Elementary Principles 218 - 6.71 (cont'd) yE3 = * 0.450 pE...

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6- 52 6.71 (cont’d) () * E 4 E3 0.450 40 C 0.450(0.360) 2.13 10 kmol E kmol 760 p y P −° == = × H3 A3 E3 2 1 0.9674 kmol H kmol yy y =− = Mole balance about still: A balance about still: kmol / h recycle kmol / h ±±± ± . ± . ± .(.) . ± ± . ± . nnn n n nn n n cpr c r cr r c =+⇒= + =+ U V W = = 22 67 0550 097 2267 005 29 5 521 A balance about scrubber: A4 3 A3 3 0.03242 y n = = ±± ± (1) E balance about scrubber: 4 E4 3 E3 3 2.13 10 y n × ± (2) H balance about scrubber: 2 H4 3 H3 3 0.9764 y n = = ± (3) Overall C balance: ± . ± . ± n n n pp 0 2 2 097 2 003 2 (mol E) 2 mol C h1 m o l E A4 E4 =++ + b g bg b g bg di ± . n 044 22 67 = + + AE ( 4 ) Overall H balance: 6 2 4 6 097 4 003 6 0 ± .. n p =+++ + H4 A4 E4 b g b g (5) Solve (1)–(5) simultaneously (E-Z Solve) : 0H 4 2 23.4 kmol E/h (fresh feed), 22.7 kmol H /h (in off-gas) 3A 4 E 4 = 23.3 kmol/h, = 0.755 kmol A/h, = 0.00496 kmol E/h n ± A balance about feed mixing point: A1 0.05 1.475 kmol A h r = = E balance about feed mixing point:
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