[Scorael]

Hi guys,
I'm an electrical engineering undergrad working on a hydrogen fuel cell simulation. I have been using a book to help me with this but there are some terms that I do not understand properly. I hope someone can help out.

In the book, the author mentioned the stoichimetry and the stoichiometric ratio of air in 2 separate examples. The stoichiometric ratio is was used to calculate oxygen flow rate while the stoichiometry is for air flow rate (I'm assuming they actually mean the same because both are listed as 2.5 but I'll rather get a 2nd opinion).
I need to know the difference between the stoichiometric ratio and stoichiometry and how do I obtain them.

[ARGOS++]

[Scorael]

Hi Argos,
Those links did help my understanding of the terms. I now understand the significance of stoichiometry in the equation (turns out it meant the fuel-air ratio). There's still some confusion on the meaning of stoichiometric ratio and I hope you can help me further.

I am given these equations to calculate mass flow rate of Oxygen and Hydrogen into the cell:

m_O2 = 0.5*S_O2*M_O2*I/(2F)
m_H2 = S_H2*M_O2*I/(2F)

The flow of unused reactants out of the cell is given as:
m_O2 = 0.5*(S_O2-1)*M_O2*I/(2F)
m_H2 = (S_O2-1)*M_O2*I/(2F)

m_i is the mass flow rate of species i in g/s
S_i is termed as the stoichiometric ratio of species i
M_i is the molar mass of species i
F is the Faraday's number
I is the current drawn

What I don't really get is the stoichiometric ratio in here. From my understanding, the reaction of hydrogen and oxygen to form water is H₂+0.5O₂ = H₂O giving oxygen the stoichiometric ratio of 0.5 but clearly in the outlet flow equation S can be bigger than 1 (In 1 example S = 2).

[ARGOS++]

[Scorael]

I don't have a scanner so I can't really do that.

[ARGOS++]

[Scorael]

After studying the earlier chapters I realise the equation is I/(nF), n being the number of moles of electron, which appears to be what you have written.

Still does not explain where did stoichiometric ratio, S, came from though. I am given S in the question as 2 but I am not sure whether they arrived at that value through measurements or is it a constant.

I'll update later. Need to sleep now. Thanks for all the help though.

[ARGOS++]

[ARGOS++]

                            Copy of lost replay No. 1

Dear Scorael;

Stoichiometric ratio is an absolutely required part of the whole stoichiometry.
Stoichiometric ratio is dealing with moles whereas stoichiometry deals with both moles and masses.
For Stoichiometric ratio you may best visit Mr. Borek's:   

http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

For doing the whole stoichiometry you may then follow the simple recipe on:

"Stoichiometry Problem (#23'025)"

Therein is also an Example Diagram: How to do a “Stoichiometry Problem”.

I hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

                             Copy of lost replay No. 3

Dear Scorael;

After several tries to interpret these equations as a model, do you allow that I say it in plain language?:

They don’t make real sense to me at all; and that for several reasons!

I have Questions, three times the number of equations you have given!

Are you able to scan the few pages around your problem and send them to me via pm?
I hope this way I will have more chances to interpret the model correctly.

Sorry!, - but I think that’s the only chance we will have.
Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

                             Copy of lost replay No. 5

Dear Scorael;

In this moment I can give you only what I did for the model:

I used the following Equations:
   A.)   Ox:       O    +  2e^-    ---->    O^2-
   B.)   Red:      H                ---->    H⁺    +  1e^-
   C.)   RxN:     2H₂  +  1O₂   ---->   2H₂O
   D.)   Current :    moles e- / sec  =  I / F =  I * 1 C/sec /  96’485.3 C/mole

So I got for the mass flow(in) as a function of the current:
m_O2(in) =  I / F  / nox_O  / rxn_O  * MW_O2 = I / 96’485.3 / 2 / 2 * 32.0  g/mole = x mole O₂ / sec.
m_H2(in) =  I / F  / nred_H  / rxn_H * MW_H2 =  I / 96’485.3 / 1 / 2 *  2.0  g/mole = x mole H₂ / sec.
With:
     -    nox_O   =   2 from A.)    and  rxn_O =  1 * 2 from C.)
     -    nred_H  =   1 from B.)    and  rxn_H =  2 * 1 from C.)

Or simpler spoken:
If you combine A.) with C.) or B.) with C.) it results that with one RxN you have a generation/exchange of 4e^- (respective 4 moles) and for that 1.0 mole O₂ and 2 moles H₂ is required.
That’s all valid for a conversion efficiency of 100%.

You may think about and tell me your conclutions.

I hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

                             Copy of the lost replay No. 7

Dear Scorael;

Sorry!, -  My old mistake, because I very rarly have to deal with RedOx:
    I interchanged the terms  Red and Ox once again!

Here the corrected Equations:
   A.)   Red:     O    +  2e^-    ---->    O^2-
   B.)   Ox:       H                ---->    H⁺    +  1e^-
   C.)   RxN:     2H₂  +  1O₂   ---->   2H₂O
   D.)   Current :    moles e^- / sec  =  I / F =  I * 1 C/sec /  96’485.3 C/mole

And:
m_O2(in) =  I / F  / nred_O  / rxn_O  * MW_O2 = I / 96’485.3 / 2 / 2 * 32.0  g/mole = x mole O₂ / sec.
m_H2(in) =  I / F  / nox_H   / rxn_H   * MW_H2 =  I / 96’485.3 / 1 / 2 *  2.0  g/mole = y mole H₂ / sec.
With:
     -    nred_O  =   2 from A.)    and  rxn_O =  1 * 2 from C.)
     -    nox_H   =   1 from B.)     and  rxn_H =  2 * 1 from C.)


Be remembered that it is bad practice to use the term: I /(nF), because I/F is clearly indicating the current in mole electrons per second, as pointed out in equation D.)!

The stoichiometric ratio in your case is = mole O₂  / mole H₂   = x / y = 1/2 (from above, but it follows also from C.)).
But then using 0.5 and S_x and (nF) in the same equation doesn’t make any sense, as in your equations.

Good Luck!
                    ARGOS⁺⁺