Wastewater Calculations - Mathematics and Statistics for the Environment - page 699

( )

QS − S

o

KilogramsO perday = − 142 p x (24.70)

2

( 1000 g/kg ) f

( )

QS − S 1

o

KilogramsO perday = − 1.42 Y obs (2 4.71)

2

( 1000 g/kg ) f

1

( ) ×

Pounds Oper day = QS − S 834 . − 142 . Y (24.72)

2 o obs

f

where
BOD u = Ultimate BOD.
P x = Net waste activated sludge (VSS) (kg/day, lb/day).
Q = Influent flow (m 3 /day, MGD).
S o = Influent soluble BOD concentration (mg/L).
S = Effluent soluble BOD concentration (mg/L).
f = Conversion factor for converting BOD to BOD u .
Y obs = Observed yield (g/g, lb/lb).
8.34 = Conversion factor [(lb/MG):(mg/L)].

24.8 OXIDATION DITCH DETENTION TIME
Oxidation ditch systems may be used where the treatment of wastewater is amendable to aerobic
biological treatment and the plant design capacities generally do not exceed 1.0 MGD. The oxida-
tion ditch is a form of aeration basin where the wastewater is mixed with return biosolids. The
oxidation ditch is essentially a modification of a completely mixed activated biosolids system used
to treat wastewater from small communities. This system can be classified as an extended aeration
process and is considered to be a low loading rate system. This type of treatment facility can remove
90% or more of influent BOD. Oxygen requirements will generally depend on the maximum diurnal
organic loading, degree of treatment, and suspended solids concentration to be maintained in the
aerated channel mixed liquor suspended solids. Detention time is the length of time required for a
given flow rate to pass through a tank. Detention time is not normally calculated for aeration basins,
but it is calculated for oxidation ditches.

Note: When calculating detention time, it is essential that the time and volume units used in the
equation be consistent with each other.

Oxidationditch volume (gal)

Detentiontime(hr) = (24.73)

Flow rate (gph)

■ EXAMPLE 24.55
Problem: An oxidation ditch has a volume of 160,000 gal. If the flow to the oxidation ditch is
185,000 gpd, what is the detention time in hours?

Solution: Because detention time is desired in hours, the flow must be expressed as gph:

185,000 gpd
24 hr/day

= 7708 gph