Principles of Erosion & Sediment Control

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Transcript Principles of Erosion & Sediment Control

2011 PE Review:

Michael C. Hirschi, PhD, PE, D.WRE

Professor and Assistant Dean University of Illinois [email protected]

Acknowledgements:

Rod Huffman, PE Review coordinator Daniel Yoder (2006 presenter) Rabi Mohtar & Majdi Abu Najm (2010 presenters)

Topics

• Principles of Soil Physics (some duplication with Management session) • Soil Strength & Mechanics • Sediment Transport • Erosion Control • Slope Stabilization

Sources

• Environmental Soil Physics; Hillel; 1998

Hi

• Essentials of Soil Mechanics & Foundations, 2 nd McCarthy; 1982;

M

• USLE/RUSLE Handout:

HO

ed.; • Soil and Water Conservation Engineering – 4 th – 5 th ed. Schwab, Fangmeier, Elliott, Frevert:

S4

ed. Fangmeier, Elliott, Workman, Huffman, Schwab:

S5

• Design Hydrology & Sedimentology for Small Catchments; Haan, Barfield, Hayes:

H

Soil Physics & Mechanics

• Soil classes and particle size distributions • Soil water – Content – Potential – Flow • Soil strength & mechanics

Soil classes & particle sizes Hi.61

Soil classes & particle sizes - 2 • ISSS easiest – Sand 0.02-2.0mm (20-2000 μ) – Silt 0.002-0.02mm (2-20μ) – Clay <0.002mm (<2μ) • Soil texture classes based on USDA – Sand 0.05-2.0mm

– Silt 0.002-0.05mm

– Clay <0.002mm

Soil classes & particle sizes – 3 • Texture – EX: 50% sand, 20% silt 30% clay?

Hi.64

Soil classes & particle sizes – 4 • Particle size distribution – EX: draw in a sandy clay loam?

Hi.65

Soil water content

0 • M t = M s + M w + M a • V t = V s + V w + V a – t = total, s = solids, w = water, a = air • ρ b • ρ p = bulk density = M s /V t ≈ 1.1-1.4 g/cc = particle density = M s /V s ≈ 2.65 g/cc • Porosity = (V w • ρ w + V a ) / V = water density = M t w ≈ 25-60% /V w = 1.0 g/cc

Soil water content – 2 • w w = M w / (M s + M w ) – Water content wet basis • w = mass wetness = M w / M s – Water content dry basis • θ = V w /V t = V w / (V s + V w – Volumetric water content + V a )

Calc.: soil water content • • Given: – Soil with 30% water content dry basis Find: – Best guess at equivalent inches of water in the top foot of soil?

Calc.: soil water content – 2 • M w – / M s M w = V w = 0.30

* ρ w – ρ b = M s / V t ; M s = V t * ρ b

Calc.: soil water content – 3 • M w – / M s = 0.30

M w = V w * ρ w – ρ b = M s / V t ; M s = V t * ρ b – – – – – M w / M s = (V w * ρ w )/(V t * ρ b ) = (V w / V t )(ρ w / ρ b ) θ = V w /V t θ *(ρ w / ρ b ) = 0.3; θ = (ρ b / ρ w ) * 0.3

θ = 0.3 *(1.3/1.0) = 0.39

0.39 * 1 ft * 12”/ft = 4.7”

Soil water potential Hi.157

Cuenca.58

Soil water potential – 2

S5.337

Soil water potential – 3

Hi.162

Soil water potential – 4

Calc.: soil water potential • • Given: – Mercury tensiometer • SG = 13.6

– Situation as shown Find: – Total potential at C – Above or below water table?

Cuenca.64

Calc.: soil water potential - 2 • • • 1) pick datum 2) add pressures – Suction – Water depth – Gravity T = z + p + p os – = -86cm – Above water table

Soil water flow • • q = A*K*H/L – K = (q*L)/(A*H) K values S4.359; S5.261; H.430

A L q H

Calc.: soil water flow • • Given: – Need 50000 gpd through a 1-ft thick sand filter with K = 8 ft/d, and a total driving head of 3 ft Find : – Required diameter for circular tank?

Calc.: soil water flow – 2 • q = A*K*H/L; A = (q*L)/(K*H)

A

   50000

gal day

   1

ft

1  

day

8

ft

    3 1

ft

    1

ft

3 7 .

481

gal

   278

ft

2

d

 4

A

  18

ft

Soil strength and mechanics • • M.233-237,373-379 Soil stresses – Normal = F n /A = σ – Shear = F t /A = τ • F n = normal force • F t = tangential or shear force – As σ ↑ τ to cause failure = τ f ↑ – tan Φ = τ f / σ; Φ = angle of internal friction

Soil strength and mechanics – 2 M.234

Calc.: soil strength • • Given: – Well-graded sand; density 124 lb/ft 3 Find: – Ultimate shear strength 6 ft below surface?

Calc.: soil strength – 2 • From table, for well-graded sand, Φ = 32-35 o = 33.5

o • Normal stress = (124 lb/ft 3 )(6 ft) = 744 lb/ft 2 • tan Φ = τ f / σ; τ f = σ * tan Φ = τ f = 744 lb/ft 2 * tan(33.5

o ) = 492 lb/ft 2

Footing bearing loads

• M.374-379 • • q ult – = a 1 *c*N c + a 2 *B* γ 1 *N γ + γ 2 *D f *N q c = soil cohesion beneath footer – γ 1 , , γ 2 = effective soil unit weight above and below footer – B = footer size term – N c , N γ , N q – D f = capacity factors = footing depth below surface q design = q ult / FS 2 3 4 6

Length/width

1 (square) Strip Circular

B

Width Width Width Width Width Width Radius 1.2

a 1

1.12

1.07

1.05

1.03

1.00

1.2

a 2

0.42

0.45

0.46

0.47

0.48

0.50

0.60

M.375

Footing bearing loads – 2

Calc.: footing load • • Given: – Strip footing 3 ft wide – Wet soil with density of 125 lb/ft 3 – Angle of internal friction = 30 o – Cohesive strength of 400 lb/ft 2 – Use factor of safety of 3 Find: – q design in lb/ft 2

Calc.: footing load – 2 • a1 = 1.0, a2 = 0.5, B = width = 3’ • γ 1 = 125/2 = 62.5 lb/ft3; γ 2 • c = 400 lb/ft2 = 125 lb/ft3 • N c = 30, N γ = 18, N q = 20

q ult

 ( 1 .

0 )   400

lb ft

2   ( 30 )  ( 0 .

5 )( 3

ft

 )  62 .

5

lb ft

3   ( 18 )    125

lb ft

3   ( 4

ft

)( 20 )  23 , 700

lb

/

ft

2 • q design = 23,700/3 = 7900 lb/ft 2

Soil compaction and density • • Soil compaction – Greater strength – Dependent on water content Proctor test – Pack soil into mold with pounding – Modified Proctor > 56000 ft-lbs

The Erosion Process • • • Soil erosion is a multi-step process: – Soil particle/aggregate detachment – Soil particle/aggregate transport – Soil particle/aggregate deposition There must be detachment and transport for erosion to occur Deposition (sedimentation) will occur somewhere downstream

A little soils review… • • • Soil primary particles: – Sand, 0.05mm to 2mm, 2.65 g/cc density – Silt, 0.002mm to 0.05mm, 2.65 g/cc – Clay, <0.002mm, 2.6 g/cc Soil aggregates, chemically/electrically bonded sets of primary particles: – Large, in the sand range, 1.6 g/cc – Small, in the large silt range, 1.8 g/cc These aggregate sizes are approximately those used in the CREAMS model (USDA-ARS)

Detachment • There are many sources of force and energy required to detach soil particles & aggregates: – Raindrop impact – Shallow surface flow shear – Concentrated flow shear – Many more, at larger scales

Transportation • Many of the same processes contribute force and energy for soil particle & aggregate transport: – Raindrop impact – Shallow surface flow – Concentrated surface flow – Channelized flow – Others

Balancing act • Foster & Meyer (1972) proposed a balance between detachment and transport for flowing water: • 1 = (transport load/transport capacity) + (detachment load/detachment capacity)

Essentially, if the flow is using all its transport capacity transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

Example • • • In the 80’s and 90’s there was a successful push to conservation tillage as a method to reduce sediment in lakes and streams In many situations, no improvement was seen, but streambank erosion became more of a problem than it was in the past I contend that now that the streams are receiving cleaner water (because of less upland erosion), but at similar rates, from farm fields, the stream uses less of its erosive energy to transport load it receives from runoff water, so it has capacity to undercut banks and scour the streambed

Multi-stage erosion

Soil erosion and sediment yield • • • • • Hillslope erosion Channel system erosion Sediment delivery to streams Sediment transport in streams Slope stability

Hillslope soil erosion • • S4.91-111; S5.134-156; H.238-285 Background – Detachment • Raindrop impact • By turbulent overland flow – Runoff – Transport downslope • By runoff

Hillslope soil erosion background • At the top of the slope – Detachment by raindrop impact – Transport by shallow sheet flow – Sheet erosion USDA NRCS

Hillslope soil erosion background - 2 • Lower on slope – Small flow concentrations – Start to cut small channels – Rills • Roughly parallel • Head straight downslope • Random formation – Flow from sheet areas between rills – Sheet and rill erosion USDA NRCS

Hillslope soil erosion background - 3 • Bottom of hillslope – Ends at concentrated flow channel – Low area in macrotopography – “ephemeral gullies” USDA NRCS

Hillslope erosion factors • • • • Rainfall erosivity – Intensity – Total storm energy Soil erodibility Topography – Slope length – Steepness Management – Reduce local erosion – Change runoff path – Slow and spread runoff => deposition

USLE/RUSLE • • • • • • • A = R * K * LS * C * P A = average annual soil erosion (T/A/Y) R = rainfall erosivity (awful units) K = soil erodibility (awful units) – R * K gives units of T/A/Y LS = topographic factor (dimensionless) C = cover-management (dimensionless) P = conservation practice (dimensionless)

USLE/RUSLE – background • • • • • Empirical approach – >10000 plot-years of data – International use Unit Plot basis; LS = C = P = 1 – Near worst-case management R from good fit rainfall-erosion K from K = A / R C and P from studies – Subfactors in later versions

USLE/RUSLE – approach • • Lookup – Maps, tables, figures – Databases Process-based calculations – Show changes over time – Where don’t have good data

R factor – rainfall erosivity • maps – H.251; H.Appendix 8A; S4.99(SI); S5.143(SI); HO.1-5 – R(customary SI) = 17.02 * R(customary US) S4

K factor – soil erodibility • • Soil surveys, NASIS, H.261-262; HO.6

Erodibility nomograph – H.255; S4.101; S5.144; HO.7

– No short-term OM

LS – topography factor • • New tables & figures – H.264; HO.8

Know susceptibility to rilling – High for highly disturbed soils – Low for consolidated soils

C – cover-management factor • • • Part of normal management scheme Lookup: S4.102; S5.146; H.266; H.Appendix 8; HO.9

Change over time

C – cover-management factor - 2 • Subfactor approach (RUSLE) – C = PLU * CC * SC * SR * SM; all 0-1 • PLU = prior land use – roots, buried biomass, soil consolidation • • CC = canopy cover; % cover & fall height SC = exp(-b * % cover) – B = 0.05 if rills dominant; 0.035 typical; 0.025 interrill • • SR = roughness; set by tillage, reduces over time SM = soil moisture; used only in NWRR

P – conservation practice factor • • • Common practices – Contouring, stripcropping, terraces Change flow patterns or cause deposition Lookup tables – S4.103; S5.146; H.281; HO.10

Calc.: USLE/RUSLE • • Given: – – Materials in handout 3-Acre construction site near Chicago – Straw mulch applied at 4 T/A – Average 20% slope, 100’ length – Loamy sand subsoil – Fill (loose soil) Find: – Erosion rate in T/A/Y

Calc: USLE/RUSLE – 2 • • • • • • R = 150 (HO.1) K = 0.24 (HO.7) LS = 4 (HO.8-high rilling) C = 0.02 (HO.9) P = 1.0

A = R * K * LS * C * P = 2.9 T/A/Y

Calc: USLE/RUSLE – 2.1

• • Given: – Materials in handout – 16-A site near Dallas, TX – Silty clay loam subsoil – Average 50% slope, 75’ length – Cut soil Find: – By what percentage will the erosion be reduced if we increase our straw mulch cover from 40% cover to 80% cover?

Calc: USLE/RUSLE – 2.2

• • • • • Only thing different is C – Only subfactor different is SC SC = exp(-b * %cover) – For consolidated soil, b = 0.025

SC 1 = exp(-0.025 * 40%) = 0.368

SC 2 = exp(-0.025 * 80%) = 0.135

Reduction = (0.368 – 0.135)/0.368 = 63%

Sediment delivery • • • USLE/RUSLE for hillslopes – Erosion – Delivery Erosion critical for soil resource conservation Delivery critical for water quality – Movement through channel system

Sediment delivery – 2

Sediment delivery – 3 • • SDR (Sediment Delivery Ratio) – Hillslope erosion – Empirical fit for watershed delivery Channel erosion/deposition modeling – Erosion – Transport – Deposition

Sediment Delivery Ratio • • H.293-299 SDR = SY / HE – SDR = sediment delivery ratio – SY = sediment yield at watershed exit – HE = hillslope erosion over watershed

Sediment Delivery Ratio – 2 • Area-delivery relationship H.294

Sediment Delivery Ratio – 3 • Relief-length ratio – Relief = elev change along main branch – Length = length along main branch H.294

Sediment Delivery Ratio – 4 • Forest Service Delivery Index Method H.295

Sediment Delivery Ratio – 5 • MUSLE (H.298) – Y = 95(Q * q p ) 0.56

• (K a )(LS Y = storm yield in tons a )(C a )(P a ) • Q = storm runoff volume in acre-in • q = peak runoff rate in cfs • K, LS, C, P = area=weighted watershed values – SDR = 95(Q * q p ) 0.56

/(R * area) • R = storm erosivity in US units – Routing for channel delivery

Calc: SDR • • Given: – Flow path length in watershed = 4000ft – Elevation difference = 115ft Find: – SDR

Calc.: SDR – 2 • • R/L = 115/4000 = 0.029

From figure SDR = 0.45

Channel erosion-deposition modeling • Process-based small channel models – Foster-Lane model • H.285-289 • Complicated and process-based – Ephemeral Gully Erosion Model • EGEM • Fit to Foster-Lane Model results

Channel erosion-deposition modeling – 2 • Large-channel models – Sediment transport – Channel morphology

Sediment transport • Settling (H.204-209) – Stokes’ Law • V s = settling velocity • d = particle diameter • g = accel due to gravity • • ν = kinematic viscosity

V s

SG = particle specific gravity – Simplified Stokes’ Law • SG = 2.65

• Quiescent water at 68 o F • d in mm, V s in fps  1 18  

d

 2

g

SG

 1    

V s

 2 .

81 

d

2

Calc.: Stokes’ Law settling • • Given: – ISSS soil particle size classification Find: – Settling velocities of largest sand, silt, and clay particles

Calc.: Stokes’ Law settling – 2 • ISSS classification – Largest particles size • • • Clay = 0.002mm

Silt = 0.2mm

Sand = 2mm – – – V s,clay = 1.12*10 -4 fps = 0.04 ft/hr = 0.97 ft/day V s,silt = 0.11 fps = 405 ft/hr = 1.83 mi/day V s,sand = 11.24 fps = 7.66 mph = 184 mi/day

Calc.: Stokes’ Law settling • • Given: – Stokes’ Law settling Find: particles larger than what size can be assumed to settle 1 ft in one hour?

Calc.: Stokes’ Law settling – 2 • • V s = [(1 ft)/(1 hr)](1 hr/3600s) = 2.778*10 -4 fps d = (V s /2.81) 1/2 = 0.00994mm

Other settling • • Flocculant, hindered, compression – Order of increasing concentration – Slower settling with increased conc.

Settleable solids – Imhoff cone H.211

Sediment transport relationships • • • • H.223-233 Bedload – Einstein equation Suspended load Total load – Yalin total load equation • V avg , d, SG, r, т, т c – Yang stream power relationship • Stream power instead of critical shear

Channel morphology • • H.391-419 Channel cross-section – – Width B, depth D, avg. vel. U, suspended sediment load q s , flow rate q B = C a * q a , D = C b * q b , U = C c * q c , q s = C d * q d – C a , C b , C c , C d , a, b, c, d = fit coefficients & exponents

Channel morphology – 2 • • Channel-forming discharge – Greatest impact on channel itself – Return period ≈ 1.5 years Channel roughness

Channel morphology – 3 • Channel regime – H.405-419 H.416

Application of process knowledge to control • • Limit individual parts to limit whole – Limit detachment – Limit transport Enhance deposition strategically – Where damage is minimal – Where cleanup is possible

Control of Soil Erosion by Water • • Detachment limiting strategies – Reduce raindrop impact (“Stop the Drops”) – Reduce runoff – Reduce detachment capacity of runoff – Increase soil resistance to erosive forces Transport limiting strategies – Reduce runoff volume – Reduce runoff transport capacity (“Slow the Flow”)

Example – No-Till • • Detachment – Raindrop impact detachment is very low due to high surface cover percentage – Flow shear detachment is low due to low velocities caused by tortuous flow path – Soil is resistant to erosion because of low disturbance Transport – Raindrop transport is limited by surface residue – Flow transport is limited by increased infiltration, lessening runoff – Flow transport is further limited by small dams created by surface residue

Example – Mulch on newly seeded area • • Detachment – Raindrop impact detachment is very low due to high surface cover percentage – Flow shear detachment is low due to low velocities caused by tortuous flow path Transport – – Raindrop transport is limited by surface residue Flow transport is limited by increased infiltration, lessening runoff – Flow transport is further limited by small dams created by surface residue

Comparison of no-till vs. mulch • • Detachment – Likely higher with mulch for same surface cover fraction because of higher soil disturbance for seedbed preparation – Likely higher for no-till following dry years because amount of residue cover is dictated by prior year crop growth Transport – Likely higher for mulch, unless “cut” in because no-till residue is effectively “cut” in during planting, at least for a small area, hopefully across slope – Likely higher for mulch situation because seedbed prep likely reduced average aggregate diameter

Control of Sediment in Runoff • • Reduce transport capacity of flow Enhance deposition of sediment

Reduce transport capacity • • Slow the flow – Barriers • Must let water pass, though slowly • Must be flow-stable, even after use • Must be where maintenance is possible – Reduce slope steepness • Channel must be of adequate capacity Increase infiltration

Enhance deposition of sediment • Use flocculant to increase sedimentation – Usually in sedimentation ponds when other methods are not adequate – Expensive…

Slope stability & failure

Possible forms of Failure McCarthy (1982) page 437-455 McCarthy (1982) page 440 I.B Mohtar 90

Slope stability & failure – 2 • Terms – β=max. slope angle before sliding – Φ=angle of internal friction • Cohesionless soil – tan(β) = tan( Φ) – Saturated: tan( β) = (1/2)tan( Φ) I.B Mohtar 91

Slope stability & failure – 3 • Cohesive soil –γ*z*sin(β)*cos(β) = c + σ*tan(Φ) • z = assumed depth • c = cohesive force • σ = effective compressive stress –Rotational or sliding block I.B Mohtar 92

Slope stability & failure – 4

For clay soil For soil with cohesion and internal friction > 0 93

Slope stability & failure – 5 • N s = c / (γ * H max ) – c = cohesion force – γ = soil unit weight – H max = max depth without sliding I.B Mohtar 94

Calc.: slope stability • • Example 8: Given: – Cohesion strength = 500 lb/ft 2 – Unit weight = 110 lb/ft 3 – Slope steepness = 50 o – Internal friction angle = 15 o Find: – Max. slope height I.B Mohtar 95

Calc.: slope stability – 2 • Fig. b, φ = 15 o , i = 50 o • H max = c / ( γ * N s ) = (500 lb/ft 2 )(1ft 3 / 10)(1/ 0.095) = 48 ft I.B Mohtar 96

Questions?