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I'll dive right into it.

A project of mine involves a glider in a cold climate. Past the point of gliders not being efficient in cold climates in the first place, I want to keep the batteries in the glider warm, and decided that heating them up directly would be one way to tackle the issue.

I want to heat the batteries, 18650 Li Ion rated at 7.4V 2600 mAh, to 25 C (ideally) for around three hours. Is it possible to do this without drawing more than 500 mAh? I would prefer to build the "heater" and do the soldering myself but if there's a product that efficiently does this then I wouldnt be opposed to just outright buying that as well.

The actual question, or TL;DR:

1.) Is it possible to heat an aluminum surface, 3 in^2 to 25C for three hours while drawing less than 500 mAh?

2.) if so, what kind of resistors/any other items would I need?

I would of course have an insualted enclousure for the batteries to sit in with the aluminum surface. I just need the surface itself to reach 25C.

PS:

I saw this video as well: https://www.youtube.com/watch?v=xIg2R9327bg

It basically works by short circuiting the battery, and draws 1 A. Is there any way to lower the amp draw? What would I need to add to the circuit? Also, is it wiser to straight up use resistors or to use nicrome wire to heat it up? Weight is also an important factor...

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    \$\begingroup\$ It doesn't directly answer your question: but a much more reasonable way of doing this is chemical heat packs. You can insulate the battery and stuff a wadded up hand-warmer in with the insulation. \$\endgroup\$ – Bryan Boettcher Jul 19 '17 at 21:29
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    \$\begingroup\$ You say raise the temperature or maintain the temperature at 25C but what is the ambient temperature? Is there/can there be insulation? Most of the math involves delta T and thermal resistance so these are basic data points. And it isn't a question of getting a single surface to 25C, it is a volume and mass to 25C. \$\endgroup\$ – Glenn W9IQ Jul 19 '17 at 21:38
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    \$\begingroup\$ General comment: temperature is not a function of power. A perfectly insulated enclosure can heat up to any desired temperature with any arbitrarily low power source of heat (given time). So you need to look at quantities like the specific heat capacity of the air and materials and energy loss through the insulation in order to work out how much energy you need. The energy is joules. The power is joules/second. Work from there. \$\endgroup\$ – Ian Bland Jul 19 '17 at 21:48
  • \$\begingroup\$ Best case is preheat to room temp and with constant P and known thermal resistance of cells Rja design a thermal resistance to regulate T with a low power electric vent as needed. i.e. use the I^2ESR of the pack to keep it at room temp when below 25'C If you know cell T rises say 25'C and ambient is -50'C then thermal resistance needs to be increase 300%.. Rja , P loss and delta T are variables. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 19 '17 at 21:50
  • \$\begingroup\$ What you are going to want to do is put a resistor close to the batteries and wrap the whole thing in insulation. Use some kind of control to turn the resistor on and off depending on temperature. The control could also be inside the insulation with the resistor. \$\endgroup\$ – mkeith Jul 19 '17 at 21:56
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The specific heat of an 18650 LI Ion battery is 0.83 J/gK. Weight is 46.5 g. To convert mAh to Joules, you need to know the voltage. If you are working at 3.6 volts, 500 mAh = 1.8 watt-hr. Since a joule is 1 watt-second, multiply by 3600 to get 6480 joules. You need 0.83 * 46.5 or 38.6 J for each cell to raise the temperature 1 degree K (or C). So 6480 J would increase a single cell's temperature by 167°C in a perfectly insulated environment. The addition of an aluminum surface just adds specific heat, so you just need a little foil to prevent a hot spot. Just insulate the batteries well and put in whatever type of resistive load makes your batteries happy. Efficiency will be a function of the insulation; essentially all of the energy will go into heat. Of course if you use a small load it will take longer and your imperfect insulator will cause a loss of some heat, so you don't want too small of a load. Also don't let the batteries get too hot.

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Before leaping into a solution for maintaining battery temperature, it is valuable to examine the effects of temperature on the energy available from the battery. The OP did not name a specific battery so for this example, I will use a battery made up of two Panasonic NCR18650b cells. This is a 3400 mAh, 3.7 V nominal cell per the OP's specifications. The OP also did not specify an ambient temperature so I will use 0C for this analysis.

Here are the discharge characteristics based on temperature:

enter image description here

Most RC electronics powered by a battery composed of two of these cells will probably not operate at a voltage of 2.5 volts per cell so I will restrict my comments to a cut off of 3.0 volts per cell or 6 volts for the battery.

Notice the gap in discharge capacity between 0C and 25C at 3 volts. It amounts to a 500 mAh discharge differential. It quickly becomes apparent that this is the budget available to maintain the battery temperature at 25C while expending power to achieve this goal. If more energy than this is expended to achieve the temperature goal, the effort is counterproductive.

So our maximum heating budget is 500 mAh over a 3 hour flight time or 167 mA. The nominal battery voltage is 7.4 volts which yields an maximum heating budget of 1.23 watts.

Assuming the battery pack starts at 25C, a first order analysis indicates it would take a polystyrene (0.033 W/mk) box with walls over an inch thick all the way around to meet this power budget. The box would be in excess of 3.5 x 4.5 x 2.75 inches. This assumes a moderate heat transfer coefficient (50 watts / m2C) given the flight conditions and no self heating from the discharging battery. Attempting to heat a polystyrene box of less than this thickness will result in a net loss of battery capacity.

Some advantage can be gained by initially heating the battery to its maximum allowed operating temperature just prior to TO. While the heat capacity of the cell is not specified, I would estimate that preheating the battery to 40C would save ~0.2 watts.

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All heaters are 100% efficient, in the sense that all electrical watts in come out as thermal watts. Just make sure you use the heat coming out the thing that's getting hot.

If the purpose of heating the battery is to increase its delivered capacity, then it is somewhat self defeating, as the power required for heating will reduce the capacity that can be delivered to a load.

The temperature of something is a function of the power going in, and the rate of heat loss from it. Minimise the heat loss, and you'll minimise the power to reach any given temperature.

Ideally, the battery would keep itself warm with the waste heat generated in the internal resistance of the battery at normal loads, let's call this 'free heating'. This will be small.

The best insulation currently economically available is PU foam, conductivity around 0.02 SI units. Rockwool, though not as insulating, is fire-proof, which may be worth thinking about for Li batteries on a plane.

Your plan should be
a) warm the battery in an oven back at base to its maximum safe operating temperature
b) put it in the insulating box, carry it to the plane

Arrange your box to be sufficiently insulating so that by the end of the flight, given your ambient temperature, together with the free heat generated by normal operation, the temperature has not dropped below your target minimum.

A box this insulating may cause the battery to overheat if there's an excess load, you would need fuse and thermal protection on the battery pack.

Water has a much higher specific thermal capacity than batteries. You may want to make a little hot-water-bottle for your cells, which would improve the thermal time constant significantly, if you can tolerate the excess weight.

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It's unwise to use the battery's internal resistance to warm it up if it's already cold, as the reason you don't want them cold is that they degrade in annoying ways if you discharge them when cold, so then discharging them to warm up is.... defeating the purpose. You can use a pulsed current drain to keep them warm once they are, or even rely on the static current draw of the system to keep them warm, but that does involve good knowledge of your cells and system and some extra maths or very good guessing.

Any "normal people wire" has resistance. I have to add the quoted bit these days to avoid people moaning about super conductors, but if you're flying at such low temperatures, hire experts to help you ;-).

In a professional project I'd use a well controlled heating element to generate the heat, but in personal hobby projects it has happened that I just took very thin wire, such as any AWG38 enamelled copper I found somewhere, and wound that around each cell. Until I got to tens of Ohms total, maybe hundreds, depending on the energy I expected to need, bits in series or parallel, all is possible. You can then power that with a small amount of energy to warm up the cells from the outside, a much lower amount of current than you'd need to pulse through the battery itself, also reducing your losses in the control of that heating scheme. But instead of thin wire you can use resistive wire, that depends on what's the easier process for you. Lots of thin, or several rotations of resistive. When you use lots of thin wire, do try to use something like enamelled wire, else avoiding shorts is difficult, because you don't want plastic, you want good contact. With resistive, be sure you don't turn up the power too high, as the plastic on the batteries could melt where there's bad contact and that could cause shorts with the metal casing. Or add thermal paste and heat-shrink around the finished object.

The next step is either controlling the element with an Op-Amp circuit or microcontroller, of which many can be found all over the internet. I'd suspect "low voltage heater control" might be a good Google term.

Or making sure the leakage of heat to the outside and heat generated constantly is such that the system naturally balances between 10 degrees and 30 degrees depending on outside temperature, where you just switch on the heater in cold weather.

In either case, if you want low energy usage it will include run of the mill insulation. The better the insulation, the lower the amount of energy you need to keep them hot. With infinite insulation you need no energy to keep them warm. But that'll be ... difficult.

I would say with (very) good insulation, assuming you are aiming for good operation at nothing below -20°C outside, you should be able to do with 1W or less to keep warm. Maybe you'd need an extra element to initially warm them up with a bit more power. But again, warming up is best done with a small current patiently, as when cold, larger currents aren't the best thing for the Li-Ion batteries.

For power requirements and all such, you'll need to do some research into the lower levels of what's called "thermodynamics". Insulating materials have a thermal resistance, which means at a certain temperature difference they leak a certain amount of energy. The more layers you use, the more resistances in series, so the less leakage. Look up Thermal conductivity and/or thermal resistivity, to find what wire resistance (electrical in this case) you need search for things like "Ohm's law" and "current, voltage and power". Power in Watts relates to Joules through time, by E = P * t, I'll give you that as a gift. Where E is energy in Joules, P is power in Watts and t is time in seconds.

Once you've learned a little about the above and found out what kinds of insulation you could employ and what kinds of wire (or internal battery resistance tricks, that's up to you!), you can always come back here or to Physics or such Stacks with new questions.

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