Determining heat loss and required insulation

In summary,The author is trying to determine the proper material and design specifications for the enclosure he will be building to keep batteries warm in cold weather. He has come to a bit of an impasse and needs help calculating the initial R-value.
  • #1
Scratchy
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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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  • #2
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
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
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
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
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
Thanks! I did that out several times and just missed that!
 

FAQ: Determining heat loss and required insulation

1. How do you determine the heat loss in a building?

To determine heat loss in a building, you need to calculate the difference between the indoor and outdoor temperatures, the area and materials of the building envelope, and any air leaks or ventilation. This can be done using a heat loss calculator or through manual calculations using the building's specifications.

2. What factors affect the required insulation for a building?

The required insulation for a building depends on several factors, including the climate and location, building materials, energy efficiency goals, and building regulations. The size and layout of the building, as well as the type of heating and cooling systems, also play a role in determining the required insulation.

3. How is the R-value of insulation determined?

The R-value of insulation is determined by its resistance to heat flow. It is calculated by dividing the thickness of the insulation by its thermal conductivity. The higher the R-value, the more effective the insulation is at reducing heat loss.

4. What are the benefits of using insulation in a building?

Insulation helps to reduce heat loss, improve energy efficiency, and maintain a comfortable temperature inside the building. It also helps to reduce noise pollution, prevent moisture and condensation, and improve the overall durability and lifespan of the building.

5. How can I determine the most cost-effective insulation for my building?

The most cost-effective insulation for a building depends on various factors such as the climate, building materials, and energy efficiency goals. It is essential to consider the initial cost, as well as the long-term savings in energy costs, when choosing the right insulation. Consulting with a professional and comparing different types of insulation can help determine the most cost-effective option for your building.

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