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Old 07-14-2008, 08:30 PM   #1
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Default Heating element for boil kettle

electric heating of the boil kettle:
Well I was told today that a propane burner is no longer allowed..... so I'm in need of making an electrically powered boil kettle. I'm thinking something modeled closely off of this: http://home.chattanooga.net/~cdp/boilnew/boilnew.htm

Couple of questions for anyone who has done this, what size element are you using? I'm limited to 4000W @ 240 as I have a spare 240 outlet on a 20A breaker (no idea why so low current on 240). Is that big enough to boil say 10gallons in a reasonable amount of time?

Next what is the importance of the low or extra low density element? The page I linked above mentions that it will help prevent scorching, but is this really a problem?

Just wanted to add. I did some math, and someone please correct me if I'm wrong, but to heat 10 gallons of water (I'll be using less) 40*C (approx difference in temp of all the wort out of mash tun and boiling temp) using a 4000W element:
37850g H2O * 4.1855J * 40C = 6336847 / 4000W = 1584s / 60 = 26.4min
Seems like a good amount of time, especially considering that is assuming no heat loss and 100% eff.


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Old 07-15-2008, 02:51 AM   #2
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Well I am in the process of building an electric boil kettle as I type this message. I got some helpful information from brewbeemer on electric element performance. He actualy gave me a website that had a calculator. I calculated for a 13 gallon boil, and a raise in temp of 70 degrees. This doesn't sound like much, but hear me out. Wort from the sparge will be hitting the boil kettle at 150 to 170F. Boil starts at 212F for water and not much higher for wort, this would only equate to a little over 70 to 72 degree rise. The calculator gave me a number of around 4870 for thirty minutes to boil. I hope this helps. S.

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Old 07-15-2008, 02:57 AM   #3
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mind forwarding that information either via PM or here?

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Old 07-15-2008, 03:05 AM   #4
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Sorry, I was looking through my email for the link. Here it is http://www.brewheads.com/powerrequired.php any whos I think that I decided to go with a 5500 watt element for my boil kettle. This will get me a real close performance to my propane burner.

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Old 07-15-2008, 01:45 PM   #5
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Here's another one: http://suburb.semo.net/jet1024/Electric%20Heat.xls

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Old 07-15-2008, 03:33 PM   #6
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If you wanted to spped up the time to bring it to a boil, you could build a heat stick powered off a 110v circuit, use it for your mashout and other things, then stick it in the wort while bringing it to a boil with the 240v circuit. Once boiling, just remove the heatstick and boil with the 240v system.

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Old 07-15-2008, 04:59 PM   #7
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http://www.phpdoc.info/brew/boilcalc.html might help you.
I have 2x3000W elements in my kettle. I use all 6000W (actually less because the elements are rated @ 240V, but I actually get ~220V) to bring ~13gal to a boil, and then maintain boil with only 3000W.

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Old 07-15-2008, 08:30 PM   #8
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can anyone comment on the high density elements actually scorching the wort? From what I've read no one has ever actually experienced this, but most everyone warns against it.

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Old 07-16-2008, 03:05 AM   #9
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Quote:
Originally Posted by z987k View Post
can anyone comment on the high density elements actually scorching the wort? From what I've read no one has ever actually experienced this, but most everyone warns against it.
There are a couple of older, or maybe not older, but more experienced brewers warn against it, because the surface gets so hot, due to higher density , it carmelizes the wort. Higher density elements are smaller for there rated wattage because they get hotter. In the HLT high density is not so bad. In wort they are not so good. A low density element is so big because wattage per square inch is more spread out, so the element does not have to get as hot to do the same work. This make any sense? I h
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Old 07-18-2008, 12:15 AM   #10
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so I picked up an element today, it's the ultra low density 4500W @ 240 element. It also says that it has an alternate rating of 3500W at 208V opposed to 240V. Now my question is, how does it achieve this? First, I'd be willing to be I don't even get 240V, more around 220 so naturally either A) it would have to pull more amps to hit 4500 or what I would think it would do, not quite get to 4500W. Now according to my calculations, if you were only feeding it 208V it would need 21.6A to get the 4500, but what I'm sure it does is just give me like 3900W not 3500. Could someone explain to me how they come to this conclusion.

Basically what I'm getting at, is the coil can't possible have two different resistances.

Thanks
Zach

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