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authorpriyanka2015-06-24 15:03:17 +0530
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+clc
+clear
+//Initialization of variables
+d1=1 //in
+l=1 //ft
+r=0.5 //ft
+L=0.5 //in
+Ts=430 //F
+Ta=170 //F
+del=0.0125 //ft
+h=10 //Btu/hr ft^2 F
+eta=0.77
+eta2=0.94
+n=60 //fins
+thick=0.025 //in
+k2=132 //Btu/hr ft F
+//calculations
+Q=h*%pi*d1^2 *(Ts-Ta)/12
+rate=(r+L)/r
+k=26 //Btu/hr ft F
+Lt=L/12 *(h*12/(k*del))^(1/2)
+dtm=eta*(Ts-Ta)
+As=2*%pi*((2*d1)^2 -d1^2)/4
+Q1=h*n*As*dtm/144
+Q2=h*%pi*d1*(12-60*thick)*(Ts-Ta)/144
+Qt=Q1+Q2
+al=0.8
+tl=Ta+(Ts-Ta)/cosh(al)
+al2=r/12 *(h*12*2/(k2*thick))
+dtm2=eta2*(Ts-Ta)
+Q12=h*n*As*dtm2/144
+Qt2=Q12+Q2
+//results
+printf("Heat rate per foot of bare tube = %.1f Btu/hr",Q)
+printf("\n Total hourly heat loss per foot of finned tube = %.1f Btu/hr",Qt)
+printf("\n Approx. temp for tip of the fin = %d F",tl)
+printf("\n In case of Al, Total beat loss = %.1f Btu/hr",Qt2)
+disp("The answers in the textbook are a bit different due to rounding off errors")