Reiterated Mash - Unequal Grain Split

[meryvn]

My last brew i tried a reiterated mash for the first time and it worked quite well, i've been thinking about the process of it and had a few questions and wondered if anyone had answers or thoughts on the below.

My last grain bill was, for a 12L final volume at 50% efficiency

• 5kg Pale Malt
• 650g Light Crystal
• 450g Chocolate
• 450g Roasted Barley
• 260g Flaked Oats

I split the grain above in half and did a mash with the same liquor on each half but my queries are:

• Would it be better to split 60/40 and have the first mash with a higher grain weight?
• Would it be better to have all pale malt (i.e. ~3.4kg of pale in above example) in the first mash then all the 'specialty malts' plus remaining pale malt in the 2nd mash or vice versa?

 

[brewman !]

Interesting questions. I'm about to do an iterative mash as well. I look forward to the comments you receiver. @doug293cz ?

 

[Wagon_6]

50% efficiency seems low. And just to confirm you are BIAB? Do you recirculate, stir, and or sparge?

I typically split my grains down the middle with the goal of keeping the pH in line. I plug it into brunwater as a huge beer and then plug in the grains as the first 50% to make sure it still works. I’ve started using RO water so it makes it easy.

Either way you split it, you would have to account for it in the mineral additions for each mash’s pH. I think 50/50 and grain bill equally split is an easy way to do it and stay consistent.

 

[meryvn]

50% efficiency seems low. And just to confirm you are BIAB? Do you recirculate, stir, and or sparge?

I typically split my grains down the middle with the goal of keeping the pH in line. I plug it into brunwater as a huge beer and then plug in the grains as the first 50% to make sure it still works. I’ve started using RO water so it makes it easy.

Either way you split it, you would have to account for it in the mineral additions for each mash’s pH. I think 50/50 and grain bill equally split is an easy way to do it and stay consistent.

I've got a Brewdevil 50L (which is a rebaged Guten machine) my efficiency generally hovers around the 60% - 70% mark but i dropped it when doing this recipe as i didn't think it would be as high with the size of the grain bill. Actual efficiency, according to Brewfather, was 59% Mash Efficiency.

I'm quite a careless brewer when it comes to efficiency i don't bother with stirring or sparging. I usually aim low (set most recipes to 60%) and add a bit of extra grain and then liquor back if needed then end up lower OG than i planned.

 

[brewman !]

Efficiency isn't the only reason to do a reiterative mash. Mash tun sizing plays a part in it as well. The higher the gravity you try to brew, the lower the efficiency gets. So when you go bigger, the mash tun grows almost exponentially because you need more grain to go bigger and you also need more grain to compensate for the falling efficiency.

Here is a thread on the topic of reiterative brewing with some math in it.
https://www.homebrewtalk.com/threads/huge-stout-split-mash-or-sparge.660526/

 

[brewman !]

I need 8.5 gallons (into the fermentor) of the highest gravity wort I can get. My mash tun is a bit less than 8 gallons. The underlet volume is about 1 gallon. Total vessel volume is 11 gallons. I'm doing this in Thing1, see link in my sig.

Let's guess that I'll use 24 pounds of malt, divided into 2 x 12 pound mashes.

Assume that conversion efficiency is 100%, which is fair if the grind is fine enough, we mash long enough and circulate enough. This is not sparge efficiency as it does not include the extract that remains trapped in the malt post sparge.

I need 8.5 gallons into the fermenter. let's assume we lose 1 gallon during boil. That means we need 9.5 gallons after the second sparge.

Each sparge will retain 12 pounds x 0.06 gallons/lb of malt = 0.72 gallons.

The post 1st stage sparge volume will be 9.5 + 2 x .72 gallons = 10.94 gallons total water.

Mash 1. 12 pounds x 37 points/pound = 444 points total extracted. We sparge with 10.94 gallons of water, for a gravity of 444/10.94 = 1.0406. 0.72 gallons of wort stays in the grain, so post sparge we have 10.22 gallons of 1.0406 wort.

Mash 2. 12 pounds x 37 points = 444 points total extracted. We sparge with 10.22 gallons of the wort from mash 1. Ending gravity is 1.0406 + 444/10.22 = 1.084. We leave 0.72 gallons of wort in the grain, so our ending volume is 9.5 gallons @ 1.084.

Overall efficiency of these two steps is:

(9.5 gallons x 84 points) / (2 x 444) = 89.8%

Next we boil down to 8.5 gallons. The gravity going into the fermenter will be (9.5 gallons x 84 points) / 8.5 gallons = 1.0939

I would like a higher gravity than that, so let's repeat the calc for 30 pounds of malt.

Each sparge will lose 15 pounds x 0.06 = 0.90 gallons of wort. Starting sparge volume will be 8.5 gallons into the fermentor + 1 gallon boil off + 2x 0.9gallons lost to grain = 11.3 gallons.

Mash 1. 15 pounds x 37 points = 555 points. OG = 555 points/11.3 gallons = 1.0491 Post sparge volume is 11.3 gallons = 0.9 gallons = 10.4 gallons.

Mash 2. 15 pounds x 37 points = 555 points. OG = 1.0481 + 555/10.4 gallons = 1.1067 Post sparge volume is 10.4 - 0.9 = 9.5 gallons.

Overall efficiency of these two steps is:

(9.5 gallons x 106.7 points) / (2x 555 points) = 90%.

Next we boil down to 8.5 gallons. The gravity going into the fermentor will be (9.5 x1.1067)/8.5 = 1.119. This is perfect.


If we tried to mash 30 pounds at 100% conversion efficiency in one mash, the volume of wort retained would be 30 x 0.06 = 1.8 gallons. Assuming a post mash volume into the boiler of 9.5 gallons, the total sparge water would be 11.3 gallons, same as before.

30 pounds x 37 points = 1110 total points. OG = 1110 pnts/11.3 gallons = 1.0982. Final volume is 9.5 gallons. Mash efficiency = 9.5 x 98.2 / 1110 = 84% <-- This would never be attained in the real world.

This is all theoretical. Real gravity will probably be 80% of the calc numbers.

I'll be brewing this beer twice in the near future. I'll report back how it goes.

 

[meryvn]

I need 8.5 gallons (into the fermentor) of the highest gravity wort I can get. My mash tun is a bit less than 8 gallons. The underlet volume is about 1 gallon. Total vessel volume is 11 gallons. I'm doing this in Thing1, see link in my sig.

Let's guess that I'll use 24 pounds of malt, divided into 2 x 12 pound mashes.

Assume that conversion efficiency is 100%, which is fair if the grind is fine enough, we mash long enough and circulate enough. This is not sparge efficiency as it does not include the extract that remains trapped in the malt post sparge.

I need 8.5 gallons into the fermenter. let's assume we lose 1 gallon during boil. That means we need 9.5 gallons after the second sparge.

Each sparge will retain 12 pounds x 0.06 gallons/lb of malt = 0.72 gallons.

The post 1st stage sparge volume will be 9.5 + 2 x .72 gallons = 10.94 gallons total water.

Mash 1. 12 pounds x 37 points/pound = 444 points total extracted. We sparge with 10.94 gallons of water, for a gravity of 444/10.94 = 1.0406. 0.72 gallons of wort stays in the grain, so post sparge we have 10.22 gallons of 1.0406 wort.

Mash 2. 12 pounds x 37 points = 444 points total extracted. We sparge with 10.22 gallons of the wort from mash 1. Ending gravity is 1.0406 + 444/10.22 = 1.084. We leave 0.72 gallons of wort in the grain, so our ending volume is 9.5 gallons @ 1.084.

Overall efficiency of these two steps is:

(9.5 gallons x 84 points) / (2 x 444) = 89.8%

Next we boil down to 8.5 gallons. The gravity going into the fermenter will be (9.5 gallons x 84 points) / 8.5 gallons = 1.0939

I would like a higher gravity than that, so let's repeat the calc for 30 pounds of malt.

Each sparge will lose 15 pounds x 0.06 = 0.90 gallons of wort. Starting sparge volume will be 8.5 gallons into the fermentor + 1 gallon boil off + 2x 0.9gallons lost to grain = 11.3 gallons.

Mash 1. 15 pounds x 37 points = 555 points. OG = 555 points/11.3 gallons = 1.0491 Post sparge volume is 11.3 gallons = 0.9 gallons = 10.4 gallons.

Mash 2. 15 pounds x 37 points = 555 points. OG = 1.0481 + 555/10.4 gallons = 1.1067 Post sparge volume is 10.4 - 0.9 = 9.5 gallons.

Overall efficiency of these two steps is:

(9.5 gallons x 106.7 points) / (2x 555 points) = 90%.

Next we boil down to 8.5 gallons. The gravity going into the fermentor will be (9.5 x1.1067)/8.5 = 1.119. This is perfect.


If we tried to mash 30 pounds at 100% conversion efficiency in one mash, the volume of wort retained would be 30 x 0.06 = 1.8 gallons. Assuming a post mash volume into the boiler of 9.5 gallons, the total sparge water would be 11.3 gallons, same as before.

30 pounds x 37 points = 1110 total points. OG = 1110 pnts/11.3 gallons = 1.0982. Final volume is 9.5 gallons. Mash efficiency = 9.5 x 98.2 / 1110 = 84% <-- This would never be attained in the real world.

This is all theoretical. Real gravity will probably be 80% of the calc numbers.

I'll be brewing this beer twice in the near future. I'll report back how it goes.

Having read the thread you had linked above i get the 'normal' reiterative process, but what i was wondering was if you decided not to split the grain 50/50 and do 60/40 for the grain (mash tun size allowing), would that change the efficiency?

I'll have to have a play with the spreadsheet in the linked thread to see if anything changes in that with different volumes in each mas

 

[brewman !]

I haven't run those numbers. My mash tun only holds 15 pounds, so I can't split it unevenly anyway.

Run the numbers and see where it goes.

 

[meryvn]

Having read through the linked thread again and with the supposition that the second mash does not lose any efficiency from being done in wort rather than water, then there is no theoretical difference from doing 50/50 than 60/40 or 40/60.

Unfortunately i don't have time to brew the same beer a few times to check if this bears out in real life so will just stick to the 50/50 that has served me well in the past.

 

[brewman !]

Let us know how it works out for you.