Elementary Principles 107 - 4.74 (cont'd) 100 kg fuel oil...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
4- 66 4.74 (cont’d) 100 kg fuel oil 7.08 kmol C n 2 (kmol N 2 ) 12.0 kmol H n 3 (kmol O 2 ) 0.053 kmol S C + O 2 CO 2 n 4 (kmol CO 2 ) 1.3 kg NC (s) C + 1/2 O 2 CO (8/92) n 4 (kmol CO) 2H + 1/2 O 2 H 2 O n 5 (kmol SO 2 ) 20% excess air S + O 2 SO 2 n 6 (kmol H 2 O) n 1 (kmol O 2 ) 3.76 n 1 (kmol N 2 ) Theoretical O 2 : 2 2 2 2 O kmol 133 . 10 S kmol 1 O kmol 1 S kmol .053 0 H kmol 2 O kmol 5 . H kmol 2 1 C kmol 1 O kmol 1 C kmol 08 . 7 = + + 20 % excess air : n 1 = 1.2(10.133) = 12.16 kmol O 2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n 3 n 3 = 2.3102 kmol O 2 C balance: 7.08 = n 4 +8n 4 /92 n 4 = 6.514 mol CO 2 8 (6.514)/92 = 0.566 mol CO S balance: n 5 = 0.53 kmol SO 2 H balance: 12 = 2n 6 n 6 = 6.00 kmol H 2 O N 2 balance: n2 = 3.76(12.16) = 45.72 kmol N 2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online