A TRANSIENT SIMULATION OF A SYSTEM WITH MATERIAL RECYCLE

Consider the following system:

in which the following parameters are fixed:

Flow rates:

dm1/dt=100 kg/min

Compostion:

x1,A = 0.20

x1,B = 0.80

x3,A = 0.10

x3,B = 0.90

x4,A = 0.80 = x5,A = x6,A

x4,B = 0.20 = x5,B = x6,B

Others:

Purge fraction (pf) = m5/m4 = 0.20

 Working equations:

At the kth cycle (where we assume 1 cycle is 1 min)

  1. [m2,A]k = m1,A + [m6,A](k-1)
  2. [m2,B]k = m1,B + [m6,B](k-1)
  3. [m2]k = [m2,A]k+[m2,B]k
  4. [x2,A]k = [m2,A]k / [m2]k
  5. [x2,B]k = [m2,B]k / [m2]k

    6.  

     

    Balance around the unit:
    [m3]k + [m4]k = [m2]k

    [m3]k x3,A + [m4]k x4,A = [m2]k [x2,A]k

    or

  6. [m3]k = [m2]k ( [x2,A]k - x4,A)/(x3,A - x4,A)
  7. [m3,A]k=[m3]k x3,A
  8. [m3,B]k=[m3]k x3,B
  9. [m4]k= [m2]k - [m3]k
  10. [m4,A]k=[m4]k x4,A
  11. [m4,B]k=[m4]k x4,B
  12. [m5]k = pf [m4]k
  13. [m5,A]k=[m5]k x5,A
  14. [m5,B]k=[m5]k x5,B
  15. [m6]k = [m4]k - [m5]k
  16. [m6,A]k=[m6]k x6,A
  17. [m6,B]k=[m6]k x6,B
INITIAL CONDITIONS:

[m6,A]0 = [m6,B]0 = 0

BASIS: 1 min per cycle

Using Excel Spreadsheet, we get:

So, as we can see, the values have settled to a steady state value around k=70

To compare with the usual mass balance techniques,

Basis: m1 = 100 kg

Overall Balance:

m5 + m3 = m1
m5 x5,A + m3 x3,A = m1 x1,A

m5 = m1 (x1,A-x3,A)/(x5,A-x3,A)

= 100 ( 0.2 - 0.1)/( 0.8 - 0.1) = 14.286 m4 = m5/pf = 14.286/0.2 = 71.428
m6 = m4-m5 = 57.142
m2 = m1+m6 = 157.142
which are the same values obtained by the simulation.

This page is maintained by

Tomas Co
Associate Professor
Department of Chemical Engineering
Michigan Technological University
Houghton, MI 49930

(tbco@mtu.edu)

Latest revision 9/12/2002

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