Bioenergy
• Bioethanol
• Biogas
• Bio diesel ???
• Strategy in Perlis
BIOETHANOL, BIOGAS and
BIOHYDROGEN
10 May 2013
BIO ETHANOL
3
Raw materials Fermentation
Fermentation
product
separation
Microorganisms : Saccharomyces cerevisiae
Zymomonas mobilis
Raw material : Glucose containing materials
Process : Aerobic and anaerobic
ehanol
5
Saccharomyces cerevisiae
Well studied, widely used commercially
It growth is inhibited by ethanol (its own product)
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

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Raw materials (feedstocks) :
Sugar Feedstocks
7
Starch Feedstocks
:
8
Process Diagram
9
Calculation
Prove that : 1 kg glucose will produce 0.511 kg
ethanol
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Extended question :Ethanol from starch
n is the number of glucose residues in the
starch
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 :
The conversion factor approaches 1.111.
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.
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Exampl
e
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.
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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 production by starch hydrolysis: 210 g x 1.111 = 233.3 g
Starch + water --------------->> glucose
210
233.3
Means that water consumption in starch hydrolysis:233.3 – 210 =23.3 gr or
23,3 ml
Theoretical ethanol production: 233.3 x 0.511 = 119.0 g
Volume of the ethanol produced: 119.0 g / 0.79 (g/mL) = 150.8 mL
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%.
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BIOGAS
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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
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Simple oil drum batch digester
Indian ‘gobar gas’ digester
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Chinese ‘dome’ for small-scale use
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Accelerated rate farm digester with heating
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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.
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BIODIESEL
April 2013
Methods of Biodiesel Production
Raw material:
waste vegetable oil (WVO) or
straight vegetable oil (SVO).
WVO is attractive to small producers , but large-scale
plants may have difficulties with feedstock supplies because
soap manufacturers use a lot of waste oils.
Straight vegetable oil (SVO)
Important Specifications :
a. Cetane number.
b. Viscosity.
d. Freezing point / Cloud point.
Mixed triglycerides do not have a clear freezing point—some componen
Cetane number
(C16H34), also called hexadecane,
Cetane number is a measure of how quickly a fuel
will ignite in a compression ignition engine.
Cetane is very quickly ignited, and its number is set to 100.
The cetane number of the fuel should be above 41.
Most vegetable oils have an adequate cetane number,
but, if necessary, additives can be mixed with the oil.
Small amounts of acetone have been used.
Straight vegetable oils (SVO):
- Too viscous → can damage the fuel injection system of
a normal diesel engine.
Viscosity At 40C:
Petrodiesel 4 to 5 mm2/s,
vegetable oils 30 and 40 mm2/s.
What to do ?????
Solutions :
a. Blend with Diesel oil or kerosene.
b. Preheat (65 C)
c. Transesterification --> biodiesel
Transesterification
Transesterification of triglycerides with alcohol
Methanol or ethanol ?
For biodiesel, methanol is currently popular,
But ethanol is the better choice because it is
produced from biomass
Methanol is usually obtained from petroleum
although it can be produced from biomass.
Methanol is toxic
You have 1000 kg of a vegetable oil that happens
to consist of a single triglyceride
(not a mixture of triglycerides).
The three acids in each molecule
are all palmitic acid (C16:0).
What is the proper chemical name of this triglyceride?
What is the molecular mass of the triglyceride?
Estimate the number of kilograms of glycerine produced
when the above vegetable oil is transesterified with ethanol.
C16H32O2
Molar mass 256.42 g/mol
This is a triple palmitate of glycerine hence it is
glyceryl tri-palmitate.
Molecular formula C16H32O2
Molar mass 256
When make a bond, loss 1 H → 255
Tri → 255 x 3 = 765
CH2-CH-CH2
= 41
Total = 806
Glycerol C 3H5(OH)3
Molar mass
92.09382 g/mol
Each mole of triglyceride releases 1 mole of glycerine upon being
esterified.
1000 kg of oil correspond to 1000/806 = 1.24 kilomoles.
Hence
glycerine produced : 1.24 kilomoles
Weight of glycerine produced : 1.24 × 92 = 114 kg.
Example 2:
To produce biodiesel, ethyl or methyl esters from
rapeseed oil, ethanol or methanol, and
potassium hydroxide are used in the reaction:
a. Determine the theoretical conversion
efficiency of the biodiesel reaction
b. Determine the amount of ethanol required
to convert 1 L of rapeseed oil reacted with
1 percent potassium hydroxide.
Base your calculations on triolein since this is often
the predominant triglyceride in
vegetable-based oils: C57H98O6,
Oleic acid
18:1 cis-9
C18H34O2
C57H98O6, MW 878 g/gmol, density 0.88 kg/L
Ethanol: C2H5OH, MW: 46 g/mol, density 0.794 kg/L
Methanol: CH3OH, MW: 32 g/mol, density: 0.796 kg/L
Potassium hydroxide in alcohol (1 wt %/vol):
Fuel
Density (kg/L Energy (MJ/
)
L)
Biodiesel
0.88
36.6
Methanol
0.79
22.7
CH2-CH-CH2 = 41
3 Oxygen = 3 X 16 = 48
3 Carbonyl (CO) = 3 x 28 = 84
3 oleic = 3 [( 17x12 )+(33 x 1)] = 3[ 237]= 711
Total = 711+84+48+41 = 884
Molecular formula
C57H104O6
57*12+104+6*16 = 884
3H5(OOCR)3 + 3 C2H5OH → 3 RCOOC2H5 + C3H5(OH)
R= C17H33
237 + 12+32+24+5
36+5+(16+16+12+237)3
884
-->237
46
1000 gram
= 1.13 mol
36+5+3(17)
310
92
3 x 1.13
=3.39 mol
=1050 gram
a. Conversion efficiency : 105 %
b. The weight of 1 litre oil = volume x density , ------>0.88 kg
= 0.995 mol ------> ethanol = 3 x 0.99 mol
= 3 mol
= 3 x 46 g
= 138 gra
Every 1 litre ( or 880 gram) oil needs 138 gram ethanol.
Every 1 gram of ethanol will convert 880/138 = 6.4 gram 0f oil
b. Molar ratio for alcohol to triglyceride molecule is 3 to 1;
→ 3 mol EtOH × 46 g/mol = 138 g of ethanol per
gmol of triglyceride are needed.
1 gmol of triolean × 878 g/gmol = 878 g triolean;
EtOH needed for compl conversion = 878 g triolean/138 g EtOH
Fuel
Density (kg/L Energy (MJ/
)
L)
Biodiesel
0.88
36.6
Methanol
0.79
22.7
If methanol is used instead of ethanol :
a. Stoichiometrically, how many kg of methanol are required for ea
b. If the methanol, were used directly as fuel, how much energy w
c. If 1 liter of biodiesel is used as fuel, how much energy is release
d. What is the methanol-to-biodiesel energy ratio?
C3H5(OOCR)3 + 3 CH3OH → 3 RCOOCH3 + C3H5(OH)3
a. 1 litre oil = 0.88 kg
= 0.995 mol ------> Methanol = 3 x 0.99 mol
= 3 mol
= 3 x 32 g
= 96 gram
Every 1 litre ( or 880 gram) oil needs 96 gram Methanol.
Every 1 gram of Methanol will convert 880/96 = 9.16 gram 0f oil
(compare with that of ethanol)
b. ref to table in question, methanol energy = 22.7 MJ / L
Methanol density = 0.8 kg/L.
Every 1 L methanol will produce 22.7 MJ energy
Or, every 0.8 kg ( 800 gr) methanol will produce 22.7 MJ energy
For every 96 gr methanol – > 96/800 x 22.7
= 2.72 MJ
C3H5(OOCR)3 + 3 CH3OH → 3 RCOOCH3 + C3H5(OH)3
Fuel
Density (kg/L Energy (MJ/
)
L)
Biodiesel
0.88
36.6
Methanol
0.79
22.7
c.
Mw biodiesel = 237+12+32+15 = 296
1 mol oil will produce 3 mol biodesel. 3 mol biodesel= 3 x 296 gr
= 888 gr
ref to table , biodiesel energy = 36.6 MJ / L
Biodiesel density = 0.88 kg/L.
Every 1 L methanol will produce 36.6 MJ energy
Or, every 0.88 kg ( 880 gr) methanol will produce 36.6 MJ energy
For every 888 gr biodiesel – > 888/880 x 36.6
= 37 MJ
What is the main reason to prefer biodiesel to
straight vegetable oils (SVO) as fuel for diesel engines?
What should you do to make SVOs more acceptable as
diesel fuel (other than making biodiesel out of them)?