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