I did some calculations about using induction elements for brewing.

I want to use one with a built in temperature control device in it to do mashing for a 5 gallon batch, and I figured that I might as well use the same element to boil while I'm at it.

I started with the following tidbit :

"A 3.5-kilowatt (kW) Luxine induction range was shown to boil 20 pounds of water in virtually the same time as a 5.1-kW electric resistance coil (about 15 minutes)."

http://www.appliancedesign.com/CDA/A...00f932a8c0____
(conversions with help from google "convert n x to y", in USA)

20 lb water X 0.45359237 kg/lb = 9.0718474 kg

9.0718474 kg X 1 kg/L = ~9 L water (~2.4 US gal)

Not especially encouraging for fast, full boils.

However, I realized that for a full boil, with all grain, you aren't raising the water from room temperature to boiling, you are raising it from room temperature to mashing and from mashing to boiling, so to be fair, I redid all the calculations.

=== Mashing ===

Assume I have 4 gallons of water for mashing, which I could use to process 8-16 lbs of grain (2qt/lb - 1qt/lb), which seems like a reasonable amount.

Now I need to know the amount of heat needed to raise the water from tap temperature (~20C = ~ 68F) to strike temp for mashing (73C = ~163F).

4 gallons X 3.7854118 L/gallon = ~ 15.14 L

Need to change the temperature 73C - 20C = 53C.

1 calorie raises 1 gram of water 1 degree C, 1 gram = 1 ml

so,

1000 calories raises 1 L of water 1 degree C :

15.14 L X 1000 calories/L-delta_C X 53 delta_C = 802,420 calories

I don't happen to think in calories.

1 calorie = 0.00396566683 BTU.

1 calorie = 1.16222222 × 10^-6 kilowatt hours

So that's 3182 BTU, which seems ... low, more on that later. In order to accomplish that in an hour, one needs 0.93 kilowatt hours, so in order to finish in 15 minutes, one needs to apply 4 times as much energy per unit of time, so that'd be almost 4 kWh! To put that in perspective, a good 3.6 kWh element needs a 220V outlet and 16.4 amperes and they seem to cost about $1000+ online. If you can wait a half hour, the a 2kWh element would be fine (much cheaper).

Interestingly enough, the BTU figure looks incredibly low. To the best of my ability to discern, the reason why it takes a while with a 50k - 150k BTU burner is two-fold. First, propane burners only impart 30-60% of their heat to the vessel being heated. Second, the BTU ratings on burners are wildly innacurate. Induction is lauded for its efficiency, which is usually quoted at 90+%. So, technically, all those figures in the above paragraph describe the amount of energy needed, not the size of the appropriate heating element. The induction element at 90% efficiency for 15 minutes would need to be 3.72 * 1.1 = 4.09kWh (I estimated the 4 kWh in the previous paragraph). The BTU figure doesn't make a lot of sense to me, because it would indicate that only a 4772 BTU burner would be needed and that is clearly not the right size burner. A range top is something like 12,000 BTU, and I KNOW it takes more than 15 minutes to raise 4 gallons of water to 73C on that.

=== Boiling ===

We're going from about 73 C to about 100 C, so that's a 27 C difference.

I'm going to assume I did the math correctly, and I'll just say that should be possible for 6 gallons of wort in 15 minutes on a ~ 3 kWh element.

=== Propane Burner Actual Output Estimation ===

I'll do a calculation that figures out how much heat is actually getting into the liquid if someone posts the following : change in temperature, gallons water or wort heated, reported BTU output of propane burner. I think I read somewhere that a 100,000 BTU burner will boil 6 gallons in 15 minutes, but I am suspicious that it was from mash temp to boil, not from tap temperature. That's something like 10,250 BTUs required, so the efficiency is terrible.

=== Checking the Math ===

Starting with the 20 lbs of water example, lets see if my calculations are correct since the BTUs seems way off from what one might expect...

9 L X 80 delta_C X 1000 cal / L-delta_C X 1.16e^-6 kWh / 1 cal X 4 kW_15min / 1 kWh = 3.341

That seems reasonable.

So the element should be 3.341/.9 if it is 90% efficient, which is 3.712kWh. Dang that's efficient (about 95%!!!)!

*************************************

*Please check my math. I'm far from perfect. *

*************************************

=== Crazy Ideas ===

I've noticed that large brew kettles with spigots seem to be expensive, as are quality grain mills, and high output burners. So I had this insane idea that perhaps 3-4 of those 1.3 kWh induction elements, each with it's own wee boil pot (~2 gallons?) with a spigot might save big money. The 1.3 kWh elements were available for $65 (!) on the one site. Anyone know where to get smallish induction friendly brew pots with spigots for cheap?

I figure, even if those cheap elements are 80% efficient, one can take 1.5 gallons of water from 20C to 100C in 30 minutes, which is 2-3 times as fast as trying to boil 5.5 gallons on an ancient natural gas range top burner.

Then my only problem is finding a room I can draw 4 x 11amperes @ 120V without tripping a breaker or starting a fire...