Homework 12 Solution

# Homework 12 Solution - Lecture 12 MEEN 357 Homework...

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Lecture 12 MEEN 357 Homework Solution 1 Lecture-12-Homework-Solution-2009.doc RMB 1. Problem 13.31 of the textbook. Solution: The approach you must follow is to use the kind of argument given in Lectures 11 and 12. You are given an equation / a BT De μ = (12.1) The linear version of this nonlinear equation is ln ln a B D T = + (12.2) Equation (12.2) is of the linear form 01 y aa x = + (12.3) where 0 1 ln ln 1 a y aD aB x T = = = = (12.4) You are given a data table T in o C 0 5 10 20 30 40 1.787 1.519 1.307 1.002 0.7975 0.6527 The next step is to convert this table into one for x and y according to the rules (12.4). A problem that is easy to overlook is that a T is the absolute temperature and the table above is the temperature in centigrade. You must add 273.15 to the numbers in the above table in order to go forward with the regression. Of course, MATLAB can do these conversions and table creations without difficulty. The script clc,clear all T=[0,5,10,20,30,40]; mu=[1.787,1.519,1.307,1.002,0.7975,0.6527]; x=1./(T+273.15); y=log(mu); P=polyfit(x,y,1); a1=P(1) a0=P(2)

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Lecture 12 MEEN 357 Homework Solution 2 Lecture-12-Homework-Solution-2009.doc RMB %Use equation (12.4) above B=a1 D=exp(a0) Produces the output a1 = 2.1512e+003 a0 = -7.3160 B = 2.1512e+003 D = 6.6480e-004 If we use these numbers in the formula (12.1) yields the answer / 2151.2/ .00066480 aa BT T De e μ == (12.5) 2. Problem 14.3 of the textbook.
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Homework 12 Solution - Lecture 12 MEEN 357 Homework...

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