Elementary Principles 99 - 4.66 CO + 1 O 2 CO 2 2 H2 + 1 O2...

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

View Full Document Right Arrow Icon
4- 58 4.66 CO 1 2 O CO H 1 2 OH O 22 2 2 2 +→ 175 kmol/h 0.500 kmol N 2 /kmol x (kmol CO/mol) (0.500– x ) (kmol H 2 /kmol) 20% excess air Note : Since CO and H 2 each require 0 ./ 5 mol O mol fuel 2 for complete combustion, we can calculate the air feed rate without determining x CO . We include its calculation for illustrative purposes. A plot of x vs. R on log paper is a straight line through the points Rx 11 10 0 0 05 == ., . bg and 99 7 10 . . ln ln ln ln . . ln . . . ln ln . . ln . . . . exp . . .. xbR a xa R b b ax R a =+ = =− = = × =⇒ = @ 10 005 997 100 1303 997 600 249 10 31 3 0 3 6 00 2 49 10 3 38 3 0 288 b g b g moles CO mol Theoretical O 2 175 kmol kmol CO 0.5 kmol O 2 h kmol kmol CO 175 kmol kmol H 2 0.5 kmol O 2 h kmol kmol H 2 kmol O 2 h :. . . 0288 0 212 4375 += Air fed: 43.75 kmol O 2 required 1 kmol air 1.2 kmol air fed h 0.21 kmol O 2
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