Elementary Principles 226 - 6.81. Basis : 100 kg liquid...

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6- 60 6.81. Basis : 100 kg liquid feed. Assume P atm =1 atm Degree of freedom analysis : Reactor Filter 6 unknowns (n 1 , n 2 , y 2w , y 2c , m 3 , m 4 ) 2 unknowns –4 atomic species balances (Na, C, O, H) –2 balances –1 air balance 0 DF –1 (Raoult's law for water) 0 D F Na balance on reactor 100kg 0.07kgNa CO 46kgNa kg 106 kg Na CO ( 0.024 ) kg NaHCO 23 kg Na 84 kg NaHCO 3.038 0.2738( ) (1) 23 34 3 3 = + ⇒= + mm Air balance : 0 300 2 12 .( ) nn a = C balance on reactor : n nm m n n m m cc 2 (kmol) 0.700 kmol CO 12 kg C kmol 1kmol CO 100 kg 0.07 kg Na CO + =+ + + = + + () ( ) ( . ) () . . . ( . ) ( ) 2 1 2 3 4 0 024 84 840 0 7924 12 01429 3 H balance : ( . ) ( ) ( ) ( . ) ( ). ( ) 100 0 93 2 18 20 0 2 4 1 84 0976 2 18 4 4 + + m m w + + + 10 33 2 0 01190 0 024 01084 4 4 4 .. ( . ) . ( ) m m w n (kmol H O )(sat'd) n ( k m o l C O) (kmol Air) 70 C, 3 atm(absolute)
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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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