Boiloff: would this be a viable experiment?

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.

No. I"m going through this very experiment right now.

I'm figuring out what my boiloff is for a 1-gallon 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.

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

salzar said:

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 .

Click to expand...

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

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?

klyph said:

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

Click to expand...

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

Hoppus_Poppatopolis said:

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.

Click to expand...

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.

JMSetzler said:

No. I"m going through this very experiment right now.

I'm figuring out what my boiloff is for a 1-gallon 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.

Click to expand...

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

Joe Camel said:

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.

Click to expand...

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 3-7 gallons with which I'm concerned.

Two gallons would be 67% of 3 gallons, 40% of 5 gallons or 29% of 7 gallons.

Boil-off 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 boil-off 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.