Explanation Given: The length of the fin is L = 50 mm . The base thickness of the fin is t = 20 mm . The thermal conductivity of the fin is k = 25 W / m ⋅ K . The base temperature of the fin is T b = 50 °C . The temperature of the surrounding fluid is T ∞ = 20 °C . The convection heat transfer coefficient is h = 50 W / m 2 ⋅ K . The length of the grid is Δ x = 10 mm . The with of the grid is Δ y = 2 mm . Formula Used: The nodal distance along the inclined surface is given by, Δ s = ( Δ x ) 2 + ( Δ y ) 2 The finite difference equation at the node 7 is given by, k ( Δ y 2 ) ( T 7 − T 1 ) Δ x + k Δ x ( T 8 − T 7 ) Δ y + k ( Δ y 2 ) ( T 12 − T 7 ) Δ x = 0 The finite difference equation at the node 8 is given by, k Δ y ( T 2 − T 8 ) Δ x + k Δ x ( T 7 − T 8 ) Δ y + k Δ y ( T 13 − T 8 ) Δ x + k Δ x ( T 9 − T 8 ) Δ y = 0 The finite difference equation at the node 9 is given by, k Δ y ( T 3 − T 9 ) Δ x + k Δ x ( T 8 − T 9 ) Δ y + k Δ y ( T 14 − T 9 ) Δ x + k Δ x ( T 10 − T 9 ) Δ y = 0 The finite difference equation at the node 10 is given by, k Δ y ( T 4 − T 10 ) Δ x + k Δ x ( T 9 − T 10 ) Δ y + k Δ y ( T 15 − T 10 ) Δ x + k Δ x ( T 11 − T 10 ) Δ y = 0 The finite difference equation at the node 11 is given by, k Δ y ( T 5 − T 11 ) Δ x + k Δ x ( T 10 − T 11 ) Δ y + h Δ s ( T ∞ − T 11 ) = 0 The finite difference equation at the node 12 is given by, k ( Δ y 2 ) ( T 7 − T 12 ) Δ x + k Δ x ( T 13 − T 12 ) Δ y + k ( Δ y 2 ) ( T 16 − T 12 ) Δ x = 0 The finite difference equation at the node 13 is given by, k Δ y ( T 8 − T 13 ) Δ x + k Δ x ( T 12 − T 13 ) Δ y + k Δ y ( T 17 − T 13 ) Δ x + k Δ x ( T 14 − T 13 ) Δ y = 0 The finite difference equation at the node 14 is given by, k Δ y ( T 9 − T 14 ) Δ x + k Δ x ( T 13 − T 14 ) Δ y + k Δ y ( T 18 − T 14 ) Δ x + k Δ x ( T 15 − T 14 ) Δ y = 0 The finite difference equation at the node 15 is given by, k Δ y ( T 10 − T 15 ) Δ x + k Δ x ( T 14 − T 15 ) Δ y + h Δ s ( T ∞ − T 15 ) = 0 The finite difference equation at the node 16 is given by, k ( Δ y 2 ) ( T 12 − T 16 ) Δ x + k Δ x ( T 17 − T 16 ) Δ y + k ( Δ y 2 ) ( T 19 − T 16 ) Δ x = 0 The finite difference equation at the node 17 is given by, k Δ y ( T 13 − T 17 ) Δ x + k Δ x ( T 16 − T 17 ) Δ y + k Δ y ( T 20 − T 17 ) Δ x + k Δ x ( T 18 − T 17 ) Δ y = 0 The finite difference equation at the node 18 is given by, k Δ y ( T 14 − T 18 ) Δ x + k Δ x ( T 17 − T 18 ) Δ y + h Δ s ( T ∞ − T 18 ) = 0 The finite difference equation at the node 19 is given by, k ( Δ y 2 ) ( T 16 − T 19 ) Δ x + k Δ x ( T 20 − T 19 ) Δ y + k ( Δ y 2 ) ( T 21 − T 19 ) Δ x = 0 The finite difference equation at the node 20 is given by, k Δ y ( T 17 − T 20 ) Δ x + k Δ x ( T 19 − T 20 ) Δ y + h Δ s ( T ∞ − T 20 ) = 0 The finite difference equation at the node 21 is given by, k ( Δ y 2 ) ( T 19 − T 21 ) Δ x + h ( Δ s 2 ) ( T ∞ − T 21 ) = 0 The conduction heat transfer per unit depth for the fin is given by, q f = 2 [ k ( Δ y 2 ) ( T 1 − T 7 ) Δ x + k Δ y ( T 2 − T 8 ) Δ x + k Δ y ( T 3 − T 9 ) Δ x + k Δ y ( T 4 − T 10 ) Δ x + k Δ y ( T 5 − T 11 ) Δ x ] The maximum possible rate of heat transfer per unit depth for fin is given by, q max = h L 2 + ( t 2 ) 2 ( T 1 − T ∞ ) The fin efficiency is given by, η = q f q max The dimensionless fin parameter is given by, m = L c 3 / 2 h k A p Calculation: The nodal distance along the inclined surface is calculated as, Δ s = ( Δ x ) 2 + ( Δ y ) 2 = ( 10 mm ) + ( 2 mm ) = 10.2 mm The finite difference equation at the node 7 is calculated as, k ( Δ y 2 ) ( T 1 − T 7 ) Δ x + k Δ x ( T 8 − T 7 ) Δ y + k ( Δ y 2 ) ( T 12 − T 7 ) Δ x = 0 ( 25 W / m ⋅ K ) [ ( 2 mm × 1 m 1000 mm 2 ) ( 50 °C − T 7 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 8 − T 7 ) ( 2 mm × 1 m 1000 mm ) ( 2 mm × 1 m 1000 mm 2 ) ( T 12 − T 7 ) ( 10 mm × 1 m 1000 mm ) ] = 0 ( 50 °C − T 7 ) + 12.5 ( T 8 − T 7 ) + ( T 12 − T 7 ) = 0 The finite difference equation at the node 8 is calculated as, k Δ y ( T 2 − T 8 ) Δ x + k Δ x ( T 7 − T 8 ) Δ y + k Δ y ( T 13 − T 8 ) Δ x + k Δ x ( T 9 − T 8 ) Δ y = 0 ( 25 W / m ⋅ K ) [ ( 2 mm × 1 m 1000 mm ) ( 50 °C − T 8 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 7 − T 8 ) ( 2 mm × 1 m 1000 mm ) + ( 2 mm × 1 m 1000 mm ) ( T 13 − T 8 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 9 − T 8 ) ( 2 mm × 1 m 1000 mm ) ] = 0 5 ( 50 °C − T 8 ) + 125 ( T 7 − T 8 ) + 5 ( T 13 − T 8 ) + 125 ( T 9 − T 8 ) = 0 The finite difference equation at the node 9 is calculated as, k Δ y ( T 3 − T 9 ) Δ x + k Δ x ( T 8 − T 9 ) Δ y + k Δ y ( T 14 − T 9 ) Δ x + k Δ x ( T 10 − T 9 ) Δ y = 0 ( 25 W / m ⋅ K ) [ ( 2 mm × 1 m 1000 mm ) ( 50 °C − T 9 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 8 − T 9 ) ( 2 mm × 1 m 1000 mm ) + ( 2 mm × 1 m 1000 mm ) ( T 14 − T 9 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 10 − T 9 ) ( 2 mm × 1 m 1000 mm ) ] = 0 5 ( 50 °C − T 9 ) + 125 ( T 8 − T 9 ) + 5 ( T 14 − T 9 ) + 125 ( T 10 − T 9 ) = 0 The finite difference equation at the node 10 is calculated as, k Δ y ( T 4 − T 10 ) Δ x + k Δ x ( T 9 − T 10 ) Δ y + k Δ y ( T 15 − T 10 ) Δ x + k Δ x ( T 11 − T 10 ) Δ y = 0 ( 25 W / m ⋅ K ) [ ( 2 mm × 1 m 1000 mm ) ( 50 °C − T 10 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 9 − T 10 ) ( 2 mm × 1 m 1000 mm ) + ( 2 mm × 1 m 1000 mm ) ( T 15 − T 10 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 11 − T 10 ) ( 2 mm × 1 m 1000 mm ) ] = 0 5 ( 50 °C − T 10 ) + 125 ( T 9 − T 10 ) + 5 ( T 15 − T 10 ) + 125 ( T 11 − T 10 ) = 0 The finite difference equation at the node 11 is calculated as, k Δ y ( T 5 − T 11 ) Δ x + k Δ x ( T 10 − T 11 ) Δ y + h Δ s ( T ∞ − T 11 ) = 0 ( 25 W / m ⋅ K ) [ ( 2 mm × 1 m 1000 mm ) ( 50 °C − T 11 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 10 − T 11 ) ( 2 mm × 1 m 1000 mm ) + ( 50 W / m 2 ⋅ K ) ( 10.2 mm × 1 m 1000 mm ) ( 20 °C − T 11 ) ] = 0 5 ( 50 °C − T 11 ) + 125 ( T 10 − T 11 ) + 0.51 ( 20 °C − T 11 ) = 0 The finite difference equation at the node 12 is calculated as, k ( Δ y 2 ) ( T 1 − T 7 ) Δ x + k Δ x ( T 8 − T 7 ) Δ y + k ( Δ y 2 ) ( T 12 − T 7 ) Δ x = 0 ( 25 W / m ⋅ K ) [ ( 2 mm × 1 m 1000 mm 2 ) ( T 7 − T 12 ) ( 10 mm × 1 m 1000 mm ) + ( 10 mm × 1 m 1000 mm ) ( T 13 − T 12 ) ( 2 mm × 1 m 1000 mm ) ( 2 mm × 1 m 1000 mm 2 ) ( T 16 − T 12 ) ( 10 mm × 1 m 1000 mm ) ] = 0 2.5 ( T 7 − T 12 ) + 125 ( T 13 − T 12 ) + 2

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 4, Problem 4.66P

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Introduction to Heat Transfer

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The efficiency of the fin η f and compare the fin efficiency with the reported in Figure 3.19 .

Solution Summary: The author compares the fin efficiency with the reported in Figure 3.19.

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The efficiency of the fin ηf and compare the fin efficiency with the reported in Figure 3.19 .

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