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1B Clastic Sediments Lecture 27 SEDIMENT TRANSPORT Onset of motion Mode of transport Estimation of bedload transport Estimation of suspended load transport.

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Presentation on theme: "1B Clastic Sediments Lecture 27 SEDIMENT TRANSPORT Onset of motion Mode of transport Estimation of bedload transport Estimation of suspended load transport."— Presentation transcript:

1 1B Clastic Sediments Lecture 27 SEDIMENT TRANSPORT Onset of motion Mode of transport Estimation of bedload transport Estimation of suspended load transport NH

2 SEDIMENT CONTINUITY For constant sediment concentration C, (q s ) 1 < (q s ) 2 : erosion of channel bed. (q s ) 1 > (q s ) 2 : deposition on channel bed. Bed volume change in time dt: dydx = -1/(1- ) dq s dt, where dy is thickness eroded/aggraded is porosity of bed material. Bed volume change may also result from change in sediment concentration: dydx = -1(1- ) (hdC)dx, where h is flow depth. dydt = -1/(1- ) (dq s /dx + hdC/dt) Net change in bed elevation over time is related to down- stream change of sediment transport rate and the change in suspended sediment concentration in time.

3 TRESHOLD OF SEDIMENT MOVEMENT Stream power  =  0 u is rate of work done on channel bed. Below threshold shear stress  c : no sediment motion and transport. Effective shear stress:  –  c

4 FORCES ON PARTICLE ON BED Cylinder on horizontal bed below inviscid fluid. No viscosity: no drag force. Stream lines converge and then diverge over cylinder in symmetrical fashion: pressure distribution results in net upward force: lift force.

5 FORCES ON PARTICLE ON BED Cylinder on horizontal bed below inviscid fluid. No viscosity: no drag force. Stream lines converge and then diverge over cylinder in symmetrical fashion: pressure distribution results in net upward force: lift force. Cylinder under viscous flow: Flow separation behind cylinder causes drag force in addition to lift. Resultant fluid pressure force F F has upward and downstream component.

6 FORCES ON PARTICLE ON BED Resisting grain motion: Gravitational force F G ; neighbouring grains. Entrainment of grain by rotation about pivot. Angle of easiest movement: . Threshold of movement: balance of moments about pivot point: F G (sin  )a 1 = F F (cos  )a 2, where a 1 and a 2 are moment arms. F G = c 1 D 3  ’, where  ’ = (  s –  f  g, and constant c 1 accounts for flow variability grain characteristics. F F = c 2 D 2  0, where constant c 2 accounts for grain shape and packing. Ignore lift force: vanishes after entrainment.

7 FORCES ON PARTICLE ON BED Threshold of movement: balance of moments about pivot point: F G (sin  )a 1 = F F (cos  )a 2, F G = c 1 D 3  ’, F D = c 2 D 2  0. Combine and regroup: [a 1 c 1 /a 2 c 2 tan  =  0 /D  ’.  ~ grain characteristics local flow: boundary Reynolds no. Definition of threshold condition relies on experimentation.

8 DIMENSIONAL ANALYSIS OF MOTION THRESHOLD Variables: Critical bed shear stress,  c [ML -1 T -2 ]repeat Fluid density,  f [ML -3 ]repeat Fluid viscosity,  [ML -1 T -1 ] Grain diameter, D[L]repeat Submerged specific weight of grain,  ’[ML -2 T -2 ] Sought:balance of inertial and viscous forces (Reynolds number), balance of gravitational and fluid forces. Combine  ’ with repeating variables:  0 /  ’D =  (Shields’ stress). Combine  with repeating variables: (D√  f √  0 )/  = (  f u * D)/  = u * D/ = Re *, where  is the kinematic viscosity. Remember: shear velocity u * 2 =  0 / . Re * is boundary Reynolds number.  and Re * fully characterise onset of sediment motion. Their relation was constrained experimentally by Shields.

9 SHIELDS’ DIAGRAM Re * < 10: fine grain sizes: well-packed, cohesive sediment, enclosed within viscous sublayer. Entrainment more difficult than fine sand. Shields’ stress  increases with decreasing Re * Re * > 10: Non-cohesive silt and sand. Entrainment more difficult with increasing grainsize. Expected: Shields’ stress  increases with increasing Re * Experimental results: flat trend. Shear stress and grain size on both axes.

10 YALIN’S DIAGRAM  In Shields’ diagram: shear stress and grain size on both axes. Solve by combining boundary Reynolds number with Shields’  to eliminate shear stress:  = Re * 2 / . Yalin’s plot of  against √  has same general form as Shields’ curve.

11 MODES OF SEDIMENT TRANSPORT BED LOAD Sliding, rolling, saltation SUSPENDED LOAD Mode of transport depends on grain density grain size flow hydraulics Conditions vary in space and time: Modes of transport change frequently. Distinction between bed load and suspended load is not easy. Transport stage

12 TRANSPORT STAGE u * /w, where u * is the shear velocity,  0 /  w is the settling velocity (cf. Stoke’s law). With increasing shear velocity, proportion of load moving in suspension increases. Therefore dimensionless grain velocity u g /U increases with transport stage. Here, u g is the grain velocity, and U is the flow velocity. u * = w approximates saltation – suspension threshold. When u * > w, then grains move with approximately the velocity of the flow. Results shown for quartz sand in flow 48 mm deep.

13 BEDLOAD TRANSPORT Bedload transport rate ~ stream power,  0 u. conversion factor to be constrained empirically Prediction of bedload transport complicated by: bed armouring and consolidation of gravels. resistance of bedforms in sand and gravel rivers. lack of constraints on threshold of sediment motion. unsteadiness in high stage flows.  has dimensions [ML 2 T -3 ] over unit area of stream bed [L 2 ]: Proportional to the cube of (excess) flow velocity. Distribution of flow velocity in open channels.

14 FLOW RESISTANCE IN BEDLOAD RIVERS Force driving flow down inclined plane: downslope component of gravity acting on the mass of water. Flow resistance: frictional energy loss during flow on bed and banks. Parameters of the problem: Flow depth hCombine two length scales in problem in flow velocity udimensionless variable: h/k s, the relative roughness. fluid density  f fluid viscosity  Express fluid flow in a Reynolds number: Re =  f uh/ . basal shear stress  0 roughness height k s Make shear stress dimensionless by dividing by  u 2 : 8  0 /  f u 2 = ff is the friction factor. Using  0 =  gSh,  f gh sin  = 1/8  f u 2 f. Solving for u, u = √(8 gh sin  )/f. Usually written as:u = C(h sin  ) 0.5, C = (8g/f) 0.5 Chezy coefficient. Flow depth h is inadequate length scale.

15 FLOW RESISTANCE IN BEDLOAD RIVERS Only bed and banks exert friction. Together they are termed: Wetted perimeter. Hydraulic radius R H = channel cross section area wetted perimeter R H is better length scale for calculation of flow resistance. u = (R H 2/3 sin  1/2 )/n, where n is Manning’s roughness coefficient. n = 1/C R H 4/3 ; C = (8g/f)0.5 ; f = 8  0 /  f u 2 Use to estimate formative discharges in (paleo) channels: Measure channel slope from exposed geometry or terrace form. Largest clasts are assumed to represent maximum discharge conditions. Critical bed shear stress is estimated from particle size using  c =  ’D.

16 SUSPENDED LOAD The distribution of suspended sediment in a flow can be treated as diffusion problem, with high concentration at bed, and low concentration near surface. Mass flux Q is linearly proportional to concentration gradient: Q = -  ∂C/∂y, where  is a diffusivity constant. Assuming conservation of mass, ∂C/∂t = ∂Q/∂y. Combined: ∂C/∂t = ∂/∂y (  ∂C/∂y). If diffusivity constant in y, then ∂C/∂t =  (∂ 2 C/∂y 2 ). If concentration is constant in time, then ∂C/∂t = 0, and ∂ 2 C/∂y 2 = 0. Concentration profile obtained by twice integrating for boundary conditions: C = K 1 y + K 2 It can be shown that K 1 = -C 0 /h, where C 0 is the concentration at the bed. K 2 = C 0. C = C 0 (1 – y/h)This profile is linear in depth.

17 SUSPENDED LOAD Mississippi River at St. Louis C = C 0 (1 – y/h) This profile is linear in depth. Observed suspended sediment concentration profiles are not linear in depth. We have ignored the settling of grains. Concentration profile reflects balance of upward diffusion and gravitational settling of grains. When C is constant in time, then any loss of sediment due to settling is balanced by upward diffusion of sediment. Settling flux is wC. Upward diffusion flux Q = -  dC/dy. wC = -  dC/dy, ordC/C = -wdy/  t, where  t = , and  is the kinematic eddy viscosity.  ≈ 1, and  = u * [(h-y)/h]ky. Von Karman’s k = 0.4. dC/C = whdy/[  ku * (h-y)y].

18 SUSPENDED LOAD dC/C = whdy/[  ku * (h-y)y] At reference height a, C = C a. Integrate: C/C a = [h-y/y × a/h-a] w/  ku*. This gives suspended sediment concentration at any depth in flow in relation to concentration at reference depth. Only need to know a and C a. The grouping w/  ku * is the Rouse number. Since  ≈ 1, and k = 0.4, w = u * for a Rouse number of 2.5. This is the criterion for suspension. Rouse number > 2.5: w > u * with bedload transport dominant. High bed roughness → high shear velocity → high suspended sediment conc. High viscosity → low settling velocity → high suspended sediment conc.

19 SUSPENDED LOAD dC/C = whdy/[  ku * (h-y)y] At height a close to bed, C = C a Then, integration gives: C/C a = [h-y/y × a/h-a] w/  ku* This gives suspended sediment concentration at any depth in flow in relation to concentration at reference depth. Only need to know C a. Suspended sediment transport rate is product of the mass of suspended sediment m s in a column of water over a unit area of bed and the depth- averaged flow velocity U at the station: Q s = m s U Assumes that all sediment is mobilized from bed.


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