once Ive posted already the way Im calculating the efficiency, Ill do again:
first we measure the Wort in litre, but we need it in kg.
approxymately we may calculate the density of wort as follows:
0 Mass % --> density = 1
1 Mass % --> density = 1.004
2 Mass % --> density = 1.008
we may calculate:
(1) Mass [kg] = Volume[l] * (1 + Mass% * 0,004) [kg/l]
but stop, this only if the wort is at 20C
if the wort is still boiling or very hot, we have to calculate a contraction-factor of 0,96
that means: (2) Volume [l, 20] = Volume [l, 100] * 0,96
Now we may reckon the weight of the wort:
(1)+(2)
(3) Mass [kg, 20] = amount of wort[l, 100] * 0,96 * (1 + Mass% * 0,004)
In this amount of wort is contained:
(4) Extrakt [kg] = Mass [kg, 20] * Extrakt-readings[%] / 100
or easier:
(5) Extrakt [kg] = amount of wort[l, 20] * 0,96 * (1 + Mass% * 0,004) * Extrakt-readings[%] / 100
now we can conclude:
(6) brewhouse-efficiency = Extrakt [kg] / amount of grain [kg] * 100 %
I have made an simple Excel-sheet, you may just put in your datas:
take any result-cell and put in this formula (for cold wort):
=((B4*0,004+1)*B4)*A4/C4
and for hot or boiling wort:
=((B4*0,004+1)*B4*0,96)*A4/C4
You may put in your datas in Cell A4, B4 and C4,
where the cell A4 is the amount of wort in litre
and the cell B4 is the reading of extract in %, for example your reading ist 1048, so you put in 12 (%)
and the cell C4 is the total amount of grain (kg)
the result will show you the brewhouse-efficiency in %
poooohh, hard work for me writing in english....
I got approx 78% efficience on a 1092 brew on the weekend so Im a happy brewer.Insulating the mash tun better did a treat (good).
