Sediment and Erosion Design Guide - SSCAFCA design guide/Channel... · Sediment Transport & Bulking...
Transcript of Sediment and Erosion Design Guide - SSCAFCA design guide/Channel... · Sediment Transport & Bulking...
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Sediment and ErosionDesign Guide
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Sediment Transport & Bulking FactorsGoals of this Session
• Review key principals • Review basic relationships and
available tools• Review bulking factor relationships
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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
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Modes of SedimentModes of Sediment--load load TransportTransport
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Bed Load Processes
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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
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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
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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
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Jan 2008 Sample Locations
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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
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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
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Incipient Motion
τc = F*(γs-γ)DiF* = Shields Dimensionless Shear
0.03 < F* < 0.047
Equal MobilityConcept
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Bed Shear Stress
RSγτ =0
8
2
0Vfρτ =
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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
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Armoring Potential
⎟⎟⎠
⎞⎜⎜⎝
⎛−= 11
cas P
yY
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Bed Material Transport CapacityAvailable Equations/Tools
• MPM-Woo
• Zeller-Fullerton
• HEC-RAS 4.0
• SAM
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Diffusion Equation
0=+∂∂
ss CWyCε
01 =−+∂∂
ss WCCyC )(ε
General Form
@ Low Concentrations
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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
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Suspended Sediment Concentration Profilesfrom (Woo, 1983)
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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
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• 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
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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
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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
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Validation Data
Yellow Canyon (Site 15)
Coal Mine Wash
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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
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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
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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 =
+=
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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
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(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
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Sediment TransportWorkshop
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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
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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).
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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:
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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
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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
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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
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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 ==
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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:
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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)
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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 −= ,
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Bed Material and Total Sediment Load: Example Problem #4
( )( )1S10C652
10C1
1BF
6Totals
6Totals
−−−
=
/.
/
( )
141
165210
875270652
108752701
1
6
6 .
.,.
/,=
−−−
=
PbulkedP QBFQ =
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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.
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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 =
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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)