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A puzzle

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Posted by nicksanspam on January 11, 2007, 8:19 pm
 
Assuming the sun's rays are all either horizontal or vertical :-)
a 1'x1' horizontal pond collector below a 45 degree 1'x1.414' reflector
with the upper edge of the reflector directly above the south edge of
the pond with 100% reflectance and 2 R1 horizontal transparent covers
with 100% solar transmission might collect 1000 Btu of south wall sun
and 0 Btu of overhead sun and lose 6h(140-34)1ft^2/R2 = 318 on an average
January day in Phila, for a net gain of 682.

If we figure 1890 Btu/ft^2 falls on the ground and 820 falls on a south
wall on an average 80 F July day, it might collect 820 Btu of south wall
sun and 0 Btu of overhead sun and lose 10h(140-80)1ft^2/R2 = 300 on
an average July day, for a net gain of 520.

How much would we have to cut off the upper reflector edge to reduce
summer shading and make equal hot water outputs in July and January?
How much would that output be? Here's a picture (view in fixed font.)
What's d?

               620, 1890    

1|..........................
 |                    ..
 |                   . .  d
 |                  .  .
 |                 r   . ---
 |                r    .
 |               r     .
 |              r      .
 |             r       .
 |            r        .
 |           r         .
 |          r          . <--1000, 820 Btu
 |         r           .      34,  80 F
 |        r            .
 |       r             .
 |      r              .
 |     r               .
 |    r                .
 |   r                 .
 |  r                  .
 | r 45                .
  ---------------------|--------->
(0,0)                  1
                   | d |

After we cut off some reflector, we can tilt it back and extend it again
to increase the output...

Nick


Posted by nicksanspam on January 12, 2007, 1:48 pm
 

If the reflector is specular vs diffuse, so all the south-wall sun hits
the collector, 1000(1-d)+620d-6(140-34)/2 = 820(1-d)+1890d-10(140-80)/2
makes 1000 -380d -318 = 820 +1070d -300, ie -1450d = -162, so d = 0.1117,
and we cut off dsqrt(2) = 0.158' of the reflector, making w = 1.256'.


               620, 1890 Btu    

1|..........................
 |                    . .
 |                   .  .  
 |                  .   .
 - y0              r    .
 |                r     .
 |               r      .
 |              r       .
 |             r        .
 |            r         .
 |         / r          .
 |        w r           . <--1000, 820 Btu
 |       / r            .      34,  80 F
 |        r             .
 |       r              .
 |      r               .
 |     r                .
 |    r                 .
 |   r                  .
 |  r                   .
 | r beta          delta.
  -----------------|----|--------->
                   x0   1

If all the south-wall sun hits the collector, incidence angle beta equals
reflection angle (0,0)-(x0,y0)-(1,0), so angle (0,0)-(1,0)-(x0,y0) delta
= 180-2beta, and tan(delta) = -tan(2beta) = -2tan(beta)/(1-tan^2(beta))
= y0/x0, and tan(beta) = y0/x0, which makes x0^2 -2x0 -y0^2 = 0 [1].

Meanwhilst, 1000y0+620(1-x0)-6(140-34)/2 = 820y0+1890(1-x0)-10(140-80)/2
makes 180y0 -1270(1-x0) -18 = 0, ie 180y0 +1270x0 - 1288 = 0, so
y0 = 7.15555-7.05555x0 [2].

Substituting [2] into [1] makes 50.781x0^2 - 102.97x0 + 51.202 = 0, so
x0 = 0.873761 and y0 = 0.99069 and w = 1.321' and beta = 48.63 degrees,
with 752 Btu/ft^2 of hot water in December and July.

Now what happens if the sun moves normally in the sky?

Nick


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