1 Chapter 1 Sediment transport and river bed evolution 1.1 What is the sediment transport? What is the river bed evolution? System of the interaction between flow and river beds Rivers transport a variety of matters as well as water. One of the most important matters transported in rivers is sediment from an engineering point of view. A river bed determines flow in the river as a boundary while the flow causes bed evolution through sediment transport. Conceptual diagram of a system of interaction between flow and river beds. Modes of sediment transport Sediment transport Bed material load Wash load { Bedload Suspended load Bedload: Sediment transported by rolling, sliding, and saltation near the bed. Composed of relatively large sediment. Suspended load: Sediment transported in suspension in flow. Composed of relatively small sediment. Washload: Sediment transported in suspension in flow, which does not deposit on the bed. Rather small sediment with the diameter smaller than 0.2 mm.
2 Chapter 1 Sediment transport and river bed evolution River bed evolution and river channel shifting River bed evolution often causes shifting of river channels. The shifting of river channels results in the formation of fluvial fans and river mouth deltas. River bed evolution and river channel shifting are unavoidable natural processes. However, river channel shifting puts human beings at a disadvantage. Therefore, it is necessary to control river morphology by understanding physics of river bed evolution and river channel shifting. 1.2 Bed shear stress agent bringing about sediment transport Bed shear stress expressed by the flow depth If flow is in an equilibrium state (normal flow condition), the streamwise component of the gravity force is balanced with the shear stress. Therefore, the shear stress τ is a linear function in the depth direction, such that τ = ρg(h z)s (1.1) H S H - z z τdx dx ρg(h - z)sdx where H is the flow depth, z is the coordinate in the depth direction the origin of which is taken at the bed, and S is the bed slope. The bed shear stress right at the bed is τ b = ρghs (1.2) τ τ b Therefore, the bed shear stress can be obtained if the flow depth and the bed slope are known. It should be noted that this relation is not valid if flow is not in the normal flow condition. Bed shear stress expressed by the average velocity Based on the mixing length hypothesis and experimental results, it is known that the velocity distribution in the depth direction can be described by the following logarithmic function: u u = 2.5 ln z + 8.5 = 2.5 ln 30z (1.3) Here u is the friction velocity, and is the roughness height, which is known to be 1 to 3 times the bed material. According to the logarithmic velocity distribution, the velocity right at the origin is (u as x 0). The logarithmic law is H u
1.3 Threshold of sediment motion 3 not valid in the vicinity of the origin. Integrating the velocity distribution from z = to H, we obtain Because u = τ b /ρ, we obtain U = 2.5 ln H + 6 = 2.5 ln 11H (1.4) u τ b = ρc f U 2, C f = [ 2.5 ln 11H ] 2 (1.5) where C f is sometimes called Keulegan s friction coefficient. The above equation gives an expression of the bed shear stress by the average velocity. The bed shear stress can be evaluated from the average velocity, flow depth, and sediment diameter even if the flow is not in the normal flow condition. Otherwise, the following Manning s resistant law is commonly used: τ b = ρ gn2 R 1/3 U2 (1.6) Here R is the hydraulic radius, which is nearly equal to the flow depth if the channel width is sufficiently larger than the flow depth. In addition, n is Manning s roughness coefficient, which takes values between 0.01 and 0.04. Manning s roughness coefficient is determined from observations of discharge (water surface elevation and velocity) during floods in practice. Modifying the above equation, and applying to normal flow, we obtain well-known Manning s normal flow formula of the form U = 1 n R2/3 S 1/2 (1.7) an approximation of Keulegan s friction coefficient by a power function is the following Manning- Strickler formula: ( ) 1/6 U H = 7.66 (1.8) u From (1.4) (1.8), we can obtain the relation between Manning s roughness coefficient n and Keulegan s friction coefficient C f. 1.3 Threshold of sediment motion Forces exerted on sediment on the bed The four main forces exerted on sediment on the bed are the following: F r L f D f z D f : drag force of flow. Force moving sediment in the flow direction. L f : lift force of flow. Force moving sediment in the upper depth direction. F g
4 Chapter 1 Sediment transport and river bed evolution F r : friction force of the bed. Resistant force preventing sediment from moving in the streamwise direction. F g : gravity force Here the upper limit of the bed friction force F r is expressed by the Coulomb friction factor µ in the form F r = µ ( ) F g L f (1.9) If the bed shear stress is small, D f and L f are both small, so that D f cannot exceed the maximum value of friction force described above, and sediment is kept still. Meanwhile, if the bed shear stress exceeds the friction force and D f > F r, sediment starts to move. The drag force D f and the lift force L f are known to be functions of the non-dimensional bed shear stress θ (= τ/ρrgd, Shields stress) and the particle Reynolds number R p (= u d/ν). Here R is the submerged specific gravity (= ρ s /ρ 1), ρ s is the density of sediment (ρ s = 2650 kg/m 3 and R = 1.65 in the case of normal sand), d is the sand diameter, and ν is the kinematic viscosity. The critical Shields number θ c is then expressed in the form θ c = f ( ) R p (1.10) A diagram showing experimental data in the R p -θ c plane is called a Shields diagram. θ c R p Once the sediment size is given, the critical Shields number can be obtained by the use of the Shields diagram, and the critical bed shear stress can be evaluated. It should be noted, however, that θ is related to u included in R p in the following equation: θ c = u2 c Rgd (1.11) We obtain R p by guessing θ c and u c, estimate θ c by the use of the Shields diagram, calculate R p again with the use of the obtained θ c, and adjust the guessed value of θ c. We repeat this procedure several times. We can finally obtain a correct value of θ c. In order to avoid this troublesome task, we introduce a particle Reynolds number newly defined by R ep = Rgdd/ν (1.12) Equation (1.10) is written in the form θ c = f ( R ep ) (1.13)
1.3 Threshold of sediment motion 5 Brownlie (1981) proposed the following equation as an explicit form obtained by the interpolation of experimental data: θ c = 0.22Rep 0.6 + 0.06 10 7.7R 0.6 e p (1.14) As a further simple equation in which the critical shear velocity u c (cm) can be obtained directly from the sediment diameter, we have Iwagaki s formula in the form u 2 c = 80.9d (0.303 cm d) 134.6d 31/22 (0.118 cm d 0.303 cm) 55.0d (0.0565 cm d 0.118 cm) 8.41d 11/32 (0.0065 cm d 0.0565 cm) 226d (d 0.0065 cm) Because this formula has dimensions, it is necessary to be careful about dimensions when used. (1.15) Assignment 1 One of the most inexpensive methods for protect river banks from bank erosion is to cover river banks with rocks (called riprap) larger than sediment on the river bed. In the bank region, the bed shear stress is maximized at the lower end of the bank (point J in the figure below). If rocks at the point J move, whole riprap collapses. Therefore, it is important to design riprap for rocks at the point J not to move. In this reach, the channel width is 30 m, the bed slope is 0.01, the diameter of bed material is 2 cm, and the designed flow depth is 1.5 m. Estimate the discharge in the following procedure, and design the riprap in this reach. Bank Bank Flow depth J J 1) With the use of the designed flow depth, find C f from the following equation: C f = ( 2.5 ln 11H ) 2 Assume that the roughness height is 2.5 times the diameter of bed material. 2) Assuming that flow is in the uniform normal flow condition, find the average velocity U from the relation ρghs = ρc f U 2. Calculate the discharge by multiplying the average velocity by the flow depth and width. You can ignore the effect of side banks, and assume that the cross-section of the channel is a wide rectangle with the width of 30 m. 3) Find the bed shear stress and the shear velocity on the bed. 4) How large rocks should be used as riprap? Find the diameter of rocks with the use of Brownlie s equation. You can assume that the bed shear stress obtained in 3 is exerted on rocks at the lower end of the bank (point J in the figure).