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Sequencing Separation Trains. CHEN 4460 – Process Synthesis, Simulation and Optimization Dr. Mario Richard Eden Department of Chemical Engineering Auburn University Lab Lecture No. 2 – Sequencing of Separation Trains September 18, 2012
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Sequencing Separation Trains CHEN 4460 – Process Synthesis, Simulation and Optimization Dr. Mario Richard EdenDepartment of Chemical EngineeringAuburn University Lab Lecture No. 2 – Sequencing of Separation Trains September 18, 2012 Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel
Example 1: Four Components The feed to a separation process consists of the following species:
Example 1: Four Components • It is desired to separate this mixture into essentially pure species. The use of two types of separators is to be explored: • Ordinary distillation • Extractive distillation with furfural • The separation orderings are: Species number
Example 1: Four Components • Determine the number of possible separation sequences. • What splits would you forbid so as to reduce greatly the number of possible sequences?
Example 1: Solution • To determine the number of possible sequences , use Eqs. (8.9) and (A): • Combining Eqs. (8.9) and (A), the number of possible sequences is: (8.9) (A)
Example 1: Solution • For ordinary distillation, the approximate relative volatilities, a between adjacent species are as follows at 150 F:
C2 C3 1-C4= n-C4 C3 1-C4= n-C4 1-C4 n-C4= Example 1: Solution From Fig 5.3, forbid use of extractive distillation when a>2 for ordinary distillation. Thus, forbid the use of extractive distillation for the splits: C2/C3 C3/1-C4= C3/nC4 (a=2.45x1.18=2.89) Thus, the only splits to be considered for extractive distillation are those involving nC4/1-C4= and 1-C4=/furfural
C2 C3 1-C4= n-C4 C3 1-C4= n-C4 n-C4 1-C4= Furfural 1-C4= Furfural C2 C3 C2 C3 n-C4= 1-C4 Furfural n-C4 1-C4= Furfural 1-C4= Furfural Example 1: Solution
Best Sequence using Heuristics • The following guidelines are often used to reduce the number of OD sequences that need to be studied in detail: • Remove thermally unstable, corrosive, or chemically reactive components early in the sequence. • Remove final products one-by-one as distillates (the direct sequence). • Sequence separation points to remove, early in the sequence, those components of greatest molar percentage in the feed. • Sequence separation points in the order of decreasing relative volatility so that the most difficult splits are made in the absence of other components. • Sequence separation points to leave last those separations that give the highest purity products. • Sequence separation points that favor near equimolar amounts of distillate and bottoms in each column. The reboiler duty should not be excessive.
Example 2: Three Components • A mixture of benzene, toluene and biphenyl needs to be separated. The boiling points of the three components are: 80.1 C, 110.8 C, and 254.9 C, respectively. • Suggest possible separation sequences for this mixture. Select the most appropriate sequence for each of the two cases in the table below. • Simulate the two sequences you suggest using Aspen Plus, to produce each of the three components at 99% purity.
Example 2: Solution • Case 1: Direct Sequence Total cooling: 10.7 MM kcal/hr Total heating: 30.0 MM kcal/hr
Sequence separation points to remove, early in the sequence, those components of greatest molar percentage in the feed. • Sequence separation points that favor near equimolar amounts of distillate and bottoms in each column. The reboiler duty should not be excessive. Example 2: Solution • Case 1: Indirect Sequence Total cooling: 4.4 MM kcal/hr (Direct sequence: 10.7 MM kcal/hr) Total heating: 27.0 MM kcal/hr (Direct sequence: 30 MM kcal/hr)
Example 2: Solution • Case 2: Direct Sequence Total cooling: 1055 MM kcal/hr Total heating: 1064 MM kcal/hr
Sequence separation points to remove, early in the sequence, those components of greatest molar percentage in the feed. • Sequence separation points that favor near equimolar amounts of distillate and bottoms in each column. The reboiler duty should not be excessive. Example 2: Solution • Case 2: Indirect Sequence Total cooling: 14.5 MM kcal/hr (Direct sequence: 1055 MM kcal/hr) Total heating: 25 MM kcal/hr (Direct sequence: 1064 MM kcal/hr)