
01012010, 10:53 PM

#1

May 2009
Lititz, PA
Posts: 364
Liked 8 Times on 7 Posts

I want to get a better handle on the amount of boiloff I'm experiencing without introducing other factors.
However, I don't feel like boiling 6 gallons of water for an hour if I can avoid it. Would I get an accurate answer if I were to boil, say, 3 gallons of water for half an hour and then double the amount that was lost?
I seem to be getting quite a bit of boiloff, like a gallon and a half. My kettle is relatively low and wide which would make sense but I'd like to do some actual measuring.



01022010, 05:01 AM

#2

Dec 2008
Hickory, North Carolina
Posts: 901
Liked 15 Times on 15 Posts

No. I"m going through this very experiment right now.
I'm figuring out what my boiloff is for a 1gallon batch of beer. When brewing a 5.5 gallon batch, I'm boiling off 11% of my total volume during a 1 hour boil.
For a 1 gallon batch (1 gallon into the fermenter) you can't assume the same 11%. I boiled 1 gallon of water for an hour and boiled off 55% of the total volume. I"m boiling 2 gallons at this very moment figuring out what percentage of that is going to boil off.
It's best to test the actual volume probably.
*** ADDED ***
Boiling 2 gallons in my 12 quart stock pot boiled off 35% of the total volume in 1 hour.
Reason: 2gallon result



01022010, 07:53 AM

#3

Jun 2007
Riverside Ca
Posts: 99
Liked 2 Times on 2 Posts

I dont understand why boil off is expressed as a percentage of boil, instead of a mass flow rate. If the heat of vaporization doesnt change much the rate should be constant. Say you have a 1kW heater, it will vaporize about 0.5 g/s .
0.5g/s the percentage that represents depends on the volume of water.
I understand there are many variables that I did not take into acount for such as kettle shape and wind factors, these have to be empirically determined
Reason: Bad spelling and incomplete logic



01022010, 07:57 AM

#4

Dec 2009
North Pole, Alaska
Posts: 1,809
Liked 42 Times on 38 Posts

Quote:
Originally Posted by salzar
I dont undertand why boil off is expressed as a percentage of boil, instead of a mass flow rate. If the heat of vaporization doesnt change much the rate should be constant. Say you have a 1kW heater, it will vaporize about 0.5 g/s .

It's expressed as a percentage because the boil time is generally a fixed constant of 1 hour.



01022010, 03:21 PM

#5

Death by Magumba!
Aug 2007
Melbourne, Fl
Posts: 2,222
Liked 50 Times on 21 Posts

Yeah, I've been thinking about this too. There is an inverse relationship between volume and boil off. It requires more energy to sustain 212 deg with a larger volume than a smaller volume, so less energy is transfered as steam at the larger volumes. The same reason why you can boil 1 gallon quickly with 1000W but can't get 5 gallons to boil with the same element. I'm sure the correct equation is floating around on the web some where.
Differential equations any one?



01022010, 05:31 PM

#6

Mar 2008
Charlottetown, PE, Canada
Posts: 284
Liked 4 Times on 4 Posts

Quote:
Originally Posted by klyph
It's expressed as a percentage because the boil time is generally a fixed constant of 1 hour.

you would need constant time and volume to fix the percentage. Better to use L per hour or Gal per hour, these are constant
Cheers
Joe
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01022010, 05:36 PM

#7

Mar 2008
Charlottetown, PE, Canada
Posts: 284
Liked 4 Times on 4 Posts

Quote:
Originally Posted by Hoppus_Poppatopolis
I want to get a better handle on the amount of boiloff I'm experiencing without introducing other factors.
However, I don't feel like boiling 6 gallons of water for an hour if I can avoid it. Would I get an accurate answer if I were to boil, say, 3 gallons of water for half an hour and then double the amount that was lost?
I seem to be getting quite a bit of boiloff, like a gallon and a half. My kettle is relatively low and wide which would make sense but I'd like to do some actual measuring.

Yes, as long as you don't think in terms of percent loss. My pot boils off 1.5 gallons per hour whether I start with 5 gallons or 12 gallons. So if I started with 3 and boiled for 30 minutes, I'd expect to find 2.25 gallons left in the pot, give or take a bit for evaporation while cooling, etc. As long as you start the clock at the start of boil, the numbers should play out.
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01022010, 05:39 PM

#8

Mar 2008
Charlottetown, PE, Canada
Posts: 284
Liked 4 Times on 4 Posts

Quote:
Originally Posted by JMSetzler
No. I"m going through this very experiment right now.
I'm figuring out what my boiloff is for a 1gallon batch of beer. When brewing a 5.5 gallon batch, I'm boiling off 11% of my total volume during a 1 hour boil.
For a 1 gallon batch (1 gallon into the fermenter) you can't assume the same 11%. I boiled 1 gallon of water for an hour and boiled off 55% of the total volume. I"m boiling 2 gallons at this very moment figuring out what percentage of that is going to boil off.
It's best to test the actual volume probably.
*** ADDED ***
Boiling 2 gallons in my 12 quart stock pot boiled off 35% of the total volume in 1 hour.

11% of 5.5 gallons is .6 gallons per hour
55% of 1 gallon is .55 gallons per hour
35% of 2 gallons is .7 gallons per hour
Close enough for horseshoes and hand grenades
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01022010, 06:44 PM

#9

May 2009
Lititz, PA
Posts: 364
Liked 8 Times on 7 Posts

Quote:
Originally Posted by Joe Camel
Yes, as long as you don't think in terms of percent loss. My pot boils off 1.5 gallons per hour whether I start with 5 gallons or 12 gallons. So if I started with 3 and boiled for 30 minutes, I'd expect to find 2.25 gallons left in the pot, give or take a bit for evaporation while cooling, etc. As long as you start the clock at the start of boil, the numbers should play out.

I should've been more clear. I'm not talking about a percentage because that depends on the volume. I'm talking about the total amount of water lost to the atmosphere as Joe expressed above.
Say, for an easy example, that I boiled away 1 gallon in 30 minutes. That would mean, I assume, that I would lose 2 gallons per hour and that the loss would be independent of the volume or at least close to independent with the volumes of 37 gallons with which I'm concerned.
Two gallons would be 67% of 3 gallons, 40% of 5 gallons or 29% of 7 gallons.



01022010, 06:51 PM

#10

Gritty.
Jul 2006
Southwest
Posts: 14,297
Liked 814 Times on 514 Posts

Boiloff rate is just that. It's a rate of change (and it is not exactly constant). It should never be expressed in terms of percent volume loss. The variables include (but are not limited to): heat energy, volume, surface area, ambient temperature, relative humidity, and dissolved solids.
To simplify the problem, calculate a boiloff rate for any given kettle assuming that temperature, humidity, and dissolved solids are negligible. Run a full sized boil for at least an hour using the heat source exactly as you would during a brew day. Take accurate sample volume measurements as often as you like. Calculate the loss in terms of average volume per hour. Consider the rate constant.





