Algebra conversion

Relatively simple algebraic conversion will bring us to so-called Master Resolution Equation ... 

Keep this value of A/tgas in mind because it appears in the algebraic conversion that follows. The reaction quotient based on concentrations is... 

Results for this example are given in Table 9.2. The table shows that the implicit method gives better results than the explicit method of the same order. Generally this is true. However, implicit methods have an inherent disadvantage in requiring that an algebraic conversion to an explicit... 

Empirical Equations Tabular (C,t) data are easier to use when put in the form of an algebraic equation. Then necessary integrals and derivatives can be formed most readily and accurately. The calculation of chemical conversions by such mechanisms as segregation, maximum mixedness, or dispersion also is easier with data in the form of equations. 

When using a conversion factor, we treat the units just like algebraic quantities they are multiplied or canceled in the normal way. 

When using a conversion factor, we treat the units just like algebraic quantities they are multiplied or canceled in the normal way. Thus, the units in the denominator of the conversion factor cancel the units in the original data, leaving the units in the numerator of the conversion factor. 

This messy result apparently requires knowledge of five parameters k, (A )o> Poo, and po- However, conversion to dimensionless variables usually reduces the number of parameters. In this case, set Y = Na/(Na)o (the fraction unreacted) and r = t/thatch (fractional batch time). Then algebra gives... 

Since 7 , is not an experimentally measurable quantity, it is useful to insert the solution for Ts (from the Clausius-Clapeyron equation) and solve for W h as an explicit function of RH0 and RHC. VanCampen et al. showed (using sample algebraic approximations and conversion factors) that substituting for Ts in Eq. (35) gives the useful solution... 

Determine the reactor size requirements for cascades composed of one, two, and three identical CSTR s. Use an algebraic approach and assume isothermal operation at 25 °C where the reaction rate constant is equal to 9.92 m3/ kmole-ksec. Reactant concentrations in the feed are equal to 0.08 kmole/m3. The liquid feed rate is equal to 0.278 m3/ksec. The desired degree of conversion is equal to 87.5%. 

For other reaction networks, a similar set of equations may be developed, with the kinetics terms adapted to account for each reaction occurring. To determine the conversion and selectivity for a given bed depth, Ljh equations 23.4-11 and -14 are numerically integrated from x = 0 to x = Lfl, with simultaneous solution of the algebraic expressions in 23.4-12, -13, -15, and -16. The following example illustrates the approach for a series network. 

This procedure will result in the best conversion for an overall specified residence time and the proportions of the residence time in each stage. The required algebra, however, is discouraging. 

The simplest case of this parameter estimation problem results if all state variables jfj(t) and their derivatives xs(t) are measured directly. Then the estimation problem involves only r algebraic equations. On the other hand, if the derivatives are not available by direct measurement, we need to use the integrated forms, which again yield a system of algebraic equations. In a study of a chemical reaction, for example, y might be the conversion and the independent variables might be the time of reaction, temperature, and pressure. In addition to quantitative variables we could also include qualitative variables as the type of catalyst. 

There are good reasons for choosing to carry out certain operations in the Laplace domain rather than performing equivalent operations in the time domain. In particular, integration of a function with respect to time is equivalent in the Laplace domain to division by the Laplace variable s. Conversely, differentiation corresponds to multiplication by s. This latter property enables differential equations to be Laplace transformed and then solved by algebraic means. These Laplace domain operations are all more simple than their time domain counterparts. In addition convolution in the time domain is equivcdent to multiplication in the Laplace domain. Formally, this may be represented by eqn. (24), the left-hand side of which is termed the convolution integral. 

These equations are significantly more complicated to solve than those for constant density. If we specify the reactor volume and must calculate the conversion, for second-order kinetics we have to solve a cubic polynomial for the CSTR and a transcendental equation for the PFTR In principle, the problems are similar to the same problems with constant density, but the algebra is more comphcated. Because we want to illustrate the principles of chemical reactors in this book without becoming lost in the calculations, we win usually assume constant density in most of our development and in problems. 

A conversion factor simply uses your knowledge of the relationships between units to convert from one unit to another. For example, if you know that there are 2.54 centimeters in every inch (or 2.2 pounds in every kilogram or 101.3 kilopascals in every atmosphere), then converting between those units becomes simple algebra. Peruse Table 2-3 for some useful conversion factors. And remember If you know the relationship between any two units, you can build your own conversion factor to move between those units. 

Recall another algebra rule You can multiply any quantity by 1, and you ll always get back the original quantity. Now look closely at the conversion factors in the example 2.2 lb and 1 kg are exactly the same thing Multiplying by 2.2 lb/1 kg or by 1 kg/2.2 lb is really no different from multiplying by 1. 

For a CSTR the stationary-state relationship is given by the solution of an algebraic equation for the reaction-diffusion system we still have a (non-linear) differential equation, albeit ordinary rather than partial as in eqn (9.14). The stationary-state profile can be determined by standard numerical methods once the two parameters D and / have been specified. Figure 9.3 shows two typical profiles for two different values of )(0.1157 and 0.0633) with / = 0.04. In the upper profile, the stationary-state reactant concentration is close to unity across the whole reaction zone, reflecting only low extents of reaction. The profile has a minimum exactly at the centre of the reaction zone p = 0 and is symmetric about this central line. This symmetry with the central minimum is a feature of all the profiles computed for the class A geometries with these symmetric boundary conditions. With the lower diffusion coefficient, D = 0.0633, much greater extents of conversion—in excess of 50 per cent—are possible in the stationary state. 

Conversely let s assume that S is integral and that p t) p(t) is independent of sc S. By proposition (5.1) it suffices to show that p F(h) is locally free of rank p(h) for all h 0. The question being local in the neighborhood of any point s c S, we may assume that S - Spec(A), where A is a local noetherian integral k-algebra with residue field k(s) and quotient field K. Let... 

A second solution avoids the algebra in the solution by using the concept of the conversion factor and setting it up so that units cancel properly. 

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