Water Treatment Processes - Mathematics and Statistics for the Environment - page 626

Then, use Q = A × V to calculate velocity:

Q = A × V

84 cfm = (18 ft × 12 ft) × x fpm

x = 84 cfm/(18 ft × 12 ft) = 0.4 fpm

■ EXAMPLE 23.51
Problem: A rectangular sedimentation basin 50 ft long by 20 ft wide has a water depth of 9 ft. If the
flow to the basin is 1,880,000 gpd, what is the mean flow velocity 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):

1,880,000 gpd
1440 min/day

= 175 cfm

3

× 7.48 gal/ft

Then, use the Q = A × V equation to calculate velocity:

Q = A × V

175 cfm = (20 ft × 9 ft) × x fpm

x = 175 cfm/(20 ft × 9 ft) = 0.97 fpm

23.5.5 W eir o verFloW r ate
Weir overflow rate (weir loading rate) is the amount of water leaving the settling tank per linear foot
of weir. The result of this calculation can be compared with design. Normally, weir overflow rates
of 10,000 to 20,000 gal/day/ft are used in the design of a settling tank. Typically, the weir overflow
rate is a measure of the flow in gallons per minute (gpm) over each foot of weir. The weir overflow
rate is determined using the following equation:

Flow (gpm)
Weir

Weir overflow rate (gpm/ft) = (23.53)

length (ft)

■ EXAMPLE 23.52
Problem: A rectangular sedimentation basin has a total of 115 ft of weir. What is the weir overflow
rate in gpm/ft when the flow is 1,110,000 gpd?

Solution:

1,110,000 gpd
1440 min/day

Flow = = 771 gpm

Flow (gpm)
Weirlength(f

771 gpm
115 ft

Weir overflow rate = = = 6.7 gpm/ft

t)