Day4 - CE 561 Lecture Notes Fall 2009 Day 4 Laplace...

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CE 561 Lecture Notes Fall 2009 p. 1 of 14 Day 4: Laplace transform methods for solving rate equations; Stochastic (kinetic Monte Carlo) methods for modeling reacting systems Preliminary comments: Matrix methods discussed last week work only for linear differential equations – which means that they work only for 0 th or 1 st order reactions (any other order leads to non-linear terms in the rate equations), and of course only work for systems at constant temperature (where the rate “constants” are constant – the Arrhenius temperature dependence introduces even stronger nonlinearities in non-isothermal systems). The same is true for Laplace transform methods of solving rate equations, which are described below. We can occasionally, through luck or cleverness, find an analytical solution to non-linear ODE’s, but there is not a general procedure for doing so like there is for linear equations. A common practice in experimental chemical kinetics is to perform experiments under pseudo- first-order conditions. This means that all reactants except one are supplied in large excess, so that all but one concentration effectively remain constant, i.e. for A + B C + D with r = kC A C B , one might do experiments with a large excess of B so that C B is approximately constant. Then the reaction is pseudo-first-order with r=k eff C A , where the constant concentration of B is lumped into the effective rate constant ( k eff =kC B ) . If this cannot be done, and there is no “clever” solution to the non-linear equations, then the equations must be solved numerically. Laplace transform methods for solution of rate equations An alternative method for integrating the rate equations (which are a system of 1 st order ODE’s) is to use the Laplace transform. We may remember from process control class that the Laplace transform is an integral transform that converts differentiation into multiplication. The definition of the transform is: F s f t e f t dt st ( ) ( ) ( ) = = z L 0 The inverse transform is f t F s i e F s ds - st i i ( ) ( ) ( ) = = − ∞ + ∞ z L 1 1 2 π γ This is often called the complex inversion formula . We generally look up both the transform and inverse transform in tables (or use Mathematica or Maple). It is relatively easy to do the transform using the formula above, but the inverse transform is more difficult. The inverse transform requires taking an integral over a contour in the complex plane. This requires some knowledge of complex analysis, and we won’t cover it further. If we are using the transform to solve ODE’s, and we use tables to find the transform and inverse transform, then the Laplace transform method simply becomes a convenient way of tabulating solutions of ordinary
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CE 561 Lecture Notes Fall 2009 p. 2 of 14 differential equations. Applied in that way, this method is simply a systematic means of looking up the known answer to the problem.
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Day4 - CE 561 Lecture Notes Fall 2009 Day 4 Laplace...

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