chap5-3 - sediment transport

8/2/2019 Chap5-3 - Sediment Transport

1/19

Sediment Transport

The x-section and slope of true regime channel are controlled

by following three variables independent of channel:

Discharge in the channel;

Nature of sediments entering the channel, i.e. the grain

size distribution, shape of the grains and their specific

gravity; and

Quantity of the sediments entering the channel.

Regime theories account for only first two variables.

The third variable is very important and affects the slopeto very large extent and also the x-section.


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Classification of Sediment

1. Suspended Load:It is carried in the fluid away from the bed.

2. Bed Load:It moves on or near the bed.

Total load = Suspended load + Bed load


3/19

Sediment Discharge

It is rate of transportation of sediment by a channel.

Expression in terms of hydraulic parameters and sedimentproperties.

To predict amount of degradation, agradation or bankerosion.


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Suspended Load

It is the sediment that is lifted off the bed of a channel andcarried up into the body of flow by the vertical components ofthe turbulence velocities due to eddies.

Concentration of sediment decreases with distance up from thebed.

Gravity pulls down and eddies pushes up .

The steady state distribution of concentration of suspendedload is obtained by

*

where

ku

w

k

wz

aD

a

y

yD

C

C

wo

z

a

==

=


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C = sediment concentration at distance y up from bed.

Ca = known concentration at some reference level i.e. height

a above the bed.D = depth of water in the channel

w = settling or fall velocity of sediment grains in the channel

k = Von-karmans universal constant = 0.4

d = average grain size of suspended load

ks = average grain size of bed load

waterofdensity

'/'cityshear velo*

=

====

w

wo SgRSRu

k

nd

n

n

nRR

SR

s'

wo

24

and

24

,where

stressshearbed

6161

'

'

'

==

=

==


6/19

Bed Load

Moves along bottom of the channel either by rolling, slidingor jumping in small leaps.

Transmits its load to static grains below.

Grains exchange places with similar particles.

Not vertically supported but rest on bed.


7/19

Meyer-Peter-Muller Equation

Dimensionless equation for any system of units.

where

Qs/Q = actual discharge/estimated discharge assuming walls to

be frictionless 1 for wide channels.n/n = ratio between the value of Mannings coefficient as itwould be obtained on a plain bed to the actual value on ripplebed.

w = specific weight of water

= specific weight of sediment particle

S = Slope of channel

D = depth of water

d = grain diameter

g = acceleration due to gravity

( ) ( )32'

3123'

25.0047.0 sw

ww

s

ggdSDn

n

Q

Q

+=


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Meyer-Peter-Muller equation can also be written as

where, b = bed shear = (Qs/Q) x shear stress

= (Qs/Q) x w R S

= w D S

Since, for wide channels R = D and Qs/Q 1

where, ks = effective grain diameter in mm

c = critical tractive force in kg/m2

= minimum tractive force at which grains start moving.

kg/m/hr4700

23'

=

cbsn

ng

24

61

' sk

n =

( )

stresseffectiveofmeasureis

047.0

'

=

n

n

d

b

wc


9/19

Einsteins Bed Load FunctionBased on law of probability

Involves number of experimental coefficients and assumptions.

Universal relationship,

Bed load transport = f(flow intensity, sediment size)According to Einstein, the probability p for motion of a sediment particleis:

where,

iB = fraction of bed load gs of diameter dib = fraction of grains of diameter d in the bed

gs = bed load discharge per unit width of channel

w = mass density of water

= density of sedimentd = grain size or diameter

**vs.

==21

3

21

*1

gd

g

i

i

w

s

b

B

+==

1

**

**

**

**

2

1

11

B

B

t

A

Adtep lyrespective2and0.14343.5,1and,,Where *** = BA


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For uniform soils

(i)iB = ib

(ii)* =

(iii)

where, = shear intensity of particle

relationship for uniform bed material

Graph,* = *

SRd

'*

==

391.0

465.0

1 = e


11/19


12/19

Example # 1

A wide channel 4 m deep consists of uniform grain of 0.4 mm.The fall velocity of grains in still water is 0.04 m/sec.

Determine the concentration of load at 1.0 m above the bed ifthe concentration of sediment particles at 0.4 m above thebed is 400 ppm. Take specific gravity of particles as 2.67, L-slope as 1 in 4444 and representative roughness of size of bedks = 2.0 mm.

Data:D = 4 m, d = 0.4 mm, w = 0.04 m/sec, a =0.4 m

Ca = 400 ppm y = 1.0 m, G = 2.67, S = 1/4444,

ks = 2.0 mm, C = ?Solution:

*

where

ku

w

k

wz

aD

a

y

yD

C

C

wo

z

a

==

=


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Ca = 400 ppm = 400/106 = 400 x 10-6w kg/m

3

= 400 x 10-6 x 103 = 400 x 10-3 kg/m3

For 1 m width Ca = 400 x 10-3 kg/m3/m or 400 x 10-3 kg/m2

Shear velocity u* = (o/w) = (wRS/w) = (gRS), but R =

R(n/n)3/2

For wide channel, R D = 4.0 m

R = R(n/n)3/2 = D(n/n)3/2 = 4.0(0.765)3/2 =2.675 m

u* = (gRS) = (9.81 x 2.675 x1 /4444) = 0.0768 m/sec

765.02

4.0

24

246161

61

61'

=

=

==

ssk

d

k

d

n

n

3.10768.04.0

04.0

*

=

==ku

wz

ppm.95.9/mkg/m0959.0

4.04

4.0

1

1410400 2

3.1

3 ==

=

=

z

a

aD

a

y

yDCC


14/19

Example # 2

In the above problem, determine the quantity of bed loadmoved by the channel applying (a) Meyer-Peter-Muller

equation, and (b) Einsteins method.Solution:

(a) Meyer-Peter-Muller equation

b = w D S = 103 x 4 x 1/4444 = 0.9 kg/m2

Now, substituting values in the above equation, we get

kg/m/hr4700

2323'

= cbs nn

g

( ) ( ) ( ) dGdGdwwwwc

1047.0047.0047.0 ===

( ) 233 kg/m0314.0104.01067.1047.0 == c

( ){ } kg/m/hr20260314.0765.09.047002/323' ==

sg


15/19

(b) Einsteins equation

For uniform soils

From Einsteins curve

For * = = 1.11 and * = =7

From

We get

( ) ( ) 11.14444/1675.2104.0

67.11

3

'''* =

==

=

==

SR

d

GSR

dG

SR

d

w

ww

w

ws

=21

3

21

1

gd

g

g

s

( )33

21

3

21

1 gdGGgd

G

Ggdg wwgs =

=

=

( ) kg/m/hr7.21783600104.081.967.11067.27333 ==

sg


16/19

Example # 3Design an irrigation channel carrying a full supply discharge of 28cumec with a bed load concentration of 40 ppm. The average graindiameter of the bed material may be taken as 0.4 mm and its

specific gravity as 2.67. Apply Laceys regime perimeter and Meyer-Peter-Muller equation.

Given Data:Q = 28 cumec, Bed load concentration = 40 ppm,G = 2.67, d = 0.4 mm.

Required:B = ? D = ? S = ?

Solution:Quantity of bed load transported = 40 ppm = 40/106

= 40/106 x 28 m3/sec x 103 kg/m3

= 1.12 kg/sec = 1.12 x 3600 kg/hr =4032 kg/hrApplying Laceys P-Q relationship, P = 4.75 Q = 4.7528 = 25.13 mAssuming channel bed width, B = 20 mRate of bed load transport per unit width, gs = 4032/20 = 201.6kg/m/hr


17/19

According to Meyer-Peter-Muller equation

where

n= d1/6 / 24

Where d is effective diameter since the material is not uniform andthe average grain diameter is given as 0.4 mm, let us takeeffective grain diameter as 0.5 mm.

n= d1/6 / 24 = (0.5 x 10-3)1/6 / 24 = 0.0117

Since the discharge in the channel is > 12 cumec and the channelmay be taken in good shape and smoother soil, we can take thevalue ofn as 0.02.

[Note: If Q < 12 cumec, n = 0.0225 and if Q > 12 cumec, n = 0.02]

Now, critical tractive force

b = (Q/Qs) c = 1 x w D S = w D S = 103 D S [Q/Qs = 1 for wide

channels]

(1)------kg/m/hr4700

2323'

=

cbsn

ng

( ) ( ) ( ) dGdGdwwwwc

1047.0047.0047.0 ===233 kg/m028.0)104.0(1067.1047.0 ==

c


18/19

Substitution of values in equation (1) yields:

201.6 = 4700[103RS(0.585)3/2-0.028]3/2

RS = 337 x 10-6 ---------- (2)

Applying Mannings equations

R = A/P, A = P R = 25.13 R

Substituting in (3)

R = 1.167 m

S = 337 x 10-6 /1.167 = 1 / 3470

21321 SARn

Q =

( )3-------022.0.13.2502.0

1

282135

2132

==SR

SRR

( )R

-610337S,2From

=

022.010337 21632 =

RR


19/19

Taking side slope, (H):1(V)

RESULTS:

Bed width of channel = 20 mDepth of channel = 1.3 m

Slope of channel bed = 1 / 3470

m3.1

034.234.175.0

5.0206.234.23

236.220

5.020167.1

236.2205

5.0205.0

2

2

2

22

==+

+=++

+==

+=+=

+=+=

D

DD

DDD

D

DD

P

AR

DDBP

DDDBDA

chap5-3 - sediment transport

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