ANSWERS (SEDIMENT TRANSPORT EXAMPLES ......Hydraulics 3 Answers (Sediment Transport Examples) -3 Dr...
Transcript of ANSWERS (SEDIMENT TRANSPORT EXAMPLES ......Hydraulics 3 Answers (Sediment Transport Examples) -3 Dr...
Hydraulics 3 Answers (Sediment Transport Examples) -1 Dr David Apsley
ANSWERS (SEDIMENT TRANSPORT EXAMPLES) AUTUMN 2020
Q1.
Chengβs formula:
π€π π
Ξ½= [(25 + 1.2πβ2)1/2 β 5] 3/2, where πβ = π [
(π β 1)π
Ξ½2]
1/3
Densities:
Οsand = 2650 kg mβ3 Οair = 1.2 kg mβ3
Οwater = 1000 kg mβ3
Kinematic viscosities:
Ξ½air = 1.5 Γ 10β5 m2 sβ1
Ξ½water = 1.0 Γ 10β6 m2 sβ1
(a) For sand particles in air:
π =Οπ Ο =
2650
1.2 = 2208
πβ = π [(π β 1)π
Ξ½2]
1/3
= 0.001 [(2208 β 1) Γ 9.81
(1.5 Γ 10β5)2]
1/3
= 45.82
π€π π
Ξ½= [(25 + 1.2πβ2)1/2 β 5] 3/2 = [(25 + 1.2 Γ 45.822)1/2 β 5] 3/2 = 306.3
π€π = 306.3 ΓΞ½
π = 306.3 Γ
1.5 Γ 10β5
10β3 = 4.595 m sβ1
Answer: settling velocity in air = 4.60 m sβ1.
(b) For sand particles in water:
π =Οπ Ο =
2650
1000 = 2.65
πβ = π [(π β 1)π
Ξ½2]
1/3
= 0.001 [(2.65 β 1) Γ 9.81
(1.0 Γ 10β6)2]
1/3
= 25.30
π€π π
Ξ½= [(25 + 1.2πβ2)1/2 β 5] 3/2 = [(25 + 1.2 Γ 25.302)1/2 β 5]
3/2 = 111.5
π€π = 111.5 ΓΞ½
π = 111.5 Γ
1.0 Γ 10β6
10β3 = 0.1115 m sβ1
Answer: settling velocity in water = 0.112 m sβ1.
Hydraulics 3 Answers (Sediment Transport Examples) -2 Dr David Apsley
Q2.
As in Q1(b) above,
πβ = 25.30
Soulsbyβs formula,
Οcritβ =
0.30
1 + 1.2πβ+ 0.055 [1 β exp(β0.020πβ)] where Οβ =
Οπ(Οπ β Ο)ππ
Hence,
Οcritβ =
0.30
1 + 1.2 Γ 25.30+ 0.055 [1 β exp(β0.020 Γ 25.30)] = 0.03141
Οcrit = Οcritβ Γ (Οπ β Ο)ππ = 0.03141 Γ (2650 β 1000) Γ 9.81 Γ 0.001 = 0.5084 N m
β2
Answer: critical Shields parameter = 0.0314; critical shear stress = 0.508 N mβ2.
Hydraulics 3 Answers (Sediment Transport Examples) -3 Dr David Apsley
Q3.
From Q1 and Q2 above, the settling velocity and critical shear stress for incipient motion are
given, respectively, by:
π€π = 0.1115 m sβ1
Οcrit = 0.5084 N mβ2
(a) The bed shear stress is given, in general, by
Οπ = ππ (1
2Οπ2)
Rearranging for V:
π = β2
ππ
ΟπΟ
At incipient motion, Οπ = Οcrit. Hence,
π = β2
0.005Γ0.5084
1000 = 0.4510 m sβ1
Answer: for incipient motion, velocity = 0.451 m sβ1.
(b) For incipient suspended load,
π’Ο = π€π
where π’Ο is the friction velocity and π€π is the settling velocity.
By definition,
π’Ο = βΟπΟ = βππ (
1
2π2) = πβ
ππ
2
Hence,
πβππ
2= π€π
π = π€π β2
ππ = 0.1115β
2
0.005 = 2.23 m sβ1
Answer: for incipient suspended load, velocity = 2.23 m sβ1.
Hydraulics 3 Answers (Sediment Transport Examples) -4 Dr David Apsley
Q4.
π = 0.9 m2 sβ1 π = 0.06 m
Οπ = 2650 kg mβ3
πcritβ = 0.056
ππ = 0.01
π€π = 0.8 m π β1
From the critical Shields stress we can determine the critical bed stress for incipient motion:
Οcrit = Οβ(Οπ β Ο)ππ = 0.056 Γ (2650 β 1000) Γ 9.81 Γ 0.06
= 54.39 N mβ2
(a) Upstream of the gate:
h = 2.5 m
π =π
β =
0.9
2.5 = 0.36 m sβ1
Οπ = ππ Γ1
2Οπ2 = 0.01 Γ
1
2Γ 1000 Γ 0.362 = 0.648 N mβ2
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.
π§π 1 +π12
2π= π§π 2 +
π22
2π
For a flat bed:
β1 +π2
2πβ12 = β2 +
π2
2πβ22
Substituting values:
2.507 = β2 +0.04128
β22
Rearranging for the supercritical solution:
β2 = β0.04128
2.507 β β2
Iteration (from, e.g., β2 = 0) gives
β2 = 0.1318 m
Then:
π =π
β2 =
0.9
0.1318 = 6.829 m sβ1
Οπ = ππ Γ1
2Οπ2 = 0.01 Γ
1
2Γ 1000 Γ 6.8292 = 233.2 N mβ2
Hydraulics 3 Answers (Sediment Transport Examples) -5 Dr David Apsley
Οπ exceeds Οcrit; 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:
Οπ = 54.39 N mβ2
ππ Γ1
2Οπ2 = 54.39
π = β2
ππΓ54.39
Ο = β
2
0.01Γ54.39
1000 = 3.298 m sβ1
The overall depth downstream is then
β =π
π =
0.9
3.298 = 0.2729 m
Since the water level is fixed by the gate it is the same as before. The depth of scour is then
0.2729 β 0.1318 = 0.1411 m
Sluice gate assumption: total head the same on both sides of the gate.
π§π 1 +π12
2π= π§π 2 +
π22
2π
Upstream, π§π 1 = β1, but downstream there is a distinction between water level π§π 2 = 0.1318 m
and depth β2 = 0.2729 m.
β1 +π2
2πβ12 = π§π 2 +
π2
2πβ22
Substituting values:
β1 +0.04128
β12 = 0.6861
Rearranging for the subcritical solution:
β1 = 0.6861 β0.04128
β12
Iteration (from, e.g., β1 = 0.6861) gives
β1 = 0.5493 m
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).
Hydraulics 3 Answers (Sediment Transport Examples) -6 Dr David Apsley
Settling velocity:
π€π = 1.1 m sβ1 (given)
Friction velocity:
π’Ο = βΟ
Ο = β
233.2
1000 = 0.4829 m sβ1
Here, the settling velocity greatly exceeds the friction velocity, so suspended load is negligible.
Hydraulics 3 Answers (Sediment Transport Examples) -7 Dr David Apsley
Q5.
π = 5 m (width of main channel)
πmin = 3 m (width of restricted section)
π = 0.01 m
Οπ = 2650 kg mβ3 (π = 2.65)
Οcritβ = 0.05
ππ = 0.01
π = 5 m3 sβ1
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 β this is more convenient here.
The critical shear stress for incipient motion is
Οcrit = Οcritβ (Οπ β Ο)ππ = 0.05 Γ (2650 β 1000) Γ 9.81 Γ 0.01 = 8.093 N m
β2
and the corresponding velocity for incipient motion is given by
Οcrit = ππ (1
2Οπcrit
2 )
whence
πcrit = β2
ππ
ΟcritΟ = β
2
0.01Γ8.093
1000 = 1.272 m sβ1
This is a venturi so we need to determine velocities at various locations.
Upstream, π§π = β = 1 m and the velocity and total head are, respectively,
π =π
πβ =
5
5 Γ 1 = 1.0 m sβ1
π»π = π§π +π2
2π = 1 +
12
2 Γ 9.81 = 1.051 m
If critical-flow conditions occur at the throat then the total head there would be
π»π =3
2βπ =
3
2(ππ2
π)
1/3
=3
2(π2
πmin2 π
)
1/3
=3
2(
52
32 Γ 9.81)
1/3
= 0.9850 m
Since the upstream head exceeds that required for critical-flow conditions the flow remains
subcritical throughout, with total head in the restricted section, π» = 1.051 m. The depth β is
given by:
π» = β +π2
2π = β +
π2
2ππmin2 β2
Rearranging as an iterative formula for the subcritical value of β:
Hydraulics 3 Answers (Sediment Transport Examples) -8 Dr David Apsley
β = π» βπ2
2ππmin2 β2
Substituting numerical values:
β = 1.051 β0.1416
β2
Iteration (from, e.g., β = 1.051) gives
β = 0.8592 m
and corresponding velocity
π =π
πminβ =
5
3 Γ 0.8592 = 1.940 m sβ1
Hence we have:
β’ in the 5 m width, π = 1.0 m sβ1 < πcrit and the bed is stationary;
β’ in the 3 m width, π = 1.94 m sβ1 > πcrit and the bed is mobile.
(b) The bed will erode in the restricted section until π = πcrit. Then the flow depth is given by
π = πcrit Γ βπmin
β =π
πmin Γ πcrit =
5
3 Γ 1.272 = 1.310 m
If Ξπ§π is the change in height of the bed, then the total head is given by
π» = Ξπ§π + πΈ = Ξπ§π + β +π2
2π
Hence,
Ξπ§π = π» β β βπ2
2π = 1.051 β 1.310 β
1.2722
2 Γ 9.81 = β0.3415 m
Answer: depth of scour hole = 0.342 m.
Hydraulics 3 Answers (Sediment Transport Examples) -9 Dr David Apsley
Q6.
π = 12 m
π = 0.003
π = 200 m3 sβ1
Οcritβ = 0.056
Let the minimum stone diameter (corresponding to incipient motion) be π. Then the critical
shear stress for incipient motion is given by
Οcrit = Οcritβ (Οπ β Ο)ππ = 0.056 Γ (2650 β 1000) Γ 9.81 Γ π = 906.4π
For normal flow:
Οπ = Οππ βπ
For incipient motion:
π β =ΟcritΟππ
=906.4π
1000 Γ 9.81 Γ 0.003 = 30.80π (*)
The corresponding depth of flow β is given by the expression for π β in a rectangular channel,
30.80π =β
1 +2βπ
With π = 12 m this gives (after some rearrangement) an expression for β in terms of π:
β =30.8π
1 β 5.133π (**)
From Manningβs equation:
π = ππ΄ =1
ππ β2/3π1/2 Γ πβ, where π =
π1/6
21.1 (Stricklerβs equation)
Hence,
π =21.1
π1/6Γ (30.80π)2/3π1/2 Γ π Γ
30.8π
1 β 5.133π
Subtituting numerical values:
200 = 4198π3/2
1 β 5.133π
Rearranging as an iterative formula for π:
π = [200(1 β 5.133π)
4198]
2/3
Iteration (from, e.g., π = 0) gives
π = 0.08802 m
Substituting in (**) gives
Hydraulics 3 Answers (Sediment Transport Examples) -10 Dr David Apsley
β =30.8π
1 β 5.133π =
30.8 Γ 0.08802
1 β 5.133 Γ 0.08802 = 4.945 m
Answer: minimum diameter of stone = 88 mm; river depth = 4.95 m.
Hydraulics 3 Answers (Sediment Transport Examples) -11 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:
β0 =3
2(π2
π)1/3
Hence, the flow rate (per metre width of embankment) is
π = (2/3)3/2π1/2β03/2 = (2/3)3/2 Γ 9.811/2 Γ 0.153/2 = 0.09905 m2 sβ1
Critical depth:
βπ =2
3β0 =
2
3Γ 0.15 = 0.1 m
Answer: flow rate (per metre width) = 0.0990 m2 sβ1; depth over embankment = 0.1 m.
(b)
π = 0.125
π = 0.013 mβ1/3 s
Normal Depth
π = πβ, where π =1
ππ β2/3π1/2, π β = β (wide channel)
π =1
πβ5/3π1/2
β = (ππ
βπ)3/5
= (0.013 Γ 0.09905
β0.125)3/5
= 0.03442 m
Velocity
π =π
β =
0.09905
0.03442 = 2.878 m sβ1
Bed shear stress
For normal flow,
Ο = Οππ βπ, where π β = β (wide channel)
Ο = 1000 Γ 9.81 Γ 0.03442 Γ 0.125 = 42.21 N mβ2
Answer: depth = 0.034 m; velocity = 2.88 m sβ1; bed stress = 42.2 N mβ2.
Hydraulics 3 Answers (Sediment Transport Examples) -12 Dr David Apsley
(c) For the given bed material,
πβ = π [(π β 1)π
Ξ½2]
1/3
= 0.002 Γ [(2.65 β 1) Γ 9.81
(1.0 Γ 10β6)2]
1/3
= 50.59
Οcritβ =
0.30
1 + 1.2πβ+ 0.055 [1 β exp(β0.020πβ)] = 0.03987
For the actual flow down the slope,
Οβ =Ο
(Οπ β Ο)ππ =
42.21
(2650 β 1000) Γ 9.81 Γ 0.002 = 1.304
Οβ is far in excess of Οcritβ ; hence the surface will erode.
(Alternatively, one could compute the absolute critical stress Οcrit to compare with Ο).
Answer: slope erodes.
(d) Cost-effective examples include:
β’ increasing the size of bed material; e.g. rock armour;
β’ semi-immobilising by vegetating the slope or using geotextiles.
Hydraulics 3 Answers (Sediment Transport Examples) -13 Dr David Apsley
Q8.
(a)
π =π1/6
21.1 = 0.01800 mβ1/3 s
Answer: Manningβs π = 0.0180 mβ1/3 s.
(b)
π = πβ, where π =1
ππ β2/3π1/2, π β = β
Hence,
π =1
πβ5/3π1/2
or
β = (ππ
βπ)3/5
= 1.532 m
Answer: depth of flow = 1.53 m.
(c)
Οπ = Οππ βπ = 18.79 N mβ2
Answer: bed shear stress = 18.8 N mβ2.
(d)
πβ = π [(π β 1)π
Ξ½2]
1/3
= 75.89
By Soulsby,
Οcritβ =
0.30
1 + 1.2πβ+ 0.055 [1 β exp(β0.020πβ)] = 0.04620
Here,
Οβ =Οπ
(Οπ β Ο)ππ = 0.3869
As Οβ > Οcritβ the bed is mobile.
Hydraulics 3 Answers (Sediment Transport Examples) -14 Dr David Apsley
Meyer-Peter and MΓΌller
πβ = 8(Οβ β Οcritβ )3/2 = 1.591
Hence,
ππ = πββ(π β 1)ππ3 = 1.052 Γ 10β3 m2 sβ1
Van Rijn
πβ =0.053
(πβ0.3) (Οβ
Οcritβ β 1)
2.1 = 0.9604
Hence,
ππ = πββ(π β 1)ππ3 = 6.349 Γ 10β4 m2 sβ1
Answer: bed load = 1.0510β3 m2 sβ1 (Meyer-Peter and MΓΌller); 6.3510β4 m2 sβ1 (Van Rijn)
(e) By Chengβs formula,
π€π =Ξ½
π [(25 + 1.2πβ2)1/2β 5]
3/2 = 0.2309 m sβ1
By definition,
π’Ο = βΟπΟ = 0.1371 m sβ1
π’Ο < π€π , so no significant suspended load occurs.
Hydraulics 3 Answers (Sediment Transport Examples) -15 Dr David Apsley
Q9.
π = 1/800 = 0.00125
π = 0.0005 m
π = 5 m2 sβ1
Assume water has density Ο = 1000 kg mβ3 and kinematic viscosity Ξ½ = 1.0 Γ 10β6 m2 sβ1.
(a)
π =π1/6
21.1 =
(5 Γ 10β4)1/6
21.1 = 0.01335 mβ1/3 s
Answer: Manningβs π = 0.0134 mβ1/3 s.
(b)
π = πβ, where π =1
ππ β2/3π1/2, π β = β
Hence,
π =1
πβ5/3π1/2
β = (ππ
βπ)3/5
= (0.01335 Γ 5
β0.00125)3/5
= 1.464 m
Answer: depth of flow = 1.46 m.
(c)
Οπ = Οππ βπ = 1000 Γ 9.81 Γ 1.464 Γ 0.00125 = 17.95 N mβ2
Answer: bed shear stress = 18.0 N mβ2.
(d)
πβ = π [(π β 1)π
Ξ½2]
1/3
= 0.0005 Γ [(2.65 β 1) Γ 9.81
(1.0 Γ 10β6)2]
1/3
= 12.65
By Soulsby, the critical Shields stress is
Οcritβ =
0.30
1 + 1.2πβ+ 0.055 [1 β exp(β0.020πβ)]
=0.30
1 + 1.2 Γ 12.65+ 0.055 [1 β exp(β0.020 Γ 12.65)] = 0.03084
For this flow the actual Shields stress is
Hydraulics 3 Answers (Sediment Transport Examples) -16 Dr David Apsley
Οβ =Οπ
(Οπ β Ο)ππ =
17.95
(2650 β 1000) Γ 9.81 Γ 0.0005 = 2.218
As Οβ > Οcritβ the bed is mobile.
Meyer-Peter and MΓΌller
πβ = 8(Οβ β Οcritβ )3/2 = 8 Γ (2.218 β 0.03084)3/2 = 25.88
Hence,
ππ = πββ(π β 1)ππ3 = 25.88 Γ β1.65 Γ 9.81 Γ 0.00053 = 1.164 Γ 10β3 m2 sβ1
Nielsen
πβ = 12(Οβ β Οcritβ )βΟβ = 12 Γ (2.218 β 0.03084)β2.218 = 39.09
Hence,
ππ = πββ(π β 1)ππ3 = 39.09 Γ β1.65 Γ 9.81 Γ 0.00053 = 1.758 Γ 10β3 m2 sβ1
Answer: bed load = 1.1610β3 m2 sβ1 (Meyer-Peter and MΓΌller); 1.7610β3 m2 sβ1 (Nielsen).
(e) By Chengβs formula, the settling velocity is given by
π€π =Ξ½
π [(25 + 1.2πβ2)1/2β 5]
3/2 =1.0 Γ 10β6
0.0005[(25 + 1.2 Γ 12.652)1/2 β 5]3/2
= 0.06072 m sβ1
By definition, the friction velocity is
π’Ο = βΟπΟ = β
17.95
1000 = 0.1340 m sβ1
π’Ο > π€π , so suspended load occurs.
Answer: settling velocity = 0.0607 m sβ1.
(f)
πΆref = min {0.117
πβ(Οβ
Οcritβ β 1) , 0.65} = min {
0.117
12.65(2.218
0.03084β 1) , 0.65}
= min{0.6559, 0.65} = 0.65
Hydraulics 3 Answers (Sediment Transport Examples) -17 Dr David Apsley
π§ref = π Γ 0.3πβ0.7 (
Οβ
Οcritβ β 1)
1/2
= 0.0005 Γ 0.3 Γ 12.650.7 (2.218
0.03084β 1)
1/2
= 7.463 Γ 10β3 m
The concentration profile is then
πΆ
πΆref= (
βπ§ β 1
βπ§ref
β 1)
π€π ΞΊπ’Ο
or, with π = 0.41,
πΆ = 0.65(
1.464π§ β 1
195.2)
1.105
(*)
The velocity profile is
π’(π§) =π’ΟΞΊln (33
π§
ππ )
Taking the roughness length ππ equal to a diameter, i.e. ππ = 0.0005 m, gives
π’(π§) = 0.3268 ln(66000π§) (**)
The total suspended load (per unit width) is:
ππ = β« πΆπ dπ§β
π§ref
With equations (*) and (**) this becomes
ππ = β« (6.255 Γ 10β4) (1.464
π§β 1)
1.105
ln(66000π§) dπ§1.464
0.007463
and this can be evaluated using numerical integration (e.g. trapezium rule) to give
ππ = 0.044 m2 sβ1
Answer: suspended-load flux = 0.044 m3 sβ1 per metre width.
A simple Fortran code for doing the numerical integration is given on the following page. The
code can be modified for any alternative values of ππ or velocity and concentration profiles.
A significant number of trapezia is necessary (probably 200+), although this could be reduced
by using Simpsonβs 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 Answers (Sediment Transport Examples) -18 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: ', ES10.3, ' 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 Answers (Sediment Transport Examples) -19 Dr David Apsley
Q10.
(a)
π =π1/6
21.1 =
(0.0025)1/6
21.1 = 0.01746 mβ1/3 s
Answer: Manningβs π = 0.0175 mβ1/3 s.
(b)
For normal flow:
π = πβ, where π =1
ππ β2/3π1/2, π β = β
π =1
πβ5/3π1/2
β = (ππ
βπ)3/5
=
(
0.01746 Γ 3.5
β 11500 )
3/5
= 1.677 m
Average velocity:
π =π
β =
3.5
1.677 = 2.087 m sβ1
Answer: depth of flow = 1.68 m; average velocity = 2.09 m sβ1.
(c) Using the normal-flow relation:
Οπ = Οππ βπ = 1000 Γ 9.81 Γ 1.677 Γ (1
1500) = 10.97 N mβ2
The Shields stress is
Οβ =Οπ
(Οπ β Ο)ππ =
10.97
(2650 β 1000) Γ 9.81 Γ 0.0025 = 0.2711
To find the critical Shields stress:
πβ = π [(π β 1)π
Ξ½2]
1/3
= 0.0025 Γ [(2.65 β 1) Γ 9.81
(1.0 Γ 10β6)2]
1/3
= 63.24
Οcritβ =
0.30
1 + 1.2πβ+ 0.055 [1 β exp(β0.020πβ)]
=0.30
1 + 1.2 Γ 63.24+ 0.055 [1 β exp(β0.020 Γ 63.24)] = 0.04338
Hydraulics 3 Answers (Sediment Transport Examples) -20 Dr David Apsley
As Οβ > Οcritβ the bed is mobile.
Answer: bed shear stress = 11.0 N mβ2 and exceeds the critical shear stress.
(d) For the bed-load flux:
πβ = 8(Οβ β Οcritβ )3/2 = 8 Γ (0.2711 β 0.04338)3/2 = 0.8693
whence
ππ = πββ(π β 1)ππ3 = 0.8693 Γ β1.65 Γ 9.81 Γ 0.00253
= 4.372 Γ 10β4 m2 sβ1
Answer: bed load = 4.3710β4 m3 sβ1 per metre width.
(e)
By Chengβs formula:
π€π β = [(25 + 1.2πβ2)
1/2β 5]
3/2
= [(25 + 1.2 Γ 63.242)12 β 5]
3/2
= 517.5
π€π =Ξ½
πΓ π€π
β =1.0 Γ 10β6
0.0025Γ 517.5 = 0.2070 m sβ1
By definition, the friction velocity is
π’Ο = βΟπΟ = β
10.97
1000 = 0.1047 m sβ1
π’Ο < π€π so significant suspended load would not be expected to occur.
Answer: settling velocity = 0.207 m sβ1.
(f) If suspended load does occur then it may be computed by summing
concentration (πΆ) volumetric flow rate (π dπ§ per unit width)
over the water column. i.e.
ππ = β« πΆπ dπ§β
0
Hydraulics 3 Answers (Sediment Transport Examples) -21 Dr David Apsley
Q11.
ππ· =24
Re
Expanding:
drag
12 Οπ€π
2 Γ (projected area)= 24 Γ
Ξ½
π€π π
But, at terminal velocity,
drag = reduced weight = (π βππ€)π = (Οπ β Ο) Γ4
3Ο (π
2)3
Γ π
=1
6Ο(Οπ β Ο)ππ
3
whilst
projected area =Οπ2
4
Hence,
16Ο(Οπ β Ο)ππ
3
12 Οπ€π
2 Γ14Οπ
2= 24
Ξ½
π€π π
4
3
(Οπ Ο β 1)ππ
π€π 2= 24
Ξ½
π€π π
1
18
(π β 1)ππ2
Ξ½= π€π
Hydraulics 3 Answers (Sediment Transport Examples) -22 Dr David Apsley
Q12.
Consider the velocity profile
π(π§) =π’ΟΞΊln (33
π§
ππ )
The total flow per unit span is given by
π = β« π(π§) dπ§β
0
Since π = πβ (by definition of average velocity π):
πβ = β«π’ΟΞΊln (33
π§
ππ ) dπ§
β
0
Integrating by parts:
πβ =π’ΟΞΊ{[π§ ln (33
π§
ππ )]0
β
ββ« dπ§β
0
}
πβ =π’ΟΞΊ{β ln (33
β
ππ ) β β}
π =π’ΟΞΊ{ln (33
β
ππ ) β 1}
or, since 1 = ln e,
π =π’ΟΞΊln (33
π
β
ππ ) =
π’ΟΞΊln (12
β
ππ )
(with 2 sig. fig. accuracy for the constant)
Hydraulics 3 Answers (Sediment Transport Examples) -23 Dr David Apsley
Q13.
(a)
π’Ο = βΟπΟ
where Οπ is the bed shear stress and Ο is the fluid density.
(b) Linear shear stress profile with Ο = Οπ β‘ Οπ’Ο2 at π§ = 0 and Ο = 0 at π§ = β:
Ο = Οπ’Ο2 (1 β
π§
β)
From the given velocity profile:
dπ
dπ§=π’ΟΞΊπ§
By the definition of eddy viscosity (ΞΌπ‘ = Οππ‘):
Ο = ΟΞ½π‘dπ
dπ§
Rearranging for Ξ½t:
Ξ½π‘ =
ΟΟdπdπ§
=π’Ο2 (1 β
π§β)
π’ΟΞΊπ§
= ΞΊπ’Οπ§ (1 βπ§
β)
Thus, the kinematic eddy viscosity ππ‘ is a quadratic function of π§.
(c) According to Fickβs gradient-diffusion law the net upward flux of sediment volume across
a horizontal area A is
βπΎdπΆ
dπ§π΄
whilst the net downward flux due to settling is volume flux concentration, or
π€π π΄πΆ
At equilibrium these are equal and hence
βπΎdπΆ
dπ§π΄ = π€π π΄πΆ
Dividing by A and rearranging:
πΎdπΆ
dπ§+ π€π πΆ = 0
Hydraulics 3 Answers (Sediment Transport Examples) -24 Dr David Apsley
It is assumed that, as the same turbulent eddies are responsible for the transport of both
momentum and particulate material the kinematic eddy viscosity (Ξ½t) and eddy diffusivity (K)
are equal. Hence, πΎ = ΞΊπ’Οπ§ (1 βπ§
β) and so
ΞΊπ’Οπ§ (1 βπ§β)dπΆ
dπ§+ π€π πΆ = 0
(d)
(Note that all sketches below have the independent variable β here the vertical coordinate β on
the vertical axis.)
Velocity:
Eddy viscosity:
Concentration (Rouse number = 0.5 here)
(e)
Bed-load transport:
β’ set in motion by the fluid stress;
β’ main mechanisms: sliding, rolling, saltating (small jumps).
Suspended-load transport:
β’ occurs for sufficiently vigorous turbulent fluid motion;
β’ balance between net upward transport by turbulent eddies and downward settling;
β’ usually quantified by solving a diffusion equation (see the following question), then
integrating the resultant flux density (πΆπ) over the flow cross-section.
z
U
z
nt
z
C
Hydraulics 3 Answers (Sediment Transport Examples) -25 Dr David Apsley
Q14.
βπΎdπΆ
dz= π€π πΆ
Substituting the eddy diffusivity πΎ = ΞΊπ’Οπ§ (1 βπ§
β):
βΞΊπ’Οπ§ (1 β
π§β) dπΆ
dπ§= π€π πΆ
Separating variables:
dπΆ
πΆ= β
π€π ΞΊπ’Ο
dπ§
π§ (1 βπ§β)
Using partial fractions:
dπΆ
πΆ= β
π€π ΞΊπ’Ο
(1
π§+
1
β β π§) dπ§
Integrating between π§ref, πΆref and a general π§, πΆ pair:
β«dπΆ
πΆ
C
πΆref
= βπ€π ΞΊπ’Ο
β« (1
π§+
1
β β π§) dπ§
π§
π§ref
lnπΆ
πΆref= β
π€π ΞΊπ’Ο
(lnπ§
π§refβ ln
β β π§
β β π§ref)
lnπΆ
πΆref=π€π ΞΊπ’Ο
ln (π§refπ§
β β π§
β β π§ref)
On the RHS, inside the logarithm divide both numerator and denominator by π§ Γ π§ref:
lnπΆ
πΆref= ln(
βπ§ β 1
βπ§ref
β 1)
π€π ΞΊπ’Ο
πΆ
πΆref= (
βπ§ β 1
βπ§ref
β 1)
π€π ΞΊπ’Ο
Hydraulics 3 Answers (Sediment Transport Examples) -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.
Consider a horizontal surface, where the concentration is πΆ. An upward turbulent velocity π’β² for half the time carries material of concentration (πΆ β π dπΆ/dπ§), where π is a typical size of
turbulent eddy. The corresponding downward velocity for the other half of the time carries
material at concentration (πΆ + π dπΆ/dπ§). The average upward flux of sediment (volume flux
Γ concentration) through horizontal area π΄ is
1
2π’β²π΄ (πΆ β π
dπΆ
dπ§) β
1
2π’β²π΄ (πΆ + π
dπΆ
dπ§) = βπ’β²π
dπΆ
dπ§π΄
Writing π’β²π as πΎ, the net upward flux is
βπΎdπΆ
dπ§π΄
At equilibrium this is balanced by a net downward flux of material π€π π΄πΆ due to settling:
βπΎdπΆ
dπ§π΄ = π€π π΄πΆ
Dividing by A:
βπΎdπΆ
dπ§= π€π πΆ
(b)
βΞΊπ’Οπ§ (1 β
π§β) dπΆ
dπ§= π€π πΆ
Separating variables:
dπΆ
πΆ= β
π€π ΞΊπ’Ο
dπ§
π§ (1 βπ§β)
dπΆ
πΆ= β
π€π ΞΊπ’Ο
(1
π§+
1
β β π§) dπ§
Integrating between a reference height π§ref and π§:
β«dπΆ
πΆ
C
πΆref
= βπ€π ΞΊπ’Ο
β« (1
π§+
1
β β π§) dπ§
π§
π§ref
C
C-ldCdz
C+ldCdz
ws
Hydraulics 3 Answers (Sediment Transport Examples) -27 Dr David Apsley
lnπΆ
πΆref= β
π€π ΞΊπ’Ο
(lnπ§
π§refβ ln
β β π§
β β π§ref)
lnπΆ
πΆref=π€π ΞΊπ’Ο
ln (π§refπ§
β β π§
β β π§ref)
Dividing top and bottom of the last fraction by π§ and π§ref:
lnπΆ
πΆref= ln(
βπ§ β 1
βπ§ref
β 1)
π€π ΞΊπ’Ο
πΆ
πΆref= (
βπ§ β 1
βπ§ref
β 1)
π€π ΞΊπ’Ο
(c) The sediment flux can be determined by summing contributions πΆ(π dπ΄) over a cross-
section, where dπ΄ = π dπ§ and π is the width of the channel:
ππ = πβ« πΆπ dπ§β
π§πππ
Using the trapezium rule (no need to learn this formula β just sum trapezoidal areas if you
prefer) on π intervals (here, π = 3) this is approximated by
ππ = πΞπ§
2(π0 + 2β πππβ1π=1 + ππ)
(*)
where π is the integrand.
In this case,
π = 5 m
Ξπ§ =β β π§refπ
=1.5 β 0.001
3 = 0.4997 m
πΆ = πΆref(
βπ§ β 1
βπ§ref
β 1)
π€π ΞΊπ’Ο
= 0.65 (
1.5π§ β 1
1499)
0.3659
π =π’ΟΞΊln (33
π§
ππ ) = 0.4878 ln(33000π§)
π = πΆπ
Hydraulics 3 Answers (Sediment Transport Examples) -28 Dr David Apsley
π§π πΆπ ππ ππ 0.0010 0.65000 1.706 1.1089
0.5007 0.05764 4.738 0.2731
1.0004 0.03472 5.075 0.1762
1.5000 0.00000 5.273 0.0000
Using (*),
ππ = 5 Γ0.4997
2Γ (1.1089 + 2 Γ (0.2731 + 0.1762) + 0) = 2.508 m3 sβ1
Answer: ππ = 2.5 m3 sβ1.