I want a deep-freeze to keep large amounts of frozen food in. But
they're expensive; the cheapest ones cost about $1000 (US$320), and a
125-liter one I wanted cost $1159. So I've been putting off buying
one.
Our refrigerator has been malfunctioning badly enough that we've been
leaving it off much of the time. (It sounds like the compressor motor
is about to go.) So we've been keeping PET bottles full of frozen
water in the freezer to keep it cold while it's off, and putting some
in the refrigerator compartment to keep it cool as well. This would
work better if the refrigerator were better insulated and didn't lose
all of its cold air every time we opened the door.
So maybe I can freeze some ice in our small freezer and put it in a
big insulated icebox with some frozen food, and just add new colder
ice occasionally to make up for heat loss. Obviously that's feasible
if the box is arbitrarily well insulated; how feasible is it with real
materials? When I was learning about thermal storage for indoor
climate control, I learned that thermal insulators like styrofoam and
fiberglass pretty much uniformly conduct heat at about 0.04 W/m/K.
(Straw is twice as bad; silica aerogel is twice as good.).
An Absurdly Demanding Requirement.
----------------------------------
So suppose you want a 2-liter bottle of salted ice to keep the
temperature at -5°C (if you salt ice enough it will melt at as low as
0°F = -18°C; I assume without justification that the associated change
in heat of fusion is small. The solution needs to be about 10% salt
for the freezing point to depress to -6°C) for 24 hours at a time.
That means the total heat flux out has to be 2 L * 80 kcal/L over
those 24 hours, which is 6.7 kcal/h, pretty fucking low. Suppose the
freezer is 250 L in a convenient cubical shape; that's a 63-cm cube,
which has .63 * .63 * 6 = 2.3 m² of surface area to lose heat through.
0.04 W/m/K * 2.3 m² = 0.09 m*W/K. Suppose, pessimistically, that the
outside air temperature is 35°C, so we have a 40 K difference across
the wall; now we have 3.7 m*W. A kcal/h is fairly close to a watt;
6.7 kcal/h is 7.8 watts. So I'd need 48 cm of insulation on every
side of this 63-cm cube; 48 cm would lose 3.7 m*W / (.47 m) = 7.7 W.
So that's a pretty annoyingly large box --- a 1.59-meter cube, which
is 4000 liters, of which all but 250 would be made of styrofoam. You
save a little bit if you round off the corners, maybe. But remember,
this is what's necessary for a 2-liter bottle of ice to keep 250
liters cold --- more than 100 times as much.
A More Reasonable Requirement
-----------------------------
Our small freezer in the top of the fridge currently contains 16
liters of water in 10 bottles of various sizes. It's nearly full, so
if we were putting ice from it into the icebox once a day, we couldn't
transfer much more ice than this 16 liters, per day.
The fridge's compressor is 140 watts, so it can probably pump about
280 watts of heat. Maybe half of this gets lost just from normal use
of the fridge, so only 140 watts or so is available for making ice to
cool the icebox.
So these are two wattage limits on the lossage of the icebox. Which
is more stringent?
16 liters of ice melted per day is 16 L * 80 kcal/L/day = 1280
kcal/day, which is 62 watts. So that's the more stringent
requirement.
If the inside of the icebox is the same size and shape, we have our
same 3.7 m*W figure as before. If we divide that by 62 watts, we get
6.0 cm.
So if I make a styrofoam box that's 6 cm thick, I can use salted ice
from the refrigerator's freezer to keep it at -5°C if I replace the
ice once a day. The whole box will be 420 L in size, 75 cm on a side,
and be made of 170 L of styrofoam.
(Styrofoam is actually a bit better than this; I've seen numbers of
0.033 W/m/K.)
Materials, Strength, Safety
---------------------------
I'm thinking I could build this thing out of discarded styrofoam cut
into standard-size "bricks" or "tiles" and glued together. It's
probably important to build the thing so that there aren't any
non-styrofoam paths all the way from the inside to the outside ---
that would be a "thermal bridge", and the heat conducted through that
path could dwarf the heat conducted through the neighboring styrofoam.
I think I can probably get OK thermal performance by the following
approach:
- make the thing out of three or more layers of tiles, and stagger the
joints between layers so that no joint makes an uninterrupted path
from from the inside to the outside;
- glue the layers together, but don't put glue in the joints between
tiles in the same layer; just let air fill whatever space is left
open by my imprecision.
- make special non-flat "corner tiles" for the corners, like angle
iron, so that there aren't joints actually at the corners.
- cut the edges of the tiles to size using a sharp knife rather than a
hot wire to avoid any increase in density.
This approach inevitably leaves the insulating material permeable to
water, since the cracks between tiles inevitably connect with the
cracks between the tiles in the next layer, and water (e.g. from
condensation) could thus flow from one layer to the next. If water
were to saturate the cracks like this, the path between the inside and
outside wouldn't be particularly short, nor would the cross-sectional
area of the water path be particularly large, nor could the water
convect efficiently through these narrow channels, particularly since
it would be frozen toward the inside; but water's thermal conductivity
is about 0.58 W/m/K, more than fifteen times that of styrofoam, so it
pays to be careful.
If the whole box is sealed on the inside and out with some kind of
impermeable plastic liner, water won't be able to get in and cause
problems.
It's pretty important that the tiles all be the same thickness to
fairly high precision (<1mm?), but I'm not sure how to make that
happen. In wood construction, apparently you cut wood to a given
thickness using a "thickness planer", which is a huge expensive piece
of machinery. Maybe there's a less expensive approach available for
styrofoam.
The bottom will need to support most of the weight of the contents;
that means an average pressure of 63 cm of water (63 g/cm²), or about
0.90 psi, and probably commonly pressures of several times that, due
to e.g. bottles that don't have flat bottoms being dropped in.
Unevenness of pressure can be reduced to some extent by a rigid bottom
liner. I don't think the remaining pressure will be a problem; I took
a small sheet of expanded polystyrene foam sitting around the house
(from an ice cream delivery) and balanced a 2L bottle of water on its
2.55cm-diameter cap on top of the foam, and it left no detectable
dent. That's 2kg on 5.1cm², or 390 g/cm², 5.5psi. (Random web pages
suggest compressive strength of 10-30 psi, which is quite strong
enough.)
Sooner or later, no doubt, one of us will trip and fall on the thing,
or a big pile of food will fall over inside, and break it. If the
outer impermeable plastic liner mentioned earlier is strong enough to
survive this and contain the potential ensuing flood of meltwater, it
will be much less of a pain in the ass to deal with.
Inevitably as we open and close the thing, frost will form on the
inside, which eventually has to be removed. Probably the best way to
do this is to have a fairly strong removable inner liner that doesn't
stick very strongly to ice, and doesn't admit much air between it and
the nonremovable inner liner. Then we can remove it periodically,
like a garbage bag from a garbage can, and empty out the excess ice
chunks.
On weight: styrofoam's density is apparently normally specified in the
US in "pcf", pounds per cubic foot, and it ranges from about 0.9 pcf
to about 2.2 pcf. 170 liters at 1.3 pounds per cubic foot would be
3.5 kg. Probably the necessary plastic liners will weigh more.
How much would this cost if I were to buy the foam new? I ran across
some random web page with this ad:
EPS Insulation Products and Prices
SPECIAL SALE
2500 Sq. Ft., 2" Thick of Non Standard Sized Insulation Panels
$700.00 (While Supplies Last)
That works out to 11 800 liters of styrofoam for US$700, about
US$0.059 per liter, which would put the total cost of the styrofoam
for this project at US$10. The non-sale price in small units seems to
be only a little higher; 1" x 48" x 96" costs US$5.44, and that's 76
liters, US$0.072 per liter.
The Liner
---------
I folded an HDPE shopping bag in half nine times, to 512 layers thick,
and measured it at about 6mm thick, which means each layer is about 12
microns, about 0.5 mils. I think the liner needs to be considerably
more resistant to tearing than this shopping bag, so I tried a Ziploc
bag as well, which folded 7 times to 128 layers thick at about 5mm ---
about 39 microns per layer, or 1.5 mils. Probably 80 microns, or pi
mils, should be adequate. (Now I just have to find a cheap source for
plastic that thick.)
The inside of the box is 6 * .63 * .63 m² = 2.4 m², so 80 microns on
the inside of the box is 190 mL of polyethylene; the outside is
another 6 * .75 * .75 = 3.38 m², so another 270 mL, for 460 mL in all,
plus a small amount around the lip. LDPE is about 0.92 g/cm³, so 460
mL is about 423 g.
The Lid
-------
Somehow you have to be able to put things in and take them out, but
without constantly losing huge amounts of cold through the lid. And
the liner is supposed to prevent water or air from getting into the
styrofoam, so it has to be continuous between the inside and the
outside of the box, so it has to run across the lip of the lid.
How much do we lose through conduction through that liner?
Let's say the lid is the entire top side of the cube. We have
63 * 4 = 252 cm of circumference around the inside of the top of the
lid. Low-density polyethylene, the most likely material to make the
liner out of, conducts heat at about 0.33 W/m/K, about an order of
magnitude worse than styrofoam, according to some random web page. So
if our cross-sectional area is 250 cm * 80 microns * 2 (remember, the
lid is coated in a plastic liner too), we have a cross-sectional area
of about 4.0e-4 m², so we have 1.33e-4 m*W/K. Earlier I assumed 40 K
temperature difference from the inside to the outside, so we have
5.3e-3 m*W. The minimal distance across the lip is 6 cm, which would
give us 88 mW of heat loss, which is about thirty times too small to
worry about.
So, unless I fucked up my figures, the main thing to worry about is
that the lid fits well and doesn't have much airflow. It doesn't need
to have fancy interlocking shapes to lengthen the thermal path,
although some degree of interlockingness may be helpful to ensure that
the silly thing actually shuts properly.
Energy Cost of Ice
------------------
Suppose you wanted to sell a bunch of iceboxes like this to some
people who didn't have refrigerators at home. How much will they
spend on salted ice from the iceman?
They can give the iceman their melted bottles of salted ice each
morning, so there shouldn't be any significant material cost --- only
the energy cost. If it costs 60 watts of chilling to keep one of
these things cold, that will probably be about 30 watts of electricity
to run an icemaker somewhere nearby. That allows you to serve 30
households for under a kilowatt, which is 720 kWh per month ---
US$72.00 at prevailing US electricity rates, about AR$33 at prevailing
residential Argentine electricity rates, or about AR$330 at prevailing
business Argentine electricity rates. So if you can run the icemaker
in somebody's house, and they have an electric bill that they pay, it
will cost the customers about AR$1.10 a month, plus labor and profit,
which seems like a pretty reasonable cost. If you have to run it out
of a commercial property, it will cost the customers about AR$11.00
per month, which probably makes it completely infeasible in that form.
(Maybe if you tripled the thickness of the styrofoam.)
Materials Cost of Ice
---------------------
So if we have two 16kg "charges" of, say, 10% salted ice, we'll need
32 liters' worth of bottles (say, 11 3-L bottles), 32 liters of water,
and about 3.2 kg of salt. (I think. Is that 10% by molecules or by
mass? "For every mole of foreign particles dissolved in a kilogram of
water, the freezing point goes down by roughly 1.8°C", says
http://www.worsleyschool.net/science/files/saltandfreezing/ofwater.html,
down to about -21.1°C.) This should be pretty cheap.
