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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 /1244/CH4/EX4.2 | |
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Diffstat (limited to '1244/CH4/EX4.2')
-rwxr-xr-x | 1244/CH4/EX4.2/Example42.sce | 25 |
1 files changed, 25 insertions, 0 deletions
diff --git a/1244/CH4/EX4.2/Example42.sce b/1244/CH4/EX4.2/Example42.sce new file mode 100755 index 000000000..53359286e --- /dev/null +++ b/1244/CH4/EX4.2/Example42.sce @@ -0,0 +1,25 @@ +
+
+// Display mode
+mode(0);
+
+// Display warning for floating point exception
+ieee(1);
+
+clc;
+disp("Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 4 Example # 4.2 ")
+
+disp("The given example is theoretical and does not involve any numerical computation")
+
+// Local shear stress is given as:
+// tau=0.3*((rho*mu/x)^0.5)*(Uinfinity^1.5)
+
+// Using Local friction coefficient = local shear stress /
+// (0.5*rho*Uinfinity*Uinfinity), we get local friction coefficient as:
+
+//disp("Cfx = 0.6/((ReL*xstar))^0.5")
+
+//Integrating the local value of shear stress over length L and dividing by
+//area i.e. A=L*1, we get average friction coefficient as:
+
+//disp("Cfbar = 1.2/(ReL^0.5)")
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