Boil Off Rate

[Justibone]

I was going to start a similar thread but I'll ask my question here. I use a 15 gallon ss pot and the boil off rate was completely different with my last 2 ten gallon batches. I started with 12 gallons with an imperial nut brown(assuming I'd lose 2 gallons in the boil) and ended up with a little over 11 gallons of beer. Just recently I did a pale ale and started with 11 gallons of water, this time I ended up with about 8.5 gallons of beer does the different styles of beer effect boil off rate? Both batches seemed to have the same vigorous rolling boil.

You didn't mention ambient humidity... is there a chance that played a role for you?

 

[Gremlyn]

I was going to start a similar thread but I'll ask my question here. I use a 15 gallon ss pot and the boil off rate was completely different with my last 2 ten gallon batches. I started with 12 gallons with an imperial nut brown(assuming I'd lose 2 gallons in the boil) and ended up with a little over 11 gallons of beer. Just recently I did a pale ale and started with 11 gallons of water, this time I ended up with about 8.5 gallons of beer does the different styles of beer effect boil off rate? Both batches seemed to have the same vigorous rolling boil.

A few factors can affect this:

• ambient humidity (as Justibone also said): higher humidity will lead to less boil off
• wind: more wind will increase boil off
• vigour of the boil: more energy in = less final wort out, this will work in conjunction with starting volume. If you're only able to put in X amount of energy to boil, and that X amount caused you to lose 2.5 gal of wort when you started at 11 gal, you'll get significantly less boil off when starting with 12 gal since there is more mass to heat and the energy is used less efficiently.
• starting volume: you can't assume 2 gal over 1 hour under all conditions, it's best to calculate as a percentage and keep good notes on the boil off amounts you got with varying batch sizes. Eventually you can plot that data and get an equation/chart to figure out approximate boil off
• specific gravity for a few reasons: 1) higher gravity will have a higher boiling point, 2) significantly different gravity can affect the vapor-liquid equilibrium (neither gravity-based factors are enough to produce your drastic difference, but they can contribute).

I think it's a combination of both the larger surface area and the burner. The more energy you are giving to the system the more energy you an provide to evaporation.

Yes, absolutely, both play a big part in your boil off.

 

[prosper]

Don't boil *too* vigorously. Aside from wasting a lot of gas, it will have adverse affects on your beer. When I started, I took the 'vigorous' boil very seriously, and ended up with massive evaporation rates, >50%. I would up with a few beers that had absolutely zero head retention whatsoever. Decreasing the boil to a point where I boil off ~1 gallon per hour (regardless of the kettle that I'm using) greatly improved my beer, and I get a lot more brews out of a tank of propane.

A proper boil, for me, looks like a low rolling boil - just enough so that the hops and kettle gunk are moving around.

 

[RDWHAHB]

I pick crustys side. It pretty simple. It takes energy to vaporize water (enthalpy of vaporization). More BTU's=more vaporization. Surface area of water does not matter. It is not a vapor pressure equilibrium issue. As stated, pot geometry (including diameter) would have an effect on how many BTUs you can get into the water.

 

[Gremlyn]

I pick crustys side. It pretty simple. It takes energy to vaporize water (enthalpy of vaporization). More BTU's=more vaporization. Surface area of water does not matter. It is not a vapor pressure equilibrium issue. As stated, pot geometry (including diameter) would have an effect on how many BTUs you can get into the water.

OK, so tell me how my theory is wrong. How does VLE have NO affect whatsoever? You can't just say someone is wrong, you have to attempt to prove it.

I'm going to have to go home and boil some water and prove myself empirically.

 

[CrustyBrau]

OK, so tell me how my theory is wrong. How does VLE have NO affect whatsoever? You can't just say someone is wrong, you have to attempt to prove it.

I already told you. If evaporation rate exceeds energy input, the water simply stops boiling and drops below 212. If energy input exceeds evaporation, the water goes above 212 (only possible in a pressurized vessel). There really is no wiggle room here. During a boil, energy loss to evaporation + other energy loss (radiation, transfer through kettle walls, etc) = energy input. Period.

 

[CrustyBrau]

[*]vigour of the boil: more energy in = less final wort out, this will work in conjunction with starting volume. If you're only able to put in X amount of energy to boil, and that X amount caused you to lose 2.5 gal of wort when you started at 11 gal, you'll get significantly less boil off when starting with 12 gal since there is more mass to heat and the energy is used less efficiently.

That is only true because the extra volume, and therefore extra surface area through which to lose energy through the kettle walls, causes LESS energy to be lost through evaporation.

[*]starting volume: you can't assume 2 gal over 1 hour under all conditions, it's best to calculate as a percentage and keep good notes on the boil off amounts you got with varying batch sizes. Eventually you can plot that data and get an equation/chart to figure out approximate boil off

This is just crazy. You should never try to express boil off as a percentage of original volume. If you understood better how energy input and boil off rate are tied, you wouldn't be saying this.

 

[RDWHAHB]

My understanding.
Water (all liquids really) boil when their vapor pressure equals the atmopheric pressure. The phase change happens throughout the liquid, not just at the surface. This is because the vapor pressure at that temp is sufficent enough to overcome atmopheric pressure and "rise up". The phase change takes energy to happen. Around 8000 BTU/gallon if I remeber correct.

Why it isn't a partial pressure equilibrium thing (more out on a limb for my knowledge):
Essenstially at boiling the equilbrium vapor pressure of water is atmopheric pressure. So even if the air was saturated, it would be in equalibrium. Adding that 8000 BTU would still boil off a gallon.

By extension, I don't think that humidity makes a lick of diffrence (assuming that you are putting the same BTU into the liquid). In practice I would guess that a burner is less efficeint at that heat transfer in high humidity, but that is a diffrent issue much like the SA of the bottom.

If there was a lid on the pot, the same gallon would evaporate, but this time some would cool underneath the lid, and chage phase back to liquid. The cooling comes from the same 8000 BTU/gal radiating out into space.

 

[CrustyBrau]

^^^ Gets it.

This is pretty good: http://en.wikipedia.org/wiki/Boiling_point

a liquid at saturation temperature and pressure will boil into its vapor phase as additional thermal energy is applied.

 

[Gremlyn]

That is only true because the extra volume, and therefore extra surface area through which to lose energy through the kettle walls, causes LESS energy to be lost through evaporation.
[snip...]
This is just crazy. You should never try to express boil off as a percentage of original volume. If you understood better how energy input and boil off rate are tied, you wouldn't be saying this.

Almost, but not quite. Volume is irrelevant, MASS is what you're looking for. Equal amount of energy in will overall heat each molecule less when there is more mass.

My understanding.
Water (all liquids really) boil when their vapor pressure equals the atmopheric pressure. The phase change happens throughout the liquid, not just at the surface. This is because the vapor pressure at that temp is sufficent enough to overcome atmopheric pressure and "rise up". The phase change takes energy to happen. Around 8000 BTU/gallon if I remeber correct.

Why it isn't a partial pressure equilibrium thing (more out on a limb for my knowledge):
Essenstially at boiling the equilbrium vapor pressure of water is atmopheric pressure. So even if the air was saturated, it would be in equalibrium. Adding that 8000 BTU would still boil off a gallon.

By extension, I don't think that humidity makes a lick of diffrence (assuming that you are putting the same BTU into the liquid). In practice I would guess that a burner is less efficeint at that heat transfer in high humidity, but that is a diffrent issue much like the SA of the bottom.

If there was a lid on the pot, the same gallon would evaporate, but this time some would cool underneath the lid, and chage phase back to liquid. The cooling comes from the same 8000 BTU/gal radiating out into space.

You are assuming a perfect system with no external environmental factors here...

Humidity absolutely makes a difference. Atmospheric pressure in laboratory terms in 1 atm = 14 psi = 1.01 bar = 101 kpa. Atmospheric pressure in reality = whatever it happens to be that day. Atmospheric pressure actually more affects [absolute] humidity than the other way around, but more liquid in the air will reduce evaporation because it will alter the VLE... though you'd have to believe me for that to be true :cross:

This is pretty good: http://en.wikipedia.org/wiki/Boiling_point

I can link to Wikipedia articles too... please read: http://en.wikipedia.org/wiki/Evaporation#Factors_influencing_the_rate_of_evaporation

Would someone in the class like to tell me what factor #7 mentioned in this article is?

Oh yes, that's right... SURFACE AREA! And why is that? Oh right, because of what I said four pages ago.

Also if you look at factor #1 and #2, that answers about the affect of humidity too.

 

[CrustyBrau]

We're not talking about (just) evaporation, we're talking about boil off. Evaporation is only one way, out of two, that water escapes a boiling volume of liquid. Surface area is essential to the rate of evaporation, as I think we all know. Vaporization is inversely correlated to evaporation. With a larger surface area, the rate of evaporation goes up. As a result, the rate of vaporization goes down, so that the total boil off rate remains the same.


Let me ask you this: What would happen if you increased the surface area of a boiling volume indefinitely while maintaining the same energy input? Would the water continue to boil no matter how much surface area you added?

 

[RDWHAHB]

I can link to Wikipedia articles too...

Would someone in the class like to tell me what factor #7 mentioned in this article is?

Oh yes, that's right... SURFACE AREA! And why is that? Oh right, because of what I said four pages ago.

No need to cop a smart tone. We are all friends here. If I provoked that, I truly apologize.

Now, to the issue as I see it in our disagreement:
Evaporation is different than boiling. Evaporation occurs below the boiling point. I agree with all your points on evap (humidity and surface are do change evap rates). However, boiling is different, at least in my understanding. Boiling water is in equalibrium with H2O saturated water.

Take a look at this:http://commons.wikimedia.org/wiki/File:Phase-diag.svg

When you are boiling, you are located along the blue line. The atmospheric pressure (weather, altitude) that day dictates where exactly on the line you are, but that does not matter really. The only thing that is required to cross the blue line (which means changing from a liquid to a gas, or "boiling off") is a heat input (i.e. the 8000 BTU).

I'll try to see if I have a thermodynamics book for a better explaination when I get home.

Regard, Jeffro

 

[Germelli1]

Reading this thread felt like watching the special olmpics...even the winner is retarded.

Just kidding! This was pretty interesting and I enjoyed trying to learn something!

 

[arturo7]

I think maybe everyone is correct?

In order to get the desired "vigorous boil" in the larger vessel you must add more heat, right? Therefore, given the same amount of liquid, the BO amount goes up.

 

[RDWHAHB]

Well if everybody is right, can we at least take an informal poll on who is most right? I kid....I kid. Mostly because I never win popularity contests....much too ugly of a mug.

 

[Germelli1]

I think maybe everyone is correct?

In order to get the desired "vigorous boil" in the larger vessel you must add more heat, right? Therefore, given the same amount of liquid, the BO amount goes up.

That is what I thought the entire time with the surface area argument. With regards to the geometry of our pots, larger surface area at the bottom of the pot obviously means larger surface area of the top of the wort.

As I read, I just sat back and thought "obviously surface area affects is in some form, who the phuck cares if it is the bottom of the pot or top" haha

 

[Gremlyn]

Evaporation does indeed matter; as temperature increases, as does evaporation rate. Boiling doesn't prevent evaporation, so evaporation is a big factor in boil off rate, thus surface area is a big factor.

 

[Gremlyn]

OK, I'm running an experiment. I have two pans with 1/2 gal of tap water, one is 8" in diameter (SA = 50.24 sq in) and one is 9.5" in diamater (SA = 70.85 sq in). I have two identically powered burners on my gas stove, and I will boil these at the same level for 15-20 mins and then collect the left over water and see the difference.

 

[Gremlyn]

Initial results:

• Both pans reached equal temperatures, on my digital thermo both read 210.7F at their highest.
• The smaller pan had a more vigourous looking boil despite both pans being at the same temp. I attribute this to the depth of the bigger pan, which would increase the pressure at the surface of the water, thus reducing the boil and it would also serve to reduce the evaporation.
• Despite the last point, it visually appears that the larger pan lost more water.

The burners were turned off and immediately covered with their respective lids to prevent more evaporation after the boil. I'm currently waiting for them to cool down so I can measure their final volumes.

 

[CrustyBrau]

The gas is going to throw you off. The larger surface area of the bottom of the pan is going to collect more heat than the smaller one.

 

[Gremlyn]

The gas is going to throw you off. The larger surface area of the bottom of the pan is going to collect more heat than the smaller one.

If the larger surface area collected more heat, then the temperatures wouldn't be the same between the two. This is especially true in the bigger pan that boiled less vigourously. If it was higher pressure in the column above the water in the larger pan lowering the boil vigour, and your theory is correct that the larger bottom SA collects more heat, the bigger pan would have been at a higher temp, but it wasn't.

 

[Gremlyn]

Starting volume: 8 cups

Smaller Pan Final: 5.75 cups
Smaller Pan Change: 2.25 cups

Larger Pan Final: 5.66 cups
Larger Pan Change: 2.34 cups

Results: I admit, this doesn't look great in support of my theory... the results are probably not outside of my measurement abilities. I'd like to try again and go longer as I only did a 15 min boil and also make the pan masses/depths a little closer. The smaller pan was a 3qt and I have 3.5 qt pan that is wider and just a little shorter, so I'll rerun with those two pans and run longer. More to follow.

 

[CrustyBrau]

If the larger surface area collected more heat, then the temperatures wouldn't be the same between the two.

You mean... You don't mean... The temperature during the boil? Please say that's not what you mean

Edit: I was out brewing in the garage. Translated, that means I was out drinking in the garage But I went back and re-read, and it certainly appears that that's what you meant. The good news is, now we know how to direct you toward the truth

Water boils at 212 (at sea level). Water ALWAYS boils at 212 (at sea level). You could throw in one BTU, or a million. If it's boiling (at sea level, did I already say that?), it's at 212 degrees. The difference will be seen in how fast the water turns to steam, not in what your thermometer says. I think you need to read up on latent heat. Here, let me throw around another Wiki post

http://en.wikipedia.org/wiki/Latent_heat

 

[Justibone]

PURE Water ALWAYS boils at 212 (at sea level).

There, fixed it for ya!

As for why humidity affects evaporation, I believe it's been mentioned that the atmosphere can only hold a certain amount of water at a given temperature. Sweating in Georgia is not the same as sweating in New Mexico for obvious reasons, even if you are at the same ambient temperature in both places.

Inasmuch as both evaporation and vaporization are involved in boil-off volumes, humidity can be relevant. Also, if the new air flowing in to replace the steam roiling off the boil surface is already near saturation, then the already-wet replacement air will allow fewer excited water molecules to leave the liquid surface than dry air would.

 

[passedpawn]

if the new air flowing in to replace the steam roiling off the boil surface is already near saturation, then the already-wet replacement air will allow fewer excited water molecules to leave the liquid surface than dry air would.

If the water does not vaporize into the air, then what happens to the energy going into the water?

 

[Justibone]

Radiation.

 

[passedpawn]

Radiation.

Seriously (IR radiation)? I'd buy that, but I don't have any proof of it (i.e., a reference on the irrefutable interwebs).

 

[Justibone]

Radiation is how physicists usually balance these equations.

 

[RDWHAHB]

Radiation...a giant slush fund for energy balance! Well played.

I do not believe that evaporation is a big player in comparison to boiling. I could be wrong though. One way to test would be to keep the water just below boiling temps for 1 hr. See how much is lost. If you lose a gallon, and your normal boil-off (with straight H20 as well) is 2 gallons in an hour, then I would be wrong. My guess is that it is much less than that.

 

[Justibone]

My guess is that evaporation is a small but not necessarily insignificant contributor.

Totally a WAG, though... I'm no physicist!

 

[Justibone]

In air at 20°C, if the vapor pressure has reached 17.54 mm Hg, then as many water molecules are entering the liquid phase as are escaping to the gas phase, so we say that the vapor is "saturated". It has nothing to do with the air "holding" the molecules, but common usage often suggests that. As the air approaches saturation, we say that we are approaching the "dewpoint". The water molecules are polar and will exhibit some net attractive force on each other and therefore begin to depart from ideal gas behavior. By collecting together and entering the liquid state they can form droplets in the atmosphere to make clouds, or near the surface to form fog, or on surfaces to form dew.

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/relhum.html#c5

So, I'm taking from this that in the local environment of the surface of the boil, molecules will have higher density.

Higher density = more frequent collisions.

Collisions = droplets.

Droplets ==> fall back into liquid (or condense on another nearby surface)

If the air is dry (has few droplets to start) then it would permit a much greater quantity of water molecules to enter gas phase without colliding, coalescing and (potentially) dropping right back into the pot.

The convection currents may be efficient enough to carry the droplets away easily (i.e. "steam")... but drier air would presumably have more total capacity for molecules to escape water-on-water collisions.

That is my hypothesis for why humidity matters in a boil. It *absolutely* matters for evaporation, but AFAIC, for a boil the answer is more (ahem) "up in the air".

 

[RDWHAHB]

Good site. Here is a good page on evap vs boil. Nothing really to add, just says things better than I can:
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/vappre.html

Even with evap, you need to add the latent heat of voparization, so the BTUs is likely the limiting factor. Even in 100% humidity.

 

[CrustyBrau]

^^^ Awesome link.

My guess is that evaporation is a small but not necessarily insignificant contributor.

[Broken Record]It depends how much energy you're putting in![/Broken Record]

At a very slight boil, evaporation accounts for almost all the water loss. At a very hard boil, it accounts for very little. If you put enough energy into 5 gallons of water, it will boil off in five seconds. You know, if you point a Saturn V at it or something During that five seconds, very very little will escape through evaporation.

 

[daugenet]

if you would like i can perform the experiment at my house on my induction cook top. i have two elements that are 1.4kw and a couple pots of varying width. should be simple enough to do two side by side for comparison.

induction means that all the energy should be going into the pot and not to atmosphere.

 

[RDWHAHB]

Poor colors, but this site is good too:
http://www.launc.tased.edu.au/online/sciences/agsci/essoil/boiling.htm

This is really pretty clear statement:

At this pressure, the liquid abruptly changes to gas anywhere in the liquid at a rate decided by the energy ( heat ) input necessary to break the intermolecular bonds of the liquid.