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Old 02-23-2010, 06:39 PM   #31
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If I adjust the calculations that I use (based very strongly on jkarp's spreadsheet) then I get exactly the graph you have there. I am assuming the only major difference is that kaiser uses 0.19 gal/lb absorption, and jkarp uses 0.15 gal/lb. Otherwise they are almost identical graphs (except for the truncated scale on kaisers graph).

So, I am not really sure what the problem is.



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Old 02-23-2010, 06:47 PM   #32
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It's all in your point of view, or Lies, Damn Lies and Statistics.

Jkarp's graph is 0 to 100%, lamarguy's is 50 to 100%. If the graphs were plotted over the same areas they would look much closer, which is I think what jkarp was pointing out.

Love the discussion though guys, very enlightening to a new brewer.

Ron



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Old 02-23-2010, 06:51 PM   #33
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Quote:
Originally Posted by rca View Post
Jkarp's graph is 0 to 100%, lamarguy's is 50 to 100%.
Just to be clear, it's Kaiser's graph...I'm just referencing it as a good source of information.
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Old 02-23-2010, 06:56 PM   #34
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Quote:
Originally Posted by jfkriege View Post
So, I am not really sure what the problem is.
I don't believe there is a "problem" per say...But, I do believe work presented in these forums as evidence for a particular method should at least be compared to prior work. If they don't agree, why?

Like I mentioned earlier, I've grown tired of seeing misleading data thrown around like it's fact. Call it my current "soapbox".
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Old 02-23-2010, 06:59 PM   #35
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This is not mean offensively, Lamarguy, but I just realized that your avatar makes me think you are a jackass who is shouting at me. I dont think that about you, but that avatar sits right next to everything you say.

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Old 02-23-2010, 07:03 PM   #36
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Quote:
Originally Posted by jfkriege View Post
This is not mean offensively, Lamarguy, but I just realized that your avatar makes me think you are a jackass who is shouting at me.
Ha...I take it you're not a Scrubs fan.
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Old 02-23-2010, 07:06 PM   #37
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What if you doubled your water volume at 30 pounds of malt Jkarp? In your graph, would that change the eff.? Just curious.

I mean at 30 pounds of malt in a 5.5 gallon batch, youd be absorbing A LOT of your total mash wort, right? What if you doubled that water volume? What number would you have then?

Also, these graphs are cool, what is the "assumed" conversion eff. in these graphs?

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Old 02-23-2010, 07:06 PM   #38
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I have to admit that I am not. I actually don't even have a TV in the house right now. I grew up in a house with an audiophile/videophile father and I don't want to get a TV until I can get one that I really like.

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Old 02-23-2010, 07:17 PM   #39
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On my calculator I would go from 58.6% up to 72.4% assuming the same boiloff rate and time. It is a big jump, but I also would go from a 1.113 OG down to 1.069.

For the calculator that I use, I assume a 100% conversion. This is perhaps not realistic, but is good enough for playing with theory and designing a rig.

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Old 02-23-2010, 07:22 PM   #40
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Maybe we're not comparing apples and apples. Let's take a step back to the fundamentals. What are we trying to achieve? Sugar extraction.

When I think of "brewhouse efficiency" I think of how much sugar I COULD get into the wort as compared to how much sugar I ACTUALLY get into the wort. This means I could lose efficiency either through:

a) Poor Conversion
b) Dead Losses in the system (in the MLT, lines, etc.)

So how do I calculate it? Let's take 10 lbs of 2-row pilsner.

1 lb of malt should give 37 pts in 1 gal of water. This means

1.037 (SG) * 8.34 lb/gal (density of water) = 0.31 lbs of sugar

If I have 10 lbs of malt, I expect 3.1 lbs of sugar in the kettle if I achieve 100% efficiency.

If I use 5 gallons of water and have NO dead losses and 100% conversion (i.e. 100% efficiency):

5 gal * 8.34 lb/gal = 41.7 lbs of water
41.7 + 3.1 lbs of sugar = 44.8 lbs in total
44.8 lbs total / 41.7 lbs (5 gal of water) = 1.074 Specific Gravity

If 7 gallons of water is used:
7 gal * 8.34 lb/gal = 58.4 lbs of water
58.4 + 3.1 lbs of sugar = 61.5 lbs in total
61.5 lbs total / 58.4 lbs (7 gal of water) = 1.053 SG

If I get 7 gallons of 1.053 wort in my kettle and boil off 2 gallons of water, I have exactly the same numbers as the 5 gallon case.
(61.5 lbs total - 8.34 lb/gal*2 lb)= 44.8 lbs
44.8 lbs total / 41.7 lb water = 1.074 SG

Hence both show a brewhouse efficiency of 100% and it doesn't matter whether it is calculated pre or post boil.

All that matters is the volume of wort in the kettle and the gravity of that wort in conjunction together. It does not matter whether it is pre or post boil. Pre/post boil gravities only matter with respesct to your target gravity (how strong do you want the beer to be).

The way I calculate it, brewhouse efficiency doesn't care where the sugar went. It doesn't care whether it was due to conversion being less than 100% or dead losses. All that matters is the mass of sugar in the kettle.

Let's take an example. Let's say I have 100% conversion and I know I have 0.5 gallons of dead space. I use 10 lbs of 2-row and 7 gallons of water to start. However, I only collect 6.5 gallons into the kettle:

7 gallons, from above, gives me a SG of 1.053. I collect 6.5 gallons.

6.5 gal * 1.053 * 8.34lb/gal=57.1 lbs in total
6.5 gal water = 54.2 lbs of water
57.1-54.2=2.9 lbs of sugar

2.9 lbs sugar in kettle / 3.1 lbs sugar expected * 100% = 93.5% brewhouse efficiency


NOW, if I want to calculatle conversion, then I need to know my dead losses. Again, it doesn't matter where the water goes. Is it left in the MLT? In the grain? It doesn't matter. But let's say I know that I put 7.2 gallons of water IN TOTAL intot he system. Preboil I only collected 6.5 gallons. If I know this, I can approximate the conversion using a similar process.

But again, brewhouse efficiency doesn't care what the conversion is. It doesn't really care what the dead losses are. It's accounting for both at the same time.



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