Water Treatment Processes - Mathematics and Statistics for the Environment - page 625

23.5.3 s urFaCe o verFloW r ate
The surface overflow rate—similar to the hydraulic loading rate (flow per unit area)—is used to
determine loading on sedimentation basins and circular clarifiers. Hydraulic loading rate, however,
measures the total water entering the process, whereas surface overflow rate measures only the
water overflowing the process (plant flow only).

Note: Surface overflow rate calculations do not include recirculated flows. Other terms used syn-
onymously with surface overflow rate are surface loading rate and surface settling rate .

Surface overflow rate is determined using the following equation:

Flow (gpm)
Area (ft )

Surface overflow rate = (23.51)

2

■ EXAMPLE 23.48
Problem: A circular clarifier has a diameter of 80 ft. If the flow to the clarifier is 1800 gpm, what is
the surface overflow rate in gpm/ft 2 ?

Solution:

Flow (gpm)
Area (ft )

1800 gpm
0.785

0.36 gpm/ft 2 Surface overflow rate = =

×× =

2

80 ft 80 ft

■ EXAMPLE 23.49
Problem: A sedimentation basin 70 ft by 25 ft receives a flow of 1000 gpm. What is the surface
overflow rate in gpm/ft 2 ?

Solution:

Flow (gpm)
Area (ft )

1000 gpm
70 ft

0.6 gpm/ft 2

Surface overflow rate = = =

2

× 25 ft

23.5.4 m ean F loW v eloCity
The measure of average velocity of the water as it travels through a rectangular sedimentation basin
is known as mean flow velocity. Mean flow velocity is calculated using Equation 23.52:

Flow ( Q ) (ft 3 /min) = Cross-sectional area ( A ) (ft 2 ) × Volume ( V ) (ft/min) (23.52)

Q = A × V

■ EXAMPLE 23.50
Problem: A sedimentation basin is 60 ft long by 18 ft wide and has water to a depth of 12 ft. When
the flow through the basin is 900,000 gpd, what is the mean flow velocity in the basin in ft/min?

Solution: Because velocity is desired in ft/min, the flow rate in the Q = AV equation must be
expressed in ft 3 /min (cfm):

900,000 gpd
1440 min/day

= 84 cfm

3

× 7.48 gal/ft