Question 5.2: SUSPENDED SEDIMENT TRANSPORT IN A TIDAL CURRENT Calculate th......

The first stage of the solution is to determine the regime of sediment transport by calculating the Shields parameter and its critical value, the skin friction shear velocity and the fall velocity. Some of this has already been done in Example 5.1, i.e.,

\theta _{s}=0.765, \theta _{CR} =0.036   D_{*} =8.125  and  u_{*s} =0.069.

Calculate the sediment fall velocity from Equation 5.28

w_{s}=\frac{v}{D} \left[\left(10.36^{2}+1.049D^{3}_{*}\right) ^{1/2}-10.36 \right] for all D_{*} .      (5.28)
as

The regime of sediment transport can now be found as follows:

As θ_{CR} ≤ θ_{s} ≤ 0.8, then transport will occur with ripples or dunes.

As u_{*s} > w_{s} , then suspended sediment transport will occur.

Thus, we need to calculate the form drag contribution to the total bed shear stress before we can calculate the suspended sediment concentrations.

For ripples, use

For dunes, first find the critical shear stress using Equation 5.21:

\theta_{C R}=\frac{\tau_{C R}}{\left(\rho_s  –  \rho\right) g D}               (5.21)

τ_{CR}=\theta _{CR}g\left(\rho_{s}  –  \rho\right)D=0.23  N/m^{2}.

Hence, as  τ_{CR} < τ_{0s} < 26τ_{CR}, then apply Equations 5.8 and 5.10. First, find T_{s}:

T_{s}=\left\lgroup\frac{\tau _{os}  –  \tau _{CR} }{\tau _{CR} } \right\rgroup =\left\lgroup\frac{4.875  –  0.23}{0.23} \right\rgroup=20.2.

Hence,

Now find the additional roughness height due to bedforms, using Equation 5.12:

z_{of}=a_{r}\frac{\Delta ^{2}_{r} }{\lambda _{r} }.                (5.12)

Take a_{r} = 1.

For ripples,  z_{0f} = 0.057²/0.4 = 8.12 × 10^{−3}   m.

For dunes,  z_{0f} = 0.253²/73 = 0.88 × 10^{−3}   m.

It can be seen that the effect of the ripples on roughness height is much more significant than that of the dunes. However, add the two contributions to obtain the total

z_{0f} = 9 × 10^{−3}  m.

Now find the total roughness height and calculate the total drag coefficient C_D:

The total bed shear stress and total shear velocity can now be found:

We can now (finally!) calculate the suspended sediment concentrations. From Equations 5.37 & 5.38 the reference concentration and reference height are

C_{a} =\frac{0.331\left(\theta _{s}-0.045 \right) ^{1.75} }{1+ 0.72\left(\theta _{s}-0.045 \right) ^{1.75}}    (5.37)\\ =\frac{0.331\left(0.765-0.045 \right) ^{1.75}}{{1+ 0.72\left(0.765-0.045 \right) ^{1.75}}} =0.133  m^{3}   sediment/m³ sea water

From Equation 5.36 the concentration at any height (z) is given by

Finally, noting that the sediment transport rate per unit width, q(z), at any height z is given by

q(z)=u(z)c(z),

then the total suspended sediment rate (q_{s}) may be found by integration of Equation 5.39, i.e.,

q_{s} =\int_{z_{a} }^{h}{u\left(z\right)C\left(z\right)dz }     (5.39)

in combination with Equation 5.31

\frac{\varepsilon}{\rho} \frac{\mathrm{d} C}{\mathrm{~d} z}+w_{\mathrm{S}} C=0              (5.31)

The integration must be performed numerically and is shown (in part) in Table 5.1. A small increment in z (Δz = 0.1) is used and the value of C(z), u(z) is calculated midway between each pair of depths. q(z) is calculated at the midpoint and q(z)⋅Δz is summed through the water column. Using this method and depth increment the total suspended sediment transport rate is 0.00072 m³/s. The results are illustrated in Figure 5.10. It should be noted that the results are very sensitive to the method of numerical integration and the increment size (Δz), as C(z) varies very rapidly with depth. If the calculations are carried out using a computer programme, allowing choice of the increment size, then convergence of the solution may be found by sequential reduction of Δz. In this case, this does not occur until Δz < 0.001 m, for which q_{s} = 0.0006 m³/s. Also by reference to Example 5.1, it can be seen that in this case the suspended sediment transport is about 2.5 times that of the bed load transport.