Answer to Question #284220 in MatLAB for Jack

Question #284220

A typical chemical formula for aerobic growth of a microorganisms is

C2H5OH + a(O2) + (b)NH3 —> (c)CH1.7H0.15O0.4 + (d)H2O + (e)CO2

where term CH_1.7H_0.15O_0.4 represents metabolism of the microorganism. The ratio of moles of CO₂ produced per mole of O₂ consumed is called the respiratory quotient, RQ, which can be determined experimentally. Given this ratio we have 4 constants, a-d that are unknown. We can perform a mass balance on each of the four key elements

Carbon: 2 = c + (RQ)a

Hydrogen: 6 + 3b = 1.7c + 2d

Oxygen: 1 + 2a = 0.4c + d + 2(RQ)a

Nitrogen: b = 0.15c

Let RQ = 0.8 and then find the vector [a b c d] using Gauss Elimination Method.

Expert's answer

The equations may be written as

"\\begin{matrix}\n\\text{Carbon: }&\t(RQ)a + c = 2 \\\\\n\\text{Hydrogen: }& -3b + 1.7c + 2d = 6\\\\\n\\text{Oxygen: }& (2(RQ)-2)a + 0.4c + d = 1\\\\\n\\text{Nitrogen: }& b - 0.15c = 0\n\\end{matrix}"

Or substituting RQ = 0.8 in matrix form

"\\begin{bmatrix}\n 0.8 & 0 & 1 & 0 \\\\\n 0 & -3 & 1.7 & 2 \\\\\n -0.4 & 0 & 0.4 & 1 \\\\\n 0 & 1 & -0.15 & 0\n\\end{bmatrix} \n\\begin{bmatrix}\n a \\\\\n b \\\\\n c \\\\\n d \n\\end{bmatrix} =\n\\begin{bmatrix}\n 2 \\\\\n 6 \\\\\n 1 \\\\\n 0\n\\end{bmatrix}"

Adding 1/2 of the first row to the third one:

"\\begin{bmatrix}\n 0.8 & 0 & 1 & 0 \\\\\n 0 & -3 & 1.7 & 2 \\\\\n 0 & 0 & 0.9 & 1 \\\\\n 0 & 1 & -0.15 & 0\n\\end{bmatrix} \n\\begin{bmatrix}\n a \\\\\n b \\\\\n c \\\\\n d \n\\end{bmatrix} =\n\\begin{bmatrix}\n 2 \\\\\n 6 \\\\\n 2 \\\\\n 0\n\\end{bmatrix}"

Add 1/3 of the second row to the fourth:

"\\begin{bmatrix}\n 0.8 & 0 & 1 & 0 \\\\\n 0 & -3 & 1.7 & 2\\\\\n 0 & 0 & 0.9 & 1 \\\\\n 0 & 0 & 0.417 & 0.667\n\\end{bmatrix} \n\\begin{bmatrix}\n a \\\\\n b \\\\\n c \\\\\n d \n\\end{bmatrix} =\n\\begin{bmatrix}\n 2 \\\\\n 6 \\\\\n 2 \\\\\n 2\n\\end{bmatrix}"

Subtract 4.17/9 of the third row from the fourth:

"\\begin{bmatrix}\n 0.8 & 0 & 1 & 0 \\\\\n 0 & -3 & 1.7 & 2\\\\\n 0 & 0 & 0.9 & 1 \\\\\n 0 & 0 & 0 & 0.204\n\\end{bmatrix} \n\\begin{bmatrix}\n a \\\\\n b \\\\\n c \\\\\n d \n\\end{bmatrix} =\n\\begin{bmatrix}\n 2 \\\\\n 6 \\\\\n 2 \\\\\n 1.07\n\\end{bmatrix}"

Therefor

"d = 1.07 \/ 0.204 \\\\\nc = (2 - d) \/ 0.9 \\\\\nb = (6 - 1.7 c - 2 d) \/ (-3) \\\\\na = (2 - c) \/ 0.8"

"\\begin{bmatrix}\n a \\\\\n b \\\\\n c \\\\\n d\n\\end{bmatrix} =\n\\begin{bmatrix}\n 7.0 \\\\\n -0.55 \\\\\n -3.6 \\\\\n5.24\n\\end{bmatrix}"