## Conclusions

The anaerobic degradation of organics is a rather complicated process. Even if we only wish to discuss the digestion of glucose, we must take into account three different groups of bacteria. It is not easy to separate them and, in the case of conversion of propionate to acetate and hydrogen, the acetogenic and methanogenic bacteria are dependant upon each other. Although it is important to know the value of X, the bacterial concentration responsible for one interesting catabolic reaction, measurements are frequently impossible and X must be obtained by calculations using mass balances.

In many studies the concentrations:

are used in Monod or Haldane kinetics instead of cHAc and cHPr in recent publications.

Let us write:

and CH3COOH v CH3COO- + H+ Then, respectively:

SHAc as well as:

and:

After introducing Eq. (8.27) into Eq. (8.24), we obtain:

and:

Note that, in contrast to KS and Ki, KSH and K1h are not functions of pH, because of Eq. (8.25), which yields:

The next important point is to calculate how much methane is produced if 1 mol CH3COOH or 1 mol C6H12O6 is degraded anaerobically. We can compare both substrates according to their COD. Let us start with CH3COOH.

For chemical oxidation of CH3COOH we write:

and therefore 1 mol CH3COOH needs 2 mol O2; or 60 g HAc give 64 g COD. Methane production from CH3COOH can be calculated using:

giving a yield coefficient of:

rCH4 1 mol CH4 22.4 Nm3 CH4

YCh4/hac = — = --t—4 = — n— —4 = 0.35 Nm3 (kg COD)-1 (8.31)

rHAc 1 mol HAc 64 kg COD

Now we calculate the yield of CH4 from glucose: for the chemical oxidation of C6H12O6, it follows that:

Therefore, 1 mol C6H12O6 needs 6 mol O2; or 180 g C6H12O6 is equivalent to 192 g COD.

Methane production from C6H12O6 can be calculated by summarizing Eqs. (8.1), (8.2), (8.4) and (8.5)

giving a yield coefficient of:

Conclusion: from wastewater loaded with hydrocarbons (CH2O) and a given reaction rate of kg COD (m3 d)-1, which can be mineralized by anaerobic processes (digestion), the maximum rate of methane production can be calculated using Eq. (8.31):

regardless of whether polysaccharides, monosaccharides or acetate are present in the wastewater.