Question 5.3: Calculate the total load sediment transport rate in a tidal ......

First, use the Ackers and White method: From Example 5.1 D_{*}= 8.125; hence, calculate n, m, A_{gr}, C

n = 1-0.56\log D_{*} = 0.49

m = 1.67+ 6.83/D_{*} = 2.51

A_{gr} = 0.14+ 0.23/ = 0.221

\log C = 2.79\log D_{*} – 0.98\left(\log D_{*} \right)²  –  3.46 = -1.733:C = 0.0185.

Next, calculate the particle mobility number F_{gr} from Equation 5.41:

F_{gr}=\frac{u^{n}_{*} }{\left(g\left(s-1\right)D \right) ^{1/2} }\left[\frac{\bar{u} }{\sqrt{32} \log\left(10h/D\right) } \right] ^{1-n}     (5.41)

From Example 5.2 the total shear velocity was found to be 0.133  m/s; hence,

F_{gr}=\frac{u^{n}_{*} }{\left(g\left(s-1\right)D \right) ^{1/2} }\left[\frac{\bar{u} }{\sqrt{32}\log \left(10h/D\right) } \right] ^{1-n}

=\frac{0.133^{0.49} }{\left\lgroup9.81\left\lgroup\frac{2650}{1027}-1 \right\rgroup 0.0004\right\rgroup ^{1/2} }\left[\frac{2}{\sqrt{32}\log \left(10\times 10/0.0004\right) } \right] ^{1-.49}=1.177.

Now calculate the sediment transport parameter G_{gr} from Equation 5.40:

G_{gr}=\frac{q_t}{\overline{u}D }\left[\frac{u_{*} }{\overline{u} } \right] ^{n} =C\left[\frac{F_{gr} }{A_{gr} } -1\right] ^{m}    (5.40)

G_{gr} =C\left[\frac{F_{gr} }{A_{gr} } -1\right] ^{m}= 0.0185\left[\frac{1.177}{0.221}-1 \right] ^{2.51} =0.735.

Finally, calculate the sediment transport rate q_{t} from Equation 5.40:

G_{g r}=\frac{q_t}{\bar{u} D}\left[\frac{u_*}{\bar{u}}\right]^n=C\left[\frac{F_{g r}}{A_{g r}}-1\right]^m        (5.40)

q_{t} =0.735\times 2\times 0.0004\times \left\lgroup\frac{2}{0.133} \right\rgroup ^{0.49} = 2.22\times 10^{-3}m^{3}/s/m

Second, use the van Rijn method:

From Equation 5.45,

ū_{CR}=0.19\left(D_{50} \right) ^{0.1}log\left\lgroup\frac{4h}{D_{90} } \right\rgroup  for  0.1\leq D_{50} \leq 0.5  mm                      (5.45)

\begin{aligned} & \bar{u}_{C R}=0.19\left(D_{50}\right)^{0.1} \log\left\lgroup\frac{4h}{D_{90} } \right\rgroup =0.19(0.0004)^{0.1} \log \left\lgroup\frac{4\times 10}{0.0004 } \right\rgroup=0.434 \mathrm{~m} / \mathrm{s} \\ & q_b=0.005 \bar{u} h\left\lgroup\frac{\bar{u}-\bar{u}_{CR} }{\left[\left(s-1\right) gD_{50} \right] ^{1/2} } \right\rgroup ^{2.4} \left\lgroup\frac{D_{50}}{h} \right\rgroup ^{1.2} \\ & =0.005 \times 2 \times 10\left\lgroup\frac{2-0.434}{\left[\left\lgroup\frac{2650}{1027}-1\right\rgroup 9.81 \times 0.0004\right] ^{1 / 2}}\right\rgroup ^{2.4}\left\lgroup\frac{0.0004}{10}\right\rgroup ^{1.2} \\& q_b=6.9 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s} /\mathrm{m}.\end{aligned}

From Equation 5.44,

\begin{aligned} q_s & =0.012 \bar{u} h\left\lgroup\frac{\bar{u}-\bar{u}_{C R}}{\left[(s-1) g D_{50}\right]^{1 / 2}}\right\rgroup ^{2.4}\left\lgroup\frac{D_{50}}{h}\right\rgroup \left(D_*\right)^{-0.6}     (5.44)       \\ & =0.012 \times 2 \times 10\left\lgroup\frac{2-0.434}{\left[\left\lgroup\frac{2650}{1027}-1\right\rgroup 9.81 \times 0.0004\right]^{1 / 2}}\right\rgroup ^{2.4}\left\lgroup\frac{0.0004}{10}\right\rgroup8.125^{-0.6} q_s=3.57 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s} / \mathrm{m} .\end{aligned}
From Equation 5.42.

q_{t}=q_{b} +q_{s}    (5.42)

q_t=q_b+q_s=4.26 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s} / \mathrm{m} .
Finally, from Examples 5.1 and 5.2, we have alternately
q_b=2.41 \times 10^{-4}, q_s=6 \times 10^{-4}
giving
q_t=8.41 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s} / \mathrm{m} .

In summary, the three estimates of total load transport are:
Ackers and White method: q_t= 2.22 × 10^{−3} m³/s/m

van Rijn method: q_t = 4.26 × 10^{−3} m³/s/m

Bed + suspended load: q_t= 0.84 × 10^{−3}m³/s/m
Although on first sight these three estimates appear to be significantly different, all three estimates are nearly within a factor of 2 of their mean value, demonstrating their consistency within the known error bands.