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re: state of geothermal hvac
6 mar 2001
ronald herfurth wrote:
>i got a trane 2 1/2 ton - 12 seer heat pump (indoor and outdoor units) 6
>years ago for $ 4,200 but a geo-thermal rep wanted $ 10,000 and admitted
>that half of that was to drill the wells...
and you'd still end up using lots of electrical energy, with a cop of 3.
say your house needs a ton in january, ie 288k btu/day...
>...my air conditioning is pretty cheap and i think i'd come out ahead
>if i put the $ 5,800 difference toward solar thermal.
interesting thought. what kind of solar heater could you build for $5800?
nrel says 1090 btu/ft^2 of sun falls on a south wall and 475 btu/ft^2
falls on east and west walls in richmond, va on an average 35.7 f january
day, when a 1-axis ew concentrator can collect 824 btu/ft^2 over about
3 hours. with a 90% reflector and 2 layers of r1 glazing with 90% solar
transmission, that becomes 601, so you'd need 288k/601 = 479 ft^2 of
solar aperture, eg a 16'x32'x16' shed with a parabolic reflector inside.
the reflector be a 4x concentrator with a focal line f = 4' south of the
base of the north wall at dawn, with equation y^2=4fx=16x, bouncing sun
down onto a 4'x32' horizontal target at the base of the north wall. the
focus moves closer to the north wall during the day until the reflector
begins to shade the base of the wall at a 45 degree sun elevation (at
noon on april 4 and september 28, at 40 n latitude.) you would still have
lots of hot water in summertime.
with a 4' strip near the north wall dedicated to solar collection and 12x32
= 384 ft^2 of clear floorspace under the reflector, the shed could serve
some other purpose as well, eg a run-in shed for horses or a place to park
cars out of the rain. for another $1k or so, the whole cube could be glazed
to make it warm inside in the winter for use as a greenhouse or workplace.
the target might have a 0.020" clear polycarbonate cover and plastic lumber
2x4 sides and thermosyphon hot water through a 4'x8'x32' 8,192 gallon 32 ton
tank above supported by 28 2x4 studs on 2' centers, with a pressure-treated
2x4 foundation and an 8' wall at the focus with 2x4s on 4' centers to help
support the tank, with a layer of polycarbonate glazing over the wall. that
makes 384 ft^2 of polycarbonate, ie $576 at $1.50/ft^2.
we might use 1024 ft^2 of osb for the roof and north wall and 832 ft^2
more for the reflector and target bottom and 640 ft^2 more for the tank
sides and bottom. that's 2,496 ft^2, or $468 at $6 per 4x8 sheet. gluing
768 ft^2 of nielsen's reflective mylar ($0.09/ft^2 in 4'-wide rolls from
http://www.snomo.com/mylar.html) to the underside of the reflector adds
$69.12 to the materials cost ($1113.12 so far :-)
we could line the tank with a single folded $288 20'x48' piece of epdm
rubber, and cover the roof with another $154 16'x24' piece... 640 ft^2
of 3.5" fiberglass tank insulation might cost another $160. this makes
rc = 4'x8'x32'x64btu/ft^3-fxr11/704ft^2 = 1024 hours or 43 days, which
may allow storing some november solar heat for december.
we might approximate the parabolic shape with 17x24' = 408 ft of 2x4s
on 2' centers with a dozen sawkerfs in each bow. the roof might need
272 ft more 2x4s, with another 640 ft for the sides. that's 1320 ft,
$330 at 25 cents per foot. the total materials cost above is $2045.12.
if the tank water is 180 f and you want to keep the target temp 190 f
max and 93.3k btu/h of heat power falls on the target and you ignore the
heat loss to the warm air in its enclosure, you need a thermosyphoning
water flow q ft^3/s, where 3600s/hx64btu/f-ft^3q(10f) = 93.3k, so q
= 0.0405 ft^3/s, about 19 gpm.
water weighs about 63.74 - 0.0158t lb/ft^3, with t in degrees f. the
difference in density caused by the average temperature difference
between up and down pipes causes a pressure difference proportional
to the height of the water column, which makes water flow through the
resistance of the pipe loop. with 16' of height and a 10 f temperature
difference, dp = 2.53dt lb/ft^2.
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. mu is
the viscosity, about 7.26x10^-6 lb-s/ft^2 for 180 f water. we might
figure the water flows through 24' of pipe, 16' up and 8' down, which
makes q = 0.0405 = pir^4(2.53)/(8x7.26x10^-6x24'), so r = 0.0516' or
0.62", eg a 1.5" pipe.
if a total of 0.81x1090x16'x32' = 452k btu of sun enters the target
enclosure from the south wall on an average day and 0.81x475x4/3x16'x16'
= 131k enters the east and west wall area that is beneath the reflector,
a total of 583k btu enters the enclosure, including about 16'x32'x601
= 307.7k btu of beam sun.
removing this beam sun energy from the daily total, say 583k-307.7k
= 275.3k btu enters the target enclosure over 6 hours at 45.9k btu/h,
which has a thermal conductance to outdoors of about 299 ft^2/r1. if
307.7k/3 = 102.6k btu/h hits the target for 3 hours when the target
has an average temp t and 1.5x4'x32' = 192 btu/h-f of conductance to
the air in the cube which has temperature tc, we have something like
this (in courier font):
35.7 f ----www-----------------tc -------www-----tc
1/299 | | 1/299
| --- 189 f
----- | equivalent to -
|------| --> |-------| |
----- | _
45.9k/h <
< 1/192 . . . . . . . . .
<
----- |
|------| --> |-------*-------t = 190 f
----- |
102.6k/h ----- 65,536 btu/f
-----
|
_
removing the 1/192 resistor makes tc = 189 f, the thevenin equivalent
temp for that upper part. replacing the upper part with its equivalent
and replacing the 1/192 resistor, (t-189)/(1/299+1/192) flows from the
target into the air inside the enclosure, making the air temperature
tc = t + (t-189)/1+192/299) = 0.391t + 115, with no gain or loss from the
target to its enclosure if t = 189 f, ie 100% beam collection efficiency
at $2045/(102.6k/3.41) = 6.8 cents per peak watt.
you might increase the thermal collection efficiency and make electricity
at the same time by laying 4'x32' of pvs on the target floor underwater
and collecting 130 f water with 4x solar concentration.
nick
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