Are the thermal resistance values of heat sinks/etc reasonably reversible? If I wanted to dissipate an internal ambient temperature rise, could I use an internal heat sink on the enclosure wall coupled to a heat sink on the outside to pull heat out of the box?
Can I use the same thermal resistances of the heat sinks in reverse to calculate how efficient this will be at extracting heat from the air? (Assuming additional resistances of sink #1 -> wall and wall -> sink #2)
I'm developing a product that must be in a sealed enclosure. I'm also trying to have the highest operational temperature range I can (targeting -40 to 85 °C) I'm sinking some heat sources (power bricks, power op amps) directly to the metal enclosure... however I am not currently able to sink the heat from the CPU itself (~1W) directly to the case. I -could- add a wide temperature range heat pipe, but I'd like to avoid it.
We're not talking about that much heat, but I just don't have much room near the top of my targeted ambient range (105 °C Tj and 85 °C Ta)... The CPU would require a heat sink even if it wasn't enclosed.
Okay, to clarify I guess I'm asking if a heat sink's thermal resistance is simply a thermal conductance value that I can use both as a number of how much of a temperature rise a certain amount of power put into it will cause AND the exact opposite in how much power it will conduct given a specific temperature differential across it. (So, a 10 °C/watt heat sink would for 1 watt be 10 degrees above ambient, and that if ambient was 10 °C ABOVE the sink it would pull 1 watt of heat out of ambient)
Say my CPU is putting out 1 watt of power. With a heat sink, the Tjs (die to sink) is 0.5 °C/watt, and the heat sink's Tsa (sink to amibent) is 9.5 °C/watt, for 10 °C/watt total. My CPU is then 10 °C above the immediately nearby internal ambient temperature.
My internal ambient is getting about 1 watt of power sunk into it, and that power needs to go somewhere.
There is about 3" between my CPU heat sink and the enclosure wall. With a thermal conductivity of air at 0.025 W*(m^-1)/°C, there would be a rise of about 3 °C across those 3" to transfer 1 watt. (right? This seems too low to me. Also, it's ignoring any kind of internal convection effects) Call this 3 °C/watt = Tax (internal ambient transfer).
The path is then normally Tac (resistance internal ambient to case) plus TcA (case to external ambient).
If I put heat sinks on the inside and outside, I then have Tah (internal ambient to internal heat sink), Thc (internal heat sink to case), TcH (case to external heat sink), THA (external heat sink to external ambient). (although really I also have Tac and TcA still, for the case surface area not covered by heat sinks))
Say Tah is 3 °C/watt, Thc is 0.25 °C/watt, TcH is 0.1 °C/watt, and THA is 1 °C/watt.
I have 1 watt of power for Tjs, Tsa, Tax, Tah, and Thc. I additionally have up to 5 watts of power sunk into the case from other sources directly attached to the case. So, I have 6 watts of power for TcH and THA.
Can I then say that my external ambient maximum is:
Tjs + Tsa + Tax + Tah + Thc + TcH + THA, or 0.5 + 9.5 + 3.0 + 3.0 + 0.25 + 0.6 + 6.0 = 22.85 °C/watt?
With Tj of 105, my maximum external ambient would be 82.15 °C?