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

>hi nick,

hello again tim...

>> it seems reasonable to talk about "useful stored heat," ie stored
>> heat above a certain minimum usable temperature which depends on
>> the application, eg 70 f for a solar house...

>> it seems to me that 100 pounds of 150 f water contains 100(150-70)
>> = 8k btu of useful stored heat for this application, and 200 pounds
>> of 90 f water contains 200(90-70) = 4k btu.

>exactly my point. the lower mass -- higher temperature storage has 
>twice the available energy. that was my goal in setting up this example.
>i'm glad someone pushed the numbers to "discover" this  :^).

that leads to a different optimization problem involving the choice of
operating temperature for a solar heat store: is it cheaper to use
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,
including the useful energy stored. that depends on the price of tanks
(including floorspace) and collectors.

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. what's
the optimum steady-state water storage temp t, after a long string of
average days? say a sunspace keeps the house warm on average days,
with no load on the store. 

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


130           2500
140           2306.548
150           2256.944
160           2430.556
170           3333.333

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