Cost slope

Code Activity Time Cost ( ) Time cost ( ) Slope... 

Figure 7.9 The Xp parameter avoids steep slopes on the Fp curves, whereas minimum Fp does not. (Reprinted from Ahmad, Linnhoff, and Smith, Cost Optimum Heat Exchanger Networks II. Targets and Design for Detailed Capital Cost Models, Computers Chem, Engg., 7 751, 1990 with permission from Elsevier Science, Ltd.)...

Slide Conveyors Simple gravity slides and spiral chutes, while not technically conveyors, are widelv used with conveyor systems or as separate units for lowering materials from one floor to another. They are low in cost and require httle floor space if slopes are held at fairly steep angles. However, they must be used only after a careful study... 

FIG. 22-55 Typical capital-cost schematic for membrane equipment showing trade-off for membrane area and mechanical equipment. Lines shown are from families for parallel hues showing hmiting costs for membrane and for ancillary equipment. Abscissa Relative membrane area installed in a typical membrane process. Minimum capital cost is at 1.0. Ordinate Relative cost. Line with positive slope is total membrane cost. Line with negative slope is total ancillary equipment cost. Curve is total capital cost. Minimum cost is at 1.0. 

Capital Costs A typical medium-scale RO seawater plant might produce 0.25 mVs (6 MGD). For a plant with an open sea intake, seawater salinity of 38 g/1, and conversion of 45 percent, the overall cost woiild be 26.5 miUiou (1996). A capital breakdown is given in Table 22-18. Capital charges are site specific, and are sensitive to the salinity of the feed. A plant of this size would likely contain six trains. For seawater RO, the Best estimate for the slopes of the family of lines in Fig. 22-55 is —0.6 for the equipment and 0.95 for the membranes. Capital charges, shown in TaBle 22-19, usually dominate the overall economics the numbers presented are only an example. Seawater economics are based on Shields and Moch, Am. Desalination Assn. Conf. Monterey CA (1996). 

The most difficult feature of this method is that for each type of plant or plant product as well as for each type of equipment there is a break-point where the 0.6 no longer correlates the change in capacity. For small equipment or plants in reasonable pilot or semi-works size, the slope of the cost curve increases and the cost ratio is greater than 0.6, sometimes 0.75, 0.8 or 0.9. From several cost values for respective capacities a log-log plot of capacity versus cost will indicate the proper exponent by the slope of the resultant curve. Extrapolation beyond eight or ten fold is usually not too accurate. 

The criteria which would be most desirable for industrial application of a separation process involving a supercritical gas may be established by comparing Figs. 3IB, 3ID, and 32. The largest cost in such a process is likely to be that of gas compression. Therefore, the maximum separation possible of the two solvents should occur for the addition of a given amount of gas, and the total pressure required to dissolve this gas should be small. This is the case if the tie lines slope toward the 1-3 binary line and if the gas is readily soluble. In terms of the Margules parameters and Henry s constant, these favorable criteria are ... 

Program SIMCAL was expressly written to allow these sorts of what-if questions to be explored, with realistic intercepts, slopes, signal noise, digitizer characteristics, and economical factors specified, so one can get a feeling for the achieved precision and the costs this implies. 

Example 33 Assume that a simple measurement costs 20 currency units n measurements are performed for calibration and m for replicates of each of five unknown samples. Furthermore, the calibration series of n measurements must be paid for by the unknowns to be analyzed. The slope of the calibration line is > = 1.00 and the residual standard deviation is Sres = 3, cf. Refs. 75, 95. The n calibration concentrations will be evenly spaced between 50 and 150% of nominal, that is for n = 4 x, 50, 83, 117, 150. For an unknown corresponding to 130% of nominal, should be below 3.3 units, respectively < 3.3 = 10.89. What combination of n and m will provide the most economical solution Use Eq. (2.4) for S x and Eq. (2.18) for Vx-Solution since Sxx is a function of the x-values, and thus a function of n (e.g. n = 4 Sxx = 5578), solve the three equations in the given order for various combinations of n and m and tabulate the costs per result, c/5 then select the... 

E-F Toward the end of project life the rate of cash flow may tend to fall off, due to increased operating costs and falling sale volume and price, and the slope of the curve changes. 

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