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re: solar collector panel insulation
11 jul 2000
tim dierauf  wrote:

>hi nick,

hello again tim,

>i don't have the time to expand and collect the various terms in your
>equations below. what did you conclude?

i explored the choice of operating temperature for a multiple-day 
solar heat store: which is cheaper, a) a large tank of warm water with
fewer collectors, or b) a smaller higher temp tank with more collectors,
all other things being equal, eg the useful energy stored. my conclusion
was, "that depends on the price of tanks and collectors." :-)

>if you spell things out a little more, i could jam them into a few
>of the solver packages on my computer here...

we doan need no steenkeeng com, urm, oh well, if you like... 
the sum of storage and collector costs boiled down to this:
20.8k/(180-t)+125k/(t-70). the (first) collector cost rises 
as the chosen storage temp t rises, until we need umpteen 
skillion square feet of collectors as their efficiency hits
zero when t = 180 f. the (second) tank cost grows until its
volume becomes infinite as the same chosen (steady-state,
trickle-charged) storage temp t drops to 70 f. 

>economics aside, the 2nd law teaches us to match temperatures in our cycles.
>thus, if you need a process temperature of ~75f, then maintain a storage of
>75f plus whatever temperature drops occur to get the heat from the storage
>to the process (i.e. living spaces)...

plus the storage temperature drop over 5 cloudy days in a row...

>i'm sure that you would agree that heating something to 300f and
>then tempering it down to 75f before it could be used would not
>be the most effective thing to do.

yes, especially since 300 f essentially precludes water, for which 
"tempering" may not be a big deal, eg controlling a hydronic floor
circulation pump with a heating thermostat.


here's where the 20.8k/(180-t)+125k/(t-70) came from:

>> suppose we want to store 1 million btu at 70 f min to keep a solar
>> house warm for 5 cloudy days in a row, in that unlikely event...

>> we need 1mbtu/(8(t-80)) = 125k/(t-70) gallons of water. a 1500 gallon
>> poly tank costs about $400. say cheap floorspace and insulation makes
>> that $1/gallon. indoors, it loses about 24h(t-70)5x64ft^2/r20=384(t-70)
>> btu/day, about (t-70)/4 btu/gallon. if a square foot of vertical south
>> collector with r1 glazing with 90% transmission gathers 900 btu on an
>> average 30 f january day and loses 6h(t-30)1ft^2/r1, for a net gain of
>> 1080-6t btu/day, keeping g gallons of water at t (f) requires a ft^2
>> of collector, ie (1080-6t)a = (t-70)g/4, so a = (t-70)g/(4(1080-6t)).
>> with $4/ft^2 collectors, the tank+collector cost is g+(t-70)g/(1080-6t)
>> = 125k/(t-70)+125k/(1080-6t) = 20.8k/(180-t)+125k/(t-70). the collector
>> cost increases with t to 180 f, and the tank cost increases as t drops
>> to 70 f. maximizing a/(u-t)+b/(t-l), t^2(b-a)+t(2al-2bu)+bu^2-al^2 = 0,
>> and a = 104166.7 and b = -42083k and c = 3947918k in this quadratic
>> formula, so t = (-b-sqrt(b^2-4ac))/(2a) = 148.1 f, but the collector
>> plus tank cost isn't very sensitive to the choice of temperature...
>> 100 for t =130 to 170 step 10
>> 110 cc=125000!/(1080-6*t)
>> 120 tc=125000!/(t-70)
>> 130 print t,cc+tc
>> 140 next
>> run
>> 130           2500
>> 140           2306.548
>> 150           2256.944
>> 160           2430.556
>> 170           3333.333

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