References on Ammonia and Surface Area
If one is only concerned with ammonia oxidation, as little as five square feet of static submerged biomedia surface per pound of fish can be used to produce safe levels of ammonia and nitrite for tropical fish.
A reference is useful here:
“Simply put, if you do a good job of removing the solids from the system with the mechanical filter, the amount of surface area present in the biofilter will determine how much bacteria you can grow, how much you can feed the fish, and, thus, how many pounds of fish the system can hold. A general rule of thumb is that it takes at least 5 square feet of biomedia to handle 1 pound of fish being fed at 2 percent of body weight per day.”
(“Constructing a Simple and Inexpensive Recirculating Aquaculture System (RAS) for Classroom Use”, David Cline, Southern Research Aquaculture Center, Auburn University, 2005).
This article recommends 5 ft2 of effective filter media surface per pound of fish. But this is for tilapia and catfish, which are very tolerant of bad water including moderate levels of ammonia and nitrite. And it is for a submerged static filter. The important point here is that it is the square feet of surface area available which is important in biofiltration.
If a pound of fish is fed at a 2% of body weight per day of dry fish food one needs to calculate the rough amount of ammonia per day this will produce to look at further references:
- One pound is 454 grams.
- A typical feed rate is 2% of weight
- 2% of 454 grams (one pound) is 9 grams
- Fish food is typically 40% protein.
- Protein is typically 17% nitrogen. 9 x 0.4 x 0.17 = 0.612 grams nitrogen per pound of fish per day.
- NH4 to N ratio is 19/15 = 1.266
- 1.266 x 0.612 = 0.775 grams ammonia per pound of fish per day.
The University of Arizona in one of their materials (Biofiltration-Nitrification Design Overview) mentions the average ammonia removal rate on average of 0.5 gram ammonia per square meter per day. Since one square meter is roughly 11 square feet this is 0.775/0.5 = 1.55 pounds of fish per 11 square feet. 11/1.55 = 7 square feet per pound of fish per day for ONLY ammonia removal. This is quite close to the 5 square feet figure in the first reference.
Summerfelt, 2006 (“Design and Management of Conventional Fluidized-Sand Biofilters”) achieved an average removal of 0.3 grams ammonia per square meter. This is roughly 0.775/.3 = 2.58, 11/2.58 = 4 square feet per pound of fish per day for ONLY ammonia removal. Again, this is quite close to the 5 square feet figure in the first reference.
The Food and Agriculture Organization of the United Nations in its booklet available on the website www.fao.org mentions the ammonia removal rate varying between 0.2 and 2 grams per square meter of specific surface. 0.775/0.2 = 3.875, 11/3.875 = 2.8 to 28 square feet per pound of fish per day for ONLY ammonia removal. If 2.8 to 28 is considered a binomial distribution the mean will be very close to the 5 square feet figure in the first reference.
All these materials are for aquaculture operations. So one might want a reference that talks about aquariums. Grommen (“An Improved Nitrifying Enrichment to Remove Ammonium and Nitrite from Freshwater Aquaria Systems”) shows in his study of aquariums an average removal of 0.47 grams of ammonia per square meter of media in aquariums. 0.775/0.47 = 1.65, 11/1.65 = 6.7 square feet per pound of fish per day for ONLY ammonia removal. This is quite close to the 5 square feet figure in the first reference.
This makes 6 references which come in at about very roughly the 5 feet per pound figure. So Ammonia oxidation is easy! We will use the following “rule”:
One Pound Of Fish Needs 5 Square Feet of Biomedia Surface Area for Ammonia Oxidation
This is with a caveat. This is ONLY a submerged bed figure. A fluidized bed might need only two square feet while a trickle filter might need ten square feet.
Another reference gives 1.7 square foot of biomedia surface in a fluidized bed required for ammonia oxidation for one pound of fish (“Design and Operation of Moving Bed Biofilm Reactor Plants for Very Low Effluent Nitrogen and Phosphorus Concentrations”, Rusten, 2007). The difference between 1.7 and 5 reflects the increased efficiency of a fluidized bed.
Crystal Clear Water
But achieving crystal clear water is a different story than ammonia and nitrite. One pound of fish needs roughly 100 square foot of biomedia surface area to produce fairly clear water. This is a very important rule. It is paramount in fishkeeping.
This amount of biomedia gives no nitrites, no ammonia, good water clarity and good fish health. It is twenty times the level required for good ammonia oxidation in a submerged stationary media filter like a canister filter.
There is a very rough geometric progression when it comes to water parameters and surface area in a filter. This can easily be put in a table form for static submerged media:
|Square feet of
pound of fish
for 300 ppi
|0 to 0.5
Note that the 4-20-100-500 progression here is called a geometric relationship (multiply the preceding number by five) and is quite common in natural systems.
- All this is predicated on cleaning the biomedia only lightly when the flow noticeably slows down, cleaning the biomedia removes the beneficial bacteria which have slowly built up.
- This is for a canister filter (i.e. submerged static media). A fluidized bed filter is two to five times more efficient while a trickle filter is half to one fourth as efficient.
Note this surface area rule is supported in the literature (“Understanding Biological Surface Area in Aquaponics”, Michael, 2016).
“Remember: as an absolute minimum, your system needs at least: 2.5 ft2 of biological surface area/gallon of water (at low stocking densities and low feeding rates). For a healthier system, we would recommend: 10 ft2/gallon of water OR 100 square feet per pound of fish.”
We need to emphasize that this analysis is extremely simplistic. There are a lot of variables that affect how well any biofiltration media performs. Some of those variables are:
1, Flow: There is a surprisingly small relationship with the flow rate. The faster the flow rate over the media the more it can help the fish. But this flow rate relationship is not as large as common mythology suggests. Doubling the flow rate only raises the filter rate by 9%. This number came from actual testing.
2, Opening size: the larger the opening in the media the longer the media can go without clogging up. Conversely, the larger the opening the lower the surface area. So this is a somewhat two-edged sword that seems to be very roughly maximized at about one-tenth to one-eighth of an inch (2.5 to 3 millimeters).
3, Stocking: the heavier the stocking the faster the media will clog up. What is often missed is that beneficial nitrifying bacteria can be out-competed for space by non-beneficial bacteria. The heavier the stocking the more this will occur. This will reduce the media efficiency even when unclogged.
4, Amount of food: this is like the stocking variable. If a given stocking of fish is fed at a three percent of body weight per day level the media will be much less effective and clog up faster than if the feeding is at the one percent level.
5, Maturity: The longer filter media have been in place without cleaning the more “mature” the filter media and the better it will perform. This will be true until the flow slows down due to clogging.
6, Food protein levels: this interesting variable says that high-protein food is much better for the filter than a low-protein food. A 50% protein level food will give two and a half times the efficiency and the time to clog than will a 30% protein food. This desirability of high-protein food has to do with biomass buildup rate and the carbon-to-nitrogen ratio of the food. Biomass will buildup much faster and have less exposed beneficial bacteria with a low-protein food.
7, Aeration: the better the aeration the better the media will perform. The aeration in an aquarium provides both oxygen and carbon dioxide to the beneficial bacteria. Good aeration aids both in the breakdown of ammonia and in the breakdown of dissolved organic compounds and lessens the clogging.
8, Free volume: the greater the free volume the more space is available for brown gunk to form and do biofiltration
9, Temperature, pH, KH, GH and phosphates in the water all affect it in very unpredictable ways
10, Ecology: And the “ecology” of the tank will vary, often in very surprising ways. The tiny single-celled protozoan community is surprisingly important to the function of an aquarium filter and the composition of this community varies in the extreme from tank to tank.
11, pH: Biofiltration capacity is somewhat dependent on pH. If you have an acidic aquarium of pH of 6.5 triple the number. If your pH is 6.0 multiply the number by 10. Acid conditions need much larger surface areas than a 7.0 to 8.4 pH.
All these things are cumulative and can rarely make a 10 square feet per pound filter give crystal clear water. By the same token they can rarely make a 500 square feet per pound filter give cloudy water.
This somewhat unpredictable variation is why I put a downright ridiculous amounts of filtration on all my tanks. My big tanks all have large K1 fluidized bed sumps and multiple undergravel filters on them. Some of the smaller tanks have FX6 canisters on them. I almost NEVER clean a filter. And the water is crystal clear and the fish are very healthy despite being very heavily stocked.
Illustrations on How to Apply the Surface Area Numbers
Here is a comment to this web page and its answer:
I have a 75-gallon community tank with an EHEIM 2217 canister filter. I’m planning to replace all the EHEIM filter media with Swiss Tropical 20 PPI Poret foam. I’m thinking of having 5 foam pads, each 2” thick and 7” in diameter. Would this give me the 220 sq. ft of effective surface area mentioned in your chart? Using one of your other charts, I calculate that I have 0.75 lbs of fish. Using the 100 sq. ft for 1 lb. of fish metric, I would think I’ll be able to easily accommodate more fish – is that right? I’m planning to use your recommendation to squeeze liquid from the existing filter pads onto the new pads to jump start the cycling of the new foam. How long do you think I should wait before adding more fish? Thanks for your help.
OK you have a cylinder ten inches tall and 7 inches in diameter. V = π r2 h = π x 3.5 x 3.5 x 10 = 384 This will give one 384 cubic inches of 20 ppi foam. 12x12x12 is 1,728 So you will have 384/1,728 or 0.222 cubic feet. 0.222 x 220 = 49 square feet. So you can do 0.49 pounds of fish at the 100 square foot per pound of fish metric. At the current time you are at (0.49×100)/0.75 = 65 square feet per pound of fish. 65 square feet is good, but it isn’t 100 square feet. So you cannot accommodate more fish.”
And you can add fish right away and just not feed them for a few weeks or wait literally as long as you are comfortable with. I add pond mud (see the article on the “mature aquarium”) and wait at least six weeks minimum but I’m a little nuts.
Another Example of Aquarium Filter Media Calculations
This exchange is from the comments section of this website:
I’m reaching out for some clarity on ‘Square Feet of Biomedia Surface Area’. I’m considering the 5″ Poret® Foam Sheets (26 × 19.5″ Sheet) and am wondering what the calculations would be for cubic sq. ft.? I plan on executing the ‘10 ft2/gallon of water’ OR ‘100 ft2/pound of fish.’”
“To calculate the volume the math is 5 inch x 26 inch x 19.5 inch / 12 inch x 12 inch x 12inch = 1.47 cubic feet per sheet of Poret® Foam. With a 20 ppi Poret® foam (what I recommend) which has about 220 square feet of effective surface area per cubic foot, this is 1.47 x 220 = 323 square feet. So If you use the whole sheet of Poret® you can support roughly 3.23 pounds of fish in a healthy environment (100 square feet of surface area per pound of fish). And I don’t recommend “10 ft2/gallon of water”. Go ONLY with the “100 ft2/pound of fish”.
“Thank you so much for the reply but I’m still confused. Essentially I’m wondering how many – 5″ deep x 26″ long x 19.5″ high Poret® Foam Sheets, will I need for 13.2 lbs. of fish (24 × 8.8 oz Discus or 24 x 8.8/16)?”
“The calculation for 13.2 pounds of fish is 13.2 pounds divided by 3.23 pounds equals four sheets of 5″ deep x 26″ long x 19.5″ high Poret® Foam”
These examples illustrate the process very well.
If one is doing the calculations for a sump one should be aware that fluidized bed K1 is roughly 58% more efficient than static 30 ppi foam as far as filtration. The effective surface area considerations are that fluidized K1 is 900 square feet per cubic foot of media. At 60% loading this becomes 540 square feet per cubic foot of sump filter.
Foam with 30 pores per inch has a surface area of 340 square feet per cubic foot. 540/340= 1.58
Still another illustration:
How do you calculate the surface area in media used for biological filtration? For example, I have 30 pot scrubbers and some foam in a Fluval 407 canister filter on a 75 gallon tank.
OK Each pot scrubber is roughly nine cubic inches of media (large pot scrubbers???). (9×30)/(12x12x12) = 0.156 cubic feet. Pot scrubbers are 280 square feet per cubic foot media. 0.156 x 280 = 44 square feet from the pot scrubbers. If I remember right a 407 has a piece of 30 ppi foam that is 10 inches x 1 inch x 14 inches. (10x1x12)/(12x12x12) = 0.081 cubic feet. 30 ppi Foam is about 340 square feet per cubic feet 0.081×340 = 27.5 square feet. 27.5 + 44 = 71.5 square feet. That is just off the top of my head. You need to correct the numbers depending on the size of your pot scrubbers and the true size of the 407 foam.
For more on how to calculate filter volume go to these articles:
Source: Aquariumscience.org – David Bogert