1. ERT353 2014 Biofuel BIO ETHANOL What, Why, How, How much, ….

2. Biofuel : General term for fuel from bio resources What? Ethanol C 2 H 5 OH Chemical production : C 2 H 4 + H 2 O → CH 3 CH 2 OH Bioethanol : Ethanol that produced through fermentation Fermentation :?

3. Why? Properties are similar to fossil fuel Flash Point : 9 C Boiling Point : 78 C

4. Bioethanol Fermentation Glucose ----------------yeast----------  ethanol + carbondioxide Material Balance ? Mol equivalent Detail of fermentation

6. Microorganisms : Saccharomyces cerevisiae (yeast) Zymomonas mobilis

7. Saccharomyces cerevisiae Well studied, widely used commercially It growth is inhibited by ethanol (its own product) The theoretical yield is 0.511 g ethanol per gr glucose consumed. This yield can never be realized in practice since part of glucose is used for cell mass synthesis, and production of by-products such as glycerol, acetic acid, lactic acid, and succinic acid.

8. Zymomonas mobilis:  can produce ethanol at much faster rates than S. cerevisiae  high cell concentrations are not needed for high ethanol yield  high ethanol tolerance,  no requirements for aeration during the fermentation process  Not well studied

9. Sugar Raw materials (feedstocks) Sugar Sugar containing materials : Starch Cellulose

10. Typical Parameters of Batch Fermentation Process for Ethanol Production from Sugarcane Juice and Molasses

11. Starch Feedstocks : Example

12. Fermentation

13. Theoretical Ethanol Yield n is the number of glucose residues in the starch molecule. The amount of glucose produced from 1 kg of starch is 180n/ (162n + 18). When n is 2, as in maltose, the conversion factor is 1.053. When n becomes very large, this factor approaches 1.111. CONVERSION RATIO FROM STARCH OR CELLULOSE INTO GLUCOSE

14. Calculation example: Calculate the theoritical yield of ethanol that can be produced from 10 kg of glucose Calculate the theoritical yield of ethanol that can be produced from 10 kg of sugar cane Calculate the theoritical yield of ethanol that can be produced from 10 kg of molasses containing 40 % of glucose

15. Calculate theoretical ethanol yield from 1 kg of corn which has 15 percent moisture and contains 70 percent starch on a dry basis. Assume the conversion factor is 1.111.

16. Determine ethanol fermentation efficiency for the corn with 70 percent starch on dry basis in laboratory. In this experiment, a mash having 30 percent total solids on dry basis and after 72 h of fermentation. Analysis of the final liquid sample by high pressure liquid chromatography (HPLC) showed an ethanol concentration of 13.1 g/L. Example

17. Basis: 1 kg of mash. One kg of mash contains 300 g total solids and 700 g water. Starch content: 300 g x 0.70 = 210 g Glucose (MW 180) production by starch hydrolysis: 210 g x 1.111 = 233.3 g Theoretical ethanol production: 233.3 x 0.511 = 119.0 g Total liquid: (ethanol produced + volume of remaining water) Volume of the ethanol produced: 119.0 g / 0.79 (g/mL) = 150.8 mL Water (MW 18)consumption in starch hydrolysis: 210 g x 0.111 = 23.3 g Total liquid volume: 700 mL – 23.3 mL + 150.8 mL = 827.5 mL Ethanol concentration expected: 119.0 / 827.5 mL = 0.144 g/mL or 14.4 g/L. Therefore, the fermentation efficiency is: (13.1 / 14.4) x 100% = 91.0%.

18. Biomass Pretreatment High rates of hydrolysis and high yields of fermentable sugars Minimal degradation of the carbohydrate fractions No production of compounds that are inhibitory to microorganisms used in the subsequent fermentation step Inexpensive materials of construction Mild process conditions to reduce capital costs Recycle of chemicals to reduce operating costs Minimal wastes An ideal biomass pretreatment process should meet the following requirements:

19. Concentrated Sulfuric Acid Hydrolysis Dilute Sulfuric Acid Hydrolysis Steam Explosion Ammonia Treatment Lime Treatment Alkaline Peroxide Treatment Concentrated Phosphoric Acid Fractionation Enzyme Hydrolysis

20. Solubilization of cellulose can be carried out at 50°C using concentrated phosphoric acid (>82 percent). The solubilized cellulose and hemicellulose can be separated based on the insolubility of cellulose in water. \The cellulose obtained can be hydrolyzed with enzymes at very high rates. The hemicellulose can be converted to high-value products. Phosphoric acid and acetone can be recovered by simple processes and recycled. Lignin can be recovered and potentially can be used to generate high- value products.

21. BIOGAS

23. Biogas is the CH4/CO2 gaseous mix evolved from digesters, including waste and sewage pits To utilise this gas, the digesters are constructed and controlled to favour methane production and extraction

24. Indian ‘gobar gas’ digester Simple oil drum batch digester

25. Chinese ‘dome’ for small-scale use

26. Accelerated rate farm digester with heating

27. Some organic material, e.g. lignin, and all inorganic inclusions do not digested in the process.

28. Example of calculation: During anaerobic digestion, glucose is transformed into methane through a series of steps. The overall reaction is (CH2O)6 → 3CH4 + 3CO2. Calculate the percentage of methane (by both volume and mass) produced.

29. The temperature ranges (1) psicrophilic, about 20C, (2) mesophilic, about 35C (3) thermophilic, about 55C. In tropical countries, the digestion is psicrophilic, with average temperature btw 20 - 30 C. (retention times at least 14 days). In colder climates, the process is slower, the digesters is heated by using part of the biogas output. Few digesters operate at 55C (unless the purpose is to digest material rather than produce excess biogas).

30. BIOGAS biochemical processes hydrogenesis. Insoluble biodegradable materials, e.g. cellulose, polysaccharides and fats, are broken down to soluble carbohydrates and fatty acids ( about a day at 25 C in an active digester). acidogenesis Acid forming bacteria produce mainly acetic and propionic acid ( takes about one day at 25 C). methanogenesis Methane forming bacteria complete the digestion to a maximum ∼ 70%CH4 and minimum ∼ 30%CO2 with trace amounts of H2 and perhaps H2S ( about 14 -20 days at 25C)

31. The methane forming bacteria are sensitive to pH, so conditions should be mildly acidic (pH 6.6–7.0) but not more acidic than pH 6.2. Nitrogen should be present at 10% by mass of dry input and phosphorus at 2%.

32. Typical parameters for animal waste

33. Example -1 By using the table given, calculate the volume of a biogas digester suitable for the output of 6000 pigs, assuming a retention time of 20 days.

34. Vf is the flow rate of the digester fluid tr is the retention time in the digester ( ∼ 8–20 days). The volume of the digester is The volume of biogas is c is the biogas yield per unit dry mass of whole input (0.24 m3/ kg) m 0 is the mass of dry input The volume of fluid in the digester is is the density of dry matter in the fluid ( ∼ 50 kg/m3). Energy output Hb the heat of combustion per unit volume biogas (20MJ/m 3

35. ANSWER

36. (2) Calculte the power available from the digester of example 1. (assume a burner efficiency is 0.6. Biogas volume : Energy output 1 J = 2.8. 10 -7 kwh