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, Problem 9.54P

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(B) Cylindrical Water cans (diameter 6.5cm and length 15cm) are to be cooled from initial temperature of 20°C by placing theme in a cooler with ambient temperature of 1°C. Compare the initial cooling rates when the cans are laid horizontally to when the cans are laid vertically. Take v = 15.55x10° m²/s, k=0.024 W/mK and Pr=0.7. Nu = 0.53RA' a4 for horizontal orientation Nu = 0.59 Ra"4 for vertical orientation

(B) Cylindrical Water cans (diameter 6.5cm and length 15cm) are to be cooled from initial temperature of 20°C by placing theme in a cooler with ambient temperature of 1˚C. Compare the initial cooling rates when the cans are laid horizontally to when the cans are laid vertically. Take v = 15.55x10 m²/s, k=0.024 W/mK and Pr-0.7. Nu = 0.53 Ra¹ for horizontal orientation Nu = 0.59 Ra¹/4 for vertical orientation

a cylindrical can of bean puree, has a diameter of 70 mm and height of 126 mm, and is initially at a uniform temperature of 25 ° C. The cans are stacked vertically inside a retort into which steam is introduced at 120 ° C. Calculate the temperature in the center of the can after a heating time of 0.55 h at 120 ° C. Now suppose the can is in the center of a vertical stack, insulated at its two ends by the presence of the remaining cans. (The heat capacity of the metal wall of the can can be neglected.) The heat transfer coefficient of steam is estimated to be 4640 W / m2 ° K. The physical properties of the bean are k = 0.750 W / m ° K and the thermal diffusivity = 2.007 x 10-7 m2 / s.a) Calculate the temperature in the center of the product.

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