cooling shrinkage

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bengerman

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I've been reading "Designing Great Beers" and when he talks about water volumes, he incorporates cooling shrinkage into his initial volumes. I might be missing something, but this doesn't quite make sense to me, because the shrinkage should be the same as the expansion from heating. In other words, if I measure my water out at 68*f, then heat it to a boil, it will expand 4%, then I cool it and it will shrink the same 4%. So why use that figure when measuring out your total water needed?
The only place I see the cooling shrinkage mattering is if I measure the wort while it's hot to see if I'm on track to hit my volumes.

Am I missing something? Is he measuring his total water hot?
 

ajdelange

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Other than the very small discrepancy because 1/0.96 = 1.04167 (not 1.04 i.e. shrinking by 4% followed by expansion by 4% doesn't get you back to the original volume) no, I don't think you are missing anything. Perhaps he is concerned about not putting more than 9.6 gal cold water in his 10 gal pot because it would expand to more than 10 gal and overflow when heated.
 

thedartedash

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I have not read the book, so I don't know the context.

Could he be addressing topping off while the wort is hot, only to find you would need to top off again when it cools?

Or is it more of a fun fact?

ajdelange may be right as well.
 

TyTanium

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Your thought process is correct.

I measure volumes throughout (1st run to adjust sparge water, pre-boil volume for gravity readings, etc), so I'm measuring hot volume. I also know that if I want 5.5g in the carboy, I need to stop my boil around 5.75g. So just know how you're using the adjustment and you'll be right.
 
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bengerman

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LKABrewer said:
If you are measuring how long to boil and still end up with 5 gallons, you will need to take shrinkage into account.
That's evaporation/boiloff, which he addresses separately.

Where he is using this is in the calculations to determine the total water required to make a batch of beer, so you can treat all of you water at the same time if you're adding salts or diluting with distilled water, and so you can have all your water measured out so you don't have to measure in the middle of sparging.
 
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bengerman

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ajdelange said:
Other than the very small discrepancy because 1/0.96 = 1.04167 (not 1.04 i.e. shrinking by 4% followed by expansion by 4% doesn't get you back to the original volume) no, I don't think you are missing anything. Perhaps he is concerned about not putting more than 9.6 gal cold water in his 10 gal pot because it would expand to more than 10 gal and overflow when heated.
1/0.96=1.041667
1.041667*0.96 =1


You could even put it into simpler terms:
0.96*1/0.96 =1
 

ajdelange

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Take 10 gallons. Reduce by 4%: 10*.96 = 9.6 gal. Now increase that by 4%: 1.04*9.6 = 9.884 != 10. Capiche?

This simple fact is usually discovered by an investor who loses, for example 50% of his 200K portfolio in a given year and is reassured when his broker tells him that the market went up by 50% the next year and not to worry. When the statement comes he discovers he has 150K, not the original 200K.

Of course with water (ignoring evaporation) there is no loss but that's because while the water shrinks 4% with cooling it increases by 4.167% when it expands. So while perhaps my comment in #2 is interesting it isn't really relevant.
 
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bengerman

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ajdelange said:
Take 10 gallons. Reduce by 4%: 10*.96 = 9.6 gal. Now increase that by 4%: 1.04*9.6 = 9.884 != 10. Capiche?

This simple fact is usually discovered by an investor who loses, for example 50% of his 200K portfolio in a given year and is reassured when his broker tells him that the market went up by 50% the next year and not to worry. When the statement comes he discovers he has 150K, not the original 200K.

Of course with water (ignoring evaporation) there is no loss but that's because while the water shrinks 4% with cooling it increases by 4.167% when it expands. So while perhaps my comment in #2 is interesting it isn't really relevant.
Ok, we'll take a new route. You have 1 liter of water at 4* C. By definition this is 1 kilogram, which specifies the number of molecules of water. you heat it to 99* C (so you don't lose any by boiling), and it expands by a little more than 4%, so you have about 1.05 liters of water, but it still weighs the same amount. The density has changed, but you still have the same number of molecules of water, so when you return it to 4* C, it will shrink back down to 1 liter.
No way around it. It's the same amount. You didn't make any molecules disappear, until you start boiling, but that's a different calculation.
If you were to boil it, though, and re-collect the vapor, then combine it with the water that remained in the pot, then coll it all back to 4*C, it would be the same 1 liter of water.
 

ajdelange

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Ok, we'll take a new route.... The density has changed, but you still have the same number of molecules of water, so when you return it to 4* C, it will shrink back down to 1 liter.
No way around it. It's the same amount.
Of course with water (ignoring evaporation) there is no loss but that's because while the water shrinks 4% with cooling it increases by 4.167% when it expands. So while perhaps my comment in #2 is interesting it isn't really relevant.
x
 

jmf143

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Its only meaningful if you're measuring the volume of hot wort.
 

tskinsv

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It is amazing how only the shrinkage factor has been propagated and is so widely used in all these home brewing calculators and programs. If you were to start with water at 68 degrees, you boiled it, had a way to loose no evaporation, you then cooled it back to 68 degrees you would have no water loss. Just like in the radiator of a car. If you want to be fairly precise you would need to expand the wort 4%, take out your evaporation, then shrink the remaining wort 4%. That is extra work to determine a really small number. Shrinkage is only important if you are going to precisely measure your water at flame out and calculate from there. Most people should just use an evaporation number and calibrate it with real results (pre boil vs. post boil volume of wort assuming all grain brewing).
 

ajdelange

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The whole drift of the thread is that 1/0.96 != 1.04 but that as the difference (1/0.96 = 1.04167) is so small you can ignore it. To estimate evaporative loss referred to room temperature measure wort volume at boiling temperature and multiply by the factor (rho_water(100)/rho_water(T)) where T is the reference temperature. If you have the ICUMSA tables you can use the densities of sucrose solutions at the two temperatures but the water approximation is pretty good for typical wort strength. Do the same at the completion of boil. The difference is the volume loss referred to T.

I got around all this by putting load cells under the brew stand on which my kettle sits so I can see the weight loss as the boil progresses. I found it easier to just use kettle volume as determined from the distance below the lipt of the kettle to the surface of the wort. I keep a table in the brewery that lists the actual volume for a given liquid height and the equivalent volume at 20 °C.
 
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