Brewhouse Efficiency for All Grain Beer Brewing

by Brad Smith on October 26, 2008 · 30 comments

Brewhouse efficiency is a term that causes some confusion for first time all grain brewers. I previously covered how to improve your brewhouse efficiency, but we frequently see questions on our discusssion forum from brewers who don’t understand what brewhouse efficiency is or how it is used in recipe design.

Brewhouse efficiency is defined as the percent of potential grain sugars that are converted into sugar in the wort. Typically this includes losses for a given brewing setup, and these losses are taken in aggregate rather than accumulated individually. It is therefore a measure of the overall efficiency of your brewing system.

Brewhouse efficiency is a key input when designing all grain recipes, as it determines your estimated original gravity. If you don’t have an accurate brewhouse efficiency number for your particular equipment, your original gravity estimates will be way off and you will miss your target gravity.

Every grain in an all grain recipe has a potential yield, listed as the dry grain fine yield on the malt sheet. The dry grain yield is determined in laboratory conditions, by powdering the grain and extracting as much as possible and then extracting maximum potential from the sample. Yields vary from 50%-87% depending on the type of grain used. You can also express yield as a potential such as 1.038.

The actual brewhouse efficiency is measured for an entire system. Unlike the dry grain yield or potential measured in a lab, real brewers achieve only a percentage of the ideal number due to real considerations such as efficiency of the mashing process, and losses due to boiling, deadspace or trub. This percentage of the potential, as measured across the whole system into the fermenter, is the brewhouse efficiency.

A related term is mash efficiency. Unlike brewhouse efficiency, mash efficiency measures only the efficiency of the mash and sparging steps. Mash efficiency can be through of as the percent of potential fermentables extracted during the mashing process that actually make it into the boiler.

Calculating Efficiencies

Programs such as BeerSmith will calculate the brewhouse efficiency from a given recipe, volume and original gravity. However it is important to understand what’s going on under the hood. Lets look first at how to calculate the total potential of the grain for a batch of beer:

(potential_pts) = (grain_pts) * (weight lbs) / volume_gals

Each grain has a dry grain potential, which you can find from our grain listing or from the malter’s web site. The grain_pts is calculated from the grain potential by subtracting 1.000 and multiplying by 1000. For example, a grain with a potential of 1.035 becomes simply 35 points. 5 pounds of this grain in a 5 gallon batch would add 35*5/5 = 35 potential points to the beer. If we sum all of the potential points from the various grain additions we can get the overall potential. If we had no losses in the system, the 35 points above would give an ideal starting gravity for our beer of 1.035.

I mentioned that the potential points represents the gravity under ideal conditions. In practice one gets much less than this, usually around 70-80% for brewhouse efficiency overall. Therefore the actual original gravity is determined by the potential points times the gravity:

(batch_pts) = (potential_pts) * (brewhouse efficiency)

So if we consider a recipe with 40 potential points, and a 75% brewhouse efficiency we get 30 batch points or an original gravity of 1.030. This is how original gravity is estimated.

Reversing the calculation we can calculate the efficiency from an ideal recipe potential estimate (potential_pts) and actual measurement (measured_pts).

(efficiency) = (measured_pts) / (potential_pts)

So for example if we had a recipe with potential_pts of 80 and measured the wort into the fermenter 1.050 we get an efficiency of 50/80 = 62.5%. Note that this assumes we hit our target volume. If we don’t, we need to consider the target and actual volume as follows:

(efficiency) = (measured_pts * actual_vol) / (potential_pts * target_vol)

The formulas above give us the overall brewhouse efficiency, but can also be used to calculate the mash efficiency into the boiler. For efficiency into the boiler we simply use the boiler volume and measured boil specific gravity into the boiler as opposed to the fermenter. In BeerSmith you can click on the “brewhouse efficiency” button in any open recipe to perform more detailed mash or overall efficiency calculations.

Now you know how to calculate the two key all grain efficiencies: brewhouse and mash efficiency. For additional reading, consider our article on improving all grain efficiency.

Thanks for joining us again this week on the BeerSmith Home Brewing Blog. Don’t hesitate to subscribe for weekly delivery, leave a comment, or drop a vote for any of our articles on BrewPoll.com if you enjoyed today’s article.

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{ 15 comments… read them below or add one }

Jeff Brown March 17, 2012 at 10:14 am

Hello, ok I just read the efficiency article. The last paragraph says one can click on the (brewhouse efficiency) button in any open recipe.. I have the latest beersmith and I don’t find a brewhouse efficiency button anywhere? Thanks Jeff

Brad Smith March 26, 2012 at 9:33 am

In BeerSmith2 the efficiency numbers are all displayed on the mash tab and if you enter your actual volumes/gravities it will calculate the actuals here as well.

Tommy July 15, 2012 at 2:47 pm

Hi Brad,.

I’m using BeerSmith 2 and don’t see: “brewhouse efficiency” button. I see a field showing a %, but no other information? Can you point me in the right direction?

Thanks,

Tommy July 15, 2012 at 2:48 pm

oh, sorry, I just saw your reply to someone else, thanks!

Brad Smith July 18, 2012 at 11:15 pm

The efficiencies are now shown on the mash and fermentation tabs in BeerSmith 2. If you enter the actual volumes/gravities it will also estimate your actual overall brewhouse efficiency.

Boludo August 12, 2012 at 12:30 am

This equation is confusing me:
“Note that this assumes we hit our target volume. If we don’t, we need to consider the target and actual volume as follows:
(efficiency) = (measured_pts * target_vol) / (potential_pts * actual_vol)
The formulas above give us the overall brewhouse efficiency”

Is the actual_vol the amount realized to the fermenter?

Boludo September 13, 2012 at 3:58 pm

My comment was posted, although after more than a month, but there was no reply.
I believe your formula has the volumes switched. actual vol should be on top, and target vol should be below. The pts should stay in their same locations.

Brad Smith September 17, 2012 at 1:01 am

You are correct – I updated the formula.

Jeff Nagle January 5, 2013 at 2:30 pm

I believe I understand the difference between mash efficiency and brewhouse efficiency, but I’m still confused as to how Beersmith calculates them.

I just finished an English Barleywine brewday. Beersmith had my estimated mash efficiency at 101.2%, so it estimated my pre-boil gravity as 1.084. In reality my mash efficiency was 81.5% (closer to what I’d expect) and my pre-boil gravity was 1.068. What variables in Beersmith do I need to adjust so that it calculates the estimated mash efficiency closer to my actual values of around 80%?

Jerry February 26, 2013 at 9:47 pm

If I use 5.5 kilos of grain and I bottle 28 litres of beer with an OG of 1.042, what is the efficiency of that? That’s what I really want to know. Can anyone tell me this?

mplsbrewer March 29, 2013 at 11:09 am

I’m calculating efficiency slightly differently and trying to reconcile with your formula. I consider the total sugar independently of the volume. The way I look at it is:

Total extracted sugar = Measured points (pts/gal) x Boil volume (gal), which yields units of “points.” This represents the amount of sugar that made it into your boil kettle.

Total available sugar = Grain weight (lbs) x Grain Extract potential (pts/lb), which also yields units of “points.” This represents the total amount of sugar that is in your starting grain bill

Efficiency = Total extracted sugar/Total available sugar = (measured points x boil volume)/(Grain weight x Grain extract potential).

For me, this makes intuitive sense and I don’t have to worry about accounting for differences between actual and target volumes. Brewhouse efficiency would be calculated the same way, except the gravity and volume of the cooled, unfermented wort would be measured in the fermenter, rather than in the boil kettle. Brewhouse efficiency should always be lower than mash efficiency since there will be additional liquid losses in your system between the boil kettle and the fermenter.

Matt May 2, 2013 at 8:52 am

I’m confused as to why BHE is expressed in Beersmith by adjusting the OG rather than the final volume at the final gravity. I don’t really see how this is useful. On my spreadsheet I’ve been using previously, my BHE has always been calculated automatically based on my final volume and after adjusting my evap rate and mash eff to the actuals on the day
Knowing my Mash Eff allows me to know how much sugar ends up in the kettle.
Knowing my Evap rate tells me what the gravity will be at the end of my boil.
Knowing my final volume shows me the overall eff of the BH based on a combination of the mash eff, and the losses. But I already know what the gravity will be, what I’m accounting for is volume losses, not gravity point losses.

I’m not sure if this makes sense, but I really don’t see the usefulness as it’s currently operating. As mentioned above the mash eff + boil start volume + the evap rate give you all the information you need to know your OG, after that it’s all volume calcs.

Any thoughts?

MikeyS September 10, 2013 at 8:12 pm

Well, matt and Jerry… mplsbrewer has it right…this is not the best blog I think on efficiency.

The inert/insolulable parts of a grain bill weight is irrelvcant to efficiency as such, however one needs to find out what in one’s bill has as sugars.

Eg. look up moisture % and dry max yield (deduct a percent assuming coarser grind), then:

potential/ideal yield => weight of your grains x (their dry max yield -1%) x (100-moisture %). So if a base malt is 80% with about 4% moisture, one gets about 75-76% maximum yield per unit weight of the grain.

Now, By taking one’s O.G. (at any time after some process); don’t worry about solids/trop as it is not chemically dissolved and won’t affect S.G.), one can find one’s weight extract (look up Plato degrees whilst you’re at it), and this will be your efficiency. So if you figure out w.r.t. your grain weight a 66% efficiency, this would give you a brewhouse efficiency of 66/76 ~ 87%!

Note! X factor: The Plato scale accounts for sucrose as highest density of dissolvable sugar, not all sugars will be as dense, hence, the actual extraction, since we dont use pure sucrose, will be less than expected. Thus, one needs to throw in a few percent here as well, which to be honest, I think is, as I called it.. a bit of X factor… perhaps 2-5%?

So let’s adjust the 87% to ~85%. The point is, most homebrewers will usually hit these numbers within a few percent, which is far less loss in efficiency, than the already 25% we can do nothing or little about (inert material, moisture, proteins, sugar types) in a grain. This is crucial to huge breweries, not to us :)
That’s what matters. Now, S.G. is independent of volume , so dilution, evaporation but is however, dependent on correct/same/reference temperatures and thus, the efficiency should be more or less already clear after mashout/sparging. As someone said, however, one can lose some on the way due to spillage, or other leftovers. So, one usually will use the final ‘into the fermenter’ reading for accuracy in homebrewing.

So Jerry, in your case, you would need to figure out the proportions of grains you used, what their dry yield and moisture is, to do the eff… your numbers seem to be around 85% which is a typical homebrewing efficiency and looks fine :)

example to help ya: A. base malts (4kgs) assume 80% dry, 3% moisture, 1.5% loss to grind size, 500 g other malts about 76%, 4% moisture, then: [(0.785 * 0.97-0.0025(X factor sugars)]*4kg+ 0.5[*( .745*0.96)-2.5%] = 3.3kg max ideal yield in solution. Look up yor Plato degree value for 1.042 (eg. 10.5 degrees, thats 10.5% sugars in each unit volume of wort).
So this is 1.042 x 0.105 x28 liters = 3.063 kg in solution.

efficiency to raw weight thus is 55.7 % (not interesting), but to ideal solution: 3.063/3.3 ~ 93%.. so as I said, looked good… You have used plenty of water, so I am guessing you managed to mash out highly efficiently, which is why you hit that. I tend to hit 87ish more, but I dont have a big kettle atm, so use two smaller for partial boils and as you can see… can do a 5% difference. :)

Just be careful though to do temp reads very close to your indicated ref temp for your hydrometer. And homebrewing isn’t at all too bad on eff… 1-5% only is massively important in huge quantaties… A brewery can lose hundreds of thousands of pounds on a few percent over a short year.

MikeyS September 10, 2013 at 8:14 pm

Oops, and forgot to add a bit on sediments… sediments (hot/cold) in solution (but not dissolved) before being precipiated will affect S.G. slightly, but there influence is not so much on S.G. as it is in perceived total volume (for example… when your beer is done, sediment might account for 10-20%). This puts some calculations a tad off. That’s it.. enjoy.

MikeyS September 10, 2013 at 8:37 pm

Sorry, I can’t edit :p

Matt: Evaporation and dilution is not supposed to matter, given the ratios remain the same… As long as kg/L is maintained, Plato degrees remains constant. In reality yes, some sugars will follow steam, losses in each process… but they, again, are mainly important in larger scale operations.

Fret not..luckily home brewing holds high efficiencies and we are lucky like that… the important things is more about well experimentation ,)

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