Problem Statement: The following reaction stoichiometry describes the aerobic growth of yeast on ethanol.
CH3CH2OH + a O2 + b NH3 ----> c CH1.704N0.149O0.408 + d CO2 + e H2OElemental balance on C, O, N, and H gives the following four equations:
# moles in reactants = # moles in products
C: 2 = c+d (3.1)
O: 1+2a = 0.408c+2d+e (3.2)
N: b = 0.149c (3.3)
H: 6+3b = 1.704c+2e (3.4)
Furthermore, we have the following additional equation (which
is called the respiratory quotient and relates the number
of moles of carbon dioxide given off per mole of oxygen intake).
0.66 = d/a (3.5)
List the five equations (3.1)-(3.5) in the standard "Ax=b"
format. Identify the matrix "A" and the vector "b"
in the standard notation. Write minimal MATLAB codes to
solve for the five stoichiometric coefficients a through e.
Solution:
Solution:
The five equations are rearrenged as:
C: c+d =2 (3.1)
O: -2a+0.408c+2d+e=1 (3.2)
N: -b+0.149c =0 (3.3)
H: -3b+1.704c+2e =6 (3.4)
ratio: -0.66a+d =0 (3.5)
In the standard Ax=b format, the matrix A is:
a b c d e ... stoichiometric coefficients
| 0 0 1 1 0 | ... (3.1)
| -2 0 0.408 2 1 | ... (3.2)
A = | 0 -1 0.149 0 0 | ... (3.3)
| 0 -3 1.704 0 2 | ... (3.4)
| -0.67 0 0 1 0 | ... (3.5)
In the standard Ax=b format, the column vector b is:
| 2 |
| 1 |
b = | 0 |
| 6 |
| 0 |
Minimal MATLAB codes
%-----------------------------------------------------------------------
% Find the reaction stoichiometry
% CH3CH2OH + a O2 + b NH3 ----> c CH_1.704N_0.149O_0.408 + d CO2 + e H2O
% (ethanol) (yeast biomass)
% A minimal hard-coded example
% Programming note: solve Ax=b with "\"
% Instructor: Nam Sun Wang
%-----------------------------------------------------------------------
% Start fresh ----------------------------------------------------------
clear all
% Elemental balance equations ------------------------------------------
% C: c + d = 2
% O: -2a +0.408c +2d + e = 1
% N: -b +0.149c = 0
% H: -3b +1.704c +2e = 6
% ratio: -0.66a + d = 0
% The coefficients for matrix A are (in row-wise order): ---------------
% a b c d e
A = [ 0 0 1 1 0
-2 0 0.408 2 1
0 -1 0.149 0 0
0 -3 1.704 0 2
-0.66 0 0 1 0 ];
% The coefficients for column vector b are: ----------------------------
b = [ 2; 1; 0; 6; 0 ];
% Solve the linear set of equations by calling the "inv" function or predivision "\"
x = A\b;
% Print results --------------------------------------------------------
disp('The solution is ...')
disp(x)
% Print results (more elaborate version with formatted print statement
disp('The solution is ...')
stoichiometry_coeff = 'abcde';
for i=1 : size(stoichiometry_coeff,2)
fprintf( ' %s = %6.4f\n', stoichiometry_coeff(i), x(i) )
end
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