CE 573: Sediment Transport
HOME WORK 5: Chapter 6
Professor:Dr. Weiming Wu
By Cássio Rampinelli
April, 04th, 2019
OBS: This script was done in R programming language
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PROBLEM 6.4. Consider uniform flow in a sand-bed channel. The mean velocity is 0.7 m/s, the flow depth is 0.6 m, the bed slope is 0.0005, the water temperature is 20oC, and the bed sediment size distribution with \(d_{50}=0.35\) mm and \(d_{90}\)=1 mm. Predict the regime and type of bed forms using the following methods:
(a) Simons and Richardson (1961)
Solution
Simons and Richardson (1961) developed a diagram for bed form regimes using the stream power per unit area (\(\tau_0U\)), and the median sediment fall diameter as the independent variables.
First let’s compute the shear stress (\(\tau_0\)). Assuming a wide channel we can approximate the hydraulic radius (R), by the flow depth (h). Assuming the specific weight of the water 9810 N/m², then:
\[ \tau_0=\gamma R S \] \[ \tau_0=9810\cdot 0.6 \cdot 0.0005 \] \[ \tau_0=2.943 \ N/m² =2.943 \ Pa \] \[ \tau_0=2.943 \cdot 0.02 = 0.05886 \ lbs/ft^2 \]
Given the velocity U
\[ U=0.7 m/s \cdot 3.28 = 2.296 ft/s \]
The stream Power is computed as:
\[ \tau_0 \cdot U=0.05886\cdot2.296=0.135\ lbs/(ft\cdot s) \]
Considering the fall diameter as aproximately 0.93 x sieve diamter, and assuming the median sieve diameter as \(d_{50}=0.35 mm\). The one can assume the median fall diameter as \(d_{fall}=0.93 \cdot 0.35=0.32 mm\)
Checking the Simons and Richardson (1961) diagram with d=0.32 mm and \(\tau_0 \cdot U=0.135\ lbs/(ft\cdot s)\) we conclude that the predicted regime is lower flow regime and type of bed forms are dunes, as can be seen in Figure 1 below.
Figure 1:Simons and Richardson (1961) Diagram
PROBLEM 6.5. Given a mean velocity U=0.8 m/s, flow depth h=0.6 m, bed slope S=0.00075, and median diameter \(d_{50}\)=0.25 mm. Determine the regime of bed forms using the following methods:
(a) Brownlie (1983)
Solution
First, let’s check what flow regime should be used by applying Brownlie (1983) diagram. Considering the given slope of 0.00075, and computing the grain Froude number (\(F_g\)):
\[ F_g =\frac{U}{\sqrt{g\cdot d_{50} \cdot (\rho_s-\rho)/\rho}} \] \[ F_g =\frac{0.8}{\sqrt{9.81\cdot 0.00025 \cdot (2650-1000)/1000}} \] \[ F_g =12.58 \]
By using Brownlie (1983) diagram for \(F_g\)=12.58 and S=0.00075, we conclude that there is a lower regime as can be seen in Figure 2 below.
Figure 2:Brownlie (1983) Diagram
(b) Karim (1995)
Then, if we apply Karim’s method and compute the Froude number for the upper limit of the lower flow regime (\(Fr_l\)) and for the lower limit of the uper flow regime (\(Fr_u\)) and compare it with the Froude number (Fr), we conclude that the bed form regim is in the lower flow regime.
\[ Fr =\frac{U}{\sqrt{g\cdot h }} \] \[ Fr =\frac{0.8}{\sqrt{9.81\cdot 0.6 }} \] \[ Fr =0.3297 \] \[ Fr_l =2.716\Bigg(\frac{h}{d_{50}}\Bigg)^{-0.25} \] \[ Fr_l =2.716\Bigg(\frac{0.6}{0.00025}\Bigg)^{-0.25} \] \[ Fr_l =0.388 \]
\[ Fr_u =4.785\Bigg(\frac{h}{0.00025}\Bigg)^{-0.27} \] \[ Fr_u =4.785\Bigg(\frac{0.6}{d_{50}}\Bigg)^{-0.27} \] \[ Fr_u =0.585 \]
Since \(F_r\)<\(Fr_l\) \(\Rightarrow\) Lower flow regime
PROBLEM 6.7. Given a unit flow discharge of 10 m²/s in a wide channel with a bed slope of 0.001, bed-material median size of 0.5 mm, and bed-material standard deviation of 2.0. Determine the flow depth using the following methods:
(b) van Rijn (1984)
Solution
The senquence for the solution is done by iterations and is presented as follows:
First let`s assume that \(d_{90}\) \(\approx\) \(d_{84.1}\) =\(d_{50} \cdot\) \(\sigma\)=\(0.0005\cdot 2=0.001 m\) and h=10 m.
* Compute U
\[ U=\frac{Q}{A}=\frac{Q}{Bh}=\frac{q}{h}=\frac{10}{10}=1 m/s \]
* Compute \(C^{'}_{h}\)
\[ C^{'}_{h}=18log\Bigg(\frac{4h}{d_{90}}\Bigg) \] \[ C^{'}_{h}=18log\Bigg(\frac{4\cdot 10}{0.001}\Bigg) \] \[ C^{'}_{h}=82.83 \]
* Compute \(u^{'}_{*}\)
\[ u^{'}_{*}=\frac{U\cdot g^{0.5}}{C^{'}_{h}} \] \[ u^{'}_{*}=\frac{1\cdot 9.81^{0.5}}{82.83} \] \[ u^{'}_{*}=0.03781 \]
* Compute \(D^{*}\)
\[ D^{*}=d\cdot\Bigg(\Bigg(\frac{\rho_s}{\rho}-1\Bigg)\cdot \frac{g}{\nu^2}\Bigg) \] \[ D^{*}=0.0005\cdot\Bigg(\Bigg(\frac{2650}{1000}-1\Bigg)\cdot \frac{9.81}{(10^{-6})^2}\Bigg) \] \[ D^{*}=12.64 \]
* Compute \(A\)
\[ A=0.215+\frac{6.79}{D^{1.7}_*}-0.075\cdot e^{-0.00262D^*} \] \[ A=0.215+\frac{6.79}{12.64^{1.7}}-0.075\cdot e^{-0.00262\cdot 12.64}=0.2333 \] \[ A=0.233 \]
* Compute \(\omega_s\)
\[ \omega_s=\frac{\nu}{d_{50}}\Bigg(\sqrt{25+1.2D^2_{*}}-5\Bigg)^{1.5} \] \[ \omega_s=\frac{10^{-6}}{0.0005}\Bigg(\sqrt{25+1.2 \cdot 12.64^2 }-5\Bigg)^{1.5} \]
\[ \omega_s=0.0607 \]
* Compute \(u^*_c\)
\[ u^*_c=A\cdot\omega_s \] \[ u^*_c=0.233\cdot 0.0607 \] \[ u^*_c=0.0141 \]
* Compute \(T\)
\[ T=\Bigg(\frac{u_*}{u_{*cr}}\Bigg)^2-1 \] \[ T=\Bigg(\frac{0.03781}{0.014162}\Bigg)^2-1 \] \[ T=6.1278 \]
* Compute \(\Delta_d\)
\[ \Delta_d/h=0.11 \cdot \Bigg(\frac{d_{50}}{h}\Bigg)^{0.3}\cdot(1-e^{-0.5T})\cdot (25-T) \] \[ \Delta_d=\Bigg(0.11 \cdot \Bigg(\frac{d_{50}}{h}\Bigg)^{0.3}\cdot(1-e^{-0.5T})\cdot (25-T)\Bigg)\cdot h \] \[ \Delta_d=\Bigg(0.11 \cdot \Bigg(\frac{0.0005}{10}\Bigg)^{0.3}\cdot(1-e^{-0.5\cdot 6.127})\cdot (25-6.127)\Bigg)\cdot h \] \[ \Delta_d=1.014 \]
* Compute \(\lambda\)
\[ \lambda=7.3\cdot h \] \[ \lambda=7.3\cdot 10 \] \[ \lambda=73 \]
* Compute \(k_s\)* \[ k_s=3\cdot d_{90}+1.1\Delta_{d}(1-e^{-25\Delta_d/ \lambda}) \] \[ k_s=3\cdot 0.001+1.1\cdot 1.01(1-e^{-25\cdot 1.01/ 73}) \] \[ k_s=0.33 \]
* Compute \(C_h\), assuming \(R_b=h\)
\[ C_h=18\cdot log\Bigg(\frac{12R_b}{k_s}\Bigg) \] \[ C_h=18\cdot log\Bigg(\frac{12\cdot 10}{0.33}\Bigg) \] \[ C_h=46.083 \]
* Compute \(U'\) by applying Chezy Equation, assuming R=h and S_0=0.001.
\[ U'=C_h \cdot \sqrt{(R\cdot S_0)} \] \[ U'=46.083 \cdot \sqrt{(10\cdot 0.001)} \]
\[ U'=4.608 \]
* Repeat the process, assuming other value for h, until the differences between U’ and U are low. After doing that iteration process for this case we get: h= 4.24 m
PROBLEM 6.8. Given a unit flow discharge of 5 m²/s in a wide channel with a bed slope of 0.0005 and bed-material median size of 0.4 mm. Determine the Manning coefficient of channel roughness using the following methods:
(a) Wu and Wang (1999)
Solution
The senquence for the solution is done by iterations and is presented as follows:
First, let`s assume that h=3 and \(A_n\)=20
* Compute U
\[ U=\frac{Q}{A}=\frac{Q}{bh}=\frac{q}{h}=\frac{5}{3}=1.66 \]
* Compute \(D^*\)
\[ D^{*}=d\cdot\Bigg(\Bigg(\frac{\rho_s}{\rho}-1\Bigg)\cdot \frac{g}{\nu^2}\Bigg) \] \[ D^{*}=0.0004\cdot\Bigg(\Bigg(\frac{2650}{1000}-1\Bigg)\cdot \frac{9.81}{(10^{-6})^2}\Bigg) \] \[ D^{*}=10.12 \]
* Compute \(\tau_c\) by using Wu and Wang (1999) equation
\[ \tau_c=(\gamma_s-\gamma)\cdot d \cdot 0.0685\cdot D^{-0.27}_* \] \[ \tau_c=(25996.5-9810)\cdot 0.0004 \cdot 0.0685\cdot 10.12^{-0.27} \] \[ \tau_c=0.2374 \]
* Compute \(F_r\)
\[ F_r=\frac{U}{\sqrt{gh}} \] \[ F_r=\frac{1.66}{\sqrt{9.81 \cdot 3}} \] \[ F_r=0.307 \]
* Compute \(R_b\) *
\[ R_b=\frac{h}{1+\frac{0.055h}{B^2}} \] Assuming a wide channel the second fraction term of the denominater should be closer to 0
\[ R_b=\frac{3}{1}=3 \]
* Compute \(\tau_b\)
\[ \tau_b=\gamma_bR_bS \] \[ \tau_b=9810\cdot 3\cdot 0.0005 \] \[ \tau_b=14.74 \]
\[ n'=\frac{d^{1/6}_{50}}{20} \] \[ n'=\frac{0.0004^{1/6}}{20} \] \[ n'=0.0135 \]
* Compute \(\tau_b^{'}*\) by solving the following equation
\[ \frac{A_n}{g^{1/2}F_r^{1/3}}=\frac{8[1+0.0235(\tau^{'}_b/\tau_{c50})^{1.25}]}{(\tau^{'}_b/\tau_{c50})^{1/3}} \]
\[ \frac{20}{9.81^{1/2}0.307^{1/3}}=\frac{8[1+0.0235(\tau^{'}_b/0.2374)^{1.25}]}{(\tau^{'}_b/0.2374)^{1/3}} \]
\[ \tau^{'}_b=0.149 \]
* Compute n
\[ \tau^{'}_b=\Bigg(\frac{n^{'}}{n} \Bigg)^{1.5} \tau_b \]
\[ 0.149=\Bigg(\frac{0.0135}{n} \Bigg)^{1.5} 14.74 \]
\[ n=0.028 \]
* Applying Manning’s Equation to compute new value for h
\[ Q=\frac{1}{n}AR^{2/3}S^{1/2} \]
Assuming wide channel
\[ Q=\frac{1}{n}Bh h^{2/3}S^{1/2} \] \[ \frac{Q}{B}=\frac{1}{n}Bh^{5/3}S^{1/2} \] \[ q=\frac{1}{n}Bh^{5/3}S^{1/2} \]
Isolating h
\[ h=\Bigg(\frac{q\cdot n}{S^{1/2}}\Bigg)^{3/5} \]
\[ h=2.8 \]