Grain Bill Calculations - This CAN'T be right can it?

I haven't been hitting my numbers. I decided to spend some time putting a spreadsheet together to simply my process because BeerSmith isn't doing it for me. It's long and slightly technical, but I will try to make it as easy to follow as possible:

Here is what I need to make using roughly 92% 2-Row & 8% Carmel:

OG: 1.066
Volume: 5 gallons

I will lose 1.25 to trub and dead space in my brew kettle, meaning I will need a volume of 6.25g at the end of my boil. I boil off 20% (I know, it's crazy) so I will need 7.8g at the beginning of the boil.

Since I need 1.066 @ 5g at the end of the boil, I will need 1.053 @ 7.8g at the beginning of the boil (66 * 5/7.8).

Ok, so I need 7.8g of 1.053 wort in my kettle to start the brew. I have 1g of dead space in my MLT, which means I actually need 8.8g of 1.053 wort.

I need 53 PPG (points per pound per gallon) from my mash and lauter (1.053). I am averaging around 70% extraction efficiency, so I know my denominator of Maximum PPG must be 75 (53/70%). So, given my ratios, and a max PPG for Carmel 60 of 32 and a max PPG for the 2-Row of 28, here is what I need.

2-Row: 21.69 pounds (21.69 pounds / 8.8 gallons * 28 max points) = 69
Carmel: 1.81 pounds (1.81 pounds / 8.8 gallons * 32 max points) = 7

69 + 7 = 76 (roughly the denominator I need for 70% extraction efficiency)

A 23.5 pound grain bill for a 5 gallon batch of 1.066?!?

if i was you j would just use beer smith, you can figure out all of this stuff so quickly, and you can buy it on rebelbrewer.com for 17.95 one of the best investments that i've ever made

You need 1.066 at 6.25 gallons, which is

66 x 6.25 = 412.5 gravity points

Divide by 70% efficiency = 589 gravity points

Base malt is about 36 points per pound/gallon

589/36 = 16.4

You need about 16.4 lbs of grain

-edit

...Oh wait, I didn't figure the 1 gallon loss in the mash tun.

I think you need to re figure your system so that you're not making 50% extra and having it go to waste.

My knee jerk reaction is that I think you need to redesign your MLT. 1 gal of deadspace is too much. I had a similar problem when I first started and had to get a better false bottom that only left .25 gal behind. Without breaking down your numbers, I think you are leaving way too much wort in the MLT.

What are you using as a mash tun? My cooler has virtually no dead space.

Those are some big losses. If they are real, then 16.4 lbs won't be enough to yield 5 gal in the fermenter. But it won't be 23 lbs either -- you have 2-row at a max yield of 28 ppg (it should be more like 38 max ppg).

Here are my calcs:

Post-boil volume and gravity:
5 + 1.25 gal = 6.25 gal @ 1.066 = 412.5 (6.25 x 66) points total extract

Pre-boil volume and gravity:
6.25/.80 = 7.8 gal
412.5 points in 7.8 gal = 412.5/7.8 = 52.9 or about 1.053 SG

Mash tun losses of 1 gal mean you need to mash 8.8 gal @ 1.053
Total extract required = 8.8 x 53 = 466.4 points

Divide by efficiency for max extract:
466.4/.70 = 666 <------- There's your problem!!!

Base malt is 92% of grainbill
666 * 0.92 = 612.7 total max extract
Crystal is 8% of grainbill
666* 0.08 = 53 total max extract

Convert to pounds of grain
612.7 points/37 max ppg = 16.6 lbs two row
53 points/34 max ppg = 1.6 lbs crystal 60
Total grain bill: 16.6 + 1.6 = 18.2 lbs

That is still a lot of grain -- your 2.25 gals of losses are killing you, man.

Brewsmith said:

What are you using as a mash tun? My cooler has virtually no dead space.

Click to expand...

Igloo Ice Cube with a Bazooka screen.

FlyGuy said:

That is still a lot of grain -- your 2.25 gals of losses are killing you, man.

Click to expand...

No joke! Thanks for your analysis.

That's why the bazooka doesn't make for a perfect mash separator right out of the box. You should have a short run of copper tubing to get the exit close to the bottom of the cooler, then clamp your mesh to that.

Bobby_M said:

That's why the bazooka doesn't make for a perfect mash separator right out of the box. You should have a short run of copper tubing to get the exit close to the bottom of the cooler, then clamp your mesh to that.

Click to expand...

The bazooka is scraping the bottom as it is...there is no room for the copper. What you describe is exactly what I have in my BK.

Have you ever determined where the wort is sitting in your MLT after you lauter? That sounds really strange, it should siphon it right off the bottom. Do you use a small piece of tubing off the bulkhead to help with the siphon?

Doc Robinson said:

. . .
Ok, so I need 7.8g of 1.053 wort in my kettle to start the brew. I have 1g of dead space in my MLT, which means I actually need 8.8g of 1.053 wort.

I need 53 PPG (points per pound per gallon) from my mash and lauter (1.053). I am averaging around 70% extraction efficiency, so I know my denominator of Maximum PPG must be 75 (53/70%). So, given my ratios, and a max PPG for Carmel 60 of 32 and a max PPG for the 2-Row of 28, here is what I need.
. . .

Click to expand...

You mention dead space of 1G in the MLT, and then also 70% extraction efficiency. I wonder if you are double-counting here - e.g. the figure of 70% perhaps already takes into account the dead space loss.

How did you arrive at the 70% figure?

I tried to model this in BTP without much success. I was not able to figure out a direct way to model the MLT dead space in the process in such a way that the amount of dead space reduced the OG per volume of the preboil mixture.
I also tried to increase the water absorption per pound of grain and while that changes mash-in/sparge volumes to get the final volume, it doesn't do anything for the OG/Volume result.

So it seems that in BTP anyway, the efficiency figure of 70% must take into account all sugar losses, i.e. due to failure to extract and due to dead space.

I think your derivation of the 70% figure is a critical to your final answer.

carp said:

You mention dead space of 1G in the MLT, and then also 70% extraction efficiency. I wonder if you are double-counting here - e.g. the figure of 70% perhaps already takes into account the dead space loss.

How did you arrive at the 70% figure?

I tried to model this in BTP without much success. I was not able to figure out a direct way to model the MLT dead space in the process in such a way that the amount of dead space reduced the OG per volume of the preboil mixture.
I also tried to increase the water absorption per pound of grain and while that changes mash-in/sparge volumes to get the final volume, it doesn't do anything for the OG/Volume result.

So it seems that in BTP anyway, the efficiency figure of 70% must take into account all sugar losses, i.e. due to failure to extract and due to dead space.

I think your derivation of the 70% figure is a critical to your final answer.

Click to expand...

Thanks for the input. The extraction efficiency is calculated as:

(Gravity Points x Gallons of Wort Collected) / Maximum Points Per Gallon

I am between 65% and 70% always. However, my Brewhouse Efficiency is 47% because of the 2.25 gallons lost to the BK and MLT. That is pretty terrible.

Doc Robinson said:

Thanks for the input. The extraction efficiency is calculated as:

(Gravity Points x Gallons of Wort Collected) / Maximum Points Per Gallon

I am between 65% and 70% always. However, my Brewhouse Efficiency is 47% because of the 2.25 gallons lost to the BK and MLT. That is pretty terrible.

Click to expand...

It might be worth exploring this further. I'd like to expand on your equation, and rephrase it as:
(Actual gravity points in the preboil wort extracted from the MLT x Gallons of wort collected ) / maximum gravity points that could be extracted from the grain bill.

The operative term is 'Gallons of wort collected', which already accounts for the dead space in the MLT. No further consideration or adjustment for the dead space is required.
Could you provide some actual data (grain bill, input water (mash & sparge) volume, preboil volume and gravity ) from a real batch you have made that illustrates your 65-70% result.

Reading your original post again, you stated 'I have 1g of dead space in my MLT, which means I actually need 8.8g of 1.053 wort.' I think this is the key to the flaw in your logic. I think it helps to define 'wort' as the wondrous nectar that comes out of the MLT, and not think of wort as the nebulous mix of liquid and grain that exists in the MLT. Looking at it this way further illustrates why the dead space need not be considered in addition to the efficiency factor.

I would think seriously about putting a manifold in your MLT. One gallon of dead space would be hard to live with. Personally, I could never get very good efficiency with a bazooka screen on my Coleman extreme cooler. I got 5 points improvement when I used a friend's cooler with a long mesh tube in the bottom. (both batch sparge with minimal dead space)

Doc Robinson said:

OG: 1.066
Volume: 5 gallons

I will lose 1.25 to trub and dead space in my brew kettle, meaning I will need a volume of 6.25g at the end of my boil. I boil off 20% (I know, it's crazy) so I will need 7.8g at the beginning of the boil.

Click to expand...

Agreed.

Doc Robinson said:

Since I need 1.066 @ 5g at the end of the boil, I will need 1.053 @ 7.8g at the beginning of the boil (66 * 5/7.8).

Click to expand...

A little typo there. At the end of the boil, you will need 6.25g, not 5g, but 66 * 6.25 / 7.8 = 53, so your pre-boil gravity calculation is correct.

Doc Robinson said:

Ok, so I need 7.8g of 1.053 wort in my kettle to start the brew. I have 1g of dead space in my MLT, which means I actually need 8.8g of 1.053 wort.

Click to expand...

That is not quite right. As you have established, you need 7.8g of 1.053 wort in the kettle, but the gravity of the wort left in the dead space in the MLT does not have to be 1.053. If you have a good sparge, the gallon left in the MLT should be less than 1.020

Doc Robinson said:

I need 53 PPG (points per pound per gallon) from my mash and lauter (1.053). I am averaging around 70% extraction efficiency, so I know my denominator of Maximum PPG must be 75 (53/70%). So, given my ratios, and a max PPG for Carmel 60 of 32 and a max PPG for the 2-Row of 28, here is what I need.

2-Row: 21.69 pounds (21.69 pounds / 8.8 gallons * 28 max points) = 69
Carmel: 1.81 pounds (1.81 pounds / 8.8 gallons * 32 max points) = 7

69 + 7 = 76 (roughly the denominator I need for 70% extraction efficiency)

A 23.5 pound grain bill for a 5 gallon batch of 1.066?!?

Click to expand...

This is where I get completely lost.
You know that you need to collect 7.8g of 1.053 wort to end up with 5g at 1.066. You therefore need to collect 413 gravity units (7.8 * 53). If you achieve 70% mash/lauter efficiency, and each lb grain will theoretically yield ~36 gravity units at 100% efficiency, you will need 413 / (36 * .7) lbs grain = 16.4 lbs, which by a strange coincidence is exactly what FlyGuy came up with.
I must admit, I didn't understand FlyGuy's explanation with just a quick read through (because we think and express ourselves differently), but I'm sure I could have done if I took the time to try and understand what he said. The fact that we both came up with the same result plus the fact that I have a great deal of respect for his opinions was good enough for me.
If your 70% efficiency is actually brewhouse efficiency (which takes into account the losses when transferring from the kettle to the fermenter), then you would only need 16.4 * 5 / 6.25 = 13.12 lbs grain (where the 5 is the gallons in the fermenter, and the 6.25 is the volume in the kettle at the end of the boil.
If you want to improve your efficiency, then I would try to reduce the losses when transferring from the kettle to the fermenter. I drain from the kettle spigot to the fermenter, and would leave about 1g in the kettle if I left the kettle level while draining; but I tip the kettle slightly in the direction of the spigot which reduces the dead space down to less that 1 pt.

Hope this helps

-a.

Oops, I was agreeing with BrewSmith rather than FlyGuy.
Doesn't matter. I respect them both.
Perhaps I should spend a little time trying to understand what both of them actually said.

I think Brewsmith and I came up with the same numbers, but he didn't account for the 1 gal loss in the mash tun. I factored that in to my calcs.

FlyGuy said:

I think Brewsmith and I came up with the same numbers, but he didn't account for the 1 gal loss in the mash tun. I factored that in to my calcs.

Click to expand...

I think your (FlyGuy) calculation is incorrect. As observed by ajf, that extra gallon in the MLT will not contain 53 sugar points, it will be have more like 10 or 20 point.

Several of the posts above illustrate that it is not necessary (even if its possible) to factor the dead space into the calcs for grain bill / gravity. The efficiency number, e.g. 70% covers this and all other extraction sins.

Of course it is necessary to take the dead space into account when determining the amount of mash/sparge water to heat, but thats it.

carp said:

I think your (FlyGuy) calculation is incorrect. As observed by ajf, that extra gallon in the MLT will not contain 53 sugar points, it will be have more like 10 or 20 point.

Click to expand...

Unfortunately, it doesn't work like that. The calculation of extract efficiency is based on overall extraction. Also remember that the first runnings are much higher gravity, which balance out in the end.

FlyGuy said:

Unfortunately, it doesn't work like that. The calculation of extract efficiency is based on overall extraction. Also remember that the first runnings are much higher gravity, which balance out in the end.

Click to expand...

Actually I'm pretty sure that it DOES work like that. Part of your statement is correct - the calculation of efficiency is based on overall extraction, which is a function of various factors in the MLT including deadspace. Your second statement does not make any sense to me in the context of our discussion. We're talking about the last gallon left in the MLT, and its a fact, its not going to sitting there containing 53 sugar points.

I'll pick on another of the statements in your original post:
"Mash tun losses of 1 gal mean you need to mash 8.8 gal @ 1.053".

What does 'mash 8.8 gallon' mean exactly? One doesn't mash water, one mashes grain, and extracts wort. Clearly you don't extract 8.8 gallons, because its left behind in the deadspace.
Now I'm not harping on the grammar/terminology in order to be a clever dickhead, I'm just trying to highlight that careless use of the terms is leading to fundamental misunderstanding of the calculations.

Look in How to Brew, or in software (in BTP anyway) and you'll find NO mention of deadspace, or of the amount of input water (which is affected by deadspace) in either derivation of actual efficiency, or in planning the grain bill.



I really think that ajf's post is the most comprehensively correct one here.

Think of it this way -- if you wanted to increase your recipe volume by 1 gal to offset losses in your kettle, would your same thoughts still apply? No, of course not.

The same thing applies here as well. You are just over-thinking the process.

You are correct, the same thinking does not apply to losses in the brew kettle. That is exactly why that brewing software does have a place to account for such losses. For these post mash process, one is already working with extracted wort and losses have to be considered.

Do you agree that extraction efficiency is calculated as:

(Actual gravity points in the preboil wort extracted from the MLT x Gallons of wort collected ) / maximum gravity points that could be extracted from the grain bill.

Where is dead space in this equation? The reason that the BK logic is different from the mash process is because there term 'efficiency' already covers the mash process in its entirety. Think about all the imperfect things that occur in the mash - design of manifold or braid, rate of wort draining, rate of sparge water addition, temperature, pH, water absorption by grain - dead space is just another of these factors that contributes to lower efficiency.

Let me ask one more question - how do you take into account absorption of water by the grains in your calculations, in relation to grain bill and OG? Answer - YOU DON'T.

The OP originally presented this red herring of an idea about dead space being a factor in the calculation, and now you seem to be stuck on it, to the exclusion of any documented practice.

Now that I go back and look at his formula for calculating extract efficiency, you are correct -- he has already factored in his losses in the mash tun. So I am incorrect -- there is no need to add one gal of wort in the mash to offset the 1 gal loss. He was indeed double-counting a loss.

That also means that our recent back and forth about gravity of last runnings is moot. That extra gallon should never have been in there.

Regardless, I see your point now about gravity of final runnings. If one were to try to account for it, my original calculation of 1 extra gallon @ 1.053 would have overcompensated.

But that actually brings up a new point -- it will be hard now to figure out how to compensate for losses in the boil kettle because his efficiency will vary greatly with the mash volume.

carp, FlyGuy, and ajf...thank you so much. What I have taken from all this is that I should not be counting the 1 gallon of deadspace in the MLT. If so, I ordered my recipe based on an erroneous calculation. If you are correct, the extra grain I ordered should amount to a higher than planned OG. We'll see.

FlyGuy said:

. . .
That also means that our recent back and forth about gravity of last runnings is moot. That extra gallon should never have been in there.

Regardless, I see your point now about gravity of final runnings. If one were to try to account for it, my original calculation of 1 extra gallon @ 1.053 would have overcompensated.

But that actually brings up a new point -- it will be hard now to figure out how to compensate for losses in the boil kettle because his efficiency will vary greatly with the mash volume.

Click to expand...

Taking some of your points in reverse order:

'Efficiency vary with mash volume' - that is correct. If deadspace is a major contributor to mash inefficiency, then efficiency will go up as the boil volume goes up, in a way thats hard to calculate. He would see better eff for 10 gal batches than for 5 gallons. I suspect that within a normal range for say 5 gal batches, the difference won't be too big.

But . . . one has to wonder how major of a factor the deadspace is? My initial reaction was that it must be a huge problem. But when ajf highlighted that the gravity of the volume in the deadspace is pretty low, I realize it might not be such a huge deal. I have done some tracking of gravity of runnings over the course of the sparge, and once you get to that last gallon, (in my case about 13.5 gallons), the gravity is not reducing that much for each additional gallon.

Anyway it was a good discussion and going through it helped solidify the concepts in my mind.

Doc Robinson said:

I haven't been hitting my numbers. I decided to spend some time putting a spreadsheet together to simply my process because BeerSmith isn't doing it for me. It's long and slightly technical, but I will try to make it as easy to follow as possible:

Here is what I need to make using roughly 92% 2-Row & 8% Carmel:

OG: 1.066
Volume: 5 gallons

I will lose 1.25 to trub and dead space in my brew kettle, meaning I will need a volume of 6.25g at the end of my boil. I boil off 20% (I know, it's crazy) so I will need 7.8g at the beginning of the boil.

Since I need 1.066 @ 5g at the end of the boil, I will need 1.053 @ 7.8g at the beginning of the boil (66 * 5/7.8).

Ok, so I need 7.8g of 1.053 wort in my kettle to start the brew. I have 1g of dead space in my MLT, which means I actually need 8.8g of 1.053 wort.

I need 53 PPG (points per pound per gallon) from my mash and lauter (1.053). I am averaging around 70% extraction efficiency, so I know my denominator of Maximum PPG must be 75 (53/70%). So, given my ratios, and a max PPG for Carmel 60 of 32 and a max PPG for the 2-Row of 28, here is what I need.

2-Row: 21.69 pounds (21.69 pounds / 8.8 gallons * 28 max points) = 69
Carmel: 1.81 pounds (1.81 pounds / 8.8 gallons * 32 max points) = 7

69 + 7 = 76 (roughly the denominator I need for 70% extraction efficiency)

A 23.5 pound grain bill for a 5 gallon batch of 1.066?!?

Click to expand...

dang doc i knew lawyers were only good at math when it comes to money

1. if you lose 1.25g to deadspace/trub in your kettle then your final volume will be 6.25g into fermentor for calculations
2. #grain * 36ppg / final volume = max og actual gravity/max gravity = eff
3. 1.066/ x = 70% x=1.094 max eff (you drop the 1.0 from the figure for the calculations)
4. x#grain * 36ppg / 6.25 = 1.094 x=16.3 lbs grain


and no i didn't read the rest of the posts on this doc!

carp said:

Anyway it was a good discussion and going through it helped solidify the concepts in my mind.

Click to expand...

Yes me too.

And thanks to Brewsmith for getting it correct the first time around!

Doc Robinson said:

So, given my ratios, and a max PPG for Carmel 60 of 32 and a max PPG for the 2-Row of 28, here is what I need.

2-Row: 21.69 pounds (21.69 pounds / 8.8 gallons * 28 max points) = 69

Click to expand...

According to "Designing Great Beers" book, basic 2-Row has an extract potential between 1.035 & 1.037.

Your calcs for 2-Row should use 36 PPG, not 28.

Edit: Yes I replied before reading the entire thread. Now that I've read it I can see I've restated what others have already pointed out. Guess I'm lazy on xmas eve day.

Soperbrew said:

According to "Designing Great Beers" book, basic 2-Row has an extract potential between 1.035 & 1.037.

Your calcs for 2-Row should use 36 PPG, not 28.

Click to expand...

Yeah...discovered this awhile ago. Thanks though.