Please can anyone help me with tips on how to determine the mg/l content of Al in a solid substrate/feedstock (mass of 84g) for anaerobic digestion if the elemental analysis of a sample of the substrate had 145 mg/kg Al concentration.
The elemental analysis was carried out on a sample of the substrate and the result showed the substrate contains 145 mg/kg Al.
Now, I want to feed my reactor with 84g of this substrate and would like to determine the composition of Al in mg/l contained in this mass of substrate.
I will appreciate your response. Thanks.
Place container on a digital balance and tare the weight to zero. Then fill with 1 L of the substrate (ease of calculation) and take the mass reading in grams which is Y grams.
To calculate Al mass in mg in 84 g use the following conversion: 145 x Y/1000/84 mg/L.
Why? I need you to respond with scientific calculations. Simply saying wrong is not the correct approach to technical/scientific discussions.
@Selva Selvarajah.... Do you still want to modify your post? ...do it now.
And exactly.... "Why"...I was thinking the same .. why you answered the question, if you didn't understand the qurstion and might don't know the basics of anaerobic digestion.
I recommend you to read about TS/VS and the method for element analysis in substrate and its unit.
The question is nothing to do with anaerobic digestion. The question applies to any liquid or semi-liquid item, whose density is unknown. In the question the Al concentration has been assessed on the basis of mg/Kg and how this can be converted into mg/L?
Do you understand the workout in my calculation?
The elements are calculated as mg/kg dry weight of substrate. One cannot use the unitary method to find the elements, unless we know about the TS of the substrate.
@Selva Selvarajah
* Does solid means there is no water/moisture in substrate?
* How can such substrate be measured in Liter as you recommended?
*How can this substrate be used in AD if it is 100% solids without any traces of water/moisture?
* Please apply your unitary method and give a hypothetical calculation, also please try to justify the use of such substrate in AD.
Arinze should clarify whether Al was reported in a ww or dw basis. My assumption was Al concentration was reported in ww basis.
@Selva Selvarajah
You can still answer the above questions, they are now not related to the main question which Arinze asked.
2 out of 4 questions are basic physics question, 1 can be mathematics.
1. In the case of substrates, solid means solid state which may contain >15% SS
2. Sludge with the above consistency can be volumetrically measured.
3. 100% solids (i.e. dried) are not used for seeding anaerobic digesters.
4. The fundamental use of substrate is to provide sufficient seeding/microbes to accelerate the anaerobic digestion, process.
As for calculation, if Al was reported on a ww basis,
1 L of measured substrate = Y g
Therefore, the volume of 84 g substrate in L = 1 L x 84 g/Y g
Al mass in mg in 84 g substrate = 145 mg x 84 g/1000 g
Therefore Al concentration in mg/L = Al mass in mg in 84 g substrate ÷ volume of 84 g substrate in L
= 145 mg x 84 g/1000g ÷ 1 L x 84 g/Y g =145 mg x Y/1000/L
In my original answer I provided Al mass in 84 g substrate in mg/L which was wrong because the Al mass in 84 g is 12.18 mg (145 mg x 84 g/1000 g).
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