There is one question that has been bothering me for a long time.
Which grain absorption should we use in order to calculate water volumes: apparent or actual?
First let me precise what I mean by apparent absorption et actual absorption.
Almost all beer software use the grain absorption in the following way.
In fact this is only appearance, because the preboil volume is not all water but contains sugar (extract).
Similarly, weighing the grain before mash and weighing spent grain and then calculating the difference doesn't give the real water absorbed because the grain has lost sugar.
Lets take an example
With these data we can calculate the sparge water
according to (a)
Knowing the volume 24 liters and the sugar mass 3 kg and using a convergent iterative algorithm we can obtain the gravity and the water mass
In this case:
Now let's assume that we are using the actual absorption to show how the calculation works
This seems perfect but there is a rub. You probably has noticed that passing from apparent to actual varies a little bit depending on the mash efficiency and grain yield.
The question is: what is more independent from these 2 factors : apparent or actual absorption ?
Assuredly using the apparent absorption is a bit easier because it doesn't require the use of the iterative convergent algorithm.
Which grain absorption should we use in order to calculate water volumes: apparent or actual?
First let me precise what I mean by apparent absorption et actual absorption.
Almost all beer software use the grain absorption in the following way.
(here we suppose all is considered cold meaning volume quasi equal mass)(a) Mash water + Sparge water - Grain absorption = preboil volume
In fact this is only appearance, because the preboil volume is not all water but contains sugar (extract).
Similarly, weighing the grain before mash and weighing spent grain and then calculating the difference doesn't give the real water absorbed because the grain has lost sugar.
Lets take an example
Grain mass: 5 kg
Grain yield: 80 %
Mash efficiency: 75 %
Mash water 17 liters
Preboil volume (cold): 24 liters i.e. preboil volume hot 24 * 1.04 = 24.96 liters
With these data we can calculate the sparge water
according to (a)
Sparge water =Preboil volume -Mash Water + Grain absorption
Sparge water = 24 - 17 + (0.8 * 5)= 11 liters
Sugar (extract) in préboil volume= 5 kg * 0.8 * 0.75 = 3 kg
Knowing the volume 24 liters and the sugar mass 3 kg and using a convergent iterative algorithm we can obtain the gravity and the water mass
In this case:
Thus the actual absorbtion isgravity 11.927 platos (SG 1.0481) and water mass 22.15 kg (or liters)
Mash water + Sparge water - Water mass in the preboil volume
17 + 11 - 22.15 = 5.85 liters
Actual absorption is 5.85 /5 = 1.17 l/kg.
Now let's assume that we are using the actual absorption to show how the calculation works
And we can calculate sparge water using the actual absorptionThe extract produced is the same i.e. 3 kg
We the preboil volume and the sugar mass using the same algorithm we can get
the water mass in preboil volume, still 22.15 liters
Sparge water = Water in the preboil volume - Mash water + Grain absorption
Sparge water = 22.15 - 17 + (1.17 * 5) = 11 liters
This seems perfect but there is a rub. You probably has noticed that passing from apparent to actual varies a little bit depending on the mash efficiency and grain yield.
The question is: what is more independent from these 2 factors : apparent or actual absorption ?
Assuredly using the apparent absorption is a bit easier because it doesn't require the use of the iterative convergent algorithm.