Methane (CH4) and oxygen (O2) react to form formaldehyde (CH2O) (reaction #1), with a competing side reaction

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Question:

Methane (CH4) and oxygen (O2) react to form formaldehyde (CH2O) (reaction #1), with a competing side reaction (#2) in which methane and oxygen form carbon dioxide (CO2):

CH4(v) + O2(g) -> CH2O(v) + H2O(l) (reaction #1)

CH4(v) + 2O2(g) -> CO2(g) + 2H2O(l) (reaction #2)

A fresh feed of an equimolar gas mixture of CH4 and O2 is combined with a recycle stream and fed to a reactor. The single-pass conversion of CH4 is 30% and the selectivity of CH2O over CO2 is 4.5. The exiting products are sent to a condenser in which all of the CH2O and H2O are condensed. The vapor stream exiting the condenser is split into two streams: one stream is recycled and combined with the fresh feed prior to entering the reactor, while the other stream is purged. The purge stream exits the process at a rate of 55 SCMH and is composed of 50 mol% CH4, 25 mol% O2, and the balance CO2.

a) Draw and fully label a flow diagram for this process.

b) Based on this information, do you need to assume a basis? Explain briefly.

c) Assume that the purge stream is at 200°C and 15 atm. Calculate the actual volumetric flow rate of this gas stream (in m3 /h) assuming ideal gas behavior.

d) For the conditions in part (c), prove that the ideal gas law is inaccurate. Which equation of state would be better to use: (i) SRK equation of state or (ii) compressibility factor with Kay’s rule? Explain briefly.

e) Given the composition of the purge stream (50 mol% CH4, 25 mol% O2, and the balance CO2), calculate the mass fraction of each component. Molecular weight of CH4 = 16 g/mol, Molecular weight of O2 = 32 g/mol, and Molecular weight of CO2 = 44 g/mol.