re: design information
18 may 1998
brian ferris wrote:
>is it possible for a do-it-yourselfer to make a new roof out of panels.
i think so. the roof might face the equator, with a slope that equals the
latitude, for optimal overall yearly gain, or it might be more vertical,
if more winter gain is desired. it might be sheathed with a single layer
of corrugated polycarbonate plastic, like my 600 ft^2 roof, covered with
"dynaglas" or replex (800 726-5151) that costs about $1/ft^2 and comes in
4'x12' sheets, with a 10 year light transmission guarantee and an expected
mechanical lifetime of 20 years. with no plywood or shingles, it might be
less expensive than other roofs, including the labor to attach the 4x12'
sheets to rafters with a screw gun and hex-head screws.
>...have the panels serve as a roof also, just for hot water not pv?
you might think "transparent sheathing," with some way to collect solar
heat underneath, vs "panels." attach corrugated black aluminum sheets on
top of the rafters, under the polycarbonate, and trickle some water over
the sheets, as h. e. thomason did? rot the rafters, if water vapor escapes?
that seems to require good sealing. other concerns are leaks, freezing,
corrosion, pumping power, and heat loss by condensation on the inside
of the glazing.
>...i just wondered if they were actually the roof of the house without
>any air flow underneath if they would get too hot.
an opening at the soffit and some sort of vent that opens near the ridge
peak, when the rafter cavity gets too hot? polycarbonate can withstand
130 f indefinitely, although i'd guess it lasts longer if cooler. since
it is thin, i'd expect it to have a temperature equal to the average of
the air temperatures on each side.
you might put some sort of air-water heat exchanger near the peak, eg some
fin-tube pipe near the ridge, the kind used in baseboard radiators, with
some dark window screen inside the cavity to help absorb solar heat, and
fiberglass insulation under that. my roof rafter cavities are about 27" wide
x 20' long. each receives about 12k btu/h of peak sun. the us r1 sheathing
has a thermal conductance to outdoors of about 45 btu/h-f, so when outdoor
air is 30 degrees f and the average air temperature in the cavity is t
degrees, the heat loss to outdoors is (t-30)45 btu/h. if the average cavity
air temperature were 130 f, the heat loss to outdoors would be 4,500 btu/h,
leaving 7,500 btu/h of useful heat output, and the collection efficiency
would be 7.5k/12kx100 = 63%, using this linear model.
fin-tube pipe has a thermal conductance of about 5 btu/h-f-ft in still air,
and costs about $5/ft. how would that work as an air-water heat exchanger?
here's one equivalent thermal circuit, with cavity air temp t, water temp
tw, and air-water heat exchanger conductance u (btu/h-f):
|---| --> |-------*--x---www-----tw
----- | 1/u
12k btu/h |
temporarily disconnecting the air-water heat exchanger at point x
and using a (thevenin) equivalent circuit makes this simpler:
297 f----www------|------www-----tw = 130 f.
if tw = 130 f, t = 130 + (297-130)/(1/45+1/u)/u = 130 + 167/(1+u/45),
and the solar collection efficiency is (t-tw)u/12k = 167u/(1+u/45)/12k.
to make this 50% efficient, we need 167u/(1+u/45) = 0.5x12k, so u = 178,
like 35' of fin tube pipe between each rafter. ick. moving air would help,
ie venting the cavity to increase the air velocity and fin-tube pipe
conductance from 5 to about 5(2+v/2)/2 btu/h-f-ft, where v is in mph.
one might let some outdoor air flow from soffit to ridge continuously,
but that wastes heat...
if the warm air can be used to heat the house, it's thermally more
efficient to allow warm air to flow out of the top of the cavities
through a duct heat exchanger and blow it down to the house with a fan.
cooler house air could enter the cavities near the floor. in this case,
the peak solar current source (for the whole roof) becomes 600x0.9x311
= 168k btu/h, and the entire roof's thermal conductance is 600 btu/h,
so we have
|---| --> |-------*--x---www-----tw
----- | 1/u
168k btu/h |
and the thevenin equivalent temperature tt = 30 + 168k/600 = 310 f:
310 f----www------|------www-----tw = 130 f.
and the solar collection efficiency is (t-tw)u/168k = 180u/(1+u/600)/168k,
so 50% collection efficiency requires that 180u/(1+u/600) = 0.5x168k, and
u = 2,100 btu/h, something like 2 or 3 auto radiators with fans... hmmm.
with a single duct heat exchanger with a conductance of 800 btu/h-f, the
solar collection efficiency = 180(800)/(1+800/600)/168k = 37% (although the
"waste heat" heats the house), and the hydronic output power in peak sun
may be 0.37x168k = 62k btu/hour, which should be enough for showers, etc.
t = 130 + 62k/800 = 208 f, theoretically, with no air movement. with a
2,000 cfm fan and no water heating, the air temperature rise in peak sun
would be about 168k/2k = 84 f. this looks like the right ball park.