Table 5.9. Sediment Size Distribution in the St. Lawrence Seaway.
D50
Distance Downstream of Cornwall
(mm)
(km)
(mi)
0
0
28
24.1
15
0.25
40.2
25
0.018
48.3
30
0.003
56.3
35
0.001
Estimate the mean sediment size D50 10 miles (16.1 km) and 20 miles (32.2 km) downstream
of Cornwall.
The gradual decrease in sediment size with downstream distance can be approximated by the
following equation:
D50 = 28 x 10 -0.082x
(SI)
D50 = 28 x 10 -0.132x
(English)
This equation was obtained by regression analysis based on Equation 5.13. At distances of 10
and 20 miles (16.1 and 32.2 km), the expected mean sediment sizes D50 obtained by this
relationship are, respectively, 1.34 mm and 0.064 mm.
5.9.8 PROBLEM 8 Scale Ratios for Physical Models (SI)
A physical model is to be built in the Hydraulics Laboratory to simulate the flow pattern around
2
a structure in a complex multiple channel stream. About 334 m of space (with a maximum
length of 18.3 m) is available in the laboratory to model a 800 m reach. Knowing that the same
fluid (water) will be used for both the model and the prototype, determine the appropriate scale
ratios for time, discharge and force. Also, required is the flow depth in the model at the
location where flow depth reaches 6 m in the prototype.
A fixed boundary model will be used and open channel flow modeling is scaled by similarity in
Froude number. The scale ratios for γ and ρ equal unity and thus, scaling depends uniquely on
-2
the length scale L = 18.3/(800) = 2.3 x 10 . The following scale ratios for time, discharge and
force are calculated from the expressions shown in Table 5.7.
Parameter
Scale Ratio
1/2
1/2
Time
(Lρ/γ)
=L
= 0.15
5/2
1/2
5/2
-5
Discharge
L
(γ/ρ)
=L
= 8.0 x 10
3
3
-5
Force
Lγ = L
= 1.2 x 10
The flow depth of the model at h = 6.0 m is given by the product hL = 0.14 m.
5.75
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