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Dr Saad Al-Shahrani

SEPARATION POWER. The degree of separation which can be obtained with any particular separation process is indicated by “ the separation power or separation factor”.

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Dr Saad Al-Shahrani

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  1. SEPARATION POWER • The degree of separation which can be obtained with any particular separation process is indicated by “ the separation power or separation factor”. • The objective of a separation device is to produce products of differing composition. So it is logical to define the separation factor in terms of product composition. ChE 334: Separation Processes Dr Saad Al-Shahrani

  2. Product P1 Xi,1, Xj,1 (liq.) Separation device Feed Components (i, j) (e.g L-L Extraction) Product P2 Xi,2, Xj,2 (liq.) COMPONENT RECOVERIES ANDPRODUCT PURITIES Where C= Concentration (mole fraction, mass fraction or mole/ vol., mass/ vol.) ChE 334: Separation Processes Dr Saad Al-Shahrani

  3. P1 yi,, yj, (vap.) Separation device Feed Components i,j (e.g Distillation or evaporation) P2 Xi,, Xj, (liq.) COMPONENT RECOVERIES ANDPRODUCT PURITIES ChE 334: Separation Processes Dr Saad Al-Shahrani

  4. If >1: component i in product (1) more than component j and component j in product (2) more than component i. • If <1: component j in product (1) more than component i and component i in product (2) more than component j. • If =1: No separation takes place between i and j. COMPONENT RECOVERIES ANDPRODUCT PURITIES ChE 334: Separation Processes Dr Saad Al-Shahrani

  5. COMPONENT RECOVERIES ANDPRODUCT PURITIES Where SFi, SFj = split fraction of component i and j respectively. ChE 334: Separation Processes Dr Saad Al-Shahrani

  6. COMPONENT RECOVERIES ANDPRODUCT PURITIES Example Referring to example 1. Key component split Column (separator) Separation factor, SPi,j nC4H10/ iC5H12 C1 C3H8/ iC4H10 C2 iC4H10/ nC4H10 C3 ChE 334: Separation Processes Dr Saad Al-Shahrani

  7. COMPONENT RECOVERIES ANDPRODUCT PURITIES Example: A feed, F, of 100 kmol/h of air containing 21 mol% O2 and 79 mol% N2 is to be partially separated by a membrane unit according to each of the following four sets of specifications. For each case, compute the amounts, in kmol/h, and compositions, in mol%, of the two products (retentate, R, and permeate, P). The membrane is more permeable to the oxygen. Case 1:50% recovery of O2 to the permeate and 87.5% recovery of N2 to the retentate. Case 2:50% recovery of O2 to the permeate and 50 mol% purity of O2 in the permeate. Case 3:85 mol% purity of N2 in the retentate and 50 mol% purity of O2 in the permeate. Case 4:85 mol% purity of N2 in the retentate and a split ratio of O2 in the permeate to the retentate equal to 1.1. ChE 334: Separation Processes Dr Saad Al-Shahrani

  8. 50% recovery of O2 to the permeate and 87.5% recovery of N2 to the retentate F Permeate (P) Membrane ? O2 fast ? ? N2 ? slow Retentate (R) COMPONENT RECOVERIES ANDPRODUCT PURITIES Solution Th feed is: Case 1: This is the simplest case to calculate because two recoveries are given: ChE 334: Separation Processes Dr Saad Al-Shahrani

  9. F Permeate (P) ? Membrane unit ? ? ? Retentate (R) COMPONENT RECOVERIES ANDPRODUCT PURITIES Case 2: With the recovery for O2 given, calculate its distribution into the two products: 50% recovery of O2 to the permeate and 50 mol% purity of O2 in the permeate Using the purity of O2 in the permeate, the total permeate is: By a total permeate material balance, By an overall N2 material balance, ChE 334: Separation Processes Dr Saad Al-Shahrani

  10. F Permeate (P) ? Membrane unit ? ? ? Retentate (R) COMPONENT RECOVERIES ANDPRODUCT PURITIES Case 3: With two purities given, write two simultaneous material balance equations, one for each component, in terms of the total retentate and total permeate. For nitrogen, with a fractional purity of 1.00 − 0.50 = 0.50 in the permeate, 85 mol% purity of N2 in the retentate and 50 mol% purity of O2 in the permeate For oxygen, with a fractional purity of 1.00 − 0.85 = 0.15 in the retentate, Solving (1) and (2) simultaneously for the total products gives ChE 334: Separation Processes Dr Saad Al-Shahrani

  11. F Permeate (P) ? Membrane unit ? ? ? Retentate (R) COMPONENT RECOVERIES ANDPRODUCT PURITIES Therefore, the component flow rates are 85 mol% purity of N2 in the retentate and a split ratio of O2 in the permeate to the retentate equal to 1.1. Case 4: First compute the O2 flow rates using the split ratio and an overall O2 material balance, ChE 334: Separation Processes Dr Saad Al-Shahrani

  12. COMPONENT RECOVERIES ANDPRODUCT PURITIES Solving these two equations simultaneously gives Since the retentate contains 85 mol% N2 and, therefore, 15 mol% O2, the flow rates for the N2 are ChE 334: Separation Processes Dr Saad Al-Shahrani

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