Boil Off Rate

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Brett3rThanU

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So I recently upgraded my equipment to a 15 gallon SS pot and Blichmann floor standing propane burner (60,000 btu's) only to find my boil off rate has gone up to about 2 gallons an hour. I think a lot of this is due to the large surface area the wort has in my large pot as I don't have a real vigorous boil going. Is there a downside to having such a high boil off rate? I've adjusted my strike and sparge water volumes accordingly, so ending volume isn't an issue.
 
Brett3rThanU said:
So I recently upgraded my equipment to a 15 gallon SS pot and Blichmann floor standing propane burner (60,000 btu's) only to find my boil off rate has gone up to about 2 gallons an hour. I think a lot of this is due to the large surface area the wort has in my large pot as I don't have a real vigorous boil going. Is there a downside to having such a high boil off rate? I've adjusted my strike and sparge water volumes accordingly, so ending volume isn't an issue.
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I don't think there is a downside, it is what it is. As long as you're adjusting for it, and hitting your volumes. I don't think a thinner mash will hurt, and you could always top-off your pre-boil volume in the kettle if you are worried about over-sparging.
 
large diameter BK = more boil off

adjust the water as you've done

don't compromise a vigorous boil to save a little water
 
Turn the burner down once you reach 212 F, but still keep a vigorous boil at that temp. If you keep the burner on high after this point you're just wasting propane.
 
Thanks for the replies guys, I'll just live with it.

I do turn the burner way down, it's not a very vigorous boil.
 
arturo7 said:
You want a vigorous boil.
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I always read that you can overdo a boil, meaning you want a good rolling boil, but it doesn't have to be spewing everywhere. Let's just say there's no still liquid in my pot when it's boiling, but it's also not a really big boil.
 
arturo7 said:
Perhaps rolling is a better description.
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Yeah, just depends on what everyone's interpretation of a vigorous, rolling, etc boil is. No worries, I've got that aspect covered. I think I'm just losing so much because of the large surface area of my wort. I went from a 7.5 gallon pot with a 14" diameter to a 15 gallon pot with a 22" diameter, same boil, and went from losing about 1 gallon an hour to 2 gallons an hour. I'll just keep doing what I'm doing... Thanks.
 
Your boil off rate is a direct function of the amount of energy being put into the wort by the burner. It doesn't have anything to do with surface area. You put X BTU into the wort, and Y will boil off. You put 2X BTU in, and 2Y will boil off. Simple as that.

In reverse, if you actually measure the amount of boil off, you can figure out how many BTU's made it into the wort during the boil. Completely ignoring surface area.

Edit: A larger diameter pot will likely absorb more BTU from the flame, so in a sense you're right about surface area, but it's the surface area of the bottom of the pot that counts.
 
Not talking about beer, but the same principles apply.

When boil-off conditions exist, the scale of the propellant tank, which determines the surface area to volume ratio, directly effects the boil-off rate. Large tanks, with a high volume to surface area ratio loose a small percentage of their propellant in a unit of time. A small tank, with a volume to surface area ratio will loose a much higher percentage of its propellant in the same unit of time..

http://webcache.googleusercontent.c...ea"&cd=6&hl=en&ct=clnk&gl=us&client=firefox-a


That was easy. Google is my friend.
 
CrustyBrau said:
Your boil off rate is a direct function of the amount of energy being put into the wort by the burner. It doesn't have anything to do with surface area. You put X BTU into the wort, and Y will boil off. You put 2X BTU in, and 2Y will boil off. Simple as that.

In reverse, if you actually measure the amount of boil off, you can figure out how many BTU's made it into the wort during the boil. Completely ignoring surface area.

Edit: A larger diameter pot will likely absorb more BTU from the flame, so in a sense you're right about surface area, but it's the surface area of the bottom of the pot that counts.
Click to expand...

Physics was quite some time ago, but I do remember going over this stuff. Basically what you need to look at will be the vapor pressure in the boil kettle. Every liquid has a vapor-liquid equilibrium point (VLE), which means that given a liquid in a closed container at constant temperature (and given time to reach the equilibrium point), there will be a specific quantity of molecules of the liquid in the gas phase above said liquid. How many molecules is determined by the temperature of liquid, and of course the composition of the liquid.

If you also have a bigger closed container that yields double the surface area of the liquid, you'll have more molecules in the gas phase in order to reach the VLE. How many more molecules would need to be calculated based on many factors and likely involve some calculus...

Now open both containers and heat them up to boiling, and for consistency's sake assume you are doing this with an equal energy-to-mass ratio so that both containers get heated the same. The liquid will always attempt to reach the VLE, but you've heated to boiling so the VLE is higher, which means even if the container were closed you would have more molecules in the gas phase above the liquid. BUT the containers are open and the kinetic energy of the molecules is way up. So they are flying about all over the place and will zip right out of your containers very often.

When you lose a molecule out of the boiling container, another has to come out of the liquid to replace it in order to maintain the VLE. The only problem is that you very likely haven't been able to reach the VLE since it's so much higher now AND you have molecules shooting out all over the place. So more and more molecules are coming out of the liquid in a losing battle to reach VLE. As stated before: when you have a bigger container, more molecules are required to come out of the liquid phase and into the gas phase, so you end up losing vastly more molecules when boiling in a bigger container.

Long story short, surface area definitely will increase boil off rate. Hopefully that wasn't too confusing, it's been a long week for me so I'm hoping I was thinking straight...
 
arturo7 said:
When boil-off conditions exist, the scale of the propellant tank, which determines the surface area to volume ratio, directly effects the boil-off rate. Large tanks, with a high volume to surface area ratio loose a small percentage of their propellant in a unit of time. A small tank, with a volume to surface area ratio will loose a much higher percentage of its propellant in the same unit of time..
Click to expand...
There's a problem here. I believe this is referring to the surface area of the outside of the tank, rather than the surface area of the top of the liquid within. A large tank does in fact have a higher volume to surface area ratio than a smaller tank. Note that the article says that a HIGHER volume to surface area ration correlates to a SMALLER loss to boil off. When expressed as a percentage of volume, that's exactly right.

The surface area of the tank is important because heat is absorbed through that surface, causing the liquid within to boil off. That's why I said that the surface area of the bottom of the kettle matters, because it will most likely absorb more of the heat given off by the flame.


The problem with using the surface area of the top of the liquid to determine boil off rate is this. When water evaporates from the volume, it takes energy with it. If the evaporation rate is too high, the volume will stop boiling. If it is too low, the temperature of the volume will exceed the boiling point. Therefore, the total boil off rate has to match the rate of heat input, minus other heat losses (such as through the side of the kettle, electromagnetic radiation from the liquid surface, etc.) This is why water boils AT 212 (at sea level). Not near 212, AT 212. No matter how much or how little energy you pump into it, if it's boiling, it's AT 212.

Now, below the boiling point, the rate of evaporation is indeed effected heavily by surface area.
 
When water boils, some of the steam comes from the surface in what we'd call ordinary evaporation, and the rest comes from the heated surface in the kettle, where it literally explodes into vapor and then rises to the top in the form of bubbles to escape to the atmosphere.

So, one way to look at this is that the rate of ordinary evaporation is most certainly tied to surface area. But, the total energy loss to evaporation must still match the rate of energy input when boiling. This is why the boil appears less vigorous in a larger pot with the same energy input. At some point, as the surface area grows, the rate of heat loss from surface evaporation will match heat input, and the liquid will stop boiling. If the surface area is reduced, surface evaporation will slow, and the boil will appear to be more vigorous.
 
Crusty, obviously I'm no scientist, but I think you just said a larger surface area gives you a less vigorous boil?

A less vigorous boil is what we try to avoid in brewing. If the boil calms down we crank up the heat. So now we have a rolling boil again and what the Gremlin said holds true.

So, let's put it this way: When making good beer, larger surface area = more boil off. Acceptable?
 
arturo7 said:
Crusty, obviously I'm no scientist, but I think you just said a larger surface area gives you a less vigorous boil?
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Everything else being equal, yes.

arturo7 said:
A less vigorous boil is what we try to avoid in brewing. If the boil calms down we crank up the heat. So now we have a rolling boil again and what the Gremlin said holds true.
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And I'm obviously a noob brewer. Is it correct to say that we would like a minimum *percentage* of boil off? I don't really see what the appearance of the boil has to do with anything. If the wort is boiling, water and other stuff is floating away. If you boil vigorously for 10 minutes, or lightly for an hour, and lose 10% of the volume in either scenario, don't we end up with the same total loss of bad stuff?

I realize you would end up with no bitterness in a 10 minute boil, so don't kick my butt on that aspect :D

arturo7 said:
So, let's put it this way: When making good beer, larger surface area = more boil off. Acceptable?
Click to expand...
It's misleading, which is why I clarified. If some guy reads that and thinks he can get a more effective boil by putting a larger diameter pot on the same stove top, he's not going to get what he expected. He'll get the same total boil off and a less vigorous looking boil. If he wants more boil off, wider pot or not, he's going to have to get more heat into it one way or t'other.
 
Crusty, the reason we want a good, rolling boil is to drive out volatiles, specifically, DMS. We don't try to control the boil-off rate, we just deal with it. Less or more doesn't mean diddly in the end product.

The goal is a solid boil for 60-90 minutes using whatever energy it takes to achieve it. Boil-off is just a function of pot geometry. It doesn't matter if I do a 5 gallon batch or a 10 gallon batch. The amount of wort I lose to boil-off is pretty much the same for either because I use the same pot.
 
arturo7 said:
Crusty, the reason we want a good, rolling boil is to drive out volatiles, specifically, DMS. We don't try to control the boil-off rate, we just deal with it. Less or more doesn't mean diddly in the end product.
Click to expand...
I'm looking around trying to figure out the cause/effect. Note, I don't really care, I'm perfectly happy to crank up the heat for a good rolling boil like I'm told. But, I don't think I understand what the rolling boil does.

When you distill alcohol, the quantity/minute of alcohol extracted is a direct function of BTU's pumped into the mash. Since DMS has a boiling point much lower than water, like ethanol, it seems to me the same dynamic should come into play. With high DMS levels, the wort should boil ever so slightly lower than 212*, and as the DMS evaporates, the boiling temp. should come up, exactly like what happens in a still as the alcohol content drops. Not because the boil appears vigorous, but because every molecule of DMS that evaporate takes heat with it that must be replaced with energy from the burner. The faster that heat is replaced, the faster the DMS (and likewise, the water) can evaporate.

Anybody know the science behind why a vigorous boil is better than a low boil, even if overall evaporation rate is the same?
 
If the DMS is removed from the liquid by the bubbles physically passing through it, then that would explain why more vigorous boil = good.

I know I had sulphur-tasting wine at one point, and used an aquarium stone to bubble CO2 through it to help the sulfites evaporate out of the wine. It surprised the hell out of me by actually working... go figger!

More vigorous boiling would also mix the wort better, allowing the DMS molecules to get to the surface so they can get the energy to jump off the liquid surface and float away.
 
arturo7 said:
So, let's put it this way: When making good beer, larger surface area = more boil off. Acceptable?
Click to expand...

Sorry, that is not correct. Don't confuse vaporization with evaporation. The former is what happens when we boil, and it is purely an effect of how much heat energy gets into the pot (like crusy said).

So, a larger diameter pot will boil off more, but only because the BOTTOM is larger and more efficiently captures the heat from the flame. It has nothing to do with the surface area, which only contributes to evaporation.
 
passedpawn said:
So, a larger diameter pot will boil off more, but only because the BOTTOM is larger and more efficiently captures the heat from the flame. It has nothing to do with the surface area, which only contributes to evaporation.
Click to expand...

I've been lurking and I have a question. If "boil-off" is indeed water vapor escaping into the atmosphere (the same as evaporation), and surface area contributes to evaporation, isn't this argument currently configured in a circular manner?:D
 
Reelale said:
I've been lurking and I have a question. If "boil-off" is indeed water vapor escaping into the atmosphere (the same as evaporation), and surface area contributes to evaporation, isn't this argument currently configured in a circular manner?:D
Click to expand...

Evaporation on the timescale of one hour is relatively insignificant compared to vaporization. Fill your brewpot, heat it up to 200+ degrees (but don't boil) for 1 hour, and see how much is gone (as a percentage). Not much, I'd wager.
 
Damn passedD, I'd like to bust your chops but, being a bit of a details freak myself, I must concede the point.

larger diameter = more boil off

how's that?
 
Justibone said:
Evaporation on the timescale of one hour is relatively insignificant compared to vaporization. Fill your brewpot, heat it up to 200+ degrees (but don't boil) for 1 hour, and see how much is gone (as a percentage). Not much, I'd wager.
Click to expand...

You're right. Heat speeds up both physical and chemical reactions, and vaporization in this case is just evaporation at an accelerated rate. Hence my original question: heat causes more evaporation (vaporization) and evaporation is influenced by surface area. Heat and surface area=more boil-off? I experienced the same thing going from a smaller pot to a keggle.
 
CrustyBrau said:
I'm looking around trying to figure out the cause/effect. Note, I don't really care, I'm perfectly happy to crank up the heat for a good rolling boil like I'm told. But, I don't think I understand what the rolling boil does.

When you distill alcohol, the quantity/minute of alcohol extracted is a direct function of BTU's pumped into the mash. Since DMS has a boiling point much lower than water, like ethanol, it seems to me the same dynamic should come into play. With high DMS levels, the wort should boil ever so slightly lower than 212*, and as the DMS evaporates, the boiling temp. should come up, exactly like what happens in a still as the alcohol content drops. Not because the boil appears vigorous, but because every molecule of DMS that evaporate takes heat with it that must be replaced with energy from the burner. The faster that heat is replaced, the faster the DMS (and likewise, the water) can evaporate.

Anybody know the science behind why a vigorous boil is better than a low boil, even if overall evaporation rate is the same?
Click to expand...

Vaporisation (which IS evaporation) isn't the same with a low boil vs a vigorous boil. The vigorous boil has more energy in it, meaning more fast moving molecules, and as a result more boil-off.

As for DMS, it's all trapped in the 7 gal of starting wort somewhere, and the vigorous boil keeps things moving more exposing the DMS trapped deep down in the wort. It can't evaporate if it isn't at the surface. A strong, rolling boil will do the best job of exposing the DMS molecules to the surface.

passedpawn said:
Sorry, that is not correct. Don't confuse vaporization with evaporation. The former is what happens when we boil, and it is purely an effect of how much heat energy gets into the pot (like crusy said).

So, a larger diameter pot will boil off more, but only because the BOTTOM is larger and more efficiently captures the heat from the flame. It has nothing to do with the surface area, which only contributes to evaporation.
Click to expand...

I feel that arturo7 may have been the only person to read my post... your statement is incorrect. I'm sure there is a point at which it holds true, but when dealing with various brew kettle sized pots capable of handling a similarly sized batch, you can safely assume you're getting fairly consistent energy usage no matter the pot.

Surface area absolutely matters, please go back and read my last post in this thread.
 
Gremlyn1 said:
I feel that arturo7 may have been the only person to read my post... your statement is incorrect. ... please go back and read my last post in this thread.
Click to expand...

You're right, I didn't. But I just did. And you are still wrong. I have an electric system (that is boiling right now). I know how many BTU's are needed to vaporize 1 pound of water. I know how many watts in a BTU/hr. It's very mathematically predictable.

I'm going to have to leave this as "we disagree" because it has been discussed thoroughly here before.
 
passedpawn said:
You're right, I didn't. But I just did. And you are still wrong. I have an electric system (that is boiling right now). I know how many BTU's are needed to vaporize 1 pound of water. I know how many watts in a BTU/hr. It's very mathematically predictable.

I'm going to have to leave this as "we disagree" because it has been discussed thoroughly here before.
Click to expand...

You can disagree all you like, but I'm not wrong... science says so. Your math likely backs up your theory under ideal conditions, but you can't ignore vapor-liquid equilibrium and kinetic energy (i.e. motion of the molecules).
 
CrustyBrau said:
Larger diameter = less vigorous boil, where total boil off is equal.

No, really :p
Click to expand...

Still wrong. The only reason this could be right is if your bigger pot has more mass and therefor absorbs more of the energy fro your heating source.
 
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 :confused: does the different styles of beer effect boil off rate? Both batches seemed to have the same vigorous rolling boil.
 
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.
 

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