vahegan
02-02-2012, 10:28 AM
Scrubbers are known to grow the algae actively, but it is mainly plant mass. I was curious to try to find out how much of this biomass are actual waste products exported from the tank, and how much is just carbon from the atmosphere. By knowing this we could scientifically estimate the real efficiency of a scrubber.
As a starting point I am going to use the Redfield ratio. American oceanographer Alfred C. Redfield empirically determined and described in an article written in 1934 that carbon, nitrogen and phosphorus in plankton are present in the ratio: 106 : 16 : 1. Which means that in planctonic mass to 1 phosphorus atom there are 16 nitrogen and 106 carbon atoms.
For simplicity lets consider that a plant cell mainly consists of cellulose, nitrate and phosphate. Of course many other elements are present, but these are the bulk substance.
Cellulose С6(H2O)5 has a molecular weight of 6*12+5*18=162
Nitrate ion NO3- has a weight of 14+3*16=62,
and phosphate ion PO4--- has a weight of 31+16*4=95
Thus, based on the Redfield ratio, for 106*162+16*62+1*95=3949 mg of dry biomass grown by the scrubber we are exporting 16*62=992mg of nitrate and 95mg of phosphate. Since these figures are just for an estimate and can never be quite accurate, we can say roughly that
4 grams of grown dry substance would export 1000mg of nitrates and 100mg of phosphate.
Suppose we have a 100 liter (about 25 gallon) fish tank with 25ppm nitrates. This comes to 2.5 gram of nitrate ion in the system. To remove this completely the scrubber will have to grow about 2.5*4=10g of dry algae substance (at the same time it will consume about 250mg of phosphate which for the considered system is equivalent to 2.5ppm.) This sounds reasonable to me (providing that the real wet mass of algae will be probably 100 times heavier compared with the dry mass obtained through calculations above.
10 grams of dry plant substance is a lot and, having such growth in the long run, I am sure that the scrubber will soon deplete the tank of any iron and may even affect the potassium level (although it is relatively abundant in marine water). I have never seen this discussed, but it might be a good idea to start dosing iron (and maybe other microelements too, like the Plantex CSM mix which is well known to planted tank owners) to promote the growth of plant mass.
Guys, tell me your opinions on the above speculations, please. Also, did anyone ever weigh the mass of turf grown by a scrubber, both wet and dry: how much approximately is grown by each sq. inch of scrubber screen in a week?
Also, how does a scrubber behave if the proportion of available nitrates and phosphates is different from roughly 10:1. For example, if the ratio is 10ppm nitrate and 5ppm phosphate, the scrubber should "eat" 10ppm nitrates with 1ppm phosphate and the remaining 4ppm phosphates will stay in water since the nitrate will be a limiting factor to the growth of algae? Does it make sense to slowly dose some nitrate in the system (40ppm in all) to promote algae growth until the accumulated phosphate is fully consumed?
As a starting point I am going to use the Redfield ratio. American oceanographer Alfred C. Redfield empirically determined and described in an article written in 1934 that carbon, nitrogen and phosphorus in plankton are present in the ratio: 106 : 16 : 1. Which means that in planctonic mass to 1 phosphorus atom there are 16 nitrogen and 106 carbon atoms.
For simplicity lets consider that a plant cell mainly consists of cellulose, nitrate and phosphate. Of course many other elements are present, but these are the bulk substance.
Cellulose С6(H2O)5 has a molecular weight of 6*12+5*18=162
Nitrate ion NO3- has a weight of 14+3*16=62,
and phosphate ion PO4--- has a weight of 31+16*4=95
Thus, based on the Redfield ratio, for 106*162+16*62+1*95=3949 mg of dry biomass grown by the scrubber we are exporting 16*62=992mg of nitrate and 95mg of phosphate. Since these figures are just for an estimate and can never be quite accurate, we can say roughly that
4 grams of grown dry substance would export 1000mg of nitrates and 100mg of phosphate.
Suppose we have a 100 liter (about 25 gallon) fish tank with 25ppm nitrates. This comes to 2.5 gram of nitrate ion in the system. To remove this completely the scrubber will have to grow about 2.5*4=10g of dry algae substance (at the same time it will consume about 250mg of phosphate which for the considered system is equivalent to 2.5ppm.) This sounds reasonable to me (providing that the real wet mass of algae will be probably 100 times heavier compared with the dry mass obtained through calculations above.
10 grams of dry plant substance is a lot and, having such growth in the long run, I am sure that the scrubber will soon deplete the tank of any iron and may even affect the potassium level (although it is relatively abundant in marine water). I have never seen this discussed, but it might be a good idea to start dosing iron (and maybe other microelements too, like the Plantex CSM mix which is well known to planted tank owners) to promote the growth of plant mass.
Guys, tell me your opinions on the above speculations, please. Also, did anyone ever weigh the mass of turf grown by a scrubber, both wet and dry: how much approximately is grown by each sq. inch of scrubber screen in a week?
Also, how does a scrubber behave if the proportion of available nitrates and phosphates is different from roughly 10:1. For example, if the ratio is 10ppm nitrate and 5ppm phosphate, the scrubber should "eat" 10ppm nitrates with 1ppm phosphate and the remaining 4ppm phosphates will stay in water since the nitrate will be a limiting factor to the growth of algae? Does it make sense to slowly dose some nitrate in the system (40ppm in all) to promote algae growth until the accumulated phosphate is fully consumed?