ANSWERS TO EXAMPLES (SEDIMENT TRANSPORT) AUTUMN 2017...

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




Q15. (c) 2.5 m3 s

–1


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

.


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

.


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

.


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:


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.


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


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.


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.


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


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.


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


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.


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


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


.

(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



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


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,


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.


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


Q10.

(a)

sm01746.021.1

)0025.0(

1.21

3/16/16/1

d

n


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



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


12433 sm10372.40025.081.965.18693.0)1(* gdsqqb


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


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


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)


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)


(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


(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.


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


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


)(

)(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.

ANSWERS TO EXAMPLES (SEDIMENT TRANSPORT) AUTUMN 2017...

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Transcript of ANSWERS TO EXAMPLES (SEDIMENT TRANSPORT) AUTUMN 2017...