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**jkarp** · 02-23-2010, 05:48 PM

Remember we're calculating no-sparge goatcheze. Those things all matter because they alter the amount of water going into the system. More water in the system means lower pre-boil gravity and consequently fewer gravity points lost to absorption and deadloss.

The Pol · 02-23-2010, 05:49 PM

Everything matters in a no sparge calc. Because water volumes matter, boil off matters, batch size matters etc. % loss is tied to the total wort volume. The smaller that total wort volume, the larger the percentage your losses will be of that volume. Thus, reducing your eff.

Just what Jkarp said.

goatchze · 02-23-2010, 05:55 PM

Quote:

Originally Posted by jkarp

Remember we're calculating no-sparge goatcheze. Those things all matter because they alter the amount of water going into the system. More water in the system means lower pre-boil gravity and consequently fewer gravity points lost to absorption and deadloss.

Calculating efficiency is the same regarldess of whether you use 0, 1, or 2 sparges. You are correct that the amount of water "can" affect the efficiency, but I don't think it is in the way you are conveying.

We're talking about a material balance on the sugars (sorry to use engineering terms, it's what I do for a living). Let's take the example:

If i have 8 lbs of 2-row pilsner, it has the "potential" at 100% conversion to produce 37 gravity points/lb-gal. This means that at 100% conversion I could have the following, all giving me 100% efficiency:

1 gallon @ 1.296 (296 x 1 = 296)
2 gallons @ 1.148 (148 x 2 = 296)
5 gallons @ 1.0592 (59.2 x 5 = 296)
10 gallons @ 1.0296 (29.6 x 10 = 296)

All of those would give me 100% efficiency. The volume of water, by itself, doesn't matter. What matters is the gravity with respect to the volume. The boil-off and boil-time have no influence at all.

What matters is the mass of sugars extracted.

lamarguy · 02-23-2010, 05:57 PM

Quote:

Originally Posted by jkarp

I'm not buying that graph at all. The drop off in no-sparge efficiency is much steeper than it shows.

The actual drop off in efficiency is much closer to:

jfkriege · 02-23-2010, 06:15 PM

I have looked at the spreadsheet that jkarp used to make that graph. The numbers are sound and well thought out. Besides, the guy has proven his data with his own system and beer that he has actually brewed.

Lamarguy, what is the batch size, evaporation rate, dead loss, and assumed grain absorption rate to make the graph that you keep showing. Until you can tell us how you made that graph, it really doesn't mean anything.

**jkarp** · 02-23-2010, 06:16 PM

Yeah, you're still not getting it goatchez. In no-sparge, we must work backwards from the goal. Stick with me here:

Say we want 5 gallons of 1.073 finished wort and want to do a 60 min boil. We know our system will boil off 0.75 gal/hr. Let's also assume our grist absorbs 0.125 gal/lb and our MLT has 0.25 gal of deadloss.

So for total water we will need: 5 gal + (1 hr x .75 gal) + (12 lbs x .125 gal/lb) + .25 gal = 7.5 gal

12 lbs of 37 points/lb-gal = 444 total available points
444 / 7.5 gal = 59.2 or 1.059 pre-boil gravity

Now 1.75 gal of wort will be permanently "lost" in the MLT ( 12 lbs x .125 gal/lb + .25 gal )
1.75 gal * 59.2 = 103.6 gravity points lost in the MLT

(444 - 103.6) / 444 = 76.7% mash efficiency, assuming 100% conversion.

**cyberbackpacker** · 02-23-2010, 06:20 PM

Quote:

Originally Posted by goatchze

Calculating efficiency is the same regarldess of whether you use 0, 1, or 2 sparges. You are correct that the amount of water "can" affect the efficiency, but I don't think it is in the way you are conveying.

We're talking about a material balance on the sugars (sorry to use engineering terms, it's what I do for a living). Let's take the example:

If i have 8 lbs of 2-row pilsner, it has the "potential" at 100% conversion to produce 37 gravity points/lb-gal. This means that at 100% conversion I could have the following, all giving me 100% efficiency:

1 gallon @ 1.296 (296 x 1 = 296)
2 gallons @ 1.148 (148 x 2 = 296)
5 gallons @ 1.0592 (59.2 x 5 = 296)
10 gallons @ 1.0296 (29.6 x 10 = 296)

All of those would give me 100% efficiency. The volume of water, by itself, doesn't matter. What matters is the gravity with respect to the volume. The boil-off and boil-time have no influence at all.

What matters is the mass of sugars extracted.

Using your info here though goat, what you are failing to see is that we are talking more than conversion efficiency, we are talking brewhouse, and for brutus 20 and no sparge purposes, EVERYTHING matters.

If I am no-sparging a 4g batch with your example above, if I collect 5 gallons @ 1.0592 I am leaving .15 gallons per pound of 1.0592 wort behind due to grain absorption. However, if I have a higher boil off rate, longer boil, or a combination, if I collect your 10 gallons @ 1.0296, I will only be leaving behind .15 gallons per pound of 1.0296 wort due to grain absorption.

So, this will yield a greater brewhouse efficiency. It ALL matters.

EDIT: jkarp beat me to the punch, and much more eloquently... but I think my point is still valid to the conversation too!

**jkarp** · 02-23-2010, 06:23 PM

lamarguy - How about taking a closer look at those two graphs. Now, look again. Have a 3rd look for good measure.

They're NOT that far off. For a 30 lb grist, I show around 58%. The other graph shows around 54%. I'm simply saying I don't have the necessary data to understand how that other graph was determined.

lamarguy · 02-23-2010, 06:24 PM

Quote:

Originally Posted by jfkriege

I have looked at the spreadsheet that jkarp used to make that graph. The numbers are sound and well thought out. Besides, the guy has proven his data with his own system and beer that he has actually brewed.

Having done no-sparge in the past and seeing numbers posted by other folks, the real world drop off in brewhouse efficiency is simply not believable.

Keep in mind, we have folks posting in here about acheiving 110% brewhouse efficiency. Boggles the mind.

I'm not saying jkarp is making up data and I'm certainly not taking sides on this debate. I am saying the numbers do not correlate with Kaiser's data or any real world data I've seen to date.

Quote:

Originally Posted by jfkriege

Lamarguy, what is the batch size, evaporation rate, dead loss, and assumed grain absorption rate to make the graph that you keep showing.

Vb=6.5 gal, A = 0.19 gal/lb, VD=0 gal

I assume when folks engage in efficiency discussions they have at least read Kaiser's work.

lamarguy · 02-23-2010, 06:32 PM

Quote:

Originally Posted by jkarp

lamarguy - How about taking a closer look at those two graphs. Now, look again. Have a 3rd look for good measure.

No need to get touchy. If you present your work as evidence, expect comments.

Quote:

Originally Posted by jkarp

They're NOT that far off. For a 30 lb grist, I show around 58%. The other graph shows around 54%.

Correct.