Manxnorton said:
piddledribble said:
Got no choice I'm 210 mile away
It's nice to run across a Brit who still uses the old English measurement system. I can easily go back and forth between English and metric measurements; however, all-grain recipe formulation is much easier to do when using English measurements. In my humble opinion, points per pound per gallon is superior to points per kilogram because it is a non-concentrated brew house metric. For example, my average brew house extraction rate is around 30 points per pound per U.S. Gallon (25 points per pound per Imperial Gallon or an average efficiency of approximately 83%), which means that I can produce one U.S. gallon of 1.030 wort using one pound of grain. All I have to do to hit a desired original gravity (O.G.) is to divide the fractional component of the target gravity by 0.030 and multiply the result by the number of gallons that I wish to brew to know how many pounds of grain I need to use in my grist. I can easily perform this calculation in my head.
For example, we want to brew a 5 US gallon of beer with an O.G. of 1.045 with an extraction rate of 30 points per pound per U.S. gallon.
pounds_of_grist = 0.045 / 0.030 x 5 = 7.5 (In practice, I divide 45 by 30 and multiply the result by the number of gallons of wort)
Now, lets look at what we have to do when using using the metric system. My extraction rate in points per kilogram per liter is 30 x 8.35 ~= 251 points per kilogram per liter, which to me is not very intuitive (the scaling value 8.35 takes into account that there are approximately 2.20462 lbs in a kilogram and 3.78541 liters in a U.S. gallon).
Converting the example above to the metric yields a volume of 19L
kilograms_of_grist = 0.045 x 19 / 0.251 = 3.4 kilograms
With points per pound per gallon, one only needs to calculate the amount of grain that one need to make one gallon of wort and multiply that value by the number of gallons that one wishes to brew. With points per kilogram, one has to calculate the total number of gravity points in the batch with respect to a liter and then divide the result by one's brew house extraction rate in points per kilogram per liter. In my humble opinion, working with points per pound per gallon is a more natural way to brew.
Now, I know that more than one amateur brewer on this forum is more than likely asking the question, "How does one go about calculating one's average brew house extraction rate?" It's quite simple.
points_per_pound_per_gallon = (original_gravity - 1.0) x 1,000 x volume_on_gallons / pounds_of_grist
points_per_kilogram_per_liter = (original_gravity - 1.0) x 1,000 x volume_in_liters / kilograms_of_grist
We brewed a 5 Imperial gallon batch of wort with an O.G. of 1.051 batch using 11 pound of grist
points_per_pound_per_gallon = (1.051 - 1.0) x 1,000 x 5 / 11 ~= 23 points per pound
Converting to the metric system (using 23 liters as a rough equivalent of 5 Imperial gallons)
points_per_kilogram_per_liter = (1.051 - 1.0) x 1,000 x 23 / 5 ~= 235 points per kilogram
Now, all we need to do is to perform this calculation for a several batches and average the results. Points per pound and points per kilogram are much more usable brew house metrics than extraction efficiencies because maltsters quote the average extract yield in points per pound or pounds per kilogram on their malt analysis sheets.
Let's look at a hypothetical malt analysis sheet for a bag of British Pale Ale malt
Typical British Malt Analysis -- Pale Ale
Variable Typical Value
Color 2-3.5 °L (4-6.5 °EBC)
Moisture Content (MC) 2.8-3.3%
Hot Water Extract (HWE) 303-315 L°/kg 7M
Cold Water Extract (CWE) 17-20.5%
Total Nitrogen (TN) 1.4-1.7%
Soluble Nitrogen Ratio (SNR) 36-45.5%
Diastatic Power (DP) 40-65 °Lintner (124-212 °WK)
Screenings <2.2 mm 0.45%
Friability 90-100%
The key metric value above is HWE. The ratio "L°/kg" denotes points per liter per kilogram, which is another way of saying points per kilogram per liter. The reason why knowing this value is important is that without it any calculated efficiency percentage is little more than fiction.
Using the extraction rate of 235 points per pound above
HWE_listed_on_the_bag = 303 points per kilogram per liter
235 / 303 = 77.5% actual efficiency
HWE_listed_on_the_bag = 315 points per kilogram per liter
235 / 315 = 74.6% actual efficiency
The problem with using brewing software to calculate efficiency percentages is that the maximum amount of extract that can be produced by a pound of grain is usually set at around 301 points per kilogram per liter, which lures amateur brewers into believing that their brew houses are efficient than they actually are when using bags of base malt with higher extract potentials. Conversely, a bag of base malt that is incapable of producing 301 points per kilogram per liter on a good day will make amateur brewers believe that they have inefficient brew houses.
By the way, honey will not make your beer sweeter. It will make it drier. As Dennis mentioned, you need to add non-fermentable sugars to your wort.
