ERT353 2014
Biofuel
BIO ETHANOL
What, Why, How, How much, ….
Biofuel : General term for fuel from bio resources
What?
Ethanol C2H5OH
Chemical production : C2H4 + H2O → CH3CH2OH
Bioethanol : Ethanol that produced through fermentation
Fermentation :?
Why?
Properties are similar to fossil fuel
Flash Point : 9 C
Boiling Point : 78 C
Bioethanol
Fermentation
Glucose ----------------yeast---------- ethanol + carbondioxide
Material Balance ?
Mol equivalent
Detail of fermentation
Microorganisms :
Saccharomyces cerevisiae (yeast)
Zymomonas mobilis
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.
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

Raw materials (feedstocks)
Sugar
Sugar
•Sugar containing materials :
•Starch
•Cellulose
Typical Parameters of Batch Fermentation Process for
Ethanol Production from Sugarcane Juice and Molasses
Example
Starch Feedstocks :
Fermentation
CONVERSION RATIO FROM STARCH OR CELLULOSE INTO GLUCOSE
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.
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
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.
Example
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.
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%.
Biomass Pretreatment
An ideal biomass pretreatment process should
meet the following requirements:
• 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
Concentrated Sulfuric Acid Hydrolysis
Dilute Sulfuric Acid Hydrolysis
Steam Explosion
Ammonia Treatment
Lime Treatment
Alkaline Peroxide Treatment
Concentrated Phosphoric Acid Fractionation
Enzyme Hydrolysis
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 highvalue products.

BIOGAS
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
Simple oil drum batch digester
Indian ‘gobar gas’ digester
Chinese ‘dome’ for small-scale use
Accelerated rate farm digester with heating
Some organic material, e.g. lignin, and all inorganic inclusions do not
digested in the process.
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.
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).
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)
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%.
Typical parameters for animal waste
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.
The volume of the digester is
Vf is the flow rate of the digester fluid
tr is the retention time in the digester (∼8–20 days).
The volume of biogas is
c is the biogas yield per unit dry mass of whole input (0.24 m3/ kg)
m0 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/m3
ANSWER
(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