Now I have totally rewrite my post
@Dan_P your free to examine it again - I´m think that I am more near the truth this time
For the ones unfamiliar with formula - PO4 in weight is the weight of the molecule PO4-P is only the weight of the ingoing P in the molecule. The same for NO3/ NO3-N, C2H6O/C2H6O-C and so on. The first is the weight of the whole molecule the -x is the weight of the compound of interest in the molecule.
In order to test my theory that a slow and low addition of ethanol to a GFO reactor can affect the PO4 concentration in the water - start to dose 1 ml 40 % ethanol equal over the hours of the day. (in reality - I dilute 7 ml 40 % Ethanol to 1000 ml with help of RO water and dose 0.1 ml of this solution/minute)
1 ml 40 % ethanol/day correspond to 0.4 ml C2H6O/day. In g its density*ml - in this case around 0.36 g C2H6O. Hence C2H6O-C is around 0.16 g
For a long time now - 2.5 months my PO4 has been around 0.2 ppm (around 0.205 mg/L) As PO4-P is equal to around 0.07 mg/L PO4-P. My Aquarium has around 300 L water - that means an total amount of around 21 mg PO4-P
With the ratio used by
@Beuchat for heterotrophic bacteria biomass - C:N:P = 60:7:1 plus an estimate of how much of the added carbon goes into the biomass (Bio-C) and how much goes out as CO2-C, it is possible to roughly estimate how much PO4-P 1 ml of added 40% C2H6O will lock into bacterial biomass and further calculation of how much it will impact my PO3 concentrations in my aquarium
-Molecular weight C ≈ 12 g/mol
-Molecular weight P ≈ 31 g/mol
Therefore: Ratio C2H6O-C: PO4-P ≈ (60 × 12) / (1 × 31) ≈ 23:1 ≈ 4.4 %
In ecological calculations it is normally assumed that under normal conditions about 10% of the "food input" is transferred to new biomass in warm-blooded animals - the equivalent ratio for cold-blooded animals is thought to be about 20% - a higher ratio because less energy is spent heating the organism. I do not know the equivalent percentage for bacteria, but in these calculations, I use the cold-blooded animals' ratio of 20%. This means that 20% of the C2H6O-C input goes on to bacterial biomass - the rest goes out as CO2-C, so 80% is waste.
In my case I add around 0.16 g C2H6O-C a day for bacterial growth in my reactor. If I now use the ratio for cold-blooded animals - 20% of the "food" input becomes new biomass - then about 0.032 g of C2H6O-C will be bound in new bacterial biomass per day.
Now the ratio 23/1 (≈ 4.4%) for C/P in g (the same as C2H6O-C/PO4-P in g) is used on these 0.032 g/day – this corresponds to an uptake of approximately 0.0014 g PO4-P a day (1.4 mg PO4-P a day)
Molecular weight PO4 ≈ 95 g/mol
-Molecular weight P ≈ 31 g/mol
Therefore: 95/31* 1.4 mg ≈ 4,3 mg PO4 a day
As mentioned before – my aquarium is around 300 l and the calculations above show that addition of 1 ml 40% C2H6O per day can decrease my concentration with around 0,014 mg/L PO4 – in ppm - nearly the same.
It’s a rather good figure – IMO – a little too good but I am following up the experiment with tests. One thing I can add already now is that the dirt I can see in my reactor (in my synthetic filter cotton) is more corresponding to ten days run than the present run for two days - something is happen
I will move my result of my experiment to
my build thread but if any want to discuss this post in this thread - you are welcome.
Sincerely Lasse
