The stoichiometric AFR
by volume for combustion of CO is
(a) 1.19, (b) 2.38, (c)
2.45 (d) 4.76
Solution: The reaction
involved in the combustion of carbon mono-oxide (CO) is
2CO + O2 à 2CO2
2vol + 1vol à 2vol of CO2
So the combustion of 1vol
of CO needs ½ volume of O2
Now 21m3 of
O2 is present in 100 m3 of air
Therefore ½ m3
of O2 will be present in ½*100/21=2.38 m3 of air
AFR = volume of air/
volume of fuel= 2.38/1=2.38
Hence the option (b) is
correct.
No comments:
Post a Comment