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

# 0693 00555 analysis b continued exercise 625

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Unformatted text preview: 2 flow rate in exit gas = 2.22(23/737) = 0.0693 lbmol/h Molecular weight of SO2 = 64.06 SO2 flow rate in exit liquid = 0.6(1,000/(100)(64.06) = 0.0937 lbmol/h By material balance, SO2 flow rate in entering air = 0.0693 + 0.0937 = 0.1630 lbmol/h Partial pressure of SO2 in entering gas = 0.1630/(0.1630 + 2.22) = 0.0690 atm % of SO2 absorbed = 0.0937/0.1630 x 100% = 57.5% (b) At the point, the rate of mass transfer for SO2 as a flux across the gas-liquid interface, can be written by the two-film theory as, kp(pb - pi) = kL(ci - cb) (1) As shown in Fig. 6.31, the equilibrium interface composition can be determined in terms of the ratio of mass-transfer coefficients. However, here, instead of compositions in mole fractions, the gas composition is in partial pressure and the liquid is in concentration. From Eq. (1), kL 1.3 p − pi = = 6.67 = b k p 0.195 ci − cb (2) At the point (column height location), bulk liquid concentration = 0.001 lbmol SO2/lbmol H2O and the flow rate is 0.001(55.5) = 0.0555 lbmol/h for SO2 in the liquid phase. The flow of SO2 in the gas phase at that location is obtained by a material balance around the top of the column: SO2 in entering liquid + SO2 in gas at the point = SO2 in liquid at the point + SO2 in exit gas Therefore, 0 + SO2 flow rate in gas at the point = 0.0693 + 0.0555 Analysis: (b) (continued) Exercise 6.25 (continued) SO2 flow rate in gas at the point = 0.1248 lbmol/h. Therefore, the partial pressure of SO2 in the bulk gas at the point = 0.1248/(0.1248 + 2.22) x 1 atm = 0.0532 atm = pb . The density of water is 62.4 lb/ft3 or 62.4/18.02 = 3.46 lbmol/ft3. Therefore, the concentration of SO2 in the bulk liquid at the point = 0.001(3.46) = 0.00346 lbmol/ft3 = cb . We now need an algebraic equilibrium relationship between pi and ci . Convert the given equilibrium data to partial pressures and concentrations in the vicinity of the values at the point: The SO2 concentration in the liquid is obtained from, c, lbmol SO 2 lb SO 2 = 3 100 lb H 2 O ft lb SO2 / 100 lb H2O 0.30 0.50 0.70 1.00 ci , lbmol SO2 / ft3 0.00292 0.00487 0.00682 0.00974 62.4 64.06 1 100 Partial pressure SO2 , torr 14.1 26.0 39.0 59.0 pi of SO2 , atm 0.01855 0.03421 0.05132 0.07763 These equilibrium data fit the curve, pi = 5.04536ci + 511.783ci2 - 21714.4ci3 (3) From Eq. (2), (4) Solving Eqs. (3) and (4), 0.0532 - pi = 6.67(ci - 0.00346) ci = 0.00550 lbmol/ft3 and pi = 0.0396 atm The same result can be obtained by constructing a plot of SO2 partial pressure versus SO2 concentration in the liquid, similar to Fig. 6.31. First, the operating line is drawn as a straight line connecting the column end points (0.0303, 0.0) and (0.0690, 0.00584), as (p , c). Then the equilibrium curve is drawn, using Eq. (3). Then the point (pb , cb) is marked on the operating line. A straight line is extended from this point, with a slope of (-kL/kp) = -6.67, to the point of intersection on the equilibrium line, giving the same result as above Exercise 6.26 Subject: Stripping of benze...
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