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ON WIDTH VARIATIONS IN RIVER MEANDERS Luca Solari 1 & Giovanni Seminara 2 1 Department of Civil Engineering, University of Firenze 2 Department of Environmental.

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Presentation on theme: "ON WIDTH VARIATIONS IN RIVER MEANDERS Luca Solari 1 & Giovanni Seminara 2 1 Department of Civil Engineering, University of Firenze 2 Department of Environmental."— Presentation transcript:

1 ON WIDTH VARIATIONS IN RIVER MEANDERS Luca Solari 1 & Giovanni Seminara 2 1 Department of Civil Engineering, University of Firenze 2 Department of Environmental Engineering, University of Genova October 5, 2005, Urbana, USA

2 In most models developed to study morphodynamics of river meandering, river width is usually assumed constant, a reasonable first order approximation of reality Introduction Problem model river width variations in arbitrarily curved channels Method non-linear quasi-analytical model of steady hydrodynamics and morphodynamics of an arbitrarily curved fluvial channel with dominant bed- load transport. The model is developed extending a previous work (Seminara & Solari, 1998) concerning cohesionless wide channels with constant curvature. A careful inspection of the river patterns shows that river width (width of the stream free surface) undergoes spatial oscillations correlated with channel curvature

3 Formulation of the problem Assumptions Weakly meandering channel with typical measure of radius of curvature of flow streamlines >> channel width Cohesionless bottom, dominant bedload transport Wide channel with width and channel alignment varying on a longitudinal scale >> channel width Flow and bottom topography ‘slowly varying’ in both longitudinal and transverse direction Steady conditions The system curvilinear orthogonal system (s *,n *,z * ) B * is taken to oscillate around some average values B * u with oscillation correlated with channel curvature

4 Formulation of the problem 3-D Reynolds continuity fluid phase continuity solid phase no slip at the bed kinematic condition at the free surface banks impermeable to flow and sediments dynamic condition at the free surface curvature ratio width ratio Parameters and scaling Governing equationsBoundary conditions b(  ) unknown ‘slowly varying’ function to be determined

5 Formulation of the problem turbulence in equilibrium with instantaneous and local conditions fluid field the direction of bed load deviates from the direction of bottom stress due to lateral component of gravity bedload transport Closure relationships Flow and sediment discharges must keep constant in the longitudinal direction Integral constraints

6 Analysis = O(  0 )  uniform flow in straight channels with unknown shape of bottom topography + O(  1 )  correction due to secondary flow + h.o.t. The unknown functions are expanded in powers of the small parameter  Partial differential equation systems at the various order of approximation

7 Solution F 0, G 1, G 2 are vertical distributions obtained numerically and weakly dependent on the lateral and longitudinal coordinates through the relative bed roughness. Hydrodynamics: longitudinal velocity at the leading order and lateral velocity SECONDARY FLOW in a constant curvature channel (comparison with Rozovskij’s experimental observations, 1957) Log profile

8 Solution Channel geometry at the leading order: bottom profile D 0 ( ,n), free surface width b 0 (  ) Integral constraints (constant flow and sediment discharges) are satisfied determining the functions D 0 ( ,n) and b 0 (  ) by a trial and error procedure Two boundary conditions forcing the normal component of sediment flux to vanish at the side walls

9 Case study I – constant curvature channel Flow and sediment discharges must keep constant in the longitudinal direction reach, the channel is allowed to widen on point bar side BuBu BcBc

10 Case study I – constant curvature channel Integral constraints are satisfied letting the channel to widen on the point bar in the curved reach (d 0 =20, d s =0.004;  u =0.6) cross section in the straight reachcross section in the curved reach

11 Case study I – constant curvature channel In Seminara & Solari (1998) integral constraints are satisfied letting the longitudinal slope in the curved reach differs (slightly) from value in the straight reach (d 0 =20, d s =0.004;  u =0.6)

12 Flow depth at the outer bank Case study I – constant curvature channel Allowing the channel to widen implies a reduction of the scour at the outer banks B ≠const B=const

13 Case study I – constant curvature channel Position of the inner boundary For given  u and ds, the channel widens with d 0 For given d 0 and ds, the channel narrows with  u d0d0 b0b0  u =0.5 b0b0 uu d 0 =10

14 Case study II – sinuous channel Meanders show characteristic widening of the cross section close to the bend apex on the side of the point bar  =0.4, d s =0.01; d 0 =15) L s * /B u * =15 L s * /B u * =40 Sinuous channel with a sine generated curvature distribution of channel axis

15 Conclusions A non-linear quasi-analytical model to study hydrodynamics, bottom topography and river width variations in an arbitrarily curved fluvial channel with dominant bed-load transport has been developed Free surface river width appears to oscillate along the meander, being maximum in the proximity of bend apex and minimum at the inflection points. Such results are in qualitative agreement with field evidences. The process of river widening is known to promote in straight channels the formation of steady central bars (Repetto et al., 2002) A similar mechanism may be responsible for the formation of the islands (central bars) often observed in meandering channels close to the bend apex. The latter process may promote a tendency of the stream to bifurcate into an outer and inner branch leading to the mechanism of chute cutoff.

16 LINEAR DIFFERENTIAL SYSTEM O(d2)  correction due to metric variation of bottom slope and to transverse transport of longitudinal momentum

17 Flow depth at the outer bank Case study I – constant curvature channel Allowing the channel to widen implies a reduction of the scour at the outer banks B ≠const B=const d0d0 D max  u =0.5


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