Cold Vault Heat Load and Power Cut

What is modelled

A cold store loses cold in three ways: heat conducts in through the walls, warm air comes in when the doors open, and lights and equipment add heat inside. A refrigeration plant pumps that heat out, and the work it needs depends on the temperature difference it works across, so a warm climate costs much more electricity. If the power fails, only the leak through the walls warms the contents, and the large thermal mass makes this very slow.

Equations

• Wall U = 1 / (d/kins + dc/kc + 2 Rfilm), leak Q = U A (Tout − Tin).
• Door infiltration ρ cp Vdoor ΔT per opening.
• Plant COP = η Tc / (Th − Tc) and W = Q / COP.
• Power cut C dT/dt = U A (Tout − T), so T(t) = Tout − (Tout − T0) e−t/τ with τ = C / UA.

Checks run in the page

• A finite-volume solution of conduction through the layered wall against the sum of resistances.
• The second law for the refrigeration plant, with zero entropy generation for the ideal one.
• A Runge-Kutta power-cut warm-up against the exact exponential.
• The convergence of the hourly yearly electricity sum.

Honest simplifications

The vault is a lumped cube-like volume of one temperature, the contents are a single thermal mass, humidity, frost on the evaporator, ground temperature layers and defrost cycles are left out, the plant efficiency is a fixed fraction of the Carnot limit and the outside swing is a plain sine wave. A real vault needs a proper design with measured data.

• Tick Cut the power and watch the vault warm slowly, then see the days to the limit in the readout.
• Press Thin insulation to see the leak and the yearly electricity grow, and the safety time shrink.
• Press Warm climate to see how the plant COP falls and the electricity bill climbs.