Thermodynamics HW Solutions 335

Thermodynamics HW Solutions 335 - k = 0.26 Btu/h ft F and...

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Chapter 4 Transient Heat Conduction 4-58E The center temperature of meat slabs is to be lowered to 36 ° F during 12-h of cooling. The average heat transfer coefficient during this cooling process is to be determined. Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 3-in. 2 Heat conduction in the meat slabs is one-dimensional because of symmetry about the centerplane. 3 The thermal properties of the meat slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of meat slabs are given to be
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Unformatted text preview: k = 0.26 Btu/h ft F and =1.4 10-6 ft 2 /s. Air 23 F Meat 50 F Analysis The average heat transfer coefficient during this cooling process is determined from the transient temperature charts for a flat plate as follows: ) 13a 4 (Fig. 7 . 1 481 . 23 50 23 36 968 . ft) (3/12 s) 3600 ft/s)(12 10 (1.4 6 = = = = = = Bi T T T T L t i o Therefore, F Btu/h.ft. 1.5 = = = ft (3/12) F)(1/0.7) Btu/h.ft. (0.26 L kBi h Discussion We could avoid the uncertainty associated with the reading of the charts and obtain a more accurate result by using the one-term solution relation for an infinite plane wall, but it would require a trial and error approach since the Bi number is not known. 4-46...
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This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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