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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1910/CH3/EX3.1 | |
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initial commit / add all books
Diffstat (limited to '1910/CH3/EX3.1')
-rwxr-xr-x | 1910/CH3/EX3.1/Chapter31.sce | 26 |
1 files changed, 26 insertions, 0 deletions
diff --git a/1910/CH3/EX3.1/Chapter31.sce b/1910/CH3/EX3.1/Chapter31.sce new file mode 100755 index 000000000..fb58c68a5 --- /dev/null +++ b/1910/CH3/EX3.1/Chapter31.sce @@ -0,0 +1,26 @@ +// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 1")
+//Length and breadth is given as 1 unit (Gemoetry is Square)
+L = 1;//length
+//Problem can be divided into two modules
+//Solution to module 1 is given by Eq. 3.21, considering the first three terms
+//n is the looping parameter
+//theta is the non dimensional temperature defined as ((T-100)/100) where T is actual temperature in degree Celcius.
+//Initialising theta as zero
+theta = 0;
+for n = 1:3
+ theta = theta+((2/%pi)*((sin((n*%pi)/2)*sinh((n*%pi)/2))*((-1)^(n+1)+1)))/(n*sinh(n*%pi));
+end;
+//Solution to module 2 is given by Eq. 3.24, considering the first three terms
+for n = 1:3
+ theta = theta+(((3*2)/%pi)*((sin((n*%pi)/2)*sinh((n*%pi)/2))*((-1)^(n+1)+1)))/(n*sinh(n*%pi));
+end;
+//Calculating value of temperature from the value of theta
+//Temperature in degree celcius
+disp("Temperature at the centre in Degree C is")
+T = theta*100+100
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