Explanation Given: The length of the cylindrical aluminum block is L = 30 cm . The diameter of the cylindrical aluminum block is D = 15 cm . The density of the cylindrical aluminum block is ρ = 2702 kg / m 3 . The specific heat of the cylindrical aluminum block is c p = 0.896 kJ / kg . The thermal conductivity of the cylindrical aluminum block is k = 236 W / m ⋅ K . The thermal diffusivity of the cylindrical aluminum block is α = 9.75 × 10 − 5 m 2 / s . The heat transfer coefficient of cylindrical aluminum block is h = 80 W / m 2 ⋅ K . The temperature of cylindrical aluminum block is T i = 20 ° C . The temperature of furnace is T ∞ = 1200 ° C . The temperature at the center is T o = 300 ° C . Formula used: The Biot number is given as, B i = h L k Here, L c is the characteristic length, h is the heat transfer coefficient and k is the thermal conductivity. The interpolation for the coefficient of Biot number is given as, y − y 1 = y 2 − y 1 x 2 − x 1 ( x − 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. The Fourier number is given as, τ = α t L 2 Here, α is the thermal diffusivity, t is the characteristics time and L is the characteristic length. The one-term approximation for the cylindrical block is given as, θ ( 0 , 0 , t ) block = θ ( 0 , t ) wall θ ( 0 , t ) cyl = ( A 1 e − λ 2 τ ) wall ( A 1 e − λ 2 τ ) cyl T o − T ∞ T i − T ∞ = ( A e − λ 2 τ ) wall ( A e − λ 2 τ ) cyl The mass of cylinder block is given as, m = ρ π r o 2 L The maximum amount of heat transfer is given as, Q max = m c p ( T ∞ − T i ) The dimensionless heat transfer for the wall is given as, ( Q Q max ) wall = 1 − θ o,wall sin ( λ ) λ Calculation: The Biot number for long cylinder can be calculated as, B i = h L k = 80 W / m 2 ⋅ ° C × 0.3 m 236 W / m ⋅ ° C = 0.101 Refer table 4-2 “Coefficients used in the one term approximate solution” And obtain values at B i = 0.1 λ = 0.3111 A = 1.0161 The Biot number radial distance can be calculated as, B i r o = h r o k = 80 W / m 2 ⋅ ° C × 0.075 m 236 W / m 2 ⋅ ° C = 0.0254 Refer table 4-2 ‘coefficients used in the one term approximate solution’ And obtain values for B i 1 = 0.02 B i 2 = 0.04 λ 1 = 0.1995 λ 2 = 0.2814 Further, A 1 = 1.0050 A 2 = 1.0099 The interpolation for the coefficient of Biot number for cylinder length can be calculated as y = y 2 − y 1 x 2 − x 1 ( x − x 1 ) + y 1 λ = λ 2 − λ 1 B i 2 − B i 1 ( B i − B i 1 ) + λ 1 = 0.2814 − 0.1995 0.04 − 0.02 ( 0.0254 − 0.02 ) + 0.1995 = 0.2216 Solving for the constant A , A = A 2 − A 1 B i 2 − B i 1 ( B i − B i 1 ) + A 1 = 1

6th Edition

ISBN: 9781260439991

Author: CENGEL

Publisher: MCG

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Chapter 4, Problem 122P

To determine

How long the block should be kept in the furnace.

The amount of heat transfer from the aluminum block.

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Chapter 4, Problem 121P

Chapter 4, Problem 123P

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