Wastewater Calculations - Mathematics and Statistics for the Environment - page 682

■ EXAMPLE 24.32
Problem: A rotating biological contactor treats a primary effluent flow of 1,350,000 gpd. The manu-
facturer's data indicate that the media surface area is 600,000 ft 2 . What is the hydraulic loading rate
on the filter?

Solution:

Flow (gpd)
Mediaarea

1,350,000 gpd
600,000 ft

2

Hydraulic loadingrate = = = 2.3 ft

2 2

(ft )

24.6.1.2 Soluble BOD
The soluble BOD concentration of the RBC influent can be determined experimentally in the labo-
ratory, or it can be estimated using the suspended solids concentration and the K factor. The K fac-
tor is used to approximate the BOD (particulate BOD) contributed by the suspended matter. The K
factor must be provided or determined experimentally in the laboratory. The K factor for domestic
wastes is normally in the range of 0.5 to 0.7:

Soluble BOD 5 = Total BOD 5 - ( K factor × Total suspended solids) (24.23)

■ EXAMPLE 24.33
Problem: The suspended solids concentration of a wastewater is 250 mg/L. If the K value at the
plant is 0.6, what is the estimated particulate biochemical oxygen demand (BOD) concentration of
the wastewater?

Solution:

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

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

■ EXAMPLE 24.34
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

x = 72 mg/L soluble BOD

Now the lb/day soluble BOD may be determined:

SolubleBOD (mg/L) × flow (MGD) × 8.34 lb/gal = b/day l

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