Solving for Temperature in a Heat Dissipation Box

In summary, you need to calculate the thermal resistances of the various parts of the enclosure, and make alterations to the enclosure to account for the camera's 50 watt power.
  • #1
MisterCharly
2
0
Hi all

I need an some help on thermodynamics. I want to install a camera outside of my house, but the camera can only operate between 0 and +55 degrees celsius. Now I want to make an isolating box around it, and wanted to calculate how thick the material of the box should be to allow outside temperatures between -10 and 40 degrees celsius. I know that if the camera is on, its power is 50Watt. On wikipedia I found that the heat capacity of air is 20.85 J/(mol.K).

If I assume the camera is a point source of heat, located in the middle of a rectangular box, how can I calculate the temperature at the interior walls of the box, assuming the outside environment is an infinite reservoir at either -10 degrees C or +40 degrees.

Many thanks for your help!
 
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  • #2
You need to consider the thermal resistances of each part. In electronics you can have a chain of resistors in series with a voltage difference between the ends of a chain, and current will flow through the chain. At each resistor junction there will be a proportion of the voltage depending on the values of the resistors. There is an analogy here with the heat problem, the equations are the same except voltage is replaced by temperature difference, the resistance replaced by thermal resistance, and the current by heat flow. So V=IR becomes ΔT=QR[itex]_θ{}[/itex].

In your system you have a source of heat (inside the camera housing) which at equilibrium is going to be at a certain temperature, and an ambient outside temperature. Those are your end points, like the voltage across the resistor chain. Heat will flow from the hot end to the colder end, and in this case the total heat flow is 50W. You also have thermal resistances, namely internal camera heat source to camera case, camera case to box (through the air), and box to external ambient. I'm ignoring the box itself as I'm assuming it's metal with low thermal resistance, but if it's plastic there's one more junction, internal box to external box.

Your camera should have a thermal resistance specification in °C/W. The box should be calculatable, again these are treated just like heatsinks and are specified in °C/W. The more difficult one is the air, but let's assume this can be found. You should have all you need to work out the various temperatures at each junction, and if necessary alter the thermal resistances by re-designing the enclosure to meet your operating temperature criteria.

Edit: I think that since air has quite a high thermal resistance, the temperature difference between the camera case and the box is going to be quite high at 50W, and you've only got 15 deg C to work with in total. I think you might have to thermally bond the camera case to the box. Also you have two criteria, it must be greater than 10 deg C if the outside temp is -10, but it must be less than 15 deg C if the outside temp is +40.
 
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  • #3
Thanks for helping me out!
 
  • #4
MisterCharly said:
I need an some help on thermodynamics. I want to install a camera outside of my house, but the camera can only operate between 0 and +55 degrees celsius. Now I want to make an isolating box around it, and wanted to calculate how thick the material of the box should be to allow outside temperatures between -10 and 40 degrees celsius. I know that if the camera is on, its power is 50Watt.

I think it is no thermodynamics but heat engineering most likely. It is a problem of heat transfer under convection (natural convection within the box and forced one- since there may be wind- outside). The problem is very difficult for strict solution, but approximate methods of estimations are commonly used. There are a lot of books on the problem, or you may search by Google using such key words as “heat transfer”, convection, Nusselt. For the beginning you may look at http://en.wikipedia.org/wiki/Heat_transfer_coefficient .
 
  • #5


Hello there,

It seems like you are trying to solve a heat transfer problem. In order to calculate the temperature at the interior walls of the box, you will need to consider several factors such as the material and thickness of the box, the thermal conductivity of the material, and the heat transfer coefficient between the box and the outside environment.

First, you will need to determine the rate of heat dissipation from the camera. This can be calculated by multiplying the power of the camera (50W) by the time it is in operation. Then, you can use the equation Q = U*A*ΔT to calculate the heat transfer through the box, where Q is the heat dissipation rate, U is the overall heat transfer coefficient, A is the surface area of the box, and ΔT is the temperature difference between the inside and outside of the box.

Once you have the heat transfer rate, you can use the heat transfer equation Q = h*A*ΔT, where h is the heat transfer coefficient between the box and the outside environment. This equation will help you determine the temperature at the interior walls of the box. The heat transfer coefficient can be calculated using empirical correlations or by conducting experiments.

In order to determine the thickness of the box, you will need to consider the thermal conductivity of the material. The thicker the material, the slower the heat transfer will be. You can use the equation Q = k*A*ΔT/L, where k is the thermal conductivity of the material and L is the thickness of the box. This equation will help you determine the required thickness of the box to maintain the desired temperature range inside.

I hope this helps. Good luck with your project!
 

1. How does heat dissipate in a box?

Heat dissipates in a box through a process known as convection. This means that heat is transferred from the hot objects inside the box to the cooler air outside the box.

2. What factors affect the rate of heat dissipation in a box?

The rate of heat dissipation in a box is affected by the material of the box, the size and shape of the box, the temperature difference between the inside and outside of the box, and the presence of any insulating materials.

3. Can the placement of the box affect heat dissipation?

Yes, the placement of the box can affect heat dissipation. Placing the box in a well-ventilated area can allow for better air circulation and thus faster heat dissipation. Placing the box in a confined or insulated space can slow down the rate of heat dissipation.

4. How can I increase the rate of heat dissipation in a box?

There are a few ways to increase the rate of heat dissipation in a box. You can use a material with high thermal conductivity for the box, such as metal, to allow for faster heat transfer. You can also increase the surface area of the box by adding fins or using a larger box, which will allow for more contact with the surrounding air and thus faster heat dissipation.

5. Is there a limit to how much heat a box can dissipate?

Yes, there is a limit to how much heat a box can dissipate. This is determined by the size and material of the box, as well as the surrounding temperature and humidity. If the heat being generated inside the box is too high, it may exceed the box's ability to dissipate it, resulting in a buildup of heat and potential damage to the box or its contents.

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