Explanation Given: Length of can is 150 mm . Diameter of can is 60 mm . Temperature of refrigerator is 4 ° C . Concept used: Assume can is at room temperature of 300 K. Calculate the mean temperature as follows: T = 300 + ( 4 + 273 ) 2 = 288.5 K Refer to table A-4 “Thermo physical properties of air at atmospheric pressure” to obtain the kinematic viscosity of air as 14.866 × 10 − 6 m 2 /s , thermal conductivity as 0.02538 W/m ⋅ K , Thermal diffusivity as 20.982 × 10 − 6 m 2 /s and Prandtl number as 0.71 corresponding to temperature of 288.5 K by the help of interpolation. Write the expression for Rayleigh number for can in vertical position. Ra L = g β ( T s − T ∞ ) L 3 α ν ........ (1) Here, Ra L is the Rayleigh number, g is the gravitational acceleration, T s is the surface temperature, L is the length of can, α is the thermal diffusivity, ν is the kinematic viscosity and T ∞ is the temperature of ambient. Write the expression for Nusselt number as per Churchill Chu correlation for 10 4 < R a L < 10 9 . N u ¯ L = { 0.825 + 0.387 R a L 1 / 6 [ 1 + ( 0.492 Pr ) 9 / 16 ] 8 / 27 } 2 ........ (2) Here, Pr is the Prandtl number. Write the expression for heat transfer coefficient for vertical positioned can. N u ¯ L = h v ¯ L k ........ (3) Here, h v ¯ is the heat transfer coefficient for vertical positioned can and k is the thermal conductivity of air. Write the expression for heat transfer when can in vertical position. q v = h v ¯ π D L ( T s − T ∞ ) ........ (4) Here, q v is the heat transfer when can in vertical position. Write the expression for Rayleigh number for can in horizontal position. Ra D = g β ( T s − T ∞ ) D 3 α ν ........ (5) Write the expression for Nusselt number as per Morgan relation for can in horizontal position for 10 4 < R a L < 10 9 . N u ¯ D = { 0.60 + 0.387 R a D 1 / 6 [ 1 + ( 0.559 Pr ) 9 / 16 ] 8 / 27 } 2 ........ (6) Here, Pr is the Prandtl number. Write the expression for heat transfer coefficient for horizontal positioned can. N u ¯ D = h h ¯ D k ........ (7) Here, h h ¯ is the heat transfer coefficient for horizontal positioned can. Write the expression for heat transfer when can in horizontal position. q h = h h ¯ π D L ( T s − T ∞ ) ........ (8) Here, q h is the heat transfer when can in horizontal position. Calculation: Substitute 14.866 × 10 − 6 m 2 /s for ν , 20.982 × 10 − 6 m 2 /s for α , 9.81 m/s 2 for g , 1 288.5 K for β , 277 K for T ∞ , 300 K for T s and 150 mm for L in equation (1). Ra L = ( 9.81 ) ( 1 288.5 ) ( 300 − 277 ) ( 150 mm ( 1 m 1000 mm ) ) 3 ( 20.982 × 10 − 6 ) ( 14.866 × 10 − 6 ) = 2.64 × 10 − 3 311.918 × 10 − 12 = 8.46 × 10 6 Substitute 8.46 × 10 6 for R a L and 0.71 for Pr in equation (2). N u ¯ L = { 0.825 + 0.387 ( 8.46 × 10 6 ) 1 / 6 [ 1 + ( 0

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ISBN: 9780470501979

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 9, Problem 9.55P

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The position of can to maximize the cooling rate.

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Chapter 9 Solutions

Fundamentals of Heat and Mass Transfer

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The position of can to maximize the cooling rate.

Solution Summary: The author explains the position of can to maximize the cooling rate by calculating the kinematic viscosity of air, thermal conductivity, and Prandtl number.

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Chapter 9 Solutions