Document rXJ6pwEgwQk840vMmYV2BBVG

T| i ; > 5 r \ i 3 i ;2 \r . H * i i 270 CHAPTER 12 1951, Guide .. Table 7. Values of the Wall Solab Azimuth, y, fob Vabiouslt Oriented Walls and Solab Altitude Computed for 18 Deg Declination, North {August 1) 30 Deo North Latitude Sun Time Solar Altitude 0 Degrees y,Azimuth Angle Degrees AM -a i N NE E SE S SW 6 a.m. 7 8 9 6 pjn. 5 43 9.0 21.5 34.5 47.5 ; 74 81 88 shade 29 36 43 51 16 61 9 54 2 47 shade 6 39 84 10 2 - 60.0 11 1 . 72.0- 12 78.0 62 17 28 73 83 38 7 52 shade shade 90 45 0 45 - N NW W SW S SE Sun Time Solar Altitude 0 Degress AM -* ta. i 5 ajn. 6 7 8 9 10 11 12 7 PAD. 6 5 4 3 2 1 0.5 11.5 23.0 34.5 . 45.5 56.0 . 64.5 68.0 40 Deo North Latitude . Azimuth Angle y. Degrees N 66 76 85 shade NE 21 31 40 50 61 76 shade E 24 14 5 5 16 31 55 90. . 'SE ' 69 59 50 40 29 14 10 45 S 85 74 59 35 0 SW shade 80 45 pll-> N - NW W SW S SE Sun Time AM -- A 5 a.m. 6 7 8 9 10 11 12 7 pjn. 6 5. 4. 3 2 1 Solar 0Altitude Degrees . 4.5 13.5 23.5 33.0 42.0 50.0 56.0 58.0 N 67 7890 shade N y,Azimuth Angle Degrees NE . 22 33 45 57 70 87 shade NW E, 23 12 0 . 12 25 42 64 90- W- SE 68 57 45 33 20 .3 19 45 SW 8 ' SW- 90 78- 65 48 26 0 B shade 71 45 8E Table 8. Approximate Solab Declinations in Degrees Date Declination ' Date ' Declination Date Declination ' April 1 . April 15 May 1 May 15 4.5 ' 10:0 15.0 19.0 June 1 June 15 July 1 July 15 22.0 Aug. 1 . . 23.5 ' . Aug. 15 23.0 Sept. 1 .'21.5- . Sept. 15 18.0 : 14.0 8.5. 3.0 Cooling Load 271 . 00 r 14 = 104 deg, and at 7KX) ,p.m. is 90 +24 = 114 deg. By interpolation, 0 for 630 p m. is 109 deg west of south (at 5:30 a.m. <j> would be 109 deg east of south.) Example S: Find K for a wall facing 18 deg east of south at 1030 a.m. 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, 0 is 50 deg. Then K =a cos 0 cos y = cos 50 X cos 30 = 0.643 X 0.866 = 0.557. Example S: Find K for the wall in Example at 3:00 p.m.' Solution. The solar azimuth is .65 deg west. The wall solar azimuth is therefore 55 -p 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 . Solution. Use clear atmosphere solar intensities. At 50 deg altitude, the direct normal radiation is 273 Btu. per (hr)(sq ft). Then, /,, = K X / = 0.557 X 273 = 152.0 Btu per (hr) (sq ft). By linear interpolation, the diffuse irradiation is 7d = 25 + W (33 - 25) = 26.6 Btu per (hr)(sq ft). The total solar irradiation is /, = 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-