Kaiser
Well-Known Member
In order to better understand batch sparging and the efficiencies that I get with my system, I took some more measurements than usual on my last brew day. Tonight I ran the numbers and it took some time to match the brewday results. The most interesting discovery was, that the wort volume present in the mash tun after the mash is significantly larger (by about 22%) than the amount of water that was added. This has an affect on the calculation of the wort volume that is held back in the grain which affects efficiency. So far I was severely underestimating this volume by calculating it through (water used - wort volume collected). As a result my theoretical efficiency was always higher than the actual and I was wondering where I was loosing this efficiency. But now that I calculated that volume correctly, I found that I match the theory very well.
Here is the lengthy analysis:
* grist weight 5.6 kg
* total laboratory extract of that grist is 80% of 5.6 kg -> 4.5 kg
* water added to mash: 15.5 l (cold)
* extract of the first running in the kettle 22.5% (% extract is equal to Plato)
* volume of the first runnings in the kettle 9.75l at 65C -> 9.6l (cold)
* extract of the 2nd runnings: 11.75%
* volume in kettle after 2nd running: 20l at 75C -> 19.6l (cold)
* extract of the 3rd runnings: 7.4%
* volume in kettle after 3rd running (pre-boil volume): 26l at 90C -> 25l (cold)
* extract in kettle after 3rd running (pre-boil extract): 14.6%
The first analysis was for the extraction efficiency of the mash. The definition of extract percentages is:
(1) E = 100% * m_extract / ( m_water + m_extract)
If we want to know how much extract exist in a given wort of known extract content that has been created with a known amount of water we can do this by rearanging (1) to
(2) m_extract = (m_water * E / 100%) / (1 - E / 100%)
(3) m_extract = (15.5kg * 0.225) / (1 - 0.225) = 4.5 kg
This means that all of the extract available in the grain has been extracted in the mash (100% extraction efficiency). This was confirmed by a negative iodine test of the wort and the spent grain. I.e. no native starch was left.
Since batch sparging was used, a simple model can be used to calculate the lauter efficiency. lauter efficiency * extraction efficiency is the brewhouse efficiency. For that model we need the amount of wort that is held back in the lauter tun after each run-off. But this is not simply the amount of water used for the mash minus the amount of first wort collected because the volume of the wort increases when the extract is dissolved. To get that volume we can use this formula which is the weight of extract dissolved in a given volume of known gravity wort:
(4) m_extract = ( E / 100% ) * SG * V_wort
SG is the specific gravity and it will be estimated with 1+E*0.004.Rearranged to V_wort we get
(5) V_wort = m_extract / ((E/100%) * SG)
(6) V_wort = 4.5 kg / (0.225 * 1.090) = 18.3 l
This means the 15.5 l water and 4.5 kg extract from the 5.6 kg grain made 18.3 l of 22.5% wort. 9.6l of that wort were collected after the first run-off which indicates that 8.7 l are held back in the mash.
Batch sparing is a process of successive dilution of the wort remaining in the grain and running it off. This can be modeled mathematically and has bee analyzed here. But since not all run-offs were of equal size, lets just calculate the efficiency step by step:
The first run-off will extract this percentage of the extract from the mash:
(7) Eff_1st = v_1st_runoff / (v_1st_runoff + v_wort_in_grain)
(8) Eff_1st = 9.6l / (9.6l + 8.7l) = 0.52 = 52 %
If 52% were recovered by the 1st run-off, then 48% of the extract are still in the lauter tun. This extract is dilluted by the sparge water and run off. The volume of the 2nd run_off is 19.6l - 9.6l = 10l and the efficiency of that run-off is:
(9) Eff_2nd = v_2nd_run_off / (v_2nd_run_off + v_wort_in_grain)
(10) Eff_2nd = 10l / (10l + 8.7l) =0.53 = 53%
Using this and the fact that the 2nd run-off was only able to draw from 48% of the extract we can determine the combined efficiency from the 1st and 2nd run off as:
(11) Eff_1st_and_2nd = 52% + 48% * 53% = 78 %
78% of the extract are now in the boil kettle. This leaves 22% in the lauter tun. With a 3rd run off size of 5.4 l we find the efficiency of that run-off as
(12) Eff_3rd = 5.4 / (5.4 + 8.7) = 0.38 = 38%
and the combined efficiency of all 3 run-offs as:
(13) Eff_1st_2nd_3rd = 52% + 48% * 53% + 22% * 38% = 0.86 = 86%
This means that with the given run-off sizes, number of sparges and amount of wort left in the grain, an a lauter efficiency of 86% is to be expected.
The actual efficiency into the boiler is the following:
(14) Eff_kettle = V_kettle * E * SG / (m_grain * 0.8)
the 0.8 represents the 80% laboratory extract of the grain.
(15) Eff_kettle = 25l * 0.146 * 1.058 l/kg / (5.6 kg * 0.8) = 86%
Since the Efficiency is the product of extraction efficiency and lauter efficiency and the extraction efficiency was determined to be 100%, the actual lauter efficiency must have been 86%, which matches the theoretical result very well. As a result no efficiency was lost due to process inefficiencies and to increase that efficiency the following process parameters could be changed:
* more sparge water: this would lead to a larger pre boil volume and longer or stonger boils and may not be desired
* less wort kept in the grain: This mash was done with conditioned malt which makes for a"fluffier" mash. Such a mash may hold more wort and I wonder if an unconditioned mash may result in less wort being held back and thus increasing the efficiency
* equalize the run-offs: the boost expected from that is very low. Se here.
* fly sparging: this method follows a different principle and should yield better efficiencies when done properly. But in addition to more time, it also needs a better lautertun which I don't have.
So, 86% for that beer is fine with me.
Here is the lengthy analysis:
* grist weight 5.6 kg
* total laboratory extract of that grist is 80% of 5.6 kg -> 4.5 kg
* water added to mash: 15.5 l (cold)
* extract of the first running in the kettle 22.5% (% extract is equal to Plato)
* volume of the first runnings in the kettle 9.75l at 65C -> 9.6l (cold)
* extract of the 2nd runnings: 11.75%
* volume in kettle after 2nd running: 20l at 75C -> 19.6l (cold)
* extract of the 3rd runnings: 7.4%
* volume in kettle after 3rd running (pre-boil volume): 26l at 90C -> 25l (cold)
* extract in kettle after 3rd running (pre-boil extract): 14.6%
The first analysis was for the extraction efficiency of the mash. The definition of extract percentages is:
(1) E = 100% * m_extract / ( m_water + m_extract)
If we want to know how much extract exist in a given wort of known extract content that has been created with a known amount of water we can do this by rearanging (1) to
(2) m_extract = (m_water * E / 100%) / (1 - E / 100%)
(3) m_extract = (15.5kg * 0.225) / (1 - 0.225) = 4.5 kg
This means that all of the extract available in the grain has been extracted in the mash (100% extraction efficiency). This was confirmed by a negative iodine test of the wort and the spent grain. I.e. no native starch was left.
Since batch sparging was used, a simple model can be used to calculate the lauter efficiency. lauter efficiency * extraction efficiency is the brewhouse efficiency. For that model we need the amount of wort that is held back in the lauter tun after each run-off. But this is not simply the amount of water used for the mash minus the amount of first wort collected because the volume of the wort increases when the extract is dissolved. To get that volume we can use this formula which is the weight of extract dissolved in a given volume of known gravity wort:
(4) m_extract = ( E / 100% ) * SG * V_wort
SG is the specific gravity and it will be estimated with 1+E*0.004.Rearranged to V_wort we get
(5) V_wort = m_extract / ((E/100%) * SG)
(6) V_wort = 4.5 kg / (0.225 * 1.090) = 18.3 l
This means the 15.5 l water and 4.5 kg extract from the 5.6 kg grain made 18.3 l of 22.5% wort. 9.6l of that wort were collected after the first run-off which indicates that 8.7 l are held back in the mash.
Batch sparing is a process of successive dilution of the wort remaining in the grain and running it off. This can be modeled mathematically and has bee analyzed here. But since not all run-offs were of equal size, lets just calculate the efficiency step by step:
The first run-off will extract this percentage of the extract from the mash:
(7) Eff_1st = v_1st_runoff / (v_1st_runoff + v_wort_in_grain)
(8) Eff_1st = 9.6l / (9.6l + 8.7l) = 0.52 = 52 %
If 52% were recovered by the 1st run-off, then 48% of the extract are still in the lauter tun. This extract is dilluted by the sparge water and run off. The volume of the 2nd run_off is 19.6l - 9.6l = 10l and the efficiency of that run-off is:
(9) Eff_2nd = v_2nd_run_off / (v_2nd_run_off + v_wort_in_grain)
(10) Eff_2nd = 10l / (10l + 8.7l) =0.53 = 53%
Using this and the fact that the 2nd run-off was only able to draw from 48% of the extract we can determine the combined efficiency from the 1st and 2nd run off as:
(11) Eff_1st_and_2nd = 52% + 48% * 53% = 78 %
78% of the extract are now in the boil kettle. This leaves 22% in the lauter tun. With a 3rd run off size of 5.4 l we find the efficiency of that run-off as
(12) Eff_3rd = 5.4 / (5.4 + 8.7) = 0.38 = 38%
and the combined efficiency of all 3 run-offs as:
(13) Eff_1st_2nd_3rd = 52% + 48% * 53% + 22% * 38% = 0.86 = 86%
This means that with the given run-off sizes, number of sparges and amount of wort left in the grain, an a lauter efficiency of 86% is to be expected.
The actual efficiency into the boiler is the following:
(14) Eff_kettle = V_kettle * E * SG / (m_grain * 0.8)
the 0.8 represents the 80% laboratory extract of the grain.
(15) Eff_kettle = 25l * 0.146 * 1.058 l/kg / (5.6 kg * 0.8) = 86%
Since the Efficiency is the product of extraction efficiency and lauter efficiency and the extraction efficiency was determined to be 100%, the actual lauter efficiency must have been 86%, which matches the theoretical result very well. As a result no efficiency was lost due to process inefficiencies and to increase that efficiency the following process parameters could be changed:
* more sparge water: this would lead to a larger pre boil volume and longer or stonger boils and may not be desired
* less wort kept in the grain: This mash was done with conditioned malt which makes for a"fluffier" mash. Such a mash may hold more wort and I wonder if an unconditioned mash may result in less wort being held back and thus increasing the efficiency
* equalize the run-offs: the boost expected from that is very low. Se here.
* fly sparging: this method follows a different principle and should yield better efficiencies when done properly. But in addition to more time, it also needs a better lautertun which I don't have.
So, 86% for that beer is fine with me.