Subtopic: Steady state heat transfer in conduction, convection and radiation
Question Statement:
A rectangular fin of length 12 cm, width 22 cm and thickness 1.5 cm is connected to a tube at a temperature of 0 °C. The thermal conductivity of the fin material is 150 W m⁻¹ K⁻¹. The tip of the fin is not insulated. Air at a temperature of 5 °C is in contact with the fin. The heat transfer coefficient between the fin and the air is 25 W m⁻² K⁻¹. The rate of heat transfer is
Options:
(A) 3.33 W
(B) 6.63 W
(C) 9.13 W
(D) 15.23 W
Official Answer Key:
B
Step-by-Step Solution & Explanation
Fin parameter, m = √(hP/kA_c); P = 2(w+t) = 2(0.22+0.015) = 0.47 m; A_c = w×t = 0.22×0.015 = 3.3×10⁻³ m². m = √(25×0.47/(150×3.3×10⁻³)) = √23.7 = 4.872 m⁻¹. Convective (non-insulated) tip — use corrected length L_c = L + t/2 = 0.12+0.0075 = 0.1275 m. Q = √(hPkA_c)·θ_b·tanh(mL_c), θ_b = T_base−T_air = 0−5 = −5 °C (magnitude used). √(hPkA_c) = √(25×0.47×150×3.3×10⁻³) = √5.816 = 2.412; Q = 2.412×5×tanh(4.872×0.1275) = 2.412×5×0.5517 = 6.66 ≈ 6.63 W.
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