Presentation on theme: "1B Clastic Sediments Lecture 27 SEDIMENT TRANSPORT Onset of motion"— Presentation transcript:
1 1B Clastic Sediments Lecture 27 SEDIMENT TRANSPORT Onset of motion Mode of transportEstimation of bedload transportEstimation of suspended load transportNH
2 SEDIMENT CONTINUITY For constant sediment concentration C, (qs)1 < (qs)2: erosion of channel bed.(qs)1 > (qs)2: deposition on channel bed.Bed volume change in time dt:dydx = -1/(1-l) dqsdt ,where dy is thickness eroded/aggradedis porosity of bed material.Bed volume change may also resultfrom change in sediment concentration:dydx = -1(1-l) (hdC)dx ,where h is flow depth.dydt = -1/(1-l) (dqs/dx + hdC/dt)Net change in bed elevationover time is related to down-stream change of sedimenttransport rate and the changein suspended sedimentconcentration in time.
3 TRESHOLD OF SEDIMENT MOVEMENT Stream power w = t0uis rate of work done on channel bed.Below threshold shear stress tc:no sediment motion and transport.Effective shear stress: t – tc
4 FORCES ON PARTICLE ON BED Cylinder on horizontal bedbelow inviscid fluid.No viscosity: no drag force.Stream lines converge and thendiverge over cylinder in symmetricalfashion: pressure distribution resultsin net upward force: lift force.
5 FORCES ON PARTICLE ON BED Cylinder on horizontal bedbelow inviscid fluid.No viscosity: no drag force.Stream lines converge and thendiverge over cylinder in symmetricalfashion: pressure distribution resultsin net upward force: lift force.Cylinder under viscous flow:Flow separation behind cylindercauses drag force in addition to lift.Resultant fluid pressure force FF hasupward and downstream component.
6 FORCES ON PARTICLE ON BED Resisting grain motion:Gravitational force FG; neighbouring grains.Entrainment of grain by rotation about pivot.Angle of easiest movement: a.Threshold of movement: balance of momentsabout pivot point:FG(sin a)a1 = FF(cos a)a2 ,where a1 and a2 are moment arms.FG = c1D3g’ ,where g’ = (rs – rf)g, and constantc1 accounts for flow variabilitygrain characteristics.FF = c2D2t0 ,where constant c2 accounts forgrain shape and packing.Ignore lift force: vanishes afterentrainment.
7 FORCES ON PARTICLE ON BED Threshold of movement: balance of momentsabout pivot point:FG(sin a)a1 = FF(cos a)a2 ,FG = c1D3g’ ,FD = c2D2t0 .Combine and regroup:[a1c1/a2c2 tan a] = t0/Dg’ .b ~ grain characteristicslocal flow:boundary Reynolds no.Definition of threshold conditionrelies on experimentation.
8 DIMENSIONAL ANALYSIS OF MOTION THRESHOLD Variables:Critical bed shear stress, tc [ML-1T-2] repeatFluid density, rf [ML-3] repeatFluid viscosity, m [ML-1T-1]Grain diameter, D [L] repeatSubmerged specific weight of grain, g’ [ML-2T-2]Sought: balance of inertial and viscous forces (Reynolds number),balance of gravitational and fluid forces.Combine g’ with repeating variables:t0/g’D = b (Shields’ stress).Combine m with repeating variables:(D√rf√t0)/m = (rfu*D)/m = u*D/n = Re* ,where n = m/r is the kinematic viscosity.Remember: shear velocity u*2 = t0/r .Re* is boundary Reynolds number.and Re* fully characteriseonset of sediment motion.Their relation was constrainedexperimentally by Shields.
9 SHIELDS’ DIAGRAMRe* < 10: fine grain sizes: well-packed, cohesive sediment, enclosedwithin viscous sublayer. Entrainment more difficult than fine sand.Shields’ stress b increases with decreasing Re*Re* > 10: Non-cohesive silt and sand.Entrainment more difficult with increasing grainsize.Expected: Shields’ stress b increases with increasing Re*Experimental results: flat trend.Shear stress and grain size on both axes.
10 YALIN’S DIAGRAMbIn Shields’ diagram: shear stress and grain size on both axes.Solve by combining boundary Reynolds number with Shields’ b toeliminate shear stress:F = Re*2/b .Yalin’s plot of b against √F has same general form as Shields’ curve.
11 MODES OF SEDIMENT TRANSPORT BED LOADSliding, rolling, saltationSUSPENDED LOADMode of transport depends ongrain densitygrain sizeflow hydraulicsConditions vary in space and time:Modes of transport change frequently.Distinction between bed load andsuspended load is not easy.Transport stage
12 TRANSPORT STAGE u*/w , With increasing shear velocity, where u* is the shear velocity, t0/rw is the settling velocity (cf. Stoke’s law).With increasing shear velocity,proportion of load moving in suspensionincreases.Therefore dimensionless grain velocityug/U increases with transport stage.Here, ug is the grain velocity, and U is theflow velocity.u* = w approximates saltation –suspension threshold.When u* > w, then grains move withapproximately the velocity of the flow.Results shown for quartz sand in flow 48 mm deep.
13 BEDLOAD TRANSPORT Bedload transport rate ~ stream power, w = t0u . conversion factor to be constrained empiricallyPrediction 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 [ML2T-3] over unit area of stream bed [L2]: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 ofgravity acting on the mass of water.Flow resistance: frictional energy loss during flow on bed and banks.Parameters of the problem:Flow depth h Combine two length scales in problem inflow velocity u dimensionless variable: h/ks, the relative roughness.fluid density rffluid viscosity m Express fluid flow in a Reynolds number: Re = rfuh/m .basal shear stress t0roughness height ks Make shear stress dimensionless by dividing by ru2:8t0/rfu2 = f f is the friction factor.Using t0 = rgSh, rfgh sin a = 1/8 rfu2f .Solving for u, u = √(8 gh sin a)/f .Usually written as: u = C(h sin a)0.5 , C = (8g/f)0.5Chezy 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 RH =channel cross section areawetted perimeterRH is better length scale for calculation of flow resistance.u = (RH2/3 sin a1/2)/n ,where n is Manning’s roughness coefficient.n = 1/C RH4/3 ; C = (8g/f)0.5 ; f = 8t0/rfu2Use 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 tc = bg’D .
16 SUSPENDED LOADThe distribution of suspended sediment in a flow can be treated asdiffusion problem, with high concentration at bed, and low concentrationnear surface. Mass flux Q is linearly proportional to concentration gradient:Q = -k∂C/∂y , where k is a diffusivity constant.Assuming conservation of mass,∂C/∂t = ∂Q/∂y .Combined: ∂C/∂t = ∂/∂y (k∂C/∂y).If diffusivity constant in y, then ∂C/∂t = k(∂2C/∂y2).If concentration is constant in time, then ∂C/∂t = 0, and ∂2C/∂y2 = 0.Concentration profile obtained by twice integrating for boundary conditions:C = K1y + K2It can be shown that K1 = -C0/h, where C0 is the concentration at the bed.K2 = C0 .C = C0 (1 – y/h) This profile is linear in depth.
17 Mississippi River at St. Louis SUSPENDED LOADC = C0 (1 – y/h)This profile is linear in depth.Observed suspended sediment concentrationprofiles are not linear in depth.We have ignored the settling of grains.Concentration profile reflects balance ofupward diffusion and gravitational settlingof grains.When C is constant in time, then any loss of sediment due to settling isbalanced by upward diffusion of sediment.Settling flux is wC. Upward diffusion flux Q = -k dC/dy.wC = -k dC/dy ,or dC/C = -wdy/kt , where kt = be, and e is the kinematic eddy viscosity.b ≈ 1, and e = u*[(h-y)/h]ky . Von Karman’s k = 0.4 .dC/C = whdy/[bku*(h-y)y] .Mississippi River at St. Louis
18 SUSPENDED LOAD dC/C = whdy/[bku*(h-y)y] At reference height a, C = Ca . Integrate:C/Ca = [h-y/y × a/h-a]w/bku*.This gives suspended sedimentconcentration at any depth inflow in relation to concentrationat reference depth. Only need toknow a and Ca.The grouping w/bku* is the Rouse number.Since b ≈ 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/[bku*(h-y)y] At height a close to bed, C = CaThen, integration gives:C/Ca = [h-y/y × a/h-a]w/bku*This gives suspended sedimentconcentration at any depth inflow in relation to concentrationat reference depth. Only need toknow Ca.Suspended sediment transport rate is product of the mass of suspendedsediment ms in a column of water over a unit area of bed and the depth-averaged flow velocity U at the station:Qs = msUAssumes that all sediment ismobilized from bed.