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Immersion Heater calculations

http://www.thebrewingnetwork.com/forum/viewtopic.php?f=4&t=16867

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Immersion Heater calculations

Posted: Mon Sep 14, 2009 9:52 pm
by Kbar
Leave it to McMaster Carr. Years of engineering school and higher math to have most of my engineering career bottle neck down to algebraic equations. I am wanting to move to a 240V immersion heater in the HLT vs. Gas fired. I will leave the BK gas fired however. Found this equation for immersion style heaters.

V (gallons) * Delta T (F)
____________________ * 1000 = Wattage Rating of immersion heater
372 * Heat up time (Hrs)

Example - 3 Gallons of water for mash in, 60F to 168F, in 15 miniutes (0.25 hours) = 3484W or 3500W approx. Equation assumes water only. You can see their website for liquids of different densities as well. FYI

take care all!

Re: Immersion Heater calculations

Posted: Tue Sep 15, 2009 5:34 am
by codewritinfool

Re: Immersion Heater calculations

Posted: Tue Sep 15, 2009 3:28 pm
by Kbar
codewritinfool wrote:This may be of help also.
viewtopic.php?f=2&t=10822&p=103792&hilit=element#p103792


Thanks Code!

Re: Immersion Heater calculations

Posted: Wed Sep 16, 2009 5:07 pm
by ajdelange
All you have to remember is that 1 KW is equivalent to 3414 BTU/hr and that water weighs about 8.34 pounds per gallon. Thus in the example where it is desired to raise 3 gallons through a 108 °F rise the heat requirement is 3 x 8.34 x 108 = 2702.2 BTU. If this is to be accomplished in 1/4 hour then the heat input rate must be 2705.4/0.25 = 10808.6 BTU/Hr which is equivalent to 10808.6/3414 = 3.166 kw. That's all there is to it assuming 100% efficiency. The McMaster Carr formula apparently assumes 91% efficiency.

Re: Immersion Heater calculations

Posted: Wed Sep 16, 2009 8:34 pm
by codewritinfool
ajdelange wrote:All you have to remember is that 1 KW is equivalent to 3414 BTU/hr and that water weighs about 8.34 pounds per gallon. Thus in the example where it is desired to raise 3 gallons through a 108 °F rise the heat requirement is 3 x 8.34 x 108 = 2702.2 BTU. If this is to be accomplished in 1/4 hour then the heat input rate must be 2705.4/0.25 = 10808.6 BTU/Hr which is equivalent to 10808.6/3414 = 3.166 kw. That's all there is to it assuming 100% efficiency. The McMaster Carr formula apparently assumes 91% efficiency.


Excellent info, ajdelange. Of course, you just know I had to go plug your numbers into my equations and it came out 15.01 minutes, so I was pleased. I also admit that the geeky engineer in me had me considering, if only for a split second, where the rounding errors were. :)

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