Document dDRyM1dg97ojQ19pEVwrYZyzR

300 CHAPTER 13 1958 Guide Table 6. Values of the Wall Solab Azimuth, y, fob Vabiously Obiented Walls and Solab Altitude Computed for 18 Deg Declination, North (August 1) L atitude Sun Time Solab Altitude 0 Degrees y.Azimuth Angle Degrees AM-- 'i N NE E SE S SW 6 a.m. 6 p.m. s7 o8 5 4 9 3 w P 10 11 12 2 1 9.0 21.5 34.5 47.5 60.0 72.0 78.0 74 81 88 shade 29 36 43 51 62 83 shade 16 9 2 6 17 38 90 61 54 47 shade 39 84 28 73 7 52 shade 45 0 45 5 a.m. 7 p.m. e6 6 a7 5 8 4 o9 3 p 10 2 11 12 0.5 11.5 23.0 34.5 66 76 85 shade 21 31 40 50 24 s 69 14 59 5 50 shade 5 40 85 45.5 56.0 64.5 68.0 61 76 shade 16 31 55 90 29 ... 14 10 45 74 59 shade 35 80 0 45 5 a.m. 7 p.m. Eas 6 '7 6 5 z8 4 o9 3 p 10 2 11 1 12 4.5 13.5 23.5 33.0 42.0 50.0 56.0 58.0 67 78 90 shade 22 33 45 57 70 87 shade 23 12 0 12 25 42 64 90 68 57 45 90 33 78 20 65 3 48 shade 19 26 71 45 0 45 T PM - N NW W SW s SE Values of K for other seasons and latitudes may be found in the litera ture,7 or may be computed from data given in Hydrographic Office Bulletin No. 214, Tables of Computed Altitude and Azimuth8 and the Ephemeris of the Sun.9 Table 7 shows the variation of solar declination during the months ordinarily requiring cooling. Example 1: Find the solar azimuth $ at 6:30 p.m. at 40 deg north latitude on August 1st. `Solution: From Table 6 in the column of y for a wall facing west <t> for 6:00 p.m. is 90 + 14 = 104 deg, and at 7:00 p:m. is 90 +.24 = 114 deg. By interpolation, for 6:30 p.m. is 109 deg west of south (at 6:30 a.m. <f> would be 109 deg east of south.) Example 2: Find K for a wall facing 18 deg east of south at 10:00 a.m. on August 1 at SO deg north latitude. Solution: The wall azimuth is 18 deg. The solar azimuth is 48 deg east (Table 6). The wall solar azimuth is 48 -- 18 or 30 deg. From Table 6, 8 is 50 deg. Then K = cos p cos y = cos 50 X cos 30 = 0.643 X 0.866 = 0.557. Table 7. Appboximate Solab Declinations in Degbees 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 301 Example 8: Find K for 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 p 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, In = K X /on = 0.557 X 273 = 152.0 Btu per (hr) (sq ft). By linear interpolation, the diffuse irradiation is Id = 25 + if (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- 5. The rate at which the surface emits radiation to the sky and other surroundings (P1 rate at which the surface absorbs the low temperature radiation emitted by the sky and other surroundings by virtue of their temperatures and radiat ing characteristics. 7. The temperature of the surrounding air. 8- The temperature of the outer building surface. The unit convective conductance for heat transfer between the air and the building surface. The Sol-Air Temperature 'nte'relationship of the above factors can be considerably air ? Uled *^rou8^1 the use of the sol-air temperature concept. The sol- absenmPTaiilre ^ *S temperature of the outdoor air, which, in the into th6 01 r radiation exchanges, would give the same rate of heat entry radiati6 88 W0U1<1 exist with the actual combination of incident solar roundimJ' ra jnt energy exchange with the sky and other outdoor surgs, and convective heat exchange with the outdoor air. for an d^t*ie?1Perature data6' *10 as developed by Mackey and Wright showine atmosphere were used as a basis for preparing Table 8 also be <^es`Sn sol-air temperatures. Sol-air temperatures may estimated from experimental observation of surface temperatures