Explanation Given: The diameter of a burner is D = 200 mm . The thermal conductivity is k=65 W / m ⋅ K . The fin effectiveness is ε = 0.95 . The length of burner is L f = 25 mm . The length of burner is L b = 10 mm . The number of fins is N = 8 . The thickness is t = 5 mm . The width is w = 80 mm . The burner top temperature is T t = T ∞ = 20 ° C . The burner bottom temperature is T b = T ∞ = 490 ° C . The heat transfer coefficient of top burner is h t = 40 W / m 2 ⋅ K . The heat transfer coefficient of top burner is. h b = 50 W / m 2 ⋅ K Formula used: The expression for fin efficiency is given as, η f i n = tanh ( m L c ) m L c Here, m is fin parameter and L c is characteristic length. The radiation heat transfer is given as, q r o d b = ε σ A t ( T ∞ b 4 − T b 4 ) Here, A t is top burner area, σ = 5.67 × 10 − 8 W / m 2 ⋅ K 4 is Stefan’s Boltzmann constant and T b is temperature of without fin burner. Calculation: The area of top burnercan be calculate as, A t = π 4 D 2 A t = π 4 ( 0.2 m ) 2 A t = 0.031416 m 2 The area of bottom burner can be calculate as, A b = A t − N ( w t ) A b = 0.031416 m 2 − 8 × 0.08 m × 0.005 m A b = 0.028216 m 2 The radiation heat transfer at bottom of burner can be calculated as, q r o d b = ε σ A t ( T ∞ b 4 − T b 4 ) q r o d b = 0.95 × 5.67 × 10 − 8 W / m 2 ⋅ K 4 × 0.031416 m 2 × [ ( 763 K ) 4 − T b 4 ] q r o d b = 1.69 × 10 − 9 W / K 4 × [ ( 763 K ) 4 − T b 4 ] ……. (i) The fin characteristic length can be calculated as, L c = L f + t 2 L c = 0.025 m + 0.005 m 2 L c = 0.0275 m The surface area of fin can be calculated as, A f = 2 w L c A f = 2 × 0.080 m × 0.0275 m A f = 0.0044 m 2 The fin parameter can be calculated as, m = 2 h b k t m = 2 × 50 W / m 2 ⋅ K 65 W / m ⋅ K × ( 0.005 m ) m = 17.54 m − 1 The fin efficiency can be calculated as, η f i n = tanh ( m L c ) m L c η f i n = tanh ( 17.54 m -1 × 0.0275 m ) ( 17.54 m -1 × 0.0275 m ) η f i n = 0.93 The total heat transfer at base of burner can be calculated as, q b = q f i n + q c o n v b + q r o d b ……. (ii) The heat transfer for fin can be calculated as, q f i n = N [ η f i n × h b × A f × ( T ∞ b − T b ) ] q f i n = 8 [ 0.93 × 50 W / m 2 ⋅ K × 0.0044 m 2 × ( 763 K − T b ) ] q f i n = 1.6368 W / K × ( 763 K − T b ) The convection heat transfer for bottom can be calculated as, q c o n v b = h b A b ( T ∞ b − T b ) q c o n v b = 50 W / m 2 ⋅ K × 0.028216 m 2 ( 763 K − T b ) q c o n v b = 1.4108 W / K × ( 763 K − T b ) Substitute all the values in equation (ii). q b = q f i n + q c o n v b + q r o d b q b = { 1.6368 W / K × ( 763 K − T b ) + 1

6th Edition

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 3.147P

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The top surface temperature of the finned burner.

The top surface temperature of the burner without fin.

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Chapter 3, Problem 3.146P

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Introduction to Heat Transfer

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The top surface temperature of the finned burner. The top surface temperature of the burner without fin.

Solution Summary: The author compares the top surface temperature of a finned burner with that of the burner without fin.