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Partial Differential Equations (PDEs) 1Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE Daniel Baur ETH Zurich, Institut für Chemie- und Bioingenieurwissenschaften ETH Hönggerberg / HCI F128 – Zürich E-Mail: daniel.baur@chem.ethz.ch http://www.morbidelli-group.ethz.ch/education/index

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Partial Differential Equations Problem definition: In a partial differential equation (PDE), the solution depends on more than one independent variable, e.g. space and time The function is usually subject to both inital conditions and boundary conditions Examples Diffusion into semi-infinite slab: Tubular reactor with dispersion: 2Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Characterization of Second Order PDEs Second order PDEs take the general form where A, B and C are coefficients that may depend on t and z These PDEs fall in one of the following categories 1.B 2 – AC < 0: Elliptic PDE 2.B 2 – AC = 0: Parabolic PDE 3.B 2 – AC > 0: Hyperbolic PDE There are specialized solvers for some types of PDEs, hence knowing its category can be useful for solving a PDE 3Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Numerical Solution of PDEs In general, it can be very difficult to solve PDEs numerically One approach is to discretize all but one dimension of the solution; this way a system of ODEs is obtained that can be solved more easily Note that these ODE systems are usually very stiff There are different ways of discretizing a dimension, for example the finite differences method we saw earlier, or using arbitrary functions (polynomials, gaussians) Sophisticated algorithms refine the discretization in places where the solution is still inaccurate pdepe Matlab has a built-in solver for parabolic and elliptic PDEs in two dimensions, pdepe 4Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Example: Tubular reactor A tubular reactor with diffusion can be described with the following PDE model: 5Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE c in c out 0L Mass balance: Initial conditions: No discontinuity at outlet: Mass balance over inlet:

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Example: Tubular reactor Let’s take one step back: Local mass balance 6Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE c in c out 0L Convection Diffusion

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Example: Tubular reactor 7Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE Convection Diffusion

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Initial conditions and boundary conditions Implementing the initial conditions is straight forward, since every ODE requires an initial value (IVP) How do we implement the boundary conditions? One solution: Create two pseudo grid points outside the grid that guarantee that the boundary conditions are fulfilled 8Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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The discretized tubular reactor We end up with the following system of ODEs: With the «boundary conditions» 9Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Dimensionless tubular reactor We can cast the model into dimensionless form by defining Where Pe is the Peclet and Da is the Damköhler number 10Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE The numerical solution of a problem is usually much simpler if it is dimensionless (most variables will range from 0 to 1).

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Discretized dimensionless tubular reactor In dimensionless form, the equations read With the «boundary conditions» 11Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Assignment 1 1.Solve the dimensionless discretized model for the start-up of the tubular reactor for different values of Pe. Plot the concentration of A at the outlet against dimensionless time. Use Da = 1, N = 100, y start = 0, tSpan = [0,1] and assume a first order reaction n = 1. Plot the profiles for 20 different values of Pe between 10 -4 and 100, looping over Pevec = logspace(- 4, 2, 20); What do you observe? When does the reactor reach steady state, depending on the back-mixing (large Pe = low back- mixing, small Pe = strong back-mixing)? Use ode23s to solve the ODE system. In your ODE function, use a for loop going from 2 to N-1; Remember that the ODEs number 1 and N have different forms than the rest due to the «boundary conditions» 12Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Assignment 1 (continued) 2.What yields better steady-state conversion (1-y) for a first order reaction, a PFR (no back-mixing, Pe large) or a CSTR (infinite back-mixing, Pe small)? Assume steady state is reached when the change from one time point to the next is smaller than 0.1% (you might have to extend tSpan). What is the conversion (1-y)? 3.Consider a second order reaction n = 2 and plot the concentration profiles for Pe = 0.0001 and Pe = 100. How long does it take to reach steady state? What is the conversion? What is better for a second order reaction, a PFR (no back-mixing, Pe large) or a CSTR (infinite back-mixing, Pe small)? Which reaction order suffers more from choosing the wrong reactor? Calculate the ratios between the final conversions for the PFR and CSTR for first and second order reactions. 13Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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Assignment 1 (continued) 4.Now consider a reaction that is fast compared to the residence time, set Da = 5. Which reactor is better? 14Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE

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