Lesson 6 - Wort Separation Process Control
This lesson deals with wort separation process control. The different methodologies of measuring and calculating extract recovery are explained and demonstrated. A methodology is explored, as is how to investigate and resolve performance issues with extract efficiency, wort quality, and cycle time.
Calculating the Extract Recovered
As discussed earlier, to calculate the recovered extract, we need to know the Specific Gravity measured at 20˚C. We also need to know the volume of the wort at 20˚C.
Wort consists of at least 75% water. So, considering water density values as a first approximation, water at 20˚C has a density of 0.9982 kg/L and at 75˚C it has a density of 0.9748 kg/L. There is a 2.3% increase in density, and hence a 2.3% reduction in volume between wort at 75˚C and 20˚C. If we are metering wort at 75˚C from our wort separation device, then this value needs to be reduced by 2.3%, unless the meter has automatic temperature compensation.
When our wort collection to kettle is complete (sometimes called kettle full or kettle up), we can measure the volume and gravity achieved. If we measure our kettle full SG as 1.062, and our temperature corrected measured volume as 482 hL, how many kg of extract or hL˚ of extract have we recovered? To calculate the kg of extract, we can use the following equation:
kg extract in wort = hL x ˚Plato x density (˚Plato x density = kg extract/hL)
We can use the SG value to calculate the density and ˚Plato of the wort.
Density of water at 20˚C is 0.9982 kg/L (not 1.0000). We need to adjust the SG by this factor to calculate the density at 20˚C. Density is 1.062 x 0.9982 = 1.060 kg/L.
We can also use the SG to calculate ˚Plato by using the following equation:
˚Plato = (190.74 x SG3) – (800.47 x SG2) + (1,286.4 x SG) – 676.67
So by substituting SG = 1.062 into the formula, then ˚Plato = 15.14.
kg extract in wort = 482 hL x 15.14 ˚Plato x 1.060 density = 7,735 kg.
As we calculated earlier, our added extract was 7,790 kg.
Extract efficiency is the recovered extract as a percentage of the added extract:
(7,735/7,790) x 100 = 99.30%
We will now calculate the recovered hL˚ of extract:
We have recovered 482 hL at SG 1.062, or 62˚ of gravity.
So, we have collected 482 hL x 62˚ = 29,884 hL˚.
We calculated above that we have added 30,086.5 hL˚.
Therefore, our extract efficiency is 29,884/30,086.5 x 100 = 99.33%
Point of reflection: Have a go at calculating the extract efficiency from the following set of data.
8,000 kg of malt have been used. We have recovered 428 hL of wort at SG 1.055 at 13.54 ˚Plato. The malt extract dry is 80% w/w or 308.5 L˚/kg. The malt moisture content is 4 % w/w. What is the extract efficiency? Use either of the two methods outlined above. When you have an answer click ‘Next’ to continue this lesson.