Hi Andrew,

here is what I got,

Originally Posted by

**andrewmj1**
1. Heat generated from fermenting beer. I found a source that said every kg of extract was equal to 558kj of heat. This seems pretty easy to work out depending on the time of the ferment.

That is a correct ball park number for a lager beer in high kräusen, what you miss is putting that heat flow in relation to hl and time. I assume you will do mostly ales? So you need to sit down and recalculate.

To find Q you need Extract reduction kg/hl(dEs), translation from appearant to real extract (0,81), density of beer (q), Heat capacity of beer C-beer (3,98 kj/kg*k), Temperature difference in K( dT), time in days (dt) and the efficiency of your cooling system ( we assume its 0,9 ??)

Q= dEs*0,81*q*C-beer*(+/-)dT devide that by dt*0,9

and you have your heat flow in kj per hl and day

Originally Posted by

**andrewmj1**
2. Cooling a fermenter from fermenting temp say 22C down to -1.5C. The problem that I have here is using q=mxcpdt does not take into account that the load varies, with the highest load being at the start of cooling.

I don´t think that that is your problem, calculate the required Q using that formula; specific heat of wort: 3,98kj/kg K and beer: 4,06 kj/kg K, the density you will know, delta T ( in K) and V as well.

Example: cooling your lager from 9 to -1 requires 4200kj/hl. It will now be important that you decide how fast this should happen! So integrate over time : Q = V*p*c*dT/ dt

Crash cooling within one day, then your cooling unit must be able to supply that in one day! If you would not have any looses....heat radiation, heat transmission, you follow yes?

You will need just as much heat flow to cool the beer to storage temp as you need throughout the entire fermentation. Can you see what that means for your peak calculation?

Originally Posted by

**andrewmj1**
3. Cooling of chilled water to be used in wort heat exchanger cooling.

The value that I got was 148kW which seems strange considering that the system doesn't struggle too much. If someone gets a chance can they please help me out. I'd like to go through the numbers myself if you can help me out.

that one is the toughest to calculate since the setup of your wort cooling influences all calculation a lot. Basicly you could use the formula from question 2. But I have to look into this one...

Hope it was of help!

Christoph

"How much beer is in German intelligence !" - Friedrich Nietzsche