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Mesh absorber radiation calc

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Posted by nicksanspam on January 26, 2006, 8:15 pm
 
Norman Saunders says mesh absorbers can reduce reradiation loss, as well as
keeping cooler supply air near the glazing and making collector air cooler
by increasing the hot surface area and surface-to-air thermal conductance.

The calc below estimates the temperatures of a single layer of glazing,
up to 4 mesh absorbers, and an absorber plate in full sun (250 Btu/h) on
a 30 F day, with no useful heat output, assuming the R1 glazing loses heat
to the outdoors, but the rest of the surfaces only transfer heat by radiation
(we might get close to this with a small "breathing wall" airflow from south
to north through the meshes.) The equivalent circuit looks like this, viewed
in a fixed font like Courier:

              225        113        58        28
        R1  <-qr1  g2  <-qr2 g3  <-qr3 g4  <-qr4 g5       Roo
30 F ---www-------www-------www-------www-------www-------www---> useful heat
              |         |         |         |         |
              |         |         |         |         |
         q1-->|    q2-->|    q3-->|    q4-->|    q5-->|
       0 Btu/h|    113  |    56   |    28   |    28   |
              |         |         |         |         |
              T(1)      T(2)      T(3)      T(4)      T(5)
              glazing   mesh      mesh      mesh      absorber

If 1 ft^2 of glazing passes 225 Btu/h and each mesh blocks 50% of the sun,
the first mesh intercepts 112.5 Btu/h and passes on 112.5, the second mesh
intercepts 112.5 and passes on 56.25, and so on.

20 S=1.714E-09'Stefan-Boltzman constant (Btu/h-F^4-ft^2)
30 F=.5'fraction of sun blocked by mesh
40 T(0)0+460'ambient temp (R)
50 FOR N=2 TO 6'number of layers including glazing and absorber
60 FOR I=1 TO N
70 TB=(T(I-1)+T(I))/2'average temp (F)
80 G=4*S*TB^3*F^2'linearized ladder conductance (Btu/h-F)
90 Q"5*(1-F)^(I-2)*F'node current source (Btu/h)
100 QR=Q'node return current (Btu/h)
110 IF I=1 THEN G=1:Q=0:QR"5'adjustment for glazing layer (Btu/h-F)
120 IF I=2 OR I=N THEN G=G/F'adjustment for solid glazing and absorber
130 IF I=N THEN QR=0'adjustment for absorber
140 T(I)=T(I-1)+(Q+QR)/G'layer temp (F)
150 NEXT I
160 IF ABS(TP-T(N))>.01 THEN TP=T(N):GOTO 60'iterate
170 PRINT N;
180 FOR I=1 TO N-1
190 PRINT T(I)-460;
200 NEXT I
210 PRINT T(N)-460
220 NEXT N

2  255  331.7493F <--absorber, with no mesh
3  255  391.5753  416.9913  <--absorber, with 1 mesh layer
4  255  391.5767  482.5053  492.1556  ...with 2 mesh layers
5  255  391.5767  482.5061  519.4842  523.8208
6  255  391.5767  482.5061  519.4845  536.4991  538.5655

It looks like mesh layers can reduce radiation loss effective
emittance, with diminishing returns after more than 2 layers.  

The next step might be to extract some high-temp useful heat
from the absorber space with some copper or fin-tube pipe.

Nick


Posted by nicksanspam on January 27, 2006, 11:36 am
 

It looks like screens that block less sun work better, in this model...

20 S=1.714E-09'Stefan-Boltzman constant (Btu/h-F^4-ft^2)
30 F=.1'fraction of sun blocked by mesh
40 T(0)0+460'ambient temp (R)
50 FOR N=2 TO 6'number of layers including glazing and absorber
60 FOR I=1 TO N
70 TB=(T(I-1)+T(I))/2'average temp (F)
80 G=4*S*TB^3*F^2'ladder conductance (Btu/h-F)
90 Q"5*(1-F)^(I-2)*F'node current source (Btu/h)
100 QR"5*(1-F)^(I-1)'node return current (Btu/h)
110 IF I=1 THEN G=1:Q=0:QR"5'adjustment for glazing layer (Btu/h-F)
120 IF I=2 OR I=N THEN G=G/F'adjustment for solid glazing and absorber
130 IF I=N THEN QR=0'adjustment for absorber
140 T(I)=T(I-1)+(Q+QR)/G'layer temp (F)
150 NEXT I
160 IF ABS(TP-T(N))>.01 THEN TP=T(N):GOTO 60
170 PRINT N;
180 FOR I=1 TO N-1
190 PRINT T(I)-460;
200 NEXT I
210 PRINT T(N)-460
220 NEXT N

 2  255  331.7493
 3  255  671.8047  691.6501
 4  255  671.8084  1483.908  1487.517
 5  255  671.8084  1483.904  1775.256  1777.396
 6  255  671.8084  1483.904  1775.257  1964.43  1965.939

Nick


Posted by nicksanspam on January 30, 2006, 8:01 am
 
20 S=1.714E-09'Stefan-Boltzman constant (Btu/h-F^4-ft^2)
30 F=.1'fraction of sun blocked by mesh
40 C0'node cap (Btu/F)
50 T(0)0+460'ambient temp (R)
60 FOR N=2 TO 6'number of layers including glazing and absorber
70 T(N+1)0+460'water temp (R)
80 G(N+1)=5'fin-tube conductance (Btu/h-F)
90 FOR M=1 TO 50000!'relaxation
100 FOR I=1 TO N
110 TB=(T(I-1)+T(I))/2'average temp (R)
120 G(I)=4*S*TB^3*F^2'ladder conductance (Btu/h-F)
130 Q"5*(1-F)^(I-2)*F'node current source (Btu/h)
140 IF I=1 THEN G(I)=1:Q=0'adjustment for glazing layer
150 IF I=2 THEN G(I)=G(I)/F'adjustment for solid glazing
160 IF I=N THEN G(I)=G(I)/F:Q=Q/F'adjustment for solid absorber
170 DI=Q+(T(I-1)-T(I))*G(I)+(T(I+1)-T(I))*G(I+1)'net flow into node (Btu/h)
180 T(I)=T(I)+DI/C'layer temp (R)
190 NEXT I
200 NEXT M
210 PRINT N;
220 FOR I=1 TO N
230 PRINT T(I)-460;
240 NEXT I
245 QU=(T(N)-T(N+1))*G(N+1)'useful heat output (Btu/h)
250 PRINT QU
260 NEXT N

N  glazing   absorber  useful heat -->
   temp (F)  temp (F)  (Btu/h)

2  132.6884  194.4612  122.3059
3  52.45618  210.6896  210.507   202.5348
4  54.15216  219.7684  286.4629  210.1672  200.8359
5  63.30817  263.7379  519.2375  315.5369  208.3353  191.6766
6  72.3819   300.9792  653.3873  649.6066  336.7692  206.5188  182.594

N  glazing   1st mesh  2nd mesh  3rd mesh  4th mesh  absorber  useful heat
   temp (F)  temp (F)  temp (F)  temp (F)  temp (F)  temp (F)  (Btu/h)

A single 10% mesh looks best here, with N=3 and a foot of fin-tube and
202.53 Btu/h of useful heat at 170 F and the coolest glazing (at about
30+(225-203)R1 = 53 F, since all the non-useful solar heat escapes back
out the R1 glazing to the 30 F outdoors) and a 202.53/250 = 81% solar
collection efficiency.

Nick


Posted by Jeff Thies on January 30, 2006, 9:15 pm
 nicksanspam@ece.villanova.edu wrote:

I'm having some trouble visualizing the collection part of this, or for
that matter the whole concept!

I keep thinking in terms of Bill Kreamers collector where the air is
circulated from glazing side into the collection side. Do I have this
wrong in your example.

So, how does the air flow and how does it pass through the fin tube
collector?

   I'm thinking there would be a chamber at the base of the collector
connecting the front of the collector to the back. In that chanber would
be from fron to back

1: A baffle with an exhaust fan pulling heated air from the back
2: The fin tub collector (the height of the chamber) through which the
heated air is pulled

This just cycles in a loop. I would want the rear air plenum resistance
to be much greater than the rear screen, to keep air flow through that
relatively even. Hence the screens would be close together.

Perhaps your collector idea is considerably different than that, but
that is what I have been thinking about (for a few months). It's easier
to make efficient air collectors and it's easier to transport that heat
via liquids.

   Cheers,
Jeff


   Cheers,
Jeff



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