Split_00014 - 8.32 (cont'd) 95 mol M 2 mol O 2 4.76 mol air...

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8- 14 22 12 2 2 2 3 2 mol O 3.5 mol O 95 mol M 4.76 mol air 5 mol E 4.76 mol air 1.2 s 1 mol M mol O s 1 mol E mol O 1185 mol air/s 0.21 1185 249 mol O /s, 0.79 1185 936 mol N /s 1 mol CO 95 mol M s1 m air air n n nn n ⎡⎤ =+ ⎢⎥ ⎣⎦ = = = = ± ± ±± ± 2 2 4 2 5 2 62 2 mol CO 5 mol E 105 mol CO /s ol M s 1 mol E 2 mol H O 3 mol H O 95 mol M 5 mol E 205 mol H O/s s 1 mol M s 1 mol E 2 mol O 3.5 mol O 95 mol M 5 mol E 249 41.5 mol O /s s 1 mol M s 1 mol E 936 mol N n n += = =− + = == ± ± 2 /s Energy balance on air : 245 20 mol air kJ kJ ( ) 1185 6.649 7879 ( 7879 kW) sm o l a i r s air p air Qn C d T ⎛⎞ = = ⎜⎟ ⎝⎠ ± ± Energy balance on stack gas : () ( ) () () () () 6 900 3 34 5 6 900 900 900 900 7879 T ip i i TT T T pp p p CO H O v O N QH n C d T n C dT n C dT n C dT n C dT = =−Δ =− −= + + + ∫∫ ± ± ± ± Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve o T 732 C = b. 350 1000 1 434 00434 0 0434 7851 341 m (STP) h mol 22.4 L(STP) L m h 3600 s mol/s Scale factor = 4.34 mol / s 100 mol / s kW 3 3 = = . . ± . Q bg 8.33 a. Δ ² .. . . . . . HC d T p + + + + z 0 600 100 3 335 4 351 38 4 42 0 2 36 7 40 2 439 23100
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This note was uploaded on 01/07/2012 for the course CH E 210 at Pennsylvania State University, University Park.

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