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Thread: RE: calculations for peak heat load capacity

  1. #1
    Join Date
    Dec 2008
    Posts
    3

    RE: calculations for peak heat load capacity

    Hi,
    I am trying to calculate our brewery's peak refrigeration load, so that I can ensure that we have enough cooling supplied from our 3 compressors. I have 3 compressors onsite that provide a total of 121kw. I have done some rough calculations on our heat load but I am unsure if I am correct. Would some one be able to pass on/clarify some calculations for working out the following:
    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.
    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.
    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.
    Cheers
    Andrew
    Sydney

  2. #2
    Join Date
    Jul 2005
    Location
    Auburn, WA
    Posts
    229
    Hi Andrew,

    There are a number of different processes, formulas, and practices for estimating the cooling loads.

    When I do a load estimate I try to take a look at a typical to high operating scenario- if we took a snap shot of your brewery during a typical day what would each of the Fermenters and/or Brite tanks be seeing? Being based here in the United States, my data isn't metric, but hopefully this might still help.

    A typical scenario might be:

    3 EA 10 Bbl Fermenters in active Fermentation @ 70 F.
    2 EA 10 Bbl Fermenters crash cooling from 70 F to 36 F, the cooling rate is typically based on a 24 hour pull down (or whatever pull down rate the brewery might specify).
    6 EA 10 Bbl Conditioning tanks holding at 36 F

    From this data I would generate a Load estimate. Of course it is just an estimate and there will be periods the load will be a bit higher and a bit lower, but it enables a good estimate. If you would like to provide this load data, I'd be happy to run through my load estimate program and provide an estimated cooling load. You can download a survey from our website here: http://www.prochiller.com/reBrewery.html

    The calculation of the Cold Liquor Load- or chilling water to be used for Wort Cooling is pretty straightforward. Subtract your ending temperature from your starting temperature to get a temperature drop, multiply this by the total lbs of water being cooled (8.6 Lbs per gallon). Then divide by the number of hours you have to cool the water.

    For example:

    600 Gallons = 5160 lbs of water
    Starting Temp: 70 F
    Ending Temp: 35 F
    Temperature Drop: 35 F (70 - 35)

    35 F temperature drop X 5160 lbs = 180,600 BTU
    24 hour Pull Down Time: 7,525 BTU/HR

    Again sorry for the non-metric numbers, good luck.

    Jim

    Pro Refrigeration Inc.
    Last edited by jimvgjr; 12-22-2008 at 09:32 PM.

  3. #3
    Join Date
    Mar 2006
    Location
    Assens, Danmark
    Posts
    57
    Hi Andrew,

    here is what I got,


    Quote 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 krńusen, 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

    Quote 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?

    Quote 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

  4. #4
    Join Date
    Dec 2008
    Posts
    3
    Thanks Christoph,
    I haven't been back here for a while so just read your reply. Can I just clarify a few things as I'm unsure if I fully understand.
    In the fermenting beer example, dEs is this potential fermentabkle extract i.e. 13.5 plato wort (roughly 14.2kg/hl extract) bottoming out at say 3plato (roughly 3.7kg/hl extract); in this case would dEs be 14.2-3.7 = 10.5kg/hl, or is it the amount extract consumed on a day basis (i.e say around 3plato a day - roughly 3.7kg/hl extract would then be dEs ?)
    Density, I guess the same as above, is it instantaneous or end density, dT = what is the temp difference, if pitched at 15C and ferments at 15C is it the difference between glycol temp in the jacket (-5C) or is it something else?
    And the last one was days to cool, are you not mixing up the heat load from fermenting beer and the heat load for cooling the beer? Sorry mate, I'm just a bit unclear.

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