What's Your Typical Conversion Efficiency?

[bg1414]

reviving old thread. if you fly sparge how/when do you take your first runnings gravity? How does this work in the equations by not having first runnings volume?

 

[doug293cz]

reviving old thread. if you fly sparge how/when do you take your first runnings gravity? How does this work in the equations by not having first runnings volume?

The equations in the OP don't depend in any way on sparge method, nor do they require any volume measurement other than the strike volume. The defining equation for °Plato (°P) is:

°P = 100°P * Wt_of_Extract_in_Wort / (Wt_of_Extract_in_Wort + Wt_of_Water_in_Wort)​

Since Wt_of_Extract_in_Wort + Wt_of_Water_in_Wort) = Total_Wt_of_Wort, °P is just the weight percent of extract in the wort.

The weight of the water in the wort is just the weight of the strike water used in the mash. The equation in the OP assumes you are working in metric (volume in liters, and weight in kilograms.) Since 1 L of water weighs about 1 kg, volume of water in L is approximately equal to the weight of water in kg. The equation in the OP substitutes liters of strike water for kg of strike water. However, water only has a density of 1.000 kg/L at 4°C (39.2°F.) At strike temp (about 160°F) water has a density of 0.977 kg/L. So, for better accuracy, the water density at the volume measurement temp should be used.

If you are using gallons and pounds instead of liters and kilograms, the equations get a little more complicated. One gallon of water weighs 8.3304 lb at 68°F and 8.1545 lb at 160°F. The °P equation becomes the following when using lb and gal:

°P = 100°P * Extract Wt [lb] / (Extract Wt [lb] + Strike Vol [gal] * Water Density [lb/gal]), with the density at volume measurement temp being used​

So, to finally answer your question: the first runnings gravity should be taken after you have assured that the wort in the mash has been well homogenized (the wort has equal extract concentration and SG everywhere), and before any sparge water has mixed with the runnings.

Brew on

​

 

[hezagenius]

Hey @doug293cz

Why wouldn't this formula work for conversion efficiency?

(Measured_Mash_SG - 1)/(Max_SG - 1)?

If you have 10# of grain with a ppg of 37 and you mash with 10 gallons of water, your Max_SG would be 1.037. If you took a reading at the end of your mash and you got Measured_Mash_SG = 1.032, why wouldn't your conversion efficiency be 32/37 = 86.5%?

 

[doug293cz]

Hey @doug293cz

Why wouldn't this formula work for conversion efficiency?

(Measured_Mash_SG - 1)/(Max_SG - 1)?

If you have 10# of grain with a ppg of 37 and you mash with 10 gallons of water, your Max_SG would be 1.037. If you took a reading at the end of your mash and you got Measured_Mash_SG = 1.032, why wouldn't your conversion efficiency be 32/37 = 86.5%?

That's an approximate formula, and the further you are from 100% conversion, the worse the approximation. Don't know exactly how bad the approximation is (but I might run some calcs to figure it out and graph the error.)

If you mash 10 lbs of grain with 37 pts/lb potential with 10 gal of water you will not get an SG of 1.037 @ 100% conversion. Potential is defined as SG that would be obtained by creating 1 gal of wort from 1 lb of grain. Because of the volume taken up by the extract (sugar) in the wort, the volume of the water will be less than 1 gal. Also, potentials are given as dry basis, but grain contains about 4% moisture, so the "as-is" potential is only 96% of the dry potential (and unfortunately, most brewing software does not correct for dry vs. as-is, which makes your conversion efficiency look lower than it is.) If you mash 10 lb of grain with 37 pts/lb dry potential in 10 gal of water, the actual SG @ 100% conversion would be 1.0336. If we ignore the moisture content (assume the 37 pts/lb is as-is basis) the SG @ 100% conversion would be 1.0351.

Brew on

 

[hezagenius]

@doug293cz

So would using Kai's formula of FW_measured/FW_max eliminate the moisture content error?

Built into the FWmax calc is the e_grain term. Why is that 0.8? Why not 0.75 or 0.85?

 

[doug293cz]

@doug293cz

So would using Kai's formula of FW_measured/FW_max eliminate the moisture content error?

Built into the FWmax calc is the e_grain term. Why is that 0.8? Why not 0.75 or 0.85?

If you correct for moisture content when calculating FW_max, then yes. But this formula is still an approximation.

1.037 or 37 pts/lb is equal to 80.13% potential extract. Since most base grains have a potential between 1.036 and 1.038, 80% is a reasonable number to use if you don't have more detailed data on the particular lot of grain.

I don't know why Kai didn't include the moisture correction in his formulas, but did include it in the table. But, it does cause confusion.

Brew on