Hydraulics 3 A2-1 Dr David Apsley
ANSWERS TO EXAMPLES (SEDIMENT TRANSPORT) AUTUMN 2017
(Full worked answers follow on later pages)
Q1. (a) 4.60 m s–1
(b) 0.112 m s–1
Q2. τ*crit = 0.0314; τcrit = 0.508 N m–2
Q3. (a) 0.451 m s–1
(b) 2.23 m s–1
Q4. (b) 0.132 m
(c) 0.141 m; 0.549 m; 0.273 m
Q5. (b) 0.342 m
Q6. d = 88 mm; h = 4.95 m
Q7. (a) 0.0990 m2 s
–1; 0.1 m
(b) 0.034 m; 2.88 m s–1
; 42.2 N m–2
(c) erodes
Q8. (a) 0.0180 m–1/3
s
(b) 1.53 m
(c) 18.8 N m–2
(d) Meyer-Peter and Müller: 1.0510–3
m2 s
–1; Van Rijn: 6.3510
–4 m
2 s
–1
(e) uτ/ws < 1; no suspended load occurs
Q9. (a) 0.0134 m–1/3
s
(b) 1.46 m
(c) 18.0 N m–2
(d) Meyer-Peter and Müller: 1.1610–3
m2 s
–1; Nielsen: 1.7610
–3 m
2 s
–1
(e) 0.0607 m s–1
(f) 0.044 m3 s
–1 per metre width
Q10. (a) 0.0175 m–1/3
s
(b) 1.68 m; 2.09 m s–1
(c) bed shear stress = 11.0 N m–2
and exceeds the critical shear stress
(d) 4.3710–4
m3 s
–1 per metre width
(e) 0.207 m s–1
Q11. No numerical answer
Q12. No numerical answer
Q13. No numerical answer
Q14. No numerical answer
Q15. (c) 2.5 m3 s
–1
Hydraulics 3 A2-2 Dr David Apsley
Q1.
Cheng’s formula:
2/32/12 5)*2.125( dν
dws , where
3/1
2
)1(*
ν
gsdd
Densities:
ρsand = 2650 kg m–3
ρair = 1.2 kg m–3
ρwater = 1000 kg m–3
Kinematic viscosities:
νair = 1.510–5
m2 s
–1
νwater = 1.010–6
m2 s
–1
(a) For sand particles in air:
22082.1
2650
ρ
ρ ss
82.45)105.1(
81.9)12208(001.0
ν
)1(*
3/1
25
3/1
2
gsdd
3.3065)82.452.125(5)*2.125(ν
2/32/122/32/12 ddws
Hence,
1
3
5
sm595.410
105.13.306
ν3.306
d
ws
Answer: settling velocity in air = 4.60 m s–1
.
(b) For sand particles in water:
65.21000
2650
ρ
ρ ss
30.25)100.1(
81.9)165.2(001.0
ν
)1(*
3/1
26
3/1
2
gsdd
5.1115)30.252.125(5)*2.125(ν
2/32/122/32/12 ddws
Hence,
1
3
6
sm1115.010
100.15.111
ν5.111
d
ws
Answer: settling velocity in water = 0.112 m s–1
.
Hydraulics 3 A2-3 Dr David Apsley
Q2.
As in Q1(b) above,
30.25* d
Soulsby’s formula,
)]020.0exp(1[055.02.11
30.0τ *
*
* dd
crit
where gds
b
)ρρ(
τ*τ
Hence,
03141.0)]30.25020.0exp(1[055.030.252.11
30.0τ*
crit
2mN5084.0001.081.9)10002650(03141.0)ρρ(ττ * gdscritcrit
Answer: critical Shields parameter = 0.0314; critical shear stress = 0.508 N m–2
.
Hydraulics 3 A2-4 Dr David Apsley
Q3.
From Q1 and Q2 above, the settling velocity and critical shear stress for incipient motion are
given, respectively, by:
1sm1115.0 sw
2mN5084.0τ crit
(a) The bed shear stress is given, in general, by
)ρ(τ 2
21 Vc fb
Rearranging for V:
ρ
τ2 b
fcV
At incipient motion, τb = τcrit. Hence,
1sm4510.01000
5084.0
005.0
2 V
Answer: for incipient motion, velocity = 0.451 m s–1
.
(b) For incipient suspended load,
swu τ
where uτ is the friction velocity and ws is the settling velocity.
By definition,
2
)(ρ/τ 2
21
τ
f
fb
cVVcu
Hence,
s
fw
cV
2
1sm23.2005.0
21115.0
2 f
sc
wV
Answer: for incipient suspended load, velocity = 2.23 m s–1
.
Hydraulics 3 A2-5 Dr David Apsley
Q4.
q = 0.9 m2 s
–1
d = 0.06 m
ρs = 2650 kg m–3
τ*crit = 0.056
cf = 0.01
ws = 0.8 m s–1
From the critical Shields stress we can determine the critical bed stress for incipient motion:
2mN39.5406.081.9)10002650(056.0)ρρ(*ττ gdscrit
(a) Upstream of the gate:
h = 2.5 m
1sm36.05.2
9.0 h
qV
22
212
21 m N648.036.0100001.0ρτ Vc fb
The bed stress is (considerably) less than the critical value; the bed is stationary.
(b) Sluice gate assumption: total head the same on both sides of the gate.
g
Vz
g
Vz ss
22
2
22
2
11
For a flat bed:
2
2
2
22
1
2
122 gh
qh
gh
qh
Substituting values:
2
2
2
04128.0507.2
hh
Rearranging for the supercritical solution:
2
2507.2
04128.0
hh
Iteration (from, e.g., h2 = 0) gives
m1318.02 h
Then:
1
2
sm829.61318.0
9.0 h
qV
22
212
21 m N2.233829.6100001.0ρτ Vc fb
τb exceeds τcrit; the bed is mobile.
Answer: downstream depth = 0.132 m; bed stress exceeds the critical value.
(c) Scour will continue with the depth increasing and the velocity and stress decreasing, until
the stress no longer exceeds the critical value. At this point:
Hydraulics 3 A2-6 Dr David Apsley
2mN39.54τ b
39.54ρ 2
21 Vc f
1sm298.31000
39.54
01.0
2
ρ
39.542 fc
V
The overall depth downstream is then
m2729.0298.3
9.0
V
qh
Since the water level is fixed by the gate it is the same as before. The depth of scour is then
m1411.01318.02729.0
Sluice gate assumption: total head the same on both sides of the gate.
g
Vz
g
Vz ss
22
2
22
2
11
Upstream, zs1 = h1, but downstream there is a distinction between water level zs2 = 0.1318 m
and depth h2 = 0.2729 m.
2
2
2
22
1
2
122 gh
qz
gh
qh s
Substituting values:
6861.004128.0
2
1
1 h
h
Rearranging for the subcritical solution:
2
1
1
04128.06861.0
hh
Iteration (from, e.g., h1 = 0.6861) gives
m5493.01 h
Answer: depth of scour = 0.141 m; depth upstream = 0.549 m; depth downstream = 0.273 m.
(d) To determine the significance of suspended load compare the settling velocity with the
maximum friction velocity (which occurs in the initial downstream state).
Settling velocity:
ws = 1.1 m s–1
(given)
Friction velocity:
1
τ sm4829.01000/2.233ρ/τ u
Here, the settling velocity greatly exceeds the friction velocity, so suspended load is
negligible.
Hydraulics 3 A2-7 Dr David Apsley
Q5.
b = 5 m (width of main channel)
bmin = 3 m (width of restricted section)
d = 0.01 m
ρs = 2650 kg m–3
(s = 2.65)
05.0τ* crit
cf = 0.01
Q = 5 m3 s
–1
The bed is mobile if and only if the bed shear stress exceeds the critical stress for incipient
motion. Alternatively, one may compare the velocity with a critical velocity found using the
friction coefficient – this is more convenient here.
The critical shear stress for incipient motion is
2mN093.801.081.9)10002650(05.0)ρρ(ττ * gdscritcrit
and the corresponding velocity for incipient motion is given by
)ρ(τ 2
21
critfcrit Vc
whence
1sm272.11000
093.8
01.0
2
ρ
τ2 crit
f
critc
V
This is a venturi so we need to determine velocities at various locations.
Upstream, zs = h = 1 m and the velocity and total head are, respectively,
1sm0.115
5
bh
QV
m051.181.92
11
2
22
g
VzH sa
If critical-flow conditions occur at the throat then the total head there would be
m9850.081.93
5
2
3
2
3
2
3
2
33/1
2
23/1
2
min
23/1
2
gb
Q
g
qhH m
cc
Since the upstream head exceeds that required for critical-flow conditions the flow remains
subcritical throughout, with total head in the restricted section, H = 1.051 m. The depth h is
given by:
22
min
22
22 hgb
Qh
g
VhH
Rearranging as an iterative formula for the subcritical value of h:
22
min
2
2 hgb
QHh
Substituting numerical values:
2
1416.0051.1
hh
Hydraulics 3 A2-8 Dr David Apsley
Iteration (from, e.g., h = 1.051) gives
h = 0.8592 m
and corresponding velocity
1
min
sm940.18592.03
5
hb
QV
Hence we have:
in the 5 m width, V = 1.0 m s–1
< Vcrit and the bed is stationary;
in the 3 m width, V = 1.94 m s–1
> Vcrit and the bed is mobile.
(b) The bed will erode in the restricted section until V = Vcrit. Then the flow depth is given by
minhbVQ crit
m310.1272.13
5
min
critVb
Qh
If Δzb is the change in height of the bed, then the total head is given by
g
VhzEzH bb
2ΔΔ
2
Hence,
m3415.081.92
272.1310.1051.1
2Δ
22
g
VhHzb
Answer: depth of scour hole = 0.342 m.
Hydraulics 3 A2-9 Dr David Apsley
Q6.
b = 12 m
S = 0.003
Q = 200 m3 s
–1
056.0τ* crit
Let the minimum stone diameter (corresponding to incipient motion) be d. Then the critical
shear stress for incipient motion is given by
ddgdscritcrit 4.90681.9)10002650(056.0)ρρ(ττ *
For normal flow:
SgRhb ρτ
For incipient motion:
dd
gSR crit
h 80.30003.081.91000
4.906
ρ
τ
(*)
The corresponding depth of flow h is given by the expression for Rh in a rectangular channel,
bh
hd
/2180.30
With b = 12 m this gives (after some rearrangement) an expression for h in terms of d:
d
dh
133.51
8.30
(**)
From Manning’s equation:
bhSRn
VAQ h 2/13/21 where
1.21
6/1dn (Strickler’s equation)
Hence,
d
dbSd
dQ
133.51
8.30)80.30(
1.21 2/13/2
6/1
Subtituting numerical values:
d
d
133.514198200
2/3
Rearranging as an iterative formula for d:
3/2
4198
)133.51(200
dd
Iteration (from, e.g., d = 0) gives
d = 0.08802 m
Substituting in (**) gives
m945.408802.0133.51
08802.08.30
133.51
8.30
d
dh
Answer: minimum diameter of stone = 88 mm; river depth = 4.95 m.
Hydraulics 3 A2-10 Dr David Apsley
Q7.
(a) Assume rapidly-varied flow with critical conditions over the crest. Neglect upstream
dynamic head.
Measuring head relative to the top of the weir, noting that the upstream head is just the
freeboard and the head over the weir is 3/2 times critical depth:
3/12
0 )(2
3
g
qh
Hence, the flow rate (per metre width of embankment) is
122/32/12/32/3
0
2/12/3 sm09905.015.081.9)3/2()3/2( hgq
Critical depth:
m1.015.03
2
3
20 hhc
Answer: flow rate (per metre width) = 0.0990 m2 s
–1; depth over embankment = 0.1 m.
(b)
S = 0.125
n = 0.013 m–1/3
s
Normal Depth
Vhq , where hSRn
V h
2/13/21 , hRh (wide channel)
2/13/51Sh
nq
m03442.0125.0
09905.0013.05/35/3
S
nqh
Velocity
1sm878.203442.0
09905.0 h
qV
Bed shear stress
For normal flow,
SgRhρτ where hRh (wide channel)
2mN21.42125.003442.081.91000τ
Answer: depth = 0.034 m; velocity = 2.88 m s–1
; bed stress = 42.2 N m–2
.
(c) For the given bed material,
59.50)100.1(
81.9)165.2(002.0
ν
)1(*
3/1
26
3/1
2
gsdd
Hydraulics 3 A2-11 Dr David Apsley
03987.0)]020.0exp(1[055.02.11
30.0
*
*
* dd
τcrit
For the actual flow down the slope,
304.1002.081.9)10002650(
21.42
)ρρ(
τ*τ
gds
τ* is far in excess of τ
*crit, hence the surface will erode.
(Alternatively, one could compute the absolute critical stress τcrit to compare with τ).
Answer: slope erodes.
(d) Cost-effective examples include:
increasing the size of bed material; e.g. rock armour;
semi-immobilising by vegetating the slope or using geotextiles.
Hydraulics 3 A2-12 Dr David Apsley
Q8.
(a)
sm01800.0
1.213/1
6/1
d
n
Answer: Manning’s n = 0.0180 m–1/3
s.
(b)
Vhq where 2/13/21SR
nV h , hRh
Hence,
2/13/51Sh
nq
or
m532.1
5/3
S
nqh
Answer: depth of flow = 1.53 m.
(c)
2mN79.18
ρτ
SgRhb
Answer: bed shear stress = 18.8 N m–2
.
(d)
89.75
ν
)1(*
3/1
2
gsdd
By Soulsby,
04620.0
)]020.0exp(1[055.02.11
30.0τ *
*
*
dd
crit
Here,
3869.0
)ρρ(
τ*τ
gds
b
Hydraulics 3 A2-13 Dr David Apsley
As τ* > τ*crit the bed is mobile.
Meyer-Peter and Müller
591.1
)*τ*τ(8* 2/3
critq
Hence,
123
3
sm10052.1
)1(*
gdsqqb
Van Rijn
9604.0
)1*τ
*τ(
*
053.0* 1.2
3.0
critd
q
Hence,
124
3
sm10349.6
)1(*
gdsqqb
Answer: bed load = 1.0510–3
m2 s
–1 (Meyer-Peter and Müller); 6.3510
–4 m
2 s
–1 (Van Rijn)
(e) By Cheng’s formula,
1
2/32/12
sm2309.0
]5)*2.125[(ν
dd
ws
By definition,
1
τ
sm1371.0
ρ/τ
bu
uτ < ws, so no significant suspended load occurs.
Hydraulics 3 A2-14 Dr David Apsley
Q9.
S = 1/800 = 0.00125
d = 0.0005 m
q = 5 m2 s
–1
Assume the water has density ρ = 1000 kg m–3
and kinematic viscosity ν = 1.0×10–6
m2 s
–1.
(a)
sm01335.021.1
)105(
1.21
3/16/146/1
d
n
Answer: Manning’s n = 0.0134 m–1/3
s.
(b)
Vhq where 2/13/21SR
nV h , hRh
Hence,
2/13/51Sh
nq
m464.100125.0
501335.05/35/3
S
nqh
Answer: depth of flow = 1.46 m.
(c)
2mN95.1700125.0464.181.91000ρτ SgRhb
Answer: bed shear stress = 18.0 N m–2
.
(d)
65.12)100.1(
81.9)165.2(0005.0
ν
)1(*
3/1
26
3/1
2
gsdd
By Soulsby, the critical Shields stress is
03084.0)]65.12020.0exp(1[055.065.122.11
30.0
)]020.0exp(1[055.02.11
30.0τ *
*
*
dd
crit
For this flow the actual Shields stress is
218.20005.081.9)10002650(
95.17
)ρρ(
τ*τ
gds
b
As τ* > τ*crit the bed is mobile.
Hydraulics 3 A2-15 Dr David Apsley
Meyer-Peter and Müller
88.25)03084.0218.2(8)*τ*τ(8* 2/32/3 critq
Hence,
12333 sm10164.10005.081.965.188.25)1(* gdsqqb
Nielsen
09.39218.2)03084.0218.2(12τ)ττ(12 **** critq
Hence,
12333 sm10758.10005.081.965.109.39)1(* gdsqqb
Answer: bed load = 1.1610–3
m2 s
–1 (Meyer-Peter and Müller); 1.7610
–3 m
2 s
–1 (Nielsen).
(e) By Cheng’s formula, the settling velocity is given by
12/32/12
6
2/32/12
sm06072.0]5)65.122.125[(0005.0
100.1
]5)*2.125[(ν
dd
ws
By definition, the friction velocity is
1
τ sm1340.01000/95.17ρ/τ bu
uτ > ws, so suspended load occurs.
Answer: settling velocity = 0.0607 m s–1
.
(f)
65.065.0,6559.0min65.0,103084.0
218.2
65.12
117.0min
65.0,1τ
*τ
*
117.0min
*
crit
refd
C
m10463.7103084.0
218.265.123.00005.0
1τ
*τ*3.0
3
2/1
7.0
2/1
7.0
*
crit
ref ddz
The concentration profile is then
τκ
1/
1/ u
w
refref
s
zh
zh
C
C
or, with κ = 0.41,
Hydraulics 3 A2-16 Dr David Apsley
105.1
2.195
1/464.1*65.0
zC (*)
The velocity profile is
)33ln(κ
)( τ
sk
zuzu
Taking the roughness length ks equal to a diameter, i.e. ks = 0.0005 m, gives
)66000ln(3268.0)( zzu (**)
The total suspended load (per unit width) is:
h
z
s
ref
zCUq d
With equations (*) and (**) this becomes
464.1
007463.0
105.14 d)66000ln()1/464.1)(10255.6( zzzqs
and this can be evaluated using numerical integration (e.g. trapezium rule) to give
qs = 0.044 m2 s
–1
Answer: suspended-load flux = 0.044 m3 s
–1 per metre width.
A simple Fortran code for doing the numerical integration is given on the following page.
The code can readily be modified for any alternative values of ks etc., or, indeed, velocity and
concentration profiles.
Note that a significant number of trapezia is necessary (probably 200+), although this could
be reduced by using Simpson’s rule instead. For any numerical method you should always
perform the calculations with more, shorter, subintervals to confirm that numerical accuracy
is sufficient. The need to be able to vary the number of intervals is one reason why a
spreadsheet alone is not particularly good at this, as it would require intervention to change
the number of trapezia.
Hydraulics 3 A2-17 Dr David Apsley
PROGRAM SUSPENDED_LOAD
IMPLICIT NONE
DOUBLE PRECISION, PARAMETER :: KAPPA = 0.41
DOUBLE PRECISION CREF, ZREF
DOUBLE PRECISION H, KS, WS, UTAU
DOUBLE PRECISION ROUSE
DOUBLE PRECISION DZ, Z, INTEGRAL
INTEGER N, J
! Particulate and flow data
DATA CREF, ZREF, H, KS, WS, UTAU &
/ 0.65, 0.007463, 1.464, 0.0005, 0.06072, 0.1340 /
ROUSE = WS / (KAPPA * UTAU)
! Number of intervals and size
PRINT *, 'Input N'; READ *, N
DZ = (H - ZREF) / N
! Integral by trapezium rule
INTEGRAL = C(ZREF) * U(ZREF) + C(H) * U(H)
DO J = 1, N - 1
Z = ZREF + J * DZ
INTEGRAL = INTEGRAL + 2.0 * C(Z) * U(Z)
END DO
INTEGRAL = INTEGRAL * DZ / 2.0
WRITE( *, "('Suspended load: ', 1PE10.3, ' m3/s per metre')" ) INTEGRAL
! Internal functions giving profiles of concentration and velocity
CONTAINS
!--------------------------------------
DOUBLE PRECISION FUNCTION C( Z )
IMPLICIT NONE
DOUBLE PRECISION Z
C = CREF * ( (H / Z - 1.0) / (H / ZREF - 1.0) ) ** ROUSE
END FUNCTION C
!--------------------------------------
DOUBLE PRECISION FUNCTION U( Z )
IMPLICIT NONE
DOUBLE PRECISION Z
U = (UTAU / KAPPA ) * LOG( 33.0 * Z / KS )
END FUNCTION U
END PROGRAM SUSPENDED_LOAD
Hydraulics 3 A2-18 Dr David Apsley
Q10.
(a)
sm01746.021.1
)0025.0(
1.21
3/16/16/1
d
n
Answer: Manning’s n = 0.0175 m–1/3
s.
(b)
For normal flow:
Vhq where 2/13/21SR
nV h , hRh
2/13/51Sh
nq
m677.11500/1
5.301746.05/35/3
S
nqh
Average velocity:
1sm087.2677.1
5.3 h
qV
Answer: depth of flow = 1.68 m; average velocity = 2.09 m s–1
.
(c) Using the normal-flow relation:
2mN97.10)1500/1(677.181.91000ρτ SgRhb
The Shields stress is
2711.00025.081.9)10002650(
97.10
)ρρ(
τ*τ
gds
b
To find the critical Shields stress:
24.63)100.1(
81.9)165.2(0025.0
ν
)1(*
3/1
26
3/1
2
gsdd
04338.0)]24.63020.0exp(1[055.024.632.11
30.0
)]020.0exp(1[055.02.11
30.0τ *
*
*
dd
crit
As τ* > τ*crit the bed is mobile.
Answer: bed shear stress = 11.0 N m–2
and exceeds the critical shear stress.
(d) For the bed-load flux:
8693.0)04338.02711.0(8)*τ*τ(8* 2/32/3 critq
Hydraulics 3 A2-19 Dr David Apsley
12433 sm10372.40025.081.965.18693.0)1(* gdsqqb
Answer: bed load = 4.3710–4
m3 s
–1 per metre width.
(e)
By Cheng’s formula:
5.517]5)24.632.125[(
]5)*2.125[(*
2/32/12
2/32/12
dws
16
sm2070.05.5170025.0
100.1*
ν
ss wd
w
By definition, the friction velocity is
1
τ sm1047.01000/97.10ρ/τ bu
uτ < ws, so significant suspended load would not be expected to occur.
Answer: settling velocity = 0.207 m s–1
.
(f) If suspended load does occur then it may be computed by summing
concentration (C) volumetric flow rate (U dz per unit width)
over the water column. i.e.
h
s zCUq0
d
Hydraulics 3 A2-20 Dr David Apsley
Q11.
Re
24Dc
Expanding:
dwareaprojectedw
drag
ss
ν24
)(ρ 2
21
But, at terminal velocity,
3
613
34 )ρρ(π)2/(π)ρρ()( gdgdgmmweightreduceddrag ssw
whilst
4/π 2dareaprojected
Hence,
dwdw
gd
ss
s ν24
πρ
)ρρ(π2
412
21
3
61
dww
gd
ss
s ν24
)1ρ/ρ(
3
42
swgds
ν
)1(
18
1 2
Hydraulics 3 A2-21 Dr David Apsley
Q12.
Consider the velocity profile
)33ln(κ
)( τ
sk
zuzU
The total flow per unit span is given by
h
zzUq0
d)(
Since q = Vh (by definition of average velocity V):
h
s
zk
zuVh
0
τ d)33ln(κ
Integrating by parts:
hh
s
zk
zz
uVh
00
τ d)33ln(κ
hk
hh
uVh
s
)33ln(κ
τ
1)33ln(κ
τ
sk
huV
or, since 1 = ln e,
)12ln(κ
)33
ln(κ
ττ
ss k
hu
k
h
e
uV (with 2 sig. fig. accuracy for the constant)
Hydraulics 3 A2-22 Dr David Apsley
Q13.
(a)
ρ
ττ
bu
where τb is the bed shear stress and ρ is the fluid density.
(b) Linear shear stress profile with 2
τρττ ub at z = 0 and τ = 0 at z = h:
)/1(ρτ 2
τ hzu
From the given velocity profile:
z
u
z
U
κd
d τ
By the definition of eddy viscosity (μt = ρνt):
z
Ut
d
dρντ
Rearranging for νt:
)/1(κ)κ/(
)/1(
d/d
ρ/τν τ
τ
2
τ hzzuzu
hzu
zUt
Thus, the kinematic eddy viscosity νt is a quadratic function of z.
(c) According to Fick’s gradient-diffusion law the net upward flux of sediment volume across
a horizontal area A is
Az
CK
d
d
whilst the net downward flux due to settling is volume flux concentration, or
ACws
At equilibrium these are equal and hence
ACwAz
CK s
d
d
Dividing by A and rearranging:
0d
d Cw
z
CK s
It is assumed that, as the same turbulent eddies are responsible for the transport of both
momentum and particulate material the kinematic eddy viscosity (νt) and eddy diffusivity (K)
are equal. Hence, )/1(κ τ hzzuK and so
0d
d)/1(κ τ Cw
z
Chzzu s
(d)
Hydraulics 3 A2-23 Dr David Apsley
(Note that all sketches below have the independent variable – here the vertical coordinate –
on the vertical axis.)
Velocity:
z
U
Eddy viscosity:
z
nt
Concentration (Rouse number = 0.5 here)
z
C
Hydraulics 3 A2-24 Dr David Apsley
(e)
Bed-load transport:
set in motion by the fluid stress;
main mechanisms: sliding, rolling, saltating (small jumps).
Suspended-load transport:
occurs for sufficiently vigorous turbulent fluid motion;
balance between net upward transport by turbulent eddies and downward settling;
usually quantified by solving a diffusion equation (see the following question), then
integrating the resultant flux density (CU) over the flow cross-section.
Hydraulics 3 A2-25 Dr David Apsley
Q14.
Cwz
CK s
d
d
Substituting the eddy diffusivity )/1(κ τ hzzuK :
Cwz
Chzzu s
d
d)/1(κ τ
Separating variables:
)/1(
d
κ
d
τ hzz
z
u
w
C
C s
Using partial fractions:
zzhzu
w
C
C s d11
κ
d
τ
Integrating between zref, Cref and a general z, C pair:
z
z
s
Crefref
zzhzu
w
C
Cd
11
κ
d
τ
C
refref
s
ref zh
zh
z
z
u
w
C
Clnln
κln
τ
ref
refs
ref zh
zh
z
z
u
w
C
Cln
κln
τ
On the RHS, inside the logarithm divide both numerator and denominator by zzref
τκ
1/
1/lnln
u
w
refref
s
zh
zh
C
C
τκ
1/
1/ u
w
refref
s
zh
zh
C
C
Hydraulics 3 A2-26 Dr David Apsley
Q15.
(a) Due to settling, sediment concentration is greater near the bed. As a result, upward
turbulent velocity fluctuations tend to carry larger amounts of sediment than downward
fluctuations, leading to a net upward diffusive flux. An equilibrium concentration distribution
is attained when this balances the downward settling flux.
C
C-ldCdz
C+ldCdz
ws
Consider a horizontal surface, where the concentration is C. An upward turbulent velocity u′
for half the time carries material of concentration (C – l dC/dz), where l is a typical size of
turbulent eddy. The corresponding downward velocity for the other half of the time carries
material at concentration (C + l dC/dz). The average upward flux of sediment (volume flux ×
concentration) through horizontal area A is
Az
Clu
z
ClCAu
z
ClCAu
d
d)
d
d()
d
d(
21
21
Writing u′l as Ke, the net upward flux is
Az
CK e
d
d
At equilibrium this is balanced by a net downward flux of material wsAC due to settling:
ACwAz
CK s
d
d
Dividing by A:
Cwz
CK s
d
d
(b) Cwz
Chzzu s
d
d)/1(κ τ
Separating variables:
)/1(
d
κ
d
τ hzz
z
u
w
C
C s
zh-zzu
w
C
C s d11
κ
d
τ
Integrating between a reference height zref and z:
z
z
s
C
Crefref
zzhzu
w
C
Cd
11
κ
d
τ
zz
sC
C refrefzhz
u
wC )ln(ln
κln
τ
)}ln(ln)ln({lnκ
lnlnτ
refref
s
ref zhzzhzu
wCC
Hydraulics 3 A2-27 Dr David Apsley
)(
)(ln
κln
τ ref
refs
ref zhz
zzh
u
w
C
C
Dividing top and bottom of the last fraction by z and zref :
1/
1/ln
κln
τ ref
s
ref zh
zh
u
w
C
C
τκ
1/
1/lnln
usw
refref zh
zh
C
C
τκ
1/
1/ u
w
refref
s
zh
zh
C
C
(c) The sediment flux can be determined by summing contributions C(U dA) over a cross-
section, where dA = b dz and b is the width of the channel:
h
z
s
ref
zCUbQ d
Using the trapezium rule (no need to learn this formula – just sum trapezoidal areas if you
prefer) on N intervals (here, N = 3) this is approximated by
)2(2
Δ 1
1
0 N
N
i
is fffz
bQ
(*)
where f is the integrand. In this case,
b = 5 m
m4997.03
001.05.1Δ
N
zhz
ref
3659.0κ
1499
1/5.165.0
1/
1/ τ
z
zh
zhCC
u
w
ref
ref
s
)33000ln(4878.0)33ln(κ
τ zk
zuU
s
CUf
zi Ci Ui fi
0.0010 0.65 1.706 1.1089
0.5007 0.05764 4.738 0.2731
1.0004 0.03472 5.075 0.1762
1.5000 0 5.273 0
Using (*),
13 sm508.2)0)1762.02731.0(21089.1(2
4997.05 sQ
Answer: Qs = 2.5 m3 s
–1.