HOME WORK 7: Chapter 8

Professor:Dr. Weiming Wu

By Cássio Rampinelli

April, 29th, 2019

OBS: This script was done in R programming language

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PROBLEM 8.1. Derive the Rouse distribution of suspended-load concentration from the two dimensional transport equation (8.13). Make the necessary assumptions if needed.

Starting from Eq. (8.13)

\[ \frac{\partial c}{\partial t}+\frac{\partial (uc)}{\partial x}+\frac{\partial (vc)}{\partial y}-\frac{\partial (\omega_sc)}{\partial y}=\frac{\partial }{\partial x}\Bigg(\epsilon_s\frac{\partial c}{\partial x}\Bigg)+\frac{\partial }{\partial y}\Bigg(\epsilon_s\frac{\partial c}{\partial y}\Bigg) \]

Assuming steady uniform flow, Eq. (8.13) can be re-written as:

\[ -\frac{\partial (\omega_sc)}{\partial y}=+\frac{\partial }{\partial y}\Bigg(\epsilon_s\frac{\partial c}{\partial y}\Bigg) \] Eq.(8.18)

By using the sediment condition (Eq.8.7 from the book) at the water surface, Eq. (8.18) is further simplified as:

\[ \omega_sc+\epsilon_s\frac{\partial c}{\partial y}=0 \] Eq.(8.19)

The diffusion coefficient \(\epsilon_s\)  is often assumed to be proportional to the eddy viscosity of turbulent flow. The parabolic distribution of \(\epsilon_s\)  can be written as:

\[ \epsilon_s=\frac{1}{\sigma_s}ku_*y\Bigg(1-\frac{y}{h}\Bigg) \] Eq.(8.20)

Substituting Eq.(8.20) in Eq.(8.19) brings to:

\[ \omega_sc+\frac{1}{\sigma_s}ku_*y\Bigg(1-\frac{y}{h}\Bigg)\frac{\partial c}{\partial y}=0 \]

Dividing both sides by \(\omega_s\)

\[ c+\frac{1}{\sigma_s\omega_s}ku_*y\Bigg(1-\frac{y}{h}\Bigg)\frac{\partial c}{\partial y}=0 \]

Inverting both sides of the equation:

\[ \frac{\sigma_s\omega_s}{ku_*}\frac{\partial y}{y\Bigg(1-\frac{y}{h}\Bigg)}=-\frac{c}{\partial c} \]

Integrating both sides of the equation:

\[ \int \frac{\sigma_s\omega_s}{ku_*}\frac{\partial y}{y\Bigg(1-\frac{y}{h}\Bigg)}=-\int\frac{c}{\partial c} \]

\[ \frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(y)-ln(y-h)+k_1\Bigg]=-ln(c)+k_2 \]

Aggregating the constants

\[ \frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(y)-ln(y-h)\Bigg]=-ln(c)+K \]

Isolating K and re-arranging the terms

\[ K=\frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(y)-ln(y-h)\Bigg]+ln(c) \] \[ K=\frac{\sigma_s\omega_s}{ku_*}\cdot ln\Bigg[\frac{y}{y-h}\Bigg]+ln(c) \]

\[ K= ln\Bigg[\frac{y}{y-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c) \] Expression (1)

for \(y=\delta_b\) and \(c=c_b\), K can be computed as

\[ K= ln\Bigg[\frac{\delta_b}{\delta_b-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c_b) \] Expression (2)

Substituting Expression (2) in Expression (1), and rearraging the terms, brings to:

\[ ln\Bigg[\frac{\delta_b}{\delta_b-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c_b)= ln\Bigg[\frac{y}{y-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c) \]

\[ ln(c)-ln(c_b)=ln\Bigg[\frac{\delta_b}{\delta_b-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}-ln\Bigg[\frac{y}{y-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\frac{\delta_b}{\delta_b-h}}{\frac{y}{y-h}}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \] \[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\frac{\delta_b}{\delta_b-h}}{\frac{y}{y-h}}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_b}{\delta_b-h}\frac{y-h}{y}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_by-\delta_bh}{\delta_by-\delta_bh}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_by-\delta_bh}{\delta_by-yh}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_by(1-\frac{h}{y})}{\delta_by(1-\frac{h}{y})}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{(-1)(1-\frac{h}{y})}{(-1)(1-\frac{h}{y})}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\frac{h}{y}-1}{\frac{h}{\delta_b}-1}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

Finally, we get the Rouse equation

\[ \frac{c}{c_b}=\Bigg[\frac{\frac{h}{y}-1}{\frac{h}{\delta_b}-1}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]

PROBLEM 8.3. Sediment with diameter D=0.5 mm transport in a wide open channel with bed slope of 0.001 and unit flow discharge of 2 \(m^2/s\). Calculate the single-sized (uniform) suspended-load transport rate using Bagnold’s and Zhang’s formulas. Manning’s n is assumed as 0.02.

Given

\(D=0.5 \ mm\)

\(S=0.001\)

\(q=2 \ m^2/s\)

\(n=0.02\)

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

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

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

\(\nu=10^{-6} \ m^2/s\)

a) Bgnold (1966)

Solution

Bagnold (1966) established the following formula to calculate the suspended-load transport rate:

\[ q_s=0.01 \cdot\frac{\rho_s}{\rho_s-\rho}\frac{\tau_bU^2}{\omega_s} \]

  • Computing the hydraulic radius R and hydraulic depth h

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

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

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

\[ h^{5/3}=\frac{q \cdot n}{S^{1/2}} \cdot \cdot \] \[ h=\Bigg(\frac{2 \cdot 0.02}{0.001^{1/2}}\Bigg)^{3/5} \] \[ h=1.15 \ m \]

  • Computing U

\[ U=\frac{Q}{A}=\frac{Q}{Bh}=\frac{q}{h}=\frac{2}{1.15}=1.74 \ m/s \]

  • Computing \(\tau_b\)

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

\[ \tau_b = \gamma R S \] \[ \tau_b = 9810 \cdot 1.15 \cdot 0.001 \] \[ \tau_b = 11.28 \ N/m^2 \]

  • Computing \(\omega_s\) with Zhang(1961) formula

\[ \omega_{s}=\sqrt{\Bigg(13.95\cdot\frac{\nu}{d}\Bigg)^2+1.09\Bigg(\frac{\rho_s}{\rho}-1\Bigg)gd}-13.95\frac{\nu}{d} \]

\[ \omega_{s}=\sqrt{\Bigg(13.95\cdot\frac{10^{-6}}{0.0005}\Bigg)^2+1.09\Bigg(\frac{2650}{1000}-1\Bigg)\cdot 9.81 \cdot 0.0005}-13.95\frac{10^{-6}}{0.0005} \]

\[ \omega_{s}=0.07 \ m/s \]

  • Finally, we can compute the suspended-load transport rate, by plugging the calculated parameters in Bagnold (1966) formula

\[ q_s=0.01 \cdot\frac{2650}{2650-1000}\frac{11.28\cdot 1.74^2}{0.07} \] \[ q_s=7.836 \ N/(ms) \]

a) Zhang (1961)

\[ C=\frac{1}{20}\Bigg(\frac{U^3}{gR\omega_s}\Bigg)^{1.5}/\Bigg[1+\Bigg(\frac{1}{45}\frac{U^3}{gR\omega_s}\Bigg)^{1.15}\Bigg] \]

Plugging the parameters values in the previous equation

\[ C=\frac{1}{20}\Bigg(\frac{1.74^3}{9.81\cdot 1.15 \cdot 0.07}\Bigg)^{1.5}/\Bigg[1+\Bigg(\frac{1}{45}\frac{1.74^3}{9.81 \cdot 1.15 \cdot 0.07}\Bigg)^{1.15}\Bigg] \] \[ C=0.775 \ kg/m^3 \]

Plugging C in the definition of the suspended-load transporte rate expression:

\[ q_s = q\cdot C \] \[ q_s=2\cdot 0.775 \] \[ q_s=1.55 \ kg/(ms) \]

\[ q_s=1.55 \cdot g=1.54 \cdot 9.81 \] \[ q_s=15.21 \ N/(ms) \]

PROBLEM 8.5. Consider a nonuniform sediment mixture consisting of three size classes with representative diameter 0.05, 0.5, and 5 mm and fraction of 0.3, 0.4 and 0.3, respectively. The sediment mixture transports in a wide open channel with bed slope of 0.001 and flow discharge of 2 \(m^2/s\). Calculate the fractional suspended-load transport rates for the three size classes using the formula of Wu et al. (2000). Manning’s n is assumed as 0.02.

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 \(p_2=0.3 \ mm\)

\(S=0.001\)

\(q=2 \ m^2/s\)

\(n=0.03\)

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

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

  • 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 \(\gamma_s=25,996.5 N/m^2\), \(\gamma=9810 N/m^2\) the critical shear stress fro each representative diamter can be calculated as follows:

  • For k=1 (d=0.05 mm)

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

\[ \tau_{c1}=0.058\ N/m^2 \]

  • For k=2 (d=0.5 mm)

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

\[ \tau_{c2}=0.24\ N/m^2 \]

  • For k=3 (d=5 mm)

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

\[ \tau_{c3}=1.02\ N/m^2 \]

  • Then, the one can calculate the settling velocity for each size class \(\omega_{sk}\) by applying the Zhang formula

\[ \omega_{sk}=\sqrt{\Bigg(13.95\cdot\frac{\nu}{d_k}\Bigg)^2+1.09\Bigg(\frac{\rho_s}{\rho}-1\Bigg)gd_k}-13.95\frac{\nu}{d_k} \]

  • For k=1 (d=0.05 mm)

\[ \omega_{s1}=\sqrt{\Bigg(13.95\cdot\frac{10^{-6}}{0.00005}\Bigg)^2+1.09\Bigg(\frac{2650}{1000}-1\Bigg)\cdot 9.81 \cdot 0.00005}-13.95\frac{10^{-6}}{0.00005} \]

\[ \omega_{s1}=0.0016 \ m/s \]

  • For k=2 (d=0.5 mm)

\[ \omega_{s2}=\sqrt{\Bigg(13.95\cdot\frac{10^{-6}}{0.0005}\Bigg)^2+1.09\Bigg(\frac{2650}{1000}-1\Bigg)\cdot 9.81 \cdot 0.0005}-13.95\frac{10^{-6}}{0.0005} \]

\[ \omega_{s2}=0.07 \ m/s \]

  • For k=3 (d=5 mm)

\[ \omega_{s3}=\sqrt{\Bigg(13.95\cdot\frac{10^{-6}}{0.005}\Bigg)^2+1.09\Bigg(\frac{2650}{1000}-1\Bigg)\cdot 9.81 \cdot 0.005}-13.95\frac{10^{-6}}{0.005} \]

\[ \omega_{s3}=0.29 \ m/s \]

  • Then, the one can calculate \(\Phi_{sk}\) and \(q_{sk}\) by applying Wu et al. (2000)

\[ \Phi_{sk}=0.0000262\Bigg[\Bigg(\frac{\tau}{\tau_{ck}}-1\Bigg)\frac{U}{\omega_{sk}}\Bigg]^{1.74} \]

\[ \Phi_{sk}=\frac{q_{sk}}{p_{bk}\sqrt{\Bigg(\frac{\rho_s}{\rho}-1\Bigg)gd_k^3}} \]

  • For k=1 (d=0.05 mm)

\[ \Phi_{s1}=0.0000262\Bigg[\Bigg(\frac{11.28}{0.058}-1\Bigg)\frac{1.74}{0.0016}\Bigg]^{1.74} \] \[ \Phi_{s1}=47912.16 \]

\[ 47912.19=\frac{q_{s1}}{0.3\sqrt{\Bigg(\frac{2650}{1000}-1\Bigg)9.81 \cdot 0.00005^3}} \] \[ q_{s1}=0.02 \ m^2/s \]

  • For k=2 (d=0.5 mm)

\[ \Phi_{s2}=0.0000262\Bigg[\Bigg(\frac{11.28}{0.24}-1\Bigg)\frac{1.74}{0.07}\Bigg]^{1.74} \] \[ \Phi_{s2}=5.49 \]

\[ 5.49=\frac{q_{s2}}{0.4\sqrt{\Bigg(\frac{2650}{1000}-1\Bigg)9.81 \cdot 0.0005^3}} \]

\[ q_{s2}=9.87 \cdot 10^{-5} \ m^2/s \]

  • For k=3 (d=5 mm)

\[ \Phi_{s3}=0.0000262\Bigg[\Bigg(\frac{11.28}{1.02}-1\Bigg)\frac{1.74}{0.29}\Bigg]^{1.74} \] \[ \Phi_{s3}=0.033 \]

\[ 0.033=\frac{q_{s3}}{0.3\sqrt{\Bigg(\frac{2650}{1000}-1\Bigg)9.81 \cdot 0.005^3}} \]

\[ q_{s3}=1.41 \cdot 10^{-5} \ m^2/s \]