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Recovery lbmole

Feed (in 100 fresh batches) lbmol

' Recovery

Total Campaign Time, hr

Total Time for fresh feed processing (100 batches) =743.0

Total Time for off_cut 2 reprocessing (2.86 batches) A = 54.83

Total Time for off_cut 3 reprocessing (3.85 batches) = 36.0

Total Time for off_cut 4 reprocessing (2.63 batches) = 33.33

Total Time for off_cut 5 reprocessing (1.92 batches) = 13.06

Total = 880.02

Total Steam Requirement = 245.26 MMBtu, Cost = $7.8/MMBtu

Total Raw Material (Fresh Feed) = 100x78 = 7800 lbmol, Cost = $2.98/lbmol

Total Product Value = $46,909.1

Total Raw Material cost = $23,244.0 Total Operating Cost (O.C.) =$ 1,913.02

Profit = (IDistillate products x value - raw material cost) - O.C (steam) = $21,752.08/yr

A 2.86 batches means 2 batches of full capacity + 1 batch at 86% capacity

6.5. Optimisation Problem Formulation- Farhat et al.

For single separation duty Farhat et al. (1990) presented multiple criteria decisionmaking (MCDM) NLP based problem formulations for multiperiod optimisation. This involves either maximisation (Problem 1) of specified products (main-cuts) or minimisation (Problem 2) of unspecified products (off-cuts) subject to interior point constraints. These two optimisation problems are described below.

6.5.1. Problem 1 - Maximisation of Main-cuts

For a linear reflux ratio policy in each of the cuts, the problem OP1 can be written as:

where, /?is a binary variable (0,1) and determines the existence of a main-cut; Pk is the amount of main-cut with k specifying the key-component in the cut; j relates the time interval for the main-cut; n is the number of components.

The amount of main-cut in t e [tj.,, tj] can be given by:

OP1:

subject to gk = XSPECk (main-cut purity)

where, V is the vapour boilup rate; Rj is the reflux ratio; t is the time; a and b are constant parameters; A, B and T are vectors defined as:

The last production term is isolated from the general summation in Equation 6.7, because it can be expressed from the overall mass balance:

where S is the amount of off-cut.

6.5.2. Problem 2 - Minimisation of Off-cuts

For a linear reflux ratio policy in each of the cuts, the problem OP2 can be written as:

Min lYpjSA

The amount of off-cut in t z [tj.j, tj\ can be given by:

Farhat et al. (1990) used augmented Lagrangian method to solve the optimisation problem presented above. See the original reference for further details.

6.5.3. Example

Farhat et al. (1990) considered the separation of a ternary mixture with an objective to minimise two off-cuts. The operation sequence considered is shown as an STN in Figure 6.14. There are two main-cuts (P, ~ Df, P2 ~ D2), two off-cuts (S, ~ R,; S2 ~ R2) and the final specified bottom product (P3 ~ B4). Farhat et al. solved the optimisation problem OP2 which is equivalent to solving OP1. The input data for the problem is given in Table 6.11.

Main-cut 1

Off-cut 1

Main-cut 2 Off-cut 2

Main-cut 1

Off-cut 1

Main-cut 2 Off-cut 2

Initial Charge

Bottom Residue or Product

Initial Charge

Bottom Residue or Product

Figure 6.14. Operation Sequence Considered by Farhat et al. (1990)

Table 6.11. Input Data for Farhat et al. (1990) Problem

No. of Plates

(including a reboiler and a total condenser)

Total Fresh Feed, B0, mol

Feed Composition, xB0, molefraction

Column Holdup, mol

Vapour Boilup, mol/hr

Column Pressure, torr

= negligible = 110

Purity of Main-cut 1, XSPEC, (~ xlDl), molefraction = 0.95 Purity of Main-cut 2, XSPEC2 (~ x%2 X molefraction = 0.925 Purity of Bottom Product, XSPEC3 (~ x3* ), molefraction = 0.95

Total Batch Time, hr (fixed)

Farhat et al. considered both optimal constant and optimal linear reflux ratio for this problem (Figure 6.15). Final time was fixed and 4 time intervals were considered. The length of each time interval was also optimised. Table 6.12 presents the summary of the optimisation results using both options of reflux ratio profiles. A significant gain of 10.7% in specified products can be observed between the optimal linear reflux policy and the optimal constant reflux policy.

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