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A mixture of methane and air is capable of being ignited only if the mole pecent...

A mixture of methane and air is capable of being ignited only if the mole pecent of methane is between 5% and 15%. A mixture containing 9.0mole% methane in air flowing at a rate of 700.kg/h is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (note Air may be taken to consist of 21 mole % o2 and 79% N2 and to have an average molecular weight of 29.0)
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Answer #1
Concepts and reason

Material balances of the system and individual components.

Draw a fully labeled flow chart. Determine the composition of inlet stream and outlet stream in percentage of air and methane then calculate the required amount of air for desired dilution.

Fundamentals

Material balance:

66203 Input Enters through inlet Generation 1 Output [Consumption] [Accumulation Produced 1 | Leaves Consumed Buildup wiht in

To convert molar flow rate to mass flow rate, multiply the molar flow rate with molar mass.

Draw the flow sheet of the given process as shown below.

91%air 9% CH 95%air 5% CH Mixing chamber 700 kg/hr air(molar)

Molecular weight of methane (CH) is 16.04 g/mol .

Air consists of 21% oxygen and 79% nitrogen with a molecular weight of 29 g/mol .

Calculate the average molecular weight of the mixture .

M = 0.09x 16.04(8).001, 29.0(g) (mol) (mol) M = 27.83 g/mol

Determine the molar flow rate of the mixture at inlet.

700(kg) (hr) (kmol) = 25.153 kmol/hr 27.83 (kg)

Molar flow rate of methane at inlet is,

25.153 (kmol/hr) 0.09 = 2.264 kmol/hr

Molar flow rate of air at inlet is,

25.153(kmol/hr) - 2.264(kmol/hr) = 22.89(kmol/hr)

Determine the required flow rate of air for desired dilution of methane as follows:

the 0.95(kmol Air) 2.264(kmol CH,/hr)x HOLA! = 43.02 kmol Air/hr 0.05(kmol CH)

Take the difference between the original flow rate of air (43.02 kmol Air/hr) and the flow rate of air at inlet (22.89 kmol/hr) .

43.02(kmol/hr) - 22.89(kmol/hr) = 20.13(kmol/hr) = 20130(mol Air/hr)

Therefore, the required flow rate of air for desired dilution is 20130 mol Air/hr .

From the flow rate of the inlet mixture and air, the product flow rate is,

29(kgAir) - 1284 kg/hr 700 (kg/hr) +20.13(kmolAir/hr) (kmolAir)

Determine the mass flow rate of the air in mixing chamber.

22.89(kmol/hr) +20.13(kmol/hr) = 43.02(kmol/hr)

Mass of oxygen in the product gas is,

10ш 14/40 84 68682= ( )2 ( Iowa ) ico ( 7 Tour 20 er

Thus, mass percentage of oxygen in product is,

289.09 (kg 02/hr) = 0.225 kg 0,/kg 1284 (kg/hr)

Therefore, the mass percentage of oxygen in product is 22.5% kg 0,/kg .

Ans:

20130 mol Air/hr , 22.5% kg 0,/kg

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