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authorpriyanka2015-06-24 15:03:17 +0530
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+
+clc;
+clear;
+//Example 3.16
+v=20.92*10^-6 //sq m/s
+k=3*10^-2 //W/(m.K)
+Npr=0.7
+u_inf=25 //[m/s]
+d=50 //[mm]
+d=d/1000 //[m]
+Nre=u_inf*d/v //Reynold's number
+Tw=397 //[K]
+T_inf=303 //[K]
+
+//Case 1: Circular tube
+
+Nnu=0.0266*Nre^(0.805)*Npr^(1.0/3.0);
+h=Nnu*k/d //[W/sq m.K]
+A=%pi*d //Area in [sq m]
+Q=h*A*(Tw-T_inf) //[W]
+Q_by_l1=h*%pi*d*(Tw-T_inf) //[W/m]
+
+//Case 2:Square tube
+A=50*50 //Area in [sq mm]
+P=2*(50+50) //Perimeter [mm]
+l=4*A/P //[mm]
+l=l/1000 //[m]
+Nnu=0.102*(Nre^0.675)*(Npr^(1.0/3.0))
+h=Nnu*k/d //W/(sq m.K)
+A=4*l*l //[sq m]
+
+Q=h*A*(Tw-T_inf)
+Q_by_l2=Q/l //[W/m]
+printf("\nRate of heat flow from the square pipe=%f W/m \n which is more than that from the circular pipe which is equal to %f W/m",Q_by_l2,Q_by_l1);