Water Treatment Processes - Mathematics and Statistics for the Environment - page 623

To determine the drag force of the paddle on the water we use Equation 23.44:

F D = (1/2) pC D Av 2 (23.44)

where
F D = Drag force.
C D = Drag coefficient.
A = Area of the paddle.

To determine the power input imparted to the water by an elemental area of the paddle the usual
equation used is Equation 23.45:

dP = dF D v = 1/2 pC D v 3 dA (23.45)

23.5 SEDIMENTATION CALCULATIONS
Sedimentation, the solid-liquid separation by gravity, is one of the most basic processes of water
and wastewater treatment. In water treatment, plain sedimentation, such as the use of a presedimen-
tation basin for grit removal and a sedimentation basin following coagulation-flocculation, is the
most commonly used approach.

23.5.1 t anK v olume C alCulations
The two common shapes of sedimentation tanks are rectangular and cylindrical. The equations for
calculating the volume for each type tank are shown below.

23.5.1.1 Calculating Tank Volume
For rectangular sedimentation basins, we use Equation 23.46:

Volume (gal) = Length (ft) × Width (ft) × Depth (ft) × 7.48 (gal) l /f (ft) 3 (23.46)

For cylindrical clarifiers, we use Equation 23.47:

Volume (gal) = 0.785 × (Diameter) 2 × Depth (ft) × 7.48 (gal) l /f (ft) 3 (23.47)

■ EXAMPLE 23.44
Problem: A sedimentation basin is 25 ft wide by 80 ft long and contains water to a depth of 14 ft.
What is the volume of water in the basin, in gallons?

Solution:

Volume = Length (ft) × Width (ft) × Depth (ft) × 7.48 ga l /f (ft) 3

= 80 ft × 25 ft × 14 ft × 7.48 gal l /f ft 3 = 209,440 gal

■ EXAMPLE 23.45
Problem: A sedimentation basin is 24 ft wide by 75 ft long. When the basin contains 140,000 gal,
what would the water depth be?

Solution:

Volume = Length (ft) × Width (ft) × Depth (ft) × 7.48 ga l /f (ft) 3