Explanation Formula used: The expression for the maximum spacing is given as, S o p t = 2.71 ( R a s S 3 H ) − 1 4 S o p t = 2.71 [ g β ( T s − T ∞ ) α v H ] − 1 4 Here, g is the acceleration due to the gravity, β is the volumetric thermal expansion, and α is the thermal diffusivity. The expression for the Nusselt number is N u s = [ C 1 ( R a s S L ) 2 + C 2 ( R a s S L ) 1 2 ] − 1 2 The expression for the Rayleigh’s number is given as, R a L , c = g β ( T 1 − T 2 ) L c 3 α v Here v is the kinematic viscosity. The expression for the convective heat transfer is h = N u s k S The expression for the interpolation method is given as, y = y 1 + [ ( x − x 1 ) ( y 2 − y 1 ) ( x 2 − x 1 ) ] Here, y is the property, that is to be calculated, x is the given value, x 1 is the nearest lower value to the given value, x 2 is the nearest highest value to the given value, y 1 is the property of lower value and y 2 is the property of the higher value. Calculation: The following assumptions is to be made while solving the problem. 1. The fins are isothermal 2. The radiations are negligible. 3. The heat transfer from the tip of the fins are negligible. Refer to the table A-4,“ Thermo-physical properties of air”. Obtain the values at corresponding temperature of 325 K . The interpolation method can be calculated as, The interpolation method for kinematic viscosity can be calculated as, *The value of kinematic viscosity at 300 K is v 1 = ( 15.89 × 10 − 6 m 2 / s ) *The value of kinematic viscosity at 350 K is v 2 = ( 20.92 × 10 − 6 m 2 / s ) v = ( 15.89 × 10 − 6 m 2 / s ) + [ ( 325 K − 300 K ) { ( 20.92 × 10 − 6 m 2 / s − 15.89 × 10 − 6 m 2 / s ) ( 350 K − 300 K ) } ] v = 18.41 × 10 − 6 m 2 / s The interpolation method for conductive heat transfer coefficient can be calculated as, *The value of conductive heat transfer coefficient at 300 K is k 1 = ( 26.3 × 10 − 3 W / m ⋅ K ) *The value of conductive heat transfer coefficient at 350 K is k 2 = ( 30 × 10 − 3 W / m ⋅ K ) k = ( 26.3 × 10 − 3 W / m ⋅ K ) + [ ( 325 K − 300 K ) { ( 30 × 10 − 3 W / m ⋅ K − 26.3 × 10 − 3 W / m ⋅ K ) ( 350 K − 300 K ) } ] k = 0.028 W / m ⋅ K Further, The interpolation method for thermal diffusivity can be calculated as, *The value of thermal diffusivity at 300 K is α 1 = ( 22.5 × 10 − 6 W / m ⋅ K ) *The value of thermal diffusivity at 350 K is α 2 = ( 29

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ISBN: 9780470917855

Author: Bergman, Theodore L./

Publisher: John Wiley & Sons Inc

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Chapter 9, Problem 9.79P

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The optimum fin spacing.

The corresponding fin heat rate transfer.

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-1. Natural Convection from an Oven Wall. The oven wall in Example 15.5-1 is insu- lated so that the surface temperature is 366.5 K instead of 505.4 K. Calculate the natural convection heat-transfer coefficient and the heat-transfer rate per m of width. Use both Eq. (15.5-4) and the simplified equation. (Note: Radiation is being neglected in this calculation.) Use both SI and English units.

2. Losses by Natural Convection from diameter and 121.9 mm Cylinder. A vertical cylinder 76.2 mm in high is maintained at 397.1 K at its surface. It loses heat by natural convection to air at 294.3 K. Heat is lost from the cylindrical side and the flat circular end at the top. Calculate the heat loss, neglecting radiation losses. Use the simplified equations of Table 15.5-2 and those equations for the lowest range of NGr Npr. The equivalent L to use for the top flat surface is 0.9 times the diameter. Ans. q 26.0 W a 11

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Fundamentals of Heat and Mass Transfer

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The optimum fin spacing. The corresponding fin heat rate transfer.

Solution Summary: The author describes the optimum fin spacing and corresponding fin heat rate transfer. The expression for the Rayleigh’s number is given as, Nu_s=left.

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The optimum fin spacing. The corresponding fin heat rate transfer.

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The optimum fin spacing.

The corresponding fin heat rate transfer.