Methane is to be burned with 30% excess air to generate steam, 95% of the methan

Question

Methane is to be burned with 30% excess air to generate steam, 95% of the methane is consumed producing only carbon decode and water vapor Fir a basis of 10 k moles of methane determine the kg of steam generated if the combustion gates leave at a temperature of 800 degree C and methane and enter at 25 degree C and 200 degree C, respectively Liquid water enters at 5 bar, 1. Solve the mass balance problem to find the flow of the flow of each compound in the product stream. 3. Determine the amount of steam generated.

Explanation / Answer

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Methane:10 kmol

Air:20% excess

According to stoichiometry

1 kmol methane=2 kmol of O2 required

So

CH4+air------>CO2+2H2O

So if stoichiometry amount of air required then reaction

CH4+2O2+2(0.79/0.21)N2------>CO2+2H2O+2(0.79/0.21)N2

So 20% excess air means

CH4+a(O2+0.79/0.21)N2------>CO2+2H2O+bN2

if 1kmol of methane is there.

So O2 excess=1.2*2=2.4 kmol of O2

N2 excess=1.2*2*0.79/0.21=9.028 kmol

So for 10 kmole of methane

O2=24 kmol

N2=90.28 kmol

So total air=114.28 kmol

95%conversion of methane is there

So Product

1 kmol of methane=2 kmol of O2=1 kmol of CO2=2 kmol of H2O

So if 95% converted

so methane reacted=0.95*10=9.5 kmol

So methane in product=0.5 kmol

According to stoichiometry

O2 reacted=9.5*2=19 kmol

O2 remaining=5 kmol

CO2=9.5 kmol

H2O=2*9.5=19 kmol

So product moles

CH4=0.5 kmol

O2=5 kmol

CO2=9.5

H2O=19 kmol

N2=90.28 kmol

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