Evaporation rates

here my scenario:
3 cup dead space
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

I get about 29%
You might have screwed up the preboil volume
if it was 5.6875 it would be around 17%

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.

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.

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.

jkarp said:

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.

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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.

passedpawn said:

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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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.

jkarp said:

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.

Click to expand...

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).

[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.

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.

BrewBeemer said:

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.

Click to expand...

Evaporation and the water vaporized in a boil are two differentl things. Yea, they both happen at the same time, but one is affected by the surface area, and the other just isn't.

Evaporation is greater with hot water, of course, and it happens at the surface of the pot. The larger the surface area, the greater the evaporation.

However, the majority of water lost during the boil is not lost due to evaporation but to vaporization by boiling. This water is vaporized at either the element or the bottom of the pot (in the case of a propane burner). That is, it is gas well before it gets to the surface. For this water lost to the boil, only the energy applied by the element or the flame make a difference. The surface area has no effect.

I agree with jkarp that the shape of the kettle will have an effect on the heat lost to the environment through the walls of the pot (in the case of an uninsulated pot).

Beemer, you might get your wish this weekend. I have a roll of reflectix in my garage that I might put to use for an insulated graph. More later.

Do it while your still sober for good results this weekend LOL!
I realize the two different evaporations we're dealing with, mainly
the surface area is my concern. When pushing a keggle near it's max
capacity it becomes a challenge or balance for the maximum net into the fermenter is why I lean towards cutting a 10" sanke top.
It would be great to see the difference a insulated Sanke will make.
What about a vacuum liner like a Thermos coffee bottle?

Where's all the ex-nuke's at to chime in on the thermodynamics? I know there's a few lurking around here besides me.

For my two cents, I'm leaning with Passed Pawn. Evap rates would be affected by surface area, but not boil. Boiling is a straight power equation to overcome the latent heat of vaporization.

Whereas insulation would affect the boil off rates due to some heat loss, I'd think the effects would be small. Yes, heat is being lost and can't contribute to the heat required to boil off the water, but (radcon math anyone?) even at 400 watts of loss to maintain 212, we're only talking 10% (400w/4000w=10%) variation in boil-off volume due to this loss.

10% is not a huge variance for "home" testing purposes; changes 2 gallons to 1.8, for example.

To add another two cents: while heating elements are rated in "design" wattage- they will vary in and of themselves even when they are from the same batch. I wouldn't be surprised to see a 10% variation in rated and actual outputs, which introduces more error into calculations.

Ignoring that, the loss of the pot could be found from the graph PassedPawn posted- we have volume of water, and power input. It's straight calculations, and we can find how much power was lost during the heating process if we wanted. Anyone want to take a crack at it?

However, a significant heat loss would "arch" the graph visibly. As the temperature differential goes up (IE the Pot's hotter than the air around it), the heat loss would be greater, increasing the time to raise the temperature further. The effect gets worse as the temperature differential between ambient and the pot increases, so the graph would flatten more as we approached boiling point if the effect were huge.

BrewBeemer said:

What about a vacuum liner like a Thermos coffee bottle?

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yea, that'd be cool.

When I used to use propane, I really wanted to take the spear and weld it to the bottom, then cut out the hole so flames could shoot up the middle. I figured that would improve my burner's efficiency. Then I went with electrons.

Oh- missed it. Passed already ran the calcs- about 6% loss on his run.

Don't forget about thermal expansion of the liquid. Having 6.6gal at 150F is a different amount of liquid vs 6.6gal at 70F.

BrewBeemer said:

Reading this my thinking the larger exposed surface area would cause a higher evaporation rate during the boil.

Click to expand...

That's what concerns me more with my above previous reply vhampyre.

Gremlyn1 said:

Don't forget about thermal expansion of the liquid. Having 6.6gal at 150F is a different amount of liquid vs 6.6gal at 70F.

Click to expand...

About 4%.

2nd graph down:

http://www.ce.utexas.edu/prof/kinnas/319LAB/Book/CH1/PROPS/densgif.html

BrewBeemer said:

That's what concerns me more with my above previous reply vhampyre.

Click to expand...

Tougher to calculate; but I'd agree that the kettle design could vary the evaporation rate significantly.

http://www.southshoregunitepools.com/resources/htms/evaporation.htm

I think I'm following your point: an extreme example would be putting a lid on the pot- evap rate drops to nearly zero, chances of boil over go through the roof. Therefore, just simple common sense thinking with those arguments says surface area and evap rate are important, setting aside boil off rate.