Solving for Temperature in a Heat Dissipation Box

AI Thread Summary
To ensure the outdoor camera operates within its temperature range of 0 to +55 degrees Celsius, an insulating box must be designed to manage heat dissipation effectively. The camera generates 50W of heat, necessitating calculations of thermal resistance at various junctions, including the camera to its case, the case to the box, and the box to the ambient environment. Given external temperatures can range from -10 to +40 degrees Celsius, the design must account for significant temperature differences, particularly with air's high thermal resistance. It may be necessary to thermally bond the camera to the box to maintain optimal temperatures. For further understanding, resources on heat transfer and convection principles are recommended.
MisterCharly
Messages
2
Reaction score
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!
 
Science news on Phys.org
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_θ{}.

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.
 
Last edited:
Thanks for helping me out!
 
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 .
 
I was watching a Khan Academy video on entropy called: Reconciling thermodynamic and state definitions of entropy. So in the video it says: Let's say I have a container. And in that container, I have gas particles and they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles. Now, each of these particles could be in x different states. Now, if each of them can be in x different states, how many total...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top