Week 5.1 - Compact Notation Derivation for a simple Mechanism

Week 5 Challenge – Compact Notation Derivation for Simple Mechanism

Aim: - Derive the net production rate using compact notation for chemical equilibrium

Objective: -

• To study the chemical equilibrium for given chemical reactions.
• To derive stoichiometric matrix reaction rate for reactant and product.
• To derive the net reaction rate for each of the multistep reactions.
• To derive the net production rate for each of the reaction.

Introduction: -

 

Chemical equilibrium can be achieved either by considering system steady state in which the time is not participated as function of reaction. In real practice this system is not considered as it is not a one step process which will always give stoichiometric reaction of combustion process. It is well known fact that air and fuel are not get directly mixed and give proper combustion process. There are certain elementary reactions and intermediate species which give rise to the combustion process spontaneously. The study of such elementary reaction at different pressure and operating temperature is called as chemical kinetics. Chemical kinetics helps in the determining ignition delay i.e. time at which combustion process start after initial mixing of fuel and air. As the time is function of chemical reaction the reaction rate of species inside the combustion is important to determine. Introducing time zone for the chemical reaction helps to predict the reaction mechanism which further use for predicting the product side of reactions for the given properties of fuel and air.

The calculation of rates of reaction can be done manually. In this process number of ODE system will get generate. As the ODE are stiff in nature i.e. one of the species in the reaction can have drastic changes as compared with other species it is impossible to solve the reaction by hand. The chemical reaction of combustion has numbers of reaction which can’t be solved manually. The most of ODE system are solved using numerical integration method in which both explicit and implicit schemes are used. Using the numerical integration scheme the problem can be treated as initial value problem in which initial value of species at given temperature and pressure can be used to find next iterative value. To solve the ODE system, it is important to have the computer code. Using direct method of deriving reaction rate for all the species is very difficult to code as code understand the matrices rather than words. Hence, it is important to use the compact notation for the chemical reaction to arrange the problem in matrix form which can be understand by computer in simpler ways.

Reaction Mechanisms: -

`CO+O_2 ⇌ CO_2 +O`

`O+H_2 O ⇌OH+OH`

`CO+OH ⇌CO_2 +H`

`H+O_2 ⇌ OH+O`

From above four reactions it can be seen that there are 7 species which are interacting with each other. The Kf1 and Kr1 are the reaction rates which need to be calculated. Kf1 is the forward reaction rate and Kr1 is reverse reaction rates. The compact notation for the reaction can be stated as ‘i’ and compact notation for species can be stated as ‘j’.

j	Species	i	Reactions
1	CO	1	R1
2	O2	2	R2
3	CO2	3	R3
4	O	4	R4
5	H20
6	OH
7	H

This compact notation helps to understand the species term in subsequent reactions. The coefficient matrix of reactants and product are need to define in order to derive the total reaction coefficient matrix which is denoted by ‘V_ji’.

 The Reactant coefficient matrix denoted by ‘`(V_(ji))^'`’ is defined as:

`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]]`

The product coefficient matrix denoted by ‘`V_(ji)^('')`’ is defined as:

`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]]`

The net coefficient matrix for all the reactions is given by:

`(V_(ji)) = (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]]`

After finding the final net coefficient matrix, it is required to calculate the net rate of reaction (q) for each subsequent reactions. The net rate of reaction is derived as:

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

from above equation it can be seen that the net rate of reaction is substraction of species from reactant side to product side. If the qi is positive the reaction moves in forward direction and if qi is negative the reaction moves in reverse direction. The term Xj referes to the concentration per unit volume of each species in the reaction. 

The net rate of reaction for each of the multistep reaction is given as:

`q_1 = k_(f1) [CO]^1 [O_2]^1 [O]^0 [H_2O]^0 [OH]^0 [H]^0 - K_(r1) [CO]^0 [O_2]^0 [CO_2]^1 [O]^1[H_2O]^0 [OH]^0 [H]^0`

Similarly for other reaction the net rate of reaction is derived as:

`q_2 = k_(f2) [O] [H_2O] - k_(r2) [OH]^2`

`q_3 = k_(f3) [CO] [OH] - k_(r3) [CO_2][H]`

`q_4 = k_(f4) [O_2] [H] - k_(r4) [O] [OH]`

After calculating the net reaction rate for each of the reaction, the net production rate is need to defined for each of the species. The net production rate will provide the prediction of species in the forward and reverse reaction which ultimately helps to find out reaction rates of each reactions.

The net production rate is given as:

`w_j = sum_(i=1)^L (V_(ji)) (q_i)`

where L is the number of reactions in the mechanism and we can see that it is the multiplication of net coefficient of reaction rate with net reaction rate. This net production rate has to be defined for each of the species included in the reactions, which is given as:

`w_1 = -q_1 -q_3`

`w_2 = -q_1 - q_4`

`w_3 = q_1 + q_3`

`w_4 = q_1 -q_2 +q_4`

`w_5 = -q_2`

`w_6 = 2q_2 -q_3 + q_4`

`w_7 = q_3 - q_4`

The production rate can be also defined as rate of change of species with respect to time which is expressed as:

`w_j = (d[X_j]) /dt`

where subscript j is refered to number of species present in reactions. So every w1 to w7 referes to the each of the number of species from 1 to 7. This can be shown in matrix form for better understanding.

`[[(d[CO])/dt],[(d[O_2])/dt],[(d[CO_2])/dt],[(d[O])/dt],[(d[H_2O])/dt],[(d[OH])/dt],[(d[H])/dt]] = [[w_1],[w_2],[w_3],[w_4],[w_5],[w_6],[w_7]]`

From above matrix the value of each term in right hand side of matrix is determined as:

`(d[CO])/dt = k_(r1) [CO_2][O] - k_(f1) [CO][O_2] + k_(r3) [CO_2][H] - k_(f3) [CO][OH]`

`(d[O_2])/dt = k_(r1) [CO_2][O] - k_(f1) [CO][O_2] + k_(r4) [O][OH] - k_(f4) [O_2][H]`

`(d[CO_2])/dt = k_(f1) [CO][O_2] - k_(r1) [CO_2][O] + k_(f3) [CO][OH] - k_(r3) [CO_2][H]`

`(d[O])/dt = k_(f1) [CO][O_2] - k_(r1) [CO_2][O] + k_(r2) [OH]^2 - k_(f2) [O][H_2O] + k_(f4) [O_2][H] - k_(r4) [O][OH]`

`(d[H_2O])/dt = k_(r2) [OH]^2 - k_(f2) [O][H_2O]`

`(d[OH])/dt = 2k_(f2) [O][H_2O] - 2k_(r2) [OH]^2 + k_(r3) [CO_2][H] - k_(f3) [CO][OH]+ k_(f4) [O_2][H] - k_(r4) [O][OH]`

`(d[H])/dt = k_(f3) [CO][OH] - k_(r3) [CO_2][H] + k_(r4) [O][OH] - k_(f4) [O_2][H]`

In above equation we have system of ODE which are stiff in nature and now it can be evaluated by using Numerical Integration in computer code using ODE45 or other functions in different programming software.