//Variable declaration: Do = 2.5/100 //Outside diameter of tube (m) t = 1/10**3 //Thickness of fin (m) T = 25 //Fluid temperature ( C) Tb = 170 //Surface temperature ( C) h = 130 //Heat transfer coefficient (W/m^2.K) k = 200 //Thermal conductivity of fin (W/m.K) rf = 2.75/100 //Outside radius of fin (m) //Calculation: ro = Do/2 //Radius of tube (m) Ab = 2*%pi*ro*t //Area of the base of the fin (m^2) Te = Tb-T //Excess temperature at the base of the fin (K) Q1 = h*Ab*Te //Total heat transfer rate without the fin (W) Bi = h*(t/2)/k //Biot number L = rf-ro //Fin height (m) rc = rf+t/2 //Corrected radius (m) Lc = L+t/2 //Corrected height (m) Ap = Lc*t //Profile area (m^2) Af = 2*%pi*(rc**2-ro**2) //Fin surface area (m^2) Qm = h*Af*Te //Maximum fin heat transfer rate (W) A = sqrt(Lc**3*h/(k*Ap)) //Abscissa of fin efficiency C = rf/ro //Curve parameter of fin efficiency //From figure 17.4: nf = 0.86 //Fin efficiency Qf = nf*Qm //Fin heat transfer rate (W) R = Te/Qf //Fin resistance (K/W) //Result: printf("1. The heat transfer rate without the fin is : %.2f W .",Q1) printf("Or, the heat transfer rate without the fin is : %.0f Btu/h .",Q1*3.412) printf("2. The corrected length is : %.4f m .",Lc) printf("3. The outer radius is : %.3f m ",rc) printf("4. The maximum heat transfer rate from the fin is : %.2f W .",Qm) printf("5. The fin efficiency is : %.0f %%",nf*100) printf("6. The fin heat transfer rate is : %.0f %%",Qf) printf("Or, the fin heat transfer rate is : %.0f %%",Qf*3.412) printf("7. The fin thermal resistance is : %.2f K/W .",R)