Geoscience Reference
In-Depth Information
24.2.2 s Creening p it C apaCity C alCulations
Recall that detention time may be considered the time required for flow to pass through a basin or
tank or the time required to fill a basin or tank at a given flow rate. In screening pit capacity prob-
lems, the time required to fill a screening pit is being calculated. The equation used in screening pit
capacity problems is given below:
3
Volume of pit (ft )
Screeni
Fill time (days)
=
(24.3)
3
ngs removed (ft /day)
EXAMPLE 24.3
Problem: A screening pit has a capacity of 500 ft 3 . (The pit is actually larger than 500 ft 3 to accom-
modate soil for covering.) If an average of 3.4 ft 3 of screenings is removed daily from the wastewater
flow, in how many days will the pit be full?
Solution:
3
Volume of pit (ft )
Screeni
Fill time (days)
=
3
ngs removed (ft /day)
3
500 ft
3.4 ft /day
=
= 47.1 days
1
3
EXAMPLE 24.4
Problem: A plant has been averaging a screenings removal of 2 ft 3 /MG. If the average daily flow is
1.8 MGD, how many days will it take to fill the pit with an available capacity of 125 ft 3 ?
Solution: The filling rate must first be expressed as ft 3 /day:
3
2ft
×
1.8 MGD
MG
3
=
3.6 ft /day
Then,
3
Volume of pit (ft )
Screeni
Fill time (days)
=
3
ngs removed (ft /day)
3
125 ft
3.6 ft /day
=
= 4.7 days
3
3
EXAMPLE 24.5
Problem: A screening pit has a capacity of 12 yd 3 available for screenings. If the plant removes an
average of 2.4 ft 3 of screenings per day, in how many days will the pit be filled?
Solution: Because the filling rate is expressed as ft 3 /day, the volume must be expressed as ft 3 :
12 yd 3 × 27 ft 3 /yd 3 = 324 ft 3
Now calculate fill time:
3
3
Volume of pit (ft )
Screeni
324 ft
2.4 ft /day
Fill time (days)
=
=
= 35 days
1
3
3
ngs removed (ft /day)
 
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