Comprehensive Exam - Professor Wu Questions
By: Cássio Rampinelli
Date: May 20, 2020
_____________________________________________________________________
QUESTION 1:
a) What are bed load and suspended load?
SOLUTION
______________________________________________________________________________
Suspended load consist of sediment that is supported by the upward components of turbulent currents and stays in suspension for a significant length of time. They represent the mainly portion of total load in most natural rivers (Yang, 1996), and are moving in the water column above the bed rarely in contact with the bed.
Bed load is made up of particles that are moving by rolling, sliding, or jumping along the bed. It usually accounts for 5-25% of the total load for finer material, and more for coarser particles in natural rivers (Wu, 2007). They might be in either continuous or intermittent contact with the bed.
Therefore, both suspended load and bed load are terminologies used to classify sediment in motion according to its transport mechanism. The total load is the sum of the bed load and suspended load.
b) What are wash load and bed-material load? How to differentiate them?
SOLUTION
______________________________________________________________________________
Another way of classifying sediment load is by the source of the sediment. Bed-material load is comprised of particles that are found most in the channel bed. Wash load encompasses sediment particles that are derived from sources other than the bed. They are finer than the bed-material load and they are not found in appreciable quantities in the bed. Both bed-material, and wash load added also result in the total load. Bed-material load is constantly exchanging with particles in the bed and has significant contribution to the channel morphology (Wu, 2007).
Differentiating bed-material load and wash load might be subjective either on the theoretical or on the practical perspectives. From the theoretical point of view, it should be noted that the definition of wash load, for instance, relies on what is considered not found in “appreciable quantities” in the bed. The same is true for the bed-material load.
Einstein (1950) defined wash load as the grain size of which 10 percent of the bed mixture is finer. However, there is no theoretical justification for selecting \(D_{10}\) as threshold rather than some other percentile at the lower end of bed material size. Other authors, for example, have defined wash load as consisting of particles smaller than 0.063 mm (Yang and Simões, 2005; Knighton, 1998; Richards, 1982), as discussed in Biedenharn and Thorne (2006).
From a practical perspective, depending on flow and sediment conditions wash load in upstream channels may become bed-material load in downstream channels due to the weakening of flow strength, analogously, some particles that are wash load in the main channel might become bed-material load in flood plans (Wu, 2007).
c) The 3-D transport equation of sediment is given as below:
\[ \frac{\partial c}{\partial t}+\frac{\partial (uc)}{\partial x}+\frac{\partial (vc)}{\partial y}+\frac{\partial (wc)}{\partial z}-\frac{\partial (\omega_sc)}{\partial z}=\frac{\partial }{\partial x}\Bigg( \varepsilon\frac{\partial c}{\partial x}\Bigg)+\frac{\partial }{\partial y}\Bigg(\varepsilon\frac{\partial c}{\partial y}\Bigg)+\frac{\partial }{\partial z}\Bigg(\varepsilon\frac{\partial c}{\partial z}\Bigg) \]
where \(t\) is the time; \(x\), \(y\) and \(z\) are the longitudinal, transverse and vertical coordinates; \(c\) is the concentration of sediment; \(u\), \(v\) and \(w\) are the velocity components in \(x\), \(y\) and \(z\) directions;\(w_s\) is the settling velocity of sediment; and \(\varepsilon\) is the diffusivity. Derive the Rouse distribution of suspended-load concentration under a steady, uniform flow and equilibrium transport conditions. Calculate the total-load sediment transport rate by using the derived sediment concentration profile. Explain what additional empirical formulas need to be used.
SOLUTION
______________________________________________________________________________
Assuming steady uniform flow in a wide rectangular channel without any phase change, the settling of sediment particles due to the density difference between the particles and the surrounding fluid induces a downward particle settling flux. Therefore, all the terms of the 3D transport equation vanish except those describing the sediment fluxes in the vertical \(z\) direction. The equation is then simplified by:
\[ -\frac{\partial (\omega_sc)}{\partial z}=\frac{\partial }{\partial z}\Bigg(\varepsilon\frac{\partial c}{\partial z}\Bigg) \\\\\\\\(Eq.1) \]
Equilibrium is obtained when the upward turbulent flux is balanced by the downward settling flux. Assuming the net vertical velocity is zero, Eq.1 can be further simplified by: \[ \omega_sc+\varepsilon\frac{\partial c}{\partial z}=0 \\\\\\\\(Eq.2) \]
The diffusion coefficient \(\varepsilon\) is often assumed to be proportional to the eddy viscosity of turbulent flow. The parabolic distribution of \(\varepsilon\) can be written as:
\[ \varepsilon=\frac{1}{\sigma_s}ku_*z\Bigg(1-\frac{z}{h}\Bigg)\\\\\\\\(Eq.3) \] in which \(k\) is the Von Kármán constant, \(u_*\) the shear velocity, \(h\) the flow depth, and \(\sigma_s\) the Schmidt number.
Substituting Eq.3 in Eq.2 brings to:
\[ \omega_sc+\frac{1}{\sigma_s}ku_*z\Bigg(1-\frac{z}{h}\Bigg)\frac{\partial c}{\partial z}=0 \]
Dividing both sides by \(\omega_s\):
\[ c+\frac{1}{\sigma_s\omega_s}ku_*z\Bigg(1-\frac{z}{h}\Bigg)\frac{\partial c}{\partial z}=0 \]
Passing all terms related to \(c\) to the right hand side of the equation:
\[ \frac{1}{\sigma_s\omega_s}ku_*z\Bigg(1-\frac{z}{h}\Bigg)\frac{\partial c}{\partial z}=-c \]
\[ \frac{1}{\sigma_s\omega_s}ku_*z\Bigg(1-\frac{z}{h}\Bigg)\frac{1}{\partial z}=-\frac{c}{\partial c} \]
Inverting both sides of the equation:
\[ \frac{\sigma_s\omega_s}{ku_*}\frac{\partial z}{z\Bigg(1-\frac{z}{h}\Bigg)}=-\frac{c}{\partial c} \]
Integrating both sides of the equation:
\[ \int \frac{\sigma_s\omega_s}{ku_*}\frac{\partial z}{z\Bigg(1-\frac{z}{h}\Bigg)}=-\int\frac{c}{\partial c} \]
\[ \frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(z)-ln(z-h)+k_1\Bigg]=-ln(c)+k_2 \]
Aggregating the constants:
\[ \frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(z)-ln(z-h)\Bigg]+k_1\frac{\sigma_s\omega_s}{ku_*}=-ln(c)+k_2 \] \[ \frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(z)-ln(z-h)\Bigg]=-ln(c)+k_2-k_1\frac{\sigma_s\omega_s}{ku_*} \]
\[ \frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(z)-ln(z-h)\Bigg]=-ln(c)+K \]
Isolating K, and re-arranging the terms:
\[ K=\frac{\sigma_s\omega_s}{ku_*}\Bigg[ln(z)-ln(z-h)\Bigg]+ln(c) \] \[ K=\frac{\sigma_s\omega_s}{ku_*}\cdot ln\Bigg[\frac{z}{z-h}\Bigg]+ln(c) \]
\[ K= ln\Bigg[\frac{z}{z-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c) \\\\(Eq.4) \] We are interested in obtaining the concentration profile along the vertical axis \(z\), as indicated in Figure 1. Since bed load and suspended load behave differently, the water column is often divided into two zones: a bed-load zone, from the bed elevation \(z=0\) to \(\delta_b\), and a suspended-load zone from \(\delta_b\) to \(h\) (Wu, 2007). Therefore, \(\delta_b\) is the thickness of the bed-load zone.
Figure 1 Suspended load oncentration profile sketch, adapted from Wu (2007)
To compute constant K in Eq. 4 we first set \(\ z=\delta_b\), and \(c=c_b\) which results in:
\[ K= ln\Bigg[\frac{\delta_b}{\delta_b-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c_b) \]
Now we may substitute the value obtained for K in Eq.4:
\[ ln\Bigg[\frac{\delta_b}{\delta_b-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c_b)= ln\Bigg[\frac{z}{z-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}+ln(c) \] Bringing all terms related to concentration to the left hand side of the equation: \[ ln(c)-ln(c_b)=ln\Bigg[\frac{\delta_b}{\delta_b-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}-ln\Bigg[\frac{z}{z-h}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\frac{\delta_b}{\delta_b-h}}{\frac{z}{z-h}}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_b}{\delta_b-h}\frac{z-h}{z}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_bz-\delta_bh}{\delta_bz-zh}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\delta_bz(1-\frac{h}{z})}{\delta_bz(1-\frac{h}{\delta_b})}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
Multiplying numerator and denominator by (-1) \[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{(-1)(1-\frac{h}{z})}{(-1)(1-\frac{h}{\delta_b})}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
\[ ln\Bigg[\frac{c}{c_b}\Bigg]=ln\Bigg[\frac{\frac{h}{z}-1}{\frac{h}{\delta_b}-1}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \]
Finally, we get the Rouse equation
\[ \frac{c}{c_b}=\Bigg[\frac{\frac{h}{z}-1}{\frac{h}{\delta_b}-1}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \] also presented as: \[ \frac{c}{c_b}=\Bigg[\frac{(h-z)}{z}\frac{\delta_b}{(h-\delta_b)}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \] Hence, the concentration \(c\) may be expressed as a function of \(z\), as follows:
\[ c(z)=c_b\Bigg[\frac{(h-z)}{z}\frac{\delta_b}{(h-\delta_b)}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}} \\\\(Eq.5) \]
The total bed sediment discharge per unit width \(q_t\) can be calculated from the sum of the unit bed discharge \(q_b\) and the unit suspended sediment discharge \(q_s\), as follows:
\[ q_t=q_b+q_s \\\ (Eq.6) \] The unit suspended sediment discharge \(q_s\) in natural streams and canals is computed from the depth-integrated advective flux of sediment given by the \(c(z)\cdot v_x\), over the bed layer \(z>\delta_b\). Then, Eq. 6 may be updated as:
\[ q_t=q_b+\int^h_{\delta_b}c(z)v_xdz \] Substituting \(c(z)\) by Eq.5 \[ q_t=q_b+\int^h_{\delta_b}c_b\Bigg[\frac{(h-z)}{z}\frac{\delta_b}{(h-\delta_b)}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}v_xdz \\\\ (Eq.6) \]
As shown in Julien (2010), Nikuradse glued sand particles and measured velocity profiles for turbulent flow over boundaries with grain roughness height \(k^`\), resulting in the following equation for \(v_x\)
\[ v_x=\frac{u_*}{k}ln\Bigg(\frac{30z}{k^`}\Bigg) \\\ (Eq.7) \] Substituting (Eq.7) in (Eq.6): \[ q_t=q_b+\int^h_{\delta_b}c_b\Bigg[\frac{(h-z)}{z}\frac{\delta_b}{(h-\delta_b)}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}\frac{u_*}{k}ln\Bigg(\frac{30z}{k^`}\Bigg)dz \]
Assuming \(k`=d\), and rearranging:
\[ q_t=q_b+\int^h_{\delta_b}\frac{c_bu_*}{k}\Bigg[\frac{(h-z)}{z}\frac{\delta_b}{(h-\delta_b)}\Bigg]^{\frac{\sigma_s\omega_s}{ku_*}}ln\Bigg(\frac{30z}{k^`}\Bigg)dz \\\ (Eq.8) \] The reference concentration \(c_b=u_*/kv_c\) is obtained from the unit bed sediment discharge \(q_b\) transported in the bed layer of thickness \(\delta_b=2d\), given the velocity \(v_c\) at the top of the bed layer, \(v_c=(u_*/k)ln(30\delta_b/d)=4.09u_*/k\), Einstein (1950) used \(v_a=11.6u_*\). Now, rewriting Eq.8 in dimensionless form with \(z^*=z/h\) gives:
\[ q_t=q_b\Bigg[1+0.216\frac{\Big(\frac{2d}{h}\Big)^{(\frac{\sigma_s\omega_s}{ku_*}-1)}}{\Big(1-\frac{2d}{h}\Big)^\frac{\sigma_s\omega_s}{ku_*}}\int^1_{2d/h}\Big[\frac{1-z^*}{z^*}\Big]^{\frac{\sigma_s\omega_s}{ku_*}}dz^*\cdot ln\Big(\frac{30h}{d}\Big)+0.216\frac{\Big(\frac{2d}{h}\Big)^{(\frac{\sigma_s\omega_s}{ku_*}-1)}}{\Big(1-\frac{2d}{h}\Big)^\frac{\sigma_s\omega_s}{ku_*}}\int^1_{2d/h}\Big[\frac{1-z^*}{z^*}\Big]^{\frac{\sigma_s\omega_s}{ku_*}}ln(z^*)dz^*\Bigg] \]
This expression is reduced by naming part of terms with integrals as \(I_1\), and \(I_2\):
\[ q_t=q_b\Bigg[1+I_1\cdot ln\Big(\frac{30h}{d}\Big)+I_2\Bigg] \\\ (Eq.9) \] Eq.9 is the total bed load Einstein equation. The two integrals related to \(I_1\), and \(I_2\) terms have been solved with the use of nomographs prepared by Einstein (1950). The procedure to solve the equation for each size \(d\) of the considered bed material may be found in Einstein (1950), or in the Appendix A in Julien (2010).
QUESTION 2: An engineer is designing an open channel to divert flow from one small lake to another. The channel will have a \(10 \ m\) wide rectangular cross-section and the bed slope will be \(0.005\). The channel bed will be covered with uniform sediment having a diameter of \(20 \ mm\) and a density of \(2,650 \ kg/m^3\). The kinematic viscosity of water can be taken as \(10^{-6} \ m^2/s\). Assume that the channel is sufficiently long to allow the development of uniform flow conditions.
a) Using the Shields diagram, determine the discharge for which the incipient motion will take place. b) Calculate the uniform flow depth for the incipient discharge conditions.
SOLUTION
______________________________________________________________________________
The threshold relation between critical shear stress and critical Shields number for incipient motion is given by:
\[ \frac{\tau_c}{(\gamma_s-\gamma)d}=f\Bigg(\frac{u_{*c}d}{\nu}\Bigg) \] The dimensionless parameter at the left hand side of the equation is called the critical Shields number \(\Theta_c\), and the argument of the function in the right hand side is named particle shear Reynolds number \(Re_{*c}\). The relation
\[ \Theta_c=f\Bigg(Re_{*c}\Bigg) \] is empirically established by the Shields diagram for incipient motion.
Figure 2 Shields diagram adapted from Vanoni (1964), and Wu (2007)
Before using the diagram, we start by computing the channel shear stress \(\tau_0\) that is required to mobilize the bed:
\[ \tau_0= \gamma \cdot R \cdot S_e \]
Where
\(\tau_0\)= the cross-sectional average shear stress at the boundary;
\(R\)= the hydraulic radius (flow area divided by wetted perimeter); and
\(S_e\)= the friction slope. (Since we are in the steady uniform flow regime, the friction slope \(S_e\) is equivalent to the mean channel bed slope, \(S_0\)).
Therefore, by substituting the given variables in the channel shear stress equation yields:
\[ \tau_0= \gamma \cdot \frac{Area}{Perimeter} \cdot S_0 \] \[ \tau_0= 9810 \cdot \frac{10h}{2h+10} \cdot 0.005 \] \[ \tau_0=\frac{490.5h}{2h+10} \]
Where \(h\) is the flow depth associated with a discharge \(Q\). At the critical bed shear stress, at the incipient motion, \(\tau_0\) becomes equal to \(\tau_c\), and the associated \(h\), and \(Q\) will be named \(h_c\), and \(Q_c\), the variables that should be determined. Thus, one can re-write the critical Shields number \(\Theta_c\) as:
\[ \Theta_c=\frac{\tau_c}{(\gamma_s-\gamma)d}=\frac{\tau_0}{(\gamma_s-\gamma)d}=\frac{\frac{490.5h}{2h+10}}{((9.81\cdot2,650 )-9,810)\cdot 0.02}=\frac{\frac{490.5h_c}{2h_c+10}}{323.73} \] Soulsby (1997) proposed the following formula to approximate the original Shields diagram:
\[ \Theta_c=\frac{0.24}{D_*}+0.055(1-e^{-D_*/50}) \]
where
\[ D_{*}=\Bigg(\Bigg(\frac{\rho_s}{\rho}-1\Bigg)\cdot \frac{g\cdot d^3}{\nu^2}\Bigg)^{1/3} \] \[ D_{*}=\Bigg(\Bigg(\frac{2650}{1000}-1\Bigg)\cdot \frac{9.81\cdot 0.02^3}{(10^{-6})^2}\Bigg)^{1/3} \] \[ D_{*}=505.919 \]
Substituting \(D_{*}\) in Soulsby (1997) approximation for the Shields diagram:
\[ \Theta_c=\frac{0.24}{505.919}+0.055(1-e^{-505.919/50}) \]
\[ \Theta_c=0.05547217 \]
Recovering the relation between the Shields number and the shields stress:
\[ \Theta_c=0.05547217=\frac{\frac{490.5h_c}{2h_c+10}}{323.73} \] Solving for \(h_c\) to obtain:
\[ 0.05547217=\frac{\frac{490.5h_c}{2h_c+10}}{323.73} \] \[ 17.95801=\frac{490.5h_c}{2h_c+10} \]
\[ 490.5h_c=35.97602h_c+179.5801 \] \[ 454.524h_c=179.5801 \] \[ h_c=0.395 \ m \] This is the critical flow depth to incipient motion.
To obtain the associated critical discharge \(Q_c\), one should proceed by applying the Manning equation.
\[ Q_c= \frac{1}{n} \cdot Area \cdot R^{2/3} \cdot S_e^{1/2} \] with
\(S_e=S_0=0.005\) for uniform flow
\(n=((d_{50})^{1/6})/21.1\) is the Manning’s roughness coefficient based on the relation proposed by the Strickler coeeficient to the mean sediment size. In this case, we assume \(d_{50}=0.02 m\) for uniform sediment bed.
\(Area=\) flow area = \(10 \cdot h_c\)
\(R=\) hydraulic radius = \(\frac{10h_c}{2h_c+10}\)
Thus,
\[ Q_c= \frac{1}{((0.02)^{1/6})/21.1} \cdot 10 \cdot 0.395 \cdot \Bigg(\frac{10 \cdot 0.395 }{2\cdot 0.395+10}\Bigg)^{2/3} \cdot 0.005^{1/2} \] \[ Q_c=5.79 \ m^3/s \]
This is the critical flow to incipient motion.
c) What is the shear velocity under incipient flow conditions?
SOLUTION
______________________________________________________________________________
The shear velocity \(u_*\) is given by:
\[ u_*= \sqrt{\frac{\tau_0}{ \rho }} \]
For the critical shear stress:
\[ u_{*c}= \sqrt{\frac{\tau_c}{ \rho }} \]
\[ u_{*c}= \sqrt{\frac{ \tau_c=\frac{490.5h_c}{2h_c+10} }{ 1000 }} \] \[ u_{*c}= \sqrt{\frac{ \frac{490.5\cdot 0.395}{2\cdot 0.395+10} }{ 1000 }} \] \[ u_{*c}= 0.134 \ m/s \]
REFERENCES
Biedenharn, D.S, and Thorne, C.R. (2006). Wash Load/Bed Material Load Concept in Regional Sediment Management Proceedings of the Eighth Federal interagency Sedimentation Conference (8thFISC). Reno, NV, USA.
Einstein, H.A. (1950). The bed-load function for sediment transportation in open-channel flows. U.S. Department of Agriculture, Soil Conservation Service, Technical Bulletin No. 1026.
Knighton, A.D. (1998). Fluvial forms and processes: A new perspective. Arnold, London, 383 p.
Wu , W. (2007). Computational River Dynamics Taylor & Francis, London. ISBN:978-0-415-44960-I.
Julien, P.Y. (2010). Erosion and Sedimentation. Second Edition. Cambridge, UK. ISBN 978-0-521-53737-7.
Richards, K.S. (1982) Rivers: Form and process in alluvial channels Methuen, London, UK, 361 p.
Soulsby, R. (1997). Dynamics of Marine Sands: A Manual for Practical Applications. Thomas Telford Publications, London, UK.
Vanoni, V.A. (1964). Measurements of critical shear stress. Rep. No.KH-R-7, California Institute of Technology, USA.
Yang, C.T. (1996). Sediment Transport: Theory and Practice McGraw-Hill, New York. ISBN:1-57524-226-5.
Yang, C.T., and Simões, F.J.M. (2005). Wash load and bed-material load transport in the Yellow River. J. Hydraulic Engineering. Vol. 131, No. 5, pp. 413-418.