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

# 7 for f 04 analysis a example 65 in this example a

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Unformatted text preview: n. 2/3 qL From Eq. (6-51), hl = φ e hw + C Lw φ e 40,000 = 0.72 2.5 + 0.362 385 0.72 2/3 = 9.0 in. of nC8 From Eq. (6-55), with maximum bubble size of 1/4 inch = 0.00635 m, hσ = 6σ / gρ L DBmax = 6(20 / 1000) / (9.8)(703)(0.00635) = 0.00274 m = 0.11 in. nC8 From Eq. (2), ht = hd + hl + hσ = 0.045 + 9.0 + 0.11 = 9.2 in. nC8 = 0.23 psi/tray. Excessive! (c) Apply criterion of Eq. (6-64). hd + hσ = 0.045 + 0.11 = 0.155 in. < hl = 9.0 in. Therefore, weeping will occur. (d) From Fig. 6.28, since FLV = 1.64, entrainment will be very low. (e) Because entrainment is very low, EMV will not decrease. (f) From Eqs. (6-70) and (6-72), hdf =(ht + hl + hda) / 2 (3) Area of downcomer opening = Ada = Lwha Take ha = 2 in., Ada = (32.1)(2/12) = 5.4 ft2 qL From Eq. (6-71), hda = 0.03 100 Ada 2 40, 000 = 0.03 100(5.4) 2 =164 in. Very excessive Exercise 6.21 Subject: Column performance for 40% of flooding. Given: Data in Examples 6.5, 6.6, and 6.7. Find: (a) Column diameter in Example 6.5 for f = 0.4. (b) Vapor pressure drop in Example 6.6 for f = 0.4. (c) Murphree vapor-point efficiency in Example 6.7 for f = 0.4 Analysis: (a) Example 6.5: In this example, a value of f = 0.80 was used, giving DT = 2.65 ft From Eq. (6-44), by ratioing values of f, 0.80 DT = 2.65 0.40 1/ 2 = 3.75 ft = 1.15 m (b) Example 6.6: In this example, a tower diameter of 1 m gives a vapor pressure drop = 0.093 psi/tray, with a vapor hole velocity of 47.9 ft/s, a weir length of 0.73 m, an active area vapor velocity of 5.99 ft/s, and Ks = 0.265 ft/s. Vapor hole velocity varies inversely with the square of the column diameter. Therefore, uo = 14.6(1/1.15)2 = 11.0 m/s = 36 ft/s From Eq. (6-50), hd is directly proportional to hole velocity squared. Therefore, hd = 1.56(36/47.9)2 = 0.88 in. of liquid Weir length is proportional to column diameter. So, Lw = 0.73(1.15/1) = 0.84 m = 0.33 in. Vapor velocity based on active area varies inversely with the square of column diameter. Therefore, Ua = 5.99(1/1.15)2 = 4.53 ft/s. From Eq. (6-53), Ks is proportional to Ua. Thus, Ks = 0.265(4.53/5.99) = 0.20 ft/s From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.20)0.91] = 0.37 The value of C in Eq. (6-51) remains at 0.362. From Eq. (6-51), qL hl = φ e hw + C Lw φ e 2/3 12.9 = 0.37 2 + 0.362 33 0.37 2/3 = 0.88 in. hσ = 0.36 in. (no change from Example 6.6) From Eq. (6-49), ht = hd + hl + hσ = 0,88 + 0.88 + 0.36 = 2.12 in. Tray pressure drop = htρL = 2.12(0.0356) = 0.076 psi/tray (c) Example 6.7: In this example, a tower diameter of 1 m gives a EOV = 0.77. Must redo all calculations. DT = 1.15 m, A = 1.038 m2 = 10,380 cm2, Aa = 0.8(1.038) = 0.83 m2 = 8,300 cm2 Lw = 33 in. = 0.84 m, φe = 0.37, hl = 0.88 in. = 2.24 cm Ua = 4.53 ft/s = 137 cm/s, Uf = 10.2 ft/s, f = Ua/Uf = 4.53/10.2 = 0.44 Analysis: (c) (continued) Exercise 6.21 (continued) F = UaρV0..5 = 1.37(1.92)0.5 = 1.90 (kg/m)0.5/s , From Eq. (6-64), t L = From Eq. (6-65), t G = qL = 812 cm3/s hl Aa 2.24(10,380) = = 28.6 s qL 812 1 − φ e hl (1 −...
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