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re: a bubble wall?
4 may 1996
c:winsockka9qspoolmail wrote:
>> black bodies radiate heat at many wavelengths. planck's law describes
>> a curve with a peak for this, ie at wavelength lambda, a blackbody with
>> temperature t (kelvin) radiates with an intensity
>>
>> e(lambda) = c1/(lambda^5[exp(c2/(lambda t))-1]), where
>>
>> c1 = 3.74 x 10^-16 m^2w and
>> c2 = 0.0144 mk.
>something that has been bugging me for the last week, and i can't believe
>i stumbled upon it on the net - how exactly do you calculate
>[exp(c2/(lambda t))]?
i use the "e-to-the-x" button on my $20 casio fx-991h calculator, which takes
the base of natural logs, e = 2.718... to the xth power. for instance, exp(1)
is 2.71828..., exp(2) is 7.389... if t is 300 kelvin and lambda is 10 microns,
ie 10 x 10^-6 meters,
exp(c2/(lambda t)) = exp(0.0144/(10^-5 x 300)) = exp(4.8) = 121.5104...
>i've been attempting to learn about solar radiation from first principles
>from my copy of 'solar engineering of thermal processes' by duffie and
>beckman
that is not an easy thing to do. i hope you are using the second edition,
1991, i often have trouble following that good book, with bs and ms degrees
in electrical engineering.
>many thanks if you can help the mathamatically challenged :-(
glad to try, altho it seems to me that we only need arithmetic to do solar
space heating, with at most high-school algebra.
another common and more easily-understood application of exponentials
is in calculating how fast a house or some other thermal mass with
insulation around it cools... a thermal mass c surrounded by a thermal
resistance r has a natural rc "time constant" with a dimension of time,
eg in hours. if we heat the thermal mass up to some temperature dt above the
surrounding temperature ta, rc is the time it will take to the temperature
difference dt to decrease to 1/e th, ie about 1/3 of its original value.
for instance, an 8' cube of water surrounded by (us) r-20 insulation has
c = 62 x 8^3 = 31k pounds of water with a thermal mass of 31k btu/degree f,
and a thermal resistance r = 20/(8x8x6) = 0.052 f-hr/btu, so it has an rc
time constant of rc = 32k x 0.052 = 1653 hours or 68.9 days. so if we heat
the water up to an initial temperature t(0) = 100 f and sit the cube in a
ta = 70 f room, the water temperature t(d) after d days will be
t(d) = ta + (t(0)-ta) exp(-d/(rc),
so after 10 days, the water temperature would be
t(10) = 70 + (100-70) exp(-10/68.9) = 70 + 30 exp(-0.145) = 95.9 f.
after 68.9 days, the temperature would be
t(68.9) = 70 + 30 exp(-1) = 70 + 11 = 81 f.
after a year, the temperature would be
t(365) = 70 + 30 exp(-5.3) = 70.15 f.
if we made this a 16' cube with r40 insulation, we would have
rc = 62 x 16^3 x r40/(16^2 x 6) = 6613 hours or 276 days, so after a year
the water temperature would be
t(365) = 70 + 30 exp(-365/276) = 78 f.
if this cube were outside in 30 f air for a year, the water temp would be
t(365) = 30 + (100-30) exp(-365/276) = 38 f.
if we added an solar air heater with r1 glazing over one insulated side of
our 8' cube, with some simple plastic flap dampers to let the warm air into
the cube during the day and keep out the cold air at night, and collected
8' x 8' x 1000 btu/ft^2 of heat over 6 hours every day (an average december
day in philadelphia), and waited a year, ie 5 time constants, the water would
forget its original temperature, and if the average outdoor temperature over
the year were 55 f, the water would have a more or less constant temperature
t such that the energy that flowed into the cube every day would equal the
energy that flowed out of the cube:
64k btu/day = 6(t-55)64 ft^2/r1 air heater side, day
+ 18(t-55)64 ft^2/r20 air heater side, night
+ 24(t-55)64x6 ft^2/r20 other sides, 24 hours a day,
ie 64 k = (t-55)(384+57.6+460.8) = 902.4(t-55), so t = 55 + 71 = 126 f.
this might be an interesting outdoor exhibit at a science museum or some
unusual outdoor art at a college, a monolith moving even more slowly than
a very large foucault pendulum, reminding us that solar energy really can
work in partly cloudy climates, if well-applied.
we might fill up this 8' cube with 18 55-gallon plastic drums full of water,
each 3' tall and 2' in diameter, in two 3x3 drum layers, with about 300
2-liter plastic soda bottles tucked around in the spaces among them, inside
a few 2x4s and $100 worth of 6 1/2" fiberglass insulation, or strawbale/mortar
walls, with a "truth window" exposing a large thermometer inside.
i have the drums...
nick
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