|
|
re: solar heated earth
wed, 15 sep 1999
toby wrote
> a friend gave me a garagedoor opener, which i am thinking of using on a
> coldframe, to open an insulated (r10) cover during the day and close it
> at nite. the door rests on top of a 4' by 8' by 2' coldframe...
with 48 ft^2 of sidewalls?
perhaps the cover will open to reflect more sun
down into the coldframe during the day... :-)
>... which has a double layer of greenhouse (r1.8) glazing on top.
>the walls are made of strawbales (r50)...
>
> in december, here in lost wages, nevada, we receive 950 btu/sq.ft./day.
...of sun on a horizontal surface, on the average, and 1600 fall on a
south wall, in late december, ie in january, when nrel says the average
outdoor temp is 45.5 f and the average daily max is 57.3...
> suppose
> - 20% is lost for each layer of glazing
> - 50% of this is reflected by plants or leaves through the ceiling etc.
> - that absorbed by the air is negligable
> - the heat, is absorbed by the earth.
>
> are these reasonable assumptions???
maybe. do the plants mostly shade the earth?
warm air rises... so what warms the earth?
> thus, the thermal conductance is:
>
> day: 4*8/r1.8 + 2(4+8)/r50 = 18 btu/hr-f
> nite: 4*8/r10 + 2(4+8)/r50 = 4 btu/hr-f
why did you calculate these?
doubling the sidewall area makes them 18.7 and 4.2...
> the amount of heat 1 sqft of ground will absorb is:
>
> 1 sqft * 950*0.8*0.5 = 380 btu/sqft/day * 6 hours of sunlight/day
> = 60 btu/hr
why will the earth absorb this amount of heat? and 380*6 = 60?
dividing each side by 60, 6.3 = 1, which makes perfect sense,
for large values of 1...
> the thermal capacitance of 1 cuft of dirt, will absorb about 60 btu/f
ah, miracle dirt, which absorbs heat like water :-)
> thus, in the first hour, the ground temperature will rise by:
> 1 hr * 60 btu/hr * 1/60 f/btu = 1f
"thus..." so this ground is a perfect conductor, and there's a perfect
insulator 1' below the surface, and no heat is lost through the glazing? :-)
> leandre poison claimed that the temperature of the earth didn't change
> that much in 1 day, just a couple inches under the surface. thomas
> bligh recorded the grounds temperature 1/2 of an inch under the
> surface, and it's average monthly temperature was much closer to that
> at 2.6 feet deep than of that on the surface.
you might look up "skin depth" and how to calculate and use it. as i
recall, at 1 skin depth the temperature variation of the surface is
reduced by 1/e. two skin depths make this 1/e^2, and so on. but soil
is more complicated, with plants and rain and snow and water evaporating
and freezing...
> if all this energy went into 1 inch of dirt (i.e. 1 inch * 1 ft/12 inch
> * 1 sqft = 0.083 cuft), then the dirt's temperature would increase
> during the day, each hour by:
>
> 0.083 cuft * 60 btu/f/cuft * deltat = 60 btu
> deltat = 1/0.083 = 12f
but what do you suppose will actually happen to this energy?
heat doesn't flow in certain ways just because we say it does.
consider ohm's law for heatflow, and energy conservation.
> the averge monthly temperature at this depth, is 50f.
why?
>this is the same as the outdoor temperature.
incorrect. the average outdoor temp is 45.5 f in january and 45.7 in
december.
>... in the first hour the ground temperature rises to 62f
why??? the ground temperature doesn't hop up and down to simplify our
calculations. it finds its own temperature more slowly...
>and so heat leaves the ground, through it's 1 inch thermal resistance of
>0.23 hr-sqft-f/btu/inch (~1/4) into the coldframe (tf), as well as into
>the next deeper layer of dirt. remembering, from above, that:
>rc = 1.8 hr-sqft-f/btu (~2); we get...
>
> (62-tf)/(1/4) = (tf-50)/2
> 500tf - 8tf = tf-50
> tf = 500/9 = 56
nonsense... here's another way to look at it: let's put a few dark
colored containers of water among the plants, and assume the indoor
temperature t is the same day and night, and the ground has an r-value
of 10 (an ashrae approximation for downward heatflow, as i recall)
and a temperature of 45.5 f (the average monthly outdoor temp, since
this is a small structure.)
on an average day, 950x32ft^2x0.8^2 = 19..5k btu of solar energy enters
the coldframe, and 19.5k btu of heat energy flows out, so
19.5k = 6h(t-45.5)32ft^2/r1.8 btu lost from the cover during the day
+ 18h(t-45.5)32 ft^2/(r1.8+r10) " " at night
+ 24h(t-45.5)48ft^2/r50 " from the sidewalls
+ 24h(t-45.5)32ft^2/r10 " to the ground.
what's t? we can raise it by reflecting more sun in with the cover or
using a higher outdoor temp during the day, e.g. (45.5+57.3)/2...
nick
|
|
