Lecture_23 - YN+A1 [gas an] Operating line 1 {on absorbent...

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Unformatted text preview: YN+A1 [gas an] Operating line 1 {on absorbent rate} Moles solute! mole of solute-free gas, 1" 1’1 {gas out] Moles solutea’mole sotute—free liquid, X X0 XN (liquid in} ' {foerin} Figure 6.9 Opera‘ing lines for an absorber. C: . . KN W a( «a. m VAN! ' h M; W L... ‘31:. M W. 'Ku 3' Yfl+l/<l+vfl+l\ Xg/CI-a- 3(N3 W «.46... M {Axum m, X; Ll '1- Yn-HG' = ya) E'PY‘Q‘, . h L; = C’ICYM'FI "i\ CXIJ "X33 m L,“ = c-r'cvu..-Y.\ S'YA'NI /EYn+cCKdu "' )+ Kn.) -x' ok'fcu. M ' 154-031. h ‘15 ~46.) 6's ‘34. a . t - Wham—1%.. argument“ . '1' We Y“l «Kg. [a] [bl Figure 6.11 Graphical determifiation of the number of equilibrium stages for (a) absorber and (b) stripper. {bottom} {a} (bottom) {b} Figure 6.8 Continuous, steady—state operation in a counter- current cascade with equilibrium stages: (a) absorber; (b) stripper. YN+_1 (gas In) Moles solute; mole of soiute—free gas, Y Operating line ‘I too absorbent rate) Moles solutefmoie solute-free 1iquid, X X0 XN (liquid in) (foerin} Figure 6.9 Operating lines for an absorber. X XN+1 {b} Figure 6.11 Graphical determination of the number of equilibrium stages for (a) absorber and (b) stripper. 0.035 E x0 = 0 Y1: 0.0005 g: 0.03 d.) E (Y X ) “o— 1' N '5 0.025 A” .C D U E “H 0.02 E YNH = 0.0204 XN = 0.0230 Stage 7 O E $96 15 0.015 $9 ’6- 4: at E: 0Q 6 a 1 “5 °'° A 3 6 than} ‘5 2 0.005 ' 4 Jr 0 ’1 2 0 0.01 0.02 0.03 004 Mates of alcoholfmole of alcohoi—iree liquid. X Figure 6.12 Graphical determination of number of equilibrium stages for an absorber. ...
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Lecture_23 - YN+A1 [gas an] Operating line 1 {on absorbent...

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