Secondary Treatment - Wastewater Operators - page 288

 Example 7.6

Problem: The suspended solids concentration of a wastewater is 250
mg/L. If the normal k value at the plant is 0.6, what is the estimated par-
ticulate BOD concentration of the wastewater?

note: The k value of 0.6 indicates that about 60% of the suspended solids
are organic suspended solids (particulate BOD).

Solution:

250 mg/L × 0.6 = 150 mg/L particulate BOD

 Example 7.7

Problem: A rotating biological contactor receives a flow of 2.2 MGD with
a BOD content of 170 mg/L and suspended solids concentration of 140
mg/L. If the k value is 0.7, how many pounds of soluble BOD enter the
RBC daily?

Solution:

TotalBOD = ParticulateBOD + SolubleBOD

170 mg /L = (140 mg/L × + 0.7) x mg/L

170 mg/L = 98 mg/L + x mg/L

170 mg/L − 98 mg/L = = x 72 mg/L soluble BO D

Now, we can determine the lb/day soluble BOD:

Soluble BOD (lb/day) = Soluble BOD (mg/L) × Flow (MGD) × 8.34 lb/gal

= 72 mg/L × 2.2 MGD × 8.34 lb/gal = 1321 lb/day

7.4.7.2 RBC Total Media Area

Several process control calculations for the RBC use the total sur-
face area of all of the stages within the train. As was the case with the
soluble BOD calculation, plant design information or information sup-
plied by the unit manufacturer must provide the individual stage areas
(or the total train area), because physical determination of this would be
extremely difficult.

Total Area = 1st Stage Area + 2nd Stage Area + … + n th Stage Area (7.12)

7.4.7.3 RBC Organic Loading Rate

If the soluble BOD concentration is known, the organic loading on
an RBC can be determined. Organic loading on an RBC based on soluble
BOD concentration can vary from 3 to 4 lb/day/1000 ft 2 .