pset_1_sol - 2.29: Numerical Fluid Mechanics Solution of...

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2.29: Numerical Fluid Mechanics Solution of Problem Set 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.29 NUMERICAL FLUID MECHANICS— SPRING 2007 Problem Set 1 Posted 02/08/07, due Thursday 4 p.m. 02/23/07, from Lecture 1 to 3 Problem 1.1 (40 Points) Solve the below problems from “Chapara and Canale” textbook: 1.11 2.13, 2.15: Print the program and the result 3.1, 3.7, 3.9 4.5, 4.8, 4.15, 4.19 Solution of Textbook Problem 1.11: a) Note with positive “v” as upward velocity: dv dt = ! g ( x ) ! c m v = ! ( g 0 R 2 ( R + x ) 2 + c m v ) provided that v = dx dt > 0 b) For negligible drag force: dv dt = ! g 0 R 2 ( R + x ) 2 dv dx dx dt = ! g 0 R 2 ( R + x ) 2 dv dx v = ! g 0 R 2 ( R + x ) 2 dv dx v = ! g 0 R 2 ( R + x ) 2 c) 1
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2.29: Numerical Fluid Mechanics Solution of Problem Set 1 vdv v 0 v ! = " g 0 R 2 ( R + x ) 2 dx x 0 x ! 1 2 ( v 2 ! v 0 2 ) = g 0 R 2 ( 1 R + x ! 1 R + x 0 ) 1 2 ( v 2 ! v 0 2 ) = ! g 0 R 2 x ! x 0 ( R + x )( R + x 0 ) x 0 = 0 ! v 2 ( x ) = v 0 2 " 2 g 0 R x R + x d) The below plot is produced by attached file C2p29_PSET1_1p11.m; accordingly Euler’s solution stands on top of analytical solution due to: dv dx = ! g 0 R 2 v ( R + x ) 2 < v ( x ): downward convex function . BTW, as we know and we expected the Euler’s solution is only accurate for small x. 2
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2.29: Numerical Fluid Mechanics Solution of Problem Set 1 Solution of Textbook Problem 2.13: Look at C2p29_PSET1_2p13.m file. Solution of Textbook Problem 2.15: Look at C2p29_PSET1_2p15.m file. Note that a logical switch exist which checks whether within each round, any numbers are swapped or not. If not, then the numbers are already sorted and an early termination can be done. Also note after every round, we reduce the size of sort vector by one. This is because the last part of array is already sorted. Solution of Textbook Problem 3.1: a) (101101) 2 = 1 ! 2 5 + 0 ! 2 4 + 1 ! 2 3 + 1 ! 2 2 + 0 ! 2 1 + 1 ! 2 0 (101101) 2 = 1 ! 2 5 + 1 ! 2 3 + 1 ! 2 2 + 1 ! 2 0 = 45 b) (101.101) 2 = 1 ! 2 2 + 0 ! 2 1 + 1 ! 2 0 + 1 ! 2 " 1 + 0 ! 2 " 2 + 1 ! 2 " 3 (101.101) 2 = (101101) 2 ! 2 " 3 = 45 8 = 5.625 c) cos ! 3 = 0.5 (0.01101) 2 = 13 32 ! 0.4062 Solution of Textbook Problem 3.7: Look at C2p29_PSET1_3p7.m file. Note that chopping should be done at any step including the summation and the powers. For example if chop3 is the function for this work then for case a and b: 3
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2.29: Numerical Fluid Mechanics Solution of Problem Set 1 f_a=chop3(chop3(chop3(chop3(x*chop3(x*x))-chop3(7*chop3(x^2)))+chop3(8*x))-0.35); f_b=chop3(chop3(chop3(chop3(chop3(x-7)*x)+8)*x)-0.35); Note that in general chop ( x 3 ) ! chop ( x " chop ( x 2 )) or chop ( a + b + c ) ! chop ( chop ( a + b ) + c ) . For example: x = 1.37, x 2 = 1.8769, x 2 = 2.571353 chop 3( x 3 ) = 2.57 chop 3( x 2 ) = 1.88 What we really get for x 3 : chop 3( chop 3( x 2 ) ! x ) = chop 3(1.88 ! 1.37) = chop 3(2.5756) = 2.58 Apparently error for case “a” is higher (0.007 compared to 0.004). This can be due to the fact that we
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pset_1_sol - 2.29: Numerical Fluid Mechanics Solution of...

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