700 kmol/h of oxygen (O2) is fed into a reactor. The combustion of methane (CH4)

Question

700 kmol/h of oxygen (O2) is fed into a reactor. The combustion of methane (CH4) and O2 produces three compounds, namely carbon monoxide (CO), carbon dioxide (CO2) and water (H2O). The reaction can be represented by the equation as shown below:

Use the following molecular weights: C – 12 kg/kmol, H – 1 kg/kmol, O – 16 kg/kmol, N – 14 kg/kmol

Consider two possible combustion scenarios below:

1) (4 marks) The combustion involves 10% excess of CH4 (excess based on stoichiometry) and all the O2 is used up during the reaction. No CO, CO2 and H2O are found in the input stream. Determine the mass fraction of CO and CO2 in the output stream. Provide all your answers to three decimal points. (ANSWER: CO = 0.190, CO2 = 0.299)

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2). Air is now supplied in place of pure oxygen. The composition of air is 0.79-mole fraction of nitrogen gas (N2) + 0.21-mole fraction of O2. In the reaction, oxygen (O2) was consumed at the rate of 700 kmol/hr and there is no oxygen in the outlet stream. Methane is supplied at the stoichiometric rate. For some unexplained reasons, some contaminant traces of CO, CO2 and H2O up to 5 kmol/h, 10 kmol/h and 15 kmol/h, respectively, are found in the methane input stream. Determine the mass fraction of CO2 and N2 in the output stream. (ANSWERS: CO2 = 0.089, N2 = 0.713)

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Explanation / Answer

Moles of CH4 as per the reaction= 400 moles/hr. Moles of CH4 supplied= 480*1.1= 440 moles

Prpducts contains ( kmole/hr) : CH4= 440-400= 40 , CO= 200, CO2= 200 and H2O= (8/7)*700 =800 moles

Mass= moles* molar mass

Products mass ( Km/hr): CH4= 40*16= 640, CO= 200*28= 5600, CO2= 200*44= 8800, H2O= 800*18= 14400. Total mass= 640+5600+8800+14400=29440

Composition of CO2= 8800/(14400+8800+5600+640)= 0.299, CO= 5600/29440 =0.190

2.

Oxygen flow rate= 700 kmole/hr, Air supplied= 700/0.21= 3333 kmole/hr

Methane used in stoichiometric proportions is 400 kmole/hr

The combustion reaction is 4CH4+7O2---à2CO+2CO2+ 8H2O

Products contains ( kmole/hr) ; N2= 3333*0.79= 2633, CO2= 200( from combustion)+10( from methane fuel)= 210, CO= 200+5= 205, H2O= 800+15= 815

Mass = moles* molar mass

Mass flow rate ( kg/hr) : N2= 2633*28 = 73724, CO2= 210*44= 9240, CO= 205*28= 7140, H2O= 815*18=14670 , total mass flow rate= 73724+9240+7140+14670= 103374

Composition : CO2= 9240/103374= 0.089, N2= 73724/103374= 0.713

2.

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