Sediment and Erosion Design Guide - SSCAFCA design guide/Channel... · Data Requirements •...
Transcript of Sediment and Erosion Design Guide - SSCAFCA design guide/Channel... · Data Requirements •...
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
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
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
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
Bed Material Transport CapacityAvailable Equations/Tools
• MPM-Woo
• Zeller-Fullerton
• HEC-RAS 4.0
• SAM
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
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
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
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 =