## Q. 5.1

BEDLOAD SEDIMENT TRANSPORT BY A TIDAL CURRENT

Calculate the bedload sediment transport rate in a tidal current given the following data:

Depth mean current $\overline{u}$ = 2.0 m/s, grain size $D_{50}$ = 0.4 mm, water depth h = 10 m, sea water density ρ = 1027 kg/m³ (@ 10° C and salt content 35 ppt), sediment density $ρ_{s}$ = 2650 kg/m³ and kinematic viscosity $ν = 1.36 × 10^{−6} m²/s$.

## Verified Solution

First, calculate the roughness height and skin friction drag coefficient. From Equations 5.6 and 5.7,

$z_{0} =k_{s} /30$                 (5.6)

$k_{s} =2.5D_{50}$              (5.7)

$z_{0} =k_{s} /30=2.5D/30=2.5\times 0.0004/30 =3.33 \times 10^{-5}$m.

Substituting into Equation 5.5,

$C_D=\left[\frac{0.4}{1+\ln \left(z_0 / h\right)}\right]^2=\left[\frac{0.4}{1+\ln \left(3.33 \times 10^{-5} / 10\right)}\right]^2=1.187 \times 10^{-3}$ .

Now calculate skin friction shear stress from Equation 5.2 and shear velocity from Equation 5.3

$\tau _{0} =\rho C_{D}\overline{U} ^{2}$           (5.2)

$u_{*} = \sqrt{{\tau _{0} }/{\rho }}.$          (5.3)

\begin{aligned} & \tau_{0 s}=\rho C_D \bar{U}^2=1027 \times 1.187 \times 10^{-3} \times 2^2=4.875 \mathrm{~N} / \mathrm{m}^2 \\ & u_{*s}=\sqrt{\tau_{0 s} / \rho}=\sqrt{4.875 / 1027}=0.069 \mathrm{~m} / \mathrm{s} . \end{aligned}

It should be noted, at this point, that the equation for $z_{0}$ is only strictly applicable for hydraulically rough flow ($u_{*}$ $k_{s}$ /ν > 70). In this case $u_{*}$ $k_{s}$/ν = 50.7; thus, Equation 5.6

is not strictly applicable. However, the resulting error in $C_{D}$ is only about 1%, and hence Equation 5.6 is sufficiently accurate for practical purposes (see Soulsby 1997 for further details).

The Shields parameter and critical Shields parameter can now be found (from Equations 5.20, 5.22 and 5.23):

$\theta =\frac{\tau }{\left(p_{s}-p \right)gD }$.      (5.20)

$\theta _{CR}=\frac{0.3}{1+1.2D_{*} } +0.055\left[1-exp(-0.02D_{*} )\right]$    (5.22)

$D_{*} =\left[\frac{g(s-1)}{v^{2} }\right] ^{1/3} D,$    (5.23)

$\theta _{s}=\frac{\tau_{0_{s} }}{g(p_{s}-p )D} =\frac{4.875}{9.81(2650-1027)0.0004} =0.765$

$D_{*} =\left[\frac{g(s-1)}{v^{2} } \right] ^{1/3} D=\left[\frac{9.81\left\lgroup\frac{2650}{1027}-1 \right\rgroup }{(1.36\times 10^{-6} )^{2} } \right]^{1/3} \times 0.0004=8.125$

$\theta _{CR} =\frac{0.3}{1+1.2D_{*} } + 0.055\left[1-exp(-0.02D_{*} )\right]$

$=\frac{0.3}{1+1.2\times 8.125} + 0.055\left[1-exp(-0.02\times 8.125)\right]= 0.036$

Now apply Equation 5.27

$\Phi =12\theta ^{1/2}_{s}\left(\theta _{s}-\theta _{CR} \right).$                       (5.27)

$\Phi =12\theta ^{1/2}_{s}\left(\theta _{s}-\theta _{CR} \right)=12\times 0.765^{½}\left(0.765-0.036\right)=7.65$.

Substituting into Equation 5.25 gives

$\Phi =\frac{q_{b} }{\left[g\left(s-1\right)D^{3} \right] ^{1/2} }$                         (5.25)

Hence, $q_{b}=7.65[9.81((2650/1027)−1)×0.0004³]^{1/2} = 2.41 × 10^{−4} m³/m/s.$