It’s an interesting question and not one I had even considered. Prior knowledge suggests that it’s at least somewhat plausible to thermally manage an ebike battery, seeing as electric automobiles do something similar in freezing weather. However, the large packs of automobiles are unfavorable when it comes to passive heat loss, being made of a lot of metal (which has a higher thermal capacity than air) and having lots of surface area compared to volume. So the heat generated during cell discharge is being wicked away by a larger area. So an ebike battery should stand a better chance, I would think.

So I started crunching some numbers. From a different comment, you suggested a use-case of 3+ hrs in -15 C weather. For the purpose of this exercise, I considered this battery pack, because it seemed fairly representative of a common ebike battery. This is the type of pack which slides into an open slot, exposed on basically all sides to weather. I thought about considering frame-integral packs, but there’s the issue of modeling them generally, and more importantly, I think the frame would become a substantial heat sink in -15 C conditions, since there’s no way to add insulation between a frame-integrated pack and the frame itself. My working assumption is that this specimen battery pack can be wrapped in insulation on all sides, for the most optimistic results. Also, I assume the battery pack is room temperature prior to starting the sub-freezing journey.

This pack is configured as 13s5p, so we know there are 65 cells inside. I think it’s reasonable to assume they are the common 18650 cells. I found this paper regarding a model for heat generation for one particular 18650 cell during discharge, which neatly included the data they collected. As we would expect, the heat output of a single cell depends on the discharge rate, and they recorded data at 0.5C, 1C, and 1.5C (Table 6).

Since the use-case will test the endurance of an ebike battery pack to complete discharge over at least 3 hours, I will use the lowest discharge rate, since a 0.5C rate means every cell would be depleted in 2 hours. Once again, this will give us optimistic results for a 3 hr use-case.

The plot from Table 6 at 0.5C remains substantially at 25000 W/m^3 for the first 80% of time, and then quadruples by the end of the discharge. Since we are assessing the steady-state heat of the battery, we will not consider the increased tail-end heating. A single 18650 cell has a volume of 16 mL, so the entire pack has 1040 mL (0.00104 m^3) worth of heat-generating battery. Using the figure from the literature, this pack has an overall discharge heat of 26 W.

If a single hand warmer produced 26 W continuous, that would make for a very toasty hand. But we need to consider the heat loss for the battery pack’s casing, to see whether a substantial fraction remains to keep the cells warm. To do that, we can observe that this specimen pack is mostly rectilinear, so we can approximate it as a simple 6-sided shape, with dimensions 390x110x75 mm. Crunching the numbers, we arrive at a total exposed surface area of 0.1604 m^2.

For this exercise, we would have to know how much heat the casing can exchange with the environment. But since we’re trying to find an optimistic value, we can instead consider an impractically well-insulated scenario where the cells are enclosed in a thick casing of polystyrene on all sides. Specifically, I am using a value of 25 mm thickness of polystyrene, thus RSI 0.97, aka R-value 5.6. The RSI value has the units m^2*K/W. We can arrange an equation to yield K (the temperature difference in deg C) as follows:

K = (RSI * Watts)/area-in-sq-m

We have all the numbers, so we just plug them in, yielding a temperature difference of +157 C. So at -15 C ambient temperature, this pack would be at 142 C, or basically on fire. 🔥

Ok, that’s obviously way too much insulation, and while shocking, I shouldn’t be too surprised, since even 1 W continuous is enough for a wall-wart to be warm to the touch. Inside a sealed polystyrene compartment, I would also expect a wall wart to rise to a rather high steady state temperature.

What we’ve learned is that it’s actually decently plausible, although every figure here is optimistic. The cell heat output is going to be lower at lower discharge below 0.5C, possibly being half or 1/3 as much. That could sink the heat output to only 9 W. And real world insulation would likely be something like 6 mm of neoprene, so R-value of 1 (RSI 0.17). These more pessimistic numbers bring the temperature difference down to just +10 C delta, so in -15 C ambient, the pack would still be below freezing.

But I think what’s instructive here is that the cell heat can indeed be sufficient, and the insulation can be suitably thick. Other issues will be whether the insulation also absorbs heat from the sun during riding, and what should happen if the pack overheats under all that insulation whilst in slightly warmer weather.