Explanation Given: The thermal conductivity of cylinder is k = 237 W / m ⋅ K . The thermal diffusivity of cylinder is α = 9.71 × 10 − 5 m 2 / sec . The base diameter of cylindrical is D = 15 cm . The initial temperature of the cylinder is T i = 115 ° C . The temperature of water is T ∞ = 10 ° C . The convective heat transfer coefficient is h = 140 W / m 2 ⋅ K . The time of cooling is t = 8 min . Formula used: The expression for Fourier number is given as, τ = α t r 2 Here, t is the time and r is the radius of cylinder. The expression for Biot number is given as, B i = h r k Here, h is convective heat transfer coefficient and r is radius of cylinder. The dimensionless temperature 5 cm from the surface of a semi-infinite medium is given as, T − T ∞ T i − T ∞ = erfc ( x 2 α t ) − exp ( h x k + h 2 α t k 2 ) [ erfc ( x 2 α t + h α t k ) ] Here, x is the distance. The dimensionless temperature at the center of the cylinder is given as, θ cyl = T − T ∞ T i − T ∞ = A 1 e − λ 1 2 τ Here, T i is the initial temperature and A 1 , λ 1 are constants. The center temperature of semi-infinite cylinder is given as, T − T ∞ T ∞ − T i = θ semi × θ cyl Here, T is the center temperature. The expression for interpolation is given as, x 2 − x 1 x − x 1 = y 2 − y 1 y − y 1 Here, x 2 is the nearest highest value to the given value, x 1 is the nearest lower value to the given value, x is the given value, y 2 is the property of higher value, y 1 is the property of lower value, y is the property that is to be calculated. Calculation: The dimensionless temperature 5 cm from the surface of a semi-infinite medium can be calculated as, T − T ∞ T i − T ∞ = erfc ( x 2 α t ) − exp ( h x k + h 2 α t k 2 ) [ erfc ( x 2 α t + h α t k ) ] T − T ∞ T i − T ∞ = { erfc ( 5 cm × 1 cm 100 m 2 9.71 × 10 − 5 m 2 / sec × 8 ( min × 60 sec ) ) − exp ( 140 W / m 2 ⋅ K × 5 cm × 1 cm 100 m 237 W / m ⋅ K + ( 140 W / m 2 ⋅ K ) 2 × 9.71 × 10 − 5 m 2 / sec × 8 ( min × 60 sec ) ( 237 W / m ⋅ K ) 2 ) × [ erfc ( 5 cm × 1 cm 100 m 2 9.71 × 10 − 5 m 2 / sec × 8 ( min × 60 sec ) + 140 W / m 2 ⋅ K × 9.71 × 10 − 5 m 2 / sec × 8 ( min × 60 sec ) 237 W / m ⋅ K ) ] } = erfc ( 0

6th Edition

ISBN: 9781260699326

Author: CENGEL

Publisher: MCG

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Chapter 4, Problem 120P

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The center temperature of the semi-infinite aluminum cylinder.

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Chapter 4, Problem 119P

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