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MCE344 HW3...

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3-120: A plane wall with surface temperature of 350°C is attached with straight rectangular fins (k = 235 W/m·K). The fins are exposed to an ambient air condition of 25°C and the convection heat transfer coefficient is 154 W/m^2·K. Each fin has a length of 82 mm, a base of 5 mm thick and a width of 100 mm. Determine the efficiency, heat transfer rate, and effectiveness of each fin

Answer: Explanation: Given Surface temperature = Tb = 350°C Thermal Conductivity = k = 235 W/m·K Ambient air temperature = T = 25°C Convective heat transfer coefficient of air = h = 154 W/m^2·K. Length of fin = L = 82 mm Base of fin = t = 5 mm Width of fin = w = 100 mm The mass of fin is calculated by; m = √(2h/kt) m = √(2 * 154/235*0.005) m = 16.19 Then the critical length is calculated; Lc. = L + ½t Lc = 0.082 + ½ * 0.005. Lc = 0.0845m Therefore, the efficiency of the fin can be calculated as follows: tanh (mLc)/mLc = tanh(16.19 * 0.0845)/((16.19 * 0.0845) tanh(1.368055)/1.368055 = 0.878248/1.368055

= 0.6420 Eff = 64.20% Area of the fin is calculated as follows A = 2wLc A = 2 * 0.1 * 0.0845 A = 0.0169m² Rate of heat transfer is calculated by; Eff * h * A(Tb - T) = 0.642 * 154 * 0.0169 * (350 - 25) E = 543.03249W The base area of the fin is calculated by; A = wt A = 0.1 * 0.005 A = 0.0005m² Lastly, the effectiveness of each fin is calculated as follows; Eff = E/(hA(TB - T)) Eff = 543.03249/(154 * 0.0005(350 - 25)) Eff = 21.6996 Therefore, the efficiency of fin is 64.20% heat transfer rate 543.03249W and effectiveness of each fin is 21.6996...