Chper 5 Exmple 5.2-2. 6 ---------------------------------------------------------------------------------- r ower i o e deigned o or SO 2 from n ir rem uing pure wer 20 o C. The enering g conin 20 mol % SO 2 nd h leving 2 mol % ol preure of 101.3 kp. The iner g flow re i 150 kg ir/h m 2, nd he enering wer flow re i 6000 kg wer/h m 2. uming n overll r efficienc of 25%, how mn heoreicl r nd cul r re needed? ume h he ower opere 20 o C. Equilirium d for SO 2 wer em 20 o C nd 101.3 kp re given: x 0.0001403.000280.000422.000564.000842.001403.001965.00279 0.00158.00421.00763.01120.01855.0342.0513.0775 x.00420.00698.01385.0206.0273.121.212.443.682.917 Soluion ----------------------------------------------------------------------------------------- The vpor nd liquid molr flow re re clculed fir + +, 150/29 5.18 kmol iner ir/h m 2 6000/18 333 kmol iner wer/h m 2 We hve 0.20, 0.02, nd x 0. For he olue-free i x 1 x 1, x x 0 0 0 0.020 0.020 0.20 0.20 0.0204 0.250 cn e deermined from he componen lnce ( SO 2 ): + +, 6 Genkopli, C.J., Trnpor Procee nd Seprion Proce Principle, 4 h ediion, Prenice Hll, 2003, p. 663 5-9
+, 0 + 5.18 5.18 0.250 0.0204 0.00357 333 333 The opering line nd he equilirium curve cn e ploed uing he following Ml code: % Exmple 5.2-2 xe[0.0001403.000280.000422.000564.000842.001403.001965.00279.00420.00698]; e[0.00158.00421.00763.01120.01855.0342.0513.0775.121.212]; exe./(1-xe);ee./(1-e); [0.00357];[.0204.25]; plo(e,e,,,'--') legend('equilirium curve','opering line',2) xlel('');lel('') Tile('Equilirium nd Opering line on olue free coordine') grid on Figure E-1 Theoreicl numer of r. The numer of heoreicl r i deermined impl epping off he numer of r hown in Figure E-1. Thi give 2.4 heoreicl r. The cul numer of r i 2.4/.25 10 r. 5-10
Exmple 5.2-3. 6 ---------------------------------------------------------------------------------- we irrem from chemicl proce flow he re of 1.0 m 3 / 300 K nd 1 m, conining 7.4% volume of enzene vpor. I i deired o recover 85% of he enzene in he g hree-ep proce. Fir, he g i crued uing non-volile wh oil o or he enzene vpor. Then, he wh oil leving he orer i ripped of he enzene conc wih em 1 m nd 373 K. The mixure of enzene vpor nd em leving he ripper will hen e condened. Becue of he low oluili of enzene in wer, wo diinc liquid phe will form nd he enzene ler will e recovered decnion. The queou ler will e purified nd reurned o he proce oiler feedwer. The oil leving he ripper will e cooled o 300 K nd reurned o he orer. Figure E5.2-3 i chemic digrm of he proce. G ou Wh oil Cooler orer G in Condener Benzene produc Sepror Sripper To wer remen pln Sem Figure E5.2-3 Schemic digrm of he enzene-recover proce. The wh oil enering he orer will conin 0.0476 mole frcion of enzene; he pure oil h n verge moleculr weigh of 198. n oil circulion re of wice he minimum will e ued. In he ripper, em re of 1.5 ime he minimum will e ued. Compue he oil-circulion re nd he em re required for he operion. Wh oilenzene oluion re idel. The vpor preure of enzene 300 K i 0.136 m, nd i 1.77 m 373 K. Soluion ----------------------------------------------------------------------------------------- For clculion in hi exmple, ucrip will e ued o indice orer nd ucrip will e ued o indice ripper. Molr re of he g enering he orer i,o 101,300 1.0 8.314 300 40.61 mol/ 6 Beniez, J. Principle nd Modern pplicion of M Trnfer Operion, Wile, 2009, p. 183 5-11
Molr re of he crrier g i given,o (1,o ) 40.61 (1 0.074) 37.61 mol/ iquid in x,op 0.0476 G ou,op orer 85% recover iquid ou G in x,o 0.074 Convering he enering-g mole frcion o mole rio:,o,o,o,o 0.074 0.074 0.0799 Convering he enering-liquid mole frcion o mole rio:,op x,op,op 0.0476 0.0476 0.0500 Since he orer will recover 85% of he enzene in he enering g, he concenrion of he g leving will e,op 0.15 0.0799 0.0120 The equilirium d for he condiion previling in he orer cn e genered in he olue free i from he following equion: 0.136x [ From P (P enzene ) vp x (1 m) (0.136 m) x ] 1+ 0.136 1+ The following Ml code genere nd plo he equilirium d for he condiion in he orer: 0:0.02:1; 0.136*./(1+);./(1-); plo(,) xlel('_');lel('_'); grid on line([0.05 0.3], [0.012 0.0799]) line([0.05 0.88], [0.012 0.08]) 5-12
The following le how ome of he equilirium vlue genered from he Ml code. 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 digrm for he orer Figure E-1 how he equilirium curve nd he opering line for he orer. Sring wih n opering line ove he equilirium curve, uch DE, roe i owrd he equilirium curve uing D pivo poin unil he opering line ouche he equilirium curve for he fir ime. In hi ce he opering line DM ouche he equilirium curve poin P, locion eween he wo end poin of he opering line. The opering line DM correpond o he minimum olven (oil) re. From he digrm,o (mx) 0.88. Then (min),o,o,op (mx),op 0.0799 0.012 0.88 0.050 0.0818 The minimum olven re i hen: (min) 0.818 37.61 3.08 mol oil/ For n cul oil flow re which i wice he minimum concenrion of he liquid phe leving he orer i 6.16 mol/. The cul,o,op + (,o,op ) 0.050 + 37.61 6.16 (0.0799 0.012) 0.4646 We now conider he condiion in he ripper. Figure E5.2-3 how h he wh oil ccle coninuoul from he orer o he ripper, nd hrough he cooler ck o he orer. Therefore 6.16 mol/. The concenrion of he liquid enering he ripper i he me h of he liquid leving he orer (,op,o 0.4646 mol of enzene/mol oil), nd he concenrion of he liquid leving he ripper i he me h of he liquid 5-13
enering he orer (,o,op 0.05 mol of enzene/mol oil). The geou phe enering he ripper i pure em, herefore,o 0. iquid in,op 0.4646 G ou,op Sripper iquid ou G in,o 0.05,o 0 To deermine he minimum moun of em needed, we need o genere he equilirium diriuion curve for he ripper 373 K from he following equion: 1.77x [ From P (P enzene ) vp x (1 m) (1.77 m) x ] 1+ 1.77 1+ The following le how ome of he equilirium vlue genered from equilirium relion. 0 0 0.0400 0.0730 0.0600 0.1113 0.0800 0.1509 0.1000 0.1918 0.1400 0.2777 0.1600 0.3230 0.1800 0.3699 0.2000 0.4184 0.2400 0.5211 0.2600 0.5754 0.3000 0.6905 0.3400 0.8152 0.3600 0.8816 0.4000 1.0231 0.4400 1.1779 0.4600 1.2608 0.5000 1.4390 Figure E-2 digrm for he ripper Figure E-2 how he equilirium curve nd he opering line for he ripper. Sring wih n opering line elow he equilirium curve, uch DE, roe i owrd he equilirium curve uing D pivo poin unil he opering line ouche he equilirium curve for he 5-14
fir ime. In hi ce he opering line DM ouche he equilirium curve poin P, locion eween he wo end poin of he opering line. The opering line DM correpond o he minimum em re. From he digrm,op (mx) 1.13. Then (min),op (mx),op,o,o 1.13 0 0.4646 0.050 2.726 The minimum em re i hen: (min) 6.16 2.26 mol em/ 2.726 For n cul em flow re which i 1.5 ime he minimum 1.5 2.26 mol/ 3.39 mol/. The cul concenrion of he g rem leving he ripper i,op,o + (,op,o ) 0 + 6.16 3.39 (0.4646 0.05) 0.753 5-15