Elementary Principles 141 - 5.48 (cont'd) n3 = (0.85)(...

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5- 29 5.48 (cont’d) n kmol FeS kmol SO 2 kmol FeS kmol SO 3 22 2 3 == (. ) (. ). . 085 060 06833 4 0 6970 () .4646 kmol SO 5.5 kmol O n 0.21 17.08 kmol O fed 42 4 kmol SO 2 .697 kmol SO 7.5 kmol O 3 2 1.641 kmol O 2 4 kmol SO 3 −= n kmol N kmol N 52 2 = 079 1708 1349 .. . bg ( ) [ ] out V = 0.4646+0.6970+1.641+13.49 kmol 22.4 SCM (STP)/kmol 365 SCM/100 kg ore fed ⎡⎤ ⎣⎦ = 2 32 2 2 SO SO O N 0.4646 kmol SO y 100% 2.9%; y 4.3%; y 10.1%; y 82.8% 16.285 kmol = = = = b. Let (kmol) = extent of reaction ξ 2 23 3 2 2 SO SO SO 11 SO 1 O 2 1 2 N ON 1 2 n0 . 4 6 4 6 0.4646 0.697 y, y n 0.697 16.29- 16.29- n 1.641 1.641 13.49 n 13.49 y 16.29- 16.29- n=16.29- =− −+ =+ = 1 2 1 3 2 1 SO - 2 p p 1 SO O 2 Py (0.697 ) 16.29 K(T )= P ) Py (Py ) (0.4646 ) 1.641 ξξ +− ⇒⋅ = ⋅⋅ −− 1 2 - o p 2 SO SO 2 P=1 atm, T=600 C, K 9.53 atm 0.1707 kmol 0.4646 0.2939 kmol SO reacted n 0.2939 kmol f 0.367 0.4646 kmol SO fed
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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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