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+clc;
+clear all;
+disp("case 1:to find out thickness of insulation")
+r1=120/2;//mm inner radius of pipe
+r2=160/2;//mm outer radius of pipe
+kA=42;// W/(m*C) thermal conductivity of pipe material
+kB=0.8;// W/(m*C) thermal conductivity of insulation material
+thf=150;// degree C
+tcf=20;// degree C
+hcf=100;// W/(m^2*C)
+hcf=30;// W/(m^2*C)
+r=150/2;//mm mean radius
+//A=2*3.1416*r*L area for heat transfer
+//Q=2.1*2*3.1416*r*L
+//Q=2*3.1416*L*(thf-tcf)/(1000/(hhf*r1)+(log(r2/r1))/kA+(log(r3/r2))/kB+1000/(hcf*r3));
+//2.1*2*3.1416*r*L = 2*3.1416*L*(thf-tcf)/(1000/(hhf*r1)+(log(r2/r1))/kA+(log(r3/r2))/kB+1000/(hcf*r3))
+//2.1*r = (thf-tcf)/(1000/(hhf*r1)+(log(r2/r1))/kA+(log(r3/r2))/kB+1000/(hcf*r3))
+//(log(r3/r2))/kB+1/(30*r3)=0.6524
+//1.25*log(r3/r2)+1/(30*r3)=0.6524
+// By trial and error r3=0.105m= 105mm
+disp(" By trial and error, r3=0.105m= 105mm ")
+r3=105;//mm
+t=r3-r2;
+
+disp("m",t,"Thickness of insulation t =")
+
+
+
+