Second Illustrative Example

This example is similar to the previous example in Section 3.4.1, however, there is another reactor in the reaction stage as shown in Fig. 3.12. The SSN remains the same and is shown here in Fig. 3.13. The data for this example is shown in Tables 3.5 and 3.6. 

The model was solved using GAMS DICOPT, with CLPEX as the MIP solver and CONOPT as the NLP solver. The computational results are shown in Table 3.7. The resulting plant requires only one reactor as shown in Fig. 3.14. The optimal capacities of the remaining units are 75 units for the mixer (Cl), 75 units for the reactor 

Unit Capacity range Suitability Processing time Capital cost 

State Storage capacity Initial amount Price 

MILP and MINLP models have been developed to take into account the PIS operational philosophy for assessing the efficacy of this philosophy and design, respectively. The MILP model is used to determine the effectiveness of the PIS operational philosophy by, firstly, solving the model with zero intermediate storage with and without the use of latent storage. In the illustrative example a 50% increase in the throughput was achieved. 

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The second illustrative example is a modified literature example. The example was originally presented by Kim and Smith (2004). The example involves 7 water using operations with three contaminants present in the system. The example was only solved considering a central storage vessel due to the fact that the schedule used by Kim and Smith was retained for this example and there are few direct reuse opportunities within the given schedule. Due to the schedule being known, the objective was to minimise effluent. 

Table 6.4 Starting and ending times of each process in the second illustrative example...

The second illustrative example deals with wastewater minimisation in an operation involving four processes. Wastewater produced from processes 1 and 2 contain the same single contaminant Cl, while wastewater produced from processes 3 and 4 contain multiple contaminants, namely, contaminants Cl, C2 and C3. Processes 1 and 2 cannot receive water contaminated with multiple contaminants, while processes 3 and 4 can receive any type of wastewater. Table 7.3 provides the required maximum inlet and outlet concentrations for each process as well as the mass load of each contaminant in each process. 

The model for the second illustrative example was formulated in GAMS 22.0 and solved using the DICOPT2 solution algorithm. The MIP solver used was CPLEX... 

The zero effluent synthesis formulation was applied to a second illustrative example. In the example the number of processing units and the size of the central storage vessel were not known. The resulting plant required only 3 processing units and no storage vessel. The resulting schedule produced 68% less effluent than the same operation without wastewater reuse. 

Task Apply the presented mathematical formulation to the first and second illustrative examples and verify the results presented in Figs. 8.3 and 8.4. 

Solution for Second Illustrative Example with Central Storage Only... 

Table 9.2 Data for second illustrative example using inherent storage...

Fig. 9.3 Resulting schedule for second illustrative example with central storage only (Gouws and Majozi, 2009)...

Solution to Second Illustrative Example with Inherent Storage and Central Storage... 

A note must be made on the solution time of the second case in the second illustrative example. The solution time is excessive, approximately 900 CPU seconds. The long solution time is due to the solution procedure used. The solution time for the MILP accounted for more than 99% of the total solution time. Shorter solution times might have been achieved if a different solution procedure had been followed, e.g. only partially linearising the MINLP. However, the final solution found is globally optimal, which justifies the usage of the solution procedure. 

Task Apply the mathematical formulation presented in this chapter to verify the results presented for the first and second illustrative examples. In what types of problems is the proposed approach less likely to yield good results ... 

A second illustrative example of the utility of TRPES and TRCIS for studying complex molecular photodissociation dynamics that involve multiple electronic state is the case of the weakly bound cis-planar C2V nitric oxide dimer [174], The weak (Do = 710cm-1) 1 A ground-state covalent bond is formed by the pairing of two singly occupied ji orbitals, one from each NO(X2II) monomer. The very intense UV absorption spectrum of the NO dimer appears... 

However, it is inadequate to believe that any mixture of bis-tridentate ligands with will produce triple-stranded helicates. We already mentioned in the introduction how some competing coimter-ions or solvent molecules may prevent the fixation of the third strand, thus leading to unsaturated complexes (Section 1.3, Figure 13B). A second illustrative example is shown in Figure 53, whereby the use of a 1-4-disubstituted phenyl spacer in L33 makes the output of the assembly process extremely sensitive to the stoichiometric k L33 ratio (Ronson et al., 2007). 

The study of the pressure effects on the conformational inversion of cyclohexane in different solvents represents the second illustrative example of the application of high resolution, high pressure FT NMR. 

The second illustrative example is the well-known voltage-driven separately excited DC motor that drives a mechanical load against an external moment (Fig. 4.10). Figure 4.11 shows a direct bond graph model. Like the previous example, this model also has two inputs and two outputs. That is, a transfer matrix H with four transfer functions Fij can be derived ... 

The second illustrative example is the well-known voltage-driven separately excited DC motor that drives a mechanical load against an external moment (Fig. 4.10). 

The first and second illustrative examples are based on profiles synthesised through the superposition of periodie sine wave funetions with different wavelengths or phases but with equal amplitudes. The FFT of sueh profiles simply eorresponds to the amplitudes for the given sine waves, with R being equal to the sum of the power spectrum. This means that with a eonstant number of amplitudes present in the speetrum, the wavelengths and phase-shifts of the different sine waves (eorrelated to the Fourier coeffieients) do not influenee the standardised R value. 

The second illustrative example refers to a three-dimensional system of Brownian particles interacting through a strongly repulsive and short-ranged, pair potential u(f) of the form... 

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