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10-28-2011, 05:21 PM   #1
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 Correct power requirements to boil

Hello all,

I'm in the middle of a countertop brutus 20 build, and I finally broke down and did some math. I was looking for the power requirements to boil water and all I could find was the power requirements to raise water to 212 F. This is only half the story though, and not even the important half. The other part is called the latent heat of vaporization, essentially the energy required to turn water into steam. Here's the numbers I came up with:

The latent heat of vaporization of water (from wikipedia):
2257 kJ/kg

Lots of unit conversions...

(2257 kJ/kg)*(0.45 kg/lb)*(8.35 lb/g)*(0.28 Wh/kJ) = 2.37 kWh / gallon

The final unit, kilowatt-hour, is useful for us, because we usually measure boil off rate as a gallon per hour. For my system, doing 2.5g final batch size, I can assume something simple for now, like 0.5 g/h (although probably less in actuality). Putting that into the final equation,

(2.37 kWh/g) * (0.5 g/h) = 1.2 kW (rounded)

So, I need 1.2 kW to boil off a half gallon per hour. Larger batches will require higher boil-off rates, so just use the 2.37 kWh/gallon, multiplied by your expected boil off rate.

I'm trying to stick with a 1.5 kW element, due to old wiring in house. The breaker and GFCI outlet are rated for 20A, but the wiring sure as hell isn't. I rent, so I can't go tearing things up. This leaves me 300W to maintain 212 F once I'm at boiling temperatures, so I'd better insulate.

I'm a couple of weeks away from defending my PhD in mechanical engineering, so I have a fair amount of confidence in these numbers, but it is possible I did something wrong. I'm in controls engineering, not thermodynamics, and I haven't had one of those classes in years.

I just wanted to provide some more information for people trying to pick out their heating elements, and I haven't seen anybody show this yet. If it is out there on the forum already, my apologies.

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10-28-2011, 05:22 PM   #2
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I forgot to point out. While wort is not water, I dont anticipate a large difference in parameters. If somebody can figure out the latent heat of vaporization of wort, feel free to post a corrected solution

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10-28-2011, 06:07 PM   #3
jkarp
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Here's a spreadsheet that makes the calculations easy. 3KW will be plenty to maintain a boil on 5 gal. You'll need to throttle at least one of the elements, in fact.
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10-28-2011, 07:56 PM   #4
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This spreadsheet does not address the actual boil, just the energy required to raise the temperature of water up to 100C. As those boiling on wimpy stoves (such as myself), all boils are not created equal, even though the water temperature might be 100C.

Simply put, there is more to the equation than just heating water. It takes a certain amount of enthalpy exchange to transform a given unit of water into steam. My calculation was for how much energy it takes to transform 1 gallon of water into steam in 1 hour. This is after the boiling temperature has already been reached.

The more energy absorbed by the water, the more vigorous the boil, and the higher the resulting boil-off rate. If enough energy is not available, the desired boil-off rate (read: rolling boil) will not be possible.

The spreadsheet does not address this, and it does not address the increased heat transfer with ambient as the temperature of the water (and pot) increases. As the system heats up, more energy is dumped into the air, and the required energy to increase the temperature of the system another degree increases. For a weak enough heating element, it is possible that the system will stabilize at a temperature below boiling, similar to the terminal velocity of somebody sky-diving.

I'm not trying to start a fight, just help people getting into electric brewing. It seems there are a lot of people with questions on heating element sizing, and this addresses another aspect of that question

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10-28-2011, 08:18 PM   #5
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While my last class in thermodynamics is slightly greater than half a lifetime away, I can tell you that on my last batch I boiled approx. ten gallons for two hours and boiled off roughly 2 gallons with a 2000w element. I did wrap a beach towel around the kettle to minimize heat loss, but it didn't seem to make a big difference.

My hands on field experience leads me to believe that anything less than 2000w will need ideal conditions to boil; insulated narrow pot to minimize heat loss at the surface. I also have a 1675w electric turkey fryer and that will maybe barely boil five gallons without a lid. For minimal wattage boiling, a steam hood on the pot or partially covering to help to coax a better boil might help. 1500w is cutting it very close IME.

edit...with only a 2.5 gallon batch size I am much more optimistic.

edit again...I wouldn't advise brewing a larger batch w/ only 2000w, this episode was emergency practice, I had a plug break and by the time I fixed it, the batch was boiling along nice and gently so I opted to go make dinner and simply do a lengthy gentle relaxed boil.

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10-28-2011, 08:23 PM   #6
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I'm doing smaller batches, 2.5 gallons. I'm not sure how much heat loss I'll have with insulation. I'm crossing my fingers I'll be able to boil with 1.5k, considering it will be only about 3 gallons in the pot max, and it starts at 170

Your numbers seem in the ball park with my figures, so that is good enough for me.

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Last edited by BadWolfBrewing; 10-28-2011 at 08:28 PM.

10-28-2011, 08:44 PM   #7
jkarp
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Quote:
 Originally Posted by BadWolfBrewing The spreadsheet does not address this, and it does not address the increased heat transfer with ambient as the temperature of the water (and pot) increases. As the system heats up, more energy is dumped into the air, and the required energy to increase the temperature of the system another degree increases. For a weak enough heating element, it is possible that the system will stabilize at a temperature below boiling, similar to the terminal velocity of somebody sky-diving. I'm not trying to start a fight, just help people getting into electric brewing. It seems there are a lot of people with questions on heating element sizing, and this addresses another aspect of that question
The spreadsheet has % efficiency variables that one can play with to account for thermal loss. It has also been found quite accurate in its time estimates by those of us that have used it over the years in our designs.

Speaking of designs, I'm the guy who did the Countertop Brutus 20. I run my 2KW kettle at 80% with 4.5 - 5 gal and see a very good, rolling boil. Here's an old vid of mine - 5 gal of wort with 2KW at 100%. A quite vigorous boil.

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10-29-2011, 12:13 AM   #8
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Nice math OP. Your logic jives perfectly with me, (chemE)

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10-29-2011, 12:14 AM   #9
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Nice vid jkarp! Perhaps that is proof one can brew w/ 120v....I get tired everytime I read 240v, 240v, 240v, likely from those that have nver tried 120v.

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10-29-2011, 01:14 AM   #10
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Here's a couple of saved threads related to this. Maybe something helpful here.

http://www.homebrewtalk.com/f11/what-boiling-169336/index2.html#post1958160
http://www.homebrewtalk.com/f51/how-many-btus-your-hurricane-natural-gas-burner-burning-168787/index2.html#post1957831

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