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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+
+//Example2.15[Heat Loss through a Steam Pipe]
+//Given:-
+L=20;//Pipe Length[m]
+k=20;//[W/m.degree Celcius]
+r1=0.06;//Inner Radius[m]
+r2=0.08;//Outer Radius[m]
+T1=150;//Temp of inner surface[degree Celcius]
+T2=60;//Temp of outer surface[degree Celcius]
+//Solution:-
+//Integrating differential equation we get T(r)=C1logr+C2, where C1 and C2 are
+C1=(T2-T1)/log(r2/r1);
+C2=T1-((T2-T1/log(r2/r1))*log(r1));
+Q_cyl=2*%pi*k*L*(T1-T2)/(log(r2/r1));
+disp("kW",round(Q_cyl/1000),"The rate of heat conduction through the pipe is")