5 replies to this topic

[Miguel]

Hallo,

 

i have 3 different streams with the same components but different composition for example:

 

Stream A: 600 kg/h -- > mole fraction(02: 0,2; CO2: 0,7; CH4:0,1)

Stream B: 1000 kg/h ---> mole fraction(02: 0,3; CO2: 0,6; CH4:0,1)

Stream C: 800 kg/h --- > mole fraction(02: 0; CO2: 0,9; CH4:0,1)

 

how do i determine the CO2 mole fraction of the mixture?:

 

thank you!

 

Miguel

[Art Montemayor]

OK Miguel, I won't do your homework for you, but I'll tell you what you are failing ot do and think out as a promising Chemical Engineer.

 

First, you should identify the fact that you are given a MOLECULAR FRACTION with respect to a MASS flow rate.  That being said, it should be obvious that you can't work one with the other - they are not consistent units.  If you want to calculate the composition of a mass quantity, you must find the mass fraction or mass percentage.  You are not given that, so it should be obvious to you that you must convert what you have been given (the MOL fraction) to a mass fraction.   And this procedure should be easy for you because you should have already been taught how to do this.   All you do is take a basis, say 100 mols of the feed gas and break it up into it individual components and convert these into their respective mass by multiplying each by their molecular weight.  The way an engineer does this is by setting up the procedure on a spreadsheet, like Excel.

 

You do that for each of the 3 given streams and you add up the masses and re-convert them into mols.  With this all done, you then can calculate the mol percentage of the resultant mixture.

 

What year of Chemical Engineering are you in at university and where are you studying?

[Miguel]

Hello Art,

 

I resolved it as following:

 Unbenannt.PNG   2.91KB   4 downloads

 

being:  M: Mass fraction

            PM: molecular weight
           X: molar fraction

 

I am actually almost finishing my studies as chemical engineer (8th semester) which is pretty embarrassing. I was too lazy to put some thought on the calculation which is part of something for my thesis and wrote this unnecessary post. After looking again to my mass balance notes, i realized how easy it was and resolved it pretty quick.

 

Thank you again for your reply i apologize for the bad professional impression i gave you

 

Best regards,

 

Miguel

 

[Pingue2008]

Miguel,

 

Allow me!

You should follow Mr. Art Montemayor's advise. if you have used that formula, I bait you did not understand the solution. if you ever come across such problem, you will face the same challenge. Learn how to do it now. The embarrassment will be unbearable when your boss asks you to such calculation. I also want to admit that I am a bit surprise that such assignment is given at your level in the program. on the premises that you have already sold the problem, I am attaching a solution to show you that using that formula without going  through the proper steps will do you no good. if you have developed the formula, fine! but if you have blindly applied the formula ...

 

Thank you,

Attached Files

•  Miguel Homework.xlsx   110.67KB   58 downloads

[Art Montemayor]

Miguel:

 

I hope you have paid attention to what Alain (Pingue) has pointed out to you and heeded his advice.  It would be a waste of his and the Forum's time if you haven't.

 

I don't know where you obtained the formula you present as the solution to your problem.  Perhaps you derived it yourself or someone else did it.  The important thing is - as Alain points out - that you know EXACTLY how that formula is developed and can explain it in detail.  Otherwise, no professional engineer is going to accept it per se - especially a supervisor.  All professional engineers require a solid and logical basis for presenting or accepting an answer to a problem - be it in mathematical or written form.  Our profession is paid to furnish correct answers and not guesses or conjectures.  Wrong answers generate liability and we are all held accountable.

 

What Alain has said is the truth.  We old engineers are not mean, tough, jealous, or envious of young graduates.  But we are demanding and strict in enforcing that engineering solutions be logical and correct in logic and reasoning.  We also demand a means to CHECK and proof all generated engineering calculations before issuing them to a client or responsible entity.  No one is exempt from this requirement in the engineering world.   You are a young individual who is possibly about to enter the profession and it is well that you find this out now before you graduate and occupy your first job.  I have trained a lot of young graduates in the past 53 years and I have never accepted a calculation or computer print out without a full, detailed and logical explanation of how the answer was arrived at.   That is the important point that Alain points to and that is why I believe he submitted the correct answer in detail calculations - the way a professional engineer does it in industry.   Every logical step is clearly shown and can be easily followed in his spreadsheet.  Therefore the checking is accessible and easy to do.  The answers are correct because they are logical and well-explained by the detailed steps shown.  I can't do that with your equation.

 

 

[Miguel]

Dear Mr. Montemayor and Pingue2008,

 

Thank you very much for your support specially Pingue2008 who took the time to make an example. 

 

Mr.Montemayor forgive me for misunderstanding, but from the way you wrote both your answers I had the impression that i was wasting your time with basic problems of chemical engineering. I therefore simplified my answer to a short formula in order to close the topic and let you know that I had solved the problem succesfully. This however was a bad idea and resulted in a wrong deducted formula and confusion for whoever was reading this forum topic. 

 

I have uploaded as reply the excel file that Pingue2008 kindly uploaded as example with my own answers  below. You will find a step by step procedure of my calculations + the equation I intended to post corrected.

 

You will notice that the results are the same but the steps in my calculation are a bit different, that is because I considered the average molecular weight for my calculation instead of assuming a  molar flow of 100 mol/hr.

 

I apologize for any confusion and very grateful for you help

 

With kindest regards

 

Miguel

 

 

 

Attached Files

•  Miguel Homework_01.xlsx   153.1KB   21 downloads

Edited by Miguel, 29 September 2013 - 03:17 PM.