mech. eng. taking combustion class, chem equilibrium question

wallace.k · science forum beginner Joined: 09 Apr 2006 Posts: 7

Hi all, (crossposted to sci.engr.chem, sci.chemistry,
alt.mechanical.engineering)

I'm an ME student taking a graduate class in combustion. My focus
through school has been thermal and fluid sciences, but some of the
chemical equilibrium stuff I still find a bit confusing.

The following is a homework problem, so I don't want an answer, just
maybe a point in the right direction (or affirmation that I'm thinking
correctly or not).

Methane is combusted with not enough air, so the extra methane remains
in the products side. Assume no dissociation. (that means no CO, O2,
H2, etc. in products, if I am reading correctly, which is idealized I
think b/c I do know that dissociation is temperature dependent). But for
this problem, that is the assumption.

(1?)CH4 + (1/2)O2 --> (a) CO2 + (b) H2O(g) + (c)CH4.

AFAIK this is on a 1-mole-of-fuel basis. Is that an okay assumption to
make? I believe so. Otherwise it would just multiply the other #'s of
moles needed/produced, right? Also idealized to be just oxygen, not
air, though I'm sure it could be also done with (O2 + 3.76N2) also.

I need to find (psi) final, where 'psi' is the extent of reaction. I
need to find n1, n2, n3, n4, n5... for the equalities

dn1/nu1 = dn2/nu2 = dn3/nu3....= d(psi). nu is the mole number,
negative for reactants, positive for products.

I know that the reaction is 'done' when the O2 is consumed, and that
there is only 25% of the O2 required for complete combustion. I have
found that this is not possible for me to solve using a stoichiometric
balance (mass/atom balance), as then I have 3 equations and 4 unknowns.
Or thus far it doesnt' seem to work out anyway. I am thinking that I
need to use the definition of Gibbs free energy, dG being = 0 at
equilibrium, but I have some real confusion about this.

I think I'm not understanding the definition of equilibrium, maybe. Is
equilibrium attained when psi = 1 (completed reaction)? Is it when all
the O2 is gone and no more reacting can occur?

I'm now going back to my old thermodynamics book (as it's written far
more readably than this combustion text I have) to the chem. equilibrium
for mixtures chapter, and will try to understand some of this- but if
anyone here can give me some pointers on comprehending the ideas behind
what I need to do, that would be great.

TIA,
K. Wallace

protosurge@gmail.com · science forum beginner Joined: 09 Apr 2006 Posts: 4

Hi:

The problem you posted can be solved in a relatively straight foward
manner. The typical approach is to create table of reactants and
products before and after the reaction using the extent (psi) of
reaction. For ex: Number of CO2 moles Final = 0*Number of CO2 + 1*psi
(since the stoichiometric coefficient is +1). Do this for each
component and you'll get five equations.

Now add up all the right hand sides of the equations to get an
expression for the final number of moles at the end of the reaction.
Call it NTotal.

You then need an equilibrium statement. Typically it's written as K =
(P1^nu1 * P2^nu2 * .... )/(R1^nu1 * R2^nu2 * .... ) where P refers to
product, R is the reactant and nu is the stoich. coefficient. The
quantity R1 or P1 refers the activity of that component. If you assume
ideal gas behavior, you can write P1 = mole fraction of P1.

P1 = Number of CO2 final/NTotal and so on for each reactant and
product. Substitute everything back in the K equation. If you know the
initial compositions, temperature and the value of K, the only unknown
in the equation is psi. Solve for psi and you should able to retrieve
all the quantities that you need (You may have to solve the equation
iteratively or via a computer).

I leave it up to you to figure out how to determine the value of K for
a given temperature :)

If you want to see a couple good examples of this problem solved, I
recommend Chemical and Engineering Thermodynamics by Sandler. Take a
look at Chapter 9 or 10. (I think Sandler has some good examples for
this topic in his book).

Kiran

wallace.k · science forum beginner Joined: 09 Apr 2006 Posts: 7

protosurge@gmail.com wrote:

wallace.k · science forum beginner Joined: 09 Apr 2006 Posts: 7

One more question. This method seems to depend on knowing all the
stoichiometric coefficients, which aren't given; I just know that I have
only 25% of the needed air to consume the CH4. does that mean that since
for complete combustion it would generally take 2 moles of O2 to react
with 1 mole of CH4, then on the products side, there will be left 3/4 of
a mole of CH4, since with 1/2 of a mole of O2 I can only combust 1/4 of
a mole of CH4? Or is it not that straightforward? If I can get there, I
can do the rest of this.
regards,
k wallace

protosurge@gmail.com wrote:

protosurge@gmail.com · science forum beginner Joined: 09 Apr 2006 Posts: 4

If you assume complete combustion, your rxn should look like
CH4 + 2 O2 = CO2 + 2 H2O.

Create your extent equations using this reaction. The final
compositions depend on the value of K.

Kiran

wallace.k wrote:

rudnyi · science forum addict Joined: 20 May 2005 Posts: 75

Richard · science forum beginner Joined: 21 Dec 2005 Posts: 17

I don't know if you are supposed to assume only the two products
listed, but in practice when you have incomplete combustion you get a
lot of CO, as well as some hydrocarbons with more than one carbon -
C2H4, C2H6, C3H6, C3H8, C4H8, C4H10 are some examples.

(Depending on how little O2 there is relative to stoichiometric, you
can also get C2H2, C4H4, c-C5H10, C6H6 and other more "reduced"
products that can contribute to soot formation...which is another
product if you're *really* starved of O2.)

Richard

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