| The calculation isn't as simple as it seems... Typically your house air will have a dew point above 5C. That means, when it enters your fridge, dew will condense on the inside. The latent heat of condensation is really high. Just 10 grams of water could be 22 kilojoules of energy released. That water will eventually end up as ice on the evaporator (since the evaporator coil in a fridge-freezer typically has to run sub-freezing because it is shared with the freezer). Thats more energy loss (3 kJ for our 10 grams). Then the defrost mechanism will kick in to melt it into the drain - which is a resistive heater normally. So 3kJ again. The resistive heater typically heats far more than it needs though - a bunch of heat will be wasted into the metal of the coil and air in the fridge - which in turn will need more refrigeration to correct. So all in all, the energy loss of opening the door of the fridge is dominated by the water you're letting into the fridge, not the energy loss of the cold air. This analysis is tricky enough and with enough variables (house humidity, design of fridge, amount of other 'wet' food in the fridge, etc.) that I haven't seen anyone attempt to come up with a cost number, either numerically or experimentally. |
The defrost thing I guess depends on the fridge. Mine has two coils and the one in the 5 C non-freezer section generally stays above the freezing point when the compressor is not running. It does does not have a resistive heater.