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Algebra fractions

A rational algebraic fraction is the ratio of two polynomials. If the polyno-ntfal in the numerator is of a lower degree than that Of the denominator, or... [Pg.239]

Algebraic fractions are very similar to fractions in arithmetic. [Pg.171]

A rational function can also be considered as an empirical model. It is an algebraic fraction such that both the numerator and the denominator are polynomials ... [Pg.245]

It is perhaps fortunate that both versions lead to the same algebraic formulations, but we will imply a preference for the two-dimensional solution picture by expressing surface concentrations in terms of mole fractions. The adsorption process can be written as... [Pg.391]

An equation algebraically equivalent to Eq. XI-4 results if instead of site adsorption the surface region is regarded as an interfacial solution phase, much as in the treatment in Section III-7C. The condition is now that the (constant) volume of the interfacial solution is i = V + JV2V2, where V and Vi are the molar volumes of the solvent and solute, respectively. If the activities of the two components in the interfacial phase are replaced by the volume fractions, the result is... [Pg.393]

This model then leads us through a thicket of statistical and algebraic detail to the satisfying conclusion that going from small solute molecules to polymeric solutes only requires the replacement of mole fractions with volume fractions within the logarithms. Note that the mole fraction weighting factors are unaffected. [Pg.517]

Partial Fractions Rational functions are of the type f x)/g x) where /x) and g(x) are polynomial expressions of degrees m and n respectively. If the degree of/is higher than g, perform the algebraic division—the remainder will then be at least one degree less than the denominator. Consider the following types ... [Pg.446]

Sets of first-order rate equations are solvable by Laplace transform (Rodiguin and Rodiguina, Consecutive Chemical Reactions, Van Nostrand, 1964). The methods of linear algebra are applied to large sets of coupled first-order reactions by Wei and Prater Adv. Catal., 1.3, 203 [1962]). Reactions of petroleum fractions are examples of this type. [Pg.695]

Equation (8-14) shows that starts from 0 and builds up exponentially to a final concentration of Kcj. Note that to get Eq. (8-14), it was only necessaiy to solve the algebraic Eq. (8-12) and then find the inverse of C (s) in Table 8-1. The original differential equation was not solved directly. In general, techniques such as partial fraction expansion must be used to solve higher order differential equations with Laplace transforms. [Pg.720]

This formula may be used for any depth of liquid between zero and the full tank, provided the algebraic signs are obsei ved. If H is greater than R, sin Ot cos Ot will be negative and thus will add numerically to Ot/57.30. Table 10-64 gives liquid volume, for a partially filled horizontal cylinder, as a fraction of the total volume, for the dimensionless ratio H/D or H/2R. [Pg.1017]

A second shortcut allows the rate itself to be written by inspection. The net rate is written as the product of the rate at which the intermediate is produced and its success fraction. The latter represents that fraction, of the total rate constant for reactions of the intermediate, that leads to product. For the case at hand, the total rate constant for loss of R+ is fc i[X ] + t2[Y-]. But only the term 2[Y-] produces RY. Applying this prescription to the scheme gives, without any algebra, the same result derived before by a lengthier procedure ... [Pg.79]

This expression, however, is not correct. The correct one is obtained by eliminating the cyclical pair, k2k-2[X ], from the denominator. Doing so, one obtains Eq. (4-63). Since the success fraction method is so useful in general, it is advantageous to learn how to extend it to such cases rather than abandon it in favor of much tedious algebra. Once one learns to recognize the extraneous term as the one with the pair of cyclical rate constants, it is easy to apply a correction. Problem 4.5 illustrates this. The number of denominator terms is equal to the number of sequential steps in the reaction scheme, which is three in the scheme to which Eq. (4-63) applies. [Pg.81]

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... [Pg.61]

The highly interactive nature of the balance and equilibria equations for the distillation period are depicted in Fig. 3.66. An implicit, iterative algebraic loop is involved in the calculation of the boiling point temperature at each time interval. This involves guessing the temperature and calculating the sum of the partial pressures, or mole fractions. The condition required is that Zyi + yw = 1. The iterative loop for the bubble point calculation is represented by the five interconnected blocks in the lower right hand corner of Fig. 3.66. The model of Prenosil (1976) also included an efficiency term E for the steam heating, dependent on liquid depth L and bubble diameter D. [Pg.218]

The next question is how to find the partial fractions in Eq. (2-25). One of the techniques is the so-called Heaviside expansion, a fairly straightforward algebraic method. We will illustrate three important cases with respect to the roots of the polynomial in the denominator (1) distinct real roots, (2) complex conjugate roots, and (3) multiple (or repeated) roots. In a given problem, we can have a combination of any of the above. Yes, we need to know how to do them all. [Pg.18]

This is really an algebraic exercise in partial fractions. The answer hides in Table 2.1. [Pg.62]

Fourier transforms boxcar function 274 Cauchy function 276 convolution 272-273 Dirac delta function 277-279 Gaussian function 275-276 Lorentzian function 276-277 shah function 277-279 triangle function 275 fraction, rational algebraic 47 foil width at half maximum (FWHM) 55, 303... [Pg.205]

You may need to answer questions directly about the GCF or the LCM, or you will need to find these numbers in your work with fractions and algebra. [Pg.64]


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See also in sourсe #XX -- [ Pg.20 ]




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