Determining heat loss and required insulation

  • Thread starter Scratchy
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  • #1
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Howdy everyone, I'm sorry if this is in the wrong forum, but since I believe it is mostly a thermodynamics problem I think this is the right arena. My questions stems from trying to insulate a bank of batteries in cold weather. I am trying to determine the proper material and design specifications for the enclosure I will be building and I've come to a bit of an impasse. Essentially I need to keep the batteries "warm" (greater than 4.5 degrees C) for at least 24 hours with a starting temperature inside the enclosure of around 24C (this can be modified) with an outdoor temperature of -20C. I have yet to take heat transfer, but I have borrowed a textbook and I believe I need to find the allowable R-value and then work backwards from there to determine the thermal conductivity required as well as allowable thickness of the material. I just don't know how to calculate that initial R-value that I will then use to determine if a material is suitable. I believe that I might have to determine the total heat lost from 24C down to 4.5C in 24 hrs? I have a feeling I am missing a formula about the allowable heat loss perhaps? If anyone could help me out I'd really appreciate it!


These are the formulas I think will be helpful:

For R-value: R= ΔT/(q/A)
*where ΔT is change in temperature, A is area and q is the heat transfer rate.

For heat transfer rate: q=(-kA/Δx)*(T2-T1)
*where k is the thermal conductivity of the material, Δx is the thickness of the material
 
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Answers and Replies

  • #2
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So you have an equation for the rate of heat loss at a specific time. The heat loss is actually varying with time, so you will end up needing to solve a differential equation.

Let the inside temperature at time t be T(t). Let the rate of heat flow through the be ##{dQ \over dt}(t) = Q'(t)##. Your equation says ##{R \over A}Q'(t) = T_{ext} - T(t)##, where Text is the exterior temperature. It would be nice to express Q in terms of T so that we have just one variable. Over this temperature interval you can probably assume that the specific heat capacity of the battery pack is some constant c, assuming there is nothing in the batteries that would undergo a state change. So we may put ##\Delta Q = mc\Delta T##, where m is the mass of the battery pack. Differentiating, we get ##Q'(t) = mcT'(t)##. Substituting gives:
[tex]{Rmc \over A}T'(t) = T_{ext} - T(t)[/tex]
This is the differential equation. Fortunately, it is very easy to solve; the solution is ##T(t) = T_{ext} + e^{-tA \over mcR}(T_{init} - T_{ext}).##

This equation gives you the temperature at a given time, as a function of R (and other things which are presumably known). You shouldn't find it too hard to figure out the temperature at 24 hours in terms of R, and then to solve for the minimum value of R. You should include a safety margin because of various uncertainties, including the fact that the temperature distribution will not be totally uniform inside the battery pack. (You should also check my math, and assumptions.)

As a side note, you can see that the heat capacity plays an important role in this equation. Putting extra thermal mass (like bottles of water) in the battery pack will keep it warm for longer. To do this you will have to replace the term ##mc##, which assumed there was only one material in the battery pack, with the total heat capacity ##m_1c_1 + m_2c_2 + ...##, where ##m_i## is the mass of material i and ##c_i## is the specific heat capacity of material i.

Also, if you have access to a liquid that freezes between 25 and 4.5 degrees (e.g. glacial acetic acid), you can take advantage of the heat of fusion of that material; you'll have to modify the calculation accordingly.

You may find it difficult to keep the batteries warm without doing something like this (or just putting a heater).

I also didn't take into account the heat capacity of the unknown amount of insulation.
 
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  • #3
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Awesome that helps clear up a lot! Thanks so much! So with your assumptions Tinit and Text are at time 0, correct? And if I wanted the temp at T(24hrs) to be greater than or equal to 4.5 I just set T(24hrs) to 4.5C and then find the specific heat capacity of a lipo battery, and add in mass then solve for R? And as far as that heat capacity since the box will have air and the batteries I am assuming I should use m1c1 for air and m2c2 for the battery packs to be more accurate, correct?
 
  • #4
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Awesome that helps clear up a lot! Thanks so much! So with your assumptions Tinit and Text are at time 0, correct? And if I wanted the temp at T(24hrs) to be greater than or equal to 4.5 I just set T(24hrs) to 4.5C and then find the specific heat capacity of a lipo battery, and add in mass then solve for R? And as far as that heat capacity since the box will have air and the batteries I am assuming I should use m1c1 for air and m2c2 for the battery packs to be more accurate, correct?
Yeah, that's right, but I suspect the mass of the air is going to be negligible compared to the mass of the battery. (If the mass of the air is not negligible then there are actually some issues that I didn't deal with in the calculation -- gases are more difficult to deal with than liquids and solids.)
 
  • #5
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So I'm trying to perform this calculations with the actual numbers I will be using and I am getting a small, negative number for an R-value which doesn't make sense.

If:
the outside air Text = -20c
the inside air at T0 =24c
if I want the final temperature after 24hrs to be Tt = 4c
using c= 1011.8 [j/(kg/c)]
mass m=7.577kg
A = .23408 m2
and t = 86,400sec

then R = -tA/(m*c*Ln(Tt-Text) correct?

Once I punch all the numbers in I get -.6976 for the R value required to do this. Shouldn't it be a very high R value?
 
  • #6
160
21
You don't seem to have Tinit anywhere in your equation, which is a sign that something went wrong.

I get:
[tex]R = {tA \over mc}{1 \over \ln\left({T_{final} - T_{ext} \over T_{init} - T_{ext}}\right)}[/tex] and this gives a positive value for R.
 
  • #7
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Thanks! I did that out several times and just missed that!
 

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