Does anyone know how many watts are needed to maintain a 5 or 10 gallon boil?

I have just decided to go for about 4-5 kws. I have calculated how long it will take to bring the wort to a boil....but I am wondering if a single 4 or 5 kws heating element would cause a way too vigorous boil and would make a lot of water evaporate?

Maybe it would be better to get a 1,5 kws and a 2,5 kws element and turn off one of them once the boil has been reached. What do you guys think?

To answer your question only 1 watt is needed to boil water, the last one - with really good insulation. In seriousness it depends on your system, weather, etc. My poorly insulated keggle losses 767 watts to heat transfer during a 12 gallon boil, the rest converts water into steam, which is your boiloff. I think a "good" boiloff rate is quoted to be 10% per hour (or 1.2 gallons from a 12 gallon starting batch).

To boil off a gallon of water in an hour it takes about 2,400 watts delivered (not counting that lost to heat transfer) for that hour. So if you lose ~750 watts through the kettle sides you will want a 3.15kW element.

I run a 4.5kW element to boil 12.25 gallons down to ~10.70 gallons in an hour...which is on the high side for boiloff rate (12.7%). I would reccommend a 3.0 to 3.5kW, maybe pair it with a 1.5kW to get up to a boil quicker.

KegWrangler said:

I run a 4.5kW element to boil 12.25 gallons down to ~10.70 gallons in an hour...which is on the high side for boiloff rate (12.7%). I would reccommend a 3.0 to 3.5kW, maybe pair it with a 1.5kW to get up to a boil quicker.

Click to expand...

Yeah, makes perfect sense. Thanks.

The key to your boil is that you get surface deformation. This is an indication that the wort is circulating in the kettle and mixing. The boil off rate is a combination of the energy added/escaping from the kettle and the surface area of the wort. The same 15 gallon kettle with the same element should have the same boil off rate (in gal/hr) for 10 gallons or 5 gallons of wort.

I use a 5.5 kW element in my 8 gallon kettle. I usually do 3 gallons batches with just under 4 gallons of wort into the kettle. My boil off rate is 0.7 gal/hr running my PID at 42% duty cycle. If my PID is accurate, that is 2.3kW.

This spreadsheet does some of the calculations for temp, time, and power for using electric elements for heating wort.
ElectricHeat.xls

I should note that the boiloff rate I calculate comes from the following equation:

Gallons per minute (gpm evaporated) = [kW element - kW lost] / [142,568 Watts*min/gallon]

[The 142,568 comes from converting the heat enthapy of water (2,260J/g) into standard homebrewing units or SHBUs.]

Gallons per hour (gph) boiloff is ~ kW (net) / (2,376 Watts*hour/gallon).

If you don't know what your kW lost is that's OK - nobody does. Instead run a test on a batch of water the same volume as what you will be boiling in the future. Measure the amount boiled of in a hour then run that through the above equations backwards to get the kW (net). Subtract the element wattage and the resulting negative number will be your kW lost.

Some WAG numbers for reference are 800kW for a keggle. 700kW for a 10 gallon pot. Maybe 500kW for a 5 gallon pot.

KegWrangler, that is awesome info!

We aims to please...that and procrastinate from work.