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152 Chapter 7 1945 Guide ' Basement temperature, 85F.----- :------------- ----------------------- ^------.------------------- Store room temperature, 88 F. Solution. It is obvious from the shape and exposure of this store and the large glass area on the west side that the maximum cooling load will occur during the afternoon when the sun is shining on the west wall. From Fig. 1, the peak load may be expected at 4:00 p.m. The combined normal transmission and solar radiation transmission through the roof at.4:00 p.' m. is obtained from Fig. 2. While none of the roofs in Fig. 2 is exactly like this one, roof C is similar. A heat flow of 11 Btu per square foot per hour was assumed, slightly less than for roof C. The combined normal transmission and solar radiation Combined Normal and Solar Radiation Transmission: Surface S Wall W Wall Roof Total Dimensions (30 ft x 12 ft) - 155 (60 ft x 12 ft) -- 321 60 ft x 30 ft Area Sq Ft 205 399 1800 Btu per Hour per Sq Ft 3 2.5 11 Btu per Hour 615 998 19,800 21,413 Normal Transmission: Surface S .Glass Floor N Partition Total Dimensions 2 (2 ft 6 in. x 7 ft) + 2 (10 ft x 6 ft) 26 ft x 54 ft 30 ft x 12 ft Area Sq Ft 155 1404 360 u 1.13 0.34 0.34 Temp. Diff. Deg F Btu per Hour . 15 2,627; 5 2,387: 8 979 5,993 transmission through the south and west walls at 4:00 p.m. is obtained from curves TS and TW in Fig. 3. The normal heat transmission through the south glass, floor and partition is deter mined by application of Formula 1. Solar radiation transmission through the south glass can be neglected. The solar intensity / for the south side at 4:00 p.m. is 29. Apply ing a shade factor of 0.28, the calculated solar radiation transmission is 29 X 0.28 = 8 Btu per square foot per hour which is less than the normal transmission; therefore the total heat gain can be taken as that due to normal transmission. Solar radiation intensity on the west glass at 4:00 p.m. from Table 4 is 211 Btu per square foot per hour. As explained in the text, normal transmission can be neglected because it is small in comparison with solar radiation transmission. To determine the heat gain from the outside air it is necessary first to determine the volume of the outside air to be introduced. Since the windows are sealed so as not to permit infiltration and since there are only three doors in this store through which in filtration can take place, infiltration is neglected in solving this example. The dew-point within the store is, however, affected by the outside air admitted when the doors are in use. Assuming a constant apparatus dew-point, the infiltration will affect the dew-point within the space, raising it when outside air is above the line drawn on the psychrometric chart between the apparatus dew-point and the room conditions and lowering it when it is below. The volume of the store is 21,600 cu ft. Good practice indicates that in a .store of this character there should be a minimum of from 1 to 1outside air changes per hour. On a basis of 1H air changes the volume of outside air to be introduced would be 32,400 cu ft per hour. The minimum ventilation requirements as given in the Code of Minimum Requirements for Comfort Air Conditioning14 are 10 cfm per 14Code of Minimum Requirements for Comfort Air Conditioning (A.S.H.V.E. Transactions, Vol. 44. 1938. p. 27). Cooling Load 153 person. On-this basis the_ventilation requirements would be 30,000 cu ft per hour. Since this will produce approximately l'A outside air changes per hour, 30,000 cu ft per hour will be considered in this application. To determine load imposed by occupants it will be found from Chapter 2 that the average person standing at rest will dissipate 431 Btu per hour and that the moisture dissipated in 0.198 lb per hour. To determine the latent heat load, the sum of the moisture evaporated from occupants and that to be removed from outside air is multiplied by the latent heat of evaporation at the temperature at which the moisture is condensed in the conditioner. Since outside air is positively introduced, a mixture of outside and recirculated air passes through the conditioner. To remove the moisture, the air must be cooled to a temperature below the dew-point of the mixture. To obtain an approximate value of the latent heat of evapora tion, assume that the air is cooled to 55 F. At this temperature, Afg = 1062.7 Btu. per hour (steam table). Solar Radiation Through Glass: W Glass. AG = 3 (14 ft X 6 ft) + (8 ft X 6 ft) + (3 ft X 7 ft) = 321 sq ft. Hc = 321 X 0.28 X 211 = 18,965 Btu per hour (Equation 2). Outside Air: -H= (ho -- hi) (Equation 3). r = Fa + uFas (Equation 17, Chapter 1). pa = specific volume of dry air at 95 F = 13.97 cu ft per pound (Table 6, Chapter 1). Fas = difference between volume of saturated mixture and specific volume of dry air at 95 F = 0.82 cu ft per pound (Table 6, Chapter 1). ti = per cent saturation at 95 F dry-bulb and 75 F wet-bulb = 38.4 per cent (by calculation, Chapter 1). f -- 13.97 + (0.384 X 0.82) = 14.28 cu ft per pound dry air. ho = Aa + pAas (Equation 19, Chapter 1). Aa = specific enthalpy of dry air at 95 F = 22.80 Btu per pound (Table 6, Chapter 1). Aas = difference between enthalpy of saturated mixture and specific enthalpy of dry air at 95 F = 40.25 Btu per pound (Table 6, Chapter 1). ho = 22.80 + (0.384 X 40.25) = 38.26 Btu, per pound dry air. (i at 80 F dry-bulb and 67 F wet-bulb = 50.2 percent (by calculation, Chapter 1). Ai = Aa + nAas = 19.19 + (0.502 X 24.32) = 31.40 Btu per pound dry air (Table 6, Chapter 1). 'H= (38.26 - 31.40) = 14,410 Btu per hour. W0 = humidity ratio of outside air at 95 F and 75 F = 0.334 X 0.03652 = 0.01402 lb water per pound dry air. (Equation 14, Chapter 1). W\ = humidity ratio of inside air at 80 F and 67 F = 0.502 X 0.02221 = 0.01115 lb water per pound dry air.* (Equation 14, Chapter 1). Weight of water to be removed = - Wi) = 6.03 lb per hour. Occupants: 50 X 431 = 21,550 Btu per hour. 50 X 0.198 = 9.90 lb water per hour evaporated. (0.01402 - 0.01115) =