SolarCalcs  Arizona State University
SOLAR CALCULATIONS
The orbit of the Earth is an ellipse not a circle, hence the distance between the Earth and Sun varies over the year, leading to apparent solar irradiation values throughout the year approximated by [1]:
I0
=
I SC
1
+
0.033 cos
N 365
?
360?
(1)
where the solar constant, ISC = 429.5 Btu/hr?ft? (1353 W/m?). The Earth's closest point (about 146 million km) to the sun is called the perihelion and occurs around January 3; the Earth's farthest point (about 156 million km) to the sun is called the aphelion and occurs around July 4.
The Earth is tilted on its axis at an angle of 23.45?. 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 21 (vernal equinox) and September 21 (autumnal equinox). The sun reaches its northernmost latitude about June 21 (summer solstice) and its southernmost latitude near December 21 (winter solstice).
The declination is the angular distance of the sun north or south of the earth's equator. The declination angle, , for the Northern Hemisphere (reverse the declination angle's sign for the Southern Hemisphere) is [2]
=
23.45?sin N
+ 284 365
? 360?
(2)
where N is the day number of the year, with January 1 equal to 1. The Earth is divided into latitudes (horizontal divisions) and longitudes (NS divisions).
The equator is at a latitude of 0?; the north and south poles are at +90? and ?90?, respectively; the Tropic of Cancer and Tropic of Capricorn are located at +23.45? and ?23.45?, respectively. For longitudes, the global community has defined 0? as the prime 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 prime meridian. A 24hr day has 1440 mins, which when divided by 360?, means that it takes 4 mins to move each degree of longitude.
The apparent solar time, AST (or local solar time) in the western longitudes is calculated from
AST = LST + (4 min/ deg)(LSTM  Long) + ET
(3)
where LST = Local standard time or clock time for that time zone (may need to adjust for daylight savings time, DST, that is LST = DST ? 1 hr), Long = local longitude at the position of interest, and LSTM = local longitude of standard time meridian
LSTM = 15? ? Long
(4)
15? round to integer
SolarCalcs.doc
? 2007, K. E. Holbert
Page 1
The difference between the true solar time and the mean solar time changes continuously daytoday with an annual cycle. This quantity is known as the equation of time. The equation of time, ET in minutes, is approximated by [3]
ET = 9.87 sin(2 D)  7.53 cos(D) 1.5 sin(D)
where D = 360? (N  81)
(5)
365
Example 1: Find the AST for 8:00 a.m. MST on July 21 in Phoenix, AZ, which is located at 112? W longitude and a northern latitude of 33.43?.
Solution: Since Phoenix does not observe daylight savings time, it is unnecessary to make any change to the local clock time. Using Table I, July 21 is the 202nd day of the year. From
Equation (5), the equation of time is
D = 360? (N  81) = 360? (202  81) = 119.3?
365
365
ET = 9.87 sin(2 D)  7.53 cos(D) 1.5 sin(D)
= 9.87 sin(2 ?119.3?)  7.53 cos(119.3?) 1.5 sin(119.3?) = 6.05 min
From Equation (4), the local standard meridian for Phoenix is
LSTM = 15? ? Long
= 15? ? 112?
= 15? ? 7 = 105?
15? round to integer
15? round to integer
Using Equation (3), the apparent solar time (AST) is AST = LST + (4 mins)(LSTM  Long) + ET = 8 : 00 + (4 mins)(105? 112?) + (6.05 min) = 7 : 26 a.m.
The hour angle, H, is the azimuth angle of the sun's rays caused by the earth's rotation, and H can be computed from [4]
H = (No. of minutes past midnight, AST )  720 mins
(6)
4 min / deg
The hour angle as defined here is negative in the morning and positive in the afternoon (H = 0? at noon).
The solar altitude angle (1) is the apparent angular height of the sun in the sky if you are facing it. The zenith angle (z) and its complement the altitude angle (1) are given by
cos( z ) = sin(1 ) = cos(L) cos( ) cos(H ) + sin(L) sin( )
(7)
where L = latitude (positive in either hemisphere) [0? to +90?], = declination angle (negative for Southern Hemisphere) [?23.45? to +23.45?], and H = hour angle.
SolarCalcs.doc
? 2007, K. E. Holbert
Page 2
The noon altitude is N = 90?  L + . The sun rises and sets when its altitude is 0?, not
necessarily when its hour angle is ?90?. The hour angle at sunset or sunrise, HS, can be found from using Eq. (7) when 1 = 0
cos(H S ) =  tan(L) tan( )
(8)
where HS is negative for sunrise and positive for sunset. Absolute values of cos(HS) greater than unity occur in the arctic zones when the sun neither rises nor sets.
N
Zenith
W
E
Altitude
Azimuth S
The solar azimuth, 1, is the angle away from south (north in the Southern Hemisphere).
cos(1 )
=
sin(1 ) sin(L)  sin( cos(1 ) cos(L)
)
(9)
where 1 is positive toward the west (afternoon), and negative toward the east (morning), and therefore, the sign of 1 should match that of the hour angle.
If > 0, the sun can be north of the eastwest line. The time at which the sun is due east and west can be determined from
tE /W
=
4
min deg
180?
m
arccos
tan( ) tan(L)
(10)
where these times are given in minutes from midnight AST.
Example 2: For the conditions of Example 1, determine the solar altitude and azimuth angles. Solution: The hour angle is first computed using Equation (6)
H = (No. of minutes past midnight, AST )  720 mins = [60 * 7 + 26]  720 min = 68.5?
4 min / deg
4 min / deg
The declination angle is found from Equation (2)
=
23.45?
sin
N
+ 284 365
?
360?
=
23.45?
sin
202 + 284 365
? 360?
=
20.44?
The altitude angle (1) of the sun is calculated via Equation (7) sin(1) = cos(L) cos( ) cos(H ) + sin(L) sin( )
1 = arcsin[cos(33.43?) cos(20.44?) cos(68.5?) + sin(33.43?) sin(20.44?)] = 28.62?
The solar azimuth angle (1) is found from Equation (9)
SolarCalcs.doc
? 2007, K. E. Holbert
Page 3
1
=
arccos
sin(1 ) sin( cos(1 )
L)  sin( cos(L)
)
[sgn(H )]
=
arccos sin(28c.6o2s?()2s8i.n6(23?3).4co3s?()33s.4in3(?2)0.44?)
sgn(68.5?)
=
96.69?
This value indicates that the sun is north of the eastwest line. The time at which the sun is
directly east can be computed using Equation (10)
tE
=
4
min180?

arccos
tan( ) tan(L)
=
4
min180?

arccos
tan(20.44?) tan(33.43?)
=
8
:
17.5
a.m.
The AST is earlier than tE, thus verifying that the sun is still north the eastwest line.
Normal to Earth's surface
Normal to inclined surface
Inclined surface facing southwesterly
N
E 2
1
2
1
W
Horizontal projection of normal to inclined surface
Horizontal projection of sun's rays
Figure 1. Solar angles [4]
SolarCalcs.doc
? 2007, K. E. Holbert
S Page 4
The collector angle () between the sun and normal to the surface is
cos( ) = sin(1 ) cos( 2 ) + cos(1 ) sin( 2 ) cos(1  2 )
(11)
where 2 is the azimuth angle normal to the collector surface, and 2 is the tilt angle from the ground. If is greater than 90?, then the sun is behind the collector. Some collector panel angles of interest are given below.
Azimuth, 2 n/a ? 0? ?90? +90? 1
Tilt, 2 0? 90? 90? 90? 90?
90??1
90??1 varies varies varies varies
0?
Orientation Horizontal (flat) Vertical wall Southfacing Vertical East facing wall West facing wall Tracking System
The sunrise and sunset hours on the collector are different when the collector is shadowed by itself. The collector panel sunrise/sunset hours may be computed from [1]
H SRS
=
m
minH
S
,
arccos
a
b
?
a2  b2 a2 +1
+ 1
(12)
where
a = cos(L) + sin(L)
sin( 2 ) tan( 2 ) tan( 2 )
b
=
tan(
)
cos(L) tan( 2 )

sin(L) sin( 2 ) tan(
2
)
(13)
Example 3: For the conditions of Example 1, find the collector angle for a wall that faces east
southeast and is tilted at an angle equal to the location latitude.
Solution: The latitude is 33.43? which is also the tilt angle (2). Referring to Figure 1, the
collector azimuth angle (2) for an eastsoutheast direction is

90?
?
3 4
=
67.5?
.
Finally, the
collector angle is computed from Equation (11)
= arccos[sin(1 ) cos( 2 ) + cos(1 ) sin( 2 ) cos(1  2 )] = arccos[sin(28.62?) cos(33.43?) + cos(28.62?) sin(33.43?) cos(96.69?  (67.5?))] = 34.7?
Example 4: Determine the time of sunrise for the conditions of the previous examples. Solution: Using Equation (8), we find the (negative) hour angle for sunrise is
H S =  arccos[ tan(L) tan( )] =  arccos[ tan(33.43?) tan(20.44?)] = 104.2?
To find the corresponding sunrise time in AST, we rearrange Equation (6)
(Sunrise,
AST )
=
720
mins
+
HS
(4
min
/
deg)
=
720
+
(104.2?)(4
) min
deg
=
5
:
03
a.m.
SolarCalcs.doc
? 2007, K. E. Holbert
Page 5
The corresponding local time is then found using Equation (3). LST = AST  (4 mins)(LSTM  Long)  ET = 5 : 03 a.m.  (4 mins)(105? 112?)  (6.05 min) = 5 : 37 a.m.
The direct normal irradiance to the ground is [1]
I DN
=
A exp 
p p0
B sin(
1
)
(14)
where A is the apparent extraterrestrial solar intensity*, B is the atmospheric extinction coefficient (mainly due to changes in atmospheric moisture), and p/p0 is the pressure at the location of interest relative to a standard atmosphere, given by
p p0 = exp(0.0000361 z)
(15)
where z is the altitude in feet above sea level. The direct normal radiation at sealevel then is
I DN
(0 ft)
=
A
B
exp
sin(
1
)
(16)
Table I: Apparent solar irradiation (A), Atmospheric extinction coefficient (B), and Ratio of diffuse radiation on a horizontal surface to the direct normal irradiation (C) [2]
Date
Jan 21 Feb 21 Mar 21 Apr 21 May 21 June 21 July 21 Aug 21 Sept 21 Oct 21 Nov 21 Dec 21
Day of Year A (Btu/hr?ft2)
B
21
390
0.142
52
385
0.144
80
376
0.156
111
360
0.180
141
350
0.196
172
345
0.205
202
344
0.207
233
351
0.201
264
365
0.177
294
378
0.160
325
387
0.149
355
391
0.142
Note: 1 W/m2 = 0.3173 Btu/hr?ft2
C 0.058 0.060 0.071 0.097 0.121 0.134 0.136 0.122 0.092 0.073 0.063 0.057
The direct radiation flux onto the collector is
I D = I DN cos( )
(17)
The diffusescattered radiation flux onto the collector is
*
Page 69 of Ref. 1 describes the procedure for finding A for the Southern Hemisphere.
SolarCalcs.doc
? 2007, K. E. Holbert
Page 6
I DS
=C
I DN
1 + cos( 2 ) 2
(18)
where C is the ratio of diffuse radiation on a horizontal surface to the direct normal irradiation. The reflected radiation flux for a nonhorizontal surface may be approximated by
I DR
= I DN
(C
+
sin( 1 ))
1 
cos( 2 2
)
(19)
where is the foreground reflectivity with some values given below
Surroundings condition 0.2 Ordinary ground or vegetation 0.8 Snow cover 0.15 Gravel roof
The total radiation flux is then
ITot = I D + I DS + I DR
(20)
Example 5: Determine the direct and diffusescattered radiation flux to the collector of Example 3, where Phoenix is at an elevation of 1112 ft. Solution: The pressure ratio is computed using Equation (15)
p p0 = exp(0.0000361 z) = exp(0.0000361 ?1112) = 0.9607
Extracting the July 21 values of A and B from Table I, and using Equation (14) yields a direct normal radiation of
I DN
=
A exp 
p p0
B sin( 1
)
=
344
Btu hrft 2
exp

(0.9607)
0.207 sin(28.62?)
=
227
Btu hr ft 2
From Equation (17), the direct radiation onto the collector is
( ) I D = I DN cos( ) =
227
Btu hrft 2
cos(34.7?)
= 186.6
Btu hrft 2
Using Equation (18), the diffuse scattered radiation flux to the collector is
( ) I DS
=C
I DN
1
+
cos( 2 2
)
=
(0.136)
227
Btu hrft 2
1
+
cos(33.43?)
2
=
28.3
Btu hrft 2
References
1. P. J. Lunde, Solar Thermal Engineering: Space Heating and Hot Water Systems, John Wiley & Sons, 1980, pp. 62100. [TH7413.L85]
2. Chapter 30, "Solar Energy Utilization," ASHRAE Handbook, Applications, 1995. [TH7011.A12] 3. SunAngle, . 4. A.W. Culp, Principles of Energy Conversion, 2nd ed., McGrawHill, 1991, pp. 98107.
SolarCalcs.doc
? 2007, K. E. Holbert
Page 7
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