Document G6vjRQq6R6rE36gBOq3VORDkN
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CHAPTER 13
1954 Guide
Table 7. 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
of the Sun.9 Table 7 shows the variation of solar declination during the months ordinarily requiring cooling.
Cooling Load
Example 1: Find the solar azimuth <f> 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 <f> for 6:00 p.m. is 90 4* 14 = 104 deg, and at 7:00 p.m. is 90 + 24 = 114 deg. By interpolation, <f> for 6:30 p.m. is 109 deg west of south (at 5:30 a.m. 4> would be 109 deg east of south.)
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Example 2: Find K for a wall facing 18 deg east of south at 10:00 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 6). The wall solar azimuth is 48 -- 18 or 30 deg. From Table 6, & is 50deg. Then
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K -- cos 0 cos 7 -- cos 50 X cos 30 = 0.643 X 0.866 -- 0.557.
Example S: Find K for the wall in Example 2 at 3:00 p.m.
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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 S3 = 0.743 X 0.122 = 0.091.
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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,
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Id ~ K X loa -- 0.557 X 273 = 152.0 Btu per (hr) (sq ft). By linear interpolation, the diffuse irradiation is
7d = 25 4- il (33 - 25) = 26.6 Btu per (hr) (sq ft).
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The total solar irradiation is
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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 buildingi*
exposed to the weather, requires consideration of the diurnal cycles of sola||J
irradiation and air temperature. These cycles and other factors lead to
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
air conditioned space. Because of heat capacity and other factors, thesgffgl
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 ofi-vil
a building, it is necessary to know:
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1. The intensity of direct solar radiation striking the surface.
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2. The absorptivity (or reflectivity) of the surface for direct solar radiation.
3. The intensity of diffuse or sky solar radiation striking the surface.
____ _ v,* orvy ouiar raclia*>. The rate at which the surface emits radiation to the sky and other surroundings.
The 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. 9- The unit convective conductance for heat transfer between the air and the building surface.
he So,`Air Temperature
can be considerably concept. The sol-
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