2.3　Estimation and Calculation of the Amount of Methane Produced from Organic Compounds

The theoretical value of methane production can be estimated and calculated according to the chemical composition of materials. In 1939, Buswell, et al., through a series of experiments, found that all organic compounds can arise anaerobic degradation to produce methane, and the amount and composition of methane generated from complete degradation are determined by the composition of materials added. The relationship may be expressed by formula (2-1):

When,water contributes to methane production.

If the chemical compositions of the fed materials are known, the theoretical amount of gas generated may be calculated with the formula. The formula is called Buswell Formula.

Example 1: Calculate the amount and component of biogas produced from glucose.

The molecular formula of glucose is C6H12O6, and molecular weight is 180.

If n = 6, a = 12, b = 6, the Buswell Formula can be C6H12O6= 3CO2+ 3CH4, 180 g glucose can produce 3 mol CO2and 3 mol CH4.

1 kg glucose can produce 1000÷180×3≈16.67mol CO2and CH4.

Under STP (0℃ and 1 atm), gas volume is 16.67×22.4=373 (L).

So 1kg glucose can produce 373L CO2and 373L CH4. The biogas yields are about 750L, while the component of methane is 50％.

By using the formula, the amount and components generated from three major organic compounds are calculated (Table 2.3). Data listed in Table 2.3 of various organic compounds are different. It is simply because of the greater variance in chemical compositions of different polysaccharides, proteins and fats used.

From Table 2.3 it is obvious that carbohydrate and protein generate less methane, together amounting for 50％. Most of rural biogas resources, especially various straws, contain more carbohydrates, compared with industrial-waste-water which contains more lipids, therefore they produce smaller amount of methane in the biogas, around 60％.

Using formula (every kg COD generates 0.35m3methane), men can get another important theoretical content of methane, i. e. 0.25mg methane (0.35L under STP) by removing 1mg COD. Because of the difficulty in determining the components of materials used, this theoretical production of gas is of practical value.

In the processes of practical fermentation, the ferments can't be completely degraded and part of the ferments are utilized by microbes which produce other products. The net gas production is less than the theoretical amount mentioned above.

Example 2: 500t of waste water from winery, with its COD value of 34500mg/L, is treated by anaerobic digestion. Calculate the amount of methane produced from waste water.

The discharge amount of COD value per day = 500m3×0.0345t/m3= 17.25t

The methane yields = 17.25t×350m3=6037.5m3