Wastewater Calculations - Mathematics and Statistics for the Environment - page 683

■ EXAMPLE 24.35
Problem: The wastewater entering a rotating biological contactor has a BOD content of 210 mg/L.
The suspended solids content is 240 mg/L. If the K value is 0.5, what is the estimated soluble BOD
(mg/L) of the wastewater?

Solution:

( ) +

TotalBOD (mg/L) = ParticulateBOD × K Solubl eeBOD

210 = (240 × 0.5) + x

210 =+ 120

90 mg/L = x

x

SolubleBOD =

24.6.1.3 Organic Loading Rate
The organic loading rate can be expressed as total BOD loading in pounds per day per 1000 ft 2 of
media. The actual values can then be compared with plant design specifications to determine the
current operating condition of the system:

SolubleBOD × × 8.34 lb/gal

Mediaarea(1000 ft )

flow (MGD)

Organicloading rate = (24.24)

2

■ EXAMPLE 24.36
Problem: A rotating biological contactor has a media surface area of 500,000 ft 2 and receives a flow
of 1,000,000 gpd. If the soluble BOD concentration of the primary effluent is 160 mg/L, what is the
organic loading on the RBC in lb/day/1000 ft 2 ?

Solution:

SolubleBOD × × 8.34 lb/gal

Mediaarea(1000 ft)

Flow (MGD)

Organicloading rate =

2

160 mg/ L × 1.0 MGD × 8.34 lb/gal

=

2

500 × 1000 ft

ay/1000 ft 2 = 2.7 lb/d

■ EXAMPLE 24.37
Problem: The wastewater flow to an RBC is 3,000,000 gpd. The wastewater has a soluble BOD
concentration of 120 mg/L. The RBC consists of six shafts (each 110,000 ft 2 ), with two shafts com-
prising the first stage of the system. What is the organic loading rate in lb/day/1000 ft 2 on the first
stage of the system?

Solution:

SolubleBOD (lb/day)
Me

Organicloading rate =

2

diaarea(1000 ft )

120 mg/L × 3.0MGD × 8.34 lb/gal

2

= = 13.6lb/day/1000 ft

2

220 × 1000 ft