Home Distillation of Alcohol (Homemade Alcohol to Drink)
Mash Effficiency and Yield
I have stolen the following, in its entirety, from a posting from
John Palmer which was sent to the
rec.crafts.brewing discussion group. It explains how to
estimate the resulting yield from various starting brews
Date: Thu, 27 Jan 2000 15:27:21 -0800
From: John Palmer
Subject: Mash Effficiency and Yield (long)
The following is an excerpt from my book. Which is still not done yet. But
will be soon. I will post when it is available. Should be next month. :-}
John
Extraction and Maximum Yield
All of these grains can be used to produce the fermentable sugars that make
up the wort. But to brew the same beer recipe consistently, we need to be
able to quantify how much yield we can expect from each type of grain. Under
laboratory conditions, each grain will yield a typical amount of fermentable
and non-fermentable sugars that is referred to as its percent extraction or
maximum yield. This number ranges from 50 - 80% by weight, with some wheat
malts hitting as high as 85%. This means that 80% (for example) of the
malt's weight is soluble in the laboratory mash. (The other 20% represents
the husk and insoluble starches.) In the real world, we brewers will never
hit this target, but it is useful for comparison.
The reference for comparison is pure sugar (sucrose) because it yields
100% of its weight as soluble extract when dissolved in water. (One pound
of sugar will yield a specific gravity of 1.046 when dissolved in 1 gallon
of water.) To calculate the maximum yield for the malts and other
adjuncts, the percent extraction for each is multiplied by the reference
number for sucrose-46 points/pound/gallon (ppg).
For example, let's look at a typical pilsner base malt. Most light base
malts have a maximum yield of 80% by weight of soluble materials. So, if
we know that sugar will yield 100% of its weight as soluble sugar and that
it raises the gravity of the wort by 46 ppg, then the maximum increase in
gravity we can expect from pilsner base malt, at 80% solubility, is 80% of
46 or 37 ppg.
The typical maximum yields for the malts are listed in Table 9. You may be
wondering how useful the maximum yield number of a malt can be if you can
never expect to hit it. The answer is to apply a scaling factor to the
maximum yield and derive a number we will usually achieve - a typical
yield.
Extract Efficiency and Typical Yield
The maximum yield is just that, a value you might get if all the mash
variables (e.g. pH, temperature, time, viscosity, grind, phase of the
moon, etc.) lined up and 100% of the starches where converted to sugars.
But most brewers, even commercial brewers, don't get that value in their
mashes. Most brewers will approach 80 - 90% of the maximum yield (i.e. 90%
of the maximum 80%). This percentage is referred to as a brewer's extract
efficiency and the resulting yield is the typical yield from our mash. The
extract efficiency is dependent on the mash conditions and the lautering
system. This will be discussed further in the chapters to follow: Chapter
13 - What is Mashing? and Chapter 14 - The Methods of Mashing.
For the purposes of our discussion of the typical yields for the various
malts and adjuncts, we will assume an extract efficiency of 85%, which is
considered to be very good for homebrewers. A few points less yield (i.e.
80 or 75% extraction efficiency), is still considered to be good
extraction. A large commercial brewery would see the 10% reduction as
significant because they are using thousands of pounds of grain a day. For
a homebrewer, adding 10% more grain per batch to make up for the
difference in extraction is a pittance.
Table 9 - Typical Malt Yields in Points/Pound/Gallon
Malt Type | Max Yield
| Max PPG | Typical PPG (85%)
| PPG Steep |
2 Row Lager Malt | 80 | 37 | 31 | -- |
6 Row Base Malt | 76 | 35 | 30 | -- |
2 Row Pale Ale Malt | 81 | 38 | 32 | -- |
Biscuit/Victory Malt | 75 | 35 | 30 | -- |
Vienna Malt | 75 | 35 | 30 | -- |
Munich Malt | 75 | 35 | 30 | -- |
Brown Malt | 70 | 32 | 28 | 8* |
Dextrin Malt | 70 | 32 | 28 | 4* |
Light Crystal (15L) | 75 | 35 | 30 | 14* |
Pale Crystal (40L) | 74 | 34 | 29 | 22 |
Medium Crystal (60L) | 74 | 34 | 29 | 18 |
Dark Crystal (120L) | 72 | 33 | 28 | 16 |
Special B | 68 | 31 | 27 | 16 |
Chocolate Malt | 60 | 28 | 24 | 15 |
Roast Barley | 55 | 25 | 22 | 21 |
Black Patent Malt | 55 | 25 | 22 | 21 |
Wheat Malt | 79 | 37 | 31 | -- |
Rye Malt | 63 | 29 | 25 | -- |
Oatmeal (Flaked) | 70 | 32 | 28 | -- |
Corn (Flaked) | 84 | 39 | 33 | -- |
Barley (Flaked) | 70 | 32 | 28 | -- |
Wheat (Flaked) | 77 | 36 | 30 | -- |
Rice (Flaked) | 82 | 38 | 32 | -- |
Malto-DextrinPowder | 100 | 40 | (40) | (40) |
Sugar (Corn, Cane) | 100 | 46 | (46) | (46) |
Malt % Yield data obtained and averaged from several sources. Steeping
data is experimental and was obtained by steeping 1 lb. in 1 gal at 160'F
for 30 minutes. All malts were crushed in a 2 roller mill at the same
setting.
* The low extraction from steeping is attributed to unconverted, insoluble
starches as revealed by an iodine test.
Mash Efficiency
There are two different original gravities (OG) that matter to a brewer: one
is the pre-boil or extraction OG, and the other is the post-boil or pitching
OG. And, ninety percent of the time, the pitching OG is what people are
referring to because it determines the strength of the beer. When brewers
plan recipes, they think in terms of the pitching OG, which assumes that the
wort volume is the final size of the batch, e.g. 5 gallons.
But, when it comes to the efficiency of the mash and lauter, we want to
think in terms of the pre-boil gravity. The Extract Efficiency section and
table gave us the typical malt yields that allows us to evaluate our
mashing process.
When all-grain homebrewers get together to brag about their brewing
prowess or equipment and they say something like, "I got 30 (ppg) from my
mash schedule", they are referring to the overall yield from their mash in
terms of the amount of wort they collected.
It is important to realize that the total amount of sugar is constant, but
the concentration (i.e. gravity) changes depending on the volume. To
understand this, let's look at the unit of points/pound/gallon. This is a
unit of concentration, so the unit is always expressed in reference to 1
gallon ("per gallon"). In mashing, you are collecting "x" number of
gallons of wort that has a gravity of "1.0yy" that was produced from "z"
pounds of malt. To calculate your mash extraction in terms of ppg, you
need to multiply the number of gallons of wort you collected by its
gravity and divide that by the amount of malt that was used. This will
give you the gravity (points per gallon) per pound of malt used. Let's
look at an example.
Palmer's Short Stout (target OG = 1.050)
Malts
6.5 lbs. of 2 Row
0.5 lb. of Chocolate Malt
0.5 lb. of Crystal 60
0.5 lb. of Dextrin Malt
0.5 lb. of Roast Barley
(8.5 lbs. total)
For our example batch, we will assume that 8.5 pounds of malt was mashed
to produce 6 gallons of wort that yielded a gravity of 1.038. The brewer's
total sugar extraction for this batch would be 6 gallons multiplied by 38
points/gallon = 230 points. Dividing the total points by the pounds of
malt gives us our mash extraction in points/pound e.g. 230/8.5 = 27 ppg.
This value is good, if not great; 30 ppg is basically what everyone shoots
for. Comparing these numbers to lager malt's 37 ppg maximum gives us a
good approximation of our mash efficiency: 27/37 = 73%, while 30/37 =
81%.
If we look at the maximum ppg numbers from Table 9 for each of the
recipe's malts, we can calculate our actual mash efficiency:
Malts OG based on Max. PPG
6.5 lbs. of 2 Row 37 x 6.5 / 6 = 40.1
0.5 lb. of Chocolate Malt 28 x .5 / 6 = 2.3
0.5 lb. of Crystal 60 34 x .5 / 6 = 2.8
0.5 lb. of Dextrin Malt 32 x .5 / 6 = 2.6
0.5 lb. of Roast Barley 25 x .5 / 6 = 2.1
Total 49.9 points
In this case, our mash extraction of 1.038 means our percent efficiency
was 38/49.9 = 76%. Usually I think you will find that your efficiency will
be 80% or better.
Planning Malt Quantities for a Recipe
We use the efficiency concept in reverse when designing a recipe to achieve
a targeted OG. Let's go back to our Short Stout example.
To produce a 1.050 wort, how much malt will we need?
1. First, we need to assume an anticipated yield (e.g. 30 ppg), for the
recipe volume (e.g. 5 gallons).
2. Then we multiply the target gravity (50) by the recipe volume (5) to
get the total amount of sugar. 5 x 50 = 250 pts.
3. Dividing the total points by our anticipated yield (30 ppg) gives the
pounds of malt required. 250 / 30 = 8.3 lbs. (I generally round up to
the nearest half pound, i.e. 8.5)
4. So, 8.5 lbs. of malt will give us our target OG in 5 gallons. Using the
malt values for 85% Efficiency in Table 9, we can figure out how much
of each malt to use to make up our recipe. Malts OG based on PPG (85%)
6.5 lbs. of 2 Row 31 x 6.5 / 5 = 40.3
0.5 lb. of Chocolate Malt 24 x .5 / 5 = 2.4
0.5 lb. of Crystal 60 29 x .5 / 5 = 2.9
0.5 lb. of Dextrin malt 28 x .5 / 5 = 2.8
0.5 lb. of Roast Barley 22 x .5 / 5 = 2.2
8.5 lbs. total 50.6 points total
Remember though that this is the post-boil gravity. When you are
collecting your wort and are wondering if you have enough, you need to
ratio the measured gravity by the amount of wort you have collected to see
if you will hit your target after the boil. For instance, to have 5
gallons of 1.050 wort after boiling, you would need (at least):
6 gallons of 1.042 (250 pts/6g)
or 7 gallons of 1.036 (250 pts/7g)
So, when planning to brew with grain, you need to be able to figure how
much malt to use if you are going to collect 6-7 gallons of wort that will
boil down to 5 gallons at a target OG. (Actually you need 5.5 gallons if
you plan for fermentation losses from the hops and trub.) These
considerations are taken into account in Chapter 19 - Designing Recipes.
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