Sediment and ErosionDesign Guide

Sediment Transport & Bulking FactorsGoals 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 SedimentModes of Sediment--load load TransportTransport

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

0 200 400 600 800 1000 1200 1400 1600 18001.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

DonFelipeDam Plan: Existing Geom Future Flows 6/28/2007

Main Channel Distance (ft)

Vel

Chn

l (ft/

s)

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

PajaritoNorth Main

Bed Material SamplingBed Material Sampling

Maximum Particle Size

Minimum Weight of Sample

g lb3-inch 6,000 132-inch 4,000 91-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

BOULDERS COBBLESSAND

VC C M F VFGRAVEL

SILT or CLAYVC C M F VF

0.010.1110100 0.030.3330300

Grain Size in Millimeters

0

20

40

60

80

100

Perc

ent F

iner

S6, D50=1.93mmS7, D50=1.13mmS8, D50=0.81mmS9, D50=0.39mmS10, D50=0.46mmS11, D50=0.96mmS12, D50=0.29mm

Bed Material Size Trendsin SSCAFCA Area

0.01

0.1

1

10

100S5 S4 S3 S2 S1 S6 S7 S8 S9 S1

0S1

1S1

2S1

4S1

3S1

8S1

7S1

6S1

5S2

4S1

9S2

0S2

1S2

2S2

3

Part

icle

Siz

e (m

m)

D84D16D50

Calabacillas Montoyas

Lom

itas

Neg

ras

Barranca Venada

Upstream to downstream order by arroyo

Incipient Motion

τc = F*(γs-γ)DiF* = Shields Dimensionless Shear

0.03 < F* < 0.047

Equal MobilityConcept

Bed Shear Stress

RSγτ =0

8

2

0Vfρτ =

Shear PartitioningShear Partitioning

• Grain Shear: τ’ = γ R’ SR’ from fluid mechanicsusing Equation B.3

• Shear due to form resistance:τ” = γ R” S

• Total Shear: τ = γ R S ≅ γ (R’+R”) S

Armoring Potential

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 11

cas P

yY

Bed Material Transport CapacityAvailable Equations/Tools

• MPM-Woo

• Zeller-Fullerton

• HEC-RAS 4.0

• SAM

Diffusion Equation

0=+∂∂

ss CWyCε

01 =−+∂∂

ss WCCyC )(ε

General Form

@ Low Concentrations

Suspended Sediment Concentration Profiles

z

a ada

yyd

CC

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛ −=Rouse (1937):

*uWz s

βκ=

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Susp. Sediment Load/Max. Susp Load

Dep

th/T

otal

Dep

th

0

0.2

0.4

0.6

0.8

1

0.4 0.6 0.8 1 1.2Velocity/Mean Velocity

Dep

th/T

otal

Dep

th

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6Concentration/

Reference Concentration

Dep

th/T

otal

Dep

th

Suspended Sediment Concentration Profilesfrom (Woo, 1983)

Application of Woo (1983) Solution• Bed load from Meyer-Peter, Muller• Fall velocity modified by Maude and

Whitmore (1958):

• Viscosity modified by O’Brien and Julien (1988):

vCeβαμ =

αω )( CW pp −= 1

• Power-law velocity profile• Reference concentration at bed layer

from Karim and Kennedy (1983):

– Limitations in computation procedure:• Reference concentration <650,000 ppm• Total concentration < 510,000-65,000D50

cbl u

uD*

*50=τ

Application of Woo (1983) Solution

The MPM-Woo Equation

0 0.5 1 1.5 2 2.5 3 3.5 4D50 (mm)

1.0E-6

1.0E-5

1.0E-4

1.0E-3

1.0E-2

1.0E-1

a

1

2

3

4

5

6

b

-1

-0.5

0

0.5

1

c

-4

-2

0

d

Qs=aVbDc(1-Cf)d

The MPM-Woo EquationUnit

Discharge(cms/m)

Velocity(m/s)

Depth (m)

Gradient Fine Sediment Concentration

(ppm)

D50 (mm)

0.9 0.6 0.1 0.005 0 0.27.4 6.3 2.2 .04 60,000 4.0

1

1.1

1.2

1.3

1.4

0 0.05 0.1

Cf

(1-C

f)d

0.2

0.3

1.0

1.5

4.0

D50(mm)

Range of Applicability

Effect of Cf

Validation Data

Yellow Canyon (Site 15)

Coal Mine Wash

Validation Data

Yellow Canyon (Site 15) Coal Mine Wash

0 1 10 100Water Discharge (CMS)

1000

10000

100000

1000000

10000000

Bed

Mat

eria

l Dis

char

ge (M

etric

Ton

s/D

ay)

EstimatedTransport Rate

Range ofMeasured Data

0 1 10 100Water Discharge (CMS)

1000

10000

100000

1000000

Bed

Mat

eria

l Dis

char

ge (M

etric

Ton

s/D

ay)

EstimatedTransport Rate

Range ofMeasured Data

Sediment BulkingSediment Bulking

Mudflow (2)

Mudflow (1)

Mud Flood (3)

Mud Flood (2)

Mud Flood (1)

Water flood

1.0

1.2

1.4

1.6

1.8

2.0

0% 20% 40% 60% 80%Concentration (%)

Bul

king

Fac

tor

0 200 400 600 800ThousandsConcentration (ppm)

by Weightby Volume

Typical conditions in SSCAFCA Arroyos

Bulking Factors

where Bf = bulking factorQ = clear-water dischargeQstotal = total sediment load (i.e., combination of bed

material and wash load)Cs = total sediment concentration by weightSg = specific gravity of the sediment

( )( )110/10/1

1

6

6

- - -

gsg

s

s

f

SCSCQ

QQB total =

+=

Recommended Bulking Factors for Q100

1.00

1.05

1.10

1.15

1.20

0.1 1 10Median (D50) Bed Material Size (mm)

Bul

king

Fac

tor

50 cfs100 cfs250 cfs500 cfs1000 cfs

Dominant Discharge (Qd)

Upper size-limitof applicability

(Table 3.5) (Table 3.5) Estimated Estimated Sediment Sediment Bulking Bulking

Factors for Factors for Other EventsOther Events

Dominant Discharge (cfs) Recurrence Interval (yrs) 50 100 250 500 1,000

D50 (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 D50 (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 D50 (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 D50 (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 D50 (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 D50 (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

Sediment TransportWorkshop

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 YieldExample Problem #1

1. Compute fine sediment yield from the watershed using MUSLE:

( ) PCLSKQV95CY 560pws

.=

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 YieldExample Problem #1

Estimate K from Table A.2.1 (see also SCS, 1992): for Te use 0.1 - very gravelly, sandy loam and loamy sandfor ROF use 0.0 (Rock outcrop 40%, Orthids 30%) Compute weighted K:

( ) 048.00.1

)1.0(48.0)0(52.0=

+=

+=′

total

TTROFROF

AKAKAK θθ

Estimate C value from Table A.2.2:

C = 0.32

P = 1.0 (no terracing)

Estimate LS using Equation A.3:

Fine Sediment YieldExample Problem #1

( )20065.00454.0065.06.72

SSLSn

++⎟⎠⎞

⎜⎝⎛=

λ

[ ]2

5.0

)45.9(0065.0)45.9(0454.0065.06.72

100++⎟

⎠⎞

⎜⎝⎛=

= 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 YieldExample Problem #2

( )sw

sf WW

WppmC+

= 610

( )( ) ( ) ( ) ( ) ( )[ ]200019464.62435605.40

20001946106

+=

2. Compute average fine sediment concentration from watershed for the 100-year storm:

wppm −= 147,34

Bed Material and Total Sediment Load: Example Problem #1

Compute the bed material transport capacity, total sediment loadand 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 = 5.8c = -0.7d = -1.9

From hydraulics example problem: V100 = 13.4 fpsy100 = 2.0 feetW = 39 feet

Assume constant fine sediment yield throughout the storm:Cf = 34,147 ppm-w

Bed Material and Total Sediment Load: Example Problem #1

( )d

f

cb

s CyVaq 610/1 −′=

( ) ( ) ftcfsx /40.310

147,3410.24.13105.19.1

6

7.08.56 =⎟⎠⎞

⎜⎝⎛ −=

−

−−

WqQ ss =

Apply Equation C.3:

( ) ( ) cfs6.1323940.3 ==

Bed Material and Total Sediment Load: Example Problem #2

2. Compute the bed material concentration.

( )s

s6

s Q652QQ10x652C

..

+=

( )( ) wppm642251

61326521045613210x652 6

−+

= ,....

From Equation C.5:

Bed Material and Total Sediment Load: Example Problem #3

3. Compute the total sediment load. :) Q( f

⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎠⎞

⎜⎝⎛=

f6

ff

C10C

652QQ.

cfs9131473410

14734652

10456 .

,,

.=⎟

⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎠⎞

⎜⎝⎛=

fsTotals QQQ +=

cfs51469136132 ... =+=

Compute the wash load discharge

(rearranging C.7)

Bed Material and Total Sediment Load: Example Problem #4

4. Compute the total sediment concentration, bulking factor, and bulked peak discharge.

( )Totals

Totals6

Totals Q652QQ10x652

C.

.+

=

( )( )51466521045

514610x652 6

....

+=

wppm875270 −= ,

Bed Material and Total Sediment Load: Example Problem #4

( )( )1S10C652

10C1

1BF

6Totals

6Totals

−−−

=

/.

/

( )

141

165210

875270652

108752701

1

6

6 .

.,.

/,=

−−−

=

PbulkedP QBFQ =

Annual Sediment YieldThe 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 16.650 33.8 4,863 13.125 27.5 3,774 10.210 19.4 2,413 6.55 13.7 1,559 4.22 7.1 668 1.8

Compute the mean annual water and sediment yields.

Annual Sediment YieldExample Problem

Y0.40 + Y0.20 + Y0.08 + Y0.04 + Y 0.015 + Y 0.015 = Y 2 x5 x10 x25 x50 x100 xAnnual x

(7.1)0.4 + (13.7)0.2 + (19.4)0.08 + (27.5)0.04 + (33.8) 0.015 + (40.2) 0.015 = Y a w

ftac349 −= .

(668)0.4 + (1559)0.2 + (2413)0.08 + (3774)0.04 + (4863) 0.015 + (6142) .015 = Y as

From Equation 3.26:

where x = Either the total sediment or water yield.

(1) Water yield:

Sediment yield:

tons1088 =

Annual Sediment YieldExample Problem

tons/acre2.94 = 370

1088 = Y as

2*4 ft/mi-ac 0.86 = 2.94/3. =

Unit sediment yield:

(*assuming bulked unit weight of 100 pcf, see Constants and Conversions)