1/3 Beer Evaporated After Boil!!

[Djaniefer]

Just finished creating my second batch of 1 gallon beer (a Cream Ale).

Everything went well until I poured my cooled boil into my 1 gallon jug and about 1/3 had evaporated during the boil (also lost some while transferring between stockpots during the sparge stage). See photo.

I topped the gallon off with tap water. What should I be expecting in the coming days/end product?

FYI: I used 26 cups of water, total. 10 cups for the mash and 16 cups for the sparge.

 

[jtratcliff]

You need to calibrate your boil-off and account for it in your pre-boil volume. It depends on on your pot size but is usually pretty constant once you know what it is... Plus all your other volume losses

In my 10G kettle, I lose about 1 gallon per hour during the boil, some to grain absorption during the mash (squeezing the bag minimizes this), some losses to trub, etc...

A calculator will help you to estimate these losses and figure out your strike and sparge volumes...
e.g. https://pricelessbrewing.github.io/BiabCalc/#Advanced

 

[brandonlovesbeer]

As long as you did the pre-math correctly, after adding the top up water (due to boil off) everything should be fine.

Hopefully you boiled the top up water first.

 

[CoreyD]

I used to have that problem, before I learned the right way to do it. Always topped it off, with no issues. It should turn out fine.
I never boiled my top up water with no ill effects. Although now that I see a post saying you should, it's probably a good idea.

 

[RPh_Guy]

I've been there, done that. No big deal.
From my understanding it's best to use distilled water (or reverse osmosis purified water) to top off. Otherwise your beer ends up with concentrated amounts of the minerals in your tap water.

 

[madscientist451]

The decision to add water after the boil should depend on a gravity reading, not a concern for volume. If your wort was at 1.050, your beer would be about 5% ABV, but if you add 1/3 of the final volume of water to that wort, your gravity will only be about 1.033 which will give you about a 3% beer.
There are many online calculators to help determine how much water to add to reach a desired gravity:
https://www.brewersfriend.com/dilution-and-boiloff-gravity-calculator/
But before you decide on your next brew, do a simple boil test. If you are doing one gallon batches, start with 1.5 gallons of water and boil for 1/2 hour, let it cool and measure what you have left. You can use the results to determine your boil off rates for 60-90 minute boils.
Once you get a bunch of brews done, you will figure out how much water you need to start with and not really even have to think about it all that much. Good Luck!

 

[Smellyglove]

madscientist451 gave you a great tip, do a test boil.

However. On that small scale I'd do a full 60 minute boil to extract data. Your evaporation will rise over time since the more you evaporate, the lower volum you'll have, and if you add the same amount of energy into the wort over the course of a full boil, that energy is distributed into a smaller volume over time, which will give you more evaporation than at the start of the boil.

 

[CudaJoe]

Know what your equipments water loss rate is. My keggle looses about 1 gallon maybe alittle more over the course of an hour. I keep the lid off which could prevent a lot of that evaporation. Better to be safe from boil overs then sorry. if your volume is lower than expected, check your FG and adjust the amount of water to add. It probably should be to your normal fill line but Id alway check. You never know if your FG is off without checking.

 

[RPh_Guy]

-off topic?-

....Your evaporation will rise over time since the more you evaporate, the lower volum you'll have, and if you add the same amount of energy into the wort over the course of a full boil, that energy is distributed into a smaller volume over time, which will give you more evaporation than at the start of the boil.

Maybe you are correct due to other variables, but from a physics standpoint that explanation doesn't make any sense. A particular amount of heat applied to any volume of liquid at boiling point will cause the same amount of phase change.

Either way, I agree; A 60-min test boil still is best to eliminate more variables.

 

[Smellyglove]

-off topic?-


Maybe you are correct due to other variables, but from a physics standpoint that explanation doesn't make any sense. A particular amount of heat applied to any volume of liquid at boiling point will cause the same amount of phase change.

Either way, I agree; A 60-min test boil still is best to eliminate more variables.

That extra energy added pr volume unit must go somewhere, it goes into more evaporation.

 

[RPh_Guy]

That extra energy added pr volume unit must go somewhere, it goes into more evaporation.

It might be confusing because it takes longer (more heat) to raise the temperature of a larger volume of liquid.

However, once the liquid reaches boiling point additional heat will change a constant amount of liquid to gas phase regardless of total fluid volume (the law of conservation of energy).
One gram of water requires 2257 Joules to vaporize under standard conditions (heat of vaporization). Whether that one gram is in a pot with 100g or 100,000g of water, vaporization rate will be (more-or-less) the same per amount of heat applied.... the temperature of the water doesn't change.

I hope I'm explaining this adequately!

 

[Smellyglove]

It might be confusing because it takes longer (more heat) to raise the temperature of a larger volume of liquid.

However, once the liquid reaches boiling point additional heat will change a constant amount of liquid to gas phase regardless of total fluid volume (the law of conservation of energy).
One gram of water requires 2257 Joules to vaporize under standard conditions (heat of vaporization). Whether that one gram is in a pot with 100g or 100,000g of water, vaporization rate will be (more-or-less) the same per amount of heat applied.... the temperature of the water doesn't change.

I hope I'm explaining this adequately!

Hmm.

If you have 100l of water and add 2257kJ, one gram of water will evaporate. The water stille holds 100c. Then toss out 99L of water. The setting on your burner remains the same. Where does that extra heat go if you dont count losses. It takes 2257kJ to evaporate that gram of water, but now its happening faster, since «more» energy is applied.

If you'd have a perfect system with no heat loss to the environment you are saying that evaporation rate will be the same if you add 2257kJ into an amount of water, then reduce that water to maybe 1/10th and apply the same amount of heat. Doesn't that mean energy is just magically dissapearing?

The way I see it that it takes that given amount of energy to evaporate one gram of water, but if that energy is applied "faster"(more heat) then you'll get "faster" (bigger) evaporation.

Edit: I'm terrible at explaining things in english.

You have a given amount of water. It takes 10 seconds to apply a total of 2257kJ into that water. Poof, one gram of water evaporates every ten seconds.

Then you toss out half of that water, now it takes 5 seconds (to simplify) to add those 2257kJ.. Evaporation can't be the same.

 

[RPh_Guy]

Hmm.

If you have 100l of water and add 2257kJ, one gram of water will evaporate. The water stille holds 100c. Then toss out 99L of water. The setting on your burner remains the same. Where does that extra heat go if you dont count losses. It takes 2257kJ to evaporate that gram of water, but now its happening faster, since «more» energy is applied.

If you'd have a perfect system with no heat loss to the environment you are saying that evaporation rate will be the same if you add 2257kJ into an amount of water, then reduce that water to maybe 1/10th and apply the same amount of heat. Doesn't that mean energy is just magically dissapearing?

The way I see it that it takes that given amount of energy to evaporate one gram of water, but if that energy is applied "faster"(more heat) then you'll get "faster" (bigger) evaporation.

Emphasis added.

The same amount of heat is being applied to the smaller volume as the larger volume.
Why are you thinking a smaller volume would have extra heat applied to it from the same heat source at the same setting? Your reasoning hinges on this faulty assumption.

Also 2257J/g = 2.257kJ/g. SI units

 

[Smellyglove]

Emphasis added.

The same amount of heat is being applied to the smaller volume as the larger volume.
Why are you thinking a smaller volume would have extra heat applied to it from the same heat source at the same setting? Your reasoning hinges on this faulty assumption.

Also 2257J/g = 2.257kJ/g. SI units

Im thinking that if you have your burner turned to 11, and you boil 10L. Then boil 1L, also turned to 11, you are applying more energy into that lower amount of water if you look at kJ/water.

And sice water doesn't get any warmer than "boiling temp" that "extra energy" is converted into more steam.

 

[RPh_Guy]

Suppose you have a heat source dialed to deliver 2257J/s (2257 watts). We've established water's heat of vaporization is 2257J, so each second it vaporizes 1g of water (1 mL) at 100°C.

If 2257J/s is delivered to 1000mL of water at 100°C, 1mL is vaporized every second.
If 2257J/s is delivered to 10mL of water at 100°C, 1mL is vaporized every second.... same amount of energy, same amount changes phase.

Since the heat source is 2257 watts it vaporizes 1mL each second at a constant rate. The volume of water being heated is not significant because each mL of water still in liquid form remains unchanged (no heat is gained or lost).

The energy output of the heat source would need to change in order to change the rate of vaporization (boil-off rate).

The 1000mL same would take 100 times longer than the 10mL sample to go from 25°C to 100°C, given the same heat source. The volume matters here because each mL of water increases in energy.

Hope this clears it up!

 

[Smellyglove]

Suppose you have a heat source dialed to deliver 2257J/s (2257 watts). We've established water's heat of vaporization is 2257J, so each second it vaporizes 1g of water (1 mL) at 100°C.

If 2257J/s is delivered to 1000mL of water at 100°C, 1mL is vaporized every second.
If 2257J/s is delivered to 10mL of water at 100°C, 1mL is vaporized every second.... same amount of energy, same amount changes phase.

Since the heat source is 2257 watts it vaporizes 1mL each second at a constant rate. The volume of water being heated is not significant because each mL of water still in liquid form remains unchanged (no heat is gained or lost).

The energy output of the heat source would need to change in order to change the rate of vaporization (boil-off rate).

The 1000mL same would take 100 times longer than the 10mL sample to go from 25°C to 100°C, given the same heat source. The volume matters here because each mL of water increases in energy.

Hope this clears it up!

This seems pretty logical, thanks. I know since I'm the only one believing this I'm the stupid one

It just seems very odd to me that if the burner is at a constant setting, the volume of water doesn't matter, since to me it seems logical that all that energy from the burner is distributed into a smaller amount of water, thus providing a more vigorous boil, but you did clarify it with saying that the energy into the water is constant, maybe I was thinking backwards.

 

[RPh_Guy]

Hey, we're all here to learn from each other!

 

[TasunkaWitko]

I'm no scientist or engineer; however, based on my experience:

a) Surface area (width) of the boiling kettle should be as narrow as practicable with a 1-gallon batch; I use an 8-quart stock pot.

b) It is not necessary to go balls-to-the-wall with the boil; a slow, gentle boil gets the job done.