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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+
+clc;
+clear;
+//Example 3.9
+v=17.36*10^-6 //Viscosity for air [sq m./s]
+k=0.0275 //for air ..[W/(m.K)]
+Cp=1.006 //[kJ/(kg.K)]
+Npr=0.7 //for air
+u_inf=2 //[m/s]
+x=0.2 //[m]
+Nre_x=u_inf*x/v //Reynolds number at x=0.2 m
+//Since this is less than 3*10^5
+Nnu_x=0.332*sqrt(Nre_x)*(Npr^(1.0/3.0))
+hx=Nnu_x*k/x //[W/(sq m.K]
+//Average value of heat transfer coeff is twice this value
+h=2*hx //[W/sq m.K)]
+h=12.3 //Approximation
+w=1 //width in [m]
+A=x*w //[sq m] Area of plate
+Tw=333 //[K]
+T_inf=300 //[K]
+Q=h*A*(Tw-T_inf) //Heat flow in [W]
+printf("\nANSWER:\nHeat flow is :%f W\n",Q)
+//From both sides of plate:
+Q=2*Q //[W]
+printf("\nANSWER\n Heat flow from both sides of plate is %f W",Q);
+