I Calculating the thermal coefficient between the insulated interior of a telescope instrument package and the cold ambient air outside

[Erwinux]

I've built an insulated chamber to protect a sensitive instrument at freeze temperatures in the winter. The instrument is mounted on a telescope, so the heat inside the chamber will slowly dissipate in the ambient. A digital PID thermostat is used to keep the temperature at a safeguard level. The outside and inside temperature and the duty cycle of the thermostat are recorded on a remote server, so I know the power consumed at a certain difference of temperature, anytime.

How to calculate the outside minimum temperature the heater will maintain the safeguard temperature inside insulated chamber?

 

[DaveE]

A simple model will work pretty well here for steady state solutions. The power your heater makes will be proportional to the temperature difference it creates. This should be a linear relationship in normal stable systems (not true if you have things that change state like water freezing/melting) so twice the heat flow (power) will create twice the temperature difference. Also, the temperature difference won't depend much on the temperature values within "normal" temperature ranges.

So, you should be able to get a good estimate by collecting some power-temperature data and then extrapolate to maximum heater power and the minimum desired temperature.

 

[erobz]

So you have something like that above.

I think you can apply surface energy balance to the interior/exertion walls to get in the ball park for a steady state solution. Its a bit hand wavy, but it shouldn't be that bad:

Take $A_s$ to be the exterior surface area of the enclosure
$k$ is the thermal conductivity of the insulation
$L$ is the insulation thickness
$h_{int}$ is the average convection coefficient inside the enclosure maybe $10 \rm{ \frac{W}{m^2 K}}$
$h_{ext}$ is the average convection coefficient outside the enclosure could be as high as $200\rm{ \frac{W}{m^2 K}}$
$q$ is the maximum output power rating for the heater

Then you get $T_{int}$ from solving the following relation:

$$ q = A_s \frac{T_{int} - T_{\infty}}{ \frac{1}{h_{int}} + \frac{L}{k} + \frac{1}{h_{ext}} } $$

let me know if you have any questions.