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heatflow, heatflow, heatflow
13 jun 1996
william r stewart wrote:
>> >nick pine wrote:
>> >> william r stewart wrote:
>> >> >nick pine wrote:
>> >> >> >see solar today sept/oct 1995.
>
>> my copy says "published by the american solar energy society."
>
>do you have any problems with this organization, or are they all ignorant
>and criminal, too?
they are good guys with their hearts in the right places, like yours, however,
we haven't seen many 100% solar houses like those of norman saunders, pe,
described in their pages lately. bill yanda and steve baer figure nicely
in their publications. their technical journal is somewhat clogged with
academics, imo.
>perhaps an understanding of thermodynamics should be your foundation for
>hvac calculations, as is the normal in engineering schools.
been there, done that. ashrae calcs are more useful and straightforward here.
>>>you fail to realize that the thermal store can be much warmer than the
>>>ambient room temperature. since the water wall will be painted (or papered)
>>>a light color, most of the heat transfer will be from convection.
btw, i think the light color is irrelevant here, unless you are talking about
a shiny metallic surface. dark colors are good for solar absorption, but ir
emission at eg 80 f is pretty much independent of color. perhaps you believe
in painting radiators dark colors... i don't think that matters.
>> sure. i just use an r-value of 2/3. i don't add anything for paint or paper.
>> other people go on about nusselt and grashof numbers, but the results are
>> similar. r-values are actual physical measurements that include convection.
>convection equivalents are not so cut and dry. newton's law of cooling is;
>qc = hc*a(ts - tf)
right. ohm's law for heatflow :-)
>where
> hc= average convection heat-transfer coefficient over the surface a
> in w/m2*k (btu/hr*ft^2*f). hc for air, free convection is
> 1-5 btu/hr*ft^2*f
that's a u (=1/r) value, and it's hard to get below 1.5 in still air. it
increases if the air is moving to about u = 1.5 + v/5 for smooth surfaces,
u = 2 + v/2 for rougher ones like stucco, where v is in mph. see ashrae hof
fig 1, p 22.1. you might be able to get u < 1 if you make the water wall faces
shiny metal (u ~ 0.5 for upward heatflow, ~ 0.25 for sideways heatflow), but
that won't work for the side facing the sun, which will impose a lower limit
on the overall u-value of at least 1.5, unless it's a selective surface.
> ts= surface temp
> tf= undisturbed fluid temperature
natch.
>because the thermal store wall will be resting on a perpendicular floor, and
>will have a perpendicular table top extending 18 inches into the room, i will
>assume an hc value of 1 for purposes of discussion.
1.5 is probably closer to the truth, especially since the floor and table top
will act as conductive fins. gluing foil-faced foamboard to the non-south
surfaces would help...
>the wall design is not completely set; one design i am looking at is
>10'x2.5'x8", which is slightly different from 10'x4'x4".
ok.
>this design would give a storage per wall at 16.6 ft^3 x 7.48 gal/ft^3 x
>8.33 lb/gal = 1041 btu storage for each degree f delta t.
ok.
>if they are exposed to 700 btus/hr ft^2 solar insolation
that was 700 btu/day-ft^2. full sun is 311 btu/hr-ft^2, max, on earth.
^^^
>through dual pane windows, then;
>each storage wall will receive 25 x 700 = 17500 btus during the day.
agreed, for a 25 ft^2 water wall face...
>with 4 walls, that translates to 17,500 x 4 = 70,000 btus.
ok.
>of course each window group has an addition 50 ft^2 of window space that
>bypasses the thermal storage walls, so another 140,000 btus will enter
>those windows.
ok. but irrelevant.
>two more south window groups do not have thermal storage walls (will
>likely have storage floors, such as slate or dark tile on concrete),
>so 2 x 70 x 700 = 98,000 btus additional input.
ok. but irrelevant.
>70k + 140k + 98k = ~310,000 btus of solar insolation input on an average
>december day (january and february are higher).
ok. but irrelevant.
>if we want to add in the thermal storage of other aspects of the house, such as
>drywall and flooring, we will find that the thermal mass of the house is
>considerably more, though it is at, or close to, room ambient temperature.
that's what we will find, huh? :-) how much? numbers please.
>if the room temperature is higher in the daytime, then this thermal mass
>will rise to (almost) meet the ambient temperature high.
sure. or higher, if it's in the sun, eg the floor.
>you calculated 4000 btu/f for internal secondary storage in an earlier post,
that was a guess. your house might have 4,000 ft^2 of 1/2" drywall with a
thermal mass of 0.5 btu/f, ie 2,000 btu/f. how much will your solar fireplace
weigh? that would add about 0.16 btu/lb...
>so if the indoor daytime temperature is 75 f and the base winter room
>temperature is 65 f, then internal secondary storage is 40,000 btus.
ok, go ahead, double that number, but that's still a drop in the bucket
compared to a typical day's energy need of 24(70-40)150 = 108k.
>if the ts = 85 f and tf = 75f (after a day of average sunlight in december),
is ts the water wall temp, and tf the room and wall temp?
>qc = 1*50*(85-75) = 500 btu/hr per storage wall
let's see, you said above,
>qc = hc*a(ts - tf)
and you said you wanted to use hc = 1, altho 1.5 is a more reasonable min...
so where did the 50 come from? oh, 2 sides to each water wall, each 25 ft^2.
i guess you'd want to add the perimeter area of 8"/12" x 25' = 17 ft^2, and
something for the floor and table fins? so i'd make this 1.5x67x(85-75) =
1,000 btu/hr/wall, altho you seem to have pulled the 85 f out of a hat.
perhaps you'd like to recalculate that 109 f with these new shapes.
>which, according to your calculation of my house design at 150 btu/hr f
>(which you admit needs significant rework, i'll address this in another post),
perhaps you would like to recalculate that. it's your house.
>would raise the room ambient temperature, conversely lowering the qc.
right. until you are parading naked :-)
>as the room and thermal storage walls starts to cool down to
>ts = 73 and ts =68 f, for example, then qc = 125 btu/hr per storage wall
what difference does that make? you might rework that too.
>with 65 f being the base winter room temperature, the storage walls would
>still have 32000 btus of energy left, not counting the wall, ceiling, and floor
>thermal storage.
ok. i get 33,312 btu, using your numbers above.
but what's the point of this calculation, will?
>> >>the heat goes in one side of those window-blocking water walls and out
>> >>both sides, almost immediately, heating the room air.
>> >by what transfer mechanism, convection? please show me the calculation.
>>my nrel data book says about 1050 btu/ft^2 would shine in your double-glazed
>>windows over a 6 hour winter day, of which they estimate 710 btu/ft^2 would
>>get through the window, and shine on the water wall in your 70 f room, with
>>a still air film r-value of 2/3 for each side (see ashrae hof, fig 1, p 22.1),
>>heating it up to a temperature t, where 710 = 6(t-70)2/(2/3), so t = 109 f.
>>if the wall contains 4" of water, ie about 21 pounds per square foot of wall,
>>the rc time constant of the wall is about 2/3/2x21 = 7 hours. these walls
>>might keep your house warm for several hours, not just a half-hour...
>i'm glad to see that you admit you can be wrong.
sure. i miscalculated before. you are building a whole house wrong :-)
>your understanding of convection still needs work, though, because we haven't
>even touched on nusselt, prandl, reynolds, or grashof. you'll find fluid
>dynamics plays a large role in convection, free or forced. thermal
>conductivity physics only takes you so far in the real world.
r-values are sufficient here.
so let's see. your new water walls have a thermal resistance of 2/3/67 ft^2
and a thermal mass of 1041 btu/f, so their rc time constant is 10.35 hours,
and your house with the 4,000 btu/f of inherent thermal mass and 150 btu/f-hr
with all the curtains drawn has a time constant of 26.6 hours, so if i'm doing
this right, we might figure the rc time constant of your entire house with
all those ritual curtains drawn as the rms value sqrt(26.6^2+10.35^2)=28.5 hr.
suppose we say the average temp of the thermal mass in your house is 80 f
to start with. then when it's 30 f outside, with all the thermal curtains
drawn in the dark, 24 hours a day, the indoor house temp will be:
t(t) = 30 + (80-30) exp(-t/28.5)
80 f after 0 hours, 52 f after 24 hours, and 39 f after 48 hours.
still not a solar house by a long shot, i'd say. altho it could be
dramatically improved by rearranging the same materials...
nick
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