Optimization of food-to-microorganism ratio and addition of Tween 80 to enhance biohythane production via two-stage anaerobic digestion of food waste

BACKGROUND: Food waste (FW) is easily acidified to produce acid inhibition, which limits the treatment efficiency of anaerobic digestion (AD). To improve the AD efficiency of FW, two-stage anaerobic digestion (TSAD) technique was used to test the effect of food-to-microorganism (F/M) ratio on hydrogen production and the effect of Tween 80 addition on methane production. Changes in volatile fatty acids during FW TSAD were analyzed. Furthermore, the energy recovery of different TSAD conditions was compared. RESULTS: Under the condition of F/M = 10, hydrogen production of FW was the highest (102.79 mL g(-1) volatile solids (VS)). In the methanogenesis stage of FW, the methane production of different conditions was 351.97-407.11 mL g(-1) VS. Kinetic analysis showed that the hydrogen production could be increased by appropriately increasing the F/M ratios, and the time required to reach 90% methane production (t(90)) in the methanogenesis stage could be shortened by adding Tween 80. Analysis of energy recovery showed that the energy yield of TSAD was 13.01-15.68 kJ g(-1) VS, and the average daily energy recovery of hydrogen production stage F/M = 10, methane production stage with 0.1% Tween 80 (A-10 and M-10 T) were the highest at t(90), which was 9.23 and 4.60 kJ d(-1). CONCLUSION: F/M ratio affects hydrogen production, and a suitable F/M ratio can improve the hydrogen production of the TSAD of FW. The addition of Tween 80 can shorten the t(90) of the methanogenesis stage. Analysis of energy recovery calculation showed that A-10 + M-10 T was more economically rational for TSAD. (c) 2023 Society of Chemical Industry.