a. Explanation Given: t = 1mm , w = 20 mm , T b = 85°C Calculation: Calculating the film temperature by the following formula: T f = T b + T ∞ 2 Temperature = T b , T 8 = ambient temperature. Putting T b = 85°C , T 8 = 20°C in the equation : T f = 85 + 20 2 = 52.5 ∘ C ≈ 325 K From table A.4 (Thermophysical properties of gases at atmospheric pressure) , given below are the properties of air at T = 325 K and p = 1 atm : The kinematic viscosity of air is v = 18.4 × 10 − 6 m 2 / s The thermal conductivity of air is k = 0.0.282 W / m ⋅ K The prandtl of air is Pr = 0.704 (a) For the rectangular fin: Calculating the Reynolds number by following formula Re w = u ∞ w v Here Re w = Reynolds number , u ∞ = flow velocity , w = width of fin , v = kinematic viscosity Putting 5m/s for u ∞ , 20 × 10 − 3 m for w , 18.4 × 10 − 6 m 2 / s for v in equation : Re w = 5 m / s × 20 × 10 − 3 m 18.4 × 10 − 6 m 2 / s = 5435 Calculating the average convection heat transfer coefficient by using the following formula: h ¯ = k w [ 0.664 Re w 1 / 2 ] Pr 1 / 3 Here , h ¯ = average convection heat transfer coefficient , k = thermal conductivity , D = diameter , C = constant , Re = Reynolds number , Pr = Prandtl number Putting k = 0.0282 W / m ⋅ K , w = 20 × 10 − 3 m , Re = 5435 , Pr = 0.704 in equation h ¯ = 0.0282 W / m ⋅ K 20 × 10 − 3 m [ 0.664 × 5435 1 / 2 ] 0.704 1 / 3 = 61.4 W / m 2 ⋅ K The periphery of the fin is calculated by the following formula: P =2(w+t) Here, w = width of the fin , t = thickness of the fin Putting w = 20 × 10 − 3 m , t = 1 × 10 − 3 m in equation: P = 2 × ( 20 × 10 − 3 m + 1 × 10 − 3 m ) = 42 × 10 − 3 m Calculating the cross-sectional area of the fin by the following formula: A c = w t Hence , w = width of fin , t = thickness of the fin Putting w = 20 × 10 − 3 m , t = 1 × 10 − 3 m in equation: A c = 20 × 10 − 3 m × 1 × 10 − 3 m = 20 × 10 − 6 m 2 Calculating the parameter of fin M by using the following formula : M = h P k A c ⋅ θ b ..... ( 6 ) Here M = fin parameter , h = local convection heat transfer coefficient , p = perimeter , k = thermal conductivity of heat transfer , A c = area of the fin , θ b = temperature difference. Putting , h = 61.4 W / m 2 ⋅ K , P= 42 × 10 − 3 m , k = 237 W / m ⋅ K , A c = 20 × 10 − 6 m 2 , θ b = ( 85 − 20 ) ∘ C M = 61.4 W / m 2 . K × 42 × 10 − 3 m × 20 × 10 − 6 m × ( 85 − 20 ) ∘ C = 7.19 W Now for the cylinder fin: Calculating the diameter from the following formula: D = 4 π t w D = diameter of the fin W= width of the fin T= thickness of the fin Now putting the value of w, as 20 × 10 − 3 and 1 × 10 − 3 for t D = 4 π 1 × 10 − 3 × 2 × 10 − 3 = 0 (B). To determine Diameter cylindrical fin would be needed to provide the same fin heat transfer rate as the rectangular cross-section fin

7th Edition

ISBN: 9780470501979

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 7, Problem 7.63P

To determine

The heat loss observed from the rectangular and cylinder fin.

(B).

To determine

Diameter cylindrical fin would be needed to provide the same fin heat transfer rate as the rectangular cross-section fin

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Chapter 7, Problem 7.62P

Chapter 7, Problem 7.64P

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Chapter 7 Solutions

Fundamentals of Heat and Mass Transfer

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The heat loss observed from the rectangular and cylinder fin.

Solution Summary: The author calculates the thermal conductivity of air at T = 325 K and p = 1 atm.

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The heat loss observed from the rectangular and cylinder fin.

Diameter cylindrical fin would be needed to provide the same fin heat transfer rate as the rectangular cross-section fin

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Chapter 7 Solutions