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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/CH2/EX2.9 | |
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-rwxr-xr-x | 1910/CH2/EX2.9/Chapter29.sce | 56 |
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diff --git a/1910/CH2/EX2.9/Chapter29.sce b/1910/CH2/EX2.9/Chapter29.sce new file mode 100755 index 000000000..1fdf90849 --- /dev/null +++ b/1910/CH2/EX2.9/Chapter29.sce @@ -0,0 +1,56 @@ +//Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 09")
+//A thin walled copper tube of outside metal radius r=0.01m carries steam at temprature, T1=400K.It is inside a room where the surrounding air temprature is Tinf=300K.
+T1=400;
+Tinf=300;
+r=0.01;
+//The tube is insulated with magnesia insulation of an approximate thermal conductivity of k=0.07W/(m*K)
+k=0.07;
+//External convective Coefficient h=4W/(m^2*K)
+h=4;
+//Critical thickness(rc) is given by k/h
+disp("The critical thickness of insulation in metre is")
+rc=k/h
+//We use the rate of heat transfer per metre of tube length as Q=(Ti-Tinf)/((ln(r2/r1)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L))) where length,L=1m
+L=1;
+//When 0.002m thick layer of insulation r1=0.01m,r2=0.01+0.002=0.012m
+r1=0.01;//inner radius
+r2=0.012;//outer radius
+//Let ln(r2/r1)=X
+X=log(r2/r1)/log(2.718);
+//The heat transfer rate per metre of tube length is Q
+disp("The heat transfer rate Q per metre of tube length in W/m is ")
+Q=(T1-Tinf)/(((X)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L)))
+//When critical thickness of insulation r1=0.01m,r2=0.0175m
+r2=0.0175;//outer radius
+r1=0.01;//inner radius
+//Let ln(r2/r1)=X
+X=log(r2/r1)/log(2.718);
+//The heat transfer rate per metre of tube length is Q
+disp("The heat transfer rate per metre of tube length Q in W/m is ")
+Q=(T1-Tinf)/(((X)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L)))
+//When there is a 0.05 m thick layer of insulation r1=0.01m,r2=.01+0.05=0.06m
+r1=0.01;//inner radius
+r2=0.06;//outer radius
+//Let ln(r2/r1)=X
+X=log(r2/r1)/log(2.718);
+//The heat transfer rate per metre of tube length is Q
+disp("The heat transfer rate per metre of tube length Q in W/m is ")
+Q=(T1-Tinf)/(((X)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L)))
+//It is important to note that Q increases by 5.2% when the insulation thickness increases from 0.002m to critical thickness.
+//Addition of insulation beyond the critical thickness decreases the value of Q (The heat loss).
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