This is an experiment that I wanted to try for a while: Sparge a mash with cold instead of hot water.

Based
on my understanding of the lauter process sparging with cold water should have no or only little
impact on the efficiency if all the sugars, that will be dissolved, are
dissolved during the mashing process. While a colder sparge could slow
the speed of the run-off by causing the wort to be more viscous and
flocks of coagulated protein be smaller it should not affect how many
sugars are left behind. Especially in batch sparging where there is no
concern about channeling through the grain bed.

I decided to give
the cold water sparge a try on one of my Schwarzbier recipes. But since
I also wanted to change the grain bill slightly it is not a true
side-by-side where only the temperature of the sparge water changed.
Here is what I did for the two beers:

The things to note is that the conversion efficiency was very high on both batches. Almost all of the extract potential was realized in the mash which is an indication for good and complete mashing. The lauter efficiencies (percentage of dissolved extract that made it into the kettle) for both beers were very similar and as a result the efficiencies in the kettle were very similar as well. The differences that can be seen are easily within measurement errors.

This shows that a cold water sparge does not necessarily lower your efficiency. 

It should also be noted that the 2nd runnings, which were the cold runnings, never cleared up. The remained hazy throughout the sparge. 

Tonight I tasted the beers. Here are pictures that show the color and clarity of the beer

And the taste notes:

The cold sparged beer is definitely a slightly more hazy than the hot sparged version. This may actually have been the result of the cold sparge although I don't have a solid explanation for this. If the haze results from an increased protein content it may also explain the slightly better head retention and fuller mouthfeel.

Conclusions:

• Cold sparging does not have strong adverse effects on efficiency and beer quality
• when a mash-out is performed it has no apparent effect on the fermentability of the wort. I don't know if this is still the case when no mash-out is done.
• it may make the beer more prone to haze
• it does not really save time since the wort at the end of the lauter will be colder and require more time to be heated to boiling temperatures
• it can save the need for a pot for heating the sparge water
• Since the spent grain temperature is lower at the end of a cold sparge less energy is wasted.

While this was an interesting experiment I don't plan to repeat it in the near future. At this point I don't see any benefit in this practice except for cases were I forget to heat the sparge water.

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I have to start pushing this
test more. It seems as if it provides an answer to one of the most
common brewing forum questions: Why is my FG higher than expected?
Interesting enough, most of the very experienced American home brewers
don't use this test either. Might be that their process is refined
enough that the information given by this test is just redundant. But especially for beginning homebrewers, this test can provide invaluable information regarding the FG that can be expected. Almost as important ad taking an original extract (OG) reading. To many of them are just hung up on the attenuation numbers that are given for the yeasts at White Labs and Wyeast. When I asked them about the procedure that is used to get these numers, they told me that they don't even use a standard wort for all the yeasts.

I
certainly swear by it. How else can you find out if you met your
targeted fermentability during mashing before the beer fermentation is
done. It has become very important to brewing lager beers as they seem
to slow down significantly towards the end with a risk of being to
sweet before going to lagering temps. But even with Ales this test is
useful as it actually allows me to take residual fermentable sugar in
the beer into account when calculating priming sugar additions.

A few weeks back I decided to write another brewing water calculation spread sheet. The formulas were mostly taken from the literature and existing spread sheets. Then I decided to add a cation (positively charged ions) to anion (negatively charged ions) balance check just to see if the water profile that I created made sense. This is when I noticed an imbalance when creating brewing water from scratch by using distilled water and salts. The resulting water should not show an imbalance and every cation should have matching anion. But it was showing an imbalance when chalk was used. So I gave the fomulas used for chalk a closer look.

 And found that 1 mol (a unit that is proportional to the amount of molecules/ions of a particular substance) of CaCO3 was assumed to add one mol of bicarbonate to the water. And that in most spreadsheets and calculators the bicarbonate contribution was later used to calculate the alkalinity as CaCO3. But that didn't seem right. If CaCO3 adds only one bicarbonate, it also needs to add one hydroxyl ion (OH-):

(1)  CaCO3 + H20 -> Ca2+ + HCO3- + OH-

Since this would liberate hydroxyl the pH of the water would need to rise. If that is not happening then chalk can also be dissolved in the presence of CO2

(2)   CaCO3 + H2O + CO2 -> Ca2+ + HCO3- + HCO3-

In this case each mol of chalk would add 2 moles of bicarbonate. Yet another reaction is possible in the presence of acid and free protons

(3)  CaCO3 + H+ -> Ca2+ + HCO-

(4)  HCO- + H+ -> H2O + CO2

If neither of these reactions hapen the chalk won't dissolve. And that is clearly happening in brewing: If you add chalk to the brewing water it just turns the water cloudy and it will eventually settle. 

But does it really matter if the chalk dissolves or not? No. Because the bigger picture is that we added the chalk to give the water+chalk mixture more "alkalinity" I.e. acid buffering capacity. That acid buffering capacity is needed to reach a targeted mash pH once the malt, and with it acid buffers, has been added. At that point reactions (3) and (4) can take place. Whichever reaction is happening (1)..(4), chalk can neutralize 2 equivalents of acid and for all intents and purposes 1 ppm of chalk should therefore raise the alkalinity by 1 ppm as CaCO3. 

But that is not what most water treatment spreadsheets assume. They assume that 1 mmol/l CaCO3 adds 1 mmol/l HCO3- (bicarbonate) which drops one negative charge on the floor and caused the imbalance that I noticed. And then they go ahead and convert the ppm HCO3- to alkalinity as ppm CaCO3 by multiplying with the factor 50/60. In the end the addition of 1 ppm CaCO3 raises the alkalinity by only 0.5 ppm as CaCO3. This certainly seems wrong and I thought I had it all figured out until I decided to confirm this theory with an experiment.

The experiment is seemingly simple. Make small mashes with 3 different waters that are supposed to have the same residual alkalinity and test their pH. The first water (A) would be reverse osmosis water and serve as the control. The second water (B) would be reverse osmosis water with chalk and calcium chloride added such that the added residual alkalinity is 0 if the chalk contributes 2 alkalinity equivalents. The 3rd water (C) has chalk and calcium chloride added such that the added residual alkalinity is 0 if chalk contributes only one alkalinity equivalent. Whichever water that causes a mash pH to match the RO water mash pH the closest would have used the correct formula for alkalinity contributions by chalk. Here is a summary of the waters used:

• water A: reverse osmosis tap water
• water B: RO water + 80 ppm CaCO3 + 290 ppm CaCl2*2H2O; this increases the Ca2+ content by ~110 ppm
  • if 1ppm CaCO3 adds 1 ppm alkalinity as CaCO3 then the water's residual alkalinity (RA) increases by 0.0 over the RO water's RA
  • if 1 ppm CaCO3 adds 0.5 ppm alkalinity as CaCO3 then the water's RA decreases by 2.2 dH (German Hardness) or 40 ppm as CaCO3
• water C: RO water + 150 ppm CaCO3 + 150 ppm CaCl2*2H2O; this increases the Ca2+ content by ~110 ppm
  • if 1ppm CaCO3 adds 1 ppm alkalinity as CaCO3 then the water's residual alkalinity (RA) decreases by ~4.4 dH or 80 ppm as CaCO3
  • if 1 ppm CaCO3 adds 0.5 ppm alkalinity as CaCO3 then the water's RA remains unchanged compared to the RO water

200ml of each water were taken and heated to ~64C in the microwave. Then 50g of crushed pilsner malt were added to each water sample and stirred in. The mashes were occasionally stirred and a 15ml sample was taken from each mash after 5 min and cooled to 22C when it was measured with a pH meter. The results were surprising:

• mash A : pH = 5.76
• mash B : pH = 5.69
• mash C : pH = 5.77

According to these results the chalk added only 0.5 ppm alkalinity as CaCO3. And the pH shift for mash B is even in the range that would have been expected from the 2.2 dH RA drop. According to Kolbach the shift is 0.03 pH units for each dH which would be 0.066 and the results show ~0.07.

I couldn't believe it and started to ponder why that would be the case. Why is the added CaCO3 only neutralizing 1 equivalent of acid and not 2? Maybe it has something to do with the chalk not being dissolved.

So I conducted another similar experiment. This time between a control, water with suspended chalk and water with dissolved chalk. The chalk would be dissolved with CO2 which is brought into solution through shaking. Here is what I did. I added 0.24 g chalk and 0.88g calcium chloride to 1.5 l of reverse osmosis water. This is twice the salts added to water B in the previous experiment because I wanted to pronounce the effect of the residual alkalinity difference. I then shook this water and the added salts in a 2l soda bottle until the calcium chloride was dissolved. Immediately after shaking, without giving the chalk a chance to settle, I poured off 200ml for sample B. I then removed another 300ml in order to increase the head space. This headspace was then filled with CO2 and the bottle closed. When I started shaking the bottle, it immediately contracted which was a sign of the CO2 going into solution. After some shaking I let the bottle sit until the water became crystal clear again. This was not the result of the chalk settling but it being dissolved in the water. I then took 200ml of that water for samle C:

• water A: reverse osmosis
• water B: RO + 160 ppm CaCO3 + 580 ppm CaCl2*2H2O
  • RA = -4.4 dH or 80 ppm alkalinity as CaCO3 if chalk adds 1 alkalinity equivalent
  • RA = 0 dH or 0 ppm alkalinity as CaCO3 if chalk adds 2 alkalinity equivalents
• water C: water B + CO2
  • RA = -4.4 dH or 80 ppm alkalinity as CaCO3 if chalk adds 1 alkalinity equivalent
  • RA = 0 dH or 0 ppm alkalinity as CaCO3 if chalk adds 2 alkalinity equivalents

I then heated both samples to 68C, added 50g crushed pilsner malt to each and rested (with occasional stirring) them for 10 min. After that I took 15 ml samples and cooled them to 20-21C:

• mash A : pH = 5.67
• mash B : pH = 5.47
• mash C : pH = 5.66

So it appears that dissolving the chalk in the mash water changes its alkalinity potential. undissolved chalk has less alkalinity potential than dissolved chalk since mash B showed a much lower mash pH which could only have been the result of a lower RA than the 2 other mashes.

But why is this? Does not all the chalk dissolve in the mash as commonly assumed? And if yes why is that? And would it always be 50%? Shouldn't there be enough acid for this to happen via reactions (3) and (4)?

For now I don't have an answer to this. 

This weekend I took the time to take extensive
extract and volume measurements during a 2 sparge batch sparging
process here is the data and an analysis of that data:

• grist weight 5.6 kg
• total laboratory extract of that grist is  80% of 5.6 kg -> 4.5 kg
• water added to mash: 15.5 l (cold)
• extract of the first running in the kettle 22.5% (% extract is equal to Plato)
• volume of the first runnings in the kettle 9.75l at 65C -> 9.6l (cold)
• extract of the 2nd runnings: 11.75%
• volume in kettle after 2nd running: 20l at 75C -> 19.6l (cold)
• extract of the 3rd runnings: 7.4%
• volume in kettle after 3rd running (pre-boil volume): 26l at 90C -> 25l (cold)
• extract in kettle after 3rd running (pre-boil extract): 14.6%

The first analysis was for the extraction efficiency of the mash. The definition of extract percentages is:

(1)  E = 100% * m_extract / ( m_water + m_extract)

If
we want to know how much extract exist in a given wort of known extract
content that has been created with a known  amount of water we can do
this by rearanging (1) to

(2) m_extract = (m_water * E / 100%) / (1 – E / 100%)

(3) m_extract = (15.5kg * 0.225) / (1 – 0.225) = 4.5 kg

This
means that all of the extract available in the grain has been extracted
in the mash (100% extraction efficiency). This was confirmed by a
negative iodine test of the wort and the spent grain. I.e. no native
starch was left.

Since batch sparging was
used, a simple model can be used to calculate the lauter efficiency.
lauter efficiency * extraction efficiency is the brewhouse efficiency. 
For that model we need the amount of wort that is held back in the
lauter tun after each run-off. But this is not simply the amount of
water used for the mash minus the amount of first wort collected
because the volume of the wort increases when the extract is dissolved.
To get that volume we can use this formula which is the weight of
extract dissolved in a given volume of known gravity wort:

 (4) m_extract = ( E / 100% ) * SG * V_wort

SG is the specific gravity and it will be estimated with 1+E*0.004.Rearranged to V_wort we get

(5) V_wort = m_extract /  ((E/100%) * SG)

(6) V_wort = 4.5 kg / (0.225 * 1.090) = 18.3 l

This
means the 15.5 l water and 4.5 kg extract from the 5.6 kg grain made
18.3 l of 22.5% wort. 9.6l of that wort were collected after the first
run-off which indicates that 8.7 l are held back in the mash.

Batch
sparing is a process of successive dilution of the wort remaining in
the grain and running it off. This can be modeled mathematically and
has bee analyzed here. But since not all run-offs were of equal size, lets just calculate the efficiency step by step:

The first run-off will extract this percentage of the extract from the mash:

(7) Eff_1st = v_1st_runoff / (v_1st_runoff + v_wort_in_grain)

(8) Eff_1st = 9.6l / (9.6l + 8.7l) = 0.52 = 52 %

If
52% were recovered by the 1st run-off, then 48% of the extract are
still in the lauter tun. This extract is dilluted by the sparge water
and run off. The volume of the 2nd run_off is 19.6l – 9.6l = 10l and
the efficiency of that run-off is:

(9) Eff_2nd = v_2nd_run_off / (v_2nd_run_off + v_wort_in_grain)

(10) Eff_2nd = 10l / (10l + 8.7l) =0.53 = 53%

Using
this and the fact that the 2nd run-off was only able to draw from 48%
of the extract we can determine the combined efficiency from the 1st
and 2nd run off as:

(11) Eff_1st_and_2nd = 52% + 48% * 53% = 78 %

78%
of the extract are now in the boil kettle. This leaves 22% in the
lauter tun. With a 3rd run off size of 5.4 l we find the efficiency of
that run-off as

(12) Eff_3rd = 5.4 / (5.4 + 8.7) = 0.38 = 38%

and the combined efficiency of all 3 run-offs as:

(13) Eff_1st_2nd_3rd =  52% + 48% * 53% + 22% * 38% = 0.86 = 86%

This
means that with the given run-off sizes, number of sparges and amount
of wort left in the grain, an a lauter efficiency of 86% is to be
expected.

The actual efficiency into the boiler is the following:

(14) Eff_kettle = V_kettle * E * SG / (m_grain * 0.8)

the 0.8 represents the 80% laboratory extract of the grain.

(15) Eff_kettle = 25l * 0.146 * 1.058 l/kg / (5.6 kg * 0.8) = 86%

Since
the Efficiency is the product of extraction efficiency and lauter
efficiency and the extraction efficiency was determined to be 100%, the
actual lauter efficiency must have been 86%, which matches the
theoretical result very well. As a result no efficiency was lost due to
process inefficiencies and to increase that efficiency the following
process parameters could be changed:

• more sparge water: this would lead to a larger pre boil volume and longer or stonger boils and may not be desired
• less wort kept in the grain: This mash was done with conditioned
  malt which makes for a"fluffier" mash. Such a mash may hold more wort
  and I wonder if an unconditioned mash may result in less wort being
  held back and thus increasing the efficiency
• equalize the run-offs: the boost expected from that is very low. Se here.
• fly
  sparging: this method follows a different principle and should yield
  better efficiencies when done properly. But in addition to more time,
  it also needs a better lautertun which I don't have.

So, 86% for that beer is fine with me.

This experiment was designed to evaluate different Weissbier yeasts. The following yeasts were used:

• 351-1 (This yeast came from a WLP351 vial, but I think it is not the WLP351 strain anymore)
• WY3068 – Supposedly the W68 strain from yeast bank Weihenstephan. A very popular strain among German brewers
• WY3333
• WY3056 – Initially a blend of yeast, but I cultured this one from a single cell colony

The wort was a simple Helles Weissbier wort:

• 70% Weyermann light wheat, 30% Weyermann Bohemian Pils
• Step mash (55 C for 30 min -> infusion of boiling water -> 65 C for 45 min -> thin decoction boiled for 10 min -> 72C mash-out)
• 3.7g 10% Target and 7.5g 8% Northern Brewer hops boiled for 60 min
• Boiled for 60 min in a 2 stage boil: 1st stage just a simmer, 2nd stage with a 12 %/hr boil-off. I wanted to see if that type of boil, which is done by many commercial brewers, actually works for avoiding DMS. No noticable DMS was later found in the beer
• Cast-out wort: 16l @ 11.5 Plato

 4 one galon glass jugs were filled with 3l wort each. They were oxygenated with pure O2, but I did not take ones on how long (30s are likely). The following amounts of yeast were pitched

• 351-1: 10 ml sediment, propagated from an agar culture
• WY3068: 50 ml loose sediment from a Wyeast activator pack
• WY3333: 35 ml thin slurry from a Wyeast activator pack
• WY3056: 10 ml sediment propagated from an agar culture

It was noted that the pitching rates were rather different, but time and availability didn't allow for all yeasts to be grown the same way to the same amounts.

The yeast was pitched at 18C and since all growlers sat in the same water bath, it was assumed that they would have the same temperature. The temperature measured is the temperature of that water bath and because of the good heat conductivity the actual fermentation temperature was not expected to be different.

Over the next 2 days the temperature rose to 21C (70F) before it fell down to 20 C. The 2nd day after pitching the following extract values were measured:

• 351-1: 7 Plato
• WY3068: 6.5 Plato
• WY3333: 6.5 Plato
• WY3056: 6.0 Plato

Alongside the primary Fermentation, a number of fast ferment tests were done:

• dry bread yeast (1/4 tsp to 150 ml) : 2.5 Plato
• dry bread yeast (1/2 tst to 150 ml) : 2.5 Plato
• WY3056 : 2.5 Plato
• WY3333 : 2.6 Plato

The beers were bottles with residual extract. This means that the beer was simply filled into bottles once the extract level reached 3.7 – 3.8 Plato, which leaves enough residual fermentable extract to properly carbonate the beers. A practice called Gruenschlauchen in German Brewing.

During bottling a strong banana aroma was noticed for WY3056 and WY3333.

After one month (I didn't get to it earlier) the 4 beers were tasted together:

WY3086:

The beer pours a very strong head and is well carbonated. It's aroma shows moderately yeasty notes with some sulfur. The taste shows a little of the Weissbier clove spiciness but hardly any banana even though the beer smelled like banana juice at bottling time. The final extract was 2.7 Plato.

WY3333:

The beer is highly carbonated. It's aroma is yeasty with some banana/bubble gum character. But that fruit was very strong and came out later when the head subsided. The taste shows a restraint spiciness but no fruit. It is also a little yeasty, but more in a good way. Final extract 2.7 Plato. 

351-1:

The beer was not as well carbonated as the others and didn't pour a strong head. This is odd since this yeast is actually able to ferment below the 2.6 Plato of the other beers and was bottled with at the same extract level as the other beers. As a result more fermentable sugars must have been fermented that should have resulted in more CO2. The aroma spots some solvent notes (ethyl acetate). Later, the aroma is more clove dominated. It's taste is more spicy than all the other ones with less yeasty character. Final extract 1.6 Plato (!!)

WY3056:

The beer is highly carbonated. The aroma is clean initially, but once the head fell it showed a slight yeasty character. The taste is bready-yeasty (in a good way) without any signigficant spiciness. This character might make this yeast ideal for a Dunkles Weissbier. Final extract 2.8 Plato.