COUNTERCURRENT LIQUID- LIQUID EXTRACTION

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COUNTERCURRENT LIQUID- LIQUID EXTRACTION Powered By Docstoc
					COUNTERCURRENT LIQUID-
  LIQUID EXTRACTION
       SOLVENT REQUIREMENTS
• PFD FOR MULTISTAGE COUNTERCURRENT
  EXTRACTION




• STAGE TO STAGE MASS BALANCE
• Xr = Xf * (mS/F – 1)/ ( [mS/F] n+1 -1)

        http://www.cheresources.com/extraction.shtml
DEFINITION OF EXTRACTION FACTOR
            EXTRACTION FACTOR
• FOR A SPECIFIC NUMBER OF STAGES ,n, A
  HIGHER E FACTOR RESULTS IN MORE
  EFFECTIVE EXTRACTION
• LOWER PRACTICAL LIMIT FOR E IS ABOUT 1.3
  FOR MOST OPERATIONS
• SPREADSHEET EXAMPLE FOLLOWS


  http://www.cheresources.com/extraction.shtml
    COMPARISON FOR CROSS AND
     COUNTERCURRENT TRAINS




http://www.cheresources.com/extraction.shtml
            SPREADSHEET SIMULATION
http://terpconnect.umd.edu/~nsw/ench250/extract3.htm
    Process Specification (Change the number in yellow cells. Other cells in this
    worksheet contain formula.)
                  Flow rate of the raffinate
        R = 300phase
                  Flow rate of the extract
        E = 200phase
                  Inlet composition of the raffinate
       xin = 0.03phase
                  Outlet composition of the raffinate
      xout = 0.001phase
                  Inlet composition of the extract
       yin =     0phase
        K=       3Partition coefficient
    Calculated Variable:
            S     2=K*E/R
    Find the number of equilibrium stages required by forcing the following
    f(N)=0.
                    = (xin-xout)/(xin-yin/K) - (S^(xN+1)-S)/(S^(xN+1)-
      f(xN) =    -01)
                    ¬ via |Goal Seek| by setting f(xN) to a value of 0 by changing
         xN = 3.954xN.
    Since a fractional stage does not make physical sense, set the number to be an
    integer.
          N=      4
                 SPREADSHEET SIMULATION
http://terpconnect.umd.edu/~nsw/ench250/extract3.htm
Find the raffinate composition. (We prepared a matrix A and a vector b to handle up to 14 stages.)
Define the yellow cells to be A and b. As N changes, we need to manually redefine A and b to match
their size.
The following matrix A and vector b describe the material balance equation around each
stage.
       xi-1 - (1+S)*xi + S*xi+1 = 0 for i=1,2,...,N
                                            -3 2 0     0    0    0    0    0    0    0    0    0    0 0       -0.03
                                             1 -3 2    0    0    0    0    0    0    0    0    0    0 0           0
                                             0 1 -3    2    0    0    0    0    0    0    0    0    0 0           0
                                             0 0 1    -3    2    0    0    0    0    0    0    0    0 0           0
                                             0 0 0     1   -3    2    0    0    0    0    0    0    0 0           0
                                             0 0 0     0    1   -3    2    0    0    0    0    0    0 0           0
                                       A= 0 0 0        0    0    1   -3    2    0    0    0    0    0 0    b=     0
Summary                                      0 0 0     0    0    0    1   -3    2    0    0    0    0 0           0
 Stage Raffinate x Extract y                 0 0 0     0    0    0    0    1   -3    2    0    0    0 0           0
      0 0.03000 0.09000                      0 0 0     0    0    0    0    0    1   -3    2    0    0 0           0
      1 0.01452 0.04355                      0 0 0     0    0    0    0    0    0    1   -3    2    0 0           0
      2 0.00677 0.02032                      0 0 0     0    0    0    0    0    0    0    1   -3    2 0           0
      3 0.00290 0.00871                      0 0 0     0    0    0    0    0    0    0    0    1   -3 2           0
      4 0.00097 0.00290                      0 0 0     0    0    0    0    0    0    0    0    0    1 -3          0

				
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posted:12/12/2011
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