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/PO...43?cm_sp=IO-_-Home-_-VTV70300505&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

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.

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.

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

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.

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.

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.

TheModestMeader said:

Energy loss would be calculated with:

Click to expand...

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.

audger said:

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.

Click to expand...

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. :rockin: Why do you have to be such and a$$ when I was just trying to help him?

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.

lemy said:

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.

Click to expand...

10 gallons 45F to 170F with 1500 watts = 128 minutes

There's a spreadsheet floating around somewhere here with the formulas.

tektonjp said:

10 gallons 45F to 170F with 1500 watts = 128 minutes

There's a spreadsheet floating around somewhere here with the formulas.

Click to expand...

Thanks for the response, but does the equation count for heat loss ? I understand 128 minutes in the perfect world where there is no heat loss, but in a 50 degree garage with my feeble attempt at insulation...

Just looking to see if anyone has experience with it.

BMGfan said:

Hey folks,
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/PO...43?cm_sp=IO-_-Home-_-VTV70300505&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

Click to expand...

They do sell 240V gfci cords, links have been posted here before, just dont remember where, maybe someone else can??

if you got one of those you could get some long SO cord and run a line from your drier or stove outlet and use that to power a 4500 or 5500 element, which will make sure you cane boil/mash in a reasonable amount of time

lemy said:

Thanks for the response, but does the equation count for heat loss ? I understand 128 minutes in the perfect world where there is no heat loss, but in a 50 degree garage with my feeble attempt at insulation...

Just looking to see if anyone has experience with it.

Click to expand...

Sorry, that's all I know. You said you didn't care how long it takes. Well, it will take over 2 hours, best case senario.If you insulate the HLT and put a lid on it, then it should be pretty darn close. You should you didn't care how long it takes. Well, it will take over 2 hours, best case scenario.

tektonjp said:

Sorry, that's all I know. You said you didn't care how long it takes. Well, it will take over 2 hours, best case senario.If you insulate the HLT and put a lid on it, then it should be pretty darn close. You should you didn't care how long it takes. Well, it will take over 2 hours, best case scenario.

Click to expand...

2 hours is fine, my only concern is that, due to heat loss, 10-12 gallons would never actually get to 170F.

Throw a heat stick in an igloo.

Do you have a dryer? That uses probably a 30 amp circuit. Replace breaker with 30amp gfci and build yourself a 220v heatstick 4500-5000 watts with a replacment dryer cord. that is exactly what i did.

tektonjp said:

Throw a heat stick in an igloo.

Click to expand...

+1 on the igloo setup.

I tried a 1500W in a 7.5 gallon SS pot and it never gets to a boil. I can get it there with help from the stove top but it loses so much heat that if I turn the range off the boil stops.

There's a spreadsheet floating around somewhere here with the formulas.

Click to expand...

We have a page on our website that allows you to calculate various things related to electric brewing.

Electric Brewing Calculator

It is basically an online version of the spreadsheet referenced above.

I have been using a Butterball Turkey Fryer, which uses a 1650 watt direct immersion element. Getting to a boil with a full pot of 6 gallons of water was marginal at best, as you near boiling, the heat loss from evaporation and radiation through the uninsulated tub was substantial. I decided that the sleek stainless steel look of the turkey fryer had to take a back seat to functionality. I had a big roll of Mylar faced foam insulation leftover from building my house. I wrapped the fryer and lid with this, and with the lid closed, I could reach a boil with relatively little heat loss, but it still took what seemed like forever.

Next step was to insulate the cavity between the tub and the housing. Since the heating element and all electrical wiring were detachable as a unit for cleaning, I could just stuff fiberglass into the space inside the cavity. Back of the envelope calculations for boiling times were reasonable, about an hour to Mash Temp, and another hour to boil. Once boiling, it would boil over if the lid was completely closed, and peter out if fully opened, so I left my stirring spoon across the top to keep the lid open an inch or so, which allows a gentle rolling boil. The basket also come in handy to keep the grain bag off the heating element.

I did an all-grain Porter with the setup, it just doesn't quite have the muscle to heat the Strike water as rapidly as I would have liked, but I managed to make some pretty good beer, despite its limitations. I have also done a couple of extract/partial mash brews with it.