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re: sunman
20 jun 2004
morris dovey   wrote:

>...desoto (20 mi w of des moines) is noticably warmer 
>with less sunny weather than mason city.

nrel says 930 (vs 870) btu/ft^2 falls on a south wall on an average
24.4 (vs 18.3) f december day in des moines (vs mason city), ie they
say des moines is warmer and sunnier :-) the average daily max is 32.6,
so the average daytime temp is about (24.4+32.6)/2 = 28.5 f. 

>the hanger/shop has an uninsulated concrete slab floor that (i'm 
>told) is at least 6" thick to support the heaviest aircraft that 
>might ever be stored inside. it is the only semblance of storage 
>available...

perhaps you can add ceiling mass, eg 30" round poly film greenhouse
air ducts laid flat with a few inches of water inside over plywood
platforms with foil underneath... 1/2" plywood on 2' centers can
support 100 psf (19" of water :-) with less than 24"/240 deflection.

>i'm fairly well resigned to the fact that the shop will be chilly 
>at night without some conventional heat source to make up the 
>difference.

ceiling mass can stay warm at night while the shop stays cool, which
makes night setbacks effective.

with 3 r20 walls and an r20 ceiling and 1" "r7.2" double-foil foamboard
over the other wall and some air sealing, a 50'x50'x14' hangar might have
a thermal conductance of about 50x14/r20 = 35 btu/h-f for each r20 wall
and 50x14/r8 = 88 for the opening wall and 50x50/r20 = 125 for the ceiling.

if we store heat in a t (f) massy ceiling and the hangar is 65 f for 8
hours on an average december day and the south wall has 50x14 = 700 ft^2
of solar siding, so it needs no heat on an average day, and the other two
r20 walls need 8h(65-28.5)70 = 20.4k btu, the hangar needs 24h(t-24.4)125
+ 20.4k = 3000t - 52.8k btu on an average day. if the south wall collects
0.9x700x930 = 585.9k btu on an average day and loses 6h(t-28.5)700/r1 btu
to the outdoors and 585.9k-6(t-28.5)700 = 3000t-52.8k, t = 105.3 f. 

if the hangar needs (65-28.5)(3x35+88) = 7045 btu/h on a cloudy day, we
might keep it warm with the help of a slow ceiling fan in series with
a motion detector and room temp thermostat until the ceiling temp reaches
65+7045x0.5/2500 = 66.4 f, so the average ceiling temp over 5 cloudy days
would be about (105.3+66.4)/2 = 85.9 f. 

storing heat in d" of water (c = d/12x64x2500 = 13.3kd btu/f) makes 
5d(24h(85.9-24.4)125+8hx7045) = (105.3-66.4)c, so d = 2.3", if i did
that right. 

nick




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