Question

1500 pound per hour of a waste mixture of 40% benzene, 50% toluene and 10% of water is burned with 25% excessive air. Determine the total heat released and the percent by volume of each component in the flue gas.
thank you.

Answer 1

(1)

Benzene has a heating value of 18150 Btu/lb , toulene has value 18129 Btu/lb , and water has 0 value.

On a per pound basis, heat released is :

By toulene = 18129*0.5 = 9064.5 Btu/lb

By benzene = 18150*0.4 = 7260 Btu/lb

By water = 0

So total heat release = 16324.5 Btu/lb

Total heat release in the reactor = 16324.5 * 1500 = 24,486,750 Btu/hr

(2)

Toulene burns accoring to the reaction :

C₆H₅CH₃ + 9O₂ ---> 7CO₂ + 4H₂O

MW of toulene = 0.202 lb , Mass of toulene in recator = 50% = 750 lb/h, so moles of toulene = 3712.87 moles

Stoichiometry of the reaction says that these many moles of toulene will require 9*3712.87 ( = 33415.8) moles of O₂ and produce 7*3712.87 ( = 25990 ) moles CO₂ and 4*3712.87 ( = 14851.5) moles H₂O per hour

Since 25% excess air has been supplied, this much O₂ will be present extra, So extra O₂ = 0.25 * 33415.8 = 8354 moles. Overall O₂ = 33415.8 + 8354 = 41769.8 moles/h

Since air is 21% O₂ and 79% N₂, so moles of N₂ = 79/21*41769.8 = 157134 moles/h

So, for toulene we have :

O₂ = 41769.8 moles

CO₂ = 25990 moles

H₂O = 14851.5 moles

N₂ = 157134 moles

per hour.

Similarly you can calculate for benzene and add to get overall results.

Treating all gases as ideal, mole% can be treated as volume%, so calculate mole% for each component and that will be the % by volume, because in PV = nRT, keeping P and T constant , n is directly proportional to V.