HOMEBREW Digest #4861 Tue 04 October 2005

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  Re: Low Final Gravities (Bob Tower)
  Low Final Gravities (leavitdg)
  Wort Chiller Efficiency (Pete Limosani)
  Re: And another thing ! ("Mike Dixon")
  Viscosity ("Spencer W. Thomas")
  Corn, Corn, Corn (Scott Birdwell)
  efficiency, 1/ batch sparge analysis ("steve.alexander")
  efficiency, 2/ continuous sparge analysis ("steve.alexander")
  many things ("steve.alexander")

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---------------------------------------------------------------------- Date: Tue, 4 Oct 2005 01:14:23 -0700 From: Bob Tower <tower at cybermesa.com> Subject: Re: Low Final Gravities Allan J. Horn of Orinda, California wrote about his troubles with high final gravities (at least that's what he meant to say in his subject line, either that or low attenuation). I had a similar situation a few years back. I'd scaled my brew operations from 10 gallons to 20 and had built my dream fermentation system only to find that my beers all finished in the upper teens and low 20s (specific gravity) just as Allan is experiencing. Also as Allan is doing is tried repitching, adding yeast nutrients, rousing the yeast also to no avail. I also was using pure oxygen added to the wort. But still my beers were finishing high. Then a fellow brewer recommended a different oxygenation regimen. He told me to try adding the oxygen in stages rather than just once at pitching as I had been doing. On my next batch I added oxygen right before pitching, then four hours later and again another four hours later and yet again after 12 hours from pitching the yeast. This immediately corrected the problem and all my beers since have finished exactly where I expect them to (generally in the 1.008 to 1.014 range depending on the beer). My current oxygenation schedule for beers under 1.060 goes as follows (times listed are hours from pitching): 0 hours 2 hours 6 hours 12 hours I am using the red 40.1 gram bottles of oxygen intended for small torches available at hardware stores, a brass oxygen regulator (no gauges) and a 0.5 micron stainless steel aeration stone. I turn the regulator on until I just get bubbles out of the stone. At each stage I run the oxygen for 2 minutes. For beers over 1.060 I will add oxygen at the 24 hour mark and if it's really strong (say over 1.080) I will also oxygenate at the 36 hour mark. I don't have an explanation of why this works (or even if this is truly the solution to my problem) but to my knowledge it was the only variable that I changed. I was using the same yeast, in the same quantity, handling it exactly the same way as I had before. Also, I didn't change my recipes nor did I change my mashing technique in any way that I am aware of. But from the time I went from fermenting 10 gallon batches in either 2 plastic buckets or glass carboys to fermenting 20 gallon batches in a single fermentation tank I had problems with the first 4 batches finishing with high gravities. Spreading out the oxygenation seemed to bring the gravities down to where I had specified in the recipes. The only thing I can guess is that it could have something to do with the fermentation volume and/ or the fermenter geometry or some other unknown variable. Previously, fermenting in five gallon volumes I was running the oxygen for about 45 seconds (as specified in the instructions that came with the stone and regulator) right before pitching and I had no problems with my finishing gravities. Assuming that I would need more oxygen for more beer, on my first four 20 gallon batches I had increased the time to 4 minutes. It would be interesting to see if Allan tries altering his oxygenation as I have and whether or not it lowers his final gravities. I hope this information helps. Bob Tower / Los Angeles, CA Return to table of contents
Date: Tue, 04 Oct 2005 06:21:18 -0400 From: leavitdg at plattsburgh.edu Subject: Low Final Gravities Thanks to Al and Jeff...agreed that taste IS more important than color, and I am headed up to the NC Malt Supply to pick up some more Maris Otter.. === Allan; I have also been trying to get the final gravities down, and think that a lower rest (mid 140's) followed by a slow ramp up, stopping in the mid 50's, slowly ramping to mashout has helped me. The grain bill is, of course, important, too. Ane real healthy yeast, of sufficient quantity. Please let us know what does it for you. Darrell Return to table of contents
Date: Tue, 04 Oct 2005 07:29:10 -0400 From: Pete Limosani <peteLimo at comcast.net> Subject: Wort Chiller Efficiency Fellow Brewers, I'm trying cut down on my water usage during a brew session. I boil about 7.5 gallons down to 6 and use an immersion chiller to cool it. It generally takes 20-25 minutes to get from boiling to 70-75*. When I first start cooling I open up the valve full throttle and the water escaping the chiller is steaming. The difference in temperature between water entering the chiller and exiting the chiller is great. As the wort cools down, the difference in temperature closes and it takes longer to drop that next degree. As the difference in temperature decreases, I am tempted to slow the amount of water running through the chiller to save water. My thinking is that less water will be used, but the water may extract a couple more degrees while it is travelling. Is my logic flawed? Will I use less water over the whole cooling cycle by keeping it wide open? Or will I use less water by slowing the flow and waiting a little longer? How do I find the best balance between the rate of flow and the rate of heat extraction? My chiller is home made with about 20' of 3/8" O.D. copper tubing. I just measured the temperature of my tap water at 65*. Probably gets a little colder on brew days because of the volume of draw prior to chilling. Thank you in advance for your thoughts. Pete Return to table of contents
Date: Tue, 4 Oct 2005 09:00:36 -0400 From: "Mike Dixon" <mpdixon at ipass.net> Subject: Re: And another thing ! From: "steve.alexander" <-s at adelphia.net> > If anyone thinks the brewing science articles are irrelevant > should take a look at the most recent issue of JIB where there > is an article on mash viscosity which has relevance to the > discussion of extraction efficiency. > > For many years I have preached the your HB books are wrong > and that the increased temperature of mashout has a negligible > impact on mash viscosity. There are graphs to prove it in > the several articles on mash rheology. > > http://www.scientificsocieties.org/jib/contents/current.htm I guess I am not seeing the same thing. Using this article: http://www.scientificsocieties.org/jib/papers/2005/G-2005-0718-276.pdf Figure 2c shows a definite decrease in viscosity from 65C at 60 min to the time they go to 78C. Same for this article: http://www.scientificsocieties.org/jib/papers/2005/G-2005-0816-281.pdf Figure 5 What I also noticed is they made grain dust. In the first article the grain was ground to 0.06 mm. In the second article the grind was 0.02 mm. Perhaps this is a standard they are to use but contrast that with the homebrewing malt mill gaps of 0.039" - 0.050" (1 - 1.27 mm). I saw a probrewer forum where the gentlemen mentioned 0.033" (0.8382 mm), but that still is quite a coarse grist compared to the experimental method used with the viscometer. What I do know is that stirring the mash I create at home becomes easier as the temperature increases from after dough in to mashout. That indicates to me that something has changed and with the varibility that would be noticed from a standard mash setup to a grain dust slurry in a laboratory, I'm fine with atrributing it to a lowered viscosity of the mash due to the higher temperature. I suspect that could be tested by measuring the amperage draw on a continuously stirred mash from start to finish. A homemade viscometer of sorts. If the amperage drops then the mash is become less "tough" to stir, or some way to measure the torque. Anyone got a Compu-Brew system and time on your hands? Cheers, Mike Dixon Wake Forest, NC www.ipass.net/mpdixon Return to table of contents
Date: Tue, 04 Oct 2005 10:24:27 -0400 From: "Spencer W. Thomas" <hbd at spencerwthomas.com> Subject: Viscosity Mike Dixon writes: >Using this article: >http://www.scientificsocieties.org/jib/papers/2005/G-2005-0718-276.pdf >Figure 2c shows a definite decrease in viscosity from 65C at 60 min to the >time they go to 78C. > When I look at their Figure 1d, I see mash continuing at 70C until 111 minutes, during which time the viscosity decreases from about 2.85 to 2.7 (5% decrease). I don't see any significant change in viscosity due to the mashout at 78C. I have to admit that I don't quite understand why point P is labeled "end viscosity at 70C" when it's at the beginning of the 70C segment of the mash, and point R is labelled "end viscosity at 78C" when it's before the ramp-up to 78C even starts. Same thing with the second article Mike points to. At the end of the 72C mash step, the viscosity is down to about 320 mPas, and in the 78C mashout, it drops to 300 mPas (6% drop) (Figure 2d.) =S Return to table of contents
Date: Tue, 4 Oct 2005 10:20:33 -0500 From: Scott Birdwell <defalcos at sbcglobal.net> Subject: Corn, Corn, Corn Jeff Renner spake: > Brewers us corn syrup especially made for brewing that has the same > kinds of sugars as mashed malt, or at least a similar ratio of > fermentable sugars (see above). I am not sure if these kinds of > syrups are available in the HB trade. This kind of corn syrup is definitely available to the homebrew trade. We have stocked this item for over 12 years. It is our alternative to the "tin & kilo" brewing that the malt extract kit manufacturers have been trying to foist upon us for decades. A can of decent quality hopped malt extract and a couple of pounds of this stuff can make a very palatable, if not exciting, batch of beer. Unlike "sugar beers," it lacks that cidery, winey twang in the finish. I'm not ready to trade in my mash tun, but it's a good way to start, especially if you prefer a lighter-bodied beer. Thought you'd like to know. Scott Birdwell DeFalco's Home Wine & Beer Supplies Houston TX www.defalcos.com Return to table of contents
Date: Tue, 04 Oct 2005 12:22:21 -0400 From: "steve.alexander" <-s at adelphia.net> Subject: efficiency, 1/ batch sparge analysis Well my face may red, but it's others acting like baboons here. Continuous sparge is LESS efficient than batch in the general case. But it requires some detailed analysis to see why. Let's consider batch sparge first. What happens in any sparge is that we have grist which initially contains all the soluble extract and we have water which will carry of part of the extract. [Well goods becomes more soluble throughout the mash, but we assume the mash is over and all solubles are in their soluble state at beginning of mash]. Let's say we have X volume units of soluble extract (when in solution) and let's just estimate that this has some density in solution around 1.55 times that of water - close to mash extract, so the mass of extract is then X * 1.55. The grist also retains a certain volume of solution which is never removed by any reasonable sparging/draining process The volume of unremoved stuff is U. Now all mashes start with a volume of mash water M, and we assume that after the relatively long mash period that the extract both in and outside the grist are near equilibrium, same concentration of extract, and the total volume of this solution is (M+X) and the amount of spargable liquid is (M+X-U). So if we just drained the mash water we would obtain (M+X-U)/(M+X). Multiply by 100 for percent efficiency. We'll also need to reference the volume of sparge water S. I'll note without proof that it's simpler to consider the total amount of extract left behind in the grist ,and in this case U/(M+X) is the fraction left behind. Practical Example: I could quibble just slightly over the numbers, but let's go to ChadT's recent example. 10# of grist, assuming 65% of the mass is soluble extract means 6.5# of extract(2.95kg of extract mass). The extract volume is about 2.95/1.55 = 1.90L = 0.5 gallons of extract volume {X=0.5gal}. Chad also assumes the grist hold 0.1gal/lb, so {U = 1 gal}. Chad chooses an initial mash water addition of 3.1gal of mash water (1.24qt/lb) so just draining the obtainable fraction of the mash water would yield (3.1+0.5 - 1.0) / (3.1+0.5) = 0.722 => 72.2% extraction with no sparge water at all. Now in Chad's example he compares two cases, first adding all the (3.4gal) of sparge water with the mash water, obtaining equilibrium and draining and also the continuous case. I'd like to expand this it a more detailed analysis Chad's all-water together case is simple. It's the same extract result as increasing the mash water addition to (3.1+2.4=) 6.5 gallons. So it's (6.5+0.5-1)/(6.5+0.5) => 85.7% efficient. Let's also consider a 1-sparge case (1S/) where we drain the mash water and add all the 3.4gal of sparge water, achieve equilibrium then drain that too. As above the initial mash water carries off 72.2% of the extract. The retained grist contains 1 gallon of water with only 27.77% of the original extract. After the 3.4gal of sparge water reaches equilibria we can drain 3.4gal of solution and this will contain 3.4/(3.4+1.0) fraction of the 27.77% of original extract. Put into notational terms we have: E = (M+X-U)/(M+X) + [1-((M+X-U)/(M+X)] * S/(S+U) or more simply the amount left behind is: (U/(M+X)) * (U/(S+U)) For our example this is: E = 0.722 + [0.2777]*0.773 => 93.7%; 6.3 %left behind. Now consider the 2-sparge case, where we drain the mash water, add half the total sparge water(S/2), equilibrate, drain add the second half of the sparge, equilibrate drain. In our example for this case the mash drain removes 72.2% of extract leaving 27.8%, the second removes [[ (1.7/(1.7+1.0)) 62.9% of the remaining 27.8%]] or 17.5% leaving behind only 10.3% of the original; then the second sparge also removes the 62.9% of the remaining 10.3% leaving only 3.8% behind for a 96.2% efficiency in the 2-sparge case using the same amount of water. As it turns out we can re-write the fraction of the extract removed more concisely as: E = 1 - { U/(M+X) * {U/((S/N)+U) } ^N}, where N is the number of batches in the sparge and ^ is exponentiation operation. In our example above, U = 1gal ,X = 0.5gal, M=3.1gal ,S = 3.4gal, so for N = 1; E = 93.7% for N = 2; E = 96.2 % for N = 3; E = 97.1% for N = 4; E = 97.6% for N = 5; E = 97.9% for N = 10; E = 98.5% for N = 100; E = 99.0% and in the all-water at once case we got 85.7% efficiency. So as you can see dividing up the batch sparge water into smaller units we can approach 100% efficiency with batch sparging but after splitting the water into 2 batches there is typically rather little additional gain for the added effort of more smaller batches. == Next the continuous sparge case will be examined. Return to table of contents
Date: Tue, 04 Oct 2005 12:40:56 -0400 From: "steve.alexander" <-s at adelphia.net> Subject: efficiency, 2/ continuous sparge analysis In a similar vein let's examine the continuous sparge by an analysis of extraction efficiency as a series of small water 'replacements' followed by a final 'drain'; then sweep this to the limit of infinitesimal replacement volumes. Again we have a volume of solute(extract) X, a retained volume of unrecoverable solution in the grist of U, a volume of initial mash water M, and a total volume of sparge water S. We'll also refer to Chad Tchantz's example of X = 0.5gal, U = 1 gal, M = 3.1gal and S = 3.4gal. Again at the end-of-mash we have the condition where all the extract is presumed in solution of the mash water with a total volume of M+X, but only (M+X-U) of this liquid is recoverable by 'draining'. Now instead of draining this entire amount as in batch sparging, consider as a first approach to the continuous case replacing S/2 amount of wort with sparge water equilibrating, then repeating this step with the other S/2 of sparge water, then finally draining the whole. [[ note: I would have suggested removing/replacing S amount, but that exceeds the available liquid]]. After we calculate this for two S/2 replacements we'll see three S/3 replacement case then extrapolate to the general N by S/N, and finally take the limit as N approaches infinity. As N approaches infinity we have the case of the input trickle = output trickle continuous sparge. S/2 case: Step 1: remove S/2 of the original mash water. This removes (S/2) / (M+X) fraction of the extract and leaves (M+X-(S/2))/(M+X) behind. Step 2: After equilibrium, we again replace S/2 of the liquid. The removes, (S/2)/(M+X) * (M+X-(S/2))/(M+X) in this step, leaving ((M+X-(S/2))/(M+X))^2 of the total extract. Drain: Finally we drain the (M+X-U) volume of liquid which leaves ((M+X-(S/2))/(M+X))^2 * (U/(M+X)) fraction of the original extract behind and so the efficiency is E = 100 * { 1 - ((M+X-(S/2))/(M+X))^2 * (U/(M+X)) } For Chad's example this means, 1.7gal containing 47.2% of total extract is removed at step 1. At step 2, again 47.2% of the liquid is removed, but by then the tun only contains 52.8% of the original extract. So step 2 extracts 24.9% of the original, but leaves 27.9% behind. The final 'drain' step removes (M+X-U) volume of liquid leaving U behind. The spent grist still contains U/(M+X) of the 27.9% left after step 2. The bottom line is that this S/2 case leaves ((M+X-(S/2))/(M+X))^2 * (U/(M+X)) of the extract behind which is 7.7% lost or 92.3% efficient, not very good. Now the weirdness begins ... S/3 case: The S/3 case is straightforward, as above and finally leaves: ((M+X-(S/3))/(M+X))^3 * (U/(M+X)) fraction of the original extract so the efficiency is E = 100 * { 1 - (M+X-(S/3))/(M+X))^3 * (U/(M+X)) } In the example this means 8.9% lost extract and only 91.1% extracted !!! That's *less* extracted than in the S/2 case; is this possible ? Is this an error ? No, it's correct. If you add the sparge water 1 teaspoonful at a time and remove it at the same rate then the extraction is less efficient than if you replaced it 1 quart at a time (assuming equilibrium applies). If you remove a quart, then it's 1qt at the higher gravity, while if 1tsp at a time, you are constantly diluting the wort over that period of a quart of teaspoons. You are are in part removing the same water you just added and that means lower efficiency for the continuous sparge. This problem gets worse the farther you are from equilibrium, but that's for another day. So generally, if you divide the sparge water into N equal parts of volume S/N ,then you leave ( ((M+X-(S/N))/(M+X))^N * (U/(M+X)) ) fraction of extract behind and efficiency is: E = 100 * { 1 - (M+X-(S/N))/(M+X))^N * (U/(M+X)) } In the example case, for N = 2; E = 92.3% for N = 3; E = 91.1% for N = 4; E = 90.5% For N = 100; E = 89.25% extracted. For N = 1000; E = 89.20% extracted. The term ((M+X-(S/N))/(M+X))^N can be written: (1-(S/(M+X)/N)^N = (1-K/N)^N, for K = S/(M+X) and, as some of you may recognize from a calculus course, the limit of this term as N approaches infinity is e^-K, So the continuous sparge case we have that the fraction of extract left in the grist is (U/(M+X)) * e^-(S/(M+X)) and the efficiency percent is 100 * { 1 - ( U/(M+X) * e^-(S/(M+X)) ) } In the example, efficiency is 100 * { 1 - 0.2777 * e^-0.9444444 } or E = 89.19% Note that this is LOWER efficiency than the batch sparge case for even one drain and one sparge ! === My initial thinking was that Chad's all-water example at 85.7 % efficiency was a reasonable lower bound for batch sparge, and that continuous would remain around 92% (like the N=2 case above) or a bit better. That's not true. Intuition and poorly chosen examples here conspired to support the wrong conclusion. The correct take away is that continuous sparge is LESS efficient, than batch. -S Return to table of contents
Date: Tue, 04 Oct 2005 14:06:57 -0400 From: "steve.alexander" <-s at adelphia.net> Subject: many things Dave Burley says ... >SteveA what have you been drinking? {8^) > > My drink hasn't inhibited critical reading skills, as yours has, Dave. >Let's see. You believe that batch sparging is more efficient yet you aim for >inefficiency to get better tasting beer?? So what do you use? Batch or >continuous sparging? > I've done both over the years and there isn't much difference in practice. I currently use batch sparge in a recirculating pumped system and it's hard keeping the efficiency down to 75% where it belongs without restricting the sparge water volume or bypassing. > And that as you raise the temperature the mash >viscosity doesn't go down? > > PLEASE learn to read. The viscosity certainly decreases with increasing temperature. I said the change in viscosity due to mashout is microscopic , totally beyond the concern of any HBer. Many HB books play up the decreased viscosity as the reason for the 72C->78C mashout boost. That's nonsense. >I think the main reason I do continuous sparging is the uniformity in my >sparge and added complexity. This lets me calculate the grist and all sorts >of nice things. I also get a graininess I prefer that is missing with most >batch sparging I have tasted. Batch sparged beer reminds me of an extract >beer, probably for a good reason. > Ah yes, that grainy continuous sparge flavor ! The inscrutable and incalculable batch sparge. You certainly must be joking. I personally think there are good flavors in the middle runnings that should not be avoided, but everyone in the know is convinced of increasingly poor flavor from late runnings and from over-extraction. My personal taste aside, several triangle taste studies show a preference for beers made from no-sparge wort (including two informal ones I performed). In any case this is certainly all related to efficiency, amount of water and pH and not the piddling differences in sparge methods. >I think Charley P's comment long ago that "just add another pound of malt" >is right up there with most of his other silliness like a 15 minute mash >based on an incorrect iodine test procedure and passing hot wort through the >air, as in his pictures. Remember, Charlie is Type A+, graduated as a >Nuclear Physicist to become an Unclear Physicist, in his own words. {8^) > > Does anyone else recall that ad_hominem = logical fallacy ? "Charlie is a bonehead, therefore any argument he supports is wrong" is childish name calling; grow-up Dave. Address the issue not this personality muck-racking. >I am puzzled by your explanation of why batch sparging can ever be more >efficient than continuous sparging. > I DIDN'T provide an explanation before, so your puzzlement in attempting to understand the non-existent explanation is even more puzzling. It's now laid out in detail - straightforward I think. >In a batch sparge, the "sparge" water is >a high concentration sugar solution, which reduces its ability to remove the >malt sugar from the interior of the grain. > You are wrong. A batch sparge begins by draining most of the sugar with the first runnings at T=0, then at T+epsilon is at a lower equilibrium gravity than the continuous sparge will be for a long time - actually until about the time the next batch or final drain occurs. Continuous begins and finishes with more extract in the tun than a batch sparge - so it operates at a higher eq.SG and therefore with less ability to remove sugars (slower diffusion). >One test of the extraction efficiency would be to do an extraction both ways >on the same batch and the take the remaining brewer's grains and soak them >overnight (to eliminate kinetic issues) in a known volume of water to cover, >or more. and determine the concentration of the remaining sugar with >Clinitest. > > Impossible to control the two sparges with sufficient accuracy to produce meaningful and comparable results Dave. Why don't you propose the experiment you think would prove the point in detail, and I'll point out all the problems with the methodology (IOW we'll switch roles). >And, Chad. I have made this argument before, but modern usage of the "-ster" >suffix is asexual [...] > You go girl ! -S Return to table of contents
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