another 12' non-cube SOLAR.KnowledgePublications.com

another 12' non-cube
16 mar 2001
nrel says a 1-axis ew concentrator can collect 729 btu/ft^2 over 3 hours
in phila on an average 30.4 f january day. with a 90% reflector and a 90%
absorptive target, we might collect 0.9x0.9x729x12'x12' = 85k btu/day with
a 12'x12'x16' tall open-sided box with a parabolic reflective north wall and
roof (y^2=12x, with y=0 on top of a 4' tall tank, with the upper reflector
edge 12' above that. the focus moves closer to the north wall as the sun
rises until the upper reflector edge shades the target north edge at
atn(14/12) = 49.4 degree elevation. still lots of summertime hot water...

we might reflect dawn sun down into a 3' wide x 12' long x 2' deep trench
with 48 ft^2 of reflective mylar sides. grafix plastics at (800) 447-2349
http://www.graphixplastics.com sells 54"x50'x0.002" rolls of one-surface
metalized mylar for $48.31, about 22 cents/ft^2, with a $100 minimum order.
(anyone want to split an order?) it has a 740k psi modulus of elasticity
up to 5% in the 54" direction and a tensile strength of 34k psi. dupont
says it loses less than 10% of its strength at 100 c and softens at 250 c. 

the target might be a mylar film sandwich stretched over a slightly curved
3'x12' piece of osb, 2' off the ground and 1' below the tank water surface,
with some welded-wire fence over the top of the mylar to resist the head.
tractor supply sells $39.99 4'x100' rolls this fence with a 2"x4" 0.081"
diameter galvanized steel mesh. with 50k psi strength, each wire can hold 
50kxpi0.081^2/4 = 258 pounds, so 1' of fencing with 6 wires on 2" centers
can support 1546 pounds. an alternative target might have clear mylar on
top with underwater pvs inside. 

a linear foot of 3'-wide collector containing 64 psf water (a 1' head)
needs to contain 1'x3'x64psf = 192 lb in the vertical direction, so each
side of the upper surface needs to support 96 pounds. if the middle bulges
d inches above the edges with tension t (in pounds), td/18 = 96, roughly,
so t = 1728/d < 1546 when d > 1.1".

we might screw 3'x12' of 1/2" osb to a 12' vertical 2x4 down the middle,
then screw 2 12' 1x3s flat to the edges of the other side, then stretch 2
3'x12' layers of mylar over that (with silicone caulk near the edges) and
around the back and screw them to the back with 2 more 1x3s, then wrap
the fence over the mylar and around the back of the osb and screw it on
with 2 more 1x3s, then add some 1x3 tension straps under the 2x4 to make
the osb bulge up 2" in the middle. the target could sit on a bench made
>from  6 flat stacks of 4 tires and 6 tire pairs standing up.

we might paint the upper aluminized mylar flat black or selectively
(does solec paint dissolve mylar? they have no idea, but say it can
make e = 0.24 and alpha = 0.9 on aluminum, and would like to sell it
for $80 per gallon, including hazardous shipping. anyone care to
split a gallon?)

to make a 6' wide x 8' long x 3' tall (id) 1152 gallon polyethylene-
film-lined tank under the reflector, we might put 10'x12' of black
plastic film on the ground, then put a layer of 18 2' diameter x 6"
tall tires around the perimeter, flat on the ground, filled with
sawdust or mulch or compost, and tied together with rope, with 2
more ropes that cross in the center.

then put 6 more layers of tires on top, tied together and filled
with sawdust, then pile about 1' of dry leaves, mulch, sawdust,
etc. on the black plastic film in the middle, then line the tank
with a $22 18'x24' piece of 6 mil poly film while filling it with
3' of water and tucking more sawdust down into the triangular gaps
where the film meets the tires.

then lay the 2' film edges flat on top of the tire perimeter and
put another layer of tires over that, tied together and filled
with sawdust, then string 7 galvanized wires from opposite tires
across the width and length of the tank and lay 10'x12' of plastic
film over the wires and lay 48 ft^3 of dry leaves in bags over the
film, inside the tires.

we might keep the tank full with a hose and an $8 stock tank float
valve, and empty it with a 12v rv pump with a built-in pressure
switch in the basement. 

r2 per inch for the leaves, etc., compacting to 6" under the tank,
make its thermal conductance about 180ft^2/r24 = u7.5, with
rc = 1152x8btu/f/7.5btu/h-f = 1229 hours, or 51 days.

the reflector needs (sqrt(4x^2+y^2)+y^2/(2x)ln((2x+sqrt(4x^2+y^2))/y)/2
= 17.75' long curved rafters, plus 2' more at the bottom. they might
be 4 white-painted kerfed 2x4 bows with 12' 2x4s at the top and
bottom edges. we could tie these edges to the bottom tires with a
pair of 2' ropes at the bottom and a pair of 16' ropes at the top,
with twitch sticks near the top, and bolt 2 pressure-treated vertical
10' 2x4s to the sides of the reflector between the point above
the 3' focal line and the ground as compression struts.

then we'd attach 216 ft^2 of mylar ($46) to an $11 sheet of uv greenhouse
polyethylene with axle grease (to make smoothing and replacement easier)
and stretch the poly it over the upper side of the bows and staple it to
the reflector sides with 1" vinyl batten tape. 

the target has about 1.5x3'x12' = 54 btu/h-f of thermal conductance to
still outdoor air. we might collect 28.3k btu/h of beam sun over 3 hours
in 190 f water and lose (190-30)54 = 8.6k btu/h, for an average hot water
production of 59.2k btu/day.

with 180 f tank water and a 190 f max target temp and 22.9k btu/h of
target heat, we need a thermosyphoning water flow q ft^3/s, where
3600s/hx64btu/f-ft^3q(10f) = 19.7k, so q = 0.0086 ft^3/s, about 4 gpm.  

water weighs about 63.74 - 0.0158t lb/ft^3, so a 1' height difference
and 10 f temp difference makes a pressure difference of 0.158 lb/ft^2.

we might have about 20' of pipe with radius r in feet running mostly
under the target in a thermosyphoning loop, which makes q = 0.00856
= pir^4(0.158)/(8mu20') ft^3/s for laminar flow, using a formula from
one of william shurcliff's books, where mu = 7.41x10^-6 lb-s/ft^2 for
180 f water, so r = 0.067' or 0.81", eg a 2" pipe (or larger, with
some fitting head losses), which makes the total materials cost -$54
(counting 180 tires at -$1.25 each), or -$54/(19.7k/3.41) = -1 cent
per peak watt.

at t degrees f, the tank loses heat at (t-30)7.5btu/h to the outdoors.
if hot water use removes heat at a constant 59.2k/24h = 2467 btu/h,
dt/dt = -((t-30)7.5+2467)/(1152x8) = -0.0008138t -0.2433 = -ct +d, and
d/c = -299 and t = -299+(180+299)exp(-0.0008138t) = 120 f when
t = 164 hours, after 6.9 cloudy 30 f days in january.

i've heard that a typical tire contains the fuel equivalent of about
4 gallons of oil. this structure could make an interesting fire, if
something goes wrong.

nick