HOME WORK 6: Chapter 7

Professor:Dr. Weiming Wu

By Cássio Rampinelli

March, 18th, 2019

OBS: This script was done in R programming language

__________________________________________________________________________________________________________________________

PROBLEM 7.4. Uniform sediment with diameter D=0.5 mm transports in a wide open channel with a bed slope of 0.001 and a unit flow discharge of 2 m²/s. Assume Manning’s n as 0.03. Calculate the corresponding bed load transport rate using the following formulas:

a)Meyer-Peter and Muller (1948)

Solution

Given

\(D=0.5 \ mm\)

\(S=0.001\)

\(q=2 \ m²/s\)

\(n=0.03\)

\(\gamma= 9,810 N/m^3\)

\(\gamma_s= 24,525 N/m^3\)

The Meyer-Peter and Muller (1948) is given by:

\[ \frac{q_{b}}{\gamma_{s}\cdot \sqrt{\Bigg(\frac{\gamma_{s}}{\gamma}-1\Bigg)gD_{m}^3}}=8\cdot\Bigg(\frac{(k/k^{'})^{3/2} \cdot \gamma \cdot R \cdot S}{(\gamma_s -\gamma)\cdot D_m}-0.047\Bigg)^{3/2} \]

  • Computing the hydraulic radius R

Assuming a wide retangular channel (R=h) and applying Manning equation

\[ Q=\frac{1}{n} \cdot A \cdot h^{2/3} \cdot S^{1/2} \] \[ Q=\frac{1}{n} \cdot B\cdot h \cdot h^{2/3} \cdot S^{1/2} \] \[ \frac{Q}{B}=\frac{1}{n} \cdot h^{5/3} \cdot S^{1/2} \]

\[ q=\frac{1}{n} \cdot h^{5/3} \cdot S^{1/2} \]

\[ h^{5/3}=\frac{q \cdot n}{S^{1/2}} \cdot \cdot \] \[ h=\Bigg(\frac{2 \cdot 0.03}{0.001^{1/2}}\Bigg)^{3/5} \] \[ h=1.47 \ m \]

  • Computing \(k^{'}\) and \(k\)

Assuming uniforme bed gradation \(D_{50}=D_{90}=D_m=0.5 mm\)

\[ k^{'}=\frac{26}{D_{90}^{1/6}} =\frac{26}{0.0005^{1/6}}=92.29 \] \[ k=\frac{1}{n} =\frac{1}{0.03}=33.33 \]

  • Plugging the computed values on the Meyer-Peter and Muller (1948) formula

\[ \frac{q_{b}}{24525\cdot \sqrt{\Bigg(\frac{25996.5}{9810}-1\Bigg)9.81 \cdot (0.0005)^3}}=8\cdot\Bigg(\frac{(33.33/92.29)^{3/2} \cdot 9810 \cdot 1.47 \cdot 0.001}{(25996.5 -9810)\cdot 0.0005}-0.047\Bigg)^{3/2} \]

\[ q_{b}=1.85 \ (N/m\cdot s) \]

b)van Rijn (1984)

\[ U=\frac{Q}{A}=\frac{Q}{Bh}=\frac{q}{h}=\frac{2}{1.47}=1.36 m/s \]

\[ C^{'}_{h}=18log\Bigg(\frac{4h}{d_{90}}\Bigg) \] \[ C^{'}_{h}=18log\Bigg(\frac{4\cdot 1.47}{0.0005}\Bigg) \] \[ C^{'}_{h}=73.27 \]

\[ u^{'}_{*}=\frac{U\cdot g^{0.5}}{C^{'}_{h}} \]

\[ u^{'}_{*}=\frac{1.36\cdot 9.81^{0.5}}{73.27} \]

\[ u^{'}_{*}=0.0581 \]

\[ D^{*}=d\cdot\Bigg(\Bigg(\frac{\rho_s}{\rho}-1\Bigg)\cdot \frac{g}{\nu^2}\Bigg)^{1/3} \] \[ D^{*}=0.0005\cdot\Bigg(\Bigg(\frac{2650}{1000}-1\Bigg)\cdot \frac{9.81}{(10^{-6})^2}\Bigg)^{1/3} \] \[ D^{*}=12.65 \]

\[ A=0.215+\frac{6.79}{D^{1.7}_*}-0.075\cdot e^{-0.00262D^*} \]

\[ A=0.215+\frac{6.79}{12.65^{1.7}}-0.075\cdot e^{-0.00262\cdot 12.65}=0.2333 \]

\[ A=0.233 \]

\[ \omega_s=\frac{\nu}{d_{50}}\Bigg(\sqrt{25+1.2D^2_{*}}-5\Bigg)^{1.5} \]

\[ \omega_s=\frac{10^{-6}}{0.0005}\Bigg(\sqrt{25+1.2 \cdot 12.65^2 }-5\Bigg)^{1.5} \]

\[ \omega_s=0.0607 \ m/s \]

\[ u^*_c=A\cdot\omega_s \]

\[ u^*_c=0.233\cdot 0.0607 \]

\[ u^*_c=0.0141 \]

\[ T=\Bigg(\frac{u_*^{'}}{u_{*cr}}\Bigg)^2-1 \]

\[ T=\Bigg(\frac{0.0581}{0.0141}\Bigg)^2-1 \] \[ T=15.85 \]

\[ \Phi_b=0.053\cdot \frac{T^{2.1}}{D_*^{0.3}} \] \[ \Phi_b=0.053\cdot \frac{15.85^{2.1}}{12.65^{0.3}} \] \[ \Phi_b=8.2 \]

\[ \Phi_b=\frac{q_b}{\rho_s \sqrt{(\frac{\rho_s}{\rho}-1)\cdot g\cdot{D^3}}} \]

\[ 8.2=\frac{q_b}{2650 \sqrt{(\frac{2650}{1000}-1)\cdot 9.81\cdot{0.0005^3}}} \]

\[ q_b=0.978 \ (N/m\cdot s) \]

PROBLEM 7.6. Consider a nonuniform sediment mixture consisting of three size classes with representative diameters 0.05, 0.5, and 5 mm and fractions 0.3, 0.4 and 0.3, respectively. The sediment mixture transports in a wide open channel with a bed slope of 0.001 and a flow discharge of 2 m2/s. Assume Manning’s n as 0.03. Calculate the fractional transport rates for the three size classes using the following formulas:

a)Wu et al. (2000)

Solution

Given

\(d_1=0.05 \ mm\) and \(p_1=0.3 \ mm\)

\(d_2=0.5 \ mm\) and \(p_2=0.4 \ mm\)

\(d_3=5 \ mm\) and \(p_2=0.3 \ mm\)

\(S=0.001\)

\(q=2 \ m²/s\)

\(n=0.03\)

\(\rho= 1,000 \ kg/m^3\)

\(\rho_s= 2,650 kg \ /m^3\)

  • First we should compute \(p_{hk}\) and \(p_{ek}\) for each of the 3 classes regarding the three representative diameters given by:

\[ p_{hk}=\sum_{j=1}^Np_{bj}\frac{d_j}{d_k+d_j} \]

\[ p_{ek}=\sum_{j=1}^Np_{bj}\frac{d_k}{d_k+d_j} \]

For k=1 (d=0.05 mm)

\[ p_{h1}=0.3\frac{0.05}{0.05+0.05}+0.4\frac{0.5}{0.05+0.5}+0.3\frac{5}{0.05+5} \] \[ p_{h1}=0.81 \]

\[ p_{e1}=0.3\frac{0.05}{0.05+0.05}+0.4\frac{0.05}{0.05+0.5}+0.3\frac{0.05}{0.05+5} \]

\[ p_{e1}=0.19 \]

For k=2 (d=0.5 mm)

\[ p_{h2}=0.3\frac{0.05}{0.5+0.05}+0.4\frac{0.5}{0.5+0.5}+0.3\frac{5}{0.5+5} \] \[ p_{h2}=0.5 \]

\[ p_{e2}=0.3\frac{0.5}{0.5+0.05}+0.4\frac{0.5}{0.5+0.5}+0.3\frac{0.5}{0.5+5} \]

\[ p_{e2}=0.5 \]

For k=3 (d=5 mm)

\[ p_{h3}=0.3\frac{0.05}{5+0.05}+0.4\frac{0.5}{5+0.5}+0.3\frac{5}{5+5} \] \[ p_{h3}=0.19 \]

\[ p_{e3}=0.3\frac{5}{5+0.05}+0.4\frac{5}{5+0.5}+0.3\frac{5}{5+5} \]

\[ p_{e3}=0.91 \]

  • Then, the one can calculate the critical shear stress for each representative diameter (or class k) using the Wu et al. (2000) formula proposed for the Shields sediment incipient motion:

\[ \frac{\tau_{ck}}{(\gamma_s-\gamma)d_k}=\Theta_c\Bigg(\frac{p_{ek}}{p_{hk}}\Bigg)^{-m} \]

Assuming \(\Theta_c=0.03\) and m=0.6, obteined from calibrated data, and \(\gamma_s=25,996.5 N/m³\), \(\gamma=9810 N/m³\) the critical shear stress fro each representative diamter can be calculated as follows:

  • For k=1 (d=0.05 mm)

\[ \frac{\tau_{c1}}{(25996.5-9810)0.00005}=0.03\Bigg(\frac{0.19}{0.81}\Bigg)^{-0.6} \]

\[ \tau_{c1}=0.058\ N/m² \]

  • For k=2 (d=0.5 mm)

\[ \frac{\tau_{c2}}{(25996.5-9810)0.0005}=0.03\Bigg(\frac{0.5}{0.5}\Bigg)^{-0.6} \]

\[ \tau_{c2}=0.24\ N/m² \]

  • For k=3 (d=5 mm)

\[ \frac{\tau_{c3}}{(25996.5-9810)0.005}=0.03\Bigg(\frac{0.81}{0.19}\Bigg)^{-0.6} \]

\[ \tau_{c2}=1.02\ N/m² \]

  • Computing \(\Phi_{bk}\) for each size class

\[ \Phi_{bk}=0.0053 \Bigg( \Bigg(\frac{n^{'}}{n}\Bigg)^{3/2} \frac{\tau_b}{\tau_{ck}}-1\Bigg)^{2.2} \]

OBS: Considering the given sediment classes and fractions

\[ D_{50} \approx 0.275 \ mm \] \[ n^{'} = \frac{D_{50}^{1/6}}{20}=\frac{0.000275^{1/6}}{20}=0.013 \]

For a wide channel, as done in the previous problem the hydraulic depth h is computed as:

\[ h^{5/3}=\frac{q \cdot n}{S^{1/2}} \cdot \cdot \]

\[ h=\Bigg(\frac{2 \cdot 0.03}{0.001^{1/2}}\Bigg)^{3/5} \]

\[ h=1.47 \ m \]

For a wide channel, the hydraulic depth can be assumed to be equivalent to the hydraulic Radios. Then:

\[ \tau_b = \gamma R S \] \[ \tau_b = 9810 \cdot 1.47 \cdot 0.001 \] \[ \tau_b = 14.42 \ N/m^2 \]

  • For k=1 (d=0.05 mm)

\[ \Phi_{b1}=0.0053 \cdot \Bigg(\Bigg(\frac{0.013}{0.03}\Bigg)^{3/2}\frac{14.42}{0.058}\Bigg)^{2.2} \] \[ \Phi_{b1}=62.513 \]

  • For k=2(d=0.5 mm)

\[ \Phi_{b2}=0.0053 \cdot \Bigg(\Bigg(\frac{0.013}{0.03}\Bigg)^{3/2}\frac{14.42}{0.24}\Bigg)^{2.2} \]

\[ \Phi_{b2}=2.748 \]

  • For k=3(d=5 mm)

\[ \Phi_{b3}=0.0053 \cdot \Bigg(\Bigg(\frac{0.013}{0.03}\Bigg)^{3/2}\frac{14.42}{1.02}\Bigg)^{2.2} \]

\[ \Phi_{b3}=0.114 \]

  • ** Finally, one can compute the bed load for each size class applying**

\[ \Phi_{bk}=\frac{q_{bk}}{p_{bk} \cdot \rho_s \cdot \sqrt{(\rho_s/ \rho-1)\cdot g \cdot d_k^3}} \]

  • For k=1 (d=0.05 mm)

\[ 62.513=\frac{q_{bk}}{0.3 \cdot 2650 \cdot \sqrt{(2650/ 1000-1) \cdot 9.81 \cdot 0.00005^3}} \]

\[ q_{b1}=0.071 \ (N/m \cdot s) \]

  • For k=2 (d=0.5 mm)

\[ 2.748=\frac{q_{b2}}{0.4 \cdot 2650 \cdot \sqrt{(2650/ 1000-1) \cdot 9.81 \cdot 0.0005^3}} \]

\[ q_{b2}=0.131 \ (N/m \cdot s) \]

  • For k=3 (d=5 mm)

\[ 0.114=\frac{q_{b3}}{0.3 \cdot 2650 \cdot \sqrt{(2650/ 1000-1) \cdot 9.81 \cdot 0.005^3}} \]

\[ q_{b3}=0.129 \ (N/m \cdot s) \]

PROBLEM 7.11. If you are asked to design a pit trap for measuring bed load in a sand-bed stream, how wide is the pit trap in the streamwise direction?

Solution

The pit trap length (or wide in the streamwise direction) should consider the expected range of the bed load grains.

According to Einstein (1944) experiments demonstrated that samplers having streamwise slot lengths of 100 to 200 grain diameters collect nearly 100% of the bed load.

For grain sizes between 1.88 and 4.5 mm, Poreh et al. (1970) indicated that streamwise slot length of about 35 grain diameters would have an efficiency close to 100%.