If you are given an efficiency value, you should be able to predict the strength of the wort. Well actually, a software or online spreadsheet should be able to calculate it for you. However, many of you might have realized that you cannot keep this efficiency value constant over every batch you make; i.e., you get a lower efficiency value when making a barley wine then when making an English bitter. This is why in the second article I explained how to model the batch sparging process.

Using this knowledge, you should now be able to correctly predict the strength of any wort you make. Problem is you have to go through all that math every time you create a recipe, which can be tedious to say the least. You can write your own brewing spreadsheet to avoid having to go through this more than once. Alternatively, I have created a table, which shows approximate efficiency values for varying wort strength produced. It won't be as precise as modeling your own system, but it will definitely get you in the right ballpark.

The remainder of this article will explain this table in detail and what those values mean to a brewer. First, I need to explain how to calculate kettle and brewhouse efficiency. I will still be working with the example of producing 21L (5.5 gal) of beer using 5 kg (11 lb.) of grain with 4% moisture and 80% dry weight extract content. I will assume the mash tun has a dead space of 1L (1qt), that the boiler evaporates 4L (1 gal) of water per hour and that you lose 2L (0.5 gal) of wort to trub. Again, for more details on the gory calculations, go read the second article.

Note: If you are not interested in the math, you can skip this part and go directly to the next one

**Calculating efficiency**We know that the maximum extract content is:

SI 5kg*(1-0.04)*0.8=3.84 kg

US 11lb*(1-0.04)*0.8=8.45 lb.

I assumed a conversion efficiency of 95%. Furthermore, if we use a grain to water ratio of 3L/kg (1.44 qt/lb), I showed you should collect 8.5L (2.25 gal) of 1.078 wort. For the second sparge, you would need to add 18.9L (5 gal) of water to reach the desired final volume. To calculate efficiency, we need to calculate the total amount of extract collected.

The first running has a strength of 1.078 or 19.5 wt% (aka Plato value). We therefore collected;

SI 8.5L*1kg/L*1.078*0.195=1.79 kg

US 2.25 gal*8.33lb/gal*1.078*0.195=3.94 lb.

Note 8.33 is a constant to transform gallons of water to pounds of water.

We therefore left behind the following amount of extract

SI 3.84kg*0.95-1.79kg=1.9 kg

US 8.45lb*.95-3.94lb=4.1 lb.

To find the amount of liquid left behind, we use the dead volume and grain absorption of 1.56 L/kg (0.187 gal/lb.)

SI 5kg*1.56L/kg+1L=8.8L

US 11lb*0.187gal/lb+0.26gal=2.32 gal

So if we add 18.9L (5gal) of water for the sparge, we end up with 1.9 kg (4.1lb) of extract dissolved in 27.7 L (7.32 gal) of liquid. To find the strength of this in Plato, we need to account for the volume displaced by extract, which is

0.63L/kg (0.0755 gal/lb.).

SI 27.7L-0.63L/kg*1.9kg=26.5L

US 7.32gal-0.0755gal/lb*4.1lb=7 gal

SI 1.9kg/(1.9kg+26.5L*1kg/L)=6.5 wt%

US 4.1 lb/(4.1 lb+7 gal*8.33 lb/gal)=6.5 wt%

, which represents a density of 1.026. We know we should collect 18.9 L (5 gal) of wort, so to find the amount of extract collected by the second running;

SI 18.9L*1kg/L*1.026*0.065=1.3kg

US 5gal*8.33 lb/gal*1.026*0.065=2.78 lb

By adding up all the extract collected and dividing by the maximum extract, we can finally get the kettle efficiency;

SI (1.79kg+1.3kg)/3.84kg=0.80

US (4.1lb+2.78lb)/8.45lb=0.81

The small difference between the two unit systems comes from rounding off errors. A computer should have no problem getting an accurate answer. All this work was to find an efficiency value for our system. You can now go back to the first article and use the same method outlined there to find the OG of the beer. Realize however that the derived efficiency value is for all the wort found in the kettle, even the one which is lost to the trub. In other words your beer volume is 21+2=23L (5.5+0.5=6 gal). Calculating the brewhouse efficiency requires you to know the OG of the wort to calculate the extract left behind with the trub.

I will leave the final calculations to those interested. The OG should be 1.051.

**Cool, now what?!**So this is a lot of math and I feel that some of you (aka the vast majority) do not feel like going through it and don't feel like spending hours to write your own brewing spreadsheet to account for changes in efficiency based on the amount of grain used. The few of you who do should now have enough tools to get the job done. I will now try to give a more practical insight into the use of efficiency.

I ran the batch simulation 4 times to produce 4 different worts. The first one used 3 kg (6.6lb) of grain to make a 1.033 wort. This could be an English bitter, a Berliner Weisse or god forbid a light American lager . The second was the example I used to demonstrate the calculations. This is a 1.051 normal strength beer. The third one is stronger at 1.085 and required 10 kg (22 lb) of grains. The last one is a monster with an OG of 1.103. This requires 14 kg (30.9 lb) of grains and if you use a water to grain ratio of 2.1 L/kg (1.0 qt/lb), this will take up about 39L (10.3 gal) of space in a mash tun.

The charts represent how the mass from the grain used is distributed during the brewing process. The water content (dark blue) and non-extract content (red) are the same among all worts. This is based on the malster specifications so we don't have any control over it. The green part is the amount of starches, which were not converted by the enzymes during mashing. This is accounted for with the conversion efficiency and should not change between mashes.

The purple section is the extract left behind during sparging. This is what varies a lot between grain bills. The more grain present, the more liquid absorbed by the grain, the less extract you collect. In a 1.103 you could sparge once again to perform a party gyle. The light blue section is the amount of extract lost with the trub.

I kept it constant with every simulation; however, this might not always be true. Trub is formed of coagulated proteins and hop debris. So the higher the protein content of the wort (wheat beer or simply a whole lot of grain), the more trub formed. However, this should not be a very significant change. If you use kettle efficiency instead of brewhouse efficiency, you won't have to worry about this value. You might just end up with slightly less beer (maybe half a six pack less).

The practical implication of efficiency is to realize it is highly dependent on grain bill size. By running multiple simulations, I created the efficiency table, which I presented at the beginning of this article. So how do you use it?

Say you want to make a doppelbock with an OG of 1.084, look into which range this value falls into and get your efficiency. You should get a kettle efficiency of 66% and a brewhouse efficiency of 60%. Now just plug these values into your favorite program and start making your recipe. If you are making a dry stout at 1.042 OG, use a kettle efficiency value of 83% or a brewhouse efficiency of 75%.

If you have previously calculated efficiency during brew day, you should have enough data points to find a correction factor for future brews. As an example, say you brewed five batches between 1.050 and 1.060 and got a kettle efficiency closer to 85% and you are getting ready to brew a 1.093 beer, use a kettle efficiency of 67% instead of 61%. Since the loss of efficiency comes from grain absorption, you can therefore still use these values if you fly sparge. A little bit of mucking around might be required at first, but grain remains grain no matter how you wash it.

You have to realize these values are approximate and will vary between systems. I tried to use "average" values for mash tun dead volume, kettle evaporation, and losses to trub. If you

use kettle efficiency instead of brewhouse efficiency, you only have to use two assumed values instead of three, which could be better.

Furthermore, efficiency generally increases in the brewhouse if you make bigger batches, so 10-gallon brewers might have one or two extra efficiency points compared to 5-gallon brewers.

In addition, the efficiency values were derived based on a 60 minute boil. A 90-minute boil will lead to higher efficiency (but longer brew days and more energy consumption). A table of efficiencies for 90 minutes can be found below. For longer boils, use a correction factor based on experience.

**The water to grain ratio**

I have alluded to the water to grain ratio, i.e. the amount of water you add to your mash tun at first based on the size of the grain bill, but never explained how to choose it. In theory, a thicker mash can produce a less fermentable beer than a more dilute mash, but the experimental evidence is contradictory. It would seem that thicker mashes does enhance beta-amalyse activity, but thinner mashes cause beta-amalyse to denature faster.

This is all to say that as long as you stay between 2 and 4 l/kg (0.96 to 1.92 qt/lb), the enzymatic conversion should not be greatly affected. Therefore, you could pick any of these values at random and be fine. However, if you choose a very small ratio, you might collect close to no wort with your first running since all the liquid will get absorbed by the grain. On the other hand, a very big ratio will lead to almost no water being left for the sparging process. This means an optimum value for the grain to water ratio can be found.

It just so happens that you should sparge with the same volume as you collected after the mash. In other words if you want to make 21L (5.5 gal) of beer and know you will evaporate 4L (1 gal) and loose 2L (0.5 gal) to trub, you need a total volume of water of

SI 21L+4L+2L=27L

US 5.5gal+1gal+0.5gal=7 gal

So you should collect:

27/2=13.5L (7/2=3.5 gal) of wort from the first running and add 13.5L (3.5 gal)of water to sparge with to get the ideal grain to water ratio. A general formula to find the optimum grain to water ratio is:

SI R=((FV+L+Evap)/n+DV+GM*1.56-0.63*E)/GM

US R=((FV+L+Evap)/n+DV+GM*0.75-0.30*E)/GM

R=water to grain ration in L/kg or qt/lb

FV=Final volume of beer in L or qt

L=Losses to trub in L or qt

Evap=Water evaporated in L or qt

n= number of sparges

DV=Dead volume in L or qt

GM=grain mass in kg or lb

E= extract in mash

The extract in mash is calculated as follows:

E=GM*CGDB*(1-M)*CE

CGDB=coarse grind dry basis; the extract contentof the grain find from malt analysis sheets

M= moisture content of grain

CE=conversion efficiency

These formulas are not 100% accurate. The water to grain ratio optimum is actually a much uglier formula. If you are interested, it can be derived by taking a derivative of efficiency with respect to the water to grain ratio. For a single sparge (ie n=2 because you first empty you mash and then sparge once) the R value in SI units is actually

R=(0.98*(FV+L+Evap)/n+DV+GM*1.56-0.63*E)/GM

If you add a second sparge, the solved ratio is:

R=(0.96*(FV+L+Evap)/n+DV+GM*1.56-0.63*E)/GM

And if you perform 9 sparges (n=10)

R=(0.47*(FV+L+Evap)/n+DV+GM*1.56-0.63*E)/GM

In other words as you increase the number of sparges, you deviate from my simplified formula. However, since it is not logical to perform more than 3 sparges, this is good enough.

**Concluding remarks**I hope you have a better understanding of efficiency now. These articles were meant to show you on one hand the details behind all the brewing software out there. In addition, they were meant to help you prepare for brew day.

The important aspects to understand are that conversion efficiency should remain constant no matter what. If it is low, they are many things you can do to improve your efficiency during mashing. On the other hand, lautering efficiency depends highly on the amount of grist used. The table I outlined in this article should give you a good way to account for this when designing a recipe.

Finally, there are ways to optimize the water to grain ratio to get the most out of every batch. Just make sure it stays in a logical range. The derived formula only accounts for the physical parameters and not the enzymatic conversions taking place during the mash.