Chiller cooling output calculation in KW
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  1. #1
    Join Date
    Apr 2009
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    9

    Chiller cooling output calculation in KW

    Dear all,

    I have water-cooled chiller, I need to calculate its cooling output in KW. I have chilled water flow in GPM and supply & return temperature in C.

  2. #2
    Join Date
    Apr 2005
    Location
    Split, Croatia
    Posts
    37
    You need to have very precise thermometer to calculate correctly cooling output.
    Since dT is usually around 5K your imprecise temperature reading could lead to 30% (or even more) error.

    Q=m*c(T2-T1)

    m=mass of water
    c=specific heat capacity of water

  3. #3
    Join Date
    Aug 2003
    Location
    Fort Worth, TX
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    11,279
    You can estimate cooling output in British Thermal Units per Hour (BTUH) via:

    500 x GPM x deltaT = BTUH

    Keep your units consistent. If you know the Gallons Per Minute, use Fahrenheit for the temperature scale.

    Once you determine the BTUH, divide that number by 12,000 to get tonnage.

    To get kilowatts consumed per ton, you measure what the machine is using in kilowatts to deliver the measured tonnage.

    Example: process chiller flowing about 50 GPM with a 10 degree split between return and supply water...

    500 x 50 x 10 = 250,000 BTUH

    250,000/12,000 = 20 tons

    Assume machine is drawing 12 kilowatts to achieve this output:

    12 kilowatts/20 tons = .6 kilowatts per ton


    Remember that there are 3.413 BTUs per watt, 3413 BTUs per kilowatt. So, on a one to one basis, you would need about 3.515 kilowatts per ton (12,000 BTU), but we can achieve a ton of refrigeration with less kilowatts per ton consumed than that, by far.
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  4. #4
    Join Date
    Feb 2006
    Posts
    224
    The formula for KW of refrigeration is L/s x Delta T(*C) x 4.19kj/kg (for Water)
    To convert GPM to L/s GPM (div by) 60= GPS
    GPS x 3.785= L/s
    60 seconds in a minute
    3.785 litres in a US Gallon
    I have it as 3.415 BTU/ Watt

  5. #5
    Join Date
    Apr 2009
    Posts
    9
    Thanks to all of you...

  6. #6
    Join Date
    Jul 2009
    Posts
    3

    Thumbs up Clarification...

    Quote Originally Posted by shophound View Post
    You can estimate cooling output in British Thermal Units per Hour (BTUH) via:

    500 x GPM x deltaT = BTUH

    Keep your units consistent. If you know the Gallons Per Minute, use Fahrenheit for the temperature scale.

    Once you determine the BTUH, divide that number by 12,000 to get tonnage.

    To get kilowatts consumed per ton, you measure what the machine is using in kilowatts to deliver the measured tonnage.

    Example: process chiller flowing about 50 GPM with a 10 degree split between return and supply water...

    500 x 50 x 10 = 250,000 BTUH

    250,000/12,000 = 20 tons

    Assume machine is drawing 12 kilowatts to achieve this output:

    12 kilowatts/20 tons = .6 kilowatts per ton


    Remember that there are 3.413 BTUs per watt, 3413 BTUs per kilowatt. So, on a one to one basis, you would need about 3.515 kilowatts per ton (12,000 BTU), but we can achieve a ton of refrigeration with less kilowatts per ton consumed than that, by far.
    Dear Shophound

    Good calculations. Please clarify what is the '500' used for? I mean what does it denote?
    regards
    RK

  7. #7
    Join Date
    Nov 2006
    Location
    North Carolina
    Posts
    130
    500 is a constant

  8. #8
    Join Date
    Jul 2009
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    3
    Quote Originally Posted by chad_nc View Post
    500 is a constant
    Hi Chad
    I know it is a constant. What I want to know is that how is it derived? it is just not a number which is assumed.

  9. #9
    Join Date
    Feb 2006
    Posts
    224
    500 is a constant for water. The reason I believe is, you are dealing in GPM(minutes) and BTU/HR (hours)
    60 Min. in an hour and the weight of 1 US gallon of water 8.31 lbs.
    60 x 8.31 = 498.6
    rounded to 500.
    Process system with glycol depends on weight of 1 gall of dilluted glycol.

  10. #10
    Join Date
    Jul 2009
    Posts
    3

    Thumbs up

    Quote Originally Posted by bertoh View Post
    500 is a constant for water. The reason I believe is, you are dealing in GPM(minutes) and BTU/HR (hours)
    60 Min. in an hour and the weight of 1 US gallon of water 8.31 lbs.
    60 x 8.31 = 498.6
    rounded to 500.
    Process system with glycol depends on weight of 1 gall of dilluted glycol.
    Thanks Bertoh, I got my answer.

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