Elementary Principles 147 - 5.53(cont'd b Overall C balance...

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Unformatted text preview: 5.53 (cont'd) b. Overall C balance: n1 FG kmol PX IJ 8 kmol C = 100 kmol TPA H h K kmol PX h 8 kmol C n1 = 100 kmol PX / h kmol TPA c. O 2 consumed = 100 kmol TPA 3 kmol O 2 = 300 kmol O2 /h h 1 kmol TPA Overall O 2 balance: 0.21n 2 = 300 kmol O 2 +0.04n 4 n 2 = 1694 kmol air/h h n 4 = 1394 kmol/h Overall N 2 balance: 0.79n 2 = 0.96n 4 100 kmol TPA 2 kmol H 2 O Overall H 2 O balance: n 3W = = 200 kmol H 2 O / h h 1 kmol TPA V2 = V3 = n 2 RT 1694 kmol 0.08206 m3 atm 298 K = = 6.90 103 m3 air/h P h kmol K 6.0 atm ( n 3W + n 4 ) RT = ( 200+1394 ) kmol P h 0.08206 m3 atm 378 K = 8990 m3 /h kmol K 5.5 atm V3W = 200 kmol H 2 O (l) 18.0 kg 1 m3 = 3.60 m3 H 2 O(l) / h leave condenser h kmol 1000 kg d. 90% single pass conversion n 3p = 0.10 n1 + n 3p ====> n 3p = 111 kmol PX / h . d i n1 =100 mrecycle = mS + m3 P = (100 + 11.1) kg PX 106 kg PX 3 kg S 11.1 kmol PX 106 kg PX + h 1 kmol PX kg PX h 1 kmol PX = 3.65 104 kg/h e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the air. The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX. f. The stream can be allowed to settle and separate into water and PX layers, which may then be separated. n 1 (kmol CO / h), n 3 (kmol H 2 / h), 0.10n 2 (kmol H 2 / h) 5.54 Separator n 6 (kmol CO / h) n 7 (kmol H 2 / h) n 8 (kmol CO 2 / h) 2 kmol N 2 / h 0.90 n 2 2 kmol N 2 / h n1 , n 2 , n 3 2 kmol N 2 / h 0.300 kmol CO / kmol 0.630 kmol H 2 / kmol 0.020 kmol N 2 / kmol 0.050 kmol CO 2 / kmol Reactor n 1 (kmol CO / h) n 2 (kmol H 2 / h) n 3 (kmol CO 2 / h) n 4 (kmol M / h) n 5 (kmol H 2 O / h) 2 kmol N 2 / h Separator n 4 (kmol M / h) n 5 (kmol H 2 O / h) 5-35 ...
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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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