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Topic: How to calculate the mols to drain in a mixture  (Read 4212 times)

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Offline rmurray

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How to calculate the mols to drain in a mixture
« on: March 17, 2012, 08:29:48 PM »
I have an initial mixture total of 8.3613 mols.
O2 25% mols: 2.0903
He 70%mols: 5.829
N2 5% Mols: 0.4181

Final gas mols of 8.3654
O2 18% mols: 1.5058
He 45%mols: 3.7644
N2 37% Mols: 3.0952

I need to figure out how to remove X mols of initial mix and topoff with air (20.9% O2, 79.1% N2) for the O2 component. I know that removing the initial mix to 5.377 based on He component will work for the He component, but the O2 component is too high, coming out at 2.0912 mols. Reverse engineering I find the removal amount is to about 3.7294 mols and adding the air mixture of 2.9055 mols works for the O2 component and N2 component. How do I setup the equations for this? I've been working on this for a couple days and have hit a brick wall for some reason.

Also, this mixture is at about 230 bar, so I can not drain each component separately.

Thank you,
Rob
« Last Edit: March 17, 2012, 09:18:30 PM by rmurray »

Offline Borek

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Re: How to calculate the mols to drain in a mixture
« Reply #1 on: March 18, 2012, 05:34:22 AM »
Standard approach would be to describe the mixtures composition using system of equations with amount removed as an unknown. However, looks to me like there is no solution to this problem - you can't get below 20.9% of oxygen no matter what volumes of both gases you will mix.
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Offline rmurray

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Re: How to calculate the mols to drain in a mixture
« Reply #2 on: March 18, 2012, 07:31:05 AM »
Standard approach would be to describe the mixtures composition using system of equations with amount removed as an unknown. However, looks to me like there is no solution to this problem - you can't get below 20.9% of oxygen no matter what volumes of both gases you will mix.

What will happen is a certain amount of pure He will be added once the proper amount is removed, then air will be added to bring it to the final mols. I know it can be done, I'm just stuck at this step.

Online billnotgatez

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Re: How to calculate the mols to drain in a mixture
« Reply #3 on: March 18, 2012, 10:28:57 AM »
Quote
and topoff with air (20.9% O2, 79.1% N2) for the O2 component.

I assume you are not really using air, but are using a mixture that is close to air.
below derived from internet search excluding water the air is
Nitrogen       -- N2  --  78.084%
Oxygen         -- O2  --  20.9476%
Argon          -- Ar  --   0.934%
Carbon Dioxide -- CO2 --   0.0314%
Neon           -- Ne  --   0.001818%
All Other      --     --   0.001182%


Offline rmurray

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Re: How to calculate the mols to drain in a mixture
« Reply #4 on: March 18, 2012, 10:44:06 AM »
Bill,
Actually I will use air, but for simplistic sake, I have it set to 20.9% O2 and 79.1% N2. The rest is negligible for my calculations. I have an excel sheet I setup with all the calculations except where I got stuck. I then reverse engineered the answer. Works fine this way, but I'm incorporating this into a program so all the gas mixtures will change. I can send you the excel file if that helps

Offline fledarmus

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Re: How to calculate the mols to drain in a mixture
« Reply #5 on: March 18, 2012, 11:31:19 AM »
So you want a final mixture of z moles of gas, with total moles being z(1.5058 molO2+3.7644 molHe+3.0952 molN2) final mix from three initial mixtures w(2.0903 molO2+5.829 molHe+0.4181 molN2)current mix + x(20.9%O2+79.1%N2 (you still need to put these in mole fraction form))air + y(molHe)helium. So you separate these into three equations for oxygen, nitrogen, and helium, decide what your total amount z should be, and with three equations in three unknowns (w, x, and y) you should be able to solve the problem.

Is this what you are asking?

Offline rmurray

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Re: How to calculate the mols to drain in a mixture
« Reply #6 on: March 18, 2012, 12:22:41 PM »
I just figured it out. Had to think back to the good ole days of high school algebra, solving systems of equations works like a charm on this.

Thanks everyone for the help
Rob

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