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A Simple Question. Mike? Tony? Anyone?

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  • Harry
    Further to that discussion about low-power continuous hobby stills without large boilers... I think I have the answer to this on paper, but it would be nice if
    Message 1 of 3 , May 3, 2005
      Further to that discussion about low-power continuous hobby stills
      without large boilers...

      I think I have the answer to this on paper, but it would be nice if
      someone with the right training could say yea or nay. Anyway,
      here's the question and guidelines...

      Q) Calculate the amount of additional heat required to separate
      this binary mixture into azeotrope and bottoms: 20 litres of 10:90
      ethanol-water @ 70°C.

      Working parameters (assumptions):

      Assume an ideal lossless system (for simplicity).

      The specific heat capacity of water is 4.18 kJ.kg-1.K-1.
      The specific heat of vaporization of water is 2260 kJ.kg-1.

      The specific heat capacity of ethanol is 2.44 kJ.kg-1.K-1.
      The specific heat of vaporization of ethanol is 838 kJ.kg-1.

      Given that a 10:90 ETOH:H2O mixture boils at 93°C, MY rough
      calculations determine…
      The specific heat capacity of 10:90 ETOH:H2O is 3.63 kJ.kg-1.K-1.
      The specific heat of vaporization of 10:90 ETOH:H2O is 1807 kJ.kg-1.

      If any of the above is off, let me know.


      Slainte!
      regards Harry
    • Mike Nixon
      Harry wrote: Subject: [Distillers] A Simple Question. Mike? Tony? Anyone? Further to that discussion about low-power continuous hobby stills without large
      Message 2 of 3 , May 3, 2005
        Harry wrote:
        Subject: [Distillers] A Simple Question. Mike? Tony? Anyone?


        Further to that discussion about low-power continuous hobby stills
        without large boilers...

        I think I have the answer to this on paper, but it would be nice if
        someone with the right training could say yea or nay. Anyway,
        here's the question and guidelines...

        Q) Calculate the amount of additional heat required to separate
        this binary mixture into azeotrope and bottoms: 20 litres of 10:90
        ethanol-water @ 70°C.

        Working parameters (assumptions):

        Assume an ideal lossless system (for simplicity).

        The specific heat capacity of water is 4.18 kJ.kg-1.K-1.
        The specific heat of vaporization of water is 2260 kJ.kg-1.

        The specific heat capacity of ethanol is 2.44 kJ.kg-1.K-1.
        The specific heat of vaporization of ethanol is 838 kJ.kg-1.

        Given that a 10:90 ETOH:H2O mixture boils at 93°C, MY rough
        calculations determine.
        The specific heat capacity of 10:90 ETOH:H2O is 3.63 kJ.kg-1.K-1.
        The specific heat of vaporization of 10:90 ETOH:H2O is 1807 kJ.kg-1.

        If any of the above is off, let me know.
        ===================
        Try working it with kJ/mol ... it's much easier!
        Latent heat of vaporization of water is 40.639 kJ.mol and of ethanol is
        39.33 kJ/mol, so they are both pretty well much the same for any mixture ...
        take a mean of 40 kJ/mol.

        In volume, 1 mol of water is 18.016 ml, and ethanol is 58.24 ml, so the same
        amount of heat is needed to vaporize either volume. Just take it from there
        and the answers drop out fairly painlessly.

        What .... you want ME to do it????? :-))

        All the best,
        Mike N
        (Waddya mean? The dog ate your homework!!!!!)
      • Harry
        ... ethanol is ... mixture ... ... so the same ... from there ... Many thanks, Teech! :-)) Hmmm...guess I ll have to brush up on my self-taught Chem.Eng. Eh
        Message 3 of 3 , May 3, 2005
          --- In Distillers@yahoogroups.com, "Mike Nixon" <mike@s...> wrote:

          > Try working it with kJ/mol ... it's much easier!
          > Latent heat of vaporization of water is 40.639 kJ.mol and of
          ethanol is
          > 39.33 kJ/mol, so they are both pretty well much the same for any
          mixture ...
          > take a mean of 40 kJ/mol.
          >
          > In volume, 1 mol of water is 18.016 ml, and ethanol is 58.24 ml,
          so the same
          > amount of heat is needed to vaporize either volume. Just take it
          from there
          > and the answers drop out fairly painlessly.
          >
          > What .... you want ME to do it????? :-))
          >
          > All the best,
          > Mike N
          > (Waddya mean? The dog ate your homework!!!!!)


          Many thanks, Teech! :-))
          Hmmm...guess I'll have to brush up on my self-taught Chem.Eng.
          Eh bien, I have many books & such. So much to do, so little time.
          And right now it's WHISKY TIME! :-)
          Thanks again.

          Slainte!
          regards Harry
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