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.

Please show work so I understand this, thank you!! Will RATE!!

Answer 1

Given reactions:

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

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

Equimolar mixture of CH4 and O2 = 50mol% each

Conversion of methane is X_CH4 = 30%

P is the purge stream

Qp is the flow rate of the purge stream

R is the recycle stream

Selectivity of CH2O over CO2 is 4.5

A) FLOW DIAGRAM:

B)

• For any calculation basis is nothing but the estimated assumption of one of the variable which helps us in making the calculation easier.
• In the above problem if it is asked to calculate the flow rate of recycle stream then the basis is required, else it can be solved without basis as it plays no part in this particular problem provided.
• It is always recommended to use basis in all kinds of problems.
• In this problem the basis can be taken as 100SCMH

C) Assuming that purge stream is at temperature of 200^o C and pressure of 15 atm, the actual volumetric flow rate in m³ /h is calculated as follows:

SCMH means standard cubic meter per hour of gas stream

Assuming ideal gas laws to be valid we get

From PV= nRT for one mole

$V2= V1*\frac{P1}{P2}*\frac{T2}{T1}$

At state 1: let the Pressure be P1 = 1 atm; Temperature = 15^o C= 288.15K

V1 = 55 smch

At state 2: P2 =15atm; T2= 200^o C = 473.15K

V2 =?

$V2= 55*\frac{1}{15}*\frac{473.15}{288.15}$

V2 = 6.021 m3/hr

D) The ideal gas law as its name states , is valid for gassy fluids but not liquids. When the pressure is high and temperature is low the behaviour of the gas is much similar to real gas and has a certain compressibility which is neglected in ideal gas law. The feed contains CH4 and CO2 whose behaviour is similar to the real fluids.

In such cases the cubic equation of state and the virial equations come into play, such as Soave- Redlich- Kwong equation and Kays rule of compressibility.

Using the SRK equation makes the situation difficult to solve and hence we use the kays rule.

Kay's Rule uses pseudo-critical properties to calculate pseudo-reduced quantities that are then used in the generalized compressibility charts.

E )Composition of purge gas:

Component	Mol %	Molecular weight
methane CH4	50	16
oxygen O2	25	32
carbon dioxide	25	44

let us assume there are 100 moles in the purge stream :

CH4 = 50 moles

O2 = 25 moles

CO2 = 25 moles

$moles = \frac{weight}{molecular weight}$

weight of CH4 = 16*50=800 grams

weight of oxygen = 32*25= 800 grams

weight of carbon dioxide = 44*25 = 1100 grams

total weight = 800+800+1100 = 2700 grams

weight fraction = weight of component /(total weight) *100

weight of CH4 = (800/2700 ) *100 =29.6%

weight of O2 = (800/2700)*100= 29.6%

weight of carbon dioxide CO2 = (1100/2700)*100 = 40.8%

P.S : mass fraction and weight fraction are same.