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re: a 16' cube
8 mar 2001
joelncaryn wrote:
>okay, someone correct me if i mistranslate this.
thanks for the exegesis, caryn ("mr. reagan really meant to say..." :-)
>>> nrel says 1000 btu/ft^2 of sun falls on a south wall and 425 falls
>>> on east and west walls in phila, pa on an average 30.4 f january day,
>>> when a 1-axis ew concentrator can collect 729 btu/ft^2 over 3 hours.
>>> with a 90% reflector and 2 layers of r1 clear polycarbonate glazing
>>> with 90% solar transmission, we might collect 0.9x0.9x0.9x729x16'x16'
>>> = 136k btu/day with a 16' cube with a parabolic reflector inside.
>...we could collect 136,000 btu of solar energy per day even somewhere
>as north as philadelphia, even on a cold and not-sunny january day.
on an average january day, with an average amount of sun. nrel had tracking
collectors in mind, but this fixed solar trough version could have about
the same performance in january, with a larger target and a focal line that
moves over the target (a compromise between efficiency and complexity.)
>whether or not you'd find it possible (wrt homeowner's associations and
>local building codes etc.) to build a 16' cube onto your house is not an
>issue nick usually addresses.
i don't belong to a homeowner's association (a rebellious friend near
boston does, and often commits "condo crimes" like hanging laundry on
a line or making a compost heap behind the house.) my boca township
building inspector has been very cooperative. i've applied for permits
for gizmos attached to the house, and built two 100' greenhouses and
some freestanding lawn art without permits.
>to make the parabolic reflector reflect, glue a plasticised hard-to-tear
>kind of aluminum foil (mylar) to the bottom of the reflector.
duane johnson has used grease vs glue to attach the mylar to his outdoor
heliostats. this makes it easy to change the mylar, which only lasts a few
years when exposed to the weather.
>>> we might approximate the parabolic shape with 9x24' = 216 ft of 2x4s
>>> on 2' centers with a dozen sawkerfs in each bow...
>this is an iterative solution for how to build a parabola with the correct
>focal length by cutting 12 lines into each of a total of 9 24' long 2x4s
>to make them bendy.
dat would be 11 vs 12 cuts, and the 2x4s needn't be single 24' 2x4s. some
of the "cuts" might be angles formed when 2 2x4s are bolted together.
>...cheap. a little under $8/ sq. ft. - few houses are built for under
>$50/ sq. ft., and most of the ones i detail reach much higher numbers.
and this has a shiny curved cathedral ceiling. tres chic, non?
>to keep the tank at 150f and the target at 160f, water will be flowing
>through the system at 10 gallons per minute. of course, this ignores
>the heating of the air inside this room from the target --
well, no... that was the lower part of the thermal equivalent circuit.
>>> here's a formula for laminar flow in a pipe with radius r and length
>>> l in feet and pressure difference dp: q = pir^4dp/8mul ft^3/s... so
>>> r = 0.0447' or 0.54", eg a 1.5" pipe...
>assuming we've piped this system the easy way, we need a 1.5" (radius or
>diameter?) pipe to ensure that we have a 10 gpm flow.
the diameter needs to be greater than 1.08", eg 1.5".
>>> ...we have something like this (in courier font):
>
>>> 30.4 f ----www-----------------tc -------www-----tc
>>> 1/600 | | 1/600
>>> | --- 93 f
>>> ----- | equivalent to -
>>> |------| --> |-------| |
>>> ----- | _
>>> 37.5k/h <
>>> < 1/96 . . . . . . . . .
>>> <
>>> ----- |
>>> |------| --> |-------*-------t = 160 f
>>> ----- |
>>> 45.3k/h ----- 32,768 btu/f
>>> -----
>>> |
>>> _
>
>copy this diagram into your word processor, select all the text, and
>change it to courier font...
i just view news in fixed font using outlook express, when i'm not
using rn on this unix system.
>>> at 150 f, with a minimum usable temperature of 120 f, the tank would
>>> store about (150-120)32768 = 983k btu. with a 135 f average temp, it
>>> would lose about 24h(135-30)256ft^2/r21.6 = 30k btu to the outdoors
>>> on a cloudy day, so it could provide 76k btu of hot water for about
>>> 983k/(30k+76k) = 9 cloudy days in a row...
>it would take 9 cloudy days in a row to lose the heat stored in the tank.
there's a more accurate way to estimate this with a differential equation:
at temperature t, the tank loses heat at a rate of (t-30)256/21.6 btu/h,
and the hot water removes heat from the tank at a constant 76k/24h = 3167
btu/h, so dt/dt = -((t-30)256/21.6+3167)/32768 = -0.00036194t -0.085798,
= -ct +d, so d/c = -237.21 and t = -237.21+(150+237.21)exp(-0.00036194t)
= 120 f when t = -ln((120+237.21)/387.21)/0.00036194 = 222.8 hours, after
9.28 cloudy 30 f days in january.
now how can we make it frugaler?
nick
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