Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Air exits at 200of assumptions specific heat of

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Unformatted text preview: es = 2(15.15)(14.8) = 448 cm2 Rate of drying in constant-rate period = 0.151 g H2O/g dry wood-h or 0.15(84) = 12.6 g H2O/h-piece of wood Therefore, Rc = 12.6/448 = 0.0281 g H2O/h-cm2 (d) Estimate the effective diffusivity from a rearrangement of (18-62). From the given data, Xc = 0.968, X* = 0.066, Xavg = 0.121 at t = 22 h tc = 2 h, therefore, tf = 22 – 2 = 20 h or 72,000 s DAB = Xc − X * a2 0.3752 0.968 − 0.066 ln = ln = 1.8 × 10−6 cm2/s * X avg − X 3t f 3 ( 72, 000 ) 0.121 − 0.066 (c) Now check the Fourier number for mass transfer to see if moisture profile is parabolic at tc. N FoM −6 DABtc (1.8 × 10 ) 2 ( 3600 ) at tc = 2 = = 0.092 2 a ( 0.375 ) which is less than 0.5. Therefore, profile is not parabolic. Exercise 18.34 Subject: Falling-rate equations for drying. Given: Solid undergoing drying in the falling-rate period either by liquid diffusion or capillary movement. Find: Derive governing equations. Outline an experimental procedure for determining which mechanism prevails. Analysis: Case of Liquid diffusion controlling in falling-rate period: From (3) in Example 18.13, which is a simplification of (18-49) when N Fo M <0.1, N Fo M DABt 4 8 Xo − X * = 2 = 2 ln 2 a π π X avg − X * (1) From (18-32), the rate of drying in terms of average moisture content is R=− Rearranging (1), X avg = X * + ms dX avg A dt 8( Xo − X * ) π2 Taking the derivative of (3) with respect to t gives dX avg dt = 8( Xo − X * ) π2 − exp − (2) π2 DABt 4a 2 π2 DABt π2 DABt π2 DABt exp − = ( X avg − X * ) − 4a 2 4a 2 4a 2 (3) (4) Comparing (2) and (4), it is seen that the rate of drying varies inversely as the square of a, which is the thickness of the solid. Case of capillary flow controlling in falling-rate period: Capillary flow will be laminar. From (14-2), the rate of laminar flow is inversely proportional to the distance in the direction of flow, which is the thickness of the solid. To determine experimentally which case applies, use solids of different thickness. Exercise 18.35 Subject: Cross-circulation tray drying in the constant-rate drying period. Given: Wet solid in a tray subject to heat transfer by convection to the bottom of the tray, in addition to the usual heat convection from the hot air to the surface of the wet solid. Find: Derive the governing equation and show that the temperature of the surface of the wet solid will be higher than the wet-bulb temperature of the hot air used to dry the solid. In addition, determine the effect of radiation from the bottom of the tray to a tray below. Assumption: Neglect the conduction resistance of the tray bottom and heat transfer to the sides of the trays. Analysis: This situation is discussed in the reference: Shepherd, C.B, C. Hadlock, and R.C. Brewer, Ind. Eng. Chem., 30, 388-397 (1938), which develops the theory and provides experimental data on the drying of wet sand that supports the theory. Their results clearly show that when heat is transferred also to the bottom of the tray, the temperatur...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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