Evaporator Defrost Heat and Cycle Time
How much heat a defrost takes and how long it runs, from the three terms that make it up.
Example
You enter
- Frost mass per cycle (lb) 20
- Coil temperature at defrost start (F) -10
- Coil mass (lb) 60
- Coil specific heat (BTU/lb-F) 0.1
- Coil temperature rise during defrost (F) 60
- Defrost heater rating (BTU/hr) 10236
- Defrost efficiency (0-1; entered -- it varies widely) 0.8
You get
- Sensible heat to bring the ice to 32 F 420 BTU
- Latent (Btu) 2880
- Coil warm-up 360 BTU
- Total defrost heat 4,575 BTU
- Defrost time 26.8 min
Checked against published worked examples from: Oak Ridge National Laboratory (Fricke and Sharma).
Details, formula, and sources
ScopeThey are not the same size: melting the ice costs 144 BTU per pound against about 21 to warm that same ice from a freezer coil's temperature up to 32 F, so the latent term dominates -- but the coil warm-up term is the one that gets forgotten, and on a large coil with heavy fin stock it is real. Defrost efficiency captures everything that is NOT melting frost: heat going into the box instead of the coil, into the drain pan, and out through the insulation, and it varies so widely with coil, case and termination that it has to be entered -- the 80% default is optimistic, and ORNL's display-case study implies far lower for electric defrost. The last relation closes the loop, because frost accumulates at the coil's moisture removal rate, which is set by the box's latent load -- door openings, product respiration, infiltration. That determines how much frost is on the coil when defrost initiates, which determines how long defrost takes, so a box with heavy traffic needs more defrosts and each one is longer, and every minute of defrost is a minute of heat going into a freezer. A freezer coil at -10 F with 20 lb of frost, 60 lb of coil warmed 60 F, a 3 kW heater and 80% efficiency needs 4,575 BTU and runs 26.8 minutes, which lines up with the 20 to 30 minute terminations most controllers are set to. But halve the frost by fixing a door gasket and it falls to 14.7 minutes; double it and it climbs to 51.0. A fixed termination time is right for exactly one frost load, which is the argument for demand defrost. A sizing estimate; the manufacturer's defrost data and a measured coil temperature at termination govern.
Formula or tablesensible_btu = frost_lb x 0.5 x (32 - coil_temp_f); latent_btu = frost_lb x 144; coil_warmup_btu = coil_mass x specific_heat x rise; meltwater_btu = frost_lb x 1.0 x (drain_temp_f - 32); total_btu = (sensible + latent + melt water warmed to the drain temperature + coil warm-up) / defrost_efficiency; defrost_min = total_btu / heater_btuh x 60.
Source and editionDefrost heat from the sensible, latent, and coil warm-up terms -- 144 BTU/lb to melt ice, about 0.5 BTU/lb-F to warm it to 32 F -- divided by the defrost efficiency, by name; public thermodynamics with the efficiency entered rather than bundled. The equipment manufacturer's defrost data and a measured coil temperature at termination govern.
Free accessThe latent heat of fusion and the specific heat of ice are public physical constants; the frost load, coil mass, heater rating, and efficiency are the system's own values.
What governsEstimate. AHJ and licensed professional govern.
API field namesfrost_lb, coil_temp_f, coil_mass_lb, coil_specific_heat, coil_temp_rise_f, heater_btuh, defrost_efficiency, sensible_btu, latent_btu, coil_warmup_btu, total_btu, defrost_min
- Latent dominates 144 BTU/lb to melt against about 0.5 BTU/lb-F to warm the ice to 32 Fphysical constants
- Defrost efficiency entered because it varies widely by coil, case and termination (the 80% default is optimistic for electric display-case defrost per ORNL); the rest is heat into the boxrefrigeration practice
- Frost load drives time a fixed termination time is correct for exactly one frost load -- the argument for demand defrostrefrigeration practice