Transforms Laplace

3 METHODS FOR SOLVING KINETIC EQUATIONS 4.3.1 Laplace transforms 

Classical methods of integration are used to solve many of the linear differential equations that describe the kinetic behaviour of chemical reactions. However, the methods of solving 

A kernel frequently used in chemical kinetics is K(p, t) = q p . This kernel, integrated between the limits 0 and co, corresponds to the Laplace transform. The great value of this transform stems from the fact that the exponential dependence as a function of time is similar to that encountered in the integrals needed in chemical kinetics, while in the applications of the Laplace transform, with integration between the limits t = 0 and f = the transformed function only depends on p. Thus, the transform F(p) of the function/(O using the Laplace treatment is defined by the integral 

For example, for the function f(t) = exp (-at) in which a is constant, its Laplace transform is obtained using 

The application of Laplace transforms to the analytical solution of complicated differential kinetic laws is simplified significantly on the basis of three of its fundamental 

Accordingly, a new function f s) is defined which depends on 5 as independent variable. This transformation may also be written as 

A simple example of Laplace transformation is given now for the function e  

In order to eompute the time response of a dynamie system, it is neeessary to solve the differential equations (system mathematieal model) for given inputs. There are a number of analytieal and numerieal teehniques available to do this, but the one favoured by eontrol engineers is the use of the Laplaee transform. 

The Laplaee transform of a funetion of time f t) is given by the integral 

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Integral-Transform Method A number of integral transforms are used in the solution of differential equations. Only one, the Laplace transform, will be discussed here [for others, see Integral Transforms (Operational Methods) ]. The one-sided Laplace transform indicated by L[f t)] is defined by the equation L[f t)] = /(O dt. It has... 

Example The equation 3c/3f = D(3 c/3a. ) represents the diffusion in a semi-infinite medium, a. > 0. Under the boundary conditions c(0, t) = Cq, c(x, 0) = 0 find a solution of the diffusion equation. By taking the Laplace transform of both sides with respect to t,... 

Equations of Convolution Type The equation u x) = f x) + X K(x — t)u(t) dt is a special case of the linear integral equation of the second land of Volterra type. The integral part is the convolution integral discussed under Integral Transforms (Operational Methods) so the solution can be accomplished by Laplace transforms L[u x)] = E[f x)] + XL[u x)]LIK x)] or... 

Sufficient Conditions for the Existence of Laplace Transform Suppose/ is a function which is (1) piecewise continuous on eveiy finite intei val 0 < t exponential growth at infinity, and (3) Jo l/t)l dt exist (finite) for every finite 6 > 0. Then the Laplace transform of/exists for all complex numbers. s with sufficiently large real part. 

A short table (Table 3-1) of very common Laplace transforms and inverse transforms follows. The references include more detailed tables. NOTE F(/i -1- 1) = Iq x e dx (gamma function) /(f) = Bessel function of the first land of order n. 

Other applications of Laplace transforms are given under Differential Equations. ... 

The z -transform can also be used to solve difference equations, just like the Laplace transform can be used to solve differential equations. 

Equations readily solvable by Laplace transforms. For example ... 

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. 

This is the equation for a plug flow reactor. It can be derived directly from the rate equations with the aid of Laplace transforms. The sequences of second-order reactions of Figs. 7-5n and 7-5c required numerical integrations. 

To illustrate how Laplace transforms work, consider the problem of solving Eq. (8-2), subjec t to the initial condition that = 0 at t = 0, and Cj is constant. If were not initially zero, one would define a deviation variable between and its initial value (c — Cq). Then the transfer function would be developed using this deviation variable. Taking the Laplace transform of both sides of Eq. (8-2) gives ... 

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. 

The term in parentheses in Eq. (8-17) is zero at steady state and thus it can be dropped. Next the Laplace transform is taken, and the resulting algebraic equation solved. Denoting X s) as the Laplace transform of and X,(.s) as the transform of 4, the final transfer Function can be written as ... 

Using block diagram algebra and Laplace transform variables, the controlled variable C(.s) is given by... 

The combination of reac tor elements is facihtated by the concept of transfer functions. By this means the Laplace transform can be found for the overall model, and the residence time distribution can be found after inversion. Finally, the chemical conversion in the model can be developed with the segregation and maximum mixed models. 

Elimination of Ci and C3 from these equations will result in the desired relation between inlet Cj and outlet Co concentrations, although not in an exphcit form except for zero or first-order reactions. Alternatively, the Laplace transform could be found, inverted and used to evaluate segregated or max mixed conversions that are defined later. Inversion of a transform hke that of Fig. 23-8 is facilitated after replacing the exponential by some ratio of polynomials, a Pade approximation, as explained in books on hnear control theory. Numerical inversion is always possible. 

For operation with an inert tracer, the material balances are conveniently handled as Laplace transforms. For a stirred tank, the differential equation... 

Differential equations and their solutions will be stated for the elementary models with the main lands of inputs. Since the ODEs are linear, solutions by Laplace transforms are feasible. 

The Laplace transform technique also allows the reduction of the partial differential equation in two variables to one of a single variable In the present case. 

Table 3.1 gives further Laplace transforms of common functions (called Laplace transform pairs). 

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