Ch E 310 - Fall 10 - Lecture 26

Ch E 310 - Fall 10 - Lecture 26 - Lecture 26 December 7,...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture 26 – December 7, 2010 Agenda: Boundary Value Problems (solving ODEs) Finite Difference Methods Derivative (Neumann) Boundary Conditions Solving Non-linear ODEs HW 6 questions and help Course Evaluations (if you haven’t done this already)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Finite Difference Methods An alternative to the shooting method is to use finite difference approaches We looked at these previously for numerically approximating derivatives This method entails converting a linear ODE into a system of simultaneous algebraic equations Once formulated, we can use methods for linear systems to provide a solution The key difference in boundary value problems is that we divide the domain (space, time, etc.) into nodes and specify conditions at each node simultaneously This leads to a solution at each node and thus overall
Background image of page 2
Finite Difference Methods   T T h dx T d 2 2 0 Let’s revisit the heated rod example; the ODE is: The spatial domain is discretized into a series of nodes: The second derivative is depicted by a finite difference: 2 1 1 2 2 2 x T T T T d i i i K 200 T K 400 T K 300 T
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Forward Finite Difference Formulas
Background image of page 4
Backward Finite Difference Formulas
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Centered Finite Difference Formulas
Background image of page 6
Substituting the finite difference into the ODE, we get: Collecting terms yields: This can be written for any of the interior nodes along the length of the rod The first and last nodes ( T 0 and T n ) are subject to chosen boundary conditions (here, 300 and 400 K ) This leads to a system of n –1 linear algebraic equations and n –1 unknowns – we know how to solve this! Finite Difference Methods   0 2 2 1 1 i i i i T T h x T T T   T x h T T x h T i i i 2 1 2 1 2
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Use the finite difference method to solve the first heated rod example from Lecture 25 where x = 2 m From the finite difference form of the ODE, we can write an equation at each interior node i For node 1 we have: T 0 + (2 + 0.05×2 2 ) T 1 T 2 = 0.05×2 2 ×200 T 0 + 2.2 T 1 T 2 = 40 ( where T 0 = 300 K) 2.2 T 1 T 2 = 340 ( two unknowns )
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/21/2011 for the course CH E 310 taught by Professor Staff during the Spring '08 term at Iowa State.

Page1 / 22

Ch E 310 - Fall 10 - Lecture 26 - Lecture 26 December 7,...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online