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5-9 Chapter 5 Example 5.2-2. 6---------------------------------------------------------------------------------- A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 20oC. The entering gas contains 20 mol % SO2 and that leaving 2 mol % at a total pressure of 101.3 kPa. The inert gas flow rate is 150 kg air/h⋅m2, and the entering water flow rate is 6000 kg water/h⋅m2. Assuming an overall tray efficiency of 25%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 20oC. Equilibrium data for SO2 –water system at 20oC and 101.3 kPa are given: x 0 .0001403 .000280 .000422 .000564 .000842 .001403 .001965 .00279 y 0 .00158 .00421 .00763 .01120 .01855 .0342 .0513 .0775 x .00420 .00698 .01385 .0206 .0273 y .121 .212 .443 .682 .917 Solution ----------------------------------------------------------------------------------------- The vapor and liquid molar flow rates are calculated first L,A bX + V,A tY = L,A tX + V,A bY V = 150/29 = 5.18 kmol inert air/h⋅m2 L = 6000/18 = 333 kmol inert water/h⋅m2 We have yb = 0.20, yt = 0.02, and xt = 0. For the solute-free basis AX = 1AAxx−, AY = 1AAyy− ,A tX = ,,1A tA txx−= 01 0− = 0 ,A tY = ,,1A tA tyy−= 0.0201 0.020− = 0.0204 ,A bY = ,,1A bA byy−= 0.201 0.20− = 0.250 ,A bXcan be determined from the component balance (A = SO2): L,A bX + V,A tY = L,A tX + V,A bY 6 Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 6635-10 ,A bX = ,A tX + VL,A bY− VL,A tY ,A bX = 0 + 5.18333×0.250 − 5.18333×0.0204 = 0.00357 The operating line and the equilibrium curve can be plotted using the following Matlab codes: % Example 5.2-2 xe=[0 .0001403 .000280 .000422 .000564 .000842 .001403 .001965 .00279 .00420 .00698]; ye=[0 .00158 .00421 .00763 .01120 .01855 .0342 .0513 .0775 .121 .212]; Xe=xe./(1-xe);Ye=ye./(1-ye); X=[0 .00357];Y=[.0204 .25]; plot(Xe,Ye,X,Y,'--') legend('Equilibrium curve','Operating line',2) xlabel('X');ylabel('Y') Title('Equilibrium and Operating lines on solute free coordinates') grid on Figure E-1 Theoretical number of trays. The number of theoretical trays is determined by simply stepping off the number of trays as shown in Figure E-1. This gives 2.4 theoretical trays. The actual number of trays is 2.4/.25 = 10 trays.5-11 Example 5.2-3. 6---------------------------------------------------------------------------------- A waste airstream from a chemical process flows at the rate of 1.0 m3/s at 300 K and 1 atm, containing 7.4% by volume of benzene vapors. It is desired to recover 85% of the benzene in the gas by a three-step process. First, the gas is scrubbed using a non-volatile wash oil to absorb the benzene vapors. Then, the wash oil leaving the absorber is stripped of the benzene by contact with steam at 1 atm and 373 K. The mixture of benzene vapor and steam leaving the stripper will then be condensed. Because of the low solubility of benzene in water, two distinct liquid phases will form and the benzene layer will be recovered by decantation. The aqueous layer will be purified and returned to the process as boiler feedwater. The oil leaving the stripper will be cooled to 300 K and returned to the absorber. Figure E5.2-3 is a schematic diagram of the process. Figure E5.2-3 Schematic diagram of the benzene-recovery process. The wash oil entering the absorber will contain 0.0476 mole fraction of benzene; the pure oil has an average molecular weight of 198. An oil circulation rate of twice the minimum will be used. In the stripper, a steam rate of 1.5 times the minimum will be used. Compute the oil-circulation rate and the steam rate required for the operation. Wash oil-benzene solutions are ideal. The vapor pressure of benzene at 300 K is 0.136 atm, and is 1.77 atm at 373 K. Solution ----------------------------------------------------------------------------------------- For calculations in this example, subscript “a” will be used to indicate absorber and subscript “s” will be used to indicate stripper. Molar rate of the gas entering the absorber is Va,bot = 101,300 1.08.314 300×× = 40.61 mol/s 6 Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 183 Gas outGas inAbsorberStripperSteamCoolerWash oilCondenserSeparatorBenzene productTo watertreatment plant5-12 Molar rate of the carrier gas is given by aV = Va,bot(1 − ya,bot) = 40.61×(1 − 0.074) = 37.61 mol/s Converting the entering-gas mole fraction to mole ratio: Ya,bot = a,bota,bot1yy− = 0.0741 0.074− = 0.0799 Converting the entering-liquid mole fraction to mole ratio: Xa,top = a,topa,top1xy− = 0.04761 0.0476− = 0.0500 Since the absorber will recover 85% of the benzene in the entering gas, the concentration of the gas leaving will be Ya,top = 0.15×0.0799 = 0.0120 The equilibrium data for the conditions prevailing in the absorber can be generated in the solute free basis from the following equation: ya = 0.136xa [ From yaP = (Pbenzene)vap xa ⇒ ya(1 atm) = (0.136 atm) xa] aa1YY+ = 0.136aa1XX+ The following Matlab codes generate and plot the equilibrium data for the conditions in the absorber: X=0:0.02:1; a=0.136*X./(1+X); Y=a./(1-a); plot(X,Y) xlabel('X_a');ylabel('Y_a'); grid on line([0.05 0.3], [0.012 0.0799]) line([0.05 0.88], [0.012 0.08]) Gas outGas inAbsorber85% recoveryLiquid inLiquid outya,topya,botx = a,top 0.0476xa,bot = 0.0745-13 The following table shows some of the equilibrium values generated from the Matlab codes. Xa Ya 0 0 0.1000 0.0125 0.2000 0.0232 0.3000 0.0324 0.4000 0.0404 0.5000 0.0475 0.6000 0.0537 0.7000 0.0593 0.8000 0.0643 0.9000 0.0689 1.0000 0.0730 Figure E-1 XY diagram for the absorber Figure E-1 shows the equilibrium curve and the operating line for the absorber. Starting with any operating line above the equilibrium curve, such as DE, rotate it toward the equilibrium curve using D as a pivot point until the operating line touches the equilibrium curve for the first time. In this case the operating line DM touches the equilibrium curve at point P, a location between the two end points of the operating line. The operating line DM corresponds to the minimum solvent (oil) rate.


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Cal Poly Pomona CHE 313 - Chapter 5

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