PGE 310 - HW7 - Solution

PGE 310 - HW7 - Solution - The University of Texas at...

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1 The University of Texas at Austin PGE 310: Formulation and Solution in Geosystems Engineering Homework #7: System of Linear Eqns(Iterative Methods) - System of Nonlinear Eqns(Newton’s Method) By HosseinRoodi 1. System of Linear Equations: Iterative Methods (By HAND) The below sketch shows three reactors linked by pipes. The rate of transfer of chemicals through each pipe is equal to a flow rate ( Q , with units of cubic meters per second) multiplied by the concentration of the reactor from which the flow originates (c, with units of milligrams per cubic meter). If the system is at steady state, the transfer into each reactor will balance the transfer out . Writing mass-balance equations for the reactors one gets a system of 3 linear equations in matrix form: 12 1 2 3 120 20 450 80 100 0 40 80 130 250 cc c c c ! " !" # ! " ! a) Determine if the system is guaranteed to converge in iterative methods and if they are not, write them in a different order to guarantee its convergence. As can be seen, in all rows diagonal elements are greater than the summation of off diagonal elements. So, this matrix is guaranteed to converge in iterative method. 40 80 130 : 3 80 100 : 2 20 120 : 1 # $ $ $ Row Row Row b) Solve for the concentrations using an initial guess of c1=c2=c3=0 and the first two iterations using the following methods: Step 1: In all following iterative methods the first step is writing the equations in an explicit form: ) 80 40 250 ( 130 1 ) ( 100 80 ) 20 450 ( 120 1 2 1 3 1 2 2 1 c c c c c c c # # " " # " a. Jacobi Initial Guess: 0 ; 0 ; 0 3 2 1 " " " c c c 450 mg/s 250 mg/s Q33=130 Q13=40 Q12=80 Q23=80 Q21=20
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2 First Iteration: 9231 . 1 ) 0 80 0 40 250 ( 130 1 0 ) 0 ( 100 80 75 . 3 ) 0 20 450 ( 120 1 3 2 1 " % # % # " " " " % # " c c c ! 9231 . 1 130 250 0 75 . 3 120 450 3 2 1 " " " " " c c c Second Iteration: 0769 . 3 ) 0 80 75 . 3 40 250 ( 130 1 3 ) 75 . 3 ( 100 80 75 . 3 ) 0 20 450 ( 120 1 3 2 1 " % # % # " " " " % # " c c c ! 0769 . 3 13 40 3 75 . 3 120 450 3 2 1 " " " " " c c c b. Gauss Seidel First Iteration: 9231 . 4 ) 3 80 75 . 3 40 250 ( 130 1 3 ) 75 . 3 ( 100 80 75 . 3 ) 0 20 450 ( 120 1 3 2 1 " % # % # " " " " % # " c c c ! 9231 . 4 13 64 3 75 . 3 120 450 3 2 1 " " " " " c c c Second Iteration: 3231 . 5 ) 4 . 3 80 25 . 4 40 250 ( 130 1 4 . 3 ) 25 . 4 ( 100 80 25 . 4 ) 3 20 450 ( 120 1 3 2 1 " % # % # " " " " % # " c c c ! 3231 . 5 65 346 4 . 3 5 17 25 . 4 4 17 3 2 1 " " " " " " c c c c. Gauss Seidel with relaxation with 2 . 1 " & . First Iteration: 9662 . 5 ) 32 . 4 80 5 . 4 40 250 ( 130 1 6 . 3 ) 5 . 4 ( 100 80 75 . 3 ) 0 20 450 ( 120 1 3 2 1 " % # % # " " " " % # " c c c ! 1594 . 7 0 ) 2 . 1 1 ( 9662 . 5 2 . 1 ) 1 ( 32 . 4 0 ) 2 . 1 1 ( 6 . 3 2 . 1 ) 1 ( 5 . 4 0 ) 2 . 1 1 ( 75 . 3 2 . 1 ) 1 ( 3 3 3 2 2 2 1 1 1 " % ! # % " ! # " " % ! # % " ! # " " % ! # % " ! # " old new old new old new c c c c c c c c c Second Iteration: 4021 . 5 ) 4214 . 3 80 464 . 4 40 250 ( 130 1 5712 . 3 ) 464 . 4 ( 100 80 47 . 4 ) 32 . 4 20 450 ( 120 1 3 2 1 " % # % # " " " " % # " c c c !
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PGE 310 - HW7 - Solution - The University of Texas at...

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