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 slope to very large extent and also the x-section.

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

Sediment Discharge

It is rate of transportation of sediment by a channel.

Expression in terms of hydraulic parameters and sediment

properties.

To predict amount of degradation, aggradation or bank erosion.

Suspended Load

It is the sediment that is lifted off the bed of a channel and carried

up into the body of flow by the vertical components of the turbulence

velocities due to eddies.

Concentration of sediment decreases with distance up from the bed.

Gravity pulls down and eddies pushes up .

The steady state distribution of concentration of suspended load is

obtained by

*

where

ku

w

k

wz

aD

a

y

yD

C

C

wo

z

a

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

'/'city shear velo*

w

wo SgRSRu

k

nd

n

n

nRR

SR

s'

wo

24 and

24

, where

stressshear bed

6161

23'

'

'

*

where

ku

w

k

wz

aD

a

y

yD

C

C

wo

z

a

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 if the concentration of

sediment particles at 0.4 m above the bed is 400 ppm. Take specific

gravity of particles as 2.67, L-slope as 1 in 4444 and representative

roughness of size of bed ks = 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

Ca = 400 ppm = 400/106 = 400 x 10-6 w 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/m 0959.04.04

4.0

1

1410400 3

3.13

z

aaD

a

y

yDCC

Bed Load

Moves along bottom of the channel either by rolling, sliding or

jumping in small leaps.

Transmits its load to static grains below.

Grains exchange places with similar particles.

Not vertically supported but rest on bed.

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 it would be

obtained on a plain bed to the actual value on ripple bed.

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

gs = Volume of sediment transport (by weight) per unit width of

channel per hour.

32'3123'

25.0047.0s

w

ww

s gg

dSDn

n

Q

Q

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/hr 4700

23'

cbsn

ng

24

61

' sk

n

stress effective of measure is

047.0

'

n

n

d

b

wc

Einsteins Bed Load Function

Based 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 particle is:

where,

iB = fraction of bed load gs of diameter d

ib = 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 sediment

d = grain size or diameter

= density of fluid

** vs.

21

3

21

*

1 and ,

gd

g

i

i

w

s

b

B

1

**

**

**

**

2

1

11

B

B

t

A

Adtep lyrespective 2 and 0.143 43.5,1 and ,, Where *** BA

For uniform soils

(i)iB = ib(ii)* =

(iii)

where, = shear intensity of particle

relationship for uniform bed material

Graph, * = *

SR

d'*

391.0

465.0

1 e

Example # 2

A wide channel 4 m deep consists of uniform grain of 0.4 mm. Taking

specific gravity of particles as 2.67, L-slope as 1 in (4444+R) and

representative roughness of size of bed ks as 2.0 mm, determine the

quantity of bed load moved 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/hr 4700

2323'

cbsn

ng

dGdGdwwwwc 1047.0047.0047.0

233 kg/m 0314.0104.01067.1047.0 c

kg/m/hr 20260314.0765.09.04700 2/323' s

g

(b) Einsteins equation

For uniform soils

From Einsteins curve

For * = = 1.11 we get * = =4

From

We get

11.14444/1675.2

104.067.11

3

'''*

SR

dG

SR

dG

SR

d

w

ww

w

ws

21

3

21

1

gd

g

g

s

3321

3

21

1 gdGGgdG

Ggdgwwgs

kg/m/hr 12453600104.081.967.11067.24 333 sg

(b) Einsteins equation

For uniform soils

Putting above value in the relationship,

we get * = = 1.42

Now, From

We get

11.1*

31 gdGGg ws

kg/m/hr 4423600104.081.967.11067.242.1 333 sg

391.0

465.0

1 e

Example # 3Design an irrigation channel carrying a full supply discharge of 28 cumecwith a bed load concentration of 40 ppm. The average grain diameter ofthe 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/hr

Applying Laceys P-Q relationship, P = 4.75 Q = 4.7528 = 25.13 m

Assuming channel bed width, B = 20 m

Rate of bed load transport per unit width, gs = 4032/20 = 201.6 kg/m/hr

According to Meyer-Peter-Muller equation

where

n = d1/6 / 24

Where d is effective diameter since the material is not uniform and theaverage grain diameter is given as 0.4 mm, let us take effective graindiameter 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 channel maybe taken in good shape and smoother soil, we can take the value of n as0.02.

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

Now, critical tractive force

[Q/Qs = 1 for wide channels]

(1) ------kg/m/hr 4700

2323'

cbsn

ng

dG w1047.0

2kg/m 028.0

stressshearQQ sb / DSw1 DSw DS

310

dG ww 047.0 dwc 047.0

)104.0(1067.1047.0 33 c

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

128

2135

2132

SR

SRR

R

-610337S ,2 From

022.010337

216

32

RR

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

RESULTS:

Bed width of channel = 20 m

Depth of channel = 1.3 m

Slope of channel bed = 1 / 3470

m 3.1

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