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Solar radiation on tilted surfaces...

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Posted by nicksanspam on November 22, 2003, 10:04 pm
 
Energy-10 seems to have a lot of limits (eg no way to let warm air circulate
between a living space and a sunspace during the day and let the sunspace
get cold at night), so I've been exploring spreadsheet TMY2 simulations.
Excel seems happy enough with 8760 rows, and likes angles in radians...

If TMY2 data says Wh/m^2 of global (beam+diffuse) sun and 300 Wh/m^2 of
diffuse sun falls on a horizontal surface in Phila between 1 and 2 PM on
1/12, and the ground reflectance is 0.6, how much falls on south, west,
north, and east walls?

Equation 6.22 on page 251 of Kreider and Rabl's Heating and Cooling of
Buildings (McGraw Hill, 2nd edition, 2002--discount copies available) says
the amount of global sun that falls on a surface tilted up from the ground
at an angle of Thetap degrees is

Iglo,p = Idircos(Thetai)+IdifFsky+Iglo,horRhogFgrd,

   where Idir is the direct radiation from the solar disk, and

         Thetai is the (incidence) angle between the normal to the surface
              and a line to the sun, and

         Idif is the diffuse radiation from the isotropic sky, and

         Fsky is the fraction of sky seen by the surface, and

         Iglo,hor = Idircos(Thetas) + Idif (equation 6.23), and

         Thetas is the zenith angle of the sun (0 if directly overhead), and

     Rhog is the ground reflectance, and

         Fgrd is the fraction of isotropic radiation reflected by the ground.

Fsky = (1+cos(Thetap))/2 and Fgrd = (1-cos(Thetap))/2, so Fsky = Fgrd = 0.5
for vertical surfaces, and

Iglo,p = Idircos(Thetai)+0.5Idif+0.5Iglo,horRhog (equation 6.24.)

If Idif = 300 and Iglo,hor = 500 and Rhog = 0.6,

Iglo,p = Idircos(Thetai)+300.

Page 234 of the book says

cos(Thetas) = cos(L)cos(d)cos(w)+sin(L)sin(d) (equation 6.5),
  
   where L is the latitude (about 40 N for Phila), and

         d is the declination, and
    
         w is the hour angle.

Equation 6.4 on page 234 says

   sin(d) = -sin(23.45)cos((360(N+10))/365.25) on the Nth day of the year.
  
On 1/12, N = 12, so sin(d) = -0.3978, and d = -21.70 degrees, which makes

cos(Thetas) = 0.9291cos(40)cos(w)-0.3978sin(40)

            = 0.7660cos(w)-0.2376,

Equation 6.6 on page 234 says

w = 360(Tsol-12h)/24h, where Tsol is the solar time.

Ignoring the equation of time, and assuming Phila is near the center of
its standard time zone, Tsol = 14, so w = 30 degrees, which makes

cos(Thetas) = 0.7660cos(30)-0.2376 = 0.4258, so Thetas = 64.8 degrees,

(ie the sun is 90-64.8 = 25.2 degrees above the horizon),

If Iglo,hor = 500 = Idircos(Thetas)+Idif = 0.4258Idir+300, Idir = 469.7 Wh/m^2,

so Iglo,p = Idircos(Thetai)+300 = 469.7cos(Thetai)+300 Wh/m^2.

      (This is really irradiance Hglo,p, vs radiation Iglo,p in W/m^2.)

Equation 6.10 on page 238 says

cos(Thetai) = sin(Thetas)sin(Thetap)cos(Phis-Phip)+cos(Thetas)cos(Thetap)

            = sin(Thetas)cos(Phis-Phip) for vertical surfaces (equation 6.11),
  
   where Phip is the azimuth angle of the surface, and

         Phis is the azimuth angle of the sun.
    
Equation 6.8 on page 238 says

     sin(Phis) = cos(d)sin(w)/sin(Thetas)

               = cos(-21.7)sin(30)/sin(64.8) = 0.513, so

         Phis = +30.89 degrees (plus degrees are west of south.)

So cos(Thetai) = sin(64.8)cos(30.89-Phip)

               = 0.9048cos(30.89-Phip) for vertical surfaces.

From Figure 6.5b on page 237,

Phip = 0, 90, 180, and 270 degrees for S, W, N, and E walls, so

   cos(Thetai) = 0.7765, 0.4645, -0.7765, and -0.4645 for SWNE walls,

   and Thetai = 39.06, 62.32, 140.9, and 117.7 degrees for SWNE walls.

It looks like north and east walls get no direct sun at 2 PM (we should
really be using 1:30 PM, if the sun arrived between 1 and 2) since the
incidence angles for north and east walls are greater than 90 degrees.

so Iglo,p = 469.7cos(Thetai)+300

          = 664.7, 218.2, 300, and 300 Wh/m^2,

unless I made a mistake somewhere. How do we tell?

Maybe compare with Energy-10...

Nick


Posted by nicksanspam on November 23, 2003, 7:14 am
 

A more accurate and straightforward calculation:

10 PI=4*ATN(1)
20 IGLOHP0'global horizontal irradiance (Wh/m^2)
30 IDIF00'diffuse horizontal irradiance (Wh/m^2)
40 L=PI*39/180'Phila latitude (39 N)
50 T.5'solar time (1:30 PM EST)
60 N+T/24'day of year
70 X=-SIN(PI*23.45/180)*COS(2*PI*(N+10)/365.25)
80 D=ATN(X/SQR(-X*X+1))'sin^-1(x)
90 PRINT"Declination",180*D/PI'declination (degrees)
100 W=2*PI*(T-12)/24'hour angle (radians)
110 X=COS(L)*COS(D)*COS(W)+SIN(L)*SIN(D)
120 THETAS=-ATN(X/SQR(-X*X+1))+PI/2'cos^-1(x)
130 PRINT"Zenith angle",180*THETAS/PI'zenith angle (degrees)
140 IDIR=(IGLOH-IDIF)/COS(THETAS)'direct irradiance (Wh/m^2)
150 X=COS(D)*SIN(W)/SIN(THETAS)
160 PHIS=ATN(X/SQR(-X*X+1))cos^-1(x)
170 PRINT"Sun azimuth",180*PHIS/PI'azimuth angle of sun (degrees)
180 PRINT"Idirect",IDIR
190 PRINT" Surface       Incidence     Surface"
200 PRINT" azimuth       angle         irradiation
210 FOR PHIPD=0 TO 270 STEP 90'azimuth angle of plane (degrees)
220 PHIP=PI*PHIPD/180
230 X=SIN(THETAS)*COS(PHIS-PHIP)
240 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to surface (radians)
250 IF THETAI>=PI/2 THEN THETAI=PI/2
260 RHOG=0.6'ground reflectance
270 IGLOP=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'irradiation on surface (Wh/m^2)
280 PRINT PHIPD,180*THETAI/PI,IGLOP
290 NEXT PHIPD

Declination   -21.61381 degrees
Zenith angle   64.17039 degrees
Sun azimuth    23.28255 degrees
Idirect        459.0354 Wh/m^2

Surface       Incidence     Surface
azimuth       angle         irradiation
(degrees)     (degrees)     (Wh/m^2)

0             34.22897      679.5287
90            69.159        463.3137
180           90            300
270           90            300

Nick


Posted by Alec Chiasson on December 28, 2003, 3:16 pm
 On Sun, 23 Nov 2003 02:14:33 +0000, nicksanspa wrote:


Happy holidays! So the north wall receives almost half the energy
that the south does at that time? My gut says "how could this be?" but
judging from what I've put into it, it can't be trusted...

Alec



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