Moisture content and conversion efficiency

[element533]

Do most people account for moisture content when they calculate conversion efficiency, or ignore it?

Example:

12.5 lbs Pale 2-row
0.75 lbs Cara-pils
0.75 lbs Crystal 15L
0.5 lbs Munich malt
0.75 lbs White wheat malt

6.5 gal strike water

First wort gravity reading = 1.069

How would you calculate conversion efficiency for the above?

If I account for 4% moisture content in the grain, I get 95%. If I ignore moisture, I get 91%.

I'm less interested in the "right" answer and more interested in the "typical" answer.

 

[ajdelange]

I would note that 1.069 gravity corresponds to 16.826 °P meaning the wort is 16.826% extract w/w and that the wort weighs 0.998203*1.069 kg/L at room temperature. Thus the total extract is L*0.16826*0.998203*1.069 where L is the volume of the wort in liters at room temperature. Dividing this by the weight (in kg) of the grains used gives the fraction of the grain mass converted to extract. Multiply by 100 to get this efficiency in percent. I usually (in fact I don't think I've ever done otherwise) divide by the weight of the grain 'as is' given that I do not know the moisture content nor do I have any way to measure it but clearly if you want dry basis numbers and you have the moisture content you can subtract that from the grain weight before doing that final division.

 

[element533]

Thanks ajdelange. That approach gives me the same value, but it's more difficult to calculate because I have to account for the volume of wort kept back in the grain.

Here's my approach:

• 1.069 gravity = 16.826 °P actual
• sum of yield × weight for each grain = 5.45 kg extract available
• 5.45 kg extract ÷ (5.45 kg extract + 24.6 kg water) = 18.169 °P theoretical maximum
• 16.826 °P actual ÷ 18.169 °P maximum = 91% conversion efficiency

If I factor 4% moisture, the available extract is only 5.24 kg, which makes the efficiency 95%.

So... you use the weight of the grain as-is for your conversion numbers. What percentage do you typically get?

 

[ajdelange]

If I add up your grain weights I get almost 7 kg so if conversion were 100% you would have 7 kg of extract. Where does the 5.45 come from?

The method I described is the one used to compute conversion in a Congress mash. It does not account for the water retained by the grains as that is presumed to be very low in extract as it has been used to sparks the grain bed. The efficiency is simply the amount of extract in the kettle divided by the amount of grain used to produce it.

I usually use the final volume of the produced beer to calculate the total original extract so my efficiency number includes losses in kettle, transfer to fermenter and in emptying fermenter (extract converted to biomass, beer left in the cone...). I usually get about 69% grain to keg. Note that a typical base malt usually measures around 80% and I think that's FGDB so I'm pretty happy if my losses are only 11% or so. A commercial operation would not be happy with this kind of performance.

 

[element533]

The Congress mash computation is what I call the yield. The 5.45 kg in my example is the total extract available in the grain (~80% yield), not the weight of the grain itself. I compute conversion efficiency based on the grain's yield, not the total grain.

So you're seeing (let's say) 10 kg grain gets you 6.9 kg extract in the keg? That seems huge!! Maybe I'm misunderstanding?

Here's my typical:

10 kg grain
=> 8.0 kg extract in grain (80% yield)
=> 7.3 kg extract in mash (91% conversion efficiency)
=> 5.5 kg extract in kettle (76% lauter efficiency)
=> 5.0 kg extract in fermenter (~10% trub loss, which carries extract)
=> 4.8 kg extract in bottles (~5% secondary trub loss and spillage)

I calculate that as 63% brewhouse efficiency (from grain yield into fermenter). I batch sparge and I don't have very good filtration, but right now I'm focused on getting consistently high conversion. Hence, I'm curious when others say they're getting 95% conversion, do they mean dry-basis, or as-is?

 

[ajdelange]

The Congress mash computation is what I call the yield. The 5.45 kg in my example is the total extract available in the grain (~80% yield), not the weight of the grain itself. I compute conversion efficiency based on the grain's yield, not the total grain.

Ah. I know a lot of home brewers do it that way but I have never really understood why. You can't do that reliably unless you have the Congress mash data and that is sometimes found in the malt data sheets and sometimes not. Beyond, that, and this relates to your original question, malt picks up moisture over time and so the as is 'yield' decreases as time passes.

So you're seeing (let's say) 10 kg grain gets you 6.9 kg extract in the keg? That seems huge!! Maybe I'm misunderstanding?

That's only 69% which is 11% below the Congress value for a typical base malt. A commercial operation would want to see something in the high seventies.

Perhaps you have misinterpreted what I mean by 'extract in the keg'. It is not the weight of the beer multiplied by the true extract (TE) of the beer but rather the equivalent original extract computed from the TE and alcohol content.

Hence, I'm curious when others say they're getting 95% conversion, do they mean dry-basis, or as-is?

I can only guess that as they don't have access to malt moisture content data they are using as-is probably based on a number they got from a book or magazine article as the yield.