Document mq7qVyoB5ONONXq7mLMJKZE5B

Heating Ventilating Air Conditioning Guide 1938 Solution: The normal heat transmission through various surfaces shown in load calculations are determined by application of Formula 1. It is quite obvious from the shape and exposure of this store that the maximum sun load will exist on the west wall: Since the west wall has a large glass area with a negligible time lag, the peak load may be expected at 4:00 p.m. at which time, from Table 4, Iq " 182. Iq for south glass at 4:00 p.m. is 2. Because of the small amount of solar, -radiation transmitted through the south glass, the transmission due to temperature difference has also been included.. Assuming time lag in roof and walls to be 2 hours, the corresponding values for I for south and west walls and roof will be those shown in Table 4 for 2:00 p.m. They are respectively 77, 143 and 258. A time lag of 1 hour was assumed for the west door amounting to I = 192. By substituting these values in equations 2 and 3 the solar heat load is determined. . 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 show.windows-are sealed so as not to permit infiltration and since there are only three doors in this store through which infiltration can take place, it is obvious that infiltration of air will be a negligible quan tity. 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 1^ air changes the volume of outside air to be introduced would be 32,400 cfh. By reference to Chapter 3 it will be noted that the minimum ventilation require ments are 10 cfm per person. On this basis the ventilation requirements would be 30,000 cfh. Since this will produce approximately 1^ outside air changes per hour, 30,000 cfh will be considered in this application. 'v To determine load imposed by occupants it will be found from Table 4, Chapter. 3 that the average person standing at rest will dissipate 225 Btu sensible heat and 206 Btu latent heat per hour. Normal Transmission Load: 1 VSurface Dimensions Area SQ FT Temp. Diff. Deo F Btu per Hour 1 S Glass 2(2 ft 6 in. x 7 ft) + 1' S Wall 2(10 ft x 6 ft) (30 ft x 12 ft) -155 155 1.13 205 0.33 15 2,627 15 1,015 1 W Wall (60 ft x 12 ft)-321 399 0.33 15 1,975 W Door 3 ft x 7 ft 21 0.51 15 161 Roof 60 ft x 30 ft 1800 0.26 15 7,020 Floor 26 ft x 54 ft 1404 0.34 5 2,387 - N Partition 30 ft x 12 ft 360 0.34 8 979. Total 16.164 Sun Load: Surface S Wall S Glass W-Glass W Door W Wall Roof Total. Dimensions 3(14 ft x 6 ft)+(8 ft x 6 ft) Area BQ FT F a . I OB Shade Btu per Factor Hour 205 0.078 0.7 ` 77 155 2 0.28 862 87 300 21 0.118 399 . 0.078, 1800 0.062 0.7 0.7 0.9 182 0.28 192 143 258 15,288 333 3,113 25,914 45,597 Outside Air Heat Gain: Sensible heat, HB = 0.24 X 60 do Q (to -- t) (Formula 4). Q = 50 X 10 = 500 cfm. Density of air at 95 F dry-bulb and 75 F wet-bulb for a barometric.pressure of 29.92 in. is 0.07089 lb per cubic foot (Table 4, Chapter 1). Dew-point of outdoor air is 66 F (psychrometricchart). 162 . Chapter 8. Cooling Load Partial pressure of vapor is 0.64378 in. Hg. (Pressure of saturated vapor at 66 F, Table 6, Chapter 1). . w = 0.622 (^) - 0-622 (29= 0.0137 lb water vapor per pound dry air (Formula 5 a; Chapter 1). --------------- -- = 0.986 lb dry air per pound outside air. 1 plus 0.0137 / . - . do = 0.07089 X 0.986 = 0.0699 lb dry air per cubic foot outside air. He = 60 X 500 X 0.0699 X 0.24 (95-80) =, 7549 Btu per hour. Total heat, H = 60 d0 Q iK -- h) (Formula 5). ho = 38.46 Btu per pound, dry air at 75 F wet-bulb (Table 6, Chapter 1). h = 31.51 Btu per pound dry air at 67, F wet-bulb (Table 6, Chapter 1).- H = 60 X 0.0699 X 500 (38.46 - 31.51) = 14,574 Btu per hour. Latent heat gain from outside air = 14,574 -- 7549 = 7025 Btu per hour. People Heat Gain: 50 X 225 = 11,250 Btu per hour, sensible heat. 50 X 206 = 10,300 Btu per hour, latent heat. Light Heat Gain: 4200 X 3.413 = 14,335 Btu per hour. Summary: ; */ Component of Load ' Btu per Hour Sensible- 16,164 45,597' * 7,549 11,250 14,335 94,895 ' Latent ` ` 7,025 . 10,300 17,325 Total Load: 94,895 + 17,325 = 112,220 Btu per hour. PROBLEMS IN PRACTICE 1 The outdoor and indoor temperatures are 90 F and 78 F, respectively. What is the amount of heat transmitted per hour through a 7 ft by 4 ft north window? Ht = 28 X 1.13 (90-78) = 380 Btu per hour. (Equation 1, Chapter 8 and Table 13 A, Chapter 5). ` ... 2 a. If a restaurant has two 10 gal gas-heated coffee urns, what is the cooling load due to them? b. What is the cooling load due to four 1350 w burners on an electric range? a. 2 X 10 X 1000 = 20,000 Btu per hour (Table 10). ft. 4 X 1350 = 5400 w = 5.4 kw. 5.4 X 3413 = 18,430 Btu per hour (Table 10). 3 a. What is the maximum heat transmission for a flat roof located in Pitts burgh (latitude 40 deg) exposed to the sun with the outdoor and indoor tern-