Biodiesel information |
Simplified biodiesel process |
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* Method detailed below is different to the process we recommend for commercial production * |
General information | ||
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Biodiesel process | ||
Biodiesel chemistry | |||
Biodiesel standards | |||
Biodiesel & uk excise | |||
ATEX | |||
Chemical process | Base catalysed transesterification of triglyceride |
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Raw materials | Vegetable oil (other triglycerides can be used) |
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Methanol (or ethanol) |
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Sodium hydroxide (or potassium hydroxide) |
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Process outline |
(1) Coarse filtration of oil and drainage of any water present |
(2) Sample oil and perform titration - determine quantity of catalyst |
(3) Measure the reactants |
(4) Dissolve NaOH into methanol |
(5) Mix the reactants |
(6) Allow glycerol to settle |
(7) Drain glycerol |
(8) * Further processing e.g. washing / drying / additives |
(9) Filtration of biodiesel |
Example Process | Day 1 |
Fill tank with seived waste vegetable oil, warm to >40 °C, and leave to settle. | |||||
Day 2 |
Pour off any water which has settled to the bottom. | ||||||
waste vegetable oil |
Warm vegetable oil to 55°C | ||||||
methanol |
If waste vegetable oil is used pour off a small sample of the oil. | ||||||
sodium hydroxide |
Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) | ||||||
Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) | |||||||
Measure how much sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green | |||||||
To calculate the number of grams of pure sodium hydroxide required per litre of waste vegetable oil : Divide the number of ml solution required to neutralise by 10, and add to 3.5 | |||||||
eg. If 17ml of sodium hydroxide solution was used, amount of pure sodium hydroxide required: | |||||||
= (17 ÷ 10) + 3.5 | |||||||
= 5.2g NaOH per litre waste vegetable oil. | |||||||
(Virgin vegetable oil requires 3.5 g NaOH per litre) | |||||||
The quantity of methanol required is one fifth the volume of waste vegetable oil being used. | |||||||
Safety is paramount. Methanol is highly flammable + toxic, Sodium hydoxide is caustic. | |||||||
Add the calculated quantity of sodium hydroxide to a separate tank containing methanol, and agitate until it has completely dissolved. For example if 100 litres of waste oil was used, 20 litres of methanol will be required. | |||||||
Add this mixture (sodium methoxide) to the tank containing the waste vegetable oil, and agitate for at least an hour until completely mixed. The resultant mixture is biodiesel and glycerol and will settle into two distinct layers...Glycerol is more dense and will settle to the bottom of the tank. Leave to settle. | |||||||
Day 3 |
Pour off the glycerol which is noticably darker and more viscous than biodiesel. | ||||||
Additional processing |
The biodiesel can now be filtered to 5 microns and used as fuel. | ||||||
However, to make biodiesel to meet commercial standards such as ASTM / EN a more complex process must be employed. This involves additional steps such as oil pre-treatement, further purification / washing, drying and often addition of several additives... | |||||||
How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil |
The process outlined above is a single stage process, and is generally only recommended for vegetable oil with a low %FFA. Other processes are better suited to different types of oil | ||
The titration outlined above can also be used to determine the %FFA of the vegetable oil feedstock | ||
%FFA is the number of grams of fatty acid per 100ml of oil. There are a range of different fatty acids and it is not practical to determine the proportions of each in a particular oil sample. For this reason, %FFA is expressed as equivalent oleic acid – an “average” fatty acid. i.e. as if all the free fatty acid is oleic acid. | ||
In this titration, propan-2-ol is used to dissolve the 10ml of sample oil. Universal indicator solution (UI) is used to indicate pH of the mixture. The amount of 0.025M sodium hydroxide solution required to neutralise the dissolved oil is determined (0.025M NaOH is equivalent to 1 gram of pure NaOH crystals dissolved in 1 litre of distilled water) |
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The procedure is carried out as above – i.e. measure exactly 10ml of oil using either a graduated pipette or syringe. Dissolve this into approximately 100ml of propan-2-oil. Add a few drops of UI solution to this mixture so that a colour can be observed. While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. |
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In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) |
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Firstly, determine how many moles of NaOH have been used… |
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Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] |
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Example |
If 10ml of NaOH solution was used | |
Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 |
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Therefore, the equivalent to 0.00025 moles of OA have been neutralised | ||
mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) | ||
mass of OA in 10ml sample = 0.00025 x 282.52 | ||
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....rounding to 2 decimal places, %FFA = 0.71 | ||
Put simply... |
%FFA = number ml NaOH solution used x 0.07063 | |
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