exoenzymes of fermentative bacteria obligate anaerobes acetogens acetotrophic
實驗原理 exo-enzymes of fermentative bacteria ① ② obligate anaerobes ③ acetogens acetotrophic methanogens ④ hydrogenotrophic methanogens
表 1、2:原始數據 好氧消化 厭氧消化 初重 (g) 1. 4488 1. 4395 末重 (g) 1. 4613 1. 4527 MLSS (mg/L) 1250 1320 初始COD 負荷 (kg glucose/kg MLSS) 0. 600 0. 568 空白 5 min 60 min 90 min 好氧 厭氧 滴定液體積 (m. L) 9. 016 7. 410 3. 808 3. 932 3. 322 3. 784 3. 980 COD (mg/L) - 642. 4 2083. 2 2033. 6 2277. 6 2092. 8 2014. 4
圖 1:COD 變化曲線圖 2500 2000 COD (mg/L) 1500 Aerobic 1000 Anaerobic 500 0 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 time (min)
參考資料 1) Abedeen, Z. , “Sludge Treatment in Wastewater: Aerobic and Anaerobic Digestion”. 2) Eckenfelder, W. W. , Patoczka, J. B. and Pulliam, G. W. , “Anaerobic versus Aerobic Treatment in the USA”. 3) Haandel A. V. and Lubbe, J. V. D. , “Handbook Biological Waste Water Treatment - Design and Optimisation of Activated Sludge Systems”: 339 -343. 4) Handbook Biological Wastewater Treatment - Design of Activated Sludge Systems: “Aerobic Digestion”. 5) Michigan Dept. of Environmental Quality Operator Training and Certification Unit, “Anaerobic Sludge Digestion Process”. 6) Mountain Empire Community College Water/Wastewater Distance Learning Website, “Lesson 4: Aerobic and Anaerobic Digestion and Types of Decomposition”. 7) Portland State University, “Chapter 17 Aerobic Biological Treatment”; “Chapter 18: Anaerobic Sewage Treatment”. 8) Sivanandan, T. P. , ”Advantages of Aerobic over Anaerobic Digestion”, www. water-chemistry. in 9) Wikipedia, “Anaerobic digestion” ; “Metabolism” ; “Chemical Oxygen Demand” ; “Comparison of Anaerobic and Aerobic Digestion”. 10) 宋士武,“廢水污泥之厭氧消化”。 11) 孫立波,“現代環境微生物技術”: 202 -203. 12) 歐陽嶠暉,《下水道 程學》。 13) 行政院環境保護署環境檢驗所 NIEA W 517. 52 B,“水中化學需氧量檢測方法─密閉式重鉻酸 鉀迴流法”
附錄 如何決定稀釋倍數? C 6 H 12 O 6 + 6 O 2 → 6 CO 2 + 6 H 2 O M glucose = 180 g/mol 1 mole of glucose requires: 32 × 6 = 192 g of oxygen. ∴ COD for 1 g of glucose: 192/180 × 10000 ppm = 10667 mg/L Given tank/beaker volume: 1000 m. L ; Substrate volume: 75 m. L Volume ratio = 1000/75 = 13. 33 = 14 ∴ COD for 1 g of glucose in tank: 10667/14 = 762 mg/L ∵ Detection limit for potassium dichromate reflux COD test = 250 mg/L ∴ Dilution Factor = 4~5 Caution: DF cannot be 3 or less because 762/3 = 253 mg/L (>250 mg/L)
- Slides: 27