Today is 2018-11-16. My house here in Buenos Aires has another power outage, an event which happens regularly during the summer months. They think power will be restored in some four hours, but this is uncertain. On one occasion a few years ago, I had a three-week power outage.
(Note: in fact, as I wrote this, the power came back and stayed on.)
The first thing I did was to check the refrigerator and put ice into it. So I was thinking about the sizes of the thermal stores that would be needed for the various temperature-control applications of electrical energy that I had been suddenly denied.
In preparation for such events, I have some soft-drink bottles filled with water and frozen in the freezer, to absorb heat through their enthalpy of fusion, and more soft-drink bottles filled with water and chilled in the refrigerator, to provide thermal mass. When there’s a power outage, I move ice bottles from the freezer into the fridge.
Unfortunately, my refrigerator is not a super-efficient model like the Sun Frost line; it’s a rather normal refrigerator with a compressor, a magnetic-strip gasket, and a few centimeters of foam between the inside and outside walls. I’ve scoured it in vain for information about its energy-efficiency, except that it claims its nominal power is 230 watts, and it uses 180 watts to defrost the freezer and 15 watts for the light. Let’s assume that 230 watts is the maximum power, but that it doesn’t run the compressor and the defroster at the same time, so the compressor uses 215 watts. And, although I haven’t measured this, the compressor needs to run somewhere around half the time. (This is normal. If your compressor needs to run 80% of the time or more to keep the fridge cold at normal ambient temperatures, you’re in danger of going over the temperature setpoint and spoiling your food if it warms up a bit outside, say from 20° to 24°, and so compressors are sized to prevent this. But if your compressor only needs to run 20% of the time, you could have used a compressor half the size, and the fridge would be a lot cheaper.)
So let’s say the compressor averages 110 watts at normal ambient temperatures, and let’s suppose it has a coefficient of performance of 2, which is a normal kind of coefficient of performance for refrigeration systems. That means it draws off 200 watts of heat from the food within, on average, to compensate for the heat leakage through the walls and the seals and the losses when the door is opened and all the cold air falls out. (The bottles in the fridge also serve a bit to reduce the air losses.) And that means that those leakages are about 220 watts.
The enthalpy of fusion of ice is about 333 kJ/kg, so 220 W is about 660 mg/s of melting ice, which works out to 57 kg/day (19 MJ/day). I don’t have 57 kg of ice in there. I have the following bottles:
| N | ℓ | total |
|---+-------+--------|
| 1 | 2.25 | 2.25 |
| 3 | 1.25 | 3.75 |
| 2 | 0.6 | 1.2 |
| 8 | 0.5 | 4. |
| 1 | .375 | 0.375 |
| | total | 11.575 |
#+TBLFM: $3=$1*$2::@7$3=vsum(@2$3..@6$3)
So I have about 11.6 kg of ice in there, which works out to about 4.9 hours. Whew, that’s not much! And I guess that’s why old-fashioned iceboxes worked on a basis of daily door-to-door ice delivery.
Wrapping the refrigerator in a blanket might help the situation. But another possible solution would be a coolth reservoir where I actually could store 50 kg or more of ice, frozen when there’s power and used for cooling when there’s not. It could be relatively small, and indeed to keep the power costs acceptable it would need to be relatively small and relatively well insulated.
At Burning Man, where you need to either burn or haul away any garbage you generate, down to the smallest fern leaf, we dehydrated our garbage by hanging it out in the dry wind in a net bag; this prevented it from rotting while we were there, improved its qualities as fire fuel, and reduced its weight enormously for the trip back. Though this is rarely done in the US, the same procedure is commonly used to dry clothes after washing them, though usually without the bag. And of course this is a common way to preserve sliced produce as well.
Putting air in contact with moist things causes the humidity in some surface layer of air to reach 100% (in the case of pure water) or whatever the equilibrium relative humidity of the solution is (when it has solutes such as salts). Water can diffuse across this layer into other air, or you can blow the surface layer away and replace it with new air, and in either case you remove water from the vicinity of the moist thing and increase the drying rate.
The amount of moisture required to reach 100% humidity rises exponentially with air temperature, doubling about every 10½°, so by raising the temperature of the air a few degrees, you can get very substantial increases in the drying rate. This is complicated by the fact that the evaporation itself cools down the air and the moist thing, and so your surface temperature is always a bit lower than the incoming air temperature; after a short while, the heat thus absorbed is necessarily replaced by the heat in the incoming air.
The principal difficulty with dehydrators is that they have to work relatively fast to be useful, and this requires the air to be both hot, which requires a lot of energy, and fast-moving, which requires a bit of energy but mostly careful airflow design. There’s the additional difficulty that, if you’re heating things up with fire or electricity, you need a thermostat, and ideally one that turns the heat off if it fails, rather than unexpectedly setting fire to things and melting things.
The enthalpy of vaporization of water is 2.26 MJ/kg, an enormous number, and it swamps the specific heat of air or vapor, so nearly all of the energy that goes into the dehydator is used to evaporate water.
Here, I’ve been using my electric oven to dehydrate garbage. The oven has a thermostat that goes down to 50°, and also a fan and a shutoff timer. So far, the thermostat has been dependable, but I’m not sure that it’s designed to fail safe; the shutoff timer fails when there’s a power outage (as I said before, a frequent event), and in those cases the oven turns back on when power is restored. This is unacceptable and so I cannot depend on the shutoff timer.
(Ideally a dehydrator would be driven by a humidistat and would shut off when the desired humidity was reached.)
Depending on what you’re dehydrating, different temperatures may be acceptable, and clothes dryers are designed with a relatively sensitive thermostat for precisely this reason — the high heat that most quickly dries cotton clothes would be disastrous for delicate synthetics. 35° is a fairly safe temperature for anything other than certain exotic materials you probably shouldn’t be playing with anyway. Raw-foodists require their food not to be heated above 42° for somewhat debatable nutritional reasons. 75° is generally considered adequate for sterilizing food; although there are a few exotic thermophile microbes that can survive higher temperatures, they are not active at 37° (the temperature they are at after you eat them) and so they are generally not pathogenic. Some common plastics, especially PET — the polyester used for both cloth and Coke bottles — soften and change shape above 90°, and in my oven I’ve observed dried egg whites browning somewhere in the 70°–90° range. Water, of course, boils at 100°. Wood starts to brown around that temperature, too, though the autoignition temperature of paper is 233° according to Bradbury.
Using the psychrometic exponential rule of thumb, even at 35°, you evaporate water almost 3× as fast as at 20°, and at 75° you evaporate it 38× as fast. Furthermore, the same air heated to 75° has 38× lower relative humidity, so if the relative humidity is 70% outside, it’s 2% in the dehydrator. This means not only that you can dry things faster, but also that you can dry things that wouldn’t dry at all in room-temperature air, because they have enough salt or other hygroscopic substance in them that they’re already in equilibrium.
Predicting the exact drying rate is very complicated, because it depends not only on the rather complicated diffusion and advection processes I mentioned above, but also on airflow patterns, specimen thickness, humidity diffusion rate through the specimen, salinity, and the presence of other hygroscopic substances. But usually, in my oven at its nominal uncalibrated 75°, my tens-of-millimeters-thick food garbage dries out satisfactorily in an hour or two.
How much energy does this require? 100 g of food garbage (orange peels, chicken bones, outer onion layers, food particles washed off dishes, and whatnot) might be a meal’s worth, assuming you don’t let food rot. Say that’s 75% water, so 75 g of water. That’s 170 kJ at 2.26 MJ/kg. Air’s fairly low specific heat of 1.01 kJ/kg/K means that putting 170 kJ into heating air from 20° to 75° means you need to heat about 3 kg of air to deliver that energy; at 1.2 kg/m³, that’s 2.5 m³ of air you need to pass over the meal’s worth of food, or 510 kJ or 7.5 m³ per person per day, 5.9 W. You might need to double that if some food goes bad. It’s still a factor of 20 to 40 smaller than the reservoir you need for the refrigerator.
In a situation of a prolonged outage of municipal services, dehydration is probably also a reasonable weapon to have in your arsenal against food decay, corpse decay, and feces, although if you’re going to be dehydrating feces with hot air you probably want to use a separate dehydrator from the food dehydrator, and you might also want to use some kind of scrubber on the output air to keep the smell down. (This in turn suggests you might want to use a closed-loop system like the one modern condensing clothes dryers use.) Dehydrated feces can be burned, stored, or more easily composted than fresh feces, especially diarrhea, and if you can sterilize them in the process, all the better.
My stove is, unfortunately, electric. Cooking a meal involves, minimally, heating the food to the right temperature and holding it at that temperature until it’s cooked. The temperature is invariably between 40° and 100°. Some meats can be cooked sous vide at temperatures as low as 60°, at least once they’re sterilized, but many vegetables require 90°. Kidney beans in particular are dangerously poisonous if cooked below about 95°.XXX
The traditional means for holding the food at the right temperature is to continuously apply heat to it, while it’s losing heat to the environment by another route, so that the thermal equilibrium is around the desired cooking temperature, but other approaches are possible; for example, you can use a thermos or hooikist to maintain the temperature passively, or a thermostat together with some insulation to maintain it more actively, as crockpots, sous vide cookers, and my oven do.
Typically a meal’s worth of food for a person is about 500 g, and it’s largely water, with a specific heat in the neighborhood of water’s. Heating it from 20° to 80° thus requires 20 kilocalories, 84 kJ, which is 2 minutes in a 700-watt microwave oven at full power. Thermal losses might push the energy required higher, perhaps to 150 kJ. Three meals a day then require 450 kJ per person per day, or 5.2 W, very close to the 510 kJ required to dehydrate the food garbage.
Due to deplorable construction techniques, a refrigerative air conditioner of 2000 watts or more is necessary in many Buenos Aires apartments to maintain the temperature at bearable levels. 2000 watts of input produces about 4000 watts of heat removal (CoP ≈ 2), but typical duty cycles are about ⅓ on a 24-hour basis, meaning that you only need to remove 1300 watts on average (115 MJ/day). Typical set points are in the 18°–26° range; 24° is about the most I can take.
Workers in very hot environments, such as some parts of power plants, often use ice vests, which either have ice packets in the vest or coolant tubes in the vest running to an ice backpack containing some phase-change material at or below 20°, typically water ice at 0°. This allows you to shunt the 100 W or so of heat that your body produces into the ice rather than finding a way to reject that heat into a hostile environment. If you are adequately insulated, that 100 W, or up to 2000 W if you’re exercising hard enough, is all the ice vest needs to absorb.
Usually ice vests are used for short stints, like an hour or two, but an ice vest that could last 8–12 hours would be very useful in the Buenos Aires summer. An hour at 100 W is 360 kJ, or a bit over 1 kg of ice, so a 12-hour ice vest would require about 13 kg of ice, which would be an uncomfortably heavy backpack.
As an alternative to indoor climate control, an ice vest with easily interchangeable ice packs, or even that you could leave plugged into a flexible coolant line as you move around your apartment, could be lightweight, and it would require less energy than air conditioning, since it would only need to absorb the heat produced by your body, not the sunlight flooding in through the windows or the hot air seeping in under the door.
The building has a central electric hot-water system with a huge stainless steel tank, so my hot water didn’t actually go out during this power cut, but it would during a longer cut. My shower delivers about 250 mℓ/s of water at typically about 40°, but needs to be able to reach 45°; further temperature range is undesirable, as water above 45° can scald you before you can react. I typically shower for about half an hour a day, which is perhaps longer than most people, and works out to about 450 ℓ of fresh water that would have otherwise flowed unmolested past Buenos Aires into the Atlantic, where it turns brackish in the estuary of the Rio de La Plata.
450 ℓ of water at ΔT = 25 K is 47 MJ per person per day, or about 540 W.
So here are the controlled heat flows per person, with a per-day average. This is not the electrical energy, but the thermal energy.
| Use | W (avg) | MJ/day | target | |
| Air conditioning | 1300 | 115 | 20° | cooling |
| Refrigerator | 220 | 19 | 4° | cooling |
| Ice vests | 100 | 8.6 | 20° | cooling |
| Showering | 540 | 47 | 45° | heating |
| Cooking | 5.2 | 0.45 | 100° | heating |
| Dehydration | 5.9 | 0.51 | 75° | heating |
| total | 2171.1 | 190.56 | | |
#+TBLFM: @8$2..@8$3=vsum(@2..@7)
So the first thing to notice is that air conditioning actually accounts for 60% of the total, and the problem gets substantially easier if we can eliminate it, for example with ice vests or better building envelope design. The second thing to notice is that the high-temperature heat applications are very low-volume, while the high-volume heat application, showering, is very low-temperature.
What combination of thermal reservoir designs could we use to satisfy these needs? We can use thermal mass or phase-change materials, and we can recharge the reservoirs from different sources of heat and cold.
We can consider the hot reservoirs and cold reservoirs separately, because they are
Although ice vests could in theory use a higher-temperature phase-change material, you pretty much need water ice in even larger volumes for food refrigeration, and probably any other phase-change material would be both more massive and more expensive. So it’s probably best to just use ice for the phase-change cool reservoir. 28 MJ of ice is 84 kg of ice per day, of which 8.6 MJ or 26 kg is for your ice vest; a week’s store of ice would be 590 kg, almost a cubic meter. This is a substantial-sized home appliance, especially with the extra volume needed for cooling extra people.
The difficulty with water ice is that you need something below 0° to get it, unless perhaps you make clathrates or something at a slightly higher temperature. Although some very dry places do reach such temperatures during the night — and you might be able to do a bit better still with reflective optics and some evaporative cooling — here in Buenos Aires, this pretty much requires active refrigeration, either with a compressor or some other form of refrigeration, like a vortex tube or an ammonia-absorption refrigerator.
(An ammonia-absorption refrigerator powered by solar thermal energy is quite a reasonable possibility.)
But it’s quite common for the temperatures to reach cool and even chilly levels during the night; as I write this, for example, it’s 18° outside. If you could accumulate a large mass of 18° material during the night, for example by passing night air through it, you could use that for indoor climate control during the day, even if not directly for food refrigeration. But how much mass would you need?
Supposing that it’s in the form of water bottles at 18°, which you then allow to warm up to 23° as you pass air over them in order to sink the undesired daytime heat. To compare apples to apples, we’ll suppose we’re using this to cool your no-longer-ice vest rather than to air-condition your house, so we only need 8.6 MJ. But 8.6 MJ / ((23 - 18) calorie / g) = 410 kg instead of the 26 kg for the corresponding ice. (The other 58 kg were for cooling food down to 4°, for which the 18° water would be counterproductive.) Cooling the 410 kg of water back down at night involves passing at least 8.6 MJ / (5 K · 1.01 kJ/kg/K) = 1700 kg = 1400 m³ of 18° air over them. That’s 66 ℓ/s or 139 cfm if you do it in the 6 coolest hours, which is a quite feasible airflow rate, but one that will require forced air in any reasonably-sized system.
If you have a phase-change material that changes phase at a temperature between 18° and 23°, you could use that instead, and maybe you could get it down to 50 kg or something. Unfortunately, the only things I’m aware of that fits the bill is certain expensive grades of paraffin and some outrageously expensive metals.
The potential advantage of using a second cool reservoir at a more moderate temperature is that you can replenish it in relatively simple ways, like with a box fan running at night, rather than a compressor or vortex tube or something. But it’s probably not worthwhile, since you need the larger, colder cool reservoir for food preservation anyway.
On the other hand, the system should almost certainly use separate reservoirs for heat: one for the ≈1 MJ daily of cooking and dehydration heat at 100°, and another one for the ≈50 MJ daily of hot water at 45°. Both of these temperatures can be easily obtained from solar thermal energy; the equilibrium radiation temperature of “one sun” is 94°, which is to say that thin things on Earth’s surface only fail to reach 94° when they’re at right angles to sunlight because they have air blowing around to cool them off, and “two suns” — the intensity of sunlight you get from direct sun plus a flat mirror — gets you to 163°. Anything up to “ten suns”, which would get you to 379°, is considered a “low solar concentration ratio” and is easily achieved.
If we have to gather the 11.1 W of cooking and dehydration heat during 6 hours of sun exposure — a reasonably pessimistic figure — then we need 44 W during that time, which is 0.044 m², a 210-mm-square area. In order to be able to cook and dehydrate at other times of day, we need to transfer the heat from the absorber (which is of a size somewhere less than 0.04 m² to achieve the required temperature, and somewhere larger than 0.004 m² so that the optics aren’t too complicated) to some kind of reservoir. Bricks, for example, or sand. The easiest way to move this heat around is with some kind of coolant, and the easiest one that has no trouble with temperatures from 20° to a bit over 100° is air.
If we add a bit of pad, let’s suppose that the reservoir itself is at 120° when full and stores 3 MJ of heat that are released when it cools down to 100°, at which point we consider it “exhausted”. Storing 3 MJ in 20 K of ΔT in a material with, say, 1 J/g/K, means that it needs about 150 kg of active mass, say about 1500 ceramic tiles of 100 g each, which will occupy on the order of 0.15 m³ of volume including air spaces between them. This will have on the order of 1.5 m² of surface area around it through which it can lose heat to the environment with its ΔT of about 90 K above ambient. We’d like it to lose no more than, say, 1 MJ during 18 hours when it’s not warming up, which is a rather demanding 15 W or so, 9 W/m², or 0.1 W/m²/K. It’s probably not safe to use organic insulators like polyisocyanurate, styrofoam (33 mW/m/K; see Deep freeze), or straw (90 mW/m/K) at these temperatures, but fiberglass should be fine. I don’t have fiberglass’s insulance handy, but let’s say it’s 50 mW/m/K. Then you need 500 mm of fiberglass around your thermal reservoir to reach such a demanding specification. This is somewhat unreasonable.
So let’s add a lot more pad. Let’s let our reservoir be at 200° when full and store 20 MJ of heat by cooling down to 100°. Now it needs to be 200 kg, which would be a cube 58 cm on a side at 1 g/cc, with a surface area of 2.1 m². Say we can afford to lose 16 MJ (leaving 4 MJ) during 18 hours, which is 250 W (better not keep it indoors in the summer) at a ΔT of 130 K, which works out to 120 W/m² and 0.9 W/m²/K. (50 mW/m/K) / (250 W / 130 K / 2.1 m²) = 55 mm of fiberglass insulation, which is eminently feasible, comparable to what my refrigerator uses.
You can’t, of course, support 200 kg of tile or whatever on 580×580 mm of fiberglass and maintain the fiberglass’s insulance. But you can suspend the mass in a frame in the middle of the fiberglass, hanging it by metal wires. 1mm-diameter music wire would in theory be enough, but probably more like 8 such wires would be advisable to keep the suspended frame from shifting around and opening gaps in the fiberglass.
If you can use a phase-change material with a transition temperature between 100° and 200°, you might be able to simultaneously reduce the mass, and the volume, the working temperature of the reservoir, the variability of the temperature, and the surface area through which heat is lost. Alkali nitrates and typemetal occur to me as possibilities.
The absorber that heats up the air to heat the reservoir
Some electrical utilities, like PG&E in California, are introducing time-of-use metering, where they charge you more for electricity used during "peak" hours than during "off-peak" hours. While this is still a long shot from true demand response, it should create a market for appliances such as refrigerators
https://enphase.com/en-us/products-and-services/storage