Week 5.1 - Compact Notation Derivation for a simple Mechanism

Aim: To derive the reaction rates and production of each elementary species for the given simple elementary reations using the compact notations.

Elementary Reactions:

`CO + O_2 overset(kf1)(iff)_(kr1) CO_2 + O ----------text(R1)`

`O + H_2O overset(kf2)(iff)_(kr2) OH + OH ----------text(R2)`

`CO + OH overset(kf3)(iff)_(kr3) CO_2 + H ----------text(R3)`

`H + O_2 overset(kf4)(iff)_(kr4) OH + O ----------text(R4)`

where kf and kr 1,2,3,4 are the forward and reverse reaction rates of the elementary reactions R1,R2,R3 and R4.

Theory:

In general Fuel + Air -> Products, this is the final global equilibrium reaction that we write for a combustion reaction. But in reality there are number of elementary reactions that take place from start to end till the reation reaches equilibrium composition. The species in these elementary reactions are called elementary species. The entire set of elementary reactions are species is called Reaction Mechanism.
The study of entire Reaction Mechanism is called Chemical Kinetics. Chemical Kinetics concerns with understanding rates of chemical reactions. With the study of Chemical Kinetics we understand and solve problems like Ignition Delay, Heat realeses, Emissions, Food Decomposition, microrganism growth and many real life problems.

Formulas:

Let j be the number of elementary species in the elementary reaction

j	Species
1	CO
2	O2
3	CO2
4	O
5	H2O
6	OH
7	H

For j = 1 the species denoted is CO, for 2 is O2...

Let i be the number of elementary reaction given

i	Reactions
1	R1
2	R2
3	R3
4	R4

The Stoichiometry Coefficient matrix is calculated as difference of Stoichiometry Coefficient matrix of products and Stoichiometry Coefficient matrix of reactants.

`v_(ji) = v_(ji)'' - v_(ji)'`

where j is the elementary species and i is the reaction

Calculation of Net Reaction rate of each Reaction i:

`q_i = k_(fi)overset(N)prod_(j = 1) [X_j]^(v_(ji)') - k_(ri)overset(N)prod_(j = 1) [X_j]^(v_(ji)'')`

this calculates the net rate of the elementary reaction 'i'. The net reaction rate q tells us in which direction the reaction is taking place. If q is positive than the direction is forward direction and if the q is negative than the direction is backward.

Calculation of net production rate of particular species `omega_j`:

`omega_j = overset(L)sum_(j = 1) v_(ji)*q_i`

where L is the number of reactions.

for every j species the production rate is calculated. In the end the net production rate of the species will be equal to `(d[X_j])/(dt)`

Procedure:

1. The rate of change of an elementary species CO is given as

`(d[CO])/(dt) = -k_(f1)[CO][O2] + k_(r1)[CO2][O2] - k_(f3)[CO][OH] + k_(r3)[CO2][OH]`

Similarly for each species the rate of change of concentration of species is given from above formula. The sign convention is like, for species going forward the sign is negative as the concentration of that species is reducing and for the backward the sign is negative as the concentration is going to increase.

If we derive the rate of change of concentration for each species we would get a couple ODE system, which would be difficult to solve manaually. Therefore, we derive compact notation sign which can easily be fed into computer and solved numerically using any ODE packages or librarires.

2. For the compact notations we will denote the number of species as j and number of reactions as i. Refer the table.

3. First we derive the Stoichiometric Coefficient Matrix of reactants. The columns of this matrix is the number of species `[CO,O2,CO2,O,H2O,OH,H]` and the rows are the number of reactions i.

For this matrix only consider the species on the reactants side.

`v_(ji)' = [[1,1,0,0,0,0,0], [0,0,0,1,1,0,0],[1,0,0,0,0,1,0],[0,1,0,0,0,0,1]]`

4. Similarly we derive the Stoichiometric Coefficient Matrix of products.

`v_(ji)'' = [[0,0,1,1,0,0,0], [0,0,0,0,0,2,0],[0,0,1,0,0,0,1],[0,0,0,1,0,1,0]]`

5. Now we calculate the difference of the Stoichiometric Coefficient Matrix of products and reactants.

`v_(ji) = v_(ji)'' - v_(ji)' = [[-1,-1,1,1,0,0,0], [0,0,0,-1,-1,2,0],[-1,0,1,0,0,-1,1],[0,-1,0,1,0,1,-1]]`

6. Now we compute the Net Reaction Rate q for each i reaction. The net reaction rate of the ith reaction tells us the direction of the reaction. If q_i is positive than the rate of reaction is in forward direction and if q_i is negative the rate of reaction is in backward direction.

`q_i = k_(fi)overset(N)prod_(j = 1) [X_j]^(v_(ji)') - k_(ri)overset(N)prod_(j = 1) [X_j]^(v_(ji)'')`

Therefor for i = 1 calculating the Net Reaction Rate

`q_1 = k_(f1)[CO]^1[O2]^1 - k_(r1)[CO2]^1[O]^1`

Similarly, `q_2 = k_(f2)[O]^1[H2O]^1 - k_(r2)[OH]^1[OH]^1`

`q_3 = k_(f3)[CO]^1[OH]^1 - k_(r3)[CO2]^1[H]^1`

`q_4 = k_(f4)[H]^1[O2]^1 - k_(r4)[OH]^1[O]^1`

7. Now we compute the Production Rate of each particular elementary species ω_j.

`omega_j = overset(L)sum_(j = 1) v_(ji)*q_i`

We use the values we derived above and substitute in this equation and we will have the production rate of each species which will be equal to the rate of change of concentration of species with respect to time.

So i = 1 = CO, lets calcualte the production rate of CO

`omega_(CO) = v_(11)q_1 + v_(13)q_3`

`omega_(CO) = (-1)k_(f1)[CO]^1[O2]^1 - k_(r1)[CO2]^1[O]^1 - k_(f3)[CO]^1[OH]^1 + k_(r3)[CO2]^1[H]^1`

`omega_(CO) = -k_(f1)[CO]^1[O2]^1 + k_(r1)[CO2]^1[O]^1 - k_(f3)[CO]^1[OH]^1 + k_(r3)[CO2]^1[H]^1`

`therefore omega_(CO) = (d[CO])/(dt)`

Similarly, `omega_(O2) = -k_(f1)[CO]^1[O2]^1 + k_(r1)[CO2]^1[O]^1 - k_(f4)[H]^1[O2]^1 + k_(r4)[OH]^1[O]^1`

and `omega_(O2) = (d[O2])/(dt)`

`therefore [[omega_1],[omega_2],[omega_3],[omega_4],[omega_5],[omega_6],[omega_7]] = [[(d[CO])/(dt)], [(d[O2])/(dt)], [(d[CO2])/(dt)], [(d[O])/(dt)], [(d[H2O])/(dt)], [(d[OH])/(dt)], [(d[H])/(dt)]]`