Sediment and Erosion Design Guide

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1 Sediment and Erosion Design Guide

2 Sediment Transport & Bulking Factors Goals of this Session Review key principals Review basic relationships and available tools Review bulking factor relationships

3 Purposes of Sediment Transport Analysis Quantify capacity of the channel to transport sediment over range of anticipated flows Identify aggradation/degradation tendencies Quantify effect of transported sediment on flow volume

4 Modes of Sediment-load Transport

5 Bed Load Processes

6 Suspended Load vs Bed Load Particle size particle fall velocity (w) Hydraulic energy shear velocity (u * ) Particle motion: u * /u *c >1 Suspension: u * >w

Data Requirements Hydraulic conditions Sediment characteristics Vel Chnl (ft/s) 5.0 4.5 4.0 3.5 3.0 2.

7 Data Requirements Hydraulic conditions Sediment characteristics Vel Chnl (ft/s) DonFelipeDam Plan: Existing Geom Future Flows 6/28/2007 PajaritoNorth Main Legend Vel Chnl 100yearbulked Vel Chnl 100 year Vel Chnl 50 year Vel Chnl 25 year Vel Chnl 10 year Vel Chnl 5 year Vel Chnl 2 year Main Channel Distance (ft)

Bed Material Sampling Maximum Particle Size Minimum Weight of Sample g lb 3-inch 6,000 13

8 Bed Material Sampling Maximum Particle Size Minimum Weight of Sample g lb 3-inch 6, inch 4, inch 2, /2-inch 1,000 2 < No. 4 sieve < No. 10 sieve

9 Jan 2008 Sample Locations

MONTOYAS ARROYO 100 80 Percent Finer 60 40 20 S6, D 50 =1.93mm S7, D 50 =1.

10 MONTOYAS ARROYO Percent Finer S6, D 50 =1.93mm S7, D 50 =1.13mm S8, D 50 =0.81mm S9, D 50 =0.39mm S10, D 50 =0.46mm S11, D 50 =0.96mm S12, D 50 =0.29mm 0 BOULDERS COBBLES Grain Size in Millimeters 0.1 GRAVEL SAND VC C M F VF VC C M F VF SILT or CLAY 0.01

11 Bed Material Size Trends in SSCAFCA Area Upstream to downstream order by arroyo Calabacillas Montoyas Barranca Venada D84 D16 D50 S5 S4 S3 S2 S1 S6 S7 S8 S9 S10 S11 S12 S14 S13 S18 S17 S16 S15 S24 S19 S20 S21 S22 S23 Particle Size (mm) Lomitas Negras

12 Incipient Motion τ c = F * (γ s -γ)d i F * = Shields Dimensionless Shear 0.03 < F * < Equal Mobility Concept

13 Bed Shear Stress τ = γrs 0 τ = 0 fρv 8 2

14 Shear Partitioning Grain Shear: τ = γ R S R from fluid mechanics using Equation B.3 Shear due to form resistance: τ = γ R S Total Shear: τ = γ R S γ (R +R ) S

15 Armoring Potential Y s 1 = ya 1 Pc

16 Bed Material Transport Capacity Available Equations/Tools MPM-Woo Zeller-Fullerton HEC-RAS 4.0 SAM

17 Diffusion Equation General Form C ε + C( 1 C) W = 0 s y Low Concentrations C ε + CW s y s = 0

18 Suspended Sediment Concentration Profiles Rouse (1937): 1 1 C C 1 a = d y y a d a Ws z = βκ u * z Depth/Total Depth Depth/Total Depth Depth/Total Depth Velocity/Mean Velocity Concentration/ Reference Concentration Susp. Sediment Load/ Max. Susp Load

19 Suspended Sediment Concentration Profiles from (Woo, 1983)

20 Application of Woo (1983) Solution Bed load from Meyer-Peter, Muller Fall velocity modified by Maude and Whitmore (1958): W = ω ) p ( 1 C p Viscosity modified by O Brien and Julien (1988): μ = αe βc v α

21 Application of Woo (1983) Solution Power-law velocity profile Reference concentration at bed layer from Karim and Kennedy (1983): τ bl = D 50 u Limitations in computation procedure: u * c Reference concentration <650,000 ppm Total concentration < 510,000-65,000D 50 *

22 The MPM-Woo Equation E E E-3 d -2 c 0 b a Qs=aV b D c (1-C f )d 1.0E-4 1.0E D50 (mm) 1.0E-6

23 The MPM-Woo Equation Unit Discharge (cms/m) Range of Applicability Velocity (m/s) Depth (m) Gradient Fine Sediment Concentration (ppm) D 50 (mm) , (1-Cf)d Effect of C f D50(mm) Cf

24 Validation Data Yellow Canyon (Site 15) Coal Mine Wash

25 Validation Data Yellow Canyon (Site 15) Coal Mine Wash Bed Material Discharge (Metric Tons/Day) Range of Measured Data Estimated Transport Rate Bed Material Discharge (Metric Tons/Day) Range of Measured Data Estimated Transport Rate Water Discharge (CMS) Water Discharge (CMS)

26 Sediment Bulking 2.0 Concentration (ppm) Thousands Mudflow (2) Bulking Factor by Weight by Volume Mudflow (1) Mud Flood (3) Mud Flood (2) Mud Flood (1) Water flood 0% 20% 40% 60% 80% Concentration (%) Typical conditions in SSCAFCA Arroyos

27 Bulking Factors Q + stotal B = = f Q Q 1- S g - 1 C /10 ( /10 )( 1) 6 C S - s s 6 g where Bf = bulking factor Q = clear-water discharge Q stotal = total sediment load (i.e., combination of bed material and wash load) Cs = total sediment concentration by weight Sg = specific gravity of the sediment

28 Recommended Bulking Factors for Q Upper size-limit of applicability Bulking Factor Dominant Discharge (Qd) 50 cfs 100 cfs 250 cfs 500 cfs 1000 cfs Median (D50) Bed Material Size (mm)

29 Recurrence Interval Dominant Discharge (cfs) (yrs) ,000 D 50 (mm) = 0.5 mm D 50 (mm) = 1.0 mm D 50 (mm) = 1.5 mm D 50 (mm) = 2.0 mm D 50 (mm) = 3.0 mm D 50 (mm) = 4.0 mm (Table 3.5) Estimated Sediment Bulking Factors for Other Events

30 Sediment Transport Workshop

31 Fine Sediment Yield The watershed upstream of the location analyzed in the previous problems has the following characteristics: Drainage area = 370 acres Watershed soil type: 52% Rock outcrop, Orthids Complex (ROF) 48% Tesajo-Millett (Te) Percent impervious (roads, roofs, etc.) = 9.5% Average overland slope = 25% Average slope length = 100 feet Rangeland, grass-like plants, 10% ground cover, no canopy 100-year storm runoff volume = 40.2 ac-ft

32 Fine Sediment Yield Example Problem #1 1. Compute fine sediment yield from the watershed using MUSLE: Y s = C 95 ( ) V Q K LS C P w p where C is a calibration factor (for the SSCAFCA jurisdictional area, use 3.0, unless data are available indicating a more appropriate value).

33 Fine Sediment Yield Example Problem #1 Estimate K from Table A.2.1 (see also SCS, 1992): for Te use very gravelly, sandy loam and loamy sand for ROF use 0.0 (Rock outcrop 40%, Orthids 30%) Compute weighted K: K = ( A K + A K ) ROF ROF A total Estimate C value from Table A.2.2: C = 0.32 Tθ Tθ P = 1.0 (no terracing) Estimate LS using Equation A.3: = 0.52(0) (0.1) 1.0 = 0.048

34 Fine Sediment Yield Example Problem #1 LS = λ 72.6 n 2 ( S S ) = [ (9.45) (9.45) ] = 2150 tons fine sediment This result assumes 100% of the watershed is pervious. Adjust for given percent impervious: Ys' = (1 - % impervious) Ys = ( ) (2150) = 1946 tons

35 Fine Sediment Yield Example Problem #2 2. Compute average fine sediment concentration from watershed for the 100-year storm: C f ( ppm) = 6 10 W w W + s W s 1946 ( 2000) [( 40.5)( 43560)( 62.4) ( 1946)( 2000) ] = = 34, 147 ppm w

36 Bed Material and Total Sediment Load: Example Problem #1 Compute the bed material transport capacity, total sediment load and bulking factors for the peak of the 100-year storm, for the arroyo in the previous problems. 1. Compute bed material transport capacity at Q100 = 1,045 cfs using Equation C.3. From Figure C.2: a = 1.5x10-6 b = 5.8 c = -0.7 d = -1.9 From hydraulics example problem: V = fps y 100 = 2.0 feet W = 39 feet Assume constant fine sediment yield throughout the storm: C f = 34,147 ppm-w

37 Bed Material and Total Sediment Load: Example Problem #1 Apply Equation C.3: q s = a V b y c ( ) 6 d 1 C f /10 = 1.5 x , ( 13.4) ( 2.0) 1 = 3.40 cfs / ft 1.9 Q s = = q s W ( 3.40)( 39) = cfs

38 Bed Material and Total Sediment Load: Example Problem #2 2. Compute the bed material concentration. From Equation C.5: C s = x 10 6 Q ( Q Q ) s s 6 ( ) (. 6) = x , ppm w

39 Bed Material and Total Sediment Load: Example Problem #3 3. Compute the total sediment load. ( Q ) : Compute the wash load discharge Q f = Q Cf C 6 f (rearranging C.7) , 147 = , 147 = cfs Q = Q + stotal s Q f = = cfs

40 Bed Material and Total Sediment Load: Example Problem #4 4. Compute the total sediment concentration, bulking factor, and bulked peak discharge. C s Total = x 10 6 Q stotal ( Q Q ) s Total = x ( ) ( ) = 270, 875 ppm w

41 Bed Material and Total Sediment Load: Example Problem #4 BF = C 1 s Total / 10 ( 6 C / 10 )( S 1) s Total 6 = , 875 / , ( ) = Q = BF Pbulked Q P

42 Annual Sediment Yield The following total sediment yield results were obtained by integrating the bed-material transport capacity and adding the fine sediment for each storm. Return Period (years) Water Yield (ac-ft) Total Sediment Yield (tons) Unit Sediment Yield (tons/acre) , , , , , Compute the mean annual water and sediment yields.

43 Annual Sediment Yield Example Problem From Equation 3.26: = Y Y Y Y Y Y Y x Annual x 100 x 50 x 25 x 10 x 5 x 2 where x = Either the total sediment or water yield. (1) Water yield: Y a = (40.2) (33.8) (27.5) (19.4)+ 0.2 (13.7)+ 0.4 (7.1) w = ac ft Sediment yield: Y a =.015 (6142) (4863) (3774) (2413)+ 0.2 (1559)+ 0.4 (668) s =1088 tons

44 Annual Sediment Yield Example Problem Unit sediment yield: 1088 Y s a = 370 = 2.94 tons/acre = 2.94/3. * 2 4 = 0.86 ac - ft/mi (*assuming bulked unit weight of 100 pcf, see Constants and Conversions)

3 Theoretical Basis for SAM.sed Calculations

3 Theoretical Basis for SAM.sed Calculations Purpose Sediment transport functions can be used to calculate the bed material portion of the sediment discharge rating curve. This rating curve can then be