hmwk_7_2010f_Project

# hmwk_7_2010f_Project - UNIVERSITY OF NOTRE DAME Department...

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UNIVERSITY OF NOTRE DAME Department of Civil Engineering and Geological Sciences CE 30125 November 24, 2010 J.J. Westerink Due: November 30, 2010 Homework Set # 7 - Group Project 1 We will consider a pollutant advecting and diffusing down a river that is 500 m long. Assume that the pollutant is distributed as a Gaussian distribution with a standard deviation equal to m and an initial location equal to x 0 =100 m. We must solve the advection-diffusion equation to solve this problem. i.c.’s b.c.’s The exact solution to this problem is: We want to find out what the distribution looks like after 250 seconds from the initial time. Let us solve this problem using a second order accurate central difference approximation in space and a first order accurate forward approximation in time (Explicit in time, central in space solution), a first order backward approxima- tion in time (Implicit in time, central in space solution) and a second order accurate Crank-Nicolson approxi- mation in time (Central in time at the half step, central in space solution). We will apply the exact solution to compute the initial and boundary conditions. Corbitt has been kind enough to code this up for us in Matlab in ADDIF.m The code input parameters are grid size , time step , the standard deviation of the Gaussian distribution , the current speed , the diffusion coefficient , the implicit fraction (equal to 0 for and explicit scheme, 0.5 for a Crank-Nicolson scheme, and 1.0 for a fully implicit scheme). Note that the code also computes the Courant number, , a measure of the how far information propagates across a space step during a time step; the Peclet number, , a measure of the ratio of advection to diffusion; the simulation time on your computer; the L 2 discrete error norm, ; and the error norm, σ 5 = C t ------ U C x + D 2 C x 2 --------- = 0 x 500 ≤≤ Cxt , () xx 0 2 2 σ 2 × -------------------- ⎝⎠ ⎜⎟ ⎛⎞ exp = 0 x 500 C 0 t , σ σ 2 2 Dt × × + ------------------------------------- 0 x 0 Ut × 2 2 σ 2 4 × + × -------------------------------------- exp × = C 500 t , σ σ 2 2 × × + 500 x 0 × 2 2 σ 2 4 × + × -------------------------------------------- exp × = , σ

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## This note was uploaded on 03/02/2012 for the course CE 30125 taught by Professor Westerink,j during the Fall '08 term at Notre Dame.

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hmwk_7_2010f_Project - UNIVERSITY OF NOTRE DAME Department...

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