Evaporation rates - Home Brew Forums

 Home Brew Forums > Evaporation rates

06-26-2010, 02:14 PM   #1
jjooeell
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Mar 2007
Posts: 4

here my scenario:
Start 9:07 110 degrees 6.6875 gallons or 107 cups
9:17 150 degrees
9:28 190 degrees
9:33 200 degrees
9:44 205 degrees
9:49 212 degrees Boil Start
10:50 Finished boil 4.75 gallons or 76 cups

Detailed set-up information
Converted Sanjay keg
2 - 2000 watt water heater elements
2 - GFI inline cords attached to separate 20 amp circuits
The boil does not seam to vigorous, and when I shut off one burner, the boil diminished quite significantly. This is all done in my basement that maintains a constant 67 degrees. I guess I need to know at what temperature everyone uses to
maintain a boil?
With averages around 40%, and the norm seams to be between 10 and 15%, what am I doing wrong??
Any help would be greatly appreciated

06-26-2010, 06:52 PM   #2
dzlater
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Apr 2008
New Jersey
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You might have screwed up the preboil volume
if it was 5.6875 it would be around 17%

06-27-2010, 12:33 PM   #3
jkarp

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Percentages are meaningless for comparison as they're system specific.

1.9gal in 1.7hrs = about 1.1gal/hr. Anywhere from 1-2 gal/hr is considered good so you're fine.

06-27-2010, 12:57 PM   #4
passedpawn
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I doubt there was much evaporation at all up to 212. That happens from the top surface of the wort.

Vaporization from boiling happens right at the surface of your heating element (or the bottom of the pot for propane burner users). This vaporization is a very calculable thing. Once your water reaches 212, you are ready to boil. Now, for every 2370 watts you put into your pot, you will boil 1 gallon of water per hour. So, your boiloff rate will be exactly

4000 / 2370 gallons/hr.

That's about 1.7 gallons per hour.
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06-27-2010, 02:29 PM   #5
jkarp

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While 2.37kW h is indeed the latent heat of vaporization for a gallon of water at sea level, it doesn't take into account heat losses due to kettle geometry, ambient temp, evaporation, etc.

06-27-2010, 03:32 PM   #6
passedpawn
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Quote:
 Originally Posted by jkarp While 2.37kW h is indeed the latent heat of vaporization for a gallon of water at sea level, it doesn't take into account heat losses due to kettle geometry, ambient temp, evaporation, etc.
Please, adjust my numbers to be more accurate. Of course you are right, but I am afraid this subject becomes a bit of a bore if it gets any more technical. Maybe I already crossed that line.

Because some of these things tend to cancel, I still think this simple equation I posted is sufficient. I.e., heat loss from the kettle walls works against the boil, but some evaporation will occur to increase the apparent boil-off.

Also, I feel like a broken record but kettle geometry will have no effect on the boil in an electric keggle. It will effect evaporation, but the boil-off from the element is determined solely by the wattage. The only way the shape the keggle could have on the boil is heat loss to the air.
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06-28-2010, 03:37 AM   #7
jkarp

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Quote:
 Originally Posted by passedpawn Also, I feel like a broken record but kettle geometry will have no effect on the boil in an electric keggle. It will effect evaporation, but the boil-off from the element is determined solely by the wattage. The only way the shape the keggle could have on the boil is heat loss to the air.
Sorry passedpawn but your record very definitely is broken.

Kettle geometry makes a significant difference because it alters the surface area that is radiating heat. Energy radiated out of the kettle is energy not available for heat of vaporization.

I ran extensive tests on this and I can assure you, geometry is quite important when designing electric systems.

06-28-2010, 02:18 PM   #8
passedpawn
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Quote:
 Originally Posted by jkarp Sorry passedpawn but your record very definitely is broken. Kettle geometry makes a significant difference because it alters the surface area that is radiating heat. Energy radiated out of the kettle is energy not available for heat of vaporization. I ran extensive tests on this and I can assure you, geometry is quite important when designing electric systems.
I'm truly interested in seeing the results of those tests. Please, if there is a link to those, let me know. Or, if you get some moments, post the setup and results here.

I have done some testing myself and thought it was insignificant (even on an uninsulated keggle).
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06-28-2010, 02:28 PM   #9
passedpawn
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[sorry for large pic, small one is useless here]

Here's a test I did in March (ambient around 60F)

I raised 10 gallons of water from 70 to 210 in 40 minutes. Uninsulated keggle. 5500W element.

Theoretically, this takes about 37.3 minutes. Since it took 40, I concluded that at most I lost about 6%. I'm likely not delivering the whole 5500W to the keggle due to losses in my SSRs, house wiring, etc., so I concluded that the heat lost was even less than 6%, and not signficant.

I could have made a much more elaborate study of this, but it's all I have.

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07-01-2010, 11:08 PM   #10
BrewBeemer
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Reading this my thinking the larger exposed surface area would cause a higher evaporation rate during the boil. Example a Sanke with the lid completly cut out allows 15.5" ID diameter for 188.69 sq/in of area, 20" ID pot of 314.16 sq/in of area. This difference comes to 1.66 times more exposed surface area allowing a higher evaporation rate. Unless i'm completly wrong take a Sanke with a 10" vs 12" cut opening will also reduce the evaporation rate more yet.
Using wild boil rates will about throw my above opinions out the window I bet.
I prefer 10" cut Sankes plus this allows for a wider top area left for fittings to be installed.

passedpawn, thanks for posting your chart as this can be used with different volumes and wattage elements as a fairly accurate guide with heating times and projects. I would like to see this same above chart done with a insulated Sanke, repeated then posted. Thanks.