Using the equation for thermal resistance
R_{thermal} = L/(k*A)
where k is the thermal conductivity of a material, L is the length, and A is the surface area through which the heat passes, I could find the thermal resistance of the copper and stainless steel on the reference cavity. To reduce temperature gradients across the vacuum chamber, the thermal resistance of the copper must be the same or less than that of the stainless steel. Since the copper is directly on top of the stainless steel, the length and width will be the same for both, just the thickness will be different (for ease of calculation, I assumed flat, rectangular strips of the metal). Assuming we wish to have a thermal resistance of the copper n times less than that of the stainless steel, we have
R_{Cu} = R_{SS}/n
or
L/(k_{Cu}*w*t_{Cu}) = L/(k_{SS}*w*t_{SS}*n)
so that
t_{Cu}/t_{SS} = n*k_{SS}/k_{Cu}
We know that k_{SS} = 401 W/m*K and K_{Cu} = 16 W/m*K, so
t_{Cu}/t_{SS} = 0.0399*n
By using the drawings for the short reference cavity vacuum chamber (the only one I could find drawings for online) I found a thickness of the walls of 0.12 in or 0.3048 cm. So for the same thermal resistance in both metals, the copper must be 0.0122 cm thick and for a thermal resistance 10 times less, it must be 0.122 cm thick. So we will have to keep wrapping the copper on the vacuum chamber! |