Horsefly
Solar Wizard
I’ve looked / searched a bit and didn’t find anyone addressing this specifically, so I thought I would put down some thoughts and results of my computations and hopefully others will correct and/or contribute to a better picture.
We all know that LFP batteries need to be kept above 32°F/0°C for charging, and in fact they behave much better if kept closer to 50°F or more. There are numerous great threads here (and YouTube videos) of people building insulated boxes with heating pads. It got me thinking: Is there a way to know - or at least estimate - the specifics of what you NEED in a given installation?
That’s where my quest began a few days ago. I’m a retired engineer, so I started down the rabbit hole. The below is my initial take on calculating some requirements.
I think there are really two aspects to keeping cells warm in a battery box: Heating up the LFP pack, and the heat loss to the outside of the box.
Item 1: Heating an LFP pack
I realize there are lots of secondary items to consider here. For example: How much air / dead-space there is in the box around the pack. I’m happy to hear some thoughts about how to incorporate such items into a solution, but for now I’m going to concentrate on the heating of the pack.
After lots of Google searches, I found a source that said an LFP battery has a specific heat capacity of 1306 Joules / (kilogram x °C). (I can provide the source if someone really wants it, but it was a doctoral dissertation from the University of Waterloo) This is clearly a pretty inaccurate estimate, since any given battery is made up of varying proportions of several different materials. Each of those materials has a specific heat capacity, and the specific heat capacity of the cell depends on the proportions of each of the materials. The real number will be different for larger cells than it will be for smaller cells, and it will be different for aluminum case vs. plastic. However, for this rough estimate, lets use 1306. (Again - Please jump in with suggestions)
Now we can take the mass of the cells. For example, the aluminum-cased 280Ah cells weigh about 5.6kg each. A pack of 4 (for a 12.8V battery) would then weigh 4 x 5.6 = 22.4kg. Now if we multiply this weight by the specific heat capacity of 1306, we get:
22.4 x 1306 = 29,254.4 Joules per °C
A Watt is one Joule per second, so a Joule is one “Watt second”. To get to Watt-hours, we need to divide this number by 60 x 60 = 3600 seconds per hour:
29,254.4 / 3600 = 8.126 Wh per °C
So, ignoring all the inefficiencies that may be present, this means:
To heat one 4x280Ah pack one degree Celsius requires 8.126 watt-hours of energy
With no inefficiencies, you can raise the temperature of one pack from 32°F (0°C) to 50°F (10°C) in one hour if you use about 80W of heaters. You can use 40W of heaters and do the same in two hours. Because we know our box, heating pads, metal sheet etc. are not the perfect theoretical world, we could bump the number up by say 25% (?), so we could use 50W of heating pads, using around 4 amps continuously from our pack, until the thermostat turned them off in about two hours.
Thoughts?
Item 2: Heat loss from a battery box
We also know that the warm insides of our battery box will over time cool down to approach the temperature outside the box. How fast this happens depends on temperatures inside and outside the box, the material we use to build and insulate the box and how much it resists the transfer of heat, and the exposed area of this material. The degree something resists the transfer of heat is the “R-value” (or RSI-value in metric) that is used to rate insulation. All material has an R-value.
The imperial / American R-value has the units of
Square feet (area) x °F x hours / BTU
I find this much less handy to use than the metric RSI value, which has the units of
Square meters x °K / Watts (note that in this context °K is the same as °C)
Either one is really expressing how much a material resists the transfer of power (watts) between two different temps (°C) for a given area (square meters). You can convert an R value to an RSI value by dividing the R value by 5.678.
As an example, I’ve been planning out a battery box which is built from ¾” plywood on the outside, with the walls, top, and bottom lined with 2” XPS insulation. 2” XPS insulation has an R value of 10. Plywood has an R-value of about 1.25 per inch of thickness, so my ¾” plywood would have an R value of 0.75 x 1.25 = 0.9375. Combining the two layers of my box, the R value would be 10.9375. The RSI value would then be 10.9375 / 5.678 = 1.92629.
Now I need to add up the surface areas on the inside of the box, including the top, bottom, and all four sides (again, I’m ignoring some subtleties of the bottom losing less heat than the top, etc). Lets say my box had inside dimensions of 15” left-to-right x 12” front-to-back x 12” top-to-bottom. The total surface area would then be (2 x 12 x 12)(left and right side) + (2 x 12 x 15)(front and back) + (2 x 12 x 15)(top and bottom) for a total of about 1000 square inches.
1 inch is 0.0254 meters, so 1 sq inch = 0.0254 x 0.0254 = 0.00064516 square meters
Thus my 1000 sq inches of inside surface area is 0.64516 square meters.
Now let's assume that we want the inside of our box at 50°F (10°C), and outside the box it will be 32°F (0°C). The difference between the inside and the outside is then 10°C. Then it’s just algebra:
Rsi = sq meters x °C / Watts
Watts = sq meters x °C / Rsi
Watts = 0.64516 x 10 / 1.92629 = 3.3492
This means that a constant “leak” of 3.3492W of power is escaping the box. If the temperature difference is larger between the inside and outside of the box, the watts go up.
If I chose to use the cheaper, thinner 1” XPS (with an R value of 5 instead of 10), the watts escaping the box almost doubles.
Other Considerations?
My intent here was to define a way to at least estimate things. We don’t want to use heating pads that are too small, or the batteries will never really get / stay warm. On the other hand we don’t want them too big either. We should insulate our battery box, but how much?
I’ve admitted that what I’ve provided is theoretical, and in the real world there are other losses and inefficiencies. At the very least, what one builds should use these formulas only for a baseline, to which you add a fudge factor of +20% or +50% or ?
Hopefully I’ve got the formulas right, and hopefully they will help. What else can you guys think of to consider?
We all know that LFP batteries need to be kept above 32°F/0°C for charging, and in fact they behave much better if kept closer to 50°F or more. There are numerous great threads here (and YouTube videos) of people building insulated boxes with heating pads. It got me thinking: Is there a way to know - or at least estimate - the specifics of what you NEED in a given installation?
That’s where my quest began a few days ago. I’m a retired engineer, so I started down the rabbit hole. The below is my initial take on calculating some requirements.
I think there are really two aspects to keeping cells warm in a battery box: Heating up the LFP pack, and the heat loss to the outside of the box.
Item 1: Heating an LFP pack
I realize there are lots of secondary items to consider here. For example: How much air / dead-space there is in the box around the pack. I’m happy to hear some thoughts about how to incorporate such items into a solution, but for now I’m going to concentrate on the heating of the pack.
After lots of Google searches, I found a source that said an LFP battery has a specific heat capacity of 1306 Joules / (kilogram x °C). (I can provide the source if someone really wants it, but it was a doctoral dissertation from the University of Waterloo) This is clearly a pretty inaccurate estimate, since any given battery is made up of varying proportions of several different materials. Each of those materials has a specific heat capacity, and the specific heat capacity of the cell depends on the proportions of each of the materials. The real number will be different for larger cells than it will be for smaller cells, and it will be different for aluminum case vs. plastic. However, for this rough estimate, lets use 1306. (Again - Please jump in with suggestions)
Now we can take the mass of the cells. For example, the aluminum-cased 280Ah cells weigh about 5.6kg each. A pack of 4 (for a 12.8V battery) would then weigh 4 x 5.6 = 22.4kg. Now if we multiply this weight by the specific heat capacity of 1306, we get:
22.4 x 1306 = 29,254.4 Joules per °C
A Watt is one Joule per second, so a Joule is one “Watt second”. To get to Watt-hours, we need to divide this number by 60 x 60 = 3600 seconds per hour:
29,254.4 / 3600 = 8.126 Wh per °C
So, ignoring all the inefficiencies that may be present, this means:
To heat one 4x280Ah pack one degree Celsius requires 8.126 watt-hours of energy
With no inefficiencies, you can raise the temperature of one pack from 32°F (0°C) to 50°F (10°C) in one hour if you use about 80W of heaters. You can use 40W of heaters and do the same in two hours. Because we know our box, heating pads, metal sheet etc. are not the perfect theoretical world, we could bump the number up by say 25% (?), so we could use 50W of heating pads, using around 4 amps continuously from our pack, until the thermostat turned them off in about two hours.
Thoughts?
Item 2: Heat loss from a battery box
We also know that the warm insides of our battery box will over time cool down to approach the temperature outside the box. How fast this happens depends on temperatures inside and outside the box, the material we use to build and insulate the box and how much it resists the transfer of heat, and the exposed area of this material. The degree something resists the transfer of heat is the “R-value” (or RSI-value in metric) that is used to rate insulation. All material has an R-value.
The imperial / American R-value has the units of
Square feet (area) x °F x hours / BTU
I find this much less handy to use than the metric RSI value, which has the units of
Square meters x °K / Watts (note that in this context °K is the same as °C)
Either one is really expressing how much a material resists the transfer of power (watts) between two different temps (°C) for a given area (square meters). You can convert an R value to an RSI value by dividing the R value by 5.678.
As an example, I’ve been planning out a battery box which is built from ¾” plywood on the outside, with the walls, top, and bottom lined with 2” XPS insulation. 2” XPS insulation has an R value of 10. Plywood has an R-value of about 1.25 per inch of thickness, so my ¾” plywood would have an R value of 0.75 x 1.25 = 0.9375. Combining the two layers of my box, the R value would be 10.9375. The RSI value would then be 10.9375 / 5.678 = 1.92629.
Now I need to add up the surface areas on the inside of the box, including the top, bottom, and all four sides (again, I’m ignoring some subtleties of the bottom losing less heat than the top, etc). Lets say my box had inside dimensions of 15” left-to-right x 12” front-to-back x 12” top-to-bottom. The total surface area would then be (2 x 12 x 12)(left and right side) + (2 x 12 x 15)(front and back) + (2 x 12 x 15)(top and bottom) for a total of about 1000 square inches.
1 inch is 0.0254 meters, so 1 sq inch = 0.0254 x 0.0254 = 0.00064516 square meters
Thus my 1000 sq inches of inside surface area is 0.64516 square meters.
Now let's assume that we want the inside of our box at 50°F (10°C), and outside the box it will be 32°F (0°C). The difference between the inside and the outside is then 10°C. Then it’s just algebra:
Rsi = sq meters x °C / Watts
Watts = sq meters x °C / Rsi
Watts = 0.64516 x 10 / 1.92629 = 3.3492
This means that a constant “leak” of 3.3492W of power is escaping the box. If the temperature difference is larger between the inside and outside of the box, the watts go up.
If I chose to use the cheaper, thinner 1” XPS (with an R value of 5 instead of 10), the watts escaping the box almost doubles.
Other Considerations?
My intent here was to define a way to at least estimate things. We don’t want to use heating pads that are too small, or the batteries will never really get / stay warm. On the other hand we don’t want them too big either. We should insulate our battery box, but how much?
I’ve admitted that what I’ve provided is theoretical, and in the real world there are other losses and inefficiencies. At the very least, what one builds should use these formulas only for a baseline, to which you add a fudge factor of +20% or +50% or ?
Hopefully I’ve got the formulas right, and hopefully they will help. What else can you guys think of to consider?