Subject: Solar oven improvements?

From: [email-masked] (Nick Pine)
Date: 3 Oct 1996 13:56:51 -0400

Marge Wood <[email-masked]> forwards from Barbara Kerr:

>It gets about 250 degrees F. by 9:30 a.m. rises to around
>300F at noon and slowly drops until by 4 p.m cooking is over...
>Bread takes 2 hours.  A chicken or a beef roast takes 2 to 3 hours.

How long does bread usually take in an oven? A half-hour? How about
a 20 lb turkey, or 5 gallons of rice or beans or potatoes?
Sounds like higher temperatures might be welcome. 

Or a higher thermal conductivity between the air and the food, as in a
convection oven? The $39 DecoSonic circular glass one, with a mechanical
timer and thermostat and electric heater and fan, dries and cooks some
things very quickly, especially in small pieces, compared to a regular oven.
Warm air-food conductivity might be about 2+V/2 Btu/hr-ft^2 F, where V is
in mph, or 10+V W/m^2-K, where V is in m/s. A small PV powered circulation
fan in the oven might increase V from 0 to 8 mph and reduce the time to
heat up a pint of soup by 3:1, were the pot not exposed to the sun, altho
the the pot lids are usually transparent, with some sun shining in... 
 
>If I am going to be away from home, I may put the pots in before 8 a.m.
>and let them wait for the oven to heat up. And since the oven turns
>itself off, things do not have to be taken out promptly.

Storing some heat in the oven might be an advantage, if the pots did not
have to wait, and a disadvantage if the oven did not turn itself off.
But since solar energy is free, the oven might have a timer. Most ovens do.
A hanging metal cup weight filled with water, inside the oven, that holds
down a vent flap with a spring at the top, that opens when food is done?
A flap in the bottom that opens to let heat rise up into the oven, and 
somehow closes when the food is done?
 
>The outside portion of the Solar Wall Oven is comprised of the oven proper
>and the passageway ending in a frame to hold the oven securely in place.
>If the oven is very big, there may be cantilever props for additional support.

Sounds like the space under the oven is mostly empty. If the space below
needs props, how about putting some glazing under there, and heating the
oven from below too, or instead? Most people think of the sun as heating
things from above, or at an angle, but if we want to use the sun all year,
even in winter, when the sun is lower in the sky, and the days are shorter
and colder, it seems better to have lots of vertical or almost-vertical
glazing, since the highest altitude of the sun at noon on the shortest day
in winter is only 90-23.5-latitude, 26.5 degrees above the horizon at noon
on 12/21, where I live.

Also, with a low-thermal-mass "sunspace" below, and the thing to be heated
above it, warm air rises, so the thing above the house or oven or water tank
can be heated from below during the day, and if the thing is insulated above,
it can trap the heat at the top (like a cave or an igloo, with a floor above
the entrance) and lose little heat at night. If there is glazing above, vs.
an insulated roof, most of the heat in the oven is likely to disappear at
night. Maybe the pots could have a head start in the morning...

>The insulated oven box is lined with galvanized metal, approx. 26 gauge,
>is shaped and spaced inside the insulation with 1/2 inch or so clearance
>from the insulation and 1/4 in.to 1/2 in. from the glazing. 

I can't quite imagine how this looks.
I wonder how thick the insulation is, and what kind it is. 
 
>With the south facing ovens I have experienced six to seven hours of
>temperatures above 250F daily when the sun is out.
 
I wonder how well this works in the winter. Or in cloudier places.
Could we store solar heat for a cloudy day in the oven?

>My current oven size slants from 15 to 12 inches high, 18 inches
>north to south and 30 inches wide.

Aha, numbers! :-) Let's see: the oven itself has a surface area of about
15(2(18+30)) = 1440 in^2 or 10 ft^2, and it might be at counter height, 30"
above the floor, with about 3' to the ground on the outside? and 30" wide,
another 3' or so. And 18" deep, with a volume underneath of about 10 ft^3,
and a potential vertical solar glazing area underneath of about 10 ft^2, or
more if the glass or polycarbonate glazing slopes, or it is wider near the
ground. Glazing the space underneath might collect another 10,000 Btu/day
of sun where I live, in December. 
 
>The exterior end of the passageway and the base for the oven are
>set out from the wall sufficently to allow noonday sun at mid-summer
>to strike the top of the oven and any reflector.

And in mid-winter, when the sun just slides in almost parallel to the lid?

>If this is not possible, place the oven out as far as convenient and
>compensate by the angle and size of a South reflector.

A south reflector on the ground, white paint, or a shallow frozen reflecting
pool, might reflect 60% of the sun that falls on the ground to the oven (as
well as keep weeds and lawn mowers away from the glass), vs the 20% ground 
reflection in NREL tables, so it might increase the solar input by 1.6/1.2
= 1.33, 33% more, 13K vs 10K Btu/day. Collecting more sun by reflection is
harder to do with overhead glazing.

How about a movable insulating reflective cover on top of the oven, to cover
the glazing on top at night, with an R-value of 10, say, like the walls of
the oven (?). Hinge the cover along the edge near the house wall, and run a
string into the house that raises the cover when you pull on the string, so 
the oven can gather more heat on a sunny day in December, with the cover
raised to about a 45 degree angle. Can we attach a small motor and pulley
to the string, to lower the cover at night, or when a cloud passes over,
or when cooking is done? Sure.

How hot will the oven get on a December day, without much thermal mass inside?
Say we are cooking bread, with an RC time constant of less than an hour. RC
is product of the thermal resistance and capacitance. A loaf of bread with a
surface area of 1 ft^2 and a thermal resistance of 1/(2+8/2) = 1/6 (in a
convection oven) containing a pint of water might have an RC time constant
of 1/6 x 1 = 1/6 hour or 10 minutes. Let's assume the oven does not have to
evaporate a lot of water from the food as it cooks (this is very expensive
energy-wise, at 1000 Btu/pint of water.)

An average December day where I live near Philadelphia might be 43 F outside
for 6 hours, during which 1000 Btu/ft^2 of sun falls on a south wall, if
there is no ground reflector. About 30"x18"/144in^2/ft^2x 1000 Btu/ft^2 =
3750 Btu/day of sun might fall on top of this oven, (if the lower surface of
the cover reflected 100% of the sun that fell on it) of which 90% might enter
the oven, if it had polycarbonate glazing, ie 3375 Btu, and the oven with 
internal temperature T would lose heat through its walls and ceiling over
6 hours to the 43 F outside world (let's assume the kitchen is also 43 F in
December, like mine, making the calculation simpler and the refrigerator
more efficient :-) If the conductivity of the oven were, say R1 for the
4 ft^2 of upper glazing and R10 for the 6 ft^2 of walls and bottom, ie
a total of 4/1 + 6/10 = 4.6 Btu/hr-F, we would have

3375 = 6 hoursx(T-43)x4.6 = 27.6 (T-43), or T = 43 + 3375/27.6 = 165 F. 

Not very warm, on an average December day with an average amount of sun, 
so what do we do, just cook on sunnier winter days? Add an electric heater
and thermostat in the oven and cook with the top lid closed on cloudier days?
How big an electric heater? With the lid closed, we have a conductivity of
10 ft^2/R10 = 1 Btu/hr-F :-) So raising the oven to 343 F, say, takes
(343F-43F)x10 Btu/hr-F = 3000 Btu/hr or 3000/3.41 = 880 watts. A $20 hair
dryer or small electric room heater could do this, for about 10 cents/hr,
if the electrical insulation, etc, didn't melt in the heat... Or 9 100 watt 
light bulbs in the bottom of the oven. (But it's better to have more airflow
around the food.)

Double glaze the oven? This reduces the amount of sun, but it also lowers the 
heat loss. Suppose double glazing is R2. Then the glazing conductivity drops
>from  4ft^2/R1 to 4 ft^2/R2, and the total conductivity of the oven decreases
>from  to 2.6 vs 4.6 Btu/hr-F, and the amount of sun that enters the oven might
decrease by another 10% to 3000 Btu/day. So we have 

3000 = 6 hoursx(T-43)x2.6 = 15.6 (T-43), or T = 43 + 3000/15.6 = 235 F. 

Better...

Let's add another 10 ft^2 of double glazing underneath, with some sort of
solar absorber behind it, and let the hot air rise up or better, be blown up
with some sort of fan into the oven. The oven temp T would go up as the
total becomes 4 ft^2/R2 for the upper glazing, 2 ft^2/R10 for the walls,
nothing for the floor, since it is warmer underneath, and 10 ft^2/R2 for
the lower glazing, for a total of 2 + 0.2 + 5 = 7.2. Solar power increases
>from  3K Btu/day to 3K + 10 ft^2 x 1333 x 0.9 x 0.9 = 14K Btu/day. So

14K = 6 hrx(T-43)x7.2, and T = 43 + 14K/(6x7.2) = 367 F.

How about cloudy days? Can we somehow store heat in the volume underneath, 
in say, a collection of scrap iron, with insulation around it, and a small
fan to heat the iron with solar hot air during the day? Iron and steel store
about the same amount of heat as water per cubic foot, 64 Btu/degree F, but
at a higher temperature. And it can be cheap at a junkyard, a penny a pound,
maybe. We'd need to cut it or break it up into smallish pieces and put it
inside some sort of box, or weld it together into a chunk with lots of holes
for fast-moving air circulation near the outside, and bigger hunks inside, eg
engine blocks or ingots or a stack of civil war cannonballs in a public park.

Say we mount the oven at eye level, and the heat store is 4' tall x 4' wide
x 2' thick, on the average, so we can store 4x4x2x64 = 2048 Btu/F inside.
The iron might be completely surrounded by R10 fiberglass insulation, with
some sort of oxidixed iron (a somewhat selective surface) eg stucco mesh over
that, an air gap, and two layers of glazing. What would the average temp of
this box be, ignoring the oven? The average outdoor temperature in December
over a 24 hour day where I live is 36 F. At this point, the box would lose
heat day and night.

Suppose now that the kitchen is 70 F, and the house wall is R20, so the heat
loss on the house side is not large. The 8 ft^2 bottom might have 2" R10
Styrofoam insulation exposed to R10 soil, ie a conductivity of 8 ft^2/R20
= 0.4, and the top of the box might have 8 ft^2/R10 = 0.8 Btu/hr-F, with
16 ft^2/R20 for the north side, exposed to the 70 F kitchen, with a total
of 16 ft^2/R10 for the east and west sides, and 16 ft^2/R10 for the south
side, at night, and 16 ft^2/R2 during the day. The box might collect
16 ft^2 x 1333 Btu/ft^2/day x 0.9 x 0.9 = 17K Btu/day. 

So 17K = 6 hours (T-36) 16 ft^2/R2  during the day, for the south wall,
      + 18 hours (T-36) 16 ft^2/R10 at night, for the south wall,
      + 24 hours (T-36) 16 ft^2/R10 all for the east and west walls,
      + 24 hours (T-36) 8 ft^2/R10  all day, for the top,
      + 24 hours (T-70) 16 ft^2/R20 all day, for the north wall,
      + 24 hours (T-36) 8 ft^2/R20  all day, for the bottom.

This is easier than figuring out income taxes, right? Simplifying,

   17K = 6 (T-36) 8
      + 18 (T-36) 1.6
      + 24 (T-36) (1.6 + 0.8)
      + 24 (T-70) 0.8
      + 24 (T-36) 0.4, or

   17K = (T-36) (48 + 28.8 + 57.6) (T-70) 19.2 + (T-36) 9.6
       = (T-36) 144 + (T-70) 19.2
       = 144T - 5184 + 19.2T - 1344 = 163.2T - 6528, so...

T = (17K+6528)/163.2 = 144 F. Not very warm...

Maybe we need a small air-air heat pump here, not just a heat-exchanger, but
a low-velocity vortex tube or a little jet engine to cool the block and heat
the oven to a higher temperature as needed. Compress some air with a turbine,
use the warmer air to heat the oven bottom, and return it expanded and cooler
to the block, via another small windmill attached to the same shaft, that
partially powers the input fan. We've got a fan in there anyway, right?

PV = nRT, so to increase air temp from 150 F to 250 F, ie from 610 to 710
degrees Rankine, we need to compress it by 710/610, to 86% of its original 
volume? Or push another 16% of air into that volume? Then remove some heat
at 250 F, then expand the air again, efficiently, ie slowly, to its larger
volume, and let it emerge at normal pressure from our unusual heat pump,
cooler than when it went in. We might need 14.7/0.86, ie 2.4 psi more than
atmospheric pressure. Hmmm. 67" of water. The American Supercharger II
hot tub blower, 2P681 in the Grainger catalog, $193, weighing 8.7 pounds
in the brown trunks (or the yellow ABS flame-retardant housing?) At 35 cfm,
with a ONE HORSEPOWER, 110 V, 6.6A motor... We would not need the sun if
we used that. And we might only be moving 35 cfm x 100 F, ie at most about
3500 Btu/hr or just over 1 kW, for a COP of less than 2. We need to look more,
or find someone to design a more efficient turbine.

We might double some of the insulation instead, and build some fixed
reflecting walls extending 6' or so from the south corners of the box
at about a 45 degree angle to the east and west sides of the box, to
double the sun to 34K Btu/day, and decrease the losses, so

34K = 6 hours (T-36) 16 ft^2/R2  during the day, for the south wall,
   + 18 hours (T-36) 16 ft^2/R20 at night, for the south wall,
   + 24 hours (T-36) 16 ft^2/R20 all for the east and west walls,
   + 24 hours (T-36) 8 ft^2/R20  all day, for the top,
   + 24 hours (T-70) 16 ft^2/R20 all day, for the north wall,
   + 24 hours (T-36) 8 ft^2/R20  all day, for the bottom, ie

34K = 6 (T-36) 8
   + 18 (T-36) 0.8
   + 24 (T-36) (0.8 + 0.4) 
   + 24 (T-70) 0.8
   + 24 (T-36) 0.4, or

34K = (T-36) (48 + 14.4 + 28.8) (T-70) 19.2 + (T-36) 9.6
    = (T-36) 91.2 + (T-70) 19.2
    = 91.2T - 3283 + 19.2T - 1344 = 110.4T - 4627, so...

T = (34K+4627)/110.4 = 350 F, 24 hours a day. 

But on a sunny December day, with the top oven cover open, we would have
about 300 Btu/ft^2/hr shining into the oven, ie another 8 ft^2x300x0.9x0.9
= 2000 Btu/hr shining into the oven, (the oven now being now 4' wide x 
2' deep x 2' tall) which would be losing 8 ft^2/R2 Btu/hr of heat through
the top glazing to the 43 F average outdoors, so

2000 = (T-43) 8 ft^2/R2 for the top,
     + (T-43) 16 ft^2/R20 through the sides,
     + (T-70) 8 ft^2/R20 through the north wall,
     + (T-350) 8 ft^2/R20 through the bottom, ie

2000 = (T-43) 4.8 + (T-70) 0.4 + (T-350) 0.4 = 5.6T - 374.4, so

T = (2000+374.4)/5.6 = 424 F. Almost time for solar pizza :-)

Especially if the top cover has a couple of side reflectors...

How fast will the oven heat store cool off on a cloudy day? With 2048 Btu/F
of heat storage, and the oven all buttoned up, with the movable R10 cozy
covering the R10 oven floor, the heat store might lose

24 hours x ((350-36) x (16 ft^2/R20 for the south wall, + 
                        16 ft^2/R20 for the east and west walls +
                         8 ft^2/R20 for the top + 
                         8 ft^2/R20 for the bottom) +
            (350-70) x  16 ft^2/R20 for the north wall), ie
	    
24(314(0.8+0.8+0.4+0.4)+280x0.8) = 24(754+224) = 23,472 Btu/day.

So, on the first cloudy day, it would cool 23472/2048 = 12 F to 338 F.
On the second, it would cool a bit less, since it starts off cooler, etc.

This is oversimplified, and nature is less linear at these temperatures,
so we won't really know how well it works until somebody builds one... 

Nick