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

# Assume that the thermal resistance of the metal tray

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Unformatted text preview: − 85 1045.8 = 397 min 1206.6 From the above plot, it might take an additional 10 min to go from 10.2 to 10.0 wt %. Therefore, take tf = 397 + 10 = 407 min. The total drying time = tc + tf = 123 + 407 = 530 min. Exercise 18.32 Subject: Drying time for a piece of hemlock wood Given: Hemlock wood measuring 15.15 x 14.8 x 0.75 cm is to be dried at the two large faces from an initial total moisture content of 90% to a final average total moisture content of 10%, both dry basis. Drying is in the liquid-diffusion-controlled falling-rate period with a diffusivity of 1.7 x 10-6 cm2/s. Assumptions: Equilibrium moisture content with bone-dry air is zero. Find: Drying time if bone-dry air is used. Analysis: Fig. 3.9 applies, but the coordinates are off the chart. Therefore, use (2) of Example 18.13, provided that NFo for mass transfer > 0.1 Eavg = average unaccomplished free-moisture change = N Fo M X avg − X * Xo − X * = 0.10 − 0 = 0.111 0.90 − 0 −6 DABt 1.7 × 10 ( t in s ) = 2= = 1.2 × 10 −5 ( t in s ) 2 a ( 0.75/2 ) From (2) of Example 18.13, ln X avg − X * Xo − X * 8 π2 = ln 2 − N Fo M π 4 Therfore, ln ( 0.111) = −2.197 = −0.210 − 2.467 N FoM Solving, N Fo M = 0.805 > 0.1 Therefore, 0.805 = 1.2 ×10−5 ( t in s ) and solving, t = 67,100 s or 18.6 h Exercise 18.33 Subject: Mode of drying in the falling-rate period for hemlock wood Given: A piece of hemlock wood measuring 15.15 x 14.8 x 0.75 cm, for which drying takes place only from the two largest faces. Air at 25oC is passed over the surfaces at 3.7 m/s. Dryingtime data are plotted below starting at an average total moisture content of 127 wt% dry basis and finishing at an infinite time of 6.6 wt% dry basis. Assumptions: Air is at 1 atm and a wet-bulb temperature of 17oC. Find: Whether Case 1 or Case 2 applied for the falling-rate period. If Case 1 applies, determine from the data the effective diffusivity. If Case 2 applies, determine: (a) Drying rate in g/h-cm2 for the constant-rate period assuming a dry density of 0.5 3 g/cm and no shrinkage. (b) Critical moisture content. (c) Predicted parabolic-moisture-content profile at the beginning of the falling-rate period. (d) Effective diffusivity during the falling-rate period. Analysis: Assume that the equilibrium-moisture content is that measured at infinite time. Thus, it is 6.6 wt% dry basis. Now find out if Case 1 or Case 2 applies by plotting moisture content against time with a spreadsheet. 140 % Moisture Content, dry basis 120 100 80 60 40 20 0 0 5 10 15 20 25 Time, hours From this plot, it is seen that the falling-rate period is preceded by a constant-rate period of w 2 h ours. Therefore, Case 2 applies. Exercise 18.33 continued (b) From the plot, the constant-rate period extends for 2 hours to Xc = 96.8 wt% dry basis. (a) The drying rate in the constant-rate period = (127 – 96.8)/[(100)2] = 0.151 lb H2O/lb dry wood-h. Convert this to g/h-cm2. Volume of one piece of wood = (15.15)(14.8)(0.75) = 168 cm3 Mass of dry wood in one piece = 0.5(168) = 84 g Area of 2 large fac...
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