Checking Diagram for H2, CO, CH4 Reactor Flow Ratios

[CannonSLX]

I'm about to attempt this question but beforehand I was hoping if someone could just check if my diagram is correct on the basis of the question;

A feed stream contains H2, CO and a trace of CH4. Pure CH3OH is produced by means of a reactor (CO + 2H2→CH3OH), separator and recycle with purge to remove the non-reactive CH4. Feed stream composition is 67.3 mol% H2, 32.5 mol% CO and 0.2 mol% CH4. Recycle CH4 concentration is held at 3.2 mol% CH4. What is the ratio of the feed stream to purge stream flowrates?

Diagram: https://i.gyazo.com/3552bd8e88abae971ea4686ced8a05b9.png

Would I be right in thinking that I'd also need to take a basis in order to calculate any flow rates as none have been provided.

Also does the stream coming off the separator have the same composition as the recycled one ?
In my diagram I was not sure if the arrow coming off the separator also had to have its own flow rate and compositions if it were the same as the recycled stream.

Thanks in advance for any help :)

 

[Chestermiller]

I'm about to attempt this question but beforehand I was hoping if someone could just check if my diagram is correct on the basis of the question;

The layout of the diagram looks correct.

A feed stream contains H2, CO and a trace of CH4. Pure CH3OH is produced by means of a reactor (CO + 2H2→CH3OH), separator and recycle with purge to remove the non-reactive CH4. Feed stream composition is 67.3 mol% H2, 32.5 mol% CO and 0.2 mol% CH4. Recycle CH4 concentration is held at 3.2 mol% CH4. What is the ratio of the feed stream to purge stream flowrates?

Diagram: https://i.gyazo.com/3552bd8e88abae971ea4686ced8a05b9.png

Would I be right in thinking that I'd also need to take a basis in order to calculate any flow rates as none have been provided.

Sure.

Also does the stream coming off the separator have the same composition as the recycled one ?

Yes.

In my diagram I was not sure if the arrow coming off the separator also had to have its own flow rate and compositions if it were the same as the recycled stream.

The composition is the same, but the flow rate isn't.

 

[CannonSLX]

The layout of the diagram looks correct.

Sure.

Yes.

The composition is the same, but the flow rate isn't.

Thanks :)

Would I use extent of reaction method to solve instead of multiple unit balance method ?

 

[Chestermiller]

How would you interpret what is required of you?

 

[CannonSLX]

How would you interpret what is required of you?

Well, as its asking for the ratios all I have done it calculated the flow rate for the purge stream and divided the feed stream by the purge stream.
This is what I have done;
https://i.gyazo.com/f8da4c1adac28ccb28c7aa3af095e1db.png

 

[Chestermiller]

I don't think M4 should be labeled 1 CH3OH mol, and M5, M6, and M7 shouldn't have 0.032 CH4 mol, unless you understand that these are mole fractions and not molar flows. You need to solve for the missing stream flows and mole fractions, given that, in the reactor, the reaction goes to completion, with either CO or H2 being the limiting reactant (probably CO).

 

[CannonSLX]

I don't think M4 should be labeled 1 CH3OH mol, and M5, M6, and M7 shouldn't have 0.032 CH4 mol, unless you understand that these are mole fractions and not molar flows. You need to solve for the missing stream flows and mole fractions, given that, in the reactor, the reaction goes to completion, with either CO or H2 being the limiting reactant (probably CO).

Yes I understand that they are mole fractions :)
I don't see why its necessary to solve for the mole fractions ? If I take a basis and use input=output can I not simply do as I have done in the image ?
As in Feed:purge 16:1

 

[Chestermiller]

Yes I understand that they are mole fractions :)
I don't see why its necessary to solve for the mole fractions ? If I take a basis and use input=output can I not simply do as I have done in the image ?
As in Feed:purge 16:1

I agree. I should have read over what you did more carefully. Sorry.

 

[CannonSLX]

I agree. I should have read over what you did more carefully. Sorry.

No worries, thanks anyways :)