2nhOI

where r = ratio of hourly total radiation to daily total radiation.

a = G.4G9 + G.5G16sin(hss ß = G.66G9 - G.4767 sin(hss

Example 2.15

Given the following empirical equation,

1.39G

where HD is the monthly average daily diffuse radiation on horizontal surface—see Eq. (2.105a)—estimate the average total radiation and the average diffuse radiation between 11:00 am and 12:00 pm solar time in the month of July on a horizontal surface located at 35°N latitude. The monthly average daily total radiation on a horizontal surface, H, in July at the surface location is 23.14 MJ/m2-d.

Solution

From Table 2.5 at 35° N latitude for July, Ho

40.6 MJ/m2. Therefore,

H 23.14

4G.6

Therefore, H

G.57G

^ = 1.39G - 4.G27(G.57) + 5.531(G.57)2 - 3.1G8(G.57)3 = G.316 H

From Table 2.5, the recommended average day for the month is July 17 (N = 199). The solar declination is calculated from Eq. (2.5) as

23.45sin

The sunset hour angle is calculated from Eq. (2.15) as cos(hss) = -tan(L)tan(6) ^ hss = cos_1[-tan(35)tan(21)] = 106°

The middle point of the hour from 11:00 am to 12:00 pm is 0.5 h from solar noon, or hour angle is -7.5°. Therefore, from Eqs. (2.84b), (2.84c), and (2.84a), we have a = 0.409 + 0.5016 sin(hss - 60) = 0.409 + 0.5016sin(106 - 60)

(3 = 0.6609 - 0.4767sin(hss - 60) = 0.6609 - 0.4767sin(106 - 60)

,, xx cos(h) - cos(hss) r = — (a + ß cos(h))--ss

sin(106)

2n(106)

cos(106)

n 24

sin(hss)

cos(hss)

sin(hss)

cos(hss)

cos(-7.5) - cos(106) | ||

sin(106) - |
2n(106) 360 |
0.113 Finally, Average hourly total radiation = 0.123(23.14) Average hourly diffuse radiation = 0.113(7.31) 2.85MJ/m2 or 2850kJ/m2 0.826 MJ/m2 or 826kJ/m2 2.3.8 Total Radiation on Tilted Surfaces Usually, collectors are not installed horizontally but at an angle to increase the amount of radiation intercepted and reduce reflection and cosine losses. Therefore, system designers need data about solar radiation on such titled surfaces; measured or estimated radiation data, however, are mostly available either for normal incidence or for horizontal surfaces. Therefore, there is a need to convert these data to radiation on tilted surfaces. The amount of insolation on a terrestrial surface at a given location for a given time depends on the orientation and slope of the surface. A flat surface absorbs beam (GBt), diffuse (GDt), and ground-reflected (GGt) solar radiation; that is, As shown in Figure 2.28, the beam radiation on a tilted surface is and on a horizontal surface, where GBt = beam radiation on a tilted surface (W/m2). Gb = beam radiation on a horizontal surface (W/m2). where RB is called the beam radiation tilt factor. The term cos(9) can be calculated from Eq. (2.86) and cos($) from Eq. (2.87). So the beam radiation component for any surface is In Eq. (2.88), the zenith angle can be calculated from Eq. (2.12) and the incident angle 9 can be calculated from Eq. (2.18) or, for the specific case of a south-facing fixed surface, from Eq. (2.20). Therefore, for a fixed surface facing south with tilt angle (3, Eq. (2.88) becomes _ cos(9) _ sin(L - |3)sin(6) + cos(L - |3)cos(6)cos(h) The values of RB for collector slopes equal to latitude and latitude +10°, which is the usual collector inclination for solar water-heating collectors, are shown in Figures A3.6 and A3.7 in Appendix 3. Equation (2.88) also can be applied to other than fixed surfaces, in which case the appropriate equation for cos(0), as given in Section 2.2.1, can be used. For example, for a surface rotated continuously about a horizontal east-west axis, from Eq. (2.26a), the ratio of beam radiation on the surface to that on a horizontal surface at any time is given by example 2.16 Estimate the beam radiation tilt factor for a surface located at 35°N latitude and tilted 45° at 2:00 pm solar time on March 10. If the beam radiation at normal incidence is 900 W/m2, estimate the beam radiation on the tilted surface. Solution From Example 2.14, 6 = -4.8° and h calculated from Eq. (2.90a) as 30°. The beam radiation tilt factor is sin(L)sin(6) + cos(L)cos(8)cos(h) sin(35 - 45)sin(-4.8) + cos(35 - 45)cos(-4.8)cos(30) sin(35)sin(-4.8) + cos(35)cos(-4.8)cos(30) 1.312 Therefore, the beam radiation on the tilted surface is calculated from Eq. (2.89) as Many models give the solar radiation on a tilted surface. The first one is the isotropic sky model developed originally by Hottel and Woertz (1942) and refined by Liu and Jordan (1960). According to this model, radiation is calculated as follows. Diffuse radiation on a horizontal surface, n/2 where GR = diffuse sky radiance (W/m2-rad). Diffuse radiation on a tilted surface, n/2-ß where (3 is the surface tilt angle as shown in Figure 2.28. From Eq. (2.91), the second term of Eq. (2.92) becomes GR = GD/2. Therefore, Eq. (2.92) becomes Similarly, the ground-reflected radiation is obtained by pG(GB + GD), where pG is ground albedo. Therefore, GGt is obtained as follows. Ground-reflected radiation, n/2 where Gr is the isotropic ground-reflected radiance (W/m2-rad). Ground-reflected radiation on tilted surfaces, n/2 Therefore, inserting Eqs. (2.93) and (2.96) into Eq. (2.85), we get gt - rbgb + gd The total radiation on a horizontal surface, G, is the sum of horizontal beam and diffuse radiation; that is, g - gb + gd Therefore, Eq. (2.97) can also be written as GB „ . GD [1 + cos(P) where R is called the total radiation tilt factor. |

Was this article helpful?

Do we really want the one thing that gives us its resources unconditionally to suffer even more than it is suffering now? Nature, is a part of our being from the earliest human days. We respect Nature and it gives us its bounty, but in the recent past greedy money hungry corporations have made us all so destructive, so wasteful.

## Post a comment