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.

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

Integrated Publishing, Inc. |