Problem. In an anaerobic process $($no electron acceptor$)$ a substrate is present at an initial concentration of $1\mathrm{.}5$gCO$\frac{D}{L}$ and it is totally removed from the medium.

$1500mg$

We can assume that the products in the liquid phase contain only $0\mathrm{.}1\frac{g}{L}$ of microorganisms $(C_5{H}_7{O}_2(N))$ and $0\mathrm{.}1\frac{g}{L}$ of acetic acid $(C{H}_3\frac{C}{O}OH).$ The produced gas is made only of methane and carbon dioxide.

How much methane is produced in this process, per unit volume of the liquid phase?

The COD balance under anaerobic conditions is given by Equation,

$(-S_{\text{C}\text{O}\text{D}})=\frac{x_{\text{C}\text{O}\text{D}}}{\text{P}\text{r}\text{i}\text{o}\text{m}\text{a}\text{s}\text{s}}+\frac{P_{\text{C}\text{O}\text{D}}}{\text{P}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{f}\text{o}\text{r}\text{n}}$

In this case,

$(-S_{\text{C}\text{O}\text{D}})=1\mathrm{.}5\frac{\text{g}\text{C}\text{O}\text{D}}{L},x_{\text{C}\text{O}\text{D}}=0\mathrm{.}11\mathrm{.}42\frac{\text{g}\text{C}\text{O}\text{D}}{L}=0\mathrm{.}142\frac{\text{g}\text{C}\text{O}\text{D}}{L}$

Where, $1\mathrm{.}42$ is the conversion factor between microorganisms and COD, according to Equation, $C_5{H}_7{O}_{2}N($microrganisms$)+5O_{2}5C{O}_2+2H_{2}O+N{H}_3$

The total COD of the products is$,$ therefore,

$P_{\text{C}\text{O}\text{D}}=1\mathrm{.}5-0\mathrm{.}142=1\mathrm{.}358=1\mathrm{.}36\frac{\text{g}\text{C}\text{O}\text{D}}{L}$

Sindil $(3)$ Dionisi, D$.,2017.$ Biological wastewater treatment processes: mass and heat balances. CRC press.