Water Treatment Processes - Mathematics and Statistics for the Environment - page 624

140,000 gal = 75 ft × 24 ft × x ft × 7.48 gal l /f ft 3

140,000

x ft = ××

=

75 24 7.48

x ft 10.4 ft

23.5.2 d etention t ime
Detention time for clarifiers varies from 1 to 3 hr. The equations used to calculate detention time
are shown below.

Basic detention time equation:

Volume of tank (gal)
Flo

Detentiontime(hr) = (23.48)

wrate(gph)

Rectangular sedimentation basin equation:

3

Length (ft)Width (ft) × × Depth (ft) × 7.48 gal/ft

Detentiontime(hr) = (23.49)

Flow rate (gph)

Circular basin equation:

2 3

0.785 ×× × D Depth (ft) 7.4 88gal/ft

Detentiontime(hr) = (23.50)

Flow rate (gph)

■ EXAMPLE 23.46
Problem: A sedimentation tank has a volume of 137,000 gal. If the flow to the tank is 121,000 gph,
what is the detention time in the tank (in hours)?

Solution:

Detentiontime= Volume of tank (gal)
Flow rate(gph)

137,000 gal
121,000 gph

= = 1.1 hr

■ EXAMPLE 23.47
Problem: A sedimentation basin is 60 ft long by 22 ft wide and has water to a depth of 10 ft. If the
flow to the basin is 1,500,000 gpd, what is the sedimentation basin detention time?

Solution: First, convert the flow rate from gpd to gph so the times units will match:

(1,500,000 gpd)/(24 hr/day) = 62,500 gph

Then calculate detention time:

3

Detentiontime= Volume of tank (gal)
Flow rate(gph)

60 ft × 22 ft × ×

,500 gph

10 ft 7.48 gal/ft

= = 1.6 hr

62