Summer 2001 ChE n363, Unique No. 76085 UNIT OPERATIONS II: SEPARATION PROCESSES Solution to Homework 4 Assigned: Wed, June 20, 2001 Due: Mon, June 25, 2001 Exercise 6.7 The exit gas from an alcohol fermenter consists of an air-CO2 mixture containing 10 mol% CO2 that is to be absorbed in a 5.0 N solution of triethanolamine, which contains 0.04 mol of carbon dioxide per mole of amine solution. If the column operates isothermally at 25oC, if the exit liquid contains 78.4% of the CO2 in the feed gas to the absorber, and if the absorption is carried out in a six-theoretical-plate column, calculate: (a) Moles of amine solution required per mole of feed gas (b) Exit gas composition Equilibrium Data Y 0.003 0.008 0.015 0.023 0.032 0.043 0.055 0.068 0.083 0.099 0.12 X 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 Y = moles CO2/mole air; X = moles CO2/mole amine solution Subject: Absorption of CO2 from air at 25oC by 5N aqueous triethanolamine. Given: Feed gas containing 10 mol% CO2 and 90 mol% air. Absorbent of 5N aqueous trieathanolamine containing 0.04 moles of CO2 per mole of amine solution. Column with 6 equilibrium stages. Exit liquid to contain 78.4% of the CO2 in the feed gas. Therefore, exit gas contains 21.6% of the CO2 in the entering gas. Equilibrium data for CO2 at 25oC in terms of mole ratios. Assumptions: Negligible absorption of air and stripping of amine and water. Find: (a) Moles of amine solution required per mole of feed gas. (b) Exit gas composition. Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Therefore, for CO2 X0 = 0.04 mol CO2/mol amine solution YN+1 = Y7 = 10/90 = 0.1111 mol CO2/mol air Y1 = 0.216(10/90) = 0.024 mol CO2/mol air (b) Therefore, the exit gas composition is 0.024 mol CO2/mol air or 0.024/(1 + 0.024) × 100% = 2.34 mol% CO2 and 97.66 mol% air (a) A plot of the equilibrium data as Y vs. X is given below. The operating point (X0, Y1) at the top of the column is included. A straight operating line through this point is found by trial an error to give 6 equilibrium stages, when using Y7 = 0.1111. The resulting XN = X6 = 0.0878. From Eq. (6-3), the slope of the operating line L’/G’ = (0.1111 – 0.024)/(0.0878 – 0.04) = 1.822 mol triethanolamine solution/mol air. The feed gas
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contains 9 mol air/10 mol feed gas. Therefore, moles of amine solution/mol feed gas = 1.822 (0.9) = 1.64 X6
Y7
X, mol CO2/mol amine solution
0.1
0.08
Operating line
0.06
0.04
Y1 0.02
0
Equilibrium Curve
0
0.02
X0
0.04
0.06 Y, mol CO2/mol air
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0.08
0.1
Exercise 6.11 Groundwater at a flow rate of 1,500 gpm, containing three volatile organic chemicals (VOCs), is to be stripped in a trayed tower with air to produce dinking water that will meet EPA standards. Relevant data are given below. Determine the maximum air flow rate in scfm (60oF, 1 atm) and the number of equilibrium sates required if an air flowrate of twice the minimum is used and the tower operates at 25oC and 1 atm. Also determine the composition in parts per million for each VOC in the resulting drinking water. Composition, ppm Component K-value Groundwater Max. for Drinking water 1,2- Dichloroethane (DCA) 60 85 0.005 Trichloroethylene (TCE) 650 120 0.005 1,1,1-Trichloroethane (TCA) 275 145 0.200 Note: ppm = parts per million by weight Subject: Stripping of VOCs from water by air in a trayed tower to produce drinking water. Given: 1500 gpm of groundwater containing ppm amounts of DCA, TCE, and TCA given below. Stripping at 1 atm and 25oC to reduce the ppm amounts to the very low levels below. K-values of VOCs given below. Assumptions: No stripping of water and no absorption of air. System is dilute with respect to the VOCs Find: Minimum air flow rate in scfm (60oF and 1 atm). Number of equilibrium stages for 2 times the minimum air flow. Composition in ppm for each VOC in the resulting drinking water. Analysis: Flow rate of water L = (1,500 gpm) (8.345 lb/gal)/18.02 lb/lbmol = 695 lbmol/min % stripping of a VOC = 100%× (inlet ppm – outlet ppm)/inlet ppm VOC K-value Inlet ppm Outlet ppm % stripping DCA 60 85 0.005 99.994 TCE 650 120 0.005 99.996 TCA 275 145 0.200 99.862 Because of its high % stripping and a K-value that is much lower than for the other two VOCs, the stripper is likely to be controlled by DCA. So base the calculations on DCA and then check to see that the % stripping for TCE and TCA exceed the above requirements. Because of the dilute conditions, use Kremser’s method. From Eq. (6-12), Gmin = (L/K) (fraction stripped) = (695/60) (0.99994) = 11.58 lbmol/min or mimimum flow rate = 11.58 lbmol/min (379 scf/lbmol) = 4,389 scfm at 60oF and 1 atm For 2 times the minimum value, G = 2 (11.58) = 23.16 lbmol/min K G 60(23.16 ) The stripping factor, given by Eq. (6-16), is for DCA, S DCA = DCA = = 2.0 L 695
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From Eq. (6-14), Fraction stripped = 0.99994 =
S N +1 − S 2 N +1 − 2 = Solving this S N +1 − 1 2 N +1 − 1
equation, N = 13 stages. Now, for G = 23.16 lbmol/min and N = 13 stages, compute the fraction stripped for TCE and TCA from Eq. (6-14) with the stripping factor from Eq. (6-16). The results are: VOC S Fraction stripped DCA 2.0 0.99994 TCE 21.7 1.00000 TCA 9.16 1.00000 The drinking water contains 0.005 ppm of DCA and essentially zero ppm of TCE and TCA.
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Exercise 6.13 Determine by the Kremser method the separation that can be achieved for the absorption operation indicated in Figure 6.47 for the following combinations of conditions: (a) Six equilibrium stages and 75 psia operating pressure (b) Three equilibrium stages and 150 psia operating pressure, (c) Six equilibrium stages and 150 psia operating pressure. At 90oF and 75 psia, the K-value of nC10 = 0.0011 Subject: Absorption of a light hydrocarbon gas mixture by n-decane as a function of pressure and number of equilibrium stages at 90oF. Given: Light hydrocarbon gas containing in lbmol/h, 1,660 C1, 168 C2, 96 C3, 52 nC4, and 24 nC5 for a total of 2,000 lbmol/h = V. Absorbent of L = 500 lbmol/h of nC10. K=value of nC10 = 0.0011. Find: Component flow rates in the lean gas and rich oil for: (a) N = 6 and P = 75 psia (b) N = 3 and P = 150 psia (c) N = 6 and P = 150 psia Analysis: Use the Kremser method with Eqs. (5-48) for fraction not absorbed, φA, (5-50) for fraction not stripped, φS, (5-38) for absorption factor, A, (5-51) for stripping factor, S, (5-45) and (5-53) for component flow rates in the exit gas, v1, and (5-44) and (5-52) for component flow rates in the exit liquid, lN, since no light hydrocarbons enter with the absorbent. Use Aspen Plus (or another source) to find the K-values. (a) N = 6 and P = 75 psia Component K-value A C1 29.0 0.0086 C2 6.08 0.0411 C3 2.03 0.1232 nC4 0.62 0.4032 nC5 0.21 1.1905 nC10 0.0011 Total (b) N = 3 and P = 150 psia Component K-value A C1 14.6 0.0171 C2 3.18 0.0786 C3 1.09 0.2294 nC4 0.35 0.7143 nC5 0.117 2.1368 nC10 0.00042 Total
φA 0.9914 0.9586 0.8768 0.5978 0.0797
S
0.0044
0.9956
φA 0.9829 0.9214 0.7728 0.3863 0.0573
S
0.0017
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φS
φS
0.9983
v1 1645.7 161.1 84.2 31.1 1.9 2.2 1926.2
l6 14.3 6.9 11.8 20.9 22.1 497.8 573.8
v1 1631.6 154.8 74.2 20.1 1.4 0.8 1882.9
l6 28.4 13.2 21.8 31.9 22.6 499.2 617.1
(a) N = 6 and P = 150 psia Component K-value A C1 14.6 0.0171 C2 3.18 0.0786 C3 1.09 0.2294 nC4 0.35 0.7143 nC5 0.117 2.1368 nC10 0.00042 Total
φA 0.9829 0.9214 0.7707 0.3153 0.0056
S
0.0017
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φS
0.9983
v1 1631.6 154.8 74.0 16.4 0.1 0.8 1877.7
l6 28.4 13.2 22.0 35.6 23.9 499.2 622.3
Modified Exercise 6.13 Determine using Aspen Plus, the separation that can be achieved for the absorption operation indicated in Figure 6.47 for the following combinations of conditions: (a) Six equilibrium stages and 75 psia operating pressure (b) Three equilibrium stages and 150 psia operating pressure, (c) Six equilibrium stages and 150 psia operating pressure. Subject: Same as previous exercise. Given: Same as previous exercise. Find: Same as previous exercise. Analysis: Using Aspen Plus, the following results are obtained: (a) N = 6 and P = 75 psia Component Vapor Liquid C1 1648.4 11.6 C2 161.9 6.1 C3 83.7 12.3 nC4 32.3 19.7 nC5 1.7 22.3 nC10 2.0 498.0 Total 1930.0 570 (b) N = 3 and P = 150 psia Component Vapor Liquid C1 1635.3 24.7 C2 155.6 12.4 C3 72.5 23.5 nC4 20.0 32.0 nC5 1.2 22.8 nC10 1.3 498.7 Total 1885.8 614.2 (c) N = 6 and P = 150 psia Component Vapor Liquid C1 1635.3 24.7 C2 155.7 12.4 C3 72.6 23.4 nC4 17.4 34.6 nC5 0.1 23.9 nC10 1.2 498.9 Total 1882.3 617.7
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Exercise 6.25 An SO2-air mixture is being scrubbed with water in a countercurrent-flow packed tower operating at 20oC and 1 atm. Solute-free water enters the top of the tower at a constant rate of 1,000 lb/h and is well distributed over the packing. The liquor leaving contains 0.6 lb SO2/100 lb of solute-free water. The partial pressure of SO2 in the spent gas leaving the top of the tower is 23 torr. The mole ratio of water to air is 25. The necessary equilibrium data are given below. (a) What percentage of the SO2 in the entering gases is absorbed in the tower? (b) In operating the tower it was found that the rate coefficients kg, and kL remained substantially constant throughout the tower at the following values: kL = 1.3 ft/h kg = 0.195 lbmol/h-ft2-atm At a point in the tower where the liquid concentration is 0.001 lbmol SO2 per lbmol of water, what is the liquid concentration at the gas-liquid interface in lbmol/ft3? Assume that the solution has the same density as H2O Solubility of SO2 in H2O at 20oC Partial Pressure of lb SO2 100 lb H2O SO2 in Air, torr 0.02 0.05 0.10 0.15 0.20 0.30 0.50 0.70 1.0
0.5 1.2 3.2 5.8 8.5 14.1 26.0 39.0 59
Subject: Operating data for absorption of SO2 from air into water in a packed column. Given: Column operates at 1 atm (760 torr) and 20oC. Solute-free water enters at 1,000 lh/h. Mole ratio of water to air is 25. Liquid leaves with 0.6 lb SO2/100 lb of solute-free water. Partial pressure of SO2 in exit gas is 23 torr (0.0303 atm). Equilibrium data are given as partial pressures of SO2 in air as a function of lb SO2 dissolved/100 lb H2O Assumption: No stripping of water. No absorption of air. Density of liquid taken as water. Find: (a) % of SO2 absorbed. (b) Concentration of SO2 in the liquid at the gas-liquid interface in lbmol/ft3 at a point where the bulk liquid concentration is 0.001 lbmol SO2/lbmol of water and: kL = 1.3 ft/h kg = 0.195 lbmol/h-ft2-atm Analysis: (a) Inlet water rate = 1,000/18.02 = 55.5 lbmol/hr
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Inlet air rate = water rate/25 = 55.5/25 = 2.22 lbmol/h Partial pressure of air in exit gas = 760 – 23 = 737 torr By partial pressure ratio, SO2 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) = 0.0684 atm % of SO2 absorbed = 0.0937/0.1630× 100% = 57.48% (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, kg(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 pressures and the liquid is in concentration. From Eq (1), kL/kg = 1.3/0.195 = 6.67 = (pb – pi)/(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 = 0.0555 + 0.0693 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)× 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) = 0.00246 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 SO2/ft3 = (lb SO2/100 lb H2O) (62.4/64.06) lb SO2/100 lb H2O 0.30 0.50 0.70 1.0
c i, lbmol SO2/ft3 0.00292 0.00487 0.00682 0.00974
Partial Pressure of SO2, torr 14.1 26.0 39.0 59
pi of SO2, atm 0.01855 0.03421 0.05132 0.07763
These equilibrium data fit the curve, pi = 8.6926 ci – 0.007489 From Eq. (2), 6.67 (ci – 0.00346) = (0.0532 – pi) 6.67 (ci – 0.00346) = –8.6926 ci + 0.060689 Solving this equation, we find ci = 0.0055 lbmol/ft3 and pi = 0.0403 atm
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(3)
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 en 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/kg) = -6.67, to the point of intersection on the equilibrium line, giving the same result as above.
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