
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 multipleday
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/(180t)+125k/(t70). 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 (steadystate,
tricklecharged) 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.
nick
here's where the 20.8k/(180t)+125k/(t70) 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(t80)) = 125k/(t70) 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(t70)5x64ft^2/r20=384(t70)
>> btu/day, about (t70)/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(t30)1ft^2/r1, for a net gain of
>> 10806t btu/day, keeping g gallons of water at t (f) requires a ft^2
>> of collector, ie (10806t)a = (t70)g/4, so a = (t70)g/(4(10806t)).
>>
>> with $4/ft^2 collectors, the tank+collector cost is g+(t70)g/(10806t)
>> = 125k/(t70)+125k/(10806t) = 20.8k/(180t)+125k/(t70). the collector
>> cost increases with t to 180 f, and the tank cost increases as t drops
>> to 70 f. maximizing a/(ut)+b/(tl), t^2(ba)+t(2al2bu)+bu^2al^2 = 0,
>> and a = 104166.7 and b = 42083k and c = 3947918k in this quadratic
>> formula, so t = (bsqrt(b^24ac))/(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!/(10806*t)
>> 120 tc=125000!/(t70)
>> 130 print t,cc+tc
>> 140 next
>>
>> run
>>
>> 130 2500
>> 140 2306.548
>> 150 2256.944
>> 160 2430.556
>> 170 3333.333

