Elementary Principles 388

Elementary Principles 388 - 9.29(cont'd substance CH 3 OH...

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9- 36 d. substance kmol / h kJ / kmol kmol / h kJ / kmol CH OH 75.43 O N HO HCHO H in in out out 3 2 2 2 ± ² ± ² . . .. . n H n H n nn n s −− 195220 22 63 163200 2188 3620 18410 82 29 3510 82 29 17390 237740 220920 52 80 88800 16810 22 6 5 Energy Balance : Δ Hn H ii =−= ∑∑ ² ² out in 0 ⇒+ + = × 18410 16810 220920 237704 7 406 10 25 6 6 n n s . (4) We now have four equations in four unknowns. Solve using E-Z Solve. ± n s == 58.8 kmol H O v 18.02 kg h 1 kmol kg steam fed h 2 bg 1060 ± . n 2 226 = kmol O h 2 , ± . n 5 1358 = kmol H h 2 , ± . n 6 98 00 = kmol H O h 2 Summarizing, the product gas component flow rates are 22.63 kmol CH 3 OH/h, 2.26 kmol O 2 /h, 82.29 kmol N 2 /h, 52.80 kmol HCHO/h, 13.58 kmol H 2 /h, and 98.02 kmol H 2 O/h 272 kmol h product gas 8% CH OH, 0.8% O 30% N 19% HCHO, 5% H 37% H O 32 2 2 2 ,, , Energy balance on waste heat boiler. Since we have already calculated specific enthalpies of all components of the product gas at the boiler inlet (at 600
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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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