Elementary Principles 6

# Elementary Principles 6 - 2.22 N Pr = N Pr Cp k = 0.583 J /...

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2- 6 2.22 N C k C C N p o o Pr Pr .. . () ( ) ( ) ( ) ( ) . == ⋅⋅ ×× × × ≈× × μ 0583 1936 32808 0 286 61 0 21 0 31 0 31 0 41 0 2 0 2 15 10 163 10 133 13 3 33 J / g lb 1 h ft 1000 g W / m ft h 3600 s m 2.20462 lb The calculator solution is m m 2.23 Re . . Re ( ) ( ) ( ) ( ) ( ) ( ×⋅ × −− Du ρ 048 2067 0805 043 10 51 0 281 0 1 0 34 1 01 0 4 1 0 0 3 21 0 3 11 6 34 ) 4 ft 1 m in 1 m g 1 kg 10 cm s 3.2808 ft kg / m s 39.37 in cm 1000 g 1 m the flow is turbulent 63 2.24 1/2 1/3 52 3 35 2 5 2 (a) 2.00 0.600 1.00 10 N s/m (0.00500 m)(10.0 m/s)(1.00 kg/m ) 2.00 0.600 (1.00 kg/m )(1.00 10 m / s) (1.00 10 N s/m ) (0.00500 44.426 ρμ ⎛⎞ =+ ⎜⎟ ⎝⎠ ⎡⎤ ⎢⎥ ⎣⎦ =⇒ gp p g kd y du DD k m)(0.100) 44.426 0.888 m/s 1.00 10 m /s = × g k (b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) d p (m) y D (m 2 /s) μ (N-s/m 2 ) ρ (kg/m 3 ) u (m/s) k g 0.005 0.1 1.00E-05 1.00E-05
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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