Section 2: FR&B;, Chapter 4 Combustion Processes (Section 4.8) Combustion: Rapid

Question

Section 2: FR&B;, Chapter 4 Combustion Processes (Section 4.8) Combustion: Rapid reaction of a fuel with oxygen (usually but not always in air) Complete combustion: All C in fuel forms CO, all H forms H O, all S forms SO Partial combustion: Some C forms CO Air: Actual composition given on p 161 . For most of our purposes, assume 21.0 mole% ·7%0% N: 3.76 mol Nmol O 4.76 mol air/mol O Percent excess oxygen (= percent excess air): Based on complete combustion of fuel, regardless of whether all fuel actually reacts and whether some CO is formed. Wet-basis and dry-basis product gas compositions: Mole fractions of components with water included and not included, respectively Example 20 mol CH/s 95 mol OAs 100 mol CHs 940 mol N/s 70 mol COjs 10 mol COl 160 mol H Ol 1190 mol airls 0.21 mol O/mol 0.79 mol N/mol Q: Is the flow chart balanced? Q: What reaction(s) are taking place? 2-33 Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes Copyright CJohn Wilky & Sons, Ine

Explanation / Answer

Since nitrogen does not participate in the reaction, the same amount need to be carried along with the product gas

C) total methane fed =100 mol/s for complete combustion of methane, air required= 2*100=200 mol/s   air required= 200/0.21= 952.381 moles

Air actually supplied = 1190 moles/ hr, oxygen actually supplied= 1190*0.21=250 moles/hr

Excess oxygen= 100*(250-200)/200= 100*50/200= 25%

D) This is due to incomplete combustion , some methane got converted to CO2 and some to CO.

Dry basis : CH4= 30mol/s Oxygen= 95 mol/s N2= 940 mol/S , CO2= 70 mol/s and CO= 10 mol/s

Total moles= 30+95+940+70+10= 1145 mol/s

Percentages   CH4= 100*30/1145=2.62, Oxygen= 100*95/1145=8.3% CO2= 100*70/1145=6.11%, CO= 100*10/1145=0.87, N2= 100*940/1145=82.1%

Wet basis : total moles= 1145+ moles of water= 1145+160=1305 mol/s

Percentage :CH4= 100*30/1305=2.3%, Oxygen= 7.3 CO2=5.4 mol/s CO= 0.763 N2= 100*940/1305=72.03% and H2O= 100*160/1305=12.3%

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