Dilution Refrigerator Heat Budget

What is modelled

Five plates hang one under another. N identical wiring lines connect each plate to the one above. Their metal conducts heat in proportion to temperature, k = a T. Each plate has a cooling capacity that rises with its temperature: pulse-tube lines for the two upper plates, fitted curves for the still and cold plate, and the dilution cooling power 84 n T² for the mixing chamber. Each plate settles where cooling equals the incoming wire heat plus what is dissipated on it.

Equations

• Heat through one set of lines Q = N (A/L) a (Th² - Tc²)/2.
• Balance on each plate Qcool(T) = Qwires(Tabove, T) + Qdissipated.
• Mixing chamber cooling Q = 84 ṅ3 T², so the floor is T = √(Q/(84 ṅ3)).
• Cooldown T(t) = Teq + (T0 - Teq) e-t/τ.

Checks run in the page

• The conductivity integral against its closed form.
• The mixing chamber balance residual and the quadrupling rule.
• Doubling the lines doubling the heat, and the chain of link loads adding up.
• The cooldown exponential against Euler steps.

Honest simplifications

Linear conductivity k = aT, a made-up cooling curve for each upper plate, a fixed 20 W and 0.8 W of other loads (radiation, cables) on the two upper plates, fixed time constants for the cooldown instead of heat-capacity modelling, no radiation, no thermal boundary resistance and no real attenuator thermalisation. Illustrative numbers for teaching, not a cryostat design.

• Press Full chip, steel coax then Superconducting lines to see the 4 K and mixing chamber loads fall.
• Raise the helium-3 rate and lower the base temperature.
• Press Heavy leak and see the floor rise as the square root of the heat.