Sediment and Erosion Design Guide

Sediment and Erosion Design Guide

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

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

Modes of Sediment-load Transport

Bed Load Processes

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 2.5 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 2.0 1.5 0 200 400 600 800 1000 1200 1400 1600 1800 Main Channel Distance (ft)

Bed Material Sampling Maximum Particle Size Minimum Weight of Sample g lb 3-inch 6,000 13 2-inch 4,000 9 1-inch 2,000 14 1/2-inch 1,000 2 < No. 4 sieve 200 0.5 < No. 10 sieve 100 0.25

Jan 2008 Sample Locations

MONTOYAS ARROYO 100 80 Percent Finer 60 40 20 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 300 100 COBBLES 30 10 3 Grain Size in Millimeters 0.1 GRAVEL SAND VC C M F VF VC C M F VF 1 0.3 0.03 SILT or CLAY 0.01

100 10 1 0.1 0.01 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

Incipient Motion τ c = F * (γ s -γ)d i F * = Shields Dimensionless Shear 0.03 < F * < 0.047 Equal Mobility Concept

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

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

Armoring Potential Y s 1 = ya 1 Pc

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

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

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 0.8 0.6 0.4 0.2 Depth/Total Depth 0.8 0.6 0.4 0.2 Depth/Total Depth 0.8 0.6 0.4 0.2 0 0.4 0.6 0.8 1 1.2 Velocity/Mean Velocity 0 0 0.2 0.4 0.6 Concentration/ Reference Concentration 0 0 0.2 0.4 0.6 0.8 1 Susp. Sediment Load/ Max. Susp Load

Suspended Sediment Concentration Profiles from (Woo, 1983)

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

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 *

The MPM-Woo Equation 0 1 6 1.0E-1 0.5 5 1.0E-2 4 1.0E-3 d -2 c 0 b a -0.5 3 2 Qs=aV b D c (1-C f )d 1.0E-4 1.0E-5-4 -1 1 0 0.5 1 1.5 2 2.5 3 3.5 4 D50 (mm) 1.0E-6

The MPM-Woo Equation Unit Discharge (cms/m) Range of Applicability Velocity (m/s) Depth (m) Gradient Fine Sediment Concentration (ppm) D 50 (mm) 0.9 0.6 0.1 0.005 0 0.2 7.4 6.3 2.2.04 60,000 4.0 (1-Cf)d 1.4 1.3 1.2 1.1 1 Effect of C f D50(mm) 0.2 0.3 1.0 1.5 4.0 0 0.05 0.1 Cf

Validation Data Yellow Canyon (Site 15) Coal Mine Wash

Validation Data Yellow Canyon (Site 15) Coal Mine Wash 10000000 1000000 Bed Material Discharge (Metric Tons/Day) 1000000 100000 10000 Range of Measured Data Estimated Transport Rate Bed Material Discharge (Metric Tons/Day) 100000 10000 Range of Measured Data Estimated Transport Rate 1000 0 1 10 100 Water Discharge (CMS) 1000 0 1 10 100 Water Discharge (CMS)

Sediment Bulking 2.0 Concentration (ppm) Thousands 0 200 400 600 800 Mudflow (2) Bulking Factor 1.8 1.6 1.4 by Weight by Volume Mudflow (1) Mud Flood (3) Mud Flood (2) Mud Flood (1) 1.2 1.0 Water flood 0% 20% 40% 60% 80% Concentration (%) Typical conditions in SSCAFCA Arroyos

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

Recommended Bulking Factors for Q 100 1.20 Upper size-limit of applicability Bulking Factor 1.15 1.10 1.05 Dominant Discharge (Qd) 50 cfs 100 cfs 250 cfs 500 cfs 1000 cfs 1.00 0.1 1 10 Median (D50) Bed Material Size (mm)

Recurrence Interval Dominant Discharge (cfs) (yrs) 50 100 250 500 1,000 D 50 (mm) = 0.5 mm 2 1.01 1.01 1.01 1.01 1.02 5 1.02 1.02 1.05 1.08 1.14 10 1.03 1.05 1.10 1.19 1.19 25 1.05 1.09 1.19 1.19 1.19 50 1.07 1.12 1.19 1.19 1.19 100 1.08 1.15 1.19 1.19 1.19 D 50 (mm) = 1.0 mm 2 1.01 1.01 1.01 1.01 1.01 5 1.01 1.01 1.01 1.03 1.05 10 1.01 1.01 1.03 1.07 1.16 25 1.02 1.03 1.08 1.17 1.17 50 1.02 1.04 1.12 1.17 1.17 100 1.03 1.05 1.15 1.17 1.17 D 50 (mm) = 1.5 mm 2 1.01 1.01 1.01 1.01 1.01 5 1.01 1.01 1.01 1.02 1.04 10 1.01 1.01 1.02 1.05 1.13 25 1.01 1.02 1.06 1.14 1.16 50 1.01 1.03 1.09 1.16 1.16 100 1.02 1.04 1.12 1.16 1.16 D 50 (mm) = 2.0 mm 2 1.01 1.01 1.01 1.01 1.01 5 1.01 1.01 1.01 1.01 1.03 10 1.01 1.01 1.02 1.04 1.08 25 1.01 1.01 1.04 1.09 1.15 50 1.01 1.02 1.06 1.15 1.15 100 1.01 1.03 1.08 1.15 1.15 D 50 (mm) = 3.0 mm 2 1.01 1.01 1.01 1.01 1.01 5 1.01 1.01 1.01 1.01 1.02 10 1.01 1.01 1.01 1.02 1.04 25 1.01 1.01 1.02 1.05 1.11 50 1.01 1.01 1.03 1.07 1.12 100 1.01 1.02 1.04 1.10 1.12 D 50 (mm) = 4.0 mm 2 1.01 1.01 1.01 1.01 1.01 5 1.01 1.01 1.01 1.01 1.01 10 1.01 1.01 1.01 1.02 1.03 25 1.01 1.01 1.02 1.03 1.06 50 1.01 1.01 1.02 1.04 1.10 100 1.01 1.01 1.03 1.06 1.10 (Table 3.5) Estimated Sediment Bulking Factors for Other Events

Sediment Transport Workshop

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

Fine Sediment Yield Example Problem #1 1. Compute fine sediment yield from the watershed using MUSLE: Y s = C 95 ( ) 0. 56 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).

Fine Sediment Yield Example Problem #1 Estimate K from Table A.2.1 (see also SCS, 1992): for Te use 0.1 - 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) + 0.48(0.1) 1.0 = 0.048

Fine Sediment Yield Example Problem #1 LS = λ 72.6 n 2 ( 0.065 + 0.0454S + 0.0065S ) = 100 72.6 0.5 2 [ 0.065 + 0.0454(9.45) + 0.0065(9.45) ] = 2150 tons fine sediment This result assumes 100% of the watershed is pervious. Adjust for given percent impervious: Ys' = (1 - % impervious) Ys = (1-0.095) (2150) = 1946 tons

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) ] = 10 6 + = 34, 147 ppm w

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 = 100 13.4 fps y 100 = 2.0 feet W = 39 feet Assume constant fine sediment yield throughout the storm: C f = 34,147 ppm-w

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 10 6 34,147 6 10 5.8 0.7 ( 13.4) ( 2.0) 1 = 3.40 cfs / ft 1.9 Q s = = q s W ( 3.40)( 39) = 132.6 cfs

Bed Material and Total Sediment Load: Example Problem #2 2. Compute the bed material concentration. From Equation C.5: C s = 2. 65 x 10 6 Q ( Q + 2. 65 Q ) s s 6 ( 132. 6) (. 6) = 2. 65 x 10 251 642 1045 + 2. 65 132, ppm w

Bed Material and Total Sediment Load: Example Problem #3 3. Compute the total sediment load. ( Q ) : Compute the wash load discharge Q f = Q 2. 65 10 Cf C 6 f (rearranging C.7) 1045 34, 147 = 2. 65 6 10 34, 147 = 13. 9 cfs Q = Q + stotal s Q f = 132. 6 + 13. 9 = 146. 5 cfs

Bed Material and Total Sediment Load: Example Problem #4 4. Compute the total sediment concentration, bulking factor, and bulked peak discharge. C s Total = 2. 65 x 10 6 Q stotal ( Q + 2. 65 Q ) s Total 2. 65 = 1045 6 x 10 + 2. 65 ( 146. 5) ( 146. 5) = 270, 875 ppm w

Bed Material and Total Sediment Load: Example Problem #4 BF = 1 2. 65 C 1 s Total / 10 ( 6 C / 10 )( S 1) s Total 6 = 1 2. 65 1 270, 875 / 10 270, 875 6 10 6 ( 2. 65 1) = 1. 14 Q = BF Pbulked Q P

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) 100 40.2 6,142 16.6 50 33.8 4,863 13.1 25 27.5 3,774 10.2 10 19.4 2,413 6.5 5 13.7 1,559 4.2 2 7.1 668 1.8 Compute the mean annual water and sediment yields.

Annual Sediment Yield Example Problem From Equation 3.26: = 0.015 Y + 0.015 Y + 0.04 Y + 0.08 Y + 0.20 Y + 0.40 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 = 0.015 (40.2)+ 0.015 (33.8)+ 0.04 (27.5)+ 0.08 (19.4)+ 0.2 (13.7)+ 0.4 (7.1) w = 9. 34 ac ft Sediment yield: Y a =.015 (6142)+ 0.015 (4863)+ 0.04 (3774)+ 0.08 (2413)+ 0.2 (1559)+ 0.4 (668) s =1088 tons

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)