Environmental Engineering Reference
In-Depth Information
Example 5.10
Problem: The plant is currently using two grit channels. Each channel
is 3 ft wide and has a water depth of 1.2 ft. What is the velocity when the
influent flow rate is 3.0 MGD?
Solution:
3.0MGD 1.55 cfs/MGD
2Channels
×
4.65 cfs
7.2ft
Velocity
=
=
=
.65fps
× tt1.2 ft
3f
×
2
Note: The channel dimensions must always be in feet. Convert inches to
feet by dividing by 12 inches per foot.
Example 5.11
Problem: The plant is currently using two grit channels. Each channel
is 3 ft wide and has a water depth of 1.3 ft. What is the velocity when the
influent flow rate is 4.0 MGD?
Solution:
4.0MGD 1.55 cfs/MGD
2Channels
×
6.2cfs
7.8ft
Velocity
=
=
=
0.79 fps
× tt1.3 ft
3f
×
2
Because 0.79 is within the 0.7 to 1.4 level, the operator of this unit would
not make any adjustments.
5.3.7.2 Required Settling Time
This calculation can be used to determine the time required for a
particle to travel from the surface of the liquid to the bottom at a given
settling velocity. To compute the settling time, the settling velocity in
feet per second must be provided or determined in a laboratory.
Liquid Depth (ft)
Settling
SettlingTime(s)
=
(5.9)
Velocity (fps)
Example 5.12
Problem: The grit channel of a plant is designed to remove sand that has
a settling velocity of 0.085 fps. The channel is currently operating at a
depth of 2.2 ft. How many seconds will it take for a sand particle to reach
the channel bottom?
Solution:
2.2ft
0.085 fps
SettlingTime
=
=
25.9s
 
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