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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/CH5/EX5.8/Chapter58.sce | |
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initial commit / add all books
Diffstat (limited to '1910/CH5/EX5.8/Chapter58.sce')
| -rwxr-xr-x | 1910/CH5/EX5.8/Chapter58.sce | 51 |
1 files changed, 51 insertions, 0 deletions
diff --git a/1910/CH5/EX5.8/Chapter58.sce b/1910/CH5/EX5.8/Chapter58.sce new file mode 100755 index 000000000..1ef3d4879 --- /dev/null +++ b/1910/CH5/EX5.8/Chapter58.sce @@ -0,0 +1,51 @@ +// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 5, Example 8")
+//Eletric current passes through a L=0.5m long horizontal wire of D=0.1mm diameter.
+L=0.5;
+D=0.1*10^-3;
+//The wire is to be maintained at temprature,Twire=400K and the air is at temprature,Tair=300K.
+Twire=400;
+Tair=300;
+//The resistance of the wire(R) is 0.012ohm per meter.Nusselt number(NuL) over the length of wire to be 0.4.
+NuL=0.4;
+R=0.012;
+//At mean temprature of Tf=350K, The thermal conductivity of air is k=0.03W/(m*K)
+k=0.03;
+//Nusselt number is NuL=hbar*D/k
+//hbar is the heat flux
+disp("The heat flux in W/(m^2*K) is")
+hbar=NuL*k/D
+//Q is the heat loss from the wire
+disp("The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt")
+Q=hbar*%pi*D*L*(Twire-Tair)
+//At steady state the ohmic loss in the wire equals the heat loss from its surface Therfore I^2*R=Q
+//I is the current flow.
+disp("The current in Ampere is")
+I=(Q/(R*L))^0.5
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