Heat pads to warm lithium. is the freely roaming (Dan) video missing something

Short version

I have had to relearn a lot of A level physics (50 years ago for me) to get this far . I may be wrong, I hope so. I have based by calculations on a video linked to earlier in this thread by Dan of freely roaming. He uses the same battery cells as Harry and Phil. In the third part he tested his heat pads in his camper at just above freezing. He tested how long it took to get his near frozen batteries to 7 C which he had decided was warm enough to start charging them. He did not test the temperature of his heat pads during the experiment he did not test the temperature at the top of his cells, they were hidden in his wooden enclosure box. I think he may have got some parts of his battery too hot for longevity whilst some parts were still too cold to charge . What do others think?

WHAT FOLLOWS IS COMPLICATED HEAT TRANSFER PHYSICS. DO NOT READ UNLESS YOU WILL BE GETTING DIY LITHIUM TO USE IN A SOMETIMES COLD CLIMATE. OR JUST LOVE MATHS!

I have been looking into heat movement formulae and other sources.

I found on google from several sources that the specific heat of a lifepo4 cell was 900 to 1100 j/kg/degree K. So if you want to know how long a certain weight of LiFePo4 cells will take to warm up from a cold temperature to a warmer. You can use the formula.

Time in seconds = specific heat as above x the required temperature difference between existing and a good charge temperature in degree C x the weight in kg divided by the watts.

Putting Dans video experiment with 69 watts and 21kg of cells into the formula . (He has the same size battery as Harry and one of Phil's). I took his example as 4 deg difference . Time = 1100 x 4 x 21/69 = 1339 seconds that is 22 minutes, which was about what he saw. This is not the end of it because his battery near the heat pads will be hotter than target, the temperature sensor got to target and the higher half of the battery will still be cold. So read on.

I also found a formula for heat transfer and what difference in temperature there is between pad and the coldest point. Q/t = KA dT/s

Q is the input energy in joules (a watt is a joule/second).

t is the time taken, so Q/t is the watts.

K is the coefficient of thermal conductivity for a lifepo4 cell

dT is the difference in temperature,

s is the distance from hot to cold in m.

A is the area of the cell side to be heated in m. squared.

I found this link

https://www.researchgate.net/public...on_models_for_characterization_and_simulation.

This says that when a lifepo4 battery is heated from within by self discharge it looses the heat mostly via the X axis that is out of the biggest face. The core of the battery is much slower by at least (100times) to pass heat than the Aluminium casing. Remember however the casing has a much thinner cross section.

It gives a resistance of the core out through the X axis of 0.8 Degree C/ watt. Convert that to the coefficient of thermal conductivity K and you get 1.62 W/m/degreeC. The link does not give resistances for the Y and Z directions it says they are higher.

So lets try to use Dans heating experiment to see if we can predict his battery temperature gradient.

First the Area. Dan used 4 different sized heat pads and a thin aluminium plate to try to spread the heat evenly. He used a silicon trivet with ribs to insulate to avoid some loss of heat downwards. The battery will absorb heat faster than the trivet but some heat will be lost let us assume 20% of his watts are lost downwards. Otherwise lets assume his plate spreads the heat. His battery base with 3 big cell spacers 0.174m x 0.3 = 0.0522 m sq.

Then the coefficient lets assume the x axis coefficient for the Z axis . Its the best I can find. The coefficient will be higher than that. 1.62 then.

Q/t the watts 80% of 69 is 55 watts

S the distance. The distance to the sensors about halfway up is 0.105m. To the highest point is 0.205m

dT I forget I think his experiment started at 3.7 and finished at 7 degree C dT would be 3.3 deg C

dT= Q/t x s all divided by K x A

= 55 x .105/1.62 x 0.0522

dT = 68 deg C. Starting ambient 3.7 deg C at the beginning his heat pads and aluminium plate warm up to 71.7 degree C. They then slowly produce a heat gradient that warms the cells until the sensors reach 7 C. The pads are now at 75C. Some regions of the cells will be nearly as hot. Meanwhile the temperature gradient has not started warming most of the top part of the cells.