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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
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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
'
'
'
==
=
==
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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.
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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
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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
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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
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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
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(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
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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
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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
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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
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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