Question

1. A combustion chamber is fed with 50 kmol/h of butane and 2000 kmol/h of air. Calculate the % excess air used and composition of the gases leaving combustion reactor, assuming complete combustion of butane.
2. In manufacture of chlorine, feed containing hydrochloric acid gas and air are fed to the reactor. The product gases leaving the reactor are found to contain 13.2% HCl, 6.3% 02, 42.9% N2, 30% Cl2 and 7.6% H2O (by weight). Calculate:
i) The percent excess air used.
ii) The composition of gases entering reactor
iii) The degree of completion of oxidation (conversion)
(Hint: Start solving by change the mass flow rate to mol flow rate first.)
3. The producer gas made from the coke has the following composition :
CO : 28%, CO2 : 3.5%, O2 : 0.5% and N2 : 68%.
The gas burned with such a quantity of air that the oxygen from air is 20% in excess of the net oxygen required for complete combustion. If the combustion is 98% complete, calculate the weight of the gaseous product formed. Given the oxidation reaction as below:
CO + ½ O2  CO2
4. Methanol is produced by the reaction of CO with H2 according to the equation:
CO + 2H2  CH3OH
Only 15% of the CO entering the reactor is converted to methanol. The methanol product is condensed and separated from the unreacted gases, which are recycled. The ratio feed to the reactor contains 2kmol of H2 to 1 kmol of CO. If the production of methanol is 206kmol, determine:
i) The composition of product leaving the reactor
ii) Determine the total amount of fresh feed.

Answer 1

Combustion of butane

C4H10 + 13/2 (O2 + 3.76N2) = 4CO2 + 5 H2O + 24.44 N2

From the stoichiometry of the reaction

1 kmol/h butane required = 13/2 kmol/h of O2

50 kmol/h butane required = 13*50/2 = 325 kmol/h of O2

Mol% of O2 in air = 21%

Theoretical air required = 325/0.21 = 1547.62 kmol/h

Actual air supplied = 2000 kmol/h

Excess air supplied = Actual air supplied - Theoretical air required

= 2000 - 1547.62

= 452.38 kmol/h

% excess air supplied = Excess air supplied*100 / theoretical air required

= 452.38*100/1547.62

= 29.23 %

At the outlet of the reactor

O2 = 452.38 kmol/h

N2 = actual air supplied x % of N2 in air

= 2000*0.79 = 1580 kmol/h

CO2 = 4 kmol/h CO2 x 50 kmol/h butane / 1 kmol/h butane

= 200 kmol/h

H2O = 5 kmol/h H2O x 50 kmol/h butane / 1 kmol/h butane

= 250 kmol/h