Page 244 - Soil and water contamination, 2nd edition

Sediment transport and deposition 231

The analytical solution for Equation (12.12) reads:
w s t
C ) t ( C 0 e H (12.13)
where C = initial suspended sediment concentration at t = 0.
0
If no sedimentation occurs (τ > τ ), α equals zero. In stagnant water (τ = 0), α equals
b b,e b
1. In the case of hindered deposition (0 ≤ τ ≤ τ ), α can be described as function of the
b b,d
bottom shear stress , for example a linear relation:
1 b (12.14)
b, d
-2
The critical bottom shear stress for sedimentation τ is in the order of 0.06 – 1.0 N m

b,d
(Van Rijn, 1989). In some models, α is described as a linear relation with the vertically
averaged flow velocity (which is, in turn, proportional to the square root of the bottom shear
stress ):
u x
1 (12.15)
u
x, d
where u = the critical flow velocity for sedimentation .
x,d
Example 12.3
-1
A stream 120 cm deep has an average flow velocity of 0.2 m s . The average settling
-1
velocity of the suspended sediment is 5.8 m d and the critical shear stress for deposition
-2
is 1.0 N m . Calculate the distance needed to reduce the sediment concentration in the
stream water by 75 percent if dispersion can be neglected.
Solution
First, estimate the shear stress in the stream, using Equations (11.35) and (12.2):

u 1 . 0 2 . 0 . 0 02 m s -1
*
2
1000 . 0 02 4 . 0 N m -2
b
Second, calculate the delay factor α, using Equation (12.14):
4 . 0
1 6 . 0
0 . 1
If the sediment concentration has been reduced by 75 percent of the initial value, the
C(t): C(0) ratio equals 1 – 0.75 = 0.25. Thus, according to Equation (12.13):
C (t ) w s t
e H . 0 25
C ) 0 (
The constant in the exponent is
w s 6 . 0 8 . 5 -1
9 . 2 d
H 2 . 1
Therefore
e -2.9t = 0.25
-2.9t = ln(0.25) = -1.386
-1
t = 0.478d = 0.478 d × 86400 s d = 41299 s

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