30% boil off rate for a 1 gallon batch

[lolcats]

Is this normal?

Start off with 1.5 gallon and end with 1

Thanks!

 

[MagicMatt]

It depends on your pot size and the rate of boil (i.e. rolling boil or gentle boil). Obviously a rolling boil will cause a higher evaporation rate, as will having a pot with a larger diameter.

You should expect to lose between 0.75-1.5 gallons per hour of boil depending on these factors. When you are doing 6 gallon batches and boil off 1 gallon in the hour, it equates to about 14% evaporation rate I believe (1/7). Reducing the water volume will have no effect on evaporation rate, so if you're only boiling say 1.5 gallons, you'd still lose 1gal/hour, which would mean 66% evaporation rate. Obviously this is with the same equipment.

Using a smaller pot (as you should be) would lessen the evaporation rate, so yes, I believe 30% would be completely reasonable.

 

[lolcats]

I use a 4 gallon kettle, i do a nice rolling boil with the lid half way on

Thanks for your help!

 

[thekraken]

I usual boil off about 1.4 gallons/hour in a keggle.

For the record: your boil off rate is change in vol/boil time or 0.5 gal/ how ever long you boiled. I don't know where the stating boil off rates as percentages thing started but it drives me nuts! It just ain't right! Anyway, brew on.

 

[orangehero]

It depends on your pot size and the rate of boil (i.e. rolling boil or gentle boil). Obviously a rolling boil will cause a higher evaporation rate, as will having a pot with a larger diameter. Using a smaller pot (as you should be) would lessen the evaporation rate, so yes, I believe 30% would be completely reasonable.

This is mostly false. Only if you are using a gas burner would pot diameter affect boil-off rate and that is only because you change the how much of the burner's heat gets transferred to the pot. Energy input being the same the boil-off rate will be nearly identical regardless of pot size.

Once you're boiling, boil-off rate is proportional to energy input.

 

[fosaisu]

This is mostly false. Only if you are using a gas burner would pot diameter affect boil-off rate and that is only because you change the how much of the burner's heat gets transferred to the pot. Energy input being the same the boil-off rate will be nearly identical regardless of pot size.

Once you're boiling, boil-off rate is proportional to energy input.

This is interesting (and counter to many posts on HBT that suggest exposed wort surface area and atmospheric conditions can significantly impact evaporation rate). I lack the background to have an opinion, but look forward to an interesting discussion.

In your view would the relationship between wort volume and boil off be relatively linear (assuming that the amount of energy necessary to boil wort scales linearly with wort volume?) I.e. if a 5 gallon boil results in 1 gallon of boil-off, a 10 gallon boil would result in roughly 2 gallons of boil off?

 

[Lepetitnormand]

sounds about right, I brewed 1.25 gallon Tuesday evening and ended up with .75 to .8 gallon after 45 minutes boil, it was to energetic at the beginning

 

[thekraken]

This is interesting (and counter to many posts on HBT that suggest exposed wort surface area and atmospheric conditions can significantly impact evaporation rate). I lack the background to have an opinion, but look forward to an interesting discussion.

In your view would the relationship between wort volume and boil off be relatively linear (assuming that the amount of energy necessary to boil wort scales linearly with wort volume?) I.e. if a 5 gallon boil results in 1 gallon of boil-off, a 10 gallon boil would result in roughly 2 gallons of boil off?

^^ what he said. Lay some knowledge on us orangehero.

 

[MagicMatt]

This is mostly false. Only if you are using a gas burner would pot diameter affect boil-off rate and that is only because you change the how much of the burner's heat gets transferred to the pot. Energy input being the same the boil-off rate will be nearly identical regardless of pot size.

Once you're boiling, boil-off rate is proportional to energy input.

It's most certainly not "mostly false". Regarding your last statement, it's only true because the size of the kettle doesn't change. I could also twist that around to say "once you're boiling, boil-off rate is proportional to pot diameter", and that would also be true if you assume that the constant was 'energy input'. Both are correct statements, but they are only partially correct as they omit other obvious details needed to be physically true.


To elaborate further (stop reading now, I'm warning you, thermodynamics is not for the faint of heart!), water evaporation depends on a few more things than you mention. The evaporation rate is influenced by

1) The temperature of the water at the air-water surface
2) The humidity of the air
3) The area of the air-water surface
4) The temperature of the air
5) Water current convection (heat) and the ability to keep the temperature constant at 100°C.
6) Airflow (velocity) past the water/air surface.


In a real-world situation of evaporating water, none of the first four quantities above remains constant because the process of evaporation itself changes them. Water evaporating takes quite a lot of heat away when it evaporates (540 calories per gram to be exact; that’s enough to cool down 540 grams of water by a degree). If you are not very careful to replace the lost heat energy during the evaporation, the temperature will go down. And even then the temperature right at the surface will be lower than elsewhere in the water and it will depend on #5. For a similar reason, the air near the surface of the water will become more saturated with water as the water evaporates, and thus the evaporation rate will depend on #6.


So, suppose the flame transfers h kJ/s to the water. The latent heat of evaporation of water is 2260 kJ/kg.

For energy balance, the heat given to the water must be equal to the amount of heat required to convert water into steam:

h = m˙× 2260​

Therefore:

m˙= (h/2260) kg/s​

The rate of heat transfer doesn't really matter for most of us since we get a boil going to our desired roll and then keep it there for the length of the boil. Thus, we can ignore the flame and assume the water is at a certain temperature and stays at that temperature throughout the boil. So:

m˙= (ΘA(xs−x)/2260) kg/s​

Where
Θ=(25+19v) is the evaporation coefficient (kg/m²⋅hr). This is an empirical equation, so you can't derive it from first principles.

v is the velocity of air just above the surface of the water (m/s)

A is the surface area of the water (m²)

xs is the humidity ratio in saturated air at the same temperature as the water surface (kg H2O in kg Dry Air)

x is the humidity ratio in the air (kg H2O in kg Dry Air)

It's fairly straightforward to find x and xs from the relative humidity and the Mollier chart. Also, the velocity of air at the surface (v) should remain relatively constant for all intents and purposes for the duration of the boil.


Therefore, the last remaining variable of evaporation rate is indeed the surface area (A, i.e. pot diameter). Just like with any of the other factors, if you change this number, it will directly change the evaporation rate.

So your comments:

Only if you are using a gas burner would pot diameter affect boil-off rate and that is only because you change the how much of the burner's heat gets transferred to the pot. Energy input being the same the boil-off rate will be nearly identical regardless of pot size.

are the ones that are "mostly false" (actually, completely false).

 

[orangehero]

That's all great, very detailed post and apparently you are some sort of student in a related field. I'll concede it is true that surface area affects evaporation at the surface, but we're vaporizing at the bottom of the pot (or surface of immersion element) which is the main player regarding boil-off rate (which is what I meant by mostly false).

So your comments:


are the ones that are "mostly false" (actually, completely false).

You're saying that my comment that pot diameter changes how much heat is transferred from the burner is completely false?

 

[MagicMatt]

I'll concede it is true that surface area affects evaporation at the surface, but we're vaporizing at the bottom of the pot (or surface of immersion element) which is the main player regarding boil-off rate (which is what I meant by mostly false).

Not really. Boiling and evaporation are not necessarily the same thing. It is true that boiling causes evaporation, but evaporation takes place at the surface of a liquid and can only happen when the surrounding air is not saturated with the substance being boiled. Water doesn't have to be boiled (or even heated) to evaporate.

Nucleate boiling (which is what you're referring to) is when isolated bubbles form at nucleation sites and separate from the surface of the vessel. This separation induces considerable fluid mixing near the surface (the bubbles rising), substantially increasing the convective heat transfer coefficient and the heat flux. Most of the heat transfer (i.e. vaporization) is through direct transfer from the surface to the liquid in motion at the surface and not through the vapor bubbles rising from the bottom.

 

[lolcats]

Thanks guys for your input.

Let's say I'd like to move on to a 2 gallon batch with the same 4 gallon kettle I use for the 1 gallon batch where i get a 33% boil off rate...

What should be the boil off rate for the 2 gallon batch? Should I also be losing 0.5 gallons (thus a 20% boil off rate)?

 

[thekraken]

Thanks guys for your input.

Let's say I'd like to move on to a 2 gallon batch with the same 4 gallon kettle I use for the 1 gallon batch where i get a 33% boil off rate...

What should be the boil off rate for the 2 gallon batch? Should I also be losing 0.5 gallons (thus a 20% boil off rate)?

And this is why I don't like boil off rates stated as percentages!

To answer your question: yes, all other things being equal you should boil off 0.5 gallons.