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Sizing Your Filters
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Turnover and Filtration Rate
We hear a lot about how big a filter to use, and a lot of people talk about filtering the volume of your tank four times an hour or so as a general rule. Well, while this isn't a bad rule, per se, it doesn't really tell us what's going on. If we have a 60 gallon tank, and a 240 GPH filter, we really aren't filtering the water four times an hour. Not even close. Because the water coming out of the filter mixes with unfiltered water, we are never as efficient as a turnover rate would imply.
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Say you have a 100 gallon tank with a filter that pumps 20 gallons per minute. The first minute, 20 Gallons is filtered, and the remaining water in the tank has 80 unfiltered gallons. The second minute, of the 20 gallons that you intake, only 16 gallons have not yet been filtered. Because of mixing, you re-filter four gallons. As time passes, more and more of the water you take in has already been filtered, so you never achieve 100% filtration, but you get darn close. After 31 minutes, only 0.0248 gallons of intake have been unfiltered, yielding 99.901% of total volume that has passed through the filter.
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99.9% filtration time = 6.2 x Tank Volume / Filter Flow Rate
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where filtration time is in minutes,
Tank Volume is in Gallons, and,
Filter Flow Rate is in Gallons per Minute
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This gives us a much clearer picture of what's actually going on in the tank, and while we can talk about turnover rates, we should never assume that this turnover rate has anything to do with how long it takes to actually filter the water.
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The following table shows the coefficient to use in achieving a desired filtration percentage:
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Filtration Percentage..|..Coefficient
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90%............................|..2,2
99%............................|..4,2
99,9%.........................|..6,2
99,99%.......................|..8,4
99,999%.....................|.10,4
99,9999%...................|.12,4
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Determining Media Requirements
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Mechanical:
Mechanical media, the stuff that filters out suspended solids in the water, is difficult to size mathematically. In determining how much mechanical filtration is required, we have to take into account the expected load of suspended solids, and the design of the mechanical filter, and how often we want to service the filter. Errors in sizing will simply require more frequent maintenance, and will rarely impact water clarity unless maintenance is not performed when required. The following relationships are apparent in this relationship between design characteristics and maintenance requirements. All of these factors will increase the frequency of maintenance:
Smaller average pore size of the media (media clogs more quickly)
Higher load of suspended solids (more matter available to clog the media)
Smaller surface area of the media (less surface area doing more work)
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However, not all mechanical filtration designs are as effective in removing particulates. A system that has an average pore size twice the size of the avarage suspended particle size will certainly not capture all suspended solids in one pass, and that one-pass efficiency is important to us. If we have small pore sizes, and good media depth, we have a filter that has a greater chance of capturing the particles in the first pass, and coming closer to 100% efficiency. To recap, these factors increase our filter efficiency:
Smaller pore size (captures more particles)
Greater media depth (provides more opportunities for capture)
Lower flow velocities (prevents dislodgement of captured particles)
Lower pressure (prevents dislodgement)
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We have to be somewhat cautious about our design of the mechanical filter, however, because as we increase the ability of the media to capture particles we restrict water flow through the media and increase our system head losses. The greatest effect we can have on reducing the head losses of our mechanical filter are as follows:
increasing surface area
increasing media permeability
decreasing media depth
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As much as I wanted to give you a formula for determining optimal mechanical filtration criteia, at this time there are really useful available formulas. It is possible to use a Carmen-Kozeny equation to determine the pressure drop across the media, but this depends on determining a resistance coefficient for the media and a specific surface area of the media particles, and many of the media types in use in aquariums and ponds do not have a readily determined resistance coeffficent. Many of the media are also not particulate, but screens, which would require a different formula anyways.
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Biological:
Biological media is simply a host for nitrifying bacteia. The more we can effect contact between water and biological media, the more efficient our biological filter will be. The factors important for our design are:
The load of nitrogenous wastes
The surface area of the media
The flow rate of water through the media
The availablility of oxygen in the media
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Note that we do not attempt to quantify the population of bacteria involved. First, it is nearly impossible to do, and second it is not important that we do, since bacteria will readily populate the media as long as oxygen and nitrogenous wastes are available. As the quantity of these inputs change, so will the population of bacteria, and this population will constantly be in flux with the input quantities.
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Calculating the load of nitrogenous wastes is regularly done in aquaculture. Our sources for this come from the food for the fish, and from decay of any plants or dead fish. Since only the first source, food, is what we are designing for (unless the goal of your system is to demonstrate wastewater treatment methods), we will restrict our calculations to the amount of food we intend to feed the fish. This can be calculated either by expecting to feed an amount equal to 3-5% of the fish body mass per day (for those of us planning to have so many fish we measure them in pounds), or by simply weighing the amount of food we intend to feed per day.
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Once we have this, we can look at the food we will be feeding and can usually find the protien content of the food. TetraMin, a coomon staple food, contains 45% protien, and while some of this will be metabolized into increased fish mass, for our worst-case scenario we will assume that all of this ends up entering the water column as either decayed uneaten food, or as urea from the fish, both of which are metabolized into ammonia and ammonium ions.
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This must be broken down by the bacteria, and empirical data from T. B. Lawson of Louisiana State University suggests that a biological media surface area of one square meter will host enough bacteia to process one gram of ammonia into nitrite, and then into nitrate. Thus, the following equation will give you the required biological media surface area:
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Equation 1: Nitrogen loading
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where
F = Food mass in GRAMS
P = Ratio of protien content in food
Tfw = Total Fish mass in GRAMS
A = biological filter media surface area in SQUARE METERS
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Another treatment of this problem was referenced by Spotte, focusing on an oxygen demand budget. Although designed for undergravel filtration, it gives an idea of what is possible in other scenarios:
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Equation 2: Hirayama Equation for Oxygen Demand
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where
W = media top surface area (m2)
V = filter rate (cm/min)
G = mean grain size (mm)
p = number of filters
D = media depth
B = fish biomass (g)
F = food mass (g/day)
q = number of fish
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This inequality demonstrates the oxidizing capacity of a filter bed on the left side, and the oxygen demand of the filter bed on the right hand side of the equation. This gives us pollution load estimates for fish, and an indication of oxygen demand for a filter. Although this appears to be a very detaile equation, Spotte cautions that this should be used as a guideline only for determining fish loads, and not be taken entirely literally.
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Spotte does recomend a minimum flow rate of 0.07 x 10^-3 meters per second through the media in order to maintain aerobic conditions in the biofilter and prevent anoxic conditions. In english units this is 2.3 x 10^-4 feet per second
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Chemical:
Chemical filtration, when employed, generally requires the use of granular activated carbon (GAC) or ion exchange resins, such as zeolite clay. Spotte reccomends the use of 1 gram of GAC per liter of water, to be replaced every two months (Spotte, 1979)
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If using ion exchange resins such as zeolite (used in Ammo-Carb and other "ammonia removers"), Spotte makes the following observations about use and efficiency:
Zeolite saturation is 8 m-mole NH4-N per 100 grams of zeolite.
Interionic competition is a problem when removing ammonia in hard water. A salinity of 0.5% reduced adsorbtion by 90% in one study.
Sulfate and Chloride ions are strong inhibitors of ammonia uptake by Zeolite.
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Sterilization:
I do not reccomend ozone for in-home use. UV sterilization is the preferred method of sterilization.
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When sizing a UV system, it is important to gauge the flowrate through the sterilizer to be slow enough to deliver a fatal UV dosage to the largest organism you are trying to kill. In most instances, that organism will be the parasite ichthyophthirius, which will require a dosage of between 100,000 and 336,000 microwatts per second per centimeter squared.
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The filtration rate, as opposed to the turnover rate, is particularly important for sterilization. An ichthyophthirius spore can release up to 1000 free-swimming tomites which can survive up to 48 hours before finding another host. The goal in sterilization is to mathematically reduce the probability that any of these tomites may survive long enough to find another host to as close to zero as possible. Unfortunately there is little in the way of data available that can show us what the relevant parameters in this equation are, but we can make some estimates to form the basis of our model, at least as a starting point.
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Assuming that the average time it takes a tomite to infect a host is four hours, our goal with sterilization is to remove 1000 out of 1000 tomites from a tank within four hours -- a sterilization rate of 99.99%. This rate can be achieved provided that the sterilizer flow rate is at least 8.4 x Volume / 240, and that the UV bulb wattage is sufficient to provide the minimum dosage of 100,000 microwatts per square centimeter per second. If the flow is higher, the proper dosage will not be obtained and it is possible that tomites will escape the sterilizer and still be infectious.
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To determine the size of the sterilizer, we must determine the flow velocity of the water in the sterilizer with the formula
V = .002228 x GPM/ pi x R^2,
where we will obtain the velocity in feet per second. We convert this into centimeters per second by multiplying the flow by 30.48. If we take the flow rate, and divide it by the cross section area of the sterilizer, we determine the amount of time the water is in contact with the sterilizer, and by dividing the wattage of the sterilizer by this time, we arrive at the dosage rate.
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Example:
We have a 160 gallon tank and wish to install a UV sterilizer. What wattage sterilizer do we need and what flow rate is required, assuming that the sterilizer internal diameter is four inches, and the sterilizer is two feet long, and the bulb diameter is one inch?
1. First, we need the flow rate to achieve 99.99% filtration within four hours. This is 160 x 8.4 / F = 240, or a 5.5 GPM flow rate, so we'll round up and size this for 6 GPM, or about 22.5 liters per minute.
2. The cross section area of the sterilizer will be the cross section of the 4 inch housing, minus the cross section area of the bulb within the housing. In metric units, the housing cross sectional area is 81.07 square centimeters, minus the cross section of the bulb which is 5.07 square centimeters, leaving 76 square centimeters of cross sectional area for water flow.
3. The flow velocity within the sterilizer is 22.5 liters (22,500 cc's) per minute diveded by the cross sectional area of 76 centimeters, which yields 296 centimeters per minute, or 4.93 cm/sec. This keeps the water in the housing for 61 cm / 4.93 cm/sec = 12.36 seconds.
4. The internal area available for contact is the surface area of the inside of the sterilizer, or 5.08 x pi x 61 = 973.5 sq. cm.
5. The dosage is calculated my watts x surface area / time, or 100,000 microwatts x 78.76 cm2/sec, or 7.876 x 10^6 microwatts, or 7.876 watts. An 8 watt bulb is commercially available, and is the appropriate size for this system.
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REFERENCES:
Spotte, Stephen, Fish and Invertebrate Culture, 1979, "A Wiley-Interscience Publication"
Lawson, T.B. and Scott, D., http://gumbo.bae.lse...
Escobar, Peter, Aquatic Systems Engineering, 1986 Изменено 27.1.16 автор Daxel
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