5_Shelby_Jessica_HW05A

5_Shelby_Jessica_HW05A - % Inputs: diameter of pipes d(m),...

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Jessica Shelby Section 5 A42031587 HMWK 5A Problem 1) SCRIPT: function [conversion] = FuelEfficiency (x) % Jessica Shelby, Section 5 % FuelEfficiency conversion % Input: x - beginning value (mi/G) % Output: x is converted (km/L) conversion = (x*1.609/3.785); COMMAND WINDOW: >> x=23 x = 23 >> [conversion] = FuelEfficiency (x) conversion = 9.7773 >> x=50 x = 50 >> [conversion] = FuelEfficiency (x) conversion = 21.2550 Problem 2)
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SCRIPT: function [a,v] = WaterFlow (d,q) % Jessica Shelby, Section 5 % WaterFlow is calculated due to different diameters
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Unformatted text preview: % Inputs: diameter of pipes d(m), volume flow through the pipes q(m^3/s) % Outputs: pipe's cross-sectional area a(m^2), speed of water v(m/s) a=((1/4).*(pi).*(d.^2)); v=(q./a); COMMAND WINDOW: d=[.0508,.0635,.0762,.1143,.1270] d = 0.0508 0.0635 0.0762 0.1143 0.1270 >> q=.48 q = 0.4800 [a,v] = WaterFlow (d,q) a = 0.0020 0.0032 0.0046 0.0103 0.0127 v = 236.8230 151.5667 105.2547 46.7799 37.8917...
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This note was uploaded on 11/22/2011 for the course EGR 100 taught by Professor Hinds during the Fall '08 term at Michigan State University.

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5_Shelby_Jessica_HW05A - % Inputs: diameter of pipes d(m),...

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