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re: irradiation
1 may 1998
bob stock wrote:
>the conversion factor (r) for hourly direct radiation on a surface of any
>orientation, depends on the surface tilt (b), surface azimuth (y), sun
>altitude (a), and sun azimuth (o). the conversion can be calculated by:
>
>r = cosb + sinb (cos(o-y) / tana)
that nice intuitive formula separates solar direct beam radiation into
what falls on a flat surface (r = 1 if b = 0 degrees) and what falls on
a south wall (1/tan(a) when b = 90 degrees, with an adjustment for sun
vs. wall azimuth) at a particular instant in time. so, where i live, at
40 n latitude, at noon on 12/21, with sun altitude 90-40-23.5 = 26.5
degrees and tan 26.5 = 0.499, a south wall gets about twice as much sun
as a flat surface. but finding lots of sun angles by hand isn't easy.
the second edition (wiley, 1991) of duffie and beckman's "solar
engineering of thermal processes" book has another example on
pages 16 and 26, along with some sun angle formulas:
calculate the angle of incidence of beam radiation on a surface
located in madison, wi at 10:30 (solar time) on february 13, if
the surface is tilted 45 degrees from the horizontal and pointed
15 degrees west of south, and [find] the ratio of beam radiation
to that on a horizontal surface:
10 r=atn(1)/45'degree/radian conversion factor
20 p=43'latitude (degrees)
40 st=10.5'solar time (hours)
50 w=15*(st-12)'sun hour angle (degrees)
70 n=44'julian day of year (1-365)
80 b=45'surface tilt (degrees)
100 g=15'surface azimuth (degrees)
120 d=23.45*sin(r*360*(284+n)/365)'declination (degrees)
140 ct1=sin(r*d)*(sin(r*p)*cos(r*b)-cos(r*p)*sin(r*b)*cos(r*g))
160 ct2=cos(r*d)*cos(r*w)*(cos(r*p)*cos(r*b)+sin(r*p)*sin(r*b)*cos(r*g))
180 ct3=cos(r*d)*sin(r*b)*sin(r*g)*sin(r*w)
190 ct=ct1+ct2+ct3'cosine of angle of beam radiation to surface
200 ctz=cos(r*p)*cos(r*d)*cos(r*w)+sin(r*p)*sin(r*d)'cos sun zenith angle
210 r=ct/ctz'ratio of beam sun on tilted/flat surface
220 print ct,ctz,r
.8175583 .4913945 1.663751
doing this over a representative january day, where i live, gives:
10 p=40'latitude (degrees)
20 b=90'surface tilt (degrees)
30 g=0'surface azimuth (degrees)
40 n=17'julian day of year (1-365)
50 r=atn(1)/45'degree/radian conversion factor
60 d=23.45*sin(r*360*(284+n)/365)'declination (degrees)
70 ct1=sin(r*d)*(sin(r*p)*cos(r*b)-cos(r*p)*sin(r*b)*cos(r*g))
80 cws=-tan(r*p)*tan(r*d)'cosine of sunset hour angle
90 ws=-atn(cws/sqr(1-cws^2))/r+90'sunset hour angle (degrees)
100 hs=2/15*ws'number of daylight hours
110 ts=12-hs/2'time of sunrise
120 sh=int(ts+1)'next hour after sunrise time
130 shf=sh-ts'sunrise hour fraction
140 smp=ts+shf/2'sunrise hour fraction midpoint
150 w=15*(smp-12)'hour angle (degrees)
160 ct2=cos(r*d)*cos(r*w)*(cos(r*p)*cos(r*b)+sin(r*p)*sin(r*b)*cos(r*g))
170 ct3=cos(r*d)*sin(r*b)*sin(r*g)*sin(r*w)
180 ct=ct1+ct2+ct3'cosine of angle of beam radiation to surface
190 cts=ct*shf'accumulate sun on tilted surface
200 ctz=cos(r*p)*cos(r*d)*cos(r*w)+sin(r*p)*sin(r*d)'cos of sun zenith angle
210 ctzs=ctz*shf'accumulate sun on flat surface
220 tfr=ct/ctz'ratio of beam sun on tilted/flat surface
230 print"time","t/f ratio"
240 print
250 print smp,tfr
260 for st=sh+.5 to 11.5'solar time (hours)
270 w=15*(st-12)'hour angle (degrees)
280 ct2=cos(r*d)*cos(r*w)*(cos(r*p)*cos(r*b)+sin(r*p)*sin(r*b)*cos(r*g))
290 ct3=cos(r*d)*sin(r*b)*sin(r*g)*sin(r*w)
300 ct=ct1+ct2+ct3'cosine of angle of beam radiation to surface
310 cts=cts+ct'accumulate sun on tilted surface
320 ctz=cos(r*p)*cos(r*d)*cos(r*w)+sin(r*p)*sin(r*d)'cos of sun zenith angle
330 ctzs=ctzs+ctz'accumulate sun on flat surface
340 tfr=ct/ctz'daily ratio of beam sun on tilted/flat surface
350 print st,tfr
360 next st
370 tfr=cts/ctzs'daily ratio of beam sun on tilted/flat surface
380 print
390 print"daily ratio",tfr
time t/f ratio
7.623521 7.94568
8.5 3.100142
9.5 2.217095
10.5 1.918898
11.5 1.810127
daily ratio 2.310671
nrel's "solar radiation data manual for buildings" says a horizontal
surface in phila gets an average of 620 btu/ft^2/day of sun in january,
and a south wall gets 1000 (vs 2.31x620 = 1432 with the calculation above.)
duffie and beckman mention on page 94 that treating all radiation as beam
leads to substantial overestimation of tilted surface radiation (they also
say that little solar collection happens at dawn and dusk.) i wonder why the
large overestimate. because the sun has to travel through more clouds or fog
or air in the morning and afternoon ("care must be taken that unrealistic
products of rb and beam radiation ib are not used"), or something to do
with the diffuse component of sun? seems like ground reflections would
only add to the sun that falls on a south wall...
nick
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