I thought I understood efficiency...

[Gavin C]

I use Brewer's Friend.... a simpler program that doesn't confuse the issue of efficiency. It has only one efficiency, and that's brewhouse efficiency which is everything rolled into one as far as I can tell.

My take on efficiency is that the only reason for having two efficiency ratings is the addition of fermentables to the boil. I'm talking about things like corn sugar, honey, etc, which do not have to be converted to ferment.

The ONLY efficiency rating that matters is conversion efficiency (mash efficiency), it's the only one you have control over. Nothing after the mash changes that, but if you take a gravity post boil, it will include the sugars you mashed and the ones you added to the boil. So if you add a lot of sugars to the boil, those sugars are 100% efficient.... they are already converted to sugar. That masks your mash efficiency unless you account for it mathematically. Your measured efficiency should NEVER be less than your mash efficiency if your liquid volumes are correct and your readings right.


H.W.

I thought conversion efficiency related to what percentages of starch converted to sugars in the mash.

The mash efficiency relates to how much sugars enter the boil kettle and unlike conversion efficiency is subject to additional losses including grain absorption and mash tun deadspace (if any). Is this semantics or am I way off base here.

In my experience measured BH efficiency is ALWAYS less than mash efficiency as I lose some sugars via volume losses to hop absorption, kettle true and plate chiller/hoses. (about 0.25 G)

Perhaps I am misunderstanding you. Most likely I am.

 

[Owly055]

I thought conversion efficiency related to what percentages of starch converted to sugars in the mash.

The mash efficiency relates to how much sugars enter the boil kettle and unlike conversion efficiency is subject to additional losses including grain absorption and mash tun deadspace (if any). Is this semantics or am I way off base here.

In my experience measured BH efficiency is ALWAYS less than mash efficiency as I lose some sugars via volume losses to hop absorption, kettle true and plate chiller/hoses. (about 0.25 G)

Perhaps I am misunderstanding you. Most likely I am.

I'm using conversion efficiency and mash efficiency interchangeably......... conversion efficiency is a useless figure, and one you can't really measure.

Brewhouse efficiency is a number that could be calculated in many ways.... "lies, damn like, and statistics". I recently put 2.5 gallons of wort in a fermenter, measuring the OG and using the BF program to calculate it brewhouse efficiency based on 2.5 gallons. The way I always do it. Now, I probably should factor in the quart or so I lose to trub in the fermenter (10%). The number was 93%... the highest I've hit so far......... That number really represents mash efficiency, as absorption from the small amount of pellet hops is trivial........ The real question is "does it matter".......... and it doesn't unless we are having a pissing match over who gets the highest efficiency, which would be silly. I monitor my efficiency this way every time, and use the figure to see how I'm doing with my mash. I have no way of knowing how Brewer's Friend handles the input..... weather trub loss is calculated in, etc...........nor do I really care.

Just looked up Brewhouse Efficiency at Brewer's Friend:

Brew House Efficiency - An all inclusive measure of efficiency, which counts all losses to the fermentor. This can be thought of as 'to the fermentor' efficiency. Hops absorption factors into this, and is reduced on the same equipment by ~1% in super hoppy beers.

H.W.

 

[Gavin C]

Fair enough Owly.

I would take a differing approach to the numbers but each to their own i suppose. I would disagree with you on conversion efficiency and mash efficiency equivalence even if the former is difficult but not impossible to measure.

I view the software as a tool like any other. It serves a purpose. That purpose is determined by the brewer. Putting aside their usefulness in formulating recipes, I think these tools become particularly useful when trying to find a process error or in the refinement of an existing brewing process.

I'm sure the OP will have his own take on the software and use it accordingly.

Thanks for answering my query BTW

 

[CGish]

I have been hitting my numbers pretty consistently. I use a corona mill and have Beersmith set for 68% efficiency. I am usually within a point or two of target.

Today I did a wheat.

4 lbs white wheat malt
3 lbs Rahr 2 row
2 lbs Munich light
Orange peel
.75 oz coriander.


Beersmith says I got 72% mash efficiency but the estimated OG was 1.045 and I got only 1.031.

I am not worried as I was looking for a light summer ale. Just wondering why I got relatively good mash efficiency #'s and low OG?

I usually hit 68% with a single pass through my corona mill. However, a few weeks ago I brewed a wheat beer and did not check to see if conversion was complete. It was a 90 minute mash, so I did not double check. I wound up with 40% efficiency on that brew. Three pounds of DME got me back in the ballpark, but it still came out light. I was set for a second brew that day with a similar grist. The second time I checked for conversion and it took almost 2 hours. At 90 minutes, the iodine still turned the mash black. Around 1 hour 45 minutes the color started to lighten and at 2 hours it was finally done. The second mash's numbers were right where they should be, coming in at 1.052 against the projected 1.053.

I have brew twice since then and had normal (30 - 45 minute) conversion times. It seems the bulk wheat malt I have is a slow converter.

Is it possible you just had a slow conversion and stopped the mash to early?

Cody

 

[ajdelange]

There is only one efficiency that matters and that is the effective extract delivered to the kegs on the loading dock divided by the pounds of grain that were used to produce those kegs. It accounts for
1)Slackness in the malt
2)Extract hung up in the grain bed at sparge
3)Water losses/gains due to evaporation/dilution (kettle, fermenter)
4)Extract losses due to boilovers
5)Extract hung up in the chiller or the transfer lines or the fermenter or anywhere else or spilled on the floor or retained in the yeast cake.

One can, of course, calculate the efficiency at various stages in the process. For example, if a brewer has 65% efficiency at the kettle and the maltster has told him that the FGHWE as is was 81% for a particular malt he knows he has lost 16% as unconvertible starch, husks, acrospires and equipment shortcomings. If he is used to getting 70% from this grain he knows he has lost 5% more than he should and seeks the source of the problem. If he gets 60% in the keg he knows he has lost 5% extract in the fermenter and filling process. And so on.

 

[doug293cz]

Here are the definitions of efficiency that seem most intuitive to me:

Conversion Efficiency: The percentage of grain sugar potential that is actually obtained in the mash. Conversion efficiency is affected by mash temp & time, mash pH, grain particle size, diastatic power, mash thickness, agitation, etc. Conversion efficiencies in excess of 98% are readily obtainable.

Lauter Efficiency: The percentage of the sugar in the mash that makes it into the boil kettle. Lauter efficiency is calculated as: (Mash Efficiency) / (Conversion Efficiency). It is affected by things like grain absorption and undrainable wort volume (MLT dead space.)

Mash Efficiency: The percentage of grain sugar potential that actually makes it into the boil kettle. Mash_Efficiency = Conversion_Efficiency * Lauter Efficiency

Brewhouse Efficiency: The percentage of the grain sugar potential that actually makes it into the fermenter. The difference between brewhouse efficiency and mash efficiency is caused by hop absorption, volume held in plumbing, trub volume withheld from fermenter, spills, etc.

By knowing each of your efficiencies, you can know where to look if your overall (brewhouse) efficiency is falling short.

To calculate any efficiency, we need to know the potential points of the starting grain bill. This is the sum of the grain weight * grain potential points/unit wt.

Typical pale 2 row has a potential SG of 1.036 for 1 lb of grain in 1 gal of wort, or 36 potential points/lb

10 lbs of two row would then have 10 * 36 = 360 potential points​

We also need to know the actual points in any volume of wort. Actual points is given by the volume times the points/gal, and points/gal is (SG - 1) * 1000.

Mash efficiency is easy to calculate; it is:

[(Pre-boil Volume) * (1 - Pre-boil SG) *1000)] / (Potential Grain Points)
​

With volume and SG corrected to room temp.

So if from my 10 lbs of 2 row, I obtain 6.75 gal of 1.049 SG wort @ 150°F in my boil kettle (equivalent to 6.5 gal @ 70°F), then I have obtained:

6.5 * (1.049 - 1) * 1000 = 318.5 points
​

and my mash efficiency is:

318.5 / 360 = 0.8847 => 88.5% mash efficiency
​

Brewhouse efficiency is calculated as:

[(Volume in fermenter) * (OG - 1) * 1000] / (Potential Grain Points)
​

So if I net 5.25 gal of 1.058 wort to the fermenter, my brewhouse efficiency is:

5.25 * 58 / 360 = 0.8458 => 84.6%
​

The calculations for conversion efficiency are a little trickier, as you have to know how much sugar is in a wort with a given SG. To start we need to calculate the weight of the sugar that is equivalent to the potential gravity points. The easiest way to do this is to assume a wort volume in the mash that is near the expected total wort volume. So, let's assume we have 10 gal that contain the 360 total points, which would give us an SG of 1.036. Now we convert the SG to Plato so that we know what percentage by weight of the wort is sugar, and the balance will be water.

10 gal of 1.036 wort will weigh 10 lbs * 1.036 * 8.3290 lbs/gal = 86.28844 lbs
1.036 = 9.00 Plato = 9.00 wt% sugar (I use Beersmith calculator)
Wt of sugar is 0.0900 * 86.28844 = 7.7659596 lbs
Wt of water is 0.9100 * 86.28844 = 78.5224804 lbs
Vol of water is 78.5224804 lbs / 8.3290 lbs/gal = 9.4276 gal
​

Now let's say we actually mashed with 4 gal of water. We then have:

Wt of strike water is 4.0 gal * 8.3290 lbs/gal = 33.316 lbs
Potential sugar wt = 7.7659596 lbs
Max SG = 7.7659596 lbs / (7.7659596 lbs + 33.316 lbs) = 0.1890 => 18.90 Plato
18.90 Plato = 1.079 SG of mash wort with 100% conversion
​

So, the maximum possible wort SG (in the MLT) with 10 lbs of 2 row and 4 gal of strike water is 1.079.

Now, I can measure the SG in my MLT, but I can't measure the actual wort volume, so how do I calculate how much sugar is in the wort? Let's say I measure an SG of 1.078, which is equivalent to 18.76 Plato.

Let X = the lbs of sugar created in the wort
Then 0.1876 = X / (X + 33.316) -- yes, it's algebra, which you thought you wouldn't need in real life.
X = 0.1876 * (X + 33.316)
X * (1 - 0.1876) = 6.2500816
X = 6.2500816 / (1 - 0.1876) = 7.69335499754 lbs sugar in wort in MLT
​

My Conversion efficiency is then (actual sugar in mash) / (potential sugar in mash):

Conversion efficiency = 7.69335499754 / 7.7659596 = 0.990650916796 =>99.07%
​

Finally, I can get to my lauter efficiency = (mash efficiency) / (conversion efficiency):

0.8847 / 0.99065 = 0.893050017665 => 89.31%
​

Whew

Brew on

 

[ajdelange]

It isn't really that tricky to figure out how much sugar is in a volume of wort at a given SG. The first step, of course, is to correct the wort volume to 20 °C. The next step is to compute the weight of the given volume which is simply W = SG*V*dens_water. Then find the amount of sugar in the water by multiplying by the Plato value
°P = -616.868+1111.14*SG-630.272*SG*SG+135.997*SG*SG*SG
divided by 100. For example a liter of 1.036 wort weighs 1.036*998.203 = 1034.14 grams. The w/w strength of 1.036 wort is 9.02446% so the grams of sugar are 1.036*998.203*9.02446/100 = 93.3254 grams. The rest is water.

The only thing that is at all tricky is when you put x grams of sugar into a container and add water until you have y liters and want to know what the SG is. So lets work the problem in reverse and say we put 93.3254 grams into a flask and made up to one liter. What's the SG. Well we know the weight of the sugar is
93.3254 = 1L*SG*dens_water*(-616.868+1111.14*SG-630.272*SG*SG+135.997*SG*SG*SG)/100
and we have to solve that for SG. This is actually very easy to do using the Excel Solver. It produces 1.036000024 as the solution.

Given the ASBC polynomial

°P = -616.868+1111.14*SG-630.272*SG*SG+135.997*SG*SG*SG

and modern laptops there is no reason home brewers should be worrying this potential points.... business any more than professional brewers do.

 

[doug293cz]

It isn't really that tricky to figure out how much sugar is in a volume of wort at a given SG.

"Tricky" is a relative term. Compared to the simple points computations, this problem is a little more complex. However, compared to integral equations, it's trivial. I just wanted to give readers who hadn't reached the tl:dr threshold a heads up that they needed to pay closer attention.

The first step, of course, is to correct the wort volume to 20 °C. The next step is to compute the weight of the given volume which is simply W = SG*V*dens_water. Then find the amount of sugar in the water by multiplying by the Plato value
°P = -616.868+1111.14*SG-630.272*SG*SG+135.997*SG*SG*SG
divided by 100. For example a liter of 1.036 wort weighs 1.036*998.203 = 1034.14 grams. The w/w strength of 1.036 wort is 9.02446% so the grams of sugar are 1.036*998.203*9.02446/100 = 93.3254 grams. The rest is water.

Isn't this pretty much what I did, except for using 21.1°C for a reference instead of 20°C (or did I miss something subtle)? But, I don't work with the ASBC polynomial on a regular basis, so I punted and used the Beersmith converter instead. I will be very disappointed if Brad does not use the ASBC poly. Thanks for including the actual polynomial. I will add it to my tool belt. Do you know offhand what the valid ranges of SG are for the poly?

The only thing that is at all tricky is when you put x grams of sugar into a container and add water until you have y liters and want to know what the SG is. So lets work the problem in reverse and say we put 93.3254 grams into a flask and made up to one liter. What's the SG. Well we know the weight of the sugar is
93.3254 = 1L*SG*dens_water*(-616.868+1111.14*SG-630.272*SG*SG+135.997*SG*SG*SG)/100
and we have to solve that for SG. This is actually very easy to do using the Excel Solver. It produces 1.036000024 as the solution.

This isn't really a problem you need to solve to understand your mash. You know how much potential sugar you put in, and how much water you put in, but you have no idea what the actual liquid volume is in the wort, and no way to measure it easily. I gave one way to calculate it (assuming all potential sugar has been converted, and the concentrations throughout the liquid have come to equilibrium.)

Given the ASBC polynomial

°P = -616.868+1111.14*SG-630.272*SG*SG+135.997*SG*SG*SG

and modern laptops there is no reason home brewers should be worrying this potential points.... business any more than professional brewers do.

So true. But then given the sensitivity of the calculations to small measurement errors, and the fact that most brewers don't have access to malt data for each lot that they use, how important is complete rigor in the calculations? The intent of my post was to help people understand how the different efficiencies are calculated, not how to make these measurements/calculations in a QC lab.

Brew on

 

[ajdelange]

"Tricky" is a relative term. Compared to the simple points computations, this problem is a little more complex. However, compared to integral equations, it's trivial. I just wanted to give readers who hadn't reached the tl:dr threshold a heads up that they needed to pay closer attention.

I don't know what tl:dr means but I agree that this is all very simple stuff, even if you stick to the robust solution methods. That's why I don't see any point in using the ppppg method. It makes the assumption that volume expansion is linear and it isn't. Granted that the deviation from linearity is so small that it doesn't make much practical difference. Given that you are likely to do your planning on a spreadsheet rather than a yellow legal pad I don't see any reason to accept even the small inaccuracy that comes from the approximate method.

Isn't this pretty much what I did, except for using 21.1°C for a reference instead of 20°C (or did I miss something subtle)? But, I don't work with the ASBC polynomial on a regular basis, so I punted and used the Beersmith converter instead.

I hope so or you wouldn't even get an approximately correct answer.

I will be very disappointed if Brad does not use the ASBC poly. Thanks for including the actual polynomial.

If he converts 1.036 SG to 9 °P he doesn't.

. Do you know offhand what the valid ranges of SG are for the poly?

It spans the Plato table which goes up to 1.083 20/20 SG (20.007 °P ). Above that you can use

°P = -584.6957 + 1038.2666*SG -577.9848*SG*SG + 124.5209*SG*SG*SG

which is a best fit to sucrose data in the CRC handbook (corrected to 20/20) and is continuous both in ordinate and slope with the ASBC polynomial at 1.083. Or you can use the ICUMSA polynomial which is a relative mess but is good for very concentrated solutions and at any temperature.

This isn't really a problem you need to solve to understand your mash.

It's probably the one I 'solve' the most as it's the one used to plan a brew. If I use x kg of malt that I know, from experience, has a certain percentage yield to prepare y liters of beer then I need to know how many grams of extract will be in each liter of that beer in order to calculate its °P and SG. Again, this is done in a spreadsheet so you can conveniently see the effect of swapping a kg of malt A for a kg of malt B and as you want the answers (°P and SG) to change as soon as you change a malt weight field you can't use the Solver. The answer is to precompute solutions over a reasonable range of °P and fit another polynomial to that data. The extract is

E = 0.0061289 + 0.99464*P + 0.0042888*P*P kg/hL

This is a very good fit with error about 5 mg/hL at the ends of the Plato table range and more like 2 - 3 mg/hL in the middle and, of course, is trivially solved for P if E is given without iteration (the Solver uses iterative techniques).

...most brewers don't have access to malt data for each lot that they use,

But they blindly accept tables in old brewing books that say that a malt of a given color produces a certain number of points per pound per gallon which is the same thing except less reliable than the maltsters data sheet.

....how important is complete rigor in the calculations?

Depends on what you are trying to do. The point is that with a modern spreadsheet there is no reason to accept the inaccuracies that come with the old 'points' system of calculation so why accept them? Proper calculations are just not that difficult and if the brewer's focus is on extract (rather than points) he'll have a better understanding of the science underlying the process.

The intent of my post was to help people understand how the different efficiencies are calculated, not how to make these measurements/calculations in a QC lab.

They would be better served by focusing their attention of extract levels at each stage of the process.
Yes, yes, I know that better is the enemy of good enough but I just hate that concept so I guess philosophy comes into it too.

 

[ajdelange]

Caveat - Brewing Science stuff

I will be very disappointed if Brad does not use the ASBC poly.

I got sort of intrigued here as to what he might actually be doing.

Using this as an example:

18.90 Plato = 1.079 SG of mash wort with 100% conversion

The estimated equivalent SG inverting the ASBC polynomial is 1.078093
The estimated equivalent SG interpolating into the Plato table 1.078095*
The estimated equivalent SG inverting the Lincoln equation is 1.078107
The estimated equivalent SG inverting ICUMSA polynomial is 1.078140

*Looks as if the polynomial represents the table pretty well

Those are the four methods a professional brewer would use to convert Plato to SG and all give 1.0781 to 4 decimal places so apparantly none of these is being used. I've shown 6 decimal places as that is the precision available from the best instruments, the precision of the Plato table, and the level of precision of interest to brewing scientists (recall that this is the Brewing Science thread).

I suppose it is also worth mentioning that these specific gravities are apparent (in air) and 20/20. Perhaps this calculator is calculating something else. The true specfic gravity for 18.9 °P is (ICUMSA) 1.078046 so that's not what has been done. Perhaps he is using 10/10°C as the reference temperature. That would give 1.07900 as the apparent specific gravity (ICUMSA).

All of this is, for practical purposes, moot but it is definitely of interest to brewing scientists and any one on this thread is presumably a brewing scientist (though an amateur one) and potentially interested. If not, don't read it.