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re: has anyone researched using solar for cement or pottery kilns?
3 jun 2000
anthony matonak   wrote:
>i've never really considered liquid lead a good storage medium for
>cooking. besides the possible health hazards if there were a leak of
>some kind where do you go about finding a pump to move liquid lead?

check out wave solder machines used for printed circuit electronics.

>> >steel on the other hand takes about 52.3 btu per degree f per cubic foot.
>certainly steel is pretty inexpensive these days.

our local recycling center tries to sell tin cans for $2 per ton, delivered.

>> >...let the temperature swing some 300 degrees from a low of 450 deg f
>> > a high of 750 deg f...

storing 16.5k btu per (487 lb) solid cubic foot (would they have handles?) 

>> >700 or 800 deg f should be easily produced with concentrating solar
>> >collectors like a parabolic dish, trough, or field of heliostats.

i'm picturing our outdoor oven in something like a pyramid shape, with
a field of heliostats to the north, and a secondary reflector inside
the base of the oven bouncing sun up onto the thermal mass, with the
pizzas above that. 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

>> >how much storage and how big a collector do we need to run a pizza oven?
>> let's say the oven holds 64 2' pizzas on 4 8x8' shelves, and we cook
>> 8h/20minx64 = 1536 pizzas per day at 450 f for 20 minutes and evaporate a
>> pound of water from each in your 8' cube with r40 insulation (a foot
>> of fiberglass) in a 70 f room. that's about 1.5 million btu to evaporate
>> water, plus the conductive thermal loss of 24h(450-70)6x8^2/r40 = 87k
>> btu/day (ignoring door openings.)

>i did a quick search on pizza ovens and a single oven, capable of 
>cooking a couple of pizzas at a time, have burners that put out 
>between 40,000 btu and 60,000 btu. an hour. multiply by the number 
>of ovens. i'd say the average pizza joint is probably at the two oven 
>level. say, for rough guesses, some 100,000 btu an hour or about 1
>therm an hour. it actually sounds to me like these things could be 
>designed more efficiently.

the ones i've seen seem to have less than 4 inches of insulation.

>so figure a 24 hour cooking day and we're up to 24 therms just to run
>the oven. add in 87k btu/day for thermal leakage from the storage mass 

that was leakage from the whole 8' cube.

>and we're up to 25 therms a day requirement. 

maybe 16, with more insulation.

>> >for arguments sake, say our storage unit is some 8 feet cube, with half
>> >of the internal volume composed of air for easy heat exchange...
>> maybe the bottom half, since hot air rises.

>i was thinking more along the lines of something like ball bearings
>or rough shaped chunks so as to increase surface to air inside the unit.

is it harder to get the heat into or out of the thermal mass? conductive
heat input favors solid steel, but holes and fast-moving air can help get
the heat out (and cook pizzas faster--are there convective pizza ovens?)

evaporating a pound of water from each of the 64 pizzas in 20 minutes
requires about 192k btu/h, about the same as moving 16 therms of solar heat
into the mass over 8 hours. an 8x8' surface supplying 192k btu/h through
a slowly-moving air film conductance of 1.5 btu/h-f-ft^2 would need to be
192k/(8x8x1.5) = 2,000 f degrees hotter than the air. radiation helps too. 

some copper or high temp heat pipes might help conductive heat input, but
that still seems difficult. a foot of steel conducts 26 btu/h-ft^2-f,
while copper conducts 227, with about the same heat storage per ft^3.
putting 200k btu/h into 4' of steel through a 1'x1' hot spot would seem
to make the spot 4x200k/26 = 30,769 f degrees hotter than the steel...

old iron radiators or engine blocks might be cheaper than ball bearings,
with more space for airflow, but then the volume goes up, the size goes
up, getting the heat in becomes harder, and the heat loss or the amount
of insulation goes up.

>then to circulate air using fans to transfer heat in and out.

hmmm. grainger sells a $300 2,000 cfm fan that will work at 311 f and
a $200 280 cfm power venter listed under ul standard 378 "for use on
appliances with flue gas temperatures not exceeding 600 f at the blower
inlet." maybe some kind of automobile turbocharger... 

>the idea being that a small electric circulating fan could be easily
>thermostatically controlled and the hot air stream could be plumbed
>into an existing gas appliance with minimal reworking.
interesting thought. bigger orifi... a cfm of low-temp air with a 1 f
temp difference moves about 1 btu/h of heat, so moving 200k btu/h with
a 100 f temp difference might requires about 200k/100 = 2,000 cfm, which
might move at 1000 fpm (11 mph) through a 2 ft^2 (19" diameter) hole,
using a large fan with a reasonable electrical power consumption.

>> >its volume would be some 512 cubic feet and would have 256 cubic feet of
>> >material (steel). 256 cubic feet, with 300 deg f swings, could store
>> >some (256 x 300 x 52.3) 4,016,640 btu.
>> nice. now how do we get the heat into that 63 tons of steel?

>with a 5000 sq foot array of mirrors acting as heliostats to focus
>on a collector which heats air, said air being blown into the 63 
>tons of steel with a small electric fan that doesn't mind operating
>in 800+ deg f environments. 


>i'm thinking perhaps a 1 foot in diameter air duct (or perhaps chimney)
>style metal, well insulated, attached at both ends into the storage stack
>with the thermal focus for the heliostats in the middle somewhere (with
>it's window, etc.) and some kind of insulated dampers that would close
>when the fan/collecting isn't happening. the whole collector being mounted
>directly above or to one side of those 63 tons of steel with as short a
>ducting run as possible. perhaps with the pizza oven built into one side
>of it.


>> >how big a collector to collect 40 therms in one day? figuring a
>> >parabolic reflector setup is maybe 50% efficient (wild guess) we could
>> >collect 500 watts per sq meter. say an average day of 5 sun hours would
>> >make (5 x 500) 2500 watt-hours a day per sq meter (2.5 kwh) or (x 3409)
>> >8523 btu a day. we'll need (4,000,000 / 8,523) 470 sq meters or an area
>> >21.6 meters on a side (71 feet), or over 5000 sq feet.


>...i think that about covers it. besides, with only a couple of ovens
>we'd probably not need all 40 therms a day, only 25 to run the oven
>and some extra for rainy days. unless we use gas on rainy days.

maybe on sunny days too.

>even at 100% efficient we're still talking thousands of sq feet. but
>if they could make a solar power station with a couple of acres of
>heliostats all focused on one tower i'd say it'd be possible to put
>something like this together. probably would cost a bundle though.
>might loose the occasional pigeon that flew too near the focus.

pizza 'n squab. i wonder what duane thinks about this.


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