Transporte de Sedimento

8/11/2019 Transporte de Sedimento

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Estimating Sediment TransportPeter Wilcock 9 August 2011

ore ac groun Critical and reference shear stresses

y t s ar to est mate transport rates Empirical vs formula estimates A calibrated method for estimating transport rates in

gravel-bed rivers


2/62

If a sediment rating curve is what we use in application, isnt there

Will 1,000 cfs in Silver Ck move the same amount of sediment as1,000 cfs in the Snake R?

One recent attempt:

s

sr r

Q Q

Q Q

b

s aQQ Its nice that a cancelled, but for this model to be general, therate of change of Qs with Q (i.e. the exponent b) must be thesame everywhere. But we will find that b varies from littlemore than one to more than ten!.

Basically , we are looking for a flow variable that can be

accurately scaled, such that we have a general model


3/62

A flow variable that does scale well (although it not always easy to.

transport models in common use.

What might a general transport model look like? First, lets get amore formal introduction to the la ers. The best wa to do this iswith a dimensional analysis which will produce a list of

dimensionless variables that represent the problem.This helps because* the number of dimensionless variables is (usually) 3 less than the

* the dimensionless variables often have a useful physical meaning* Any relation between the dimensionless variables should begenera , we ave s e a re evan var a es a e s ar .


4/62

( , , , , , , )s sq f h D g Our list:2L M M M M L

, L, L, , , ,

)/,*,*,(* h DsS f q

sq **

mens onaanalysis gives:

gDs

gDsgDs

)1(

)1(,

)1(

3

3

s /If s = const

& S* large


5/62

o(Flow force on bed)/(bed area): The Shields Number

3(Grain weight): ( 1)6

s g D

2

* Number of grains/area 1/

Flow Force D

e s um er:

Grain weight ( 1)

This is THE most im ortant variable in sediment trans ort

s gD

The Einstein Volumetric transport rate/width sq

TransportParameter *

Sediment fall velocity ( 1)

trans ort rate

w s gD

q

3fall velocity * grain size ( 1)

q

s gD


6/62

First, we define conditions for the onset of sediment transport;these should depend on the same variables as the transport rate

ot on o mot on , , ,s 3( 1)

* * ss gD D

( 1) /s gD h

The Shields curveIncipient motion defined in.

*c

terms of a critical value ofShields Number

Motion

* c Defined for unisize sediment No Motion

( 1)c s gD

0.0110 100 1000 10000 100000

*

ec oDimensionless plots must be

59.03.0* *35exp045.0*105.0 S S c


7/62

bcaq )*(** A typical transport model:

gDsc

c

)1( 2/3*** c-

3/ 2 3 ( 1) ( 1)( 1) s gD s gDs gD

0.1

1

10

3/ 23/ 28

( 1)s cq

s g

0.001

0.01

3/ 278.7for in (tons/day)

s c

s

q

q

1E-050.0001

0.01 0.1 1

* .and , in (Pa)c

3**q


8/62

bcaq )*(** A typical transport model:

gDsc

c

)1( 2/3*** c

)1(** s

gqsq

-

One more transport variable, W*2/32/3 )/(*)(

2/3

(sorry!)

One more stress, the reference stress r

*18*

cW

onvert t e - ormu a to anincorporate a reference shear stress. First,we divide M-PM by ( *)3/2 to get

At *= *r, W * = W *r = 0.002. Dividing by 8, raising both sides to the 2/3 power produces

** 996.0 r c *

1004.0r

c

*

*

996.018*

r W Meyer-Peter & Mller:


9/62

Reference shear stress? Critical shear stress?

3/ 210

*10

*

Meyer-Peter & Mller

** 8 1 cW

0.11

0.1

1

3/ 2*

or

.996 0.01

*W

0.01

*

.0.01 0.1 1

*

*

.0.01 0.1 1

* *c

r


10/62

How the reference shear stress gets used

100

/ ( m s

) ]

1 r a

t e W *

1 r a

t e W *

1

o r t r a t e q

b i [

0.1

s t r a n s p o r

t

0.1

s t r a n s p o r

t

0.001

0.01

.

r a v e l

t r a n s p

0.001

.

i m e n s i o n

l e

0.001

.

J06

J14 i m e n s i o n

l e

*

r W

0.00001

0.0001

G

0.00001

0.0001

0.00001

0.0001J27BOMC

0.1 1 10 100

Bed Shear Stress (Pa)

0.1 1 10 100

Bed Shear Stress (Pa)

0.1 1 10 100

r /r


11/62

The Difference Between c

and r

c : boundary between motion & no motion.Definable exactly for an individual grain;

10

*W

definable statistically for a river bed Hard to measure, tracer grains are yourbest shot

0.1

r : the stress associated with a small,agreed-upon transport rate ( W* = 0.002) 0.001

0.01

0.01 0.1 1

*r W

provides a threshold for transport estimate

Easy to determine from transport

*

* *observations c

tracers (painted rocks, magnetic rocks, radio rocks, rock scum) answer the question did the

11

, , ,flows, develop relation between %entrained, grain size, and flow)

Need large numbers of grains for reliable sample Need to place naturally


12/62

cWill sediment move? Is > ?

As increases, up go stage & (controlled by resistance and continuity).o

Q R

The incipient motion problem

For a given , we can find boundary stress using momentum =

and compare too

c

S gRS

100

1.5

2.0

2.5 Lazy River Section 2

a t i o n

( m )

10

a r s

t r e s s

( P a

)

c Motion

0.0

0.5

.

0 2 4 6 8 10 12 14 16Station (m)

E l e

1 S h e

No Motion

At what does ?o cQ

0.10.1 1 10 100

Grain Size D (mm)quartz in water

2 / 3 2 / 3Combining and :S S

Q UA U R Q ARn n

If we know and and , we can find from

and then calculate . Or, we can simply adjust untilc c c c

c c

n S R gR S

Q Q R R


13/62

Discharge ( Q ): 70.0 cfs

Enter values only in green cells Channel Velocity ( U ): 3.31 ft/sec 2 / 31.49 S

1. CALIBRATION. For the slope, flow depth, and cross

section you have specified, use Manning's eqn to find

X/S.xls illustrates the back-calculation of n

Manning's n : 0.054

Channel Slope ( S ): 0.0120 Manning's n : 0.047Flow Depth: 1.64 ft Channel Velocity ( U ): 3.82 ft/sec

Selected Flow Depth: 1.64 ft Discharge ( Q ): 80.8 cfsCross-Sectional Area ( A ): 21.1 sq.ft. 2.3 cms

Hydraulic Radius ( R ): 1.2 ft Grain Size D 60 mm nD = 0.026Critical Shields Number t*c 0 06 t' = 16 8

2 / 31.49 S Q ARn

nQ

2. FIND DISCHARGE. For the slope, flow depth, andcross section you have specified, use Manning's eqn tofind the discharge Q for a value of the roughness n incell F53. FIND CRITICAL DISCHARGE. You can also

the roughness n for a value of the discharge Q in cellF1

Apply . .

Cross section of Minebank Run stream channel, 41.4 Pa t'/tc = 0.2928 feet downstream of gage 0158397967 December 2002. c 58.3 Pa

Survey Data: c 0.71Grade Rod (ft) Distance (ft)

4.0 0.04.4 0.04.4 2.0

3

3.5Left Bank

critical stress c needed to entrain grains of size Dspecified in cell F9 (given a value of critical ShieldsNumber given in cell F10

0

*( 1)

cc

gRS

s gD

Enter inform ation only in

green cells!!No cuttin and astin !

n c

. .4.7 8.04.8 12.04.9 14.05.5 18.05.7 20.05.9 21.06.7 22.3

4

4.5

5

5.5 d e R o d ( f t )

No inserting or deleting

rows!

7.0 23.07.2 24.07.1 26.07.3 28.07.3 30.07.0 32.07.1 34.06.8 36.0

6

6.5

7

7.5

G r a

Input

7.1 36.35.7 37.0 0 10 20 30 40 50Distance (ft)

13


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Q BhU continuity

/

b B aQ

h gS

For a simple, wide,

hydraulic geometry

momentum

2 / 3

5 / 3

soS

U hn

,find Q c directly Mannings eqn.

1

or bQ aQgS n

a S Q

n gS

*

15 / 3 1*

( 1) soc c

b

s gD

a

0.045( 1)

cc s gD

7 / 6c c s

nS


15/62

Q BhU

too sure aboutsome of the values/

b B aQ

h gS

needed todetermine Q c ?2 / 3

5 / 3

soS

U hn

Like n, D , and * c

what do you do?1

or bQ aQgS n

a S Q

n gS

*

15 / 3 1*

( 1) soc c

b

s gD

a

7 / 6c c

snS


16/62

Suppose your best estimate of Mannings n is 0.035and that you are pretty sure that the real value falls n c

y0.035

0.0025n

n

e ween . an . .

We could approximate your assessment of thevalue of n with a normal distribution with

F r e q u

2 n

mean = . s an ar ev a on = . .

95% of this distribution falls between 0.03 and 0.04,as can be seen in the cumulative fre uenc lot, so

n n 0.02 0.03 0.04 0.05Manning's n

0.8

1

q u e n c y

2 n

we are saying that the real value of n is 95% likely tofall between 0.03 and 0.04 and that it is more likely

to be around the center of the distribution (0.035) 0.20.4

0.6

u m u

l a t i v e

F r e

.values of n in our Monte Carlo simulation.

How does that work? We use a random number

00.02 0.03 0.04 0.05

Manning's n

1 n c y

generator to pick a number between 0 and 1 andthen use this number to find a value of n for thecumulative frequency distribution. For example,

=0.4

0.60.8

u l a t i v e

F r e q u

. , .for 0.23, n = 0.0332 0

.

0.02 0.03 0.04 0.05

C u

Manning's n


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n D*c

1 .

The Monte Carlo simulation

*. c va ues o , , an rom spec e requency s r u ons.

2 . Calculate critical discharge and transport rate.

3 . Repeat 1000 times.

c

b

Q BhU

B a2 .

4 . Distribution of calculated values givesestimate of the effect of input uncertaintyon calculated critical discharge and transport rate.

2 / 3

/

so

h gS

S U h

n

1

5 / 3 1

or b

b

S Q aQ

gS n

a S

100200

300

400Manning's n(a)

50

100

150

200*c (b)

4 .

*

1

( 1) soc c

n gS

s gD

0

0 . 0 2 0

0 . 0 2 4

0 . 0 2 8

0 . 0 3 2

0 . 0 3 6

0 . 0 4 0

0 . 0 4 4

0 . 0 4 8

0

0 . 0 3 0

0 . 0 3 4

0 . 0 3 8

0 . 0 4 2

0 . 0 4 6

0 . 0 5 0

0 . 0 5 4

0 . 0 5 8

200

250 Grain Size D(mm)(c)

200

250

300 Critical DischargeQc (m^3/s)(d)

*7 / 6 ( 1)c caQ s DnS 0

50

100

1 4 2 0 2 6 3 2 3 8 4 4 5 0 5 60

50

100

150

0 3 6 9 1 2 1 5 1 8 2 1 Monte Carlo


18/62

LEGENDSpecify Mean St. Deviation pdf

Manning's n 0.03 0.002 normal

cMonte Carlo Simulation

. .Grain size 5 0.25 normal

D (mm) 32 lognormal95% Pred.

o Qc ca cu at on, us ng1000 trials

300

400Manning's n(a)

150

200* (b)

a cu a e ean . ev a on n erva

Qc (m^3/s) 8.1 3.4 6.7

100

200

50

100

(uncertain)

solution to

0 . 0

2 0

0 . 0

2 4

0 . 0

2 8

0 . 0

3 2

0 . 0

3 6

0 . 0

4 0

0 . 0

4 4

0 . 0

4 8

0 . 0

3 0

0 . 0

3 4

0 . 0

3 8

0 . 0

4 2

0 . 0

4 6

0 . 0

5 0

0 . 0

5 4

0 . 0

5 8

250 Grain Size D 300 Critical Discharge

thresholdchannel

roblem

100

150

200mm

(c)

150

200250 Qc (m 3/s) Need a

better

0

50

1 4

2 0

2 6

3 2

3 8

4 4

5 0

5 6 0

50

0 3 6 9 1 2

1 5

1 8

2 1 tracergravels


19/62

Application of Monte Carlo approach for assessing risk:

Given a channel with a design discharge. The channel isintended to be static. You would like to be 90% sure that the

grains in the channel bed will not move at the designdischarge.

Solution: specify possible range in n , D, and * c , use MonteCarlo to determine the range of possible discharges. Adjust

are larger than the design discharge.

300 Critical Discharge

DesignDischarge

100

150

200250 Qc (m 3/s)

0

50

0 4 8 1 2 1 6 2 0 2 4 2 8


20/62

Garcia, Vanoni Bed Load Transport Relations


21/62

100,000100,000 D = 45 mm

Sediment Transport Why its hard to estimate 0

60,000

70,000

80,000

,

t e ( k g

/ h r )

10,000

t e ( k g

/ h r )

30,000

40,000

50,000

T r a n s p o r t

R

100

1,000

T r a n s p o r t

R

0

10,000

20,000

0 20 40 60 80 100 120

D = 45 mm

D = 30 mm

1010 100

DischargeDischarge

Example calculation using Meyer-Peter and Muller for a channel= = =

Discharge (cms) Discharge (cms)

. , . , .solid curve uses = 0.045 and D = 45 mm. The dashed curve uses

= 0.045 and D = 30 mm.

*c

*c

Note that at discharge Q = 55 cms, one curve indicates zerotransport and the other a transport rate of 80,000 kg/hr.


22/62

10 M-P&M2t*c

Sediment Transport Why its hard to estimate I

1ranspor ra e s a s eep, non nearfunction of bed shear stress

0.1

q* s

Three things make it difficult to

0.001

.accura e y es ma e :

1. Accelerations in unsteady,

0.0001

. .

2. Only a portion of the total acts totrans ort sediment

1E-050.01 0.1 1

varies locally. Because transportis a nonlinear fn of estimates

* based on total will be in error


23/62

t g xg xS gR0

Three things make it difficult toh U U accura e y es ma e :

1. Accelerations in unsteady,0

0 f

g x g x

gRS

. .


NOTE:


compute S f

based on total will be in error


24/62

The Drag Partition

, 0, per area, that the flow exerts on the boundary of the channel.

But onl a ortion of the

Three things make it difficult to total 0 acts on thesediment grains to produce

accura e y es ma e :

1. Accelerations in unsteady,

.transport rates, we need to

partition 0 into the part. .


that acts only on thesediment grains. We callthis the skin friction or


grain stress .There is no direct way to

based on total will be in error .

an approximate method,using Mannings equation.


25/62

Solve the Manning eqn for the boundary stress:

2 / 3 1/ 6 2 / 3( ) ( )g S nU gRS 2 / 3SRU

0gS nU

3/ 2 1/ 4 3/ 2 '

1/ 6 .

1/ 4

Manning-Strickler Roughnessfor D in mm

.

65 65where we use 2 D D

1/6

*8.1

s

U hu k

3 21000 kg/m & = 9.81 m/sg

1/6

8.1

s D

k n

g


26/62

Using both the grain roughness and total roughness, we have anestimate of the portion of the total stess acting on the grains the

0.06

0.07

0.08

typical range of n values

. 3/ 21/ 4o gS nU

0.04

0.05

a n n i n g s n

3 21/ 4

' DgS n U

0.01

0.02

.

Manning-Strickler n

D

o

n

n

1 10 100 1000Charactersitc Bed Material Grain Size (mm)

3/ 2' D o

n

nBasically the same drag partition suggested by Meyer-Peter & Muller in the . ,application in HEC-RAS


27/62

Sediment Transport Why its hard to estimate II

W at s or a reac ?

Critical shear stress depends linearlyon grain size, but the grain size of thetransport is poorly known!

100

)

10

e a r s t r e s s

( P a

c Motion

Transport rate is a steep, nonlinearfunction of bed shear stress

1 S

h

No Motion

Small error in c can produce bigerror in qs

0.1

0.1 1 10 100Grain Size D (mm)quartz in water


28/62

Estimating Sediment Transport: Why its hard

Sediment transport is a local process: between individual grainsand the near-bed flow they encounter

The transport process is highly nonlinear: a very steep functionw w

Real streams have considerable variability in topography, flow,and bed composition The local flow actin on rains is hard to estimate The distribution of bed grain sizes is hard to determine You never know either local flow or local bed composition in

detail, so you have to estimate Errors in estimated transport rates are typically very large

(often well in excess of an order of magnitude!) Next, we examine an approach to provide reliable transport

estimates of reater accurac with acce table effort


29/62

If estimating transport rates using shear stress is so hard, remind me again why we cant find some general formula using Q.

t an est mate or , ets return tothe sediment rating curve problem. sQ Q

relation based on discharge Q can beeneral.

sr r

10 M-P&M

1

2t*c(t*)^3We approximate transport with a

power function of

* *q 3'sq 0.01.

q*

3'

0.0001

0.001

Write the relation twice, take ratio

'sr r q

1E-050.01 0.1 1

*


30/62

3

' 'sr r q

''

r

U U

We use our grain stress relation:

4.5

sq U

Which givessr r q U

.sQ U For simplicity, we assumethe width of transport does

sr r not change


31/62

We can relate U to Q using the hydraulicgeometry:

4.5 4.5 mbaQ B

r r U Q

mk U

c 4.5 msQ Q sr r

For 0.2 < m < 0.8, 1.0 < < 3.6.s no a cons an s no a su a y generavariable for a transport model.

Even larger values of found in steep mountainstreams (typically very small transport rates,

>

W find the exponent to vary between 1 and 10 (!!)


32/62

Estimating Bed-Material


33/62

Bed/Suspended Load vsWash/Bed Material Load How many sizes?


34/62

Washloadclay, silt, fine

suspension ?? uplands,banks,

true washload:(a) Transport not predictable

san , oo e n e o a ectransport of other fractions

Bed Material Fine

bed load orsuspension

interstices,stripes and

grain path in near-bed regiondominated by larger grains;

med-crs sand, pea gravel

dunes,subsurface

hard to sample & model

Bed Material bed load bed framework displacements generally rareCoarse

med-crs gravel,cobble

Hugeboulder

typical high flows

grains to exclude from thetransportable population

We focus on fine and coarse bed material


35/62

What about more fractions?

Many-fraction models available: including ones basedon the grain size of the bed surface, which allow for the

.

These models are fragile output is sensitive to the.

To simulate transport at a particular location, at aarticular time

need large amounts of detailed bed and flow data

Use these models to ask more general questions:

e.g. change in bed composition & transport witha change in flow releases from dams or

We will apply such a model later


36/62

Estimating Sediment Transport: Three overarchingconstraints bound the problem

Large spatial & temporal variabilityNeed BIG samples

Strongly nonlinear processese -mater a transport s a oca a a r

Modeling with spatial/temporal averages produces large error Sparse information available relative to that needed to model the

Modeling with high spatial/temporal resolution requires vast amounts of data Generally, we need a robust approach


37/62

Target: sediment rating curve Q s = ( Q )

Predict from a flow & transport model We cant measure ever thin but

Approaches

Models too fragile

Sample transport , Accuracy poor and a full sampling program too

burdensome

Tracer gravels and scour chains Measure volumetric change in local topography

1. Dealing with grain size2. Calculating transport rates

Issues:

3. Sampling bed material load

4. Best approach?


38/62

.

motion: What is Q c ? -

rate: What is Q se ave trou e est mat ng c, as we as y rau c parameters


39/62

Calculating cumulativeQ

h

sediment transport in asimple, prismatic channel,

b B aQ

over a hydrograph U h

nnQ

2 / 33 / 5

baQ Sha or uncer a n y nn, D , and * c

nh

a S

consequences?

3/ 23/51 0.73 *2650

bnQ S

ep ace n

ghS Replace in Meyer-Peter & Muller:

1000 ( 1)s o ca s D 0.7nQg S

a


40/62

LEGENDSpecify Mean St. Deviation pdf

Mannin 's n 0.03 0.002 normalMonte Carlo Simulation

c 0.045 0.0050 normalGrain size 5 0.25 normal

D (mm) 32 lognormal

of Qs calculation, using1000 trials

400'

200 Cumulative

.Calculate Mean St. Deviation Interval

Cumulative transport (tons) 14,997 2,806 5,509

0

100

200

300 a

0

50

100

150 c (b)

100150200250 Transport

(metric tons)(f)

0 . 0

2 0

0 . 0

2 4

0 . 0

2 8

0 . 0

3 2

0 . 0

3 6

0 . 0

4 0

0 . 0

4 4

0 . 0

4 8

0 . 0

3 0

0 . 0

3 4

0 . 0

3 8

0 . 0

4 2

0 . 0

4 6

0 . 0

5 0

0 . 0

5 4

0 . 0

5 8

200

250 Grain Size D(mm)(c)

60

80

100

( m ^ 3 / s )

(d)

050

2 0 0 0

6 0 0 0

0 0 0 0

4 0 0 0

8 0 0 0

2 0 0 0

6 0 0 0

0 0 0 0

0

50

100

1 4

2 0

2 6

3 2

3 8

4 4

5 0

5 6

0

20

40

0 20 40 60 80 100Time (hrs)

D i s c h a r g e 1 1 1 2 2 3

Useful for evaluating uncertaintyin sediment balance between

MonteCarloTransport.xls


41/62


42/62

2. Estimating Transport: Sampling

Option 1: trap all the transport in a weir, slot, or pool

-

samplers installed on the bed.

Option 1 is best, butgenerally not practical

Option 2 is practical,but involves larger

error, some r s , someluck, and lots of effort


43/62

Bi River Sam lin 1991


44/62


45/62

Big River Sampling 2007

R ll Bi Ri S li 2007


46/62

Really Big River Sampling 2007


47/62


48/62

Sampling ITransport field highly variable in space & time

Need LARGE samples!

Define Sampling Intensity for a 6.5m stream

Transport. = ,

Hand-held 13 * 0.075m X 120s = 117 msSamplingTransect

.

rapSampler . m s = ,ms (23.5%)

So, pit or net-frame traps look pretty good

Pit samplers fill up


49/62

Pit samplers fill up

Photo J Pizzuto, U Delaware, home of big samples

S d t f l


50/62

So do net-frame samplers

Bunte, K., 1998. Development and ield testing o a stationary net- rame bedload sampler ormeasuring entrainment of pebble and cobble particles, Report prepared for the StreamSystems Technology Center, USDA Forest Service, Rocky Mountain Research Station, Fort

Collins, CO, 74 pp.


51/62

3 E ti ti g S di t T t


52/62

3. Estimating Sediment TransportDirect sampling+ gives a direct relation between Q and Qs- requ res a g e or

- cannot predict future conditions- error: is it random or s stematic? - Handheld samplers: Scooping, perching, limited

grain size & TINY SAMPLES- - ,rates

+ can predict future changes- highly inaccurate

ow ar o sca eboundary conditions poorly known

The alternative? Join forces.


53/62

The alternative? Join forces.

the future is a combination of simple robustmodels and efficient measurement . A first cut, using today s technology

If pit/trap samplers can collect good samplesof small transport rates, why not use these tocalibrate a model of coarse bed-materialranspor

GOOD samples! (long duration, spatially

Two sizes: sand and gravel om ne n ro us ramewor a s

insensitive to major sources of error

T t F l (G l)


54/62

Transport Formula (Gravel)

r r 5.4

846.012.11

iW 2.14*

r r .

r Calibrate **

( 1) bs gqqW

* ( / )

Formula provides the trend,but the samples provide the accuracy

Transport formula based on shear stress so


55/62

Transport formula based on shear stress , so

stress from discharge. We return to ourMannin -Strickler formula:

2/34/16517 U SD

For Cub River: .

D65 = 90 mm = 0.42 .


56/62

0.01

0.1 ObservedTransport Funct ionW*r

W* 0.0001

0.001 ObservedTransport FunctionW*r

Qs(m^3/s)

* 0.039c

0.001 1E-05

0.00001

.

1E-07

-

0.0000010.1 1 10

/ r

1E-080.1 1 10

Q (m^3/s)

SRC.xls


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1 A B C D E F G H I J K L M N O P

Lonely R. NR Whirlpool Observed Observed23456789

s wor s ee or ranspor Slope S 0.015D65 90 (mm) D65 of bed surfaceD50 42 (mm) D50 of bed subsurfacefg 0.50 proportion fravel in bedb 15.0 (m) width of bed material, not channelrg 30 (Pa) Adjust the reference shear stress*rg 0.044 Reference Shields Number; is it reasonable?

0.001

0.01

.Transport FunctionW*r

W*

0.00001

0.0001

. Transport Function*

Qs(m^3/s)

10111213141516

Qrg 10.2 cmsObserved Observed

Q Qsg ' '/ rg qsg W*g(m^3/s) (m^3/s) (Pa) - (m^2/s) -

3.37 7.07E-09 17.0 0.57 4.71E-10 6.87E-064.90 8.80E-08 20.6 0.69 5.87E-09 6.42E-056.71 2.76E-08 24.2 0.81 1.84E-09 1.59E-05

- - - 0.000001

0.00001

0.0001

1E-08

0.0000001

0.000001

181920212223

24

. . . . . .8.78 4.35E-06 27.7 0.92 2.90E-07 2.03E-03

0.00 1.00E-20 1.00E-200.00 1.00E-20 1.00E-200.00 1.00E-20 1.00E-200.00 1.00E-20 1.00E-200.00 1.00E-20 1.00E-20

0.00 1.00E-20 1.00E-20

0.00000010.1 1 10

'/ r

1E-091 10 100

Q (m^3/s)

Lonely R. NR Whirlpool

Enter parameters of velocity rating curve herek = 0.63 m = 0.34 Qc = 0.00

( )mt U k Q Q 1.21.4

1.6

1.8

U ( m / s )

The cells below contain values of k and m

fitted to the (Q, U) values for specified Qck = 0.63 m = 0.34

0.3382 -0.1983 LINEST resultsQ U Q-Qt

(m^3/s) (m/s) (m^3/s)1.568

0.4

0.6

0.8

1

M e a n

V e l o c i

t y

. . .4.225 1.04 4.2255.946 1.17 5.9467.891 1.29 7.891

12.340 1.47 12.34013.326 1.50 13.326

0.0000.000

0

0.2

0 5 10 15 20

Discharge Q (cms)

Transport samples provide accuracy by


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p p p y y

(1) Establishing grain size D of the transport &

2 Scalin the flow

Calculate grain stress: 0.25 1.5

6517 SD U 1.5for two different flows & take the ratio:

Suppose

1 1

2 2U b

U a1.5b

1.5b

Q en

2 2Q r r

If you calibrate, the transport formula is needed only to

give the change in transport with the change in discharge.


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high flows & you base yourestimate on samples at low

, , .

in gravel-bed rivers, Earth Surface Processes & Landforms 26, 1395-1408.

Estimating bed-material transport in gravel-bedi


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rivers Conceptual basis

- fine and coarse bed material- (supply of one affects the transport of the other)

Sampling- standard needed for minimum sample size- fine bed material Helle -Smith or ?- coarse bed material pit or net frame samplers- big rivers ???

- 2-fraction model captures essential interaction between fine &coarse

& facilitates inte ral measure of reach rain size

- But it can t do everything (armoring; change in sand or gravel size) Future

,- can be done now in wadeable streams, although effort is non-trivial

SRC & BAGS


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-transport rates in gravel-bed rivers. Stream Systems Technology Center, (the StreamTeam), US Forest Service

B GS Yantao Cu , St water Sc ence : supports var ety o transport ormu as, a ows

variety of input, includes calibrated approachUsers Manual by John Pitlick

Primer by Peter Wilcock


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Transporte de Sedimento

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