Elementary Principles 220 - 6.72(cont'd b Mole fraction of...

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6- 54 6.72 (cont’d) b. Mole fraction of water in dried gas = y n nn w 4 lb - moles W / d (2.218 + 1.112 10 lb - moles / d lb - moles W(v) lb - mole = + = × ± ±± . ) . 4 34 4 2218 199 10 Henry’s law : y w P = H w x w () x wm a x = (. ) ( . 500 00170 4 × = psia)(1 atm / 14.7 psia) 0.398 atm / mole fraction lb - mole dissolved W lb - mole solution c. Solvent/solute mole ratio ± . ± .( . . ) . n n 5 24 5 37 1 4 434 44341778 222 690 == ⇒= = lb TEG lb - mole TEG 18.0 lb W lb W 150.2 lb TEG 1 lb W lb - mole TEG lb - mole W absorbed lb - moles TEG / d mm m ( x w ) in = 0.80(0.0170) 5 69.0 8 8 58 lb-mole W = 0.0136 = 0.951 lb-mole W/d lb-mole n n n = ⎯⎯⎯⎯ →= + ± ± ± Solvent stream entering absorber ± / m =+ × 0 69.0 lb .951 lb - moles W 18.0 lb d l b - m o l e - moles TEG 150.2 lb d lb - mole = 1.04 10 lb d 4 m W balance on absorber 6 (17.78 0.95 2.22) lb-moles W/d = 16.51 lb-moles W/d
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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