Help me understand something - RIMS/eHERMS Sparge

[jfkriege]

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.

 

[rca]

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

 

[lamarguy]

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.

 

[lamarguy]

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".

 

[jfkriege]

This is not mean offensively, Lamarguy, but I just realized that your avatar makes me think you are a ******* who is shouting at me. I dont think that about you, but that avatar sits right next to everything you say.

 

[lamarguy]

This is not mean offensively, Lamarguy, but I just realized that your avatar makes me think you are a ******* who is shouting at me.

Ha...I take it you're not a Scrubs fan.

 

[The Pol]

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?

 

[jfkriege]

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.

 

[jfkriege]

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.

 

[goatchze]

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.

 

[goatchze]

BTW, thanks for the debate guys. You're givine me something to do while bored in this hotel in Holland!

 

[jkarp]

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

No, no, not touchy. Just giving you a ribbing for not noticing the graphs were essentially the same.

 

[The Pol]

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.

Yup, 72% is what I thought.

 

[jkarp]

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?

I had done another worksheet at 10 gal. Efficiency goes up as the pre-boil gravity goes down and less points are lost, as expected. All these graphs were done at 37 p/p-g at 100% conversion - i.e. a congress mash. It's been a while since I read Kai's sparging paper but I suspect that's the small difference between his graphs and mine.

 

[The Pol]

Ok, your graph matches my initial statement several pages back. Thanks.

30 pounds, in a 10 gallon batch, at 72% is fine for me as that is an OG of about 1.084. I am typically in the 1.040 - 1.070 range with my beers.

I can handle 72%

 

[jkarp]

Ok, your graph matches my initial statement several pages back. Thanks.

30 pounds, in a 10 gallon batch, at 72% is fine for me as that is an OG of about 1.084. I am typically in the 1.040 - 1.070 range with my beers.

I can handle 72%

Of course, that's a theoretical max and actual results will be a few points below that. Getting the absorption rate and deadloss down makes a surprising difference in no-sparge efficiency. I recall you were doing some malt conditioning and that should really pay off for you there. With a 30 lb grist, just going from 0.15 gal/lb to 0.1 gal/lb gets you another 7-8%

I've got a little excel sheet I use for brewday volume calculations that makes grinding all these numbers easy if you're interested.

 

[The Pol]

Of course, that's a theoretical max and actual results will be a few points below that. Getting the absorption rate and deadloss down makes a surprising difference in no-sparge efficiency. I recall you were doing some malt conditioning and that should really pay off for you there. With a 30 lb grist, just going from 0.15 gal/lb to 0.1 gal/lb gets you another 7-8%

I've got a little excel sheet I use for brewday volume calculations that makes grinding all these numbers easy if you're interested.

I actually have an excel sheet which is where my #s came from too

 

[goatchze]

I'm sorry guys, but that graph is just too rosy. I've yet to hear anyone explain to me how the boil-off rate is effecting your efficiency (other than justifying the use of sparge water)? Your graph should look more like this: (should be gal/lb, not lb/gal on absorption)

 

[jkarp]

goatchze - the math was all laid out. What exactly aren't you understanding?

Boil-off must be compensated for on the front end in no-sparge. If you want 5 gallons of beer and you boil off 1 gallon, you'd better have 6 gallons pre-boil to avoid disappointment. When the mash is more dilute, the points per gallon is lower. Consequently, the actual points lost to absorption goes down. As absorption and deadloss are the ONLY sources of lost efficiency in no-sparge, the overall mash efficiency goes UP as pre-boil volume increases.

Well, of course conversion also plays a role in efficiency, but for the purposes of this discussion we're talking about theoretical best - congress mash, 100% conversion.

 

[goatchze]

goatchze - the math was all laid out. What exactly aren't you understanding?

Boil-off must be compensated for on the front end in no-sparge. If you want 5 gallons of beer and you boil off 1 gallon, you'd better have 6 gallons pre-boil to avoid disappointment. When the mash is more dilute, the points per gallon is lower. Consequently, the actual points lost to absorption goes down. As absorption and deadloss are the ONLY sources of lost efficiency in no-sparge, the overall mash efficiency goes UP as pre-boil volume increases.

Right. And this is exactly what a sparge is accomplishing? The difference being magnitude. (sorry, I left this off my post and edited it in later)

Even if you're sparging, if you want 5 gallons of beer and will boil off 1 gallon, you must use 6 gallons of water?

What are the losses if you use sparge water? Are they also ONLY absorption and deadloss?

Yes, the overal mash eff. goes UP as pre-boil volume goes down. This is always true.

 

[jkarp]

Right, but you started this thread with a Brutus 20 question. Brutus 20 is no-sparge. Batch sparging and fly sparging have different efficiency calculations entirely.

Brutus 20 is elegant in its simplicity. You've got X gallons in a closed loop, recirculating. Eventually, the wort gravity is uniform throughout the system. Assuming 100% conversion, at this point you have absolute perfect, 100% efficiency. No exceptions. No losses have occurred.

It's easy to calculate or even take an actual reading of the wort gravity at this point. Now run the wort off to the kettle. The difference in the volumes of what you started with and what's in the kettle now is your volume loss. You know the gravity, you know the volume. That's all you need to now determine the points lost and your mash efficiency.

 

[cyberbackpacker]

EDIT: again beat to the punch by jkarp!

 

[goatchze]

Right, but you started this thread with a Brutus 20 question. Brutus 20 is no-sparge. Batch sparging and fly sparging have different efficiency calculations entirely.

Brutus 20 is elegant in its simplicity. You've got X gallons in a closed loop, recirculating. Eventually, the wort gravity is uniform throughout the system. Assuming 100% conversion, at this point you have absolute perfect, 100% efficiency. No exceptions. No losses have occurred.

It's easy to calculate or even take an actual reading of the wort gravity at this point. Now run the wort off to the kettle. The difference in the volumes of what you started with and what's in the kettle now is your volume loss. You know the gravity, you know the volume. That's all you need to now determine the points lost and your mash efficiency.

...

I think this may be where the disconnect is. To me, there is no difference in the calculation, whether I have no sparge, a batch sparge, or I fly sparge.

I use:

GRAIN x PTS / [(GRAVITY-1)*VOLUME)] = EFF

where:
GRAIN = lbs of grain
PTS = gravity points per lb in one gallon
GRAVITy=specific gravity of wort, pre or post boil doesn't matter
VOLUME=volume of wort, pre or post boil doesn't matter as long as you're consistent

The above equation is what I use, so it doesn't matter the type of sparge, the length of boil, whether I'm standing on my head, holding my breath, or anything else. It only looks at how much grain you use and how much sugar you end up with.

What's simpler than that?

BTW, I'm not trying to be mean, hostile, or anything else. Just enjoying.

The real question, then, is how worthwhile is a sparge at all? That's really the difference here. How much of the residual sugar is actually recovered due to a sparge which otherwise would be unrecovered?

I'm not at home right now, so I can't look at my notes. It would be intresting to see some values for the gravities of different people's sparges. Because if you can recover just 50% of the residual sugar lost on the first run, the "max efficiency graph" would be as below, a sizable reduction in the amount of grain required. But I don't really know what those sparge gravities would look like?

 

[jkarp]

GRAIN x PTS / [(GRAVITY-1)*VOLUME)] = EFF

where:
GRAIN = lbs of grain
PTS = gravity points per lb in one gallon
GRAVITy=specific gravity of wort, pre or post boil doesn't matter
VOLUME=volume of wort, pre or post boil doesn't matter as long as you're consistent

So, 10 lbs, 37 points/lb-g, 1.050 gravity, and 5 gal:

10 x 37 / ((1.050 - 1) * 5) = 1480.

What's the units of efficiency here? I'm not familiar with numbers like that.

 

[goatchze]

Sorry, it's getting late over here in Holland and I've had one too many Hertog Jans!

Flip that equation. The numerator should be the sugar you recovered, the denominator the "potential sugar".

If we work in "points", Gravity-1 is the thousandths of SG above 1, so

GRAIVTY-1 = 1.050-1 = 50 points
50 pts/gal * 5 gal = 250 pts

10 lb*37 pts/lb-gal = 370 pts

250 pts / 370 pts = 0.68 or 68% efficiency

 

[goatchze]

The funny thing is, the use of "points" is supposed to simplify the calculation. however, since the specific gravity can change pre-post boil, I really think it makes it more complicated.

It would be easier to say that 1 lb of 2-row should yield 0.31 lbs of sugar.

10 lbs should yield 3.1 lbs.

5 gallons of wort at 1.05 SG means you have

(1.05-1) * 8.34 lb/gal * 5 gal = 2.085 lbs of sugar

NOTE: 8.34 lb/gal is the density of water, or a SG of 1.0.

2.085/3.1 = 0.67 or 67% (the difference from above is due to rounding)

But the '1.05-1' part is where someone got the idea to introduce "points", so I can see where it's coming from.

 

[jkarp]

Ah, OK. I understand. So let's apply your formula to my prior example:

GRAIN = 12
PTS = 37
GRAVITY = 1.0592 (pre-boil; you said it doesn't matter)
VOLUME = 5.75 gal (again, pre-boil, per your requirement to be consistent)

"GRAVITY-1" = 59.2
59.2 * 5.75 = 340.4
12 * 37 = 444
340.4 / 444 = 76.7% efficiency.

Holy smokes! How about that? We match exactly.

 

[jkarp]

Now, to help you understand why boil time / evaporation is important, let's do the same example, adjusted for a 90 minute boil.

(Recall our original conditions: Say we want 5 gallons of 1.073 finished wort and want to do a 90 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.)

Full Brutus 20 system volume will be 5 gal + (1.5 hr x .75 gal) + (12 lbs x .125 gal/lb) + .25 gal = 7.875 gal

444 / 7.875 = 56.4 points pre-boil

Now, back to your formula:

GRAIN = 12
PTS = 37
GRAVITY = 1.0564 (pre-boil; you said it doesn't matter)
VOLUME = 6.125 gal (again, pre-boil, per your requirement to be consistent)

"GRAVITY-1" = 56.4
56.4 * 6.125 = 345.45
12 * 37 = 444
345.45 / 444 = 77.8% efficiency.

So, simply by going from a 60 minute to 90 minute boil, NO other changes, our theoretical max Brutus 20 efficiency went up by a little over 1%.

 

[jkarp]

It would be easier to say that 1 lb of 2-row should yield 0.31 lbs of sugar.

Where do you get your data? 2-row is 37 p/p-g. Sugar is 46 p/p-g. Therefore, 1 lb of 2-row is equivalent to .8 lbs of sugar.

 

[goatchze]

Now, to help you understand why boil time / evaporation is important, let's do the same example, adjusted for a 90 minute boil.

(Recall our original conditions: Say we want 5 gallons of 1.073 finished wort and want to do a 90 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.)

Full Brutus 20 system volume will be 5 gal + (1.5 hr x .75 gal) + (12 lbs x .125 gal/lb) + .25 gal = 7.875 gal

444 / 7.875 = 56.4 points pre-boil

Now, back to your formula:

GRAIN = 12
PTS = 37
GRAVITY = 1.0564 (pre-boil; you said it doesn't matter)
VOLUME = 6.125 gal (again, pre-boil, per your requirement to be consistent)

"GRAVITY-1" = 56.4
56.4 * 6.125 = 345.45
12 * 37 = 444
345.45 / 444 = 77.8% efficiency.

So, simply by going from a 60 minute to 90 minute boil, NO other changes, our theoretical max Brutus 20 efficiency went up by a little over 1%.

OK OK, I think we're running in circles here. My issue wasn't your math, it was the statement that the boil directly affected your efficiency or the way your efficiency is calculated. To me it's not. It affects how much water you use, which then affects your efficiency (indirectly). I understand your point, but we're talking to different things (because we're looking at two different things). I calculate the amount of water I need seperately, thus the disconnect (a question of semantics).

People can get confused when you say that a longer boil will increase efficiency because they think "I'm increasing the gravity of my wort in the boil, thus increasing efficiency". This, of course, is not the case.

As for you last post, I'm not saying sugar as in "sucrose" or "table sugar". If we assume that wort is a mixture of only two things, water and sugar (of whatever kind), then every point of gravity above 1.0 is due to a sugar (of whatever kind).

0.31 lbs i the amount of sugar (of whatever kind) which should be produced from 1 lb of malt.

So, now that we're off of this tangent, let's go back to my original question. Your above comment that the gravity of the wort "lost" affects the overall efficiency (which I agree with completely) is of course the justification for using a sparge. The Brutus doesn't use a sparge, so the above logic means the efficiency must be lower. This means you need more grain to get the same amount of beer.

The Brutus also requires two pumps, so a higher cost to build (for me the pump was one of the largest expenses on my build).

It only uses two vessels, but if using tap water the system I drew in the first post only requires two vessels. Even if you're not using tapwater, you only need a small water bottle to hold the sparge water. (BTW, you can do the same cross circulation with the layout I've drawn).

So what I'm still not understanding, or seen shown, is what the advantage of the Brutus set up is? I'm not trying to knock it; it obviously works and is quite popular. However, the only advantage I see is the elmination of a small water bottle if not using tap water. If using tap water, I see no advantage at all?

Meanwhile there are multiple negatives, mainly a loss in efficiency and need for an additional pump?

So, again, what am I missing? Do the pros really outweight the cons, or are so many people building this system because it's been done before and is easy to duplicate?