hi brewmates,
just to make the confusion perfect
How do I calculate the brewhouse-efficiency?
well, Ill try to explain, despite of my poor english, for that I must apologize.
first I have to explain that were calculating the extract in %, not like most of Australian homebrewers do in gravity.
the percentage is approxymately 1/4 of your readings behind the 1000
for example, your reading is 1048, that means 12% extract.
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 %
pooooohhh, hard work for me writing in english....
have fun :lol:
