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 warthog 08-08-2012 08:06 PM

power! a practical question

i am planning my own electric brewery. my question refers to power. the place i plan on brewing already has a 30A circuit installed, i plan on using that. i really don't want to go higher, the house is old, it has 150A service, and i find all sorts of scary wiring errors whenever i dig into something.

i am planning a 5 gallon system, but really only because of the 30A number, and i honestly would rather go 10 gallon.

so here's the question for all of yous guys who have built an electric brewery:

how long does it take to boil? my calcs tell me with 6000W max (25A at 240V) it will take 11:15 (about 1 min longer for a 5500W element) to bring my 6 gallons of wort to boil ( dT 80F - i live at 7000' boiling is ~200F and my wort comes out of my m/l tun at ~120-130).

that said, 100% of my power does not go to heating the wort, or does it? what sort of heating efficiency should i expect?

i plan on insulating my hlt, m/lt and kettle, but the tops will be open, so i expect some heavy losses there. any thought?

i guess the next question has to do with the specific heat of wort, but i think i need to assume that its greater than 1.00, i assume it depends somewhat on the o.g. but does anyone know of a formula or rule of thumb?

 DustBow 08-08-2012 08:53 PM

I don't have exact numbers for you, but using a 5500w element for 5 gallon batches will be pretty painless, your temps will rise quickly. People easily brew 10 gallon batches with 5500w elements

 passedpawn 08-08-2012 09:15 PM

Effectively 100% of the energy will go into the wort. A bit is lost on in the wiring and SSR, but probably only a percent or two. Of course, you will lose some to ambient depending on insulation.

Here's a plot of a 10g boil with a 5500W element. It took about 45 minutes to go from 70F to boil. Should scale linearly to batch size, so maybe 22 minutes for a 5g batch.

http://www.homebrewtalk.com/images/3...test-38779.jpg

 warthog 08-08-2012 09:29 PM

thanks! quick number crunching tells me your system is ~85% efficient at transferring electrical power into hot wort. that's actually pretty encouraging, i may scale up. how much insulation does your system use? what was the ambient temperature? i see you are in fla.

 passedpawn 08-08-2012 09:36 PM

It was winter when I did that... probably around 65F. That was on my HLT, which has NO insulation. With an electric system, there's no reason not to insulate, but really it would save me only a few minutes and maybe 5 cents, so I don't bother. I even have a whole roll of Reflectix that I did use for my keggle mash tun.

In theory, you should see the result of heat loss as a non-linearity in that line (heat loss to ambient should increase as the temp between the water and the air increaases). It's pretty linear though (hold the straight edge of a piece of paper against the curve). Also, I should look at the voltage going into my element. It's a long run from my breaker box to where I brew (~50') so there might be significant loss in the wire.

 warthog 08-08-2012 09:53 PM

cool! thanks again. i'm an engineer (not an ee), so i love numbers :), please check that voltage, i can work that into my calcs. my basement is about 65f year round. so those numbers should jive pretty well.

 passedpawn 08-08-2012 10:00 PM

The two numbers you really need are voltage and the element resistance. Power (watts) will be V^2 / R. I'll try to measure both of those Friday and post here. (can't measure current because everything I have is fused at 10A - I need a clampon ammeter).

 warthog 08-08-2012 10:12 PM

i'm assuming i can get figure out the element resistance from its nominal wattage. v^2/w=r so at 240v^2/5500=10.47 ohm or am i missing something here?

 passedpawn 08-08-2012 10:55 PM

Quote:
 Originally Posted by warthog (Post 4315946) i'm assuming i can get figure out the element resistance from its nominal wattage. v^2/w=r so at 240v^2/5500=10.47 ohm or am i missing something here?
Reality.

That's the school answer. The school answer also says that it will take 37 minutes 52 seconds to get to boil from 70F with a 5500W element. It took longer. Time to check assumptions.

Because I can't measure the element resistance with much accuracy, I'll put a regulated voltage source across it and measure the current (I can do both at the same time with my benchtop power supply :) )

 passedpawn 08-08-2012 11:50 PM

Quote:
 Originally Posted by warthog (Post 4315946) i'm assuming i can get figure out the element resistance from its nominal wattage. v^2/w=r so at 240v^2/5500=10.47 ohm or am i missing something here?
OK, you were right. 10.00V applied, 0.958 amps measured, = 10.4 ohms.

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