Environmental Engineering Reference
In-Depth Information
7.2.2.3 Flow Rate in Acre-Feet/Day
Flow (ac-ft/day) = Flow (MGD) × 3.069 ac-ft/MG
(7.3)
Note:
Acre-feet (ac-ft) is a unit that can cause confusion, especially for
those not familiar with pond or lagoon operations; 1 ac-ft is the volume
of a box with a 1-acre top and 1 ft of depth, but the top does not have to
be an even number of acres in size to use acre-feet.
7.2.2.4 Flow Rate in Acre-Inches/Day
Flow (ac-in./day) = Flow (MGD) × 36.8 ac-in./MG
(7.4)
7.2.2.5 Hydraulic Detention Time in Days
(ac-ft)
Influent Flow (ac-ft/day)
Pond Volume
HydraulicDetention Time (Days)
=
(7.5)
Note:
Normally, hydraulic detention time ranges from 30 to 120 days for
stabilization ponds.
Example 7.1
Problem:
A stabilization pond has a volume of 53.5 ac-ft. What is the
detention time in days when the flow is 0.30 MGD?
Solution:
Flow
=
0.30 MGD3.069
×
=
0.92 ac-ft/day
53.5ac
0.92 ac-ft/day
Detentio
nTime
=
=
58.2 days
7.2.2.6 Hydraulic Loading in Inches/Day (Overflow Rate)
Influent Flow (
ac-in./day)
Pond Area (ac)
(7.6)
HydraulicLoading (in./day)
=
Population Served by Sys
tem
(7.7)
Population Loading
=
Pond Area (ac)
Note:
Population loading normally ranges from 50 to 500 people per
acre.
