Monday, August 13, 2007

Technical Tuesday: Extract Efficiency

Let's say you want to brew 5 gallons of a very basic Golden Ale. Your ingredient list looks like:

10 lbs. 2-row Pale Malt
2 ozs. Your favorite Noble Hop
Your favorite Ale Strain

You've brewed all grain once or twice before. You get the basic idea of the process, but you're not sure exactly what to expect to get OG-wise from your ten pounds of grain. You also know that variations in gravity while boiling can affect IBU utilization, and you don't want to be far off the mark. After all, this is a beer known for its subtle hopping. How can you figure out what percentage of extract you'll get from your mash?

When brewers talk about achieving a certain extract efficiency, they're talking about the amount of sugar that they coaxed out of the grain expressed as a percentage of the theoretical yield. The theoretical yield differs dependent upon the type of grain being discussed. Pale malt, Munich, Crystal, Roasted, etc. all have different theoretical yields. Luckily, there are tables that contain this information available - some are even free! [Table by Mike Uchima] We'll get back to this in a second. For right now, let's continue on with the example.

You've mashed, sparged, and collected your wort. After correcting for temperature and volume, you determine that you have a wort that will turn out to be around 1.048 in five gallons. You know that this number can be expressed in Gravity Units (GUs) as 48. Since you have five gallons, each gallon containing 48 GUs, you know that the total gravity units of your wort is 240. Since you know that the theoretical yield of two-row is 37 points per gallon, per pound of malt, you would calculate the following:

[GU*Lbs. Malt]/Gallons or, [37*10]/5 = 74. That is, if you extracted 100% of the sugars possible, your wort would have an OG of 1.074. However, it has an OG of 1.048. To determine the percentage of the sugars that you extracted, just divide Actual Extract/Potential Extract. That is, 48/74 = 64.86... ~65%. So, your system has an extract efficiency of ~65!

Of course, few recipes are this simple. Let's look at a slightly more complicated grain bill in order to calculate your efficiency.

8 lbs.2-row Pale (37 ppg)
1 lb. Munich (German) (37 ppg)
0.5 lb. Crystal 40L (37 ppg)
0.25 Chocolate 350L (30 ppg)

Since the system is known to yield around 65% efficency, we can use this to determine the resultant OG. To do this calculation by hand, you would multiply the theoretical extract by the weight of each grain used, multiply by the percentage of extract efficiency, add them all, and divide the number of gallons. It looks something like this:

T=Theoretical Extract
W=Weight of Grains
P=Percentage of Extract Efficiency
G=Total Gallons

((T1*W1*P)+(T2*W2*P)+(T3*W3*P)+(T4*W4*P))/G

For reference, you may want to download small spreadsheet that I put together to show this all in action. You can find it here. As you can see, the only big calculation here is just the calculation given above. It is the contents of cell D2. It reads:

=SUM((B2*37*C2)+(B3*37*C2)+(B4*34*C2)+(B5*30*C2))/C5

That's just the calculation above written for Excel. Armed with this knowledge, you should be able to make cells for all sorts of different fermentables, assuming that you know their theoretical yield. You can use the Uchima table linked to above, or pick up Daniels' Designing Great Beers. It is a book that is quite worthwhile, even if you already understand extract efficiency.

It is worth noting that some fermentables aren't affected by theoretical yield. Malt extracts, for example. Their efficiency is always at 100% since their fermentable sugars don't need to be extracted in any way - they're ready to go. Percentages of extract efficiency only apply to sources that you'll be extracting sugars from after all.

Go forth and be efficient!

1 comment:

v. said...

god i love it when you talk science.