Help me understand something - RIMS/eHERMS Sparge

In looking through various builds on this board, I'm still confused about the Brutus system. My confusion is centered on the way that the Brutus sparges by mixing the wort with the sparge water. Scoates has an excellent diagram on this thread.

What is the purpose of this recirculation? I understand the idea is to only need two vessels, however, I see a few big disadvantages:

1. Mixing the wort/sparge water over the grain bed will hurt efficiency
2. This system requires two pumps
3. You're supposed to recirculate until the gravity of the kettle/MLT are equal...seems like a lot of work and measuring.


I've ordered all the parts and have my stand built, just waiting on delivery to complete my build. But I still don't understand why I don't see more systems like below. I want to make sure I'm not missing something.

This configuration

1. Only uses two vessels
2. Only uses one pump
3. Allows a sparge independent of the first wort
4. Allows the first wort to start heating/approaching boil during sparge (saves time)

In this system I plan to do everything as in the Brutus up until draining the MLT. It's different from then on. I'll drain the wort to the brew kettle and fire up the heating element. Fresh water will pass through the RIMS tube to heat up to sparge temps before entering the MLT, all while the first wort is being heated to a boil. The sparge water then gets pumped/added to the kettle. (BTW, that thing in the Riddle is the RIMS tube)

I'll have the option of either recirculating the sparge water or just doing a simple fill/drain batch sparge, throttling the pump to hit a target RIMS outlet temperature. I can also do more than one sparge if I choose. (I'll find out what works best after it's built).

What is everyone's thoughts?

I just built a Brutus 20 design, I did it for simplicity and footprint.

Also, I build my water, so hooking a hose up to my 9000W RIMS heater to heat water on the go for a fly sparge, is much more complicated to do.

Brutus 20's are basically a no sparge system, your eff. loss is equal to MLT deadspace and grain absorption. The key is being mindful of your conversion eff.

I dont plan to do a lot of measuring of the gravity with my system during the "sparge". I am going to set a timer on the BCS and let the process run for a specified time, then the BCS will shut down my BK pump and continue to run the MLT pump to empty the MLT into the BK.

There is an efficiency loss, but you are also making all of your wort, first wort, which from a wort quality standpoint is about as good as you can get.

My system will achieve 70-78% depending on grain bill size. The nice thing is that since eff. losses are tied to losses in the MLT (fixed) and losses due to grain absorption (linear), I can easily determine my eff. for each brew.

Your design will work, there are guys here who have built it. You are doing a 3 vessel system where your municipal water and RIMS heater ARE your HLT.

The pol,

You're right on the RIMS heater replacing the HLT, that's the idea.

Building your own water isn't a concern. As it's drawn up, the pump will actually be able to move water from a container to feed the system. You can mix your strike water in the kettle and sprage water in a small water bottle on the side, or just use water bottles as feed. You don't necessarily need the positive pressure from your tap, although doing so would eliminate the need for any additional water container (note that the drawing doesn't really take heights into consideration...it was just a simple way to draw it)

I don't understand this statement:
"you are...making all of your wort, first wort, which from a wort quality standpoint is about as good as you can get."

What do you mean by quality? With this set up you can recirculate the sparge water until you get maximum removal. Having a lower gravity will increase the amount of sugars that you can remove, so I would expect improved effiency and good quality? What am I missing?

I too am going for a small footprint; I've got limited space in the garage. My stand is 40"x20"x20". I expect to get everything on there with the exception of the water bottle with the sparge water.

goatchze said:

The pol,

You're right on the RIMS heater replacing the HLT, that's the idea.

Building your own water isn't a concern. As it's drawn up, the pump will actually be able to move water from a container to feed the system. You can mix your strike water in the kettle and sprage water in a small water bottle on the side, or just use water bottles as feed. You don't necessarily need the positive pressure from your tap, although doing so would eliminate the need for any additional water container (note that the drawing doesn't really take heights into consideration...it was just a simple way to draw it)

I don't understand this statement:
"you are...making all of your wort, first wort, which from a wort quality standpoint is about as good as you can get."

What do you mean by quality? With this set up you can recirculate the sparge water until you get maximum removal. Having a lower gravity will increase the amount of sugars that you can remove, so I would expect improved effiency and good quality? What am I missing?

I too am going for a small footprint; I've got limited space in the garage. My stand is 40"x20"x20". I expect to get everything on there with the exception of the water bottle with the sparge water.

Click to expand...

Wort quality associated with PH and the buffering power of the malt.

My stand, with kettles, measures 48" wide x 20" deep x 40" tall... I could go larger but I wanted something less bulky too.

If you dont have positive pressure from a hose, I guess, and have another bottle or vessel that you are feeding the RIMS with, you still sort of have a three vessel system in a round about way. I assumed the design was centered on the hose providing pressure to the RIMS.

It will work, there are a couple dudes kicking around here with these setups. I had initially planned to use my RIMS for the same thing, but didn't want to create space for an additional water holding tank. As you can see in my photos, I dont have any room!!

How do you "build" water when salts don't want to dissolve in water?

Bobby_M said:

How do you "build" water when salts don't want to dissolve in water?

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I have to start with distilled, which does not flow from my tap. Meaning I would have to have another vessel and pump it from there to the RIMS heater. Thusly, not a two vessel system. That is how.

The Pol said:

My system will achieve 70-78% depending on grain bill size.

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I think you're being overly optimistic. On average, if you achieve 68% brewhouse efficiency with your no-sparge design, I'll be very impressed. I think 60% - 70%, depending upon the grain bill, is more realistic.

Realistically, to make a 1.070 beer with no-sparge at 63% efficiency requires approximately 19.25 lbs of grain. If you were at ~85% efficiency, it would only require 14.25 lbs of grain. And, I realize bulk grain is cheap (I buy in bulk too), but a 20+ percent efficiency hit is significant . Just sayin'.

Isnt your loss in eff. equal to the loss in grain absorption and MLT deadspace?

If so, it isnt hard to get in the mid 70s. If you are converting your sugars that is.

If I start with 16 gallons of water...

The grain absorbs 2 gallons (20 pounds of grist)

I lose .19 gallons to deadspace in the MLT

I have lost 2.19 gallons from my 16 gallons, right?

So I have lost about 14% of the total wort.

If I convert only 90% of my sugars, 86% (from the above wort loss calc) of 90% is still 77% right?

What am I missing?

The Pol said:

I have to start with distilled, which does not flow from my tap. Meaning I would have to have another vessel and pump it from there to the RIMS heater. Thusly, not a two vessel system. That is how.

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Sorry, what I'm asking is, how are you getting salts to dissolve in your distilled and not just precipitate out? If you stir chalk into water and walk away for 10 minutes, you come back to a layer of chalk on the bottom of the vessel. That's why most people treat their mash with salts, not the water.

20lbs of 2-row is capable of about 20 x 36 = 720 points. Assuming preboil volume of 12g for a 10g batch, your mash gravity at 100% conversion would be 1.060. Therefore the lockup of 2 gallons of 1.060 wort removes 120 GU. 600/720 = 83% theoretically with no other losses. I know that's Kaiser's chart but I don't remember what other assumptions are made about conversion and deadspace.

Bobby_M said:

Sorry, what I'm asking is, how are you getting salts to dissolve in your distilled and not just precipitate out? If you stir chalk into water and walk away for 10 minutes, you come back to a layer of chalk on the bottom of the vessel. That's why most people treat their mash with salts, not the water.

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I don't treat the sparge water... because the salts wont dissolve.

The Pol said:

Isnt your loss in eff. equal to the loss in grain absorption and MLT deadspace?

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Yes, strictly for fly sparging and, to a lesser extent, batch sparging. No-sparge does not efficiently rinse the grains, which is another significant loss.

The Pol said:

What am I missing?

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The graph I showed is based on a 6.5 gallon batch. The numbers you gave are for a ~13 gallon batch which would produce a 1.046 wort. Going back to the chart for a 6.5 gal batch, yes, you can hit ~75% efficiency with no-sparge.

My point is, no-sparge begins to impose a significant efficiency hit past 1.050 sg. If you hover around 1.050, it's acceptable, but most the brewers I know brew in the 1.050 - 1.090 range, not the 1.020 - 1.060 range.

lamarguy said:

Yes, strictly for fly sparging and, to a lesser extent, batch sparging. No-sparge does not efficiently rinse the grains, which is another significant loss.



The graph I showed is based on a 6.5 gallon batch. The numbers you gave are for a ~13 gallon batch which would produce a 1.046 wort. Going back to the chart for a 6.5 gal batch, yes, you can hit ~75% efficiency with no-sparge.

My point is, no-sparge begins to impose a significant efficiency hit past 1.050 sg. If you hover around 1.050, it's acceptable, but most the brewers I know brew in the 1.050 - 1.090 range, not the 1.020 - 1.060 range.

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The numbers I gave are for a 10 gallon batch approx at 1.060 OG. 16g of wort, loss of 2.2g to losses, loss of 2.6g to boil off and a loss of .5-.75g to trub and loss of .4g to shrinkage... 10.05g to the fermentor.

I dont get it. I know you arent rinsing the grains as cleanly as you are with a sparge. But if your wort is homogeneous, and you convert 100% of your sugars, why isnt your eff. LOSS equal to the wort LOSS?

I mean, I was under the impression, incorrect I guess... that if I converted 100% of my sugars, my wort was homogeneous, and I drained all but 20% of the wort, my eff. would equal 80%.

Woops

So if eff. loss is equal to MLT deadspace and grain absorption, with a 30 pound grist I would have...

3 gallons lost to absorption
.19 gallons to deadspace

3.19 gallons subtracted from 17 gallons total wort....

This is a 19% loss

Converting only 90% of the sugars would result in a 72% eff., but apparently I am way off on this. This would be a 10 gallon batch of 1.084 beer

Why is the eff. hit so much more significant above 1.050? What is special about that number? I thought it was pretty linear as the grain bill got larger, the absorption amount got larger since it was a steady ratio. Thus causing a pretty linear eff. loss.

The Pol said:

Why is he eff. hit so much more significant above 1.050? What is special about that number?

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It's not a magic number, just a number I threw out based on my past experience with no-sparge brewing.

Kaiser quotes the following equation:

Effbatch_sparge_step = 100% * Vrun_off / (Vrun_off + Vgrain_absorption + Vdead_space)

So, if you subtract for deadspace and grain absorption, you can get pretty close. For a 6.5 gal 1.050 batch at 75%, you're looking at ~1.4 gallons of sweet wort lost to the grain. For the same size batch at 1.070, you're looking at ~2.3 gallons lost.

Add the grain absorption loss to your deadspace loss and see where that gets you.

The Pol said:

Meaning I would have to have another vessel and pump it from there to the RIMS heater. Thusly, not a two vessel system.

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Maybe so, I guess you could say that it's technically a three vessel system because I need some water storage. However, if you're able to use tapwater, it is a two vessel system. Even if not, you only need a small container for the sparge water.

I start with RO water from my own machine, which only has a 2 gallon tank. So I start several days ahead draining the RO system every so often to build up my stock of water. In the past I've just used a water bottle, which is what I'll use in the future for sparge water.

But I guess it's a semantic argument whether this water bottle is truly a "part" of the system or if it's just a source of water. For the brief time I'm using it in the system, it will sit on top of the MLT (an Igloo Cooler) which for now will not increase the footprint of the unit.

The Pol said:

I dont get it. I know you arent rinsing the grains as cleanly as you are with a sparge. But if your wort is homogeneous, and you convert 100% of your sugars, why isnt your eff. LOSS equal to the wort LOSS?

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Your efficiency is further reduced because the concentration gradient between the water/sugar trapped in the mash and the water/sugar in the wort is less.

Just like heat exchange needs a temperature difference, mass transfer needs a concentration difference. As the difference approaches zero, your rate of exchange approaches zero.

If you have a high gravity liquid passing over the grain bed, the point at which the two reach equilbrium (ie no sugar passing into the wort) will be different than if you have fresh water.

In a nutshell, when you have a higher gravity wort passing over the grain bed, there are sugars "trapped" in the mash which you cannot remove. This is why you would have a loss in efficiency.

So it's not just losses.

Another way to look at it is IF it is losses, take your volume x your gravity. Is it more or less with a no sparge system? If it's more, that hurts your efficiency as efficiency is a measure of the sugars your produce, not just the volume.

Batch sparging and NO sparging (which the brutus 20 and some BIAB methods rely on) do NOT depend on concentration gradients. The process actually does depend on equilibrium. The wort locked in the fiber of the grist and that outside it as liquid should be the same gravity. So efficiency loss is exactly equal to absorption and deadspace in relationship to the batch size.

Of course, the higher the first wort gravity (somewhat a misnomer in no sparge), the more efficiency you lose because the gravity of the absorbed wort is higher and there is more grain doing the absorbing.

Bobby_M said:

Batch sparging and NO sparging (which the brutus 20 and some BIAB methods rely on) do NOT depend on concentration gradients. The process actually does depend on equilibrium. The wort locked in the fiber of the grist and that outside it as liquid should be the same gravity. So efficiency loss is exactly equal to absorption and deadspace in relationship to the batch size.

Of course, the higher the first wort gravity (somewhat a misnomer in no sparge), the more efficiency you lose because the gravity of the absorbed wort is higher and there is more grain doing the absorbing.

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Bobby, no offense, but what you said above doesn't make sense. If it depends on equilbrium, then it depends on concentration gradients. By definition equilibrium is when the concentration gradient is zero.

If you have, let's say 1.06 gravity potential "locked" in the grist, then you need to rinse it with water whose gravity is less than 1.06. Otherwise you're not accomplishing anything. That's my point about concentration gradients.

So I think you and I are trying to say the same thing.

But NO sparging and Batch sparging are different. In batch sparging you're staring with water whose gravity is 1. As it sits on the grain bed sugars will pass into the water, out of the grist, until equilbrium is reached (concentration gradient is zero). Then you drain the wort. If the gravity of the sparged wort is less than your "first wort", then you've picked up mores sugars than you otherwise would have been able to, thus increasing efficiency.

Yeah, there's not enough information provided to interpret that chart out of context for no-sparge. You need to have batch size, evaporation rate and boil time, absorption rate, and deadloss. For a no-sparge 5.5 gal batch of 1.070, 60 min boil, .75 gal/hr evap, .15 gal/lb absorb, and .125 gal deadloss, you'll get 74% mash efficiency, assuming good conversion.

jkarp said:

Yeah, there's not enough information provided to interpret that chart out of context for no-sparge. You need to have batch size, evaporation rate and boil time, absorption rate, and deadloss. For a no-sparge 5.5 gal batch of 1.070, 60 min boil, .75 gal/hr evap, .15 gal/lb absorb, and .125 gal deadloss, you'll get 74% mash efficiency, assuming good conversion.

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???

I guess it depends on how you calculate efficiency. The way I calculate it, it doesn't matter the batch size, evaporation rate, or boil time. Really all that matters is the conversion and dead loss. With no sparge, your "dead loss" is necessarily higher. I don't think there's any issues interpreting that chart for no sparge brewing.

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.

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.

jkarp said:

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.

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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.

jkarp said:

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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:

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.

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.

goatchze said:

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.

Click to expand...

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!

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.

jfkriege said:

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.

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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.

jfkriege said:

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

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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.

jkarp said:

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

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

jkarp said:

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

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Correct.

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.

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

rca said:

Jkarp's graph is 0 to 100%, lamarguy's is 50 to 100%.

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Just to be clear, it's Kaiser's graph...I'm just referencing it as a good source of information.

jfkriege said:

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

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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".

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.

jfkriege said:

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.

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

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 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.

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.

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.