Take the guessing out of efficiency Part 2

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Note: This is the second article in a series of 3. The link to the first one is here https://www.homebrewtalk.com/entries/take-the-guessing-out-of-efficiency.html
In this article, I will explain how to calculate extract content of wort and sparging volumes.
Conversion efficiency, as previously explained, is the amount of starch sugars that were converted to simple sugars divided by the total amount of available sugars. It should always be higher than 90%. I will be using 95% conversion to estimate the brewhouse efficiency.
The separation of sweet wort from spent grain is a physical parameter that will be taken into account with lautering efficiency. Grain absorbs 1.56L/kg (.748 qt/lb) of liquid which will not end up in your wort.Let's continue with the example of producing 21L (5.5 gal) of beer using 5kg(11lb) of grain with 80% extract CGDB and 4% moisture. For the main mash, let'ssay you use a grain to water ratio of 3L/kg (1.44 qt/lb). This means you mixed 15L (3.96 gal) water with your grain.
Maximum extractable content of mash:
SI 5*0.8*(1-0.04)=3.84 kg
US 11*0.8*(1-0.04)=8.45 lb
Taking conversion efficiency into account
SI 3.84*.95=3.65 kg
US 8.45*.95=8.03 lb
Plato value of mash
SI 3.65/(3.65+15)*100=19.6 wt%
US 8.03/(8.03+33.07)19.5 wt%
Ideal Lautering
This means the mash is at an SG of 1.078. Now you open the valve and start collecting your wort. How much will you recover? Recall dissolved sugar increases the volume by 0.63 L/kg (0.0755 gal/lb).
Volume of liquid in mash
SI 15+3.65*.63=17.3L
US 3.96+0.0755*8.03=4.57 gal
Volume absorbed by grain
SI 1.56*5=7.8L
US 0.187*11=2.06 gal
Volume collected
SI 17.3-7.8=9.5L of 1.078 wort
US 4.57-2.06=2.51 gal of 1.078 wort
Instead of using 1.56 L/kg (0.748 qt/lb) as the absorption of grain most people use a value closer to 1.1 L/kg (0.527 qt/lb) which is the apparent absorption of grain. In other words if you know you added 15L (3.96gal) of water and collected 9.5 L(2.51 gal) of liquid when you had 5 kg (11lb)of grain;
Apparent absorption
SI (15-9.5)/5=1.1 L/kg
US (3.96-2.51)/11=0.132 gal/lb=0.527qt/lb
This value is actually not dependent on the grain to water ratio, but on the extract content of the mash. Basically apparent absorption takes into account the increase in volume due to dissolved sugars.
Apparent absorption=True absorption*Max Extract*Conversion Efficiency/Grain mass
Recall that
Max Extract=Grain mass* Extract CGDB*(1-Moisture)
So grain mass cancels out and apparent absorption is dependent on the type of grain used (extract CGDB and moisture) and conversion efficiency. Since these values do not change much, using an apparent absorption of 1.1 L/kg (0.527 qt/lb) is a decent way of simplifying the math. This way you don't have to worry about correcting for the volume of dissolved sugars.
Dead volumes and stuck sparges
This is all nice and well, but the 1.56 L/kg (0.748 qt/lb)is an ideal laboratory value. You need to account for the non-idealities of your system. There is a certain amount of liquid that will remain at the bottom of the vessel and never reach the collection tube. This is known as dead
volume. People doing BIAB can just lift the bag and let it drain, which results in zero dead volume (hooray for simplicity), but if you have a mash tun there are two ways of calculating dead volume.
1.
When empty, add a nominal amount of water to your assembled mash tun, say 15L (4 gal) and open the valve. When it stops,take a measuring cup and collect the liquid at the bottom of the mash tun. This is theoretically your dead volume, ie the volume that will always remain in the
tun no matter what. In the case of a 10 gallon cylindrical with a false bottom,I got 0.3L (0.3 qt). Rectangular coolers and less efficient collection system can greatly increase this value. Don't be surprised if you get something closer to 2L (0.5 gal) or even more.
2.
The above method is simple, but we all know that when grains are added chances of stuck sparges can be high. So you can just mash using grains and calculate the amount collected. Using the above example of 5kg (11lb) in 15L (4 gal), you know you should collect 9.5L (2.51 gal). So if you just get 8 L (2.11gal), you have a dead volume of 1.5L (0.4 gal).
Note that I always use the 0.3L (0.3 qt) from above and get consistent values. You get stuck sparges in two cases. In the first case, the grain bed compacts itself too quickly and basically forms a barrier instead of a filter for liquid. You can use rice hulls to minimize this. If it still
happens, close the valve, stir up the grain, let sit for a couple of minutes and reopen the valve. Second case is harder to fix. The grain which acts as a filter is an obstacle to fluid flow. If the top of the grain runs dry, while the bottom is still wet, air can be sucked in and you lose the siphon which allowed the liquid to flow. This is why you should always sparge very slowly at
the end. However, this only represents a loss of 1-2L (1-2 qt) compared to ideality, so if you don't mind it, just take it into account when calculating efficiency. Below is an example of calculation for sparging with a 1L (1qt)dead volume.
Volume of liquid in mash
SI 15+3.65*.63=17.3L
US 3.96+0.0755*8.03=4.57 gal
Volume absorbed by grain
SI 1.56*5=7.8L
US 0.187*11=2.06 gal
Volume collected
SI 17.3-7.8-1=8.5L of 1.078wort
US 4.57-2.06-0.25=2.25 gal of 1.078 wort
Accounting forevaporation
Now comes the hard part, which is how much water we need to add for the second sparge to end up with 21L (5.5 gal) of beer at the end.First we need to know how much we will evaporate. 4L (1 gal) per hour is a good average value for people using an 8-9 gallon brew pot. Note I don't use 15%evaporation because if I want to make 12L (3 gal) of beer or 21L (5.5gal), I use the exact same kettle. Percent values are useful if you always want to run
your kettle at max volume, like in the industry. Not our case here.
So 60 minute boil means I need 25L (6.6 gal) of preboil liquid. The problem is this is wort volume and not water volume, which means I need to account for the dissolved sugars. However, I am trying to find the amount of dissolved sugars I have (calculating efficiency). This is a nonlinear equation and I can imagine the shudders of fear from people who know
what they are. Basically, I need the answer to find the answer.
Actually it really isn't that bad if you realize that there will always be the same amount of liquid left behind at every sparge. In other words;
Volume left behind =1.56 L/kg * Mass of grain Dead Volume
This means that the volume of water you sparge with should be the volume of wort you collect. You leave behind more water and less sugar.
Sparge volume = Final volume + Evaporation Volume collected from first running
SI 21+4-8.5=16.5L
US 5.5+1-2.25=4.25 gal
Non ideal transfers
The only thing missing to calculate the brewhouse efficiency is to account for losses in volume during transfer. The hops and coagulated proteins known as trub need to be separated from the boiled wort before fermentation. Also if you use counterflow chillers, some of the wort will be left behind in the chiller. This value can vary widely between systems and you need to calculate yours. A standard value is 2L (0.5 gal) losses of liquid.This needs to be accounted with extra sparge water.
Sparge volume = Final volume + Evaporation + Losses during transfer
Volume collected from first running
SI 21+4+2-8.5=18.5L
US 5.5+1+0.5-2.25=4.75 gal
Pellet hops will absorb less liquid than whole hops, however pellets hops remain in suspension and are harder to filter which means more wort is left behind. An average value I found to account for hop debris is 0.015 L/g (0.45 qt/oz) of liquid loss. Therefore if an average recipe has 50g of hops (1.75 oz), I can add or substract to the wort losses volume. However, I
only do this for very hoppy beers (think Pliny the Elder clone) since I get mixed results with this value. Also a beer made with malt high in proteins,such as wheat or rye, will form more trub debris. This is all to say that this is the one value in the modeling effort which requires some black magic to get right. However if you are consistent in the way you transfer, you should start seeing a pattern very quickly.
Last thing you need to know is how to calculate transfer loss experimentally. The obvious way is to measure volume before the boil,measure the volume in the kettle and then measure the volume in the fermenter. Since this requires a sight gauge and not everyone has access to one, we will need to get a bit more creative. You need to measure four things; the pre-boil volume
and SG (mix well before sampling) and the volume and SG in your fermenter. I will refer to these as V1, SG1, V2, SG2 respectively. P1 and P2 will be plato values of each specific density. I will use the value 0.63 L/kg, but you can just substitute 0.0755 gal/lb if you work in US units.
Water evaporated
V1-V1*SG1*P1*0.63-( V1*SG1*P1/P2 -V1*SG1*P1)Volume losses
V1-V2-Water evaporated
You should now know how to predict the extract content of the wort by taking into account grain absorption, dead volume in the lautering process and losses during transfer. On the next article I will put all of this together to obtain a brewhouse efficiency value.
Hopefully you enjoyed this article and will read the next one.

 
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