Lecture 12
●
MEEN 357 Homework Solution
1
Lecture12HomeworkSolution2009.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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View Full DocumentLecture 12
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MEEN 357 Homework Solution
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Lecture12HomeworkSolution2009.doc
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%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 =
6.6480e004
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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 Fall '07
 ANAMALAI
 Regression Analysis, Quadratic equation, TA, 0.5700 M

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