Tag Archives: reactor

Activated sludge sewage treatment bioreactors

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I ran for water commissioner of Oakland county in 2016, a county with 1.3 million people and eight sewage treatment plants. One of these plants uses the rotating disk contractor, described previously, but the others process sewage by bubbling air through it in a large tank — the so-called, activated sludge process. A description is found here in Wikipedia, but with no math, and thus, far less satisfying than it could be. I thought I might describe this process relevant mathematics, for my understanding and those interested: what happens to your stuff after you flush the toilet or turn on the garbage disposal.

Simplified sewage plant: a plug-flow reactor with a 90+% solids recycle used to maintain a high concentration of bio-catalyst material.

Simplified sewage plant: a bubbling, plug-flow bio-reactor with 90% solids recycle and a settler used to extract floc solids and bio-catalyst material.

In most of the USA, sanitary sewage, the stuff from your toilet, sink, etc. flows separately from storm water to a treatment plant. At the plant, the sewage is first screened (rough filtered) and given a quick settle to remove grit etc. then sent to a bubbling flow, plug-flow bioreactor like the one shown at right. Not all cities use this type of sludge processes, but virtually every plant I’ve seen does, and I’ve come to believe this is the main technology in use today.

The sewage flows by gravity, typically, a choice that provides reliability and saves on operating costs, but necessitates that the sewage plant is located at the lowest point in the town, typically on a river. The liquid effluent of the sewage, after bio-treatment is typically dumped in the river, a flow that is so great more than, during dry season, more than half the flow of several rivers is this liquid effluent of our plants – an interesting factoid. For pollution reasons, it is mandated that the liquid effluent leaves the plant with less than 2 ppm organics; that is, it leaves the plant purer than normal river water. After settling and screening, the incoming flow to the bio-reactor typically contains about 400 ppm of biomaterial (0.04%), half of it soluble, and half as suspended colloidal stuff (turd bits, vegetable matter, toilet paper, etc). Between the activated sludge bio-reactor and the settler following it manage to reduce this concentration to 2 ppm or less. Soluble organics, about 200 ppm, are removed by this cellular oxidation (metabolism), while the colloidal material, the other 200 ppm, is removed by adsorption on the sticky flocular material in the tank (the plug-flow tank is called an oxidation ditch, BTW). The sticky floc is a product of the cells. The rate of oxidation and of absorption processes are proportional to floc concentration, F and to organic concentration, C. Mathematically we can say that

dC/dt = -kFC

where C and F are the concentration of organic material and floc respectively; t is time, and k is a reaction constant. It’s not totally a constant, since it is proportional to oxygen concentration and somewhat temperature dependent, but I’ll consider it constant for now.

As shown in the figure above, the process relies on a high recycle of floc (solids) to increase the concentration of cells, and speed the process. Because of this high recycle, we can consider the floc concentration F to be a constant, independent of position along the reactor length.

The volume of the reactor-ditch, V, is fixed -it’s a concrete ditch — but the flow rate into the ditch, Q, is not fixed. Q is high in the morning when folks take showers, and low at night. It’s also higher — typically about twice as high — during rain storms, the result of leakage and illegal connections. For any flow rate, Q, there is a residence time for a bit of sewage flowing through a tank, τ = V/Q. We can now solve the above equation for the value of τ for an incoming concentration C° = 400 ppm, an outgoing concentration Co of 2 ppm. We integrate the equation above and find that:

ln (C°/Co) = kFτ

Where τ equals the residence time, τ = V/Q. Thus,

ln (C°/Co) = kFV/Q.

The required volume of reactor, V, is related to the flow rate, Q, as follows for typical feed and exit concentrations:

V = Q/kF ln( 400/2) = 5.3 Q/kF.

The volume is seen to be dependent on F. In Oakland county, tank volume V is chosen to be one or two times the maximum expected value of Q. To keep the output organic content to less than 2 ppm, F is maintained so that kF≥ 5.3 per day. Thus, in Oakland county, a 2 million gallon per day sewage plant is built with a 2-4 million gallon oxidation ditch. The extra space allows for growth of the populations and for heavy rains, and insures that most of the time, the effluent contains less than 2 ppm organics.

Bob Martin by the South Lyon, MI, Activated Sludge reactor

Bob Martin chief engineer the South Lyon, MI, Activated Sludge plant, 2016. His innovation was to control the air bubblers according to measurements of the oxygen content. The O2 sensor is at bottom; the controller is at right. When I was there, some bubblers were acting up.

As you may guess, the activated sludge process requires a lot of operator control, far more than the rotating disk contractor we described. There is a need for constant monitoring and tweaking. The operator deals with some of the variations in Q by adjusting the recycle amount, with other problems by adjusting the air flow, or through the use of retention tanks upstream or downstream of the reactor, or by adding components — sticky polymer, FeCl3, etc. Finally, in have rains, the settler-bottom fraction itself is adjusted (increased). Because of all the complexity. sewer treatment engineer is a high-pay, in demand, skilled trade. If you are interested, contact me or the county. You’ll do yourself and the county a service.

I’d mentioned that the effluent water goes to the rivers in Oakland county. In some counties it goes to the fields, a good idea, I think. As for the solids, in Oakland county, the solid floc is concentrated to a goo containing about 5% solids. (The goo is called unconsolidated sludge) It is shipped free to farmer fields, or sometimes concentrated to more than 5% (consolidated sludge), and provided with additional treatment, anaerobic digestion to improve the quality and extract some energy. If you’d like to start a company to do more with our solids, that would be very welcome. In Detroit the solids are burned, a very wasteful, energy-consuming process, IMHO. In Wisconsin, the consolidated sludge is dried, pelletized, and sold as a popular fertilizer, Milorganite.

Dr. Robert Buxbaum, August 1, 2017. A colleague of mine owned (owns?) a company that consulted on sewage-treatment and manufactured a popular belt-filter. The name of his company: Consolidated Sludge. Here are some sewer jokes and my campaign song.

A rotating disk bio-reactor for sewage treatment

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One of the most effective designs for sewage treatment is the rotating disk bio-reactor, shown below. It is typically used in small-throughput sewage plants, but it performs quite well in larger plants too. I’d like to present an analysis of the reactor, and an explanation of why it works so well.

A rotating disc sewage reactor.

A rotating disk sewage reactor; ∂ is the thickness of the biofilm. It’s related to W the rotation rate in radians per sec, and to D the limiting diffusivity.

As shown, the reactor is fairly simple-looking, nothing more than a train of troughs filled with sewage-water, typically 3-6 feet deep, with a stack of discs rotating within. The discs are typically 7 to 14 feet in diameter (2-4 meters) and 1 cm apart. The shaft is typically close to the water level, but slightly above, and the rotation speed is adjustable. The device works because appropriate bio-organisms attach themselves to the disk, and the rotation insures that they are fully (or reasonably) oxygenated.

How do we know the cells on the disc will be oxygenated? The key is the solubility of oxygen in water, and the fact that these discs are only used on the low biological oxygen demand part of the sewage treatment process, only where the sewage contains 40 ppm of soluble organics or less. The main reaction on the rotating disc is bio oxidation of soluble carbohydrate (sugar) in a layer of wet slime attached to the disc.

H-O-C-H + O2 –> CO2 + H2O.

As it happens, the solubility of pure oxygen in water is about 40 ppm at 1 atm. As air contains 21% oxygen, we expect an 8 ppm concentration of oxygen on the slime surface: 21% of 40 ppm = 8 ppm. Given the reaction above and the fact that oxygen will diffuse five times more readily than sugar at least, we expect that one disc rotation will easily provide enough oxygen to remove 40 ppm sugar in the slime at every speed of rotation so long as the wheel is in the air at least half of the time, as shown above.

Let’s now pick a rotation speed of 1/3 rpm (3 minutes per rotation) and see where that gets us in terms of speed of organic removal. Since the disc is always in an area of low organic concentration, it becomes covered mostly with “rotifers”, a fungus that does well in low nutrient (low BOD) sewage. Let’s now assume that mass transfer (diffusion) of sugar in the rotifer slime determines the thickness of the rotifera layer, and thus the rate of organic removal. We can calculate the diffusion depth of sugar, ∂ via the following equation, derived in my PhD thesis.

∂ = √πDt

Here, D is the diffusivity (cm2/s) for sugar in the rotifera slime attached to the disk, π = 3.1415.. and t is the contact time, 90 seconds in the above assumption. My expectation is that D in the rotifer slime will be similar to the diffusivity sugar in water, about 3 x 10-6 cm2/s. Based on the above, we find the rotifer thickness will be ∂ = √.00085 cm2 = .03 cm, and the oxygen depth will be about 2.5 times that, 0.07 cm. If the discs are 1 cm apart, we find that, about 14% of the fluid volume of the reactor will be filled with slime, with 2/5 of this rotifer-filled. This is as much as 1000 times more rotifers than you would get in an ordinary constantly stirred tank reactor, a CSTR to use a common acronym. We might now imagine that the volume of this sewage-treatment reactor can be as small as 1000 gallons, 1/1000 the size of a CSTR. Unfortunately it is not so; we’ll have to consider another limiting effect, diffusion of nutrients.

Consider the diffusive mass transfer of sugar from a 1,000,000 gal/day flow (43 liters/sec). Assume that at some point in the extraction you have a concentration C(g/cc) of sugar in the liquid where C is between 40 ppm and 1 ppm. Let’s pick a volume of the reactor that is 1/20 the normal size for this flow (not 1/1000 the size, you’ll see why). That is to say a trough whose volume is 50,000 gallons (200,000 liters, 200 m3). If the discs are 1 cm apart, this trough (or section of a trough) will have about  4×10^8 cm2 of submerged surface, and about 9×10^8 total surface including wetted disc in the air. The mass of organic that enters this section of trough is 44,000 C g/second, but this mass of sugar can only reach the rotifers by diffusion through a water-like diffusion layer of about .06 cm thickness, twice the thickness calculated above. The thickness is twice that calculated above because it includes the supernatant liquid beyond the slime layer. We now calculate the rate of mass diffusing into the disc: AxDxc/z = 8×10^8 x 3×10-6 x C/.06 cm = 40,000 C g/sec, and find that, for this tank size and rotation speed, the transfer rate of organic to the discs is 2/3 as much as needed to absorb the incoming sugar. This is to say that a 50,000 gallon section is too small to reduce the concentration to ln (1) at this speed of disc rotation.

Based on the above calculation, I’m inclined to increase the speed of rotation to .75 rpm. At this speed, the rotifer-slime layer will be 2/3 as thin 0.2 mm, and we expect an equally thinner diffusion barrier in the supernatant. At this faster speed, there is now 3/2 as much diffusion transfer per area because the thinner diffusion barrier, and we can expect a 50,000 liter reactor section to reduce the initial concentration by a fraction of 1/2.718 or C/e. Note that the mass transfer rate to the discs is always proportional to C. If we find that 50,000 gallons of tank reduces the concentration to 1/e, we find that we need 150,000 gallons of reactor to reduce the concentration of sugar from 40 ppm to 2 ppm, the legal target, ln (40/2) = 3. This 150,000 gallons is a remarkably small volume to reduce the sBOD concentration from 40 ppm to 2 ppm (sBOD = soluble biological oxygen demand), and the energy use is small too if the disc bearings are good.

The Holly sewage treatment plant is the only one in Oakland county, MI using the rotating disc contacted technology. It has a flow of 1,000,000 gallons per day, and has a contactor trough that is 215,000 gallons, about what we’d expect though their speed is somewhat higher, over 1 rpm and their input concentration is likely lower than 40 ppm. For the first stage of sewage treatment, the Holly plant use a vertical-draft, trickle-bed reactor. That is they drizzle the sewage-liquids over a slime-coated packing to reduce the sBOD concentration from 200 ppm to before feeding the flow to the rotating discs. My sense of the reason they don’t do the entire extraction with a trickle bed is that the discs use far less energy.

I should also add that the back-part of the disc isn’t totally useless oxygen storage, as it seems from my analysis. Some non-sugar reactions take place in the relatively anoxic environment there and in the liquid at the bottom of the trough. In these regions, iron reacts with phosphate, and nitrate removal takes place. These are other important requirements of sewage treatment.

Robert E. Buxbaum, July 18, 2017. As an exercise, find the volume necessary for a plug flow reactor or a stirred tank reactor (CSTR) to reduce the concentration of sugar from 40 ppm to 2 ppm. Assume 1,000,000 gal per day, an excess of oxygen in these reactors, and a first order reaction with a rate constant of dC/dt = -(0.4/hr)C. At some point in the future I plan to analyze these options, and the trickle bed reactor, too.