Page 250 - Soil and water contamination, 2nd edition

Sediment transport and deposition 237

The transport capacity as function of slope length is calculated using Equation (12.29)
with f = 170 m:
T 170 E 170 . 0 0068 l . 0 75 . 1 156l . 0 75
c r
To find the slope length at which the total cumulative erosion equals the transport
capacity , the last two equations should be equated. The slope length can then be found
by iteration or ‘trial and error ’. The equations can also be solved graphically by plotting
both the total cumulative erosion and transport capacity as function of slope length
(see Figure 12.1): the slope length corresponding to the point of intersection is the
slope length at which the total cumulative erosion equals the transport capacity. From
Figure 12.1 it can be seen that this slope length is approximately 156 m. Thus, in this
case, sediment deposition commences at 156 m from the divide and the erosion is
transport-limited at distances greater than this.

80 Total cumulative erosion
(kg -1 m -1 y -1 )
70
Transport capacity (kg -1 m -1 y -1 )
60
Cumtulative erosion/ transport capacity (kg/m/y) 50
Total cumulative erosion=Transport capacity
40
30
20
10
6642 6642 6642
0
0 20 40 60 80 100 120 140 160 180 200
Slope length (m)
Figure 12.1 Total cumulative erosion and transport capacity as function of slope length given a straight slope
of 6 percent.

EXERCISES
1. a. Calculate the bottom shear stress τ for a river with the following characteristics:
b
width = 4 m, water depth = 1 m, bed slope = 0.0001, and average flow velocity = 31
cm s-1 (see question 6 in Chapter 11).
-2
b. Given the critical shear stress for erosion τ = 1.0 N m and the critical shear stress
b,e
-2
for deposition τ = 0.5 N m , indicate whether or not there is erosion or deposition
b,d
of sediment.
2. Name three factors that control the critical shear stress for erosion or deposition.
3. Calculate the bottom shear stress τ for a shallow lake with the following characteristics:
b
-1
water depth = 1 m, wind fetch length = 300 m, wind speed = 15 m s , D90 of the bed
sediment = 70 μm.
-1
4. The water column above 1 ha of floodplain contains 125 mg l suspended solids. The
-1
-6
effective settling velocity of the sediment is 1.75 10 m s (= α·w ). Assume that the
s
water above the floodplain is continuously being refreshed and the suspended solids
concentration in the river water remains the same.

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