SOLAR CALCULATIONS (2)



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OLAR CALCULATON The orbit of the Earth is an ellise not a circle, hence the distance between the Earth and un varies over the year, leading to aarent solar irradiation values throughout the year aroximated by [1]: N = C 1 +.33cos (1) 365 the solar constant, C = 49.5 Btu/hr ft² (1353 W/m²). The Earth s closest oint (about 146 million km) to the sun is called the erihelion and occurs around January 3; the Earth s farthest oint (about 156 million km) to the sun is called the ahelion and occurs around July 4. The Earth is tilted on its axis at an angle of 3.45. As the Earth annually travels around the sun, the tilting manifests itself as our seasons of the year. The sun crosses the equator around March 1 (vernal equinox) and etember 1 (autumnal equinox). The sun reaches its northernmost latitude about June 1 (summer solstice) and its southernmost latitude near December 1 (winter solstice). The declination is the angular distance of the sun north or south of the earth's equator. The declination angle, δ, for the Northern Hemishere (reverse the declination angle's sign for the outhern Hemishere) is [] N + 84 δ = 3.45 sin 365 () N is the day number of the year, with January 1 equal to 1. The Earth is divided into latitudes (horizontal divisions) and longitudes (N- divisions). The equator is at a latitude of ; the north and south oles are at +9 and 9, resectively; the Troic of Cancer and Troic of Caricorn are located at +3.45 and 3.45, resectively. For longitudes, the global community has defined as the rime meridian which is located at Greenwich, England. The longitudes are described in terms of how many degrees they lie to the east or west of the rime meridian. A 4-hr day has 144 mins, which when divided by 36, means that it takes 4 mins to move each degree of longitude. The aarent solar time, AT (or local solar time) in the western longitudes is calculated from AT = LT + ( 4 min/ deg)( LTM Long) + ET (3) LT = Local standard time or clock time for that time zone (may need to adjust for daylight savings time, DT, that is LT = DT 1 hr), Long = local longitude at the osition of interest, and LTM = local longitude of standard time meridian Long LTM = 15 (4) 15 round to integer olarcalcs.doc 7, K. E. Holbert Page 1

The difference between the true solar time and the mean solar time changes continuously day-today with an annual cycle. This quantity is known as the equation of time. The equation of time, ET in minutes, is aroximated by [3] ET = 9.87 sin( D) 7.53 cos( D) 1.5 sin( D) D = 36 ( N 81) 365 Examle 1: Find the AT for 8: a.m. MT on July 1 in Phoenix, AZ, which is located at 11 W longitude and a northern latitude of 33.43. olution: ince Phoenix does not observe daylight savings time, it is unnecessary to make any change to the local clock time. Using Table, July 1 is the nd day of the year. From Equation (5), the equation of time is ( N 81) ( 81) D = 36 = 36 = 119.3 365 365 ET = 9.87 sin( D) 7.53 cos( D) 1.5 sin( D) = 9.87 sin( 119.3 ) 7.53 cos(119.3 ) 1.5 sin(119.3 ) = 6.5 min From Equation (4), the local standard meridian for Phoenix is Long 11 LTM = 15 = 15 = 15 7 = 15 15 round to integer 15 round to integer Using Equation (3), the aarent solar time (AT) is AT = LT + (4 mins)( LTM Long) + ET = 8 : + (4 mins)(15 11 ) + ( 6.5 min) = 7 : 6 a.m. (5) The hour angle, H, is the azimuth angle of the sun's rays caused by the earth's rotation, and H can be comuted from [4] ( No. of minutes ast midnight, AT) 7 mins H = (6) 4 min / deg The hour angle as defined here is negative in the morning and ositive in the afternoon (H = at noon). The solar altitude angle ( 1 ) is the aarent angular height of the sun in the sky if you are facing it. The zenith angle (θ z ) and its comlement the altitude angle ( 1 ) are given by cos( θ ) = sin( 1) = cos( L)cos( δ )cos( H ) sin( L)sin( δ ) (7) z + L = latitude (ositive in either hemishere) [ to +9 ], δ = declination angle (negative for outhern Hemishere) [ 3.45 to +3.45 ], and H = hour angle. olarcalcs.doc 7, K. E. Holbert Page

The noon altitude is N = 9 L + δ. The sun rises and sets when its altitude is, not necessarily when its hour angle is ±9. The hour angle at sunset or sunrise, H, can be found from using Eq. (7) when 1 = cos( ) = tan( L) tan( δ ) (8) H H is negative for sunrise and ositive for sunset. Absolute values of cos(h ) greater than unity occur in the arctic zones when the sun neither rises nor sets. W Zenith N Altitude Azimuth E The solar azimuth, α 1, is the angle away from south (north in the outhern Hemishere). sin( 1)sin( L) sin( δ ) cos( α1) = (9) cos( ) cos( L) α 1 is ositive toward the west (afternoon), and negative toward the east (morning), and therefore, the sign of α 1 should match that of the hour angle. f δ >, the sun can be north of the east-west line. The time at which the sun is due east and west can be determined from min tan( δ t E / W = 4 18 m arccos (1) deg tan( L) these times are given in minutes from midnight AT. Examle : For the conditions of Examle 1, determine the solar altitude and azimuth angles. olution: The hour angle is first comuted using Equation (6) ( No. of minutes ast midnight, AT ) 7 mins [6*7 + 6] 7 min H = = = 68. 5 4 min / deg 4 min / deg The declination angle is found from Equation () N + 84 + 84 δ = 3.45 sin = 3.45 sin =. 44 365 365 The altitude angle ( 1 ) of the sun is calculated via Equation (7) sin( ) = cos( L)cos( δ )cos( H ) + sin( L)sin( δ ) 1 1 [ cos(33.43 )cos(.44 )cos( 68.5 ) + sin(33.43 )sin(.44 )] = 8.6 1 = arcsin The solar azimuth angle (α 1 ) is found from Equation (9) olarcalcs.doc 7, K. E. Holbert Page 3

sin( 1)sin( L) sin( δ α = arccos cos( 1) cos( L) [ sgn( )] 1 H sin(8.6 )sin(33.43 ) sin(.44 = arccos sgn( 68.5 ) = 96.69 cos(8.6 ) cos(33.43 ) This value indicates that the sun is north of the east-west line. The time at which the sun is directly east can be comuted using Equation (1) tan( δ tan(.44 t E = 4 min 18 arccos = 4 min 18 arccos = 8 :17.5 a.m. tan( ) tan(33.43 ) L The AT is earlier than t E, thus verifying that the sun is still north the east-west line. Normal to Earth s surface Normal to inclined surface θ N nclined surface facing southwesterly E 1 W α α 1 Horizontal rojection of normal to inclined surface Figure 1. olar angles [4] Horizontal rojection of sun s rays olarcalcs.doc 7, K. E. Holbert Page 4

The collector angle (θ) between the sun and normal to the surface is cos( θ ) = sin( 1)cos( ) + cos( 1)sin( )cos( α1 α ) (11) α is the azimuth angle normal to the collector surface, and is the tilt angle from the ground. f θ is greater than 9, then the sun is behind the collector. ome collector anel angles of interest are given below. Azimuth, α Tilt, θ Orientation n/a 9 1 Horizontal (flat) 9 varies Vertical wall 9 varies outhfacing Vertical 9 9 varies East facing wall +9 9 varies West facing wall α 1 9 1 Tracking ystem The sunrise and sunset hours on the collector are different when the collector is shadowed by itself. The collector anel sunrise/sunset hours may be comuted from [1] a b ± a b + 1 H R = m min H,arccos a + 1 (1) cos( L) sin( L) a = + sin( α ) tan( ) tan( α ) cos( L) sin( L) b = tan( δ ) tan( α ) sin( α ) tan( ) (13) Examle 3: For the conditions of Examle 1, find the collector angle for a wall that faces eastsoutheast and is tilted at an angle equal to the location latitude. olution: The latitude is 33.43 which is also the tilt angle ( ). Referring to Figure 1, the collector azimuth angle (α ) for an east-southeast direction is 9 3 4 = 67. 5. Finally, the collector angle is comuted from Equation (11) = arccos [ sin( 1)cos( ) + cos( 1)sin( )cos( α1 α )] [ sin(8.6 )cos(33.43 ) + cos(8.6 )sin(33.43 )cos( 96.69 ( 67.5 ))] = 34.7 θ = arccos Examle 4: Determine the time of sunrise for the conditions of the revious examles. olution: Using Equation (8), we find the (negative) hour angle for sunrise is H [ tan( L) tan( δ )] = arccos[ tan(33.43 ) tan(.44 )] = 14. = arccos To find the corresonding sunrise time in AT, we rearrange Equation (6) min ( unrise, AT ) = 7 mins + H (4 min / deg) = 7 + ( 14. )(4 deg ) = 5 : 3 a.m. olarcalcs.doc 7, K. E. Holbert Page 5

The corresonding local time is then found using Equation (3). LT = AT (4 mins)( LTM Long) ET = 5 : 3 a.m. (4 mins)(15 11 ) ( 6.5 min) = 5 : 37 a.m. The direct normal irradiance to the ground is [1] B DN = A ex (14) sin( 1 ) A is the aarent extraterrestrial solar intensity *, B is the atmosheric extinction coefficient (mainly due to changes in atmosheric moisture), and / is the ressure at the location of interest relative to a standard atmoshere, given by = ex(.361 z) (15) z is the altitude in feet above sea level. The direct normal radiation at sea-level then is B DN (ft) = A ex (16) sin( 1 ) Table : Aarent solar irradiation (A), Atmosheric extinction coefficient (B), and Ratio of diffuse radiation on a horizontal surface to the direct normal irradiation (C) [] Date Day of Year A (Btu/hr ft ) B C Jan 1 1 39.14.58 Feb 1 5 385.144.6 Mar 1 8 376.156.71 Ar 1 111 36.18.97 May 1 141 35.196.11 June 1 17 345.5.134 July 1 344.7.136 Aug 1 33 351.1.1 et 1 64 365.177.9 Oct 1 94 378.16.73 Nov 1 35 387.149.63 Dec 1 355 391.14.57 Note: 1 W/m =.3173 Btu/hr ft The direct radiation flux onto the collector is = cos(θ ) (17) D DN The diffuse-scattered radiation flux onto the collector is * Page 69 of Ref. 1 describes the rocedure for finding A for the outhern Hemishere. olarcalcs.doc 7, K. E. Holbert Page 6

1+ cos( D = C DN (18) C is the ratio of diffuse radiation on a horizontal surface to the direct normal irradiation. The reflected radiation flux for a non-horizontal surface may be aroximated by 1 cos( DR = DN ρ ( C + sin( 1) ) (19) ρ is the foreground reflectivity with some values given below ρ urroundings condition. Ordinary ground or vegetation.8 now cover.15 Gravel roof The total radiation flux is then = + + () Tot D D DR Examle 5: Determine the direct and diffuse-scattered radiation flux to the collector of Examle 3, Phoenix is at an elevation of 111 ft. olution: The ressure ratio is comuted using Equation (15) = ex(.361 z) = ex(.361 111) =.967 Extracting the July 1 values of A and B from Table, and using Equation (14) yields a direct normal radiation of ex B Btu.7 Btu DN = A = 344 ex (.967) = 7 hr ft sin( 1) sin(8.6 ) hr ft From Equation (17), the direct radiation onto the collector is = cos( θ ) = D DN Btu ( 7 ) Btu cos(34.7 ) = 186. 6 Using Equation (18), the diffuse scattered radiation flux to the collector is = C D DN 1+ cos( = (.136) Btu 1+ cos(33.43 Btu ( 7 ) = 8.3 References 1. P. J. Lunde, olar Thermal Engineering: ace Heating and Hot Water ystems, John Wiley & ons, 198,. 6-1. [TH7413.L85]. Chater 3, olar Energy Utilization, AHRAE Handbook, Alications, 1995. [TH711.A1] 3. unangle, htt://www.susdesign.com/sunangle/. 4. A.W. Cul, Princiles of Energy Conversion, nd ed., McGraw-Hill, 1991,. 98-17. olarcalcs.doc 7, K. E. Holbert Page 7