Geoscience Reference
In-Depth Information
Then convert gpm/ft
2
to the in./min rise rate:
2
12.3 gpm/ft 2in./ft
7.48 gal/ft
×
Rise rate
=
=
19.7 in./min
3
23.6.6 v
olume
oF
b
aCKWash
W
ater
r
equired
(
gal
)
To determine the volume of water required for backwashing, we must know both the desired back-
wash flow rate (gpm) and the duration of backwash (min):
Backwash water volume (gal) = Backwash (gpm) × Duration of backwash (min)
(23.73)
■
EXAMPLE 23.85
Problem:
For a backwash flow rate of 9000 gpm and a total backwash time of 8 min, how many
gallons of water will be required for backwashing?
Solution:
Backwash water volume = Backwash (gpm) × Duration of backwash (min)
= 9000 gpm × 8 min = 72,000 gal
■
EXAMPLE 23.86
Problem:
How many gallons of water would be required to provide a backwash flow rate of 4850
gpm for a total of 5 min?
Solution:
Backwash water volume = Backwash (gpm) × Duration of backwash (min)
= 4850 gpm × 5 min = 24,250 gal
23.6.7 r
equired
d
epth
oF
b
aCKWash
W
ater
t
anK
(
Ft
)
The required depth of water in the backwash water tank is determined from the volume of water
required for backwashing. To make this calculation, simply use Equation 23.74:
Volume (gal) = 0.785 × (Diameter)
2
× Depth (ft) × 7.48 (gal) l /f (ft)
3
(23.74)
■
EXAMPLE 23.87
Problem:
The volume of water required for backwashing has been calculated to be 85,000 gal.
What is the required depth of water in the backwash water tank to provide this amount of water if
the diameter of the tank is 60 ft?
Solution:
Use the volume equation for a cylindrical tank, filling in known data, then solve for
x
:
Volume = 0.785 × (Diameter)
2
× Depth (ft) × 7.48 (gal) l /f (ft)
3
85,000 gal = 0.785 × (60 ft)
2
×
x
ft × 7.48 gal l /f ft
3
x
= 85,000/(0.785 × 60 × 60 × 7.48) = 4 ft
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