|
|
re: $49.95 microwave ovens at home depot
2 nov 2001
wrote:
>gig, you are a clever fellow. why not build a solar concentrator that
>would heat up some soapstone bricks during peak sun hours, and use
>them as an "instant on" oven? i think of a box, heavily insulated on
>top and sides. the bottom would be double pane glass, and reflectors
>would aim the sun into the bottom of the box and stones inside.
sounds good so far. kinda like this:
the most serious mistake was making the outer container of the receiver
of plywood. we thought that the plywood would be sufficiently insulated
from the copper panel which was the receiver proper, that it would not
get too hot. the copper panel was separated from the plywood by 4" of
fiberglass insulation. nevertheless, the plywood caught fire and the unit
was completely destroyed. we suppose this is a success, of sorts...
from "a solar collector with no convection losses," (a downward-facing
receiver over a 4:1 concentrating parabolic mirror) written by
h. hinterberger and j. o'meara of fermilab, ca 1976
we might use one layer of glass on the bottom, above a 4:1 concentrating
mirror aimed north at a larger heliostat aimed south.
>an area above the stones would serve as the oven space. it would need a
>door of course, and some way of limiting the heat from the stones for
>temperature control.
sure. this could be a convection oven with a thermostat and a blower
designed to move 450 f air, like grainger's $70.95 68 cfm 50 watt 4c940
blower. the motor is out of the airpath. it can only take 104 f.
>i'll bet nick could figure out the reflector size and insulation
>needed and weight of soapstone required to cook a 3 lb. chicken and
>couple of potatoes in the evening.
let's use iron vs soapstone, say old engine blocks. steel has about the
same specific heat by volume as water. we might make it, say, 800 f on
an average day, and let it cool to 500 over 5 cloudy 30 f days, so rc
= -120h/(ln((500-30)/(800-30)) = 243 hours.
we might surround an l' steel cube with 6" of fiberglass insulation, so
c = 64l^3 btu/f and g = 1/r = 6l^2/r19 btu/h-f (assuming the bottom is
also insulated at night and on cloudy days.) rc = 64l(19)/6 = 243 makes
l = 1.2', about 1.7 ft^3 or 780 pounds of steel. this doesn't have to
be a cube. it might be 4 side-by-side engine blocks in a rectangular
arrangement with a parabolic trough reflector hinged on the south side
that drops down underneath to collect sun.
we might somehow oxidize or nickel coat the glass-enclosed face
to make a selective surface with 0.2 emissivity, so it only loses
0.2x0.1714x10^-8((460+800)^4-(460+30)^4)1.2 = 1013 btu/h for say,
3 hours on an average day (nrel says phila gets 660 btu/ft^2 of
beam sun on a south wall on an average january day.)
with the reflector closed, we lose about 24h(800-30)6(1.2^2)/19
= 8.4k btu/day, so the steel needs to collect about 11.4k btu/day.
if the heliostat and reflector and glass each lose 10% of the sun,
we need a 11.4k/(660x0.9x0.9x0.9) = 24 ft^2 heliostat. we might
use one of duane's 4'x8' versions.
nick
|
|
