(a) Explanation Given: Fin length is L = 10 mm . Thermal conductivity is k = 180 W/ m ⋅ K . Thickness is t = 1 mm . Convection coefficient is h = 100W/ m 2 ⋅ K . Calculation: The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and (3.80), where M = ( 2 h w 2 t k ) 1 / 2 ( t b − t ∞ ) = ( 2 × 100 W/m 2 ⋅ K × 0.001 m × 180 W/m ⋅ K ) 1 / 2 ( 75 ° C ) w = 450 w W , m ≈ ( 2 h / k t ) 1 / 2 = ( 200 W/m 2 ⋅ K / 180 W/m ⋅ K × 0.001 m ) 1 / 2 = 33.3 m -1 , mL ≈ 33.3 m -1 × 0.010 m = 0.333 , and, ( h / m k ) ≈ ( 100 W/m 2 ⋅ K 33.3 m -1 × 180 W/m ⋅ K ) = 0.0167. From Table B-1, it follows that s i n h m L ≈ 0.340 , c o s h m L ≈ 1.057 , a n d t a n h m L ≈ 0.321 . From knowledge of q f , equation (3.86), (3.81) and (3.83) yield, n f = q ′ f h ( 2 L + t ) θ b , ε f ≈ q ′ f h t θ b , R ′ t , f = θ b q ′ f Case A: from equation (3.72), (3.82), (3.83) and (3.70), Heat rate per unit width is calculated as follows: q ′ f = M w ( sinh m L + ( h / m k ) cosh m L cosh m L + ( h / m k ) cosh m L ) = ( 450 W/m ) ( 0 (b) To determine Effect of length on the heat rate.

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ISBN: 9780470501962

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

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3, Problem 3.133P

(a)

To determine

Heat rate per unit width, thermal resistance, efficiency, effectiveness and tip temperature for different tip conditions.

(b)

To determine

Effect of length on the heat rate.

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Chapter 3, Problem 3.132P

Chapter 3, Problem 3.134P

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Heat rate per unit width, thermal resistance, efficiency, effectiveness and tip temperature for different tip conditions.

Solution Summary: The author explains the fin heat transfer rate for Cases A, B, and D.

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Heat rate per unit width, thermal resistance, efficiency, effectiveness and tip temperature for different tip conditions.

Effect of length on the heat rate.

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