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  1. #1

    Enthalpy conversion btu/lb to kj/kg

    Can someone explain the conversion of enthalpy from btu/lb to kj/kg?

    Every website I search says to multiply by 2.326, but when I do some checks (using web based calculator or psychrometric charts) the results do not work.

    For example;
    77F, 50%RH = 29.3 btu/lb
    25C (i.e. 77F) 50%RH= 50.29 kj/kg
    and 29.3 x 2.326 <> 50.29!

    I am using an Alerton system which has an Enthalpy block which only supports F and btu/lb and I am working in SI (metric) units.

  2. #2
    Join Date
    Apr 2007
    Location
    San Diego, CA
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    1,327
    1 Btu/lb = 2.326 kJ/kg
    http://www.engineeringtoolbox.com/un..._per_unit_mass

    Not sure where you're getting 50.29 from.

  3. #3
    Information required for proper conversion is 1 calorie is heat reqquired to raise 1 gram of water by 1 Deg C and 1 BTU is amount of heat require to raie 1 LB of water by 1 deg F . 1 calorie = 4.187 Joule . 1 Deg F = (9/5 C ) + 32. All this information gives all information to think upon.

  4. #4
    Information required for proper conversion is 1 calorie is heat reqquired to raise 1 gram of water by 1 Deg C and 1 BTU is amount of heat require to raie 1 LB of water by 1 deg F . 1 calorie = 4.187 Joule . 1 Deg F = (9/5 C ) + 32. All this information gives all information to think upon.

  5. #5
    Join Date
    Oct 2002
    Posts
    77
    Quote Originally Posted by integrationx View Post
    Can someone explain the conversion of enthalpy from btu/lb to kj/kg?

    Every website I search says to multiply by 2.326, but when I do some checks (using web based calculator or psychrometric charts) the results do not work.

    For example;
    77F, 50%RH = 29.3 btu/lb
    25C (i.e. 77F) 50%RH= 50.29 kj/kg
    and 29.3 x 2.326 <> 50.29!

    I am using an Alerton system which has an Enthalpy block which only supports F and btu/lb and I am working in SI (metric) units.
    That's a good question. Luckily for me, when I've done work where they use SI, they're still OK with BTU/LB. This explains it though. "You can not convert from metric to imperial enthalpy or vice versa directly."

  6. #6
    Join Date
    Sep 2004
    Location
    SoCal
    Posts
    446
    forget osa/ra enthalpy comparison for economizer control

    small difference in rh/temp accuracy equals large energy/capacity loss

    better to use dewpoint transmitter from rotronics or airtest

    open economizer only if,
    osa db temp is < 72 F AND
    call for cooling is present AND
    osa dewpoint is < than 52.5 F

    cheers
    lb

  7. #7
    Join Date
    Dec 2006
    Location
    Ontario, Canada
    Posts
    1
    Actually, this question is very similar to converting temperatures between Fahrenheit and Celsius. Since they cross 0 at different temperatures, you cannot simply multiple by 9/5 and 5/9. You have to move the margin by adding/subtracting 32.
    If you examine your psych charts, you will notice that enthalpy is 0 at 0 degrees for both SI and Imperial charts. You need to add/subtract the enthalpy at 32 Deg F and 0% rh. From my ASHRAE book, this would be 7.686 BTU/lb.
    Therefore 29.3 BTU/lb = (29.3 -7.686) x 2.326 = 50.27 KJ/KG

  8. #8
    Join Date
    Mar 2007
    Posts
    274
    Quote Originally Posted by integrationx View Post
    Every website I search says to multiply by 2.326, but when I do some checks (using web based calculator or psychrometric charts) the results do not work.
    you are absolutely right, ...... and the conversion factor is correct. How come ??

    In essence the enthalpy is a meaningless number. What you are interested in is what is the enthalply difference between 2 conditions

    Let's take your example
    77F, 50%RH = 29.3 btu/lb
    25C (i.e. 77F) 50%RH= 50.29 kj/kg
    We will plot this condition on a psychrometric chart and call it point "A"

    We will heat this air to 95F (or 35C)
    On the psychrometric chart, heating is a horizontal line to the right and we will call this point "B"
    On the Imperial psychrometric chart I find that RH = 28% .... Enthalpy = 34 btu/lb
    On the Metric psychrometric chart I find that RH = 28% .... Enthalpy = 62 kj/kg

    So the amount of heat required for 1 lbs of air to go from point "A" to point "B" is:
    34 btu/lb - 29.3 btu/lb = 4.7 btu/lb
    or.... 62 kj/kg - 50.29 kj/kg = 11.7 kj/kg

    4.7 btu/lb * 2.326 = pretty close to 11.7 kj/kg

  9. #9
    Join Date
    Mar 2007
    Posts
    274
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