What will be proper mixing ratio of water and cooked waste for preparation of slurry in AD? What should be the free space after introducing slurry in AD?
It depends on the water content of the cooked waste. First you need to determine the total solids TS of the cooked waste and add water so that TS of the slurry is between 8-10%. The volume of the organic mass in the AD reactor should be max. 70-80% of the reactor volume to prevent the suspension particles arrive on the gas pipe. So the free space is ca. 20-30%.
Agreed with previous answer, also take pH under control for better gas production. The optimum pH range for anaerobic digestion producing methane is 6.8–7.2 The growth rate of methanogens can be greatly reduced when the pH value is less than 6.6.
Indeed, articles indicate different S/I ratio, from 10/1 (as I used in the past, based on a local guideline and older research reports) to 1/4. After we acquired standards we noticed that the recommendation is that the S/I ratio is much lower than what we used. Also, many of the more recent articles indicate a 2-4 times higher share of inoculum than the substrate. But selecting the optimal ratio depends largely on the type of fermentation substrate, and it is done through optimization attempts, as indicated by the standard VDI4630.
I can not give you a precise indication. As mentioned above, optimal S/I ratio should be determined by testing. In case of lack of time for many fermentation experiments, I think it is good if you use the S/I ratio of 1/2. Although the reverse ratio S/I of 2/1 or higher may be as good... So you have to try. A high ratio of rice substrate can lead to acid inhibition, while too much inoculum can lead to ammonium inhibition. Good luck!
Thanks for the suggestion. Now if the TS of waste is 19%, then how much amount of water to be added to make slurry so that TS of the slurry remains 8-10% ? If the TS of slurry become more or less than 8-10%, what will be the effect on Bio-methane potential ?
If you have a waste of 19%TS, it means that 8 g TS are found in: 8/19x100 = 42,10 g waste. To prepare a slurry of 8% TS, you should add 100-42,10 = 57,9 g water for dilution. The same calculation for preparing a slurry of 10% TS: 10/19x100 = 52,63 g waste; 100-52,63 = 47,37 g water for dilution. In other words, for each 100 g waste you should add 137,5...90 g water to have a slurry of 8...10%TS. This TS range is important to facilitate bacterial mobility and metabolism, also to prevent solids deposition at the bottom of the reactor. There are many articles which indicate this range as optimal for the best biomethane potential in batch AD.