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.

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.