Document JvL38veNgK8oNQVKaB5MkqZv

[ 270 i C~~HA, PTER 12 1952 Guide Table 7. "Values of the Wall Solar Azimuth, y, fob Variously Oriented Walls and Solar Altitude Computed for 18 Deg Declination; North (August 1) Sun Time Solar ~ Altitude 0 Degrees AM-* 1 6 a.m. 7 8 9 0 p.m. 5. 4 3 .10 2 11 1 12 9.0 21.5 34.5 47.5 60.0 72.0 78.0 T PM -* 30 Deo Nortel Latitude Azimuth Angle 7, Degrees N. 74 81 88 shade N NE 29 36 43' 51 62 83 shade NW E. 16 9 2 6, ' 17 38 90 W SE 61 ` 54 47 39 28 7 .45 - SW s shade 84 73 52 0 S ' SW shade 45 SE Sun Time AM ~* i 5 a.m. 6 7 8 7 p.m. 6 5 4 '9 3 10 2 11 1 12 Solar Altitude 0 Degrees 40 Deo North Latitude Azimuth Angle 7, Degrees N NE E SE s. 0.5 11.5 23.0 34.5 45.5 56.0 64.5 68.0 66 76 85 shade 21 31 40 50 61 76 shade 24 14 5 5 16 31 55 90 69 69 50 shade 40 85 29 74 14 59 10 35 45 0 SW shade 80 45 1 PM -+ N ` NW W SW S SE Sun Time AM --* 1 5 a.m. 6 7 8 7 p.m. 6 5 4 93 10 2 .11 12 t PM -- Solar Altitude 0 Degrees 50 Deo North Latitude Azimuth Angle 7, Degrees N NE E SE S 4.5 13.5 . 23.5 33.0 42.0 50.0 56.0 58.0 67 78 90 shade N 22 33 45 57 70 87 shade NW 23 68 12 . 57 0 45 12 33 90 78 - 25 20 ' 65 42 3 48 64 19 26 90 45 0 W SW ' S _/' SW shade 71 45 SE . Table 8. Approximate Solar Declinations in Degrees Date April 1 April 15 May 1 May 15 Declination 4.5 10.0 15.0 19.0 Date June 1 June 15 July 1 July 15 Declination 22.0 23.5 23.0 21.5 Date Aug. 1 Aug. 15 . Sept. 1 Sept. 15 Declination 18.0 14.0 8.5 3.0 - Cooling Load 271 is 90 + 14 = 104 deg, and at 7:00 p.m. is 90 + 24 = 114 deg. . By interpolation, <t> for 6:30 p.m. is 109 deg west of south (at 5:30 a.m. 4> would be 109 deg east of south.) Example 2: Find K for a wall facing 18 deg east of south at 10:00 a.ni. on August 1 at 50 deg north .latitude. Solution: The wall azimuth is 18 deg. The solar azimuth is 48 deg east (Table 7). The wall solar azimuth is 48 -- 18 or 30 deg. From Table 7, f} is 50 deg. Then ' K = cos 0 cos y = cos 50 X cos 30 = 0.643 X 0.866 = 0.557. Example 3: Find K!or the wall in Example 2 at 3:00 p.m. Solution: The solar azimuth is 65 deg west. The wall solar azimuth is therefore 65 + 18 = 83 deg. The angle 0 is 42 deg. K = cos 42 X cos 83 = 0.743 X 0.122 = 0.091. Example 4: Find the total solar irradiation for the wall for the conditions of Example 2. Solution: Use clear atmosphere solar intensities. ' At 50 deg altitude, the direct normal radiation is 273 Btu per (hr) (sq ft). Then, , Id = K X Id. = 0.557 X 273 = 152.0 Btu per (hr) (sq ft). By linear interpolation, the diffuse irradiation is /d = 25 + M (33 - 25) = 26.6 Btu per (hr) (sq ft). The total solar irradiation is It = 152.0 + 26.6 = 178.6 Btu per (hr)(sq ft). PERIODIC HEAT FLOW THROUGH WALLS AND ROOFS ' The calculation of heat flow, through a structural section of a building exposed to the weather, requires consideration of the diurnal cycles of solar irradiation and air temperature. These cycles and other factors lead to a periodic variation in the instantaneous rate of heat flow into the weather surface, and a related periodic variation in the rate of heat flow into the air conditioned space. Because of heat capacity and other factors, these heat flow cycles are, in general, out of time phase and unequal in amplitude. In order to calculate the rate of heat entry into the weather surface of a building, it is necessary to know: 1. The intensity of direct solar radiation striking the surface. 2. The absorptivity (or reflectivity) of the surface for direct solar radiation. 3. The intensity of diffuse or sky solar radiation striking the surface. 4. The absorptivity (or reflectivity) of the surface for diffuse or sky solar radia tion. 5. The rate at which the surface emits radiation to the sky and other surround ings. 6. The. rate at which the surface absorbs the low temperature radiation emitted by the sky and other surroundings by virtue of their temperatures and radiating characteristics. 7. The temperature of the surrounding air. 8. The temperature of the outer building surface. 9. The unit convective conductance for heat transfer between the air and the building surface. The Sol-Air Temperature _ The complex interrelationship of the above factors can be considerably simplified through the use of the sol-air temperature concept. The sol- -3ft