00 .persons per hour entering this would be equivalent to 60 = 1130 cfm. A volume of 1130 cfm of air cooled from a wet-bulb temperature of 75 F to a wet-bulb temperature of 65 F would be equivalent to the removal of heat as follows; Total heat at 75 F wet-bulb = 37.81 Btu per hour. (See Table 2, Chapter 1). Total heat at 65 F wet-bulb = 29.62 Btu per hour 8.19 Btu per hour-to be removed II^0 cfm X 8.19 X 60 = 38,831 Btu perhaur 14.3 ' -^ Half of this, however, would be provided for by the outside air taken into the cooling unit so the heat gain to be considered under infiltration would be 19,415 Btu per,hour. Lights 4200 X 3.415 = 14,343 Btu per hour PEOPLE Fig. 4, Chapter 2, shows the heat loss from the human body at different effective temperatures. For an effective temperature of 73.4 deg for persons at rest the heat loss is 400 Btu per hour per person of which 225 Btu per hour is sensible and 175 Btu per hour is latent heat. As there are 300 occupants and 20 employees the heat gain wouldjbe (300 + 20) X 225 = 72,000 Btu per hour of sensible heat (300 + 20) X 175 = 56,000 Btu per hour of latent heat Fig. 6, Chapter 2, shows the moisture loss from human bodies and at 80 F to be about 1200 grains per hour per person at rest. The total moisture gain would then be (300 + 20) X 1200 = 384,000 grains per hour. 125