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

*>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?*

*>...Iglo,p = 469.7cos(Thetai)+300*

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

*>unless I made a mistake somewhere...*

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

>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?>...Iglo,p = 469.7cos(Thetai)+300> = 664.7, 218.2, 300, and 300 Wh/m^2,>unless I made a mistake somewhere...