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Old 02-26-2012, 05:20 PM   #1
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Default All Grain Apartment Brewing: Will a 1500W/120V element boil 6 gallons?

Hey folks,
So I'm an apartment brewer, trying to go all grain for as cheap and dirty as I possibly can. My best bet at this point is BIAB in an 8 gallon bucket with a 1500W element.

I'm not worried about the bucket, the reasearch I've done says that Food grade Polypropylene is fine to boil in.

I'm a little concerned about the element though. My apartment has 15 amp circuits so a 2000W element is out of the question. The math says that it'll take an hour to reach mash temp with a 1500W element and another 30 to hit a boil from that. The fact that it'll be slow doesn't bother me. But I want to know if anyone has done it and if it'll actually get there once losses and stuff are counted.

I do have a second circuit near enough that I could plug into with a short extension cord, so a second 1500W is possible. I'd really like to try and avoid that however as the 12.5A that each element is pulling at 120V will cause real life inconveniences around the house. I don't think that my girlfriend will enjoy not being able to use power in the living room 'cause I'm trying to boil stuff for 3 hours

The apartment does not have GFCI protection on the outlets, so I'll be relying on plug in GFCI adapters like this: http://www.grainger.com/Grainger/POW...5&cm_vc=HPPVZ3 If there's one available for 220 (I haven't seen one yet) then that may become an option, but for now I'm really only exploring the 120v option.

Thanks guys

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Old 02-26-2012, 06:33 PM   #2
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I have a small batch system that uses a 2000watt element and I can't boil 6.5 gallon batches without using my electric stove as a supplement heat source.

So I'd say that no you can't boil a 6gal batch with a 1500 watt element.

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Old 02-26-2012, 07:03 PM   #3
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I use this Formula to calculate
http://www.smartbrewer.com/tools/heat.php

deg F per min = 0.0068 * P / (gallons * LF)

0.0068 = co-efficient
P= Power
LF = loss factor, to account for heat loss. A well-insulated vessel might be 1.05 or less; uninsulated might be 1.10 to 1.15 or more depending on geometry, material, covered or open, etc.

deg F per min = 0.0068 * 1500 / (6 * 1.1)

So using the above formula.... the temperature rise per minute in your case will be approximately 1.87F

If you start with water at 60F, (212-60)/1.87 = time taken to boil which is approximately 82 minutes

But I bet in this 82 minutes the heat loss is more than what you are gaining.

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Old 03-03-2012, 01:58 AM   #4
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Energy loss would be calculated with:

btu/min = A * dT / (R * 60)

1 kW = 56.87btu / min

so:

kW = 56.87 x A x dT / (R x 60)

where:

A = square area of vessel (in sq ft)
dT = difference it temp (From outside the vessel and the liquid temp)
R = the insulating value

have fun with the math

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Old 03-04-2012, 02:16 AM   #5
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I just tested my kettle with a 5500 W element but with 120v (1375W) and I was able to do a 6 gallon batch but it took 90 min to go from mash out to boil and was hard to hold it there.

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Old 03-04-2012, 02:12 PM   #6
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My stovetop has a 1500w element, and it can and does boil 6 gallons (down to 5). However, it does take nearly an hour to reach boiling from mash-out temps and it's not the most vigorous of boils.

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Old 03-04-2012, 03:58 PM   #7
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I use 2 1500 watt high density elements and make 10 gal batches (13 gal preboil volume). I get a decent rolling boil. I've made blondes and Czech pilsners without a hint of DMS (or the notorious scorching). Takes a while to get up to a boil but I've got plenty to keep me busy during my brewday. I've never measured how long each step takes to get up to temp but my brewday takes at most 6hrs including cleaning.

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Old 03-05-2012, 05:32 PM   #8
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Quote:
Originally Posted by TheModestMeader View Post
Energy loss would be calculated with:
copy+pasting a random formula from wikipedia doesnt help anyone.

yes thats technically a "heat loss" formula, but the reason it doesnt help is that there are a dozen other variables you need to take into account for it to be useful...

what is "area"? area of the pot? does that include the top, or just the sides? what about evaporization that takes place or how much liquid the pot contains? what if you put a top on the pot? convection around the vertical sides of the pot is going to increase heatloss more than the relatively stagnant air around the bottom; 1sq inch of side area != 1 sq inch of bottom area, eventhough your forumula treats them the same...

if you want to be helpful, you can do the math yourself.
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Old 03-06-2012, 02:41 PM   #9
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Quote:
Originally Posted by audger View Post
copy+pasting a random formula from wikipedia doesnt help anyone.

yes thats technically a "heat loss" formula, but the reason it doesnt help is that there are a dozen other variables you need to take into account for it to be useful...

what is "area"? area of the pot? does that include the top, or just the sides? what about evaporization that takes place or how much liquid the pot contains? what if you put a top on the pot? convection around the vertical sides of the pot is going to increase heatloss more than the relatively stagnant air around the bottom; 1sq inch of side area != 1 sq inch of bottom area, eventhough your forumula treats them the same...

if you want to be helpful, you can do the math yourself.
it would be the total surface area and that is not a copy and paste it's a heat calc formula that I use every day (work for an industrial engineering consulting firm) . Yes it isn't going to be completly accurate because more heat does rise from the top the from the sides and more from ths sides then the bottom. Also all the points you brought up are equally pertinent.

It is a generic formula to calculate the insulating value of the pot.
I thought it would be more accurate then the "LF = loss factor, to account for heat loss. A well-insulated vessel might be 1.05 or less; uninsulated might be 1.10 to 1.15 or more depending on geometry, material, covered or open, etc."

I thought he was just trying to get a rough estimate and the post previous to me was doubting the to validity of the loss factor so I was just giving the OP another "tool for his belt" so to speak.

If YOU want to be helpful maybe you can put values to all the additionaly variables, come up with a formula and post your results instead of bashing me for nearly trying to help him. Also it would be impossible for my to produce a finite number because I didn't have the dimensions of his pot for the surface area calc and I had been drinking at the time. and FYI the formula came from my themal dynamics text book. Why do you have to be such and a$$ when I was just trying to help him?
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Last edited by TheModestMeader; 03-06-2012 at 02:43 PM. Reason: fix "broken" quote
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Old 01-07-2013, 08:25 PM   #10
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Default beating a dead horse

So I am working on putting a 1500 watt element into my keggle HLT. I only want to get 10-12 gallons of water up to mash temps (170F) I plan to plug my temp controller into a timer, so I don't care about how long it takes, but my question is, will a 1500 watt element get 10-12 gallons up to mash temps?

I have a keggle, I do plan to insulate the outside, and use a lid.


I'm only trying to save on propane costs and I do not want to go to the expense of buying 20 feet of 10 gauge wire, a 240 circuit breaker, connectors and the hassle of installation, not to mention, I would like to remain somewhat mobile if I need to, so 1500 watt is the only reasonable size.

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