Over the past few years, I’ve devoted several of these essays to analysis of first-stage sewage treatment reactors. I described and analyzed the rotating disc reactor found at the plant is Holly here, and described the racetrack,“activated sludge” plug reactor found most everywhere else here. I also described a system without a primary clarifier found near Cincinatti. All of these were effective for primary treatment; soluble organics are removed by bio-catalyzed oxidation:
2 H-C-O-H + O2 –> CO2 + H2O.
A typical plant in Oakland county treats 2,000,000 gallons per day of this stuff, with the bio-reactor receiving liquid waste containing about 200 ppm of soluble and colloidal biomass. That’s 400 dry gallons for those interested, or about 3200 dry lbs./day. About half of this will be oxidized to CO2 and water. The rest (cell bodies) are removed with insoluble components, and applied to farmers fields or buried, or burnt in an incinerator.
There is another type of reactor used in Oakland County. It’s mostly used for secondary treatment, converting consolidated sludge to higher-quality sludge that can be sold or used on farms with less restriction, but it is a type of reactor used at the South Lyon treatment plant, for primary treatment. It is a Continually stirred tank reactor, or CSTR, a design that is shown in schematic below.
As of some years ago, the South Lyon system involved a single largish pond lined with plastic with a volume about 2,000,000 gallons total. About 700,000 gallons per day of sewage liquids went into the lagoon, at 200 ppm soluble organics. Air was bubbled through the liquid providing a necessary reactant, and causing near-perfect mixing of the contents. The aim of the plant managers is to keep the soluble output to the, then-acceptable level of 10 ppm; it’s something they only barely managed, and things got worse as the flow increased. Assume as before, a value V and a flow Q.
We will call the concentration of soluble organics C, and call the initial concentration, the concentration that enters, Ci. It’s about 200 ppm. We’ll call the output concentration Co, and for this type of reactors, Co = C. The reaction is first order, approximately, so that, if there were no flow into or out of the reactor, the concentration of organics would decrease at the rate of
dC/dt = -kC.
Here k is a reaction constant, dependent on temperature oxygen and cell content. It’s typically about 0.5/hour. For a given volume of tank the rate of organic removal is VkC. We can now do a mass balance on soluble organics. Since the rate of organic entry is QCi and the rate leaving by flow is QC. The difference must be the amount that is reacted away:
QCi – QC = VkC.
We now use algebra, to find that
Co = Ci/(1 + kV/Q).
V/Q is sometimes called a residence time; for the system. At normal flow, the residence time of the South Lyon system is about 2.8 days or 68.6 hours. Plugging these numbers in, we find that the effluent from the reactor leaves at 1/35 of the input concentration, or 5.7 ppm, on average. This would be fine except that sometimes the temperature drops, or the flow increases, and we start violating the standard. A yet bigger problem was that the population increased by 50% while the EPA standard got more stringent to 2 ppm. This was solved by adding another, smaller reactor, volume = V2. Using the same algebraic analysis, as above you can show that, with two reactors,
Co = Ci/ [(1 + kV/Q)(1+kV2/Q)].
It’s a touchy system, but it meets government targets, just barely, most of the time. I think it is time to switch to a plug-flow reactor system, as used in much of Oakland county. In these, the fluid enters a channel and is reacted as it flows along. Each gallon of fluid, in a sense moves by itself as if it were its own reactor. In each gallon, we can say that dC/dt = -kC. We can thus solve for Co in terms of the total residence time, where t again is V/Q. We can rearrange this equation and integrate: ∫dC/C = – ∫kdt. We then find that,
ln(Ci/Co) = kt = kV/Q
To convert 200 ppm sewage to 2 ppm we note that Ci/Co = 100 and that V = Q ln(100)/k = Q (4.605/.5) hours. An inflow of 1000,000 gallons per day = 41,667 gal/ hour, and we find the volume of tank is 41,667 x 9.21 = 383,750 gallons. This is quite a lot smaller than the CSTR tanks at South Lyon. If we converted the South Lyon tanks to a plug-flow, race-track design, it would allow it to serve a massively increased population, discharging far cleaner sewage.
Robert Buxbaum, November 17, 2019