Water adjustment for smaller containers

Brad, this might be nice to add. I used it for strike water when container is undersize for what I need by heating to a higher temp and adding room temp water.

I looked in the back of my "Algebra for Bootleggers" teacher's edition for the answer, which looked like this:

39 qt * 164 F = 6,396
180*X + 75*Y = 6,396
X qts + Y qts = 39 qts
X = 39 - Y

substituting for X...
180*(39-Y) + 75*Y = 6396
7020 - 180*Y + 75*Y = 6396
634 = 105*Y
Y=6 quarts @ 75 F
X = 33 @ 180 F


The proper way is to set the energy to cool the 180F water equal to the energy to heat the 75F water, some thing like this

m*Cp*(180-164)=m*Cp*(164-75)
m=rho*V
where
M is mass
Cp is specific heat
rho is density
V is volume

substituting rho*V in place of m:
rho*Vhot*Cp*16=rho*Vcold*Cp*89
since rho and Cp and approximately equal, you can drop them out to get
Vhot=Vcold*89/16 = Vcold*5.56
Vhot + Vcold = 39
Vcold*5.56 + Vcold = 39
6.56*Vcold = 39
Vcold = 6 quarts
Vhot = 39 - 6 = 33 quarts

thats two ways of doing it, take your pick. I like option 1, because it is the same as calculating dilutions for gravity, but instead of points its qts*F

Thanks!
 I apologize - but exactly which situation are you trying to calculate here?  It appears to be the case where you overheat the mash and are trying to cool it down with room temperature.  This seems like a reasonable thing to do if you miss your mash temperature.

Thanks,
Brad

What I use it for example: Beersmith says I need 39 quarts at 164 F. Problem is my HLT container is only 33 quarts. So by adding 6 quarts at 75 F into mash then adding 33 quarts at 180 F it gives me the required amount Beersmith suggested 39 quarts at 164 F.

OK,
 That's what I thought - makes sense now.

Thanks!
Brad