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South window shading in Texas

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Posted by nicksanspam on May 5, 2008, 9:24 pm
 
At 31.2 N. latitude, the min/max sun elevations at noon on 12/21 and 6/21
are 90-31.2+/-23.5 = 35.3 and 82.3 degrees above the horizon. We could pass
all noon winter sun and shade all noon summer sun from a 1'-tall window
under an S' opaque vertical separation under a W' wide horizontal overhang
like this, viewed in a fixed font:
                    
                     .               .
                                  .
                    .          .
                            .    \
                   .     .      
          W           .  Aw = 35.3 degrees, and tan(Aw) = S/W,
 - ----------------.............................
  |              .                    
  |           .  .            so S = Wtan(Aw).
 S|        .            
  |     .       .
  |  .
 -|            .
  |
  |           .
  |
  |          .
  |
 1|         .
  |
  |        .    south -->
  |
  |       .
  |
  |      .
  |
  |     .
  |    
  |    .
  |
  |   .
  |  
  |  .
  |
  | .   \
  |
  |. As = 82.3 degrees, tan(As) = (1+S)/W
  |...............................

If W = (1+Wtan(Aw))/tan(As) = 0.1352+0.09573W = 0.1495', and S = 0.1059'.

NREL recommends As = 108-lat and Aw = 71-lat, based on no shading at solar
noon at 11/17 and 1/25 and complete shading at solar noon on 5/12 and 8/2,
with a modification for southern states: if the first fall month with any
heating degree days is October or later, NREL uses As = 92-lat and Aw
= 66.5-lat to avoid any shading at noon on 12/21 and completely shade
the window at noon from 3/26 to 9/18. As = 92-31.2 = 60.8 and Aw = 66.5-31.2
= 35.3 make W = 0.925' and S = 0.65'. This can be scaled. A 2'-tall window
would have W = 1.85' and S = 1.3', and so on.

We can get more winter solar gain by adding a diagonal reflector from
the south edge of the overhang to the top of the window glazing or by
eliminating separation S and tilting the overhang up to the south and making
it wider. As = 2Aw (max) will put all the reflected dawn sun into the window,
without wasting any winter sun. Aw = 35.3 makes W = 0.469 and S = 0.332.

As = 2Aw = 70.6 = 90-lat+/-delta makes declination delta = +/-11.8 degrees
= 23.5sin(360(284+N)/365) on the Nth day of the year, which makes full shade
at noon between days N = 112 (April 22) and 233 (August 21), approximately.

A perfect reflector would add 100S = 33% more dawn sun to the window.
Tilting it more than the 12/21 noon sun elevation would likely increase
the daily gain, while wasting some dawn sun, but midday sun is stronger,
since it is closer to south and it travels through less atmosphere.

Nick


Posted by nicksanspam on May 6, 2008, 7:58 pm
 

Maybe they used Cooper's 1969 equation (line 60 below):

DS#.45*SIN(2*PI*(N+284)/365)

A more accurate equation from Spencer (1971) and Iqbal (1983):

d = 0.006918 -0.399912cosB  +0.070257sinB
             -0.006758cos2B +0.000907sin2B
             -0.002697cos3B +0.00148sin3B    in radians,

20 PI=4*ATN(1)
30 DATA 25,319,320,321,322,323
40 FOR I=1 TO 6
50 READ N
60 DS#.45*SIN(2*PI*(N+284)/365)
70 B=(N-1)*2*PI/365
80 D=.006918-.399912*COS(B)+.070257*SIN(B)
90 D=D-.006758*COS(2*B)+.000907*SIN(2*B)
100 D=D-.002697*COS(3*B)+.00148*SIN(3*B)
110 D0*D/PI
120 IF N% THEN DS25=DS:D25=D
130 PRINT N,D,DS,DS-DS25,D-D25
140 NEXT I

 25          -19.16716     -19.26362      0             0

319          -18.30314     -19.14784      .1157837      .8640156
320          -18.55995     -19.37802     -.114399       .6072045
321          -18.81123     -19.60246     -.3388405      .3559265
322          -19.05686     -19.8211      -.5574723      .1102982 <--
323          -19.29675     -20.03387     -.7702427     -.129591

says day 322 (11/18) would be a better match to 1/25...

Iqbal includes the equation of time, but it doesn't look easy to find
a matching date for a given day's declination. And Cooper's equation
can be improved. It's fine for lots of purposes, but there's no fall
day that perfectly matches a given spring day's declination, and how
can we average in leap years?

If sin(360(355+C)/365) = -1 on 12/21 and 360(355+C)/365 = 270, C = -81.25,
equivalent to 365-C = 283.75. If the peak happens 365/2-10 = 172.5 days
sooner, spring day N = 25 will match fall day 2x172.5-N = 345-N = 320:

20 PI=4*ATN(1)
30 DATA 25,319,320,321
40 FOR I=1 TO 4
50 READ N
60 DS#.45*SIN(2*PI*(N+283.75)/365)
70 IF N% THEN DS25=DS
80 PRINT N,DS,DS-DS25
90 NEXT I

 25          -19.32099      0

319          -19.0894       .2315941
320          -19.32102     -2.861023E-05 :-)
321          -19.54689     -.2258988

Ah, that feels much better. Now how about leap years?

Nick


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