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+//Exa 2.7
+clc;
+clear;
+close;
+//given data
+k_cu=384;// in W/mK
+k_s=1.75;// in W/mK
+k1=k_cu;
+k2=k_s;
+hi=221;//  in W/m^2K
+ho=3605;//  in W/m^2K
+Ti=100;// in degree C
+To=125;// in degree C
+r1=0.2;// in m
+r2=0.02+0.006;// in m
+r3=0.026+0.003;// in m
+ri=0.02;// in m
+L=1;// in m
+// Part(i)
+Ao= 2*%pi*r3*L;
+Ai= 2*%pi*r1*L;
+// Formula Uo= 1/Ao*sigmaR
+Uo= 1/[ r3/(ri*hi) + r3/k1*log(r2/r1) + r3/k2*log(r3/r2) + 1/ho ];// in w/m^2K
+disp(Uo,"Overall heat transfer coefficient based on outer area in W/m^2K");
+
+//Part(ii)
+del_T= To-Ti;
+q=Uo*Ao*del_T;
+disp(q,"Water to air heat transfer rate in W/m");
+
+//Part (iii)
+// Formula q= T/(log(r3/r2)/(2*%pi*k*L)) , where T=T2-T3 and k=k_s
+k=k_s;
+T= q*log(r3/r2)/(2*%pi*k*L);
+disp(T,"Temperature drop across the scale deposited in degree C")
+
+// Note: In Part (i), they put wrong value of r2 and r1 in log(r2/r1) to calculate the value of Uo. So there is some difference in answer of coding and book
+
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