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

Analysis case of liquid diffusion controlling in

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Unformatted text preview: t material from 40 to 7 wt% on dry basis. Analysis: From (18-42), tT = tc + t f = ms ARc ( X o − X c ) + X c ln Xc X final (1) ms from given data for the 5-hour test. All moisture contents are on a freeARc moisture basis. Therefore, Xo = 0.36 – 0.05 = 0.31, Xc = 0.15 – 0.05 = 0.10, Xfinal = 0.08 – 0.05 = 0.03 Substituting into a rearrangement of (1), Solve for ms = ARc tc + t f X ( X o − X c ) + X c ln c X final = 5 0.10 ( 0.31 − 0.10 ) + 0.10 ln 0.03 = 15.13 h Use this value for the new conditions, with Xo = 0.40 – 0.05 = 0.35, Xc = 0.15 – 0.05 = 0.10, Xfinal = 0.07 – 0.05 = 0.02 tT = tc + t f = (15.13) ( 0.35 − 0.10 ) + 0.10 ln 0.10 0.02 = 6.2 h Exercise 18.31 Subject: Tunnel drying of a wet solid on trays with drying by crossflow of air from both sides. Given: Trays measuring 1.5 m long by 1.2 m wide by 25 cm deep. Initial total moisture content = 116 wt% dry basis and final average total moisture content = 10 wt% dry basis. Air is at 90oF, 1 atm, and a relative humidity of 15%. By interpolation of given data, equilibrium moisture content = 3.1 wt% dry basis. Drying-time data plotted below. Assumptions: Critical moisture content is independent of drying conditions. Drying rate is proportional to the difference between dry-bulb and wet-bulb temperatures of the air. Find: Drying time for 110 to 10 wt% dry basis using air at 125oF and 20% relative humidity. Analysis: The following is a spreadsheet plot of the drying-time data for air at 90oF and an RH of 15% starting from 116% total moisture. Two lines are shown. The upper line is the total and the bottom line is the free moisture content = total – 3.1%. 140 120 % Moisture Content, dry basis 100 80 60 40 20 0 0 100 200 300 400 500 600 700 800 900 Time, minutes From this plot, the critical free-moisture content, Xc = approximately 57.4 – 3.1 = 54.3 wt% dry basis at 211 minutes. During the constant-rate drying period, using the given drying-rate data, The rate of drying = (1.16 – 0.574)/211 = 0.00278 lb H2O/lb dry solid – min Exercise 18.31 continued In the constant-rate period, Rc = h (Tg − Tw ) ∆H vap w in lb H2O/ft2 of drying surface – min For the test data with air at 90oF, 1 atm, and RH of 15%, from Fig. 18.17, Tw = 60oF. v From the steam tables, ∆H wap = 1206.6 Btu/lb For the new conditions with air at 125oF, 1 atm, and RH of 20%, Tw = 85oF v From the steam tables, ∆H wap = 1045.8 Btu/lb Therefore, the drying time for the constant-rate period for the new conditions, by ratio using the above expression for the rate of drying is, using total moisture content differences: tc = 211 90 − 60 125 − 85 1045.8 1206.6 110 − 57.4 = 123 min 116 − 57.4 In the falling-rate period, since there is little change in the final moisture content between the test conditions and the new conditions, since both terminate at a total moisture content of approximately 10 wt%, ratio on rates using the temperature differences and the heat of vaporization. For the test data, tf = 822 – 211 = 611 minutes. Therefore to get to 10.2 wt%, t f = 611 90 − 60 125...
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