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Split_00040

# Split_00040 - 9-31(cont'd ^ ^ ^ ^ ^ ^ H = H ro(1.00 H...

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9- 40 d. e. o r 3 4 5 1 2 o ˆ ˆ ˆ Substitute for , , and through r 1 6 5 2 8 3 12 4 out out out out ˆ ˆ ˆ ˆ ˆ ˆ (1.00) (4.00) (6.00) (4.00) (6.00) (0.3479 4.28 10 0.9285 10 4.697 10 ) H H H H H H H H H H H T T T T ξ ξ Δ ⇒ Δ = Δ + + + Δ = + × + × × ± ± ± ± o out 972.24 kJ/mol = 0 E-Z Solve 2223 C T = If only the first term from Table B.2 is used, kJ / mol, kJ / mol, , , out out out ² ( ) ( ) ² . ( ) . ² . ² . ( ) ² . ( ) ² . ( ) H C dT C T H H H T H T H T i pi pi T = = = = = = = = z 25 003515 200 25 615 531 00291 25 00295 25 003346 25 25 1 2 3 4 5 o r 3 4 5 1 2 o ˆ ˆ ˆ Substitute for (=1 mol/s), ( 904.7 kJ/mol) and through r 1 6 o out out ˆ ˆ ˆ ˆ ˆ ˆ E.B. (1.00) (4.00) (6.00) (4.00) (6.00) 0 2788 0=0.3479 969.86 2788 C % error= H H H H H H H H H H T T ξ ξ Δ = − Δ = Δ + + + = = ± ± ± o o o C 2223 C 100 25% 2223 C × = If the higher temperature were used as the basis, the reactor design would be safer (but more expensive). 9.32 a. Basis : 100 lb m coke fed 84 lb C 7.00 lb - moles C fed 7.00 lb - moles CO fed m 2 400°F 7.00 lb-moles CO 2 (lb-moles CO) n 1 77°F 7.00 lb-moles(84 lb
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