Start from the water, not the timer
The only honest way into this is to find out how much water is actually going onto the coil. Put a bucket under the drain for 24 hours and weigh it. A pint is roughly a pound, so a five litre jug over a day is about 11 lb of frost, and that single measurement replaces every assumption on the page.
The defaults here — 6 lb a day on a 4 sq ft coil defrosting at an eighth of an inch — work out to 1.04 lb cleared per cycle and six defrosts a day. A timer set to four is clearing 4.17 lb of the 6, and the remaining 1.83 lb a day is the frost that quietly builds up over a week until somebody notices the box drifting.
A defrost costs twice
The heater is the obvious cost: 3,000 W for 20 minutes is 1 kWh. The part people miss is that most of that energy is now inside the box and has to be removed again by the refrigeration. At 70 percent landing in the box and 8 BTU/h per watt, that is another 0.3 kWh, so the real figure is 1.3 kWh a cycle rather than 1.
Melting the frost itself is a small part of it. On the defaults, 1.04 lb of ice at 10 °F needs 161 BTU to warm and melt, against the 3,412 BTU the heater delivers. About five percent of the heater energy goes into the job it is there for; the rest warms metal, air and the drain. That ratio is why terminating on temperature rather than running the timer out matters so much.
The hours matter as much as the kilowatt hours
Six defrosts at 20 minutes of heater plus 10 minutes of recovery is three hours a day the box is not cooling. Everything the box needs over 24 hours has to fit into the remaining 21, and that is the argument for sizing refrigeration on a run time well under 24 hours rather than on the load alone.
Where this is weakest
Two places, and both are stated on the page. Frost does not lie flat — it packs the entering face first, so airflow falls off before the nominal thickness is anywhere near uniform, which makes the calculated count a floor. And the share of heater energy that ends up in the box is genuinely uncertain; run it at 50 and at 90 percent and quote the range rather than a single figure. On the defaults that spread is 1.21 to 1.38 kWh a cycle.
Questions people ask
How many defrosts a day does a walk-in freezer need?
As many as it takes to clear the frost that actually lands on the coil, which is a measurement rather than a rule. Weigh the condensate off the drain for a day, work out what your coil carries between defrosts, and divide. On the defaults here that is 6 lb of frost against 1.04 lb a cycle, so six.
How do I measure frost on a coil?
Two ways, both cheap. Catch the drain water over a full 24 hours and weigh it — that is the total. For the density, scrape a measured patch off the coil, melt it and weigh the water; frost off a busy dock is much denser than frost off a quiet freezer, and the tables disagree wildly on this.
Why does a defrost cost more than the heater uses?
Because most of the heater energy stays in the box and has to be pulled back out by the refrigeration. On the defaults the heater uses 1 kWh and removing the heat it left costs another 0.3 kWh, so a cycle is 1.3 kWh. Cutting a cycle saves both halves.
How much of the defrost energy actually melts ice?
Less than people expect. On the defaults, warming 1.04 lb of ice from 10 °F and melting it needs 161 BTU against 3,412 BTU delivered by the heater, so about five percent. The rest goes into the coil metal, the air and the drain, which is why terminating on temperature saves more than shortening the timer does.
Should I just add more defrosts to be safe?
Each extra cycle costs the energy and takes half an hour of cooling out of the day, so extras are not free. The page prints both costs so you can see the trade. What the right schedule is for your product and your box is not something this page decides.