FLUVIAL PROCESSES: 2nd Edition - s3-eu-west-1.amazonaws.com · vii Introductory note to the 2nd...

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FLUVIAL PROCESSES: 2nd Edition

SOLUTIONS MANUAL

______________________________________________________________________

Supplement to the IAHR Monograph ‘Fluvial Processes – 2nd Edition’

Published by CRC Press, Taylor & Francis Group

© 2017 Ana Maria Ferreira da Silva, Queen’s University, Kingston,

ON, Canada

All rights reserved. No part of this publication or the information

contained herein may be reproduced, stored in a retrieval system, or

transmitted in any form or by any means, electronic, mechanical, by

photocopying, recording or otherwise, without written prior permission

from the copyright holder.

_____________________________________________________________________

FLUVIAL PROCESSES: 2nd Edition

SOLUTIONS MANUAL

______________________________________________

Ana Maria Ferreira da Silva & M. Selim Yalin

Queen’s University, Kingston, Canada

v

Table of contents

INTRODUCTORY NOTE TO THE 2nd EDITION vii

INTRODUCTORY NOTE TO THE 1st EDITION vii

1. Fundamentals

Problems 1.1 to 1.18 1

3. Flow past undulated beds


4. Regime channels and their computation


5. Formation of regime channels; meandering and braiding


6. Geometry and mechanics of meandering streams


7. Meandering-related computations


SOFTWARE INSTRUCTIONS ‒ FORTRAN 83

SOFTWARE INSTRUCTIONS ‒ MATLAB 84

vi

vii

Introductory note to the 2nd edition This revised version of the Solutions Manual reflects all updates and revisions

incorporated in the 2nd edition of the monograph. The readers are invited to read the

Introductory Note to the 1st edition included below for some further information on

this manual (disregarding the mention to the CD-ROM accompanying the

monograph).

Ana Maria Ferreira da Silva

Kingston, Ontario

August, 2017

Introductory note to the 1st edition When preparing this manual, the authors have tried to make it instructive and

reader-friendly, hence some solutions are accompanied by clarifying sketches and

explanations. The text of each question is reproduced here for the reader’s

convenience; in few cases the questions were slightly rephrased, without altering

their meaning.

The formulae and figures presented in the monograph are referred to by using their

original text-numbers; the expressions and figures that appear in the solutions have

“their own” numbers.

Before using the CD-ROM accompanying the monograph, the reader is advised to

study the Software Instructions included at the end of this manual.

M. S. Yalin and A. M. Ferreira da Silva

Kingston, Ontario

August, 2001

viii

Fundamentals

1

CHAPTER 1

Fundamentals

Problem 1.1

Prove that in the case of a uniform two-dimensional turbulent flow, the

dimensionless velocity deficit *max /)( vuu

(universal velocity distribution law) is a function of the dimensionless position

hz / only: assume that the logarithmic u-distribution is valid throughout the flow

depth.

Solution:

Consider the logarithmic u-distribution (1.6), viz

ln1

ss

k

zA

v

u

(where const 0.4 ). (1)

Assuming that Eq. (1) is valid throughout the flow depth, the maximum flow

velocity at hz is given by

ss

k

hA

v

uln

1max

. (2)

Subtracting Eq. (1) from Eq. (2), we obtain

z

h

kzA

khA

k

zA

k

hA

v

uu

ss

ss

ss

ss ln

1

)/(

)/(ln

1ln

1ln

1max

,

which indicates that

)/(max hzv

uu

.

Fluvial processes: 2nd edition – Solutions manual

2

Problem 1.2

Prove that in the case of a uniform two-dimensional turbulent flow, the velocity

deficit

*max /)( vuu

has the same constant value for any flow regime (hydraulically smooth,

transitionally rough and fully rough).

a) What is this constant value in the case of an open-channel flow?

b) What is this constant value in the case of a flow in a circular pipe?

Solution:

a) The vertically averaged flow velocity is given by (see Eq. (1.11))

ss

k

hA

v

u368.0ln

1

. (1)

The maximum flow velocity is given by Eq. (1.6), with hz :

ss

k

hA

v

uln

1max

. (2)

Subtracting Eq. (1) from Eq. (2), we obtain

ss

ss

k

hA

k

hA

v

uu368.0ln

1ln

1max

5.2)368.0/1ln(1

)/(368.0

)/(ln

1

ss

ss

khA

khA.

b) In this case, u is to be treated as the average velocity of pipe flow, h is to be

replaced by the pipe radius R and y is to be measured from the pipe wall towards

the centre (i.e. yRr ). We assume that the log-law is valid throughout the

interval Ryy 0 , where 0y is to be interpreted either as sk or as L (thickness

of the viscous sublayer) – whichever is the larger. We have

dyyRy

yA

yRdyyR

v

u

yRv

us

R

y

R

y

)(ln1

)(

2)(2

)(

1

02

02

0 00

.

Fundamentals

3

Adopting Ry / and 00 / Ry , and considering that Ry 0 , i.e. that 0 is

negligible in comparison to unity, we obtain

dy

RA

v

us

1

00

)1(ln1

2

ddy

RAs

11

000

)1(ln1

2)1(ln1

2

ddv

u

11max

00

)1(ln2

)1(2 .

But, since 0 is negligible in comparison to unity,

1212

2)1(2 200

121

00

d ,

and therefore

)( 0max v

u

v

u i.e. )( 0

max

v

uu,

where

1

0

0

)1(ln2

)(

d .

Evaluation of )( 0 . We have

1

22111

0

00

004

1ln

2

1lnlnln)(

2

dd

2

0020000

4

1ln

2

1ln

4

3 .

It can be shown (with the aid of the L’Hôpital’s rule) that if 2/1m , then the

products 00 ln m vanish in the limit 00 . Hence

75.35.12

4

3)(lim 0

00

,

and consequently,


4

75.3max

v

uu.

Problem 1.3

Consider a uniform two-dimensional fully rough turbulent open-channel flow.

Integrating the logarithmic u-distribution (1.6) between sk and h and assuming

that hks , derive the expression (1.11).

Solution:

For rough turbulent flow 5.8sB , and Eq. (1.6) can be written as

5.8ln1

sk

z

v

u

. (1)

Integrating Eq. (1) between sk and h , and bearing in mind that when hks then

hkdzhz s

h

ks

2)/ln( (2)

(see the derivation of Eq. (2) in the note below), we obtain

h

k ss

h

ksss

dzk

z

khdz

v

u

khv

u5.8ln

111

h

ks

h

k ssss

dzkh

dzk

h

h

z

kh

5.8ln

)(

1

h

ks

h

k ss

h

kssss

dzkh

dzk

h

khdz

h

z

kh

5.8ln

)(

1ln

)(

1

)(5.8

)(ln)(

1)2(

)(

1s

ss

sss

s

khkh

khk

h

khhk

kh

.

In this relation, sk is negligible in comparison to h ; hence it can be written as

5.8ln1

5.8ln11 1

ss k

he

k

h

v

u

,

i.e.

Fundamentals

5

5.8368.0ln1

sk

h

v

u

,

which is the expression (1.11) (with 5.8sB ).

Note:

Consider Eq. (2), viz

dzhz

h

ks

)/ln(I .

If hz / and ss hk / , then

sss

s

sd

h

ln10|ln|ln

11I

.

But

s

s

s

s

sss

s

20 /1

/1

)/1(

)(ln]ln[lim ,

and therefore

121 sssh

I

, i.e. hks 2I .

Problem 1.4

In addition to the dimensionless Chézy’s resistance equation (1.17), we have for the

turbulent open-channel flow also Manning’s and Darcy-Weisbach’s resistance

equations. For the case of a wide flow (R h , in which R is hydraulic radius) these

can be expressed as

2/13/21

Shn

u

and

ghSf

u8

,

respectively. Determine the interrelation between c and n, c and f, and n and f.

Solution:

Equating the right-hand side of Eq. (1.17), viz ghScu , to the right-hand side

of Manning’s equation, we obtain


6

2/13/21Sh

nghSc , and thus

g

h

nc

6/11 .

Equating the right-hand side of Eq. (1.17) to the right-hand side of Darcy-Weisbach

resistance equation, we obtain

ghSf

ghSc8

, and thus f

c8

.

Equating the right-hand side of Manning’s equation to the right-hand side of Darcy-

Weisbach resistance equation, we obtain

ghSf

Shn

81 2/13/2 , and thus 2/1

6/1

8 g

hfn .

Problem 1.5

Consider the log-log );( YX -plane in Figure 1.8, where X and Y are given by the

relations (1.24). Let 1m and 2m be two points on this plane which represent the

same granular material and fluid for two different flow stages 1)( v and 2)( v .

What is the inclination of the straight line 21mm ? What physical meaning can you

attach to this straight line?

Solution:

For the points 1m and 2m , representing the two (different) flow stages 1)( v and

2)( v , we have (see Eq. (1.24))

DvX 1

1

)( , D

vY

s

21

1

)( ,

and

DvX 2

2

)( , D

vY

s

22

2

)( ,

respectively. Introducing the notation

1

2*

)(

)(ˆ

v

vv ,

we can write, on the basis of the expressions above,

*1

2 vX

X and 2

1

2 ˆ vY

Y.

If 21mmI is the inclination of the straight line 21mm , then

Fundamentals

7

2ˆlog

ˆlog2

ˆlog

ˆlog

)/log(

)/log(

loglog

loglog 2

12

12

12

122

v

v

v

v

XX

YY

XX

YYm1m

I .

Hence the inclination of the straight line 21mm is 2/1.

Physical meaning: Consider a 2/1-inclined straight line corresponding to a given

granular material and fluid. The continual increment (decrement) of v implies the

motion of a point along this straight line in the direction of the increasing

(decreasing) values of X and Y .

Problem 1.6

Note from Figure 1.8 that the turbulent sediment transport initiation curve merges

to the straight line 3.01.0 crcr XY when the values of crX are small ( crX < 1).

Knowing this, determine (analytically) the expression of the corresponding straight

line in the plot in Figure 1.9.

Solution:

Eliminating crX between the expressions 3.01.0 crcr XY and crcr YX /23 (Eq.

(1.32)), we obtain

2/33.0/16.0/3.2 1.0

crY ,

i.e.

391.0135.0 crY ,

which is the expression sought.

Problem 1.7

The slope of a two-dimensional open-channel flow is 41015.0 S , the typical

grain size of the cohesionless bed material is mmD 18.050 . What is the value of

the flow depth which corresponds to the initiation of sediment transport?

Solution:

Consider the expression of the Y-number (second relation in (1.24)):

D

Sh

D

gSh

D

vY

sss

2* . (1)

At the stage of initiation of sediment transport, crYY , crhh , and Eq. (1)

implies


8

D

ShY cr

scr

. (2)

Here )(crY (Eq. (1.34), being given by Eq. (1.31)). Substituting

6.4101000

)1000/18.0(5.161863/1

12

33/1

2

3

Ds

in Eq. (1.34), viz

]1[045.013.0 068.0015.0392.0 2 eeYcr ,

we determine

065.0crY .

Using this value of crY in Eq. (2), we compute the critical flow depth:

mYS

Dh cr

scr 29.1065.0

1015.0

1000/18.065.1

4

.

Problem 1.8

It is intended to study the initiation of sediment transport of the open-channel flow

of Problem 1.7 with the aid of a physical model where the movable bed will be

formed by polystyrene ( 05.0/ s ). What must be the grain size D of the model

bed material and the slope S of the model flow if the model flow depth is to be

?10cmh

(Hint: model and prototype must have identical values of crX and crY for dynamic

similarity).

Solution:

We have for the prototype,

,1015.0 ,29.1 ,65.1/ , 18.0 4 SmhmmD s

smgShv /014.029.11015.081.9 4 ,

and for the model,

mhs 10.0 ,05.0/ .

Fundamentals

9

Let a ( aa / ) be the scale of a quantity a. The dynamic similarity can then be

achieved only if crcr XX and crcr YY , i.e. if 1crX and 1

crY are

provided. This means that

1vDXcr

and 1)/(2 DvY scr

(1) must be valid. From these scale relations it follows that

13 Ds

, and thus that

20.3)65.1/05.0()( 3/13/1

sD . Hence

mD 00058.0)1000/18.0(20.3 )6.0( mm . From the first equation of (1) it is clear that

31.020.3

1

v .

Equating this value of v

with

014.0

10.081.9

014.0

ShSg

v

vv

,

we determine

5

2

109.110.081.9

)014.031.0(

S .

Problem 1.9

Consider the steady and uniform two-dimensional flow in the central part of a wide

river. The flow depth and slope are mh 2 and 00067.0S . The flow bed is a

uniform sand of the grain the grain size mmD 2 : treat the mobile bed surface as

flat. Determine the specific volumetric transport rate sbq using Bagnold’s bed-load

formula.

Solution:

First we determine v , Re and :

smgShv /115.0200067.081.9 ,


10

46010

)1000/22(115.0Re

6*

skv

(fully rough flow),

6.50101000

)1000/2(5.161863/1

12

33/1

2

3

Ds .

Then we evaluate sbq from Eq. (1.45), which can be written as

)( crsb YYuDq .

Here (see Eqs. (1.34), (1.24), (1.46) and (1.11))

044.0]1[045.013.0 068.0015.0392.0 2

eeYcr ,

409.0)1000/2(5.16186

115.01000 22

D

vY

s

,

253.0

2368.0ln

11

2

1ln

2

11

D

h

B

DB

s

s

(where 5.8sB , 5.0 (fully rough flow), 5.07.0 )1/(3.0/ crYYD

5.13 (see Eq. (1.63)),

smBvD

hvu s /48.2

2368.0ln

.

Consequently,

smqsb /106.4)044.0409.0(48.2)1000/2(253.0 24 .

Problem 1.10

The Pembina River (in Alberta) has a rather large width-to-depth ratio, and

therefore in the central part of its cross-section, the flow can be treated as two-

dimensional. In a certain region of this river, the bed consists of a cohesionless sand

having the average grain size mmD 4.0 . The flow depth and slope are (at a

certain time of the year) mh 5 and 00025.0S . Treat the bed surface as flat.

a) Determine the friction factor fc .

b) Determine the flow rate Q (assuming that the average river width is

mB 100 ).

c) Determine the specific volumetric bed-load rate sbq using Bagnold’s formula.

Fundamentals

11

Solution:

a) First we determine v , Re and sB :

smghSv /111.000025.0581.9 ,

8910

)1000/4.02(111.0Re

6

skv

(fully rough flow),

5.8sB (see Figure 1.5). Then we compute the friction factor from Eq. (1.15):

9.275.81000/24.0

5368.0ln

4.0

1368.0ln

1

s

sf B

k

hc

.

b) We compute the flow rate, with the aid of Eq. (1.14), as follows:

smvBhcuBhQ f /1548111.09.275100 3 .

c) First we determine :

1.10101000

)1000/4.0(5.161863/1

12

33/1

2

3

Ds .


)( crsb YYuDq .

Here (see Eqs. (1.34), (1.24), (1.46) and (1.14))

034.0]1[045.013.0 068.0015.0392.0 2

eeYcr ,

9.1)1000/4.0(5.16186

111.01000 22

D

vY

s

,

191.0

2368.0ln

11

2

1ln

2

11

D

h

B

DB

s

s


12


2.11 (see Eq. (1.63)),

smcvu f /10.39.27111.0 . Consequently,

smqsb /104.4)034.09.1(10.3)1000/4.0(191.0 24 .

Problem 1.11

Consider the bed-load formula of H. A. Einstein, viz

*

*)/1(

)/1( 1

11

0*

0*

2

A

Ade

B

B

,

where is the reciprocal of Y -number, 50.43A , 143.0B and 5.00 .

Consider also the bed-load formula of M. S. Yalin, viz

)1ln(

11635.0 as

asYs

where

1crY

Ys and

4.0)/(45.2

s

crYa .

For which value(s) of Y do these formulae give the same value of the bed-load rate

if 043.0crY ? (Take 65.2/ s ).

Solution:

Solving , we can rewrite the formula of H.A. Einstein as

1

*

1)/1(

)/1(

* 0

0

22

2

AdeA B

B

. (1)

Equating this expression of with that given by the formula of M.S. Yalin, we

obtain

)1ln(1

1635.02

2

1*

1)/1(

)/1(

* 0

0

2

asas

YsAdeA B

B

, (2)

where

5.43* A , YB /143.0 and 2/1 0 . (3)

Fundamentals

13

Moreover, for the present 043.0crY ,

1043.0

Y

s , 344.0)65.2(

043.045.2

4.0a , (4)

and

344.081043.0

344.0

Y

Yas . (5)

Using Eqs. (3)-(5) in Eq. (2), we obtain

Y

Ye

Y

Y1

043.0635.0023.0

275.21

1)2/143.0(

)2/143.0(

2

)8656.0ln(

344.08

11 Y

Y. (6)

Solving Eq. (6) (see the note below), we determine the Y -value sought:

633.0Y .

Note:

When solving Eq. (6), the following relations were used for the evaluation of the

integral on the left-hand side:

1) 0 2

0

0

2)(

x

dexerf

)()1(1 0166

06505

404

303

20201 xxcxcxcxcxcxc ,

where 0

70 0,103)( xx , and

0705230784.01 c 0001520143.04 c

0422820123.02 c 0002765672.05 c

0092705272.03 c 0000430638.06 c

(see Romer, A. 1983: 50 BASIC-Programme: Der Computer als Hilfe in

Unterricht und Praxis. Bibliographisches Institut, Mannheim.)

2) For 012 aa :

2 1 222

1

2

0 0

12 )(erf)(erf222

a aa

a

aadedede

;


14

For 021 aa :

||

0

||

0

21

1 2 222

1

2

|)erf(||)erf(|222

a aa

a

aadedede

;

For 01 a and 02 a :

2 1 222

1

2

0

||

0

12 |)erf(|)erf(222

a aa

a

aadedede

.

Problem 1.12

Prove that Eq. (1.61), viz

m

h

zhCC

1/

1/

( vwm s /5.2 ),

implies, in terms of the basic dimensionless variables, the following relation )./,,,( DzZYXC C

Solution:

Eq. (1.61) can be written as

m

DDh

zDDhCC

1)/)(/(

1)/)(/(

, (1)

where (see Eqs. (1.62) and (1.63))

),()1(05.0 *1

CC (2) and

),()1(3.0/ /5.0

*7.0

DD . (3)

But and are the following functions of X and Y :

3/12

Y

X and

))/(()( 3/12 YX

YY

Y

Y

cr

.

Using these relations in Eqs. (2) and (3), we obtain

),())])/(([,)/(( 13/123/12 YXYXYYXCCC

(4)

Fundamentals

15

and

),())])/(([,)/(( /13/123/12

/ YXYXYYXD DD

. (5)

Consider now vwm s /5.2 . For a given shape of the grains, having the average

size D, the terminal velocity of grains, sw , is completely determined by their

specific weight in fluid )( s and by the nature of fluid ( , ). i.e.

),,,( Dfw sws .

Hence the dimensionless counterpart of sw can be expressed as

Y

XDDww

sw

s2

2

3

.

Consequently,

Y

X

Dw ws

2

,

and thus

Y

X

Xv

ww

s21

and ),(5.2 2

YXY

X

Xm mw

. (6)

Substituting Eqs. (4), (5) and (6) into Eq. (1), and taking into account that

DhZ / , we obtain

),(

1/ 1)],([

1)/(),(

YX

DC

m

YXZ

zDZYXC

,

i.e.

)/,,,( DzZYXC C , which is the relation sought.

Problem 1.13

The shape of the C -distribution curve can either be like A in Figure 1.13b or like B

(with a point of inflection P ). This depends on the numerical value of

vwm s /5.2 . Determine the range of values of m that yields the shape B.


16

Solution:

Consider Eq. (1.61), which can be written as

m

h

zh

C

C

1)/(

1)/(

. (1)

Denoting hz / and CCh m /]1)/[( by and *C , respectively, we can write

Eq. (1) as

mC )1( 1* . The first derivative of *C with respect to is

)()1( 211*

mmd

dC,

the second derivative being

)]2()1()1)(1[( 311421

2

*2

mmmmd

Cd.

The curves B have a point of inflection P, which is given by

02

*2

d

Cd.

But this means that the curves B must satisfy

0)2()1()1)(1( 311421 P

mPP

mPm , (2)

where P is the dimensionless level of the inflection point corresponding to a

given m ( 0 ).

Solving P from Eq. (2), we obtain

2

1

mP .

Observe that this relation gives the realistic P (which satisfy 1 P ) only if

m is from the range 10 m ; if 1m , then the curves are of the type A.

Note:

When 0m , then 2/1P , and when 1m , then 1P ; i.e. the increment of

m from 0 to 1 is accompanied by the increment of P from 1/2 to 1. It follows that

the boundary-curve separating the A and B curve-regions is the curve

Fundamentals

17

corresponding to 1m (and that the curve A in Figure 1.13b is, in fact, the

boundary-curve whose point of inflection is at the free surface).

Problem 1.14

In a certain river, the suspended-load concentration C at the level 75.0/ hz is

half the concentration at the level 25.0/ hz . Is the C -distribution curve like A or

B in Figure 1.13b?

Solution:

Consider Eq. (1.61), which can be rewritten as

m

hz

hz

h

h

C

C

/

/1

/1

/

. (1)

Denoting C at the levels 75.0/ hz and 25.0/ hz by 75.0C and 25.0C ,

respectively, we can write, on the basis of Eq. (1),

mmmm

h

hC

h

hCC

/1

/

75.0

25.0

75.0

75.01

/1

/75.0

(2)

and

mmmm

h

hC

h

hCC

/1

/

25.0

75.0

25.0

25.01

/1

/25.0

, (3)

which yield

mm

C

C

9

1

75.0

25.0

75.0

25.0

25.0

75.0 .

But 2/1/ 25.075.0 CC , and therefore

2

1

9

1

m

i.e. 315.0)9/1log(

)2/1log(m .

Since 10 m , the C -distribution curve is of the type B (see solution of Problem

1.13).

Problem 1.15

Consider a two-dimensional flow in a stream having a bed that consists of a

cohesionless sand having the average grain size mmD 30.0 . The flow depth and

slope are mh 80.0 and 0002.0S . Treat the bed surface as flat.


18

a) Determine the bed-load rate sbq using Bagnold’s formula.

b) Determine the thickness of the bed-load region.

c) Determine the concentration C at z .

d) Determine the specific volumetric suspended-load rate ssq . Take smws /03.0 .

Solution:

a) First we determine v , Re , sB (see Eq. (1.10)) and :

smgShv / 04.080.00002.081.9 ,

2410

)1000/30.02(04.0Re

6*

skv

,

3.9]1[5.8)5.5Reln5.2(55.2

*55.2

* )Re(ln0594.0)Re(ln0705.0*

eeBs ,

6.7101000

)1000/3.0(5.161863/1

12

33/1

2

3

Ds .


)( crsb YYuDq . Here (see Eqs. (1.34), (1.24), (1.11))

043.0]1[045.013.0 068.0015.0392.0 2

eeYcr ,

329.0)1000/3.0(5.16186

04.01000 22

D

vY

s

,

smBvD

hvu s /99.0

2368.0ln

.

is given by Eq. (1.46), viz

.

2368.0ln

11

2

1ln

2

11

D

h

B

DB

s

s

The curve in Figure 1.10 interrelates the values of D ~ and (the

proportionality factor corresponds to sand or gravel and water). From this graph, if

follows that for mmD 30.0 , we have 3.0 . Using this value of in the

Fundamentals

19

expression above, and considering that 2.3)1/(3.0/ 5.07.0 crYYD (see Eq.

(1.63)), we obtain

12.0 .

Consequently,

smqsb /100.1)043.0329.0(99.0)1000/3.0(12.0 25 .

b) From part (a), we have 2.3/ D , and thus

mmm 96.000096.0)1000/3.0(2.3 .

c) We determine C from Eq. (1.62):

044.0)1043.0/329.0(6.705.0)1/(05.0 11 crYYC .

d) We compute ssq from Eq. (1.55), viz

h

ss Cudzq

. (1)

Here C is given by Eq. (1.61), which can be rewritten as

m

z

zh

hCC

, (2)

and u is given by Eq. (1.6) (see also Eqs. (1.9) and (1.15)), viz

D

ze

vu sB

2ln

. (3)

For the present case of mh 8.0 , mmD 3.0 , smv /04.0 , 043.0C ,

m00096.0 , 29.1/5.2 vwm s , Eqs. (2) and (3) give

2

822

1102.6100096.08.0

00096.0043.0

z

h

z

hC (4)

and

)68774ln(1.01000/3.02

ln4.0

04.0 3.94.0 zz

eu

. (5)


20

Using Eqs. (4) and (5), and adopting 1.02.11)68774ln( zz , we determine

1.02

8 1)109.6( zz

hCu

,

and thus

hh

ss dzzhzzhdzzz

hq

)2(109.61109.6 1.09.09.1281.02

8

h

zzhzh

1.11.09.028

1.1

1

1.0

12

9.0

1109.6

1.11.09.0

21.11.09.0

28

1.1

1

1.0

2

9.01.1

1

1.0

2

9.0109.6

hhhh

hh

h.

For mh 8.0 and m 00096.0 , this expression gives

smqss /105.29.361109.6 258 .

Problem 1.16

At a certain stage, the flow depth and maximum velocity of the Mississippi River

(at St. Louis) are mh 12 and smu /5.1max . In the interval 6.0/5.0 hz , the

concentration distribution curve can be approximated by the straight line

)/1(0002.0 hzC . Adopting for the velocity distribution the expression 7/1

max )/( hzuu , determine the specific volumetric suspended-load rate ssq

passing through the interval 6.0/5.0 hz .

Solution:

Let us denote hz / by . The specific volumetric suspended-load rate passing

through an interval 21 is determined as follows

2

1

Cudhqss .

Substituting here )1(0002.0 C and 7/1maxuu , we obtain

2

1

2

1

7/157/8max

7/87/1max

15

7

8

7 0002.0)( 0002.0

hudhuqss

7/151

7/81

7/152

7/82max

15

7

8

7

15

7

8

7 0002.0 hu .

Fundamentals

21

For the present case of smu / 5.1max , mh 12 , 5.01 and 6.02 , this

expression gives

smqss /105.1 24 .

Problem 1.17

Prove that shU /)( , sqs / , and sq are interrelated as follows

s

hU

q

q

s

q sss

)(q

Hint: Use the open forms of sq and 0)( Uh (continuity).

Solution:

We have

s

b

b

q

s

q

s

bq

s

qb

bs

bq

b

sss

sss

1)(1q

and

01)(1

)(

s

b

b

q

s

q

s

bq

s

qb

bs

qb

bh qU .

Eliminating bsb /)/( from the equations above, we obtain

s

q

q

q

s

q sss

q ,

which is the relation sought.

Problem 1.18

Prove that Eq. (1.79) (viz qs h )( Uq , where qqsq / ) and the expression

of sq in Problem 1.17 are identical to each other.

Solution:

Eq. (1.79) can be developed as

n

qq

s

qqq

q

qh s

ns

sss

qs

)/()/()()( iiiqUq


22

s

q

q

q

s

q

q

s

qqq

s

q

qs

qqq ss

ss

s

2

)/(,

where the end-result is the expression of sq in Problem 1.17.

Flow past undulated beds

23

CHAPTER 3


Problem 3.1

Consider a two-dimensional flow in a river having a bed that consists of a

cohesionless sand with average grain size mmD 3.0 . The depth and slope of this

river are mh 0.3 , 00005.0S . Are sand waves on the bed surface ripples, dunes,

or ripples superimposed on dunes?

Solution:

First we determine v , and )(crY (Eq. (1.34)):

smgShv / 038.00.300005.081.9 ,

6.7101000

)1000/3.0(5.161863/1

12

33/1

2

3

Ds ,

043.0]1[045.013.0 068.0015.0392.0 2

eeYcr .

Then we evaluate X , and Z :

4.1110

)1000/3.0(038.06

DvX ,

9.6043.0

1

)1000/3.0(5.16186

038.010001 22

crscr YD

v

Y

Y

,

10000)1000/3.0(

3

D

hZ .

Finally, we plot the point )9.6;4.11( XP on the graph in Figure 2.29 (p. 60

of the monograph), and observe that this point lies in the common area of the R -

and D -regions ( 355.2 X ; 211 ); thus the sand waves are ripples

superimposed on dunes.

24 Fluvial processes: 2nd edition – Solutions manual

Problem 3.2

In a region of the Pembina River, we have mmD 4.0 , mh 0.5 and 00025.0S

(see Problem 1.10).

a) Are sand waves on the bed surface ripples, ripples superimposed on dunes, or

dunes? Determine also the length and height of the bed forms.

b) Determine the resistance factor c .

c) Determine the flow rate Q (assuming that the average river width is

mB 100 ).

d) Determine the specific volumetric bed-load rate sbq using Bagnold’s formula.

e) Compare the values of Q and sbq of the flow past the undulated bed

determined in c) and d) with the values of Q and sbq determined in Problem

1.10 (where the bed forms were disregarded).

Solution:

For the determination of the quantities below, which will be used throughout this

solution, see Problem 1.10:

1.10 , 034.0crY , 9.1Y ,

smv /111.0 , 9.27fc , smu f /10.3 , 191.0 .

a) Determination of the type of bed forms. First we determine X , and Z :

2.44)( 2/13 YX ,

9.55crY

Y ,

12500D

hZ .

Then we plot the point )9.55 ;2.44( XP on the graph in Figure 2.29, p. 60,

and observe that P lies in the D -region whose upper boundary (which corresponds

to 12500Z ) is not lower than the level 140 . In other words, the upper

boundary of the D -region we are dealing with is well above the point P. Hence no

ripples are present; we have only dunes.

Determination of the dune length. Observe that the present 2.44X is larger than

66.42X , beyond which the curves representing Eq. (2.17) become

indistinguishable from the straight line hd 6 (see Figure 2.20). Consequently,

md 3056 .


25

Determination of the dune height. We compute d ( dd / ) from Eq. (2.22),

viz

mddd

de )()(1

max

.

Substituting in this relation (see Eqs. (2.20), (2.23), (2.21), (2.24))

04.0)1(04.000047.0)( 002.017.02.1max

47.0

ZZd eeZ ,

99.1

6)1(35

1

1ˆ

14.0074.0

Zdd

e

,

56.1 6.016.3)log5(1.0 Zem ,

we obtain

025.0)99.1(04.0 56.199.11 ed ,

and thus

mddd 75.030025.0 .

b) We compute c from Eq. (3.11) (with d and d ):

8.175

30025.0

2

1

9.27

1

2

112/1

2

2

2/1

2

2

hcc d

d

f

.

c) We compute Q from Eq. (3.34):

smBhcvQ / 988111.08.175100 3 .

d) We compute sbq from Eq. (3.49), which can be written as

)( 2crcfcsb YYuDq .

Here 638.09.27/8.17/ fc cc , and we have

smqsb /101.1)034.09.1638.0(10.3638.0)1000/4.0(191.0 242 .


e) Q and sbq of the present flow past undulated bed compare with their flat-bed

counterparts fQ and fsbq )( determined in Problem 1.10 as follows

64.01548

988

fQ

Q ; 25.0

104.4

101.1

)( 4

4

fsb

sb

q

q.

Observe that the ratio fsbsb qq )/( could also be computed from Eq. (3.51).

Problem 3.3

The flow depth of a two-dimensional river flow is mh 3.2 . The typical grain size

of the cohesionless sand bed is mmD 8.0 . It is assumed that the sand waves are

dunes only.

a) Find for which value of the slope S the steepness of dunes has its maximum

value. And what is this maximum value?

b) For the value of S determined in (a), and knowing that the flow width is m70 ,

determine the total volumetric bed-load rate sQ . Use Bagnold’s equation.

Solution:

a) The flow intensity d for which the dune steepness acquires its maximum is

determined by Eq. (2.21). For the present 2875/ DhZ , this equation gives

2.245)1(35ˆ4.0074.0

Z

d e . Equating this result with the definition of (see Eqs. (1.24) and (1.8)), we obtain

2.2411

ˆ2

crscrscrd

YD

hS

YD

v

Y

Y

,

which gives for the slope S corresponding to maximum dune steepness,

crs Y

h

DS

2.24 .

For mmD 8.0 , we have 2.20)( 3123 Ds , and therefore (see Eq.

(1.34))

034.0]1[045.013.0 068.0015.0392.0 2

eeYcr .

Hence,

00047.0034.03.2

)1000/8.0(65.12.24 S .

The maximum dune steepness is given by the following relation (Eq. (2.20))


27

)1(04.000047.0)( 002.017.02.1max

47.0 ZZd eeZ ,

which, for the present 2875Z , gives

045.0)( max d .

b) First we determine v , Re and sB :

smgShv /103.03.200047.081.9 ,

16510

)1000/8.02(103.0Re

6

skv

(fully rough flow),

5.8sB (see Figure 1.5). Then we determine the dune length. Since 5.822/1652/Re X is larger

than 66.42X , beyond which the curves representing Eq. (2.17) become

indistinguishable from hd 6 (see Figure 2.20), we have

md 8.133.26 .

Next we compute fc (from Eq. (3.4), with Dks 2 ) and c (from Eq. (3.11), with

d and d ):

2.245.81000/8.02

3.2368.0ln

4.0

1

2368.0ln

1

sf B

D

hc

,

3.113.2

8.13045.0

2

1

2.24

1

2

112/1

2

2

2/1

2

2

hcc d

d

f

.

Finally, we compute sbq from Eq. (3.49), which can be written as

)( 2crcfcsb YYuDq .

Here (see part (a), see also Eqs. (1.35), (1.46))

034.0crY ,

823.0034.02.24 crYY ,


222.0

2368.0ln

11

2

1ln

2

11

D

h

B

DB

s

s

(where 5.0 (fully rough flow), 9.11)1/(3.0/ 5.07.0 crYYD (see Eq.

(1.63)). Moreover (see Eqs. (3.46), (3.47))

smvcu ff /49.2103.02.24 ,

467.02.24

3.11

fc

c

c .

Hence

smqsb /103)034.0823.0467.0(49.2467.0)1000/8.0(222.0 252 and

smBqQ sbsb /101.210370 335 .

Problem 3.4

The bed of the Nile River at Esna Barrage (Shalash 1983, Hartung 1987; see also

Yalin and Scheuerlein 1988) consists of a cohesionless sand having the average

grain size mmD 28.0 . The slope is 000077.0S .

a) For the flow depths mh 70.21 , mh 00.42 and mh 70.53 , determine the

type of bed forms (ripples, ripples superimposed on dunes, or dunes), the

length and steepness of the bed forms, and the resistance factor.

b) Use the information on the type and geometric characteristics of the bed forms

to explain the drop of the resistance factor with increasing flow stage.

Solution:

a) Flow depth mh 70.2 1 :

Determination of the type of bed forms. First we determine v , and )(crY

(Eq. (1.34)):

smgShv /045.07.2000077.081.9* ,

1.7101000

)1000/28.0(5.161863/1

12

33/1

2

3

Ds ,

046.0]1[045.013.0 068.0015.0392.0 2

eeYcr .


29

Then we compute X , and Z :

6.1210

)1000/28.0(045.06

DvX ,

7.9046.0

1

)1000/28.0(5.16186

045.010001 22

crscr YD

v

Y

Y

,

9643)1000/28.0(

70.2

D

hZ .

Finally, we plot the point )7.9;6.12( XP on the graph in Figure 2.29, and

observe that the point is in the common area of the R - and D -regions. Hence, we

have ripples superimposed on dunes.

Determination of the dune length. We compute d from Eq. (2.17), viz

m

d eZ

ZZh

)400)(40(01.016 .

Here

9.56.1204.09643055.004.0055.0 XZm ,

and thus

med 3.209643

)4009643)(409643(01.0170.26 9.5

.

Determination of the dune steepness. We compute d from Eq. (2.26), which can

be written as (see Eq. (2.22))

mdddd

deX )()()(1

max

.

Here (see Eqs. (2.25), (2.20), (2.23), (2.21), (2.24))

796.011)(22 )10/6.12()10/( eeX X

d ,

04.0)1(04.000047.0)( 002.017.02.1max

47.0

ZZd eeZ ,

321.0

6)1(35

1

1ˆ

14.0074.0

Zdd

e

,


54.16.016.3)log5(1.0 Zem .

Using these values in the expression of d , we determine

016.0)321.0(04.0796.0 54.1321.01 ed .

Determination of the ripple length. We determine r from Eq. (2.32):

mD

r 15.0)7.922.01(7.91.7

)1000/28.0(3000

)22.01(

300088.088.0

.

Determination of the ripple steepness. We determine r from Eq. (2.39), which,

taking into account Eq. (2.37), can be expressed as

)1.01.1( *)1(014.0)(

erXrr .

Here

594.0)(22 ]14/)5.26.12[(]14/)5.2[( eeX X

r ,

1r (since 87.0)172.9(1.0)1(1.0 r ).

Hence

082.0)17.9(1014.0594.0 )7.91.01.1( er .

Determination of the resistance factor. First we compute Re and sB (see Eq.

(1.10)):

2.2510

)1000/28.02(045.0Re

6

skv

,

2.9]1[5.8)5.5Reln5.2(55.255.2 )Re(ln0594.0)Re(ln0705.0

eeBs .

Then we compute the friction factor from Eq. (3.4):

9.272.9)1000/28.0(2

70.2368.0ln

4.0

1

2368.0ln

1

sf B

D

hc

.

Finally, we compute the resistance factor from Eq. (3.12):


31

2/1

22

2)(

2

11rrdd

fhc

c

3.20)153.0072.03.20016.0(70.22

1

9.27

12/1

22

2

.

The detailed computation of characteristics corresponding to mh 70.21 was

carried out above only for didactic purposes. Indeed, these characteristics could

have been obtained simply by using the computer program RFACTOR (available

for download on the CRC Press website as Supplementary Material to the

monograph), and so it was done for the remaining flow depths. The computed

results are displayed all together in the Table below.

h

(m)

Bed form type d d

(m) r r

(m)

c

2.70

4.00

5.70

Dunes + Ripples

Dunes + Ripples

Dunes (only)

20.3 0.016

26.5 0.026

35.5 0.035

0.15 0.082

0.25 0.049

20.3

16.7

14.1

b) The drop of the resistance factor c means the increment of the overall relative

bed roughness hKs / (see Eq. (3.28)). From the Table above it is clear that the

increment of h (from 2.70m to 5.70m) is accompanied (via decrement of c) by an

even stronger increment of sK .

Eqs. (3.28) and (3.12) indicate that hKs / is an increasing function of geometry

and skin roughness of bed forms, as embodied by the complex

hhc

rr

dd

f

22

2 2

11 .

This complex conveys that the variation of hKs / is mainly due to the variation of

the relative length hi / and/or steepness i of bed forms. In the present case, the

increment of hKs / (via decrement of c) is mainly due to the increment of d .

Problem 3.5

The Yangtze River at Da-tong, halfway between Wuhan and Nanjing, has the

following characteristics at average flood peak (Luo et al. 1980, Lin and Li 1986;

see also Yalin and Scheuerlein 1988): mmD 2.0 , 0000277.0S .

a) Determine whether the bed forms are ripples, ripples superimposed on dunes,

or dunes.

b) Determine the resistance factor c.


c) Use the resistance factor c determined in part b) to estimate the value of

Manning’s n.

Solution:

a) First we determine v , and )(crY (Eq. (1.34)):

smgShv /066.01.160000277.081.9 ,

1.5101000

)1000/2.0(5.161863/1

12

33/1

2

3

Ds ,

06.0]1[045.013.0 068.0015.0392.0 2

eeYcr .

Then we determine X , and Z :

2.1310

)1000/2.0(066.06

DvX ,

4.2206.0

1

)1000/2.0(5.16186

066.010001 22

crscr YD

v

Y

Y

,

80500)1000/2.0(

1.16

D

hZ .

Finally, we plot the point )4.22,2.13( XP on the graph in Figure 2.29.

Observe that P lies in the D -region whose upper boundary (which corresponds to

80500Z ) is well above P: thus the bed forms are dunes only.

b) First we determine the dune length and steepness. We determine d from Eq.

(2.17), viz

m

d eZ

ZZh

)400)(40(01.016 .

Here

1.162.1304.080500055.004.0055.0 XZm ,

and thus

med 6.9680500

)40080500)(4080500(01.011.166 1.16

.


33

We compute d from Eq. (2.26), which can be written as (see Eq. (2.22))

m

dddddeX )())((

1max

.

Substituting in this relation (see Eqs. (2.25), (2.20), (2.23), (2.21), (2.24))

825.011)(22 )10/2.13()10/( eeX X

d ,

04.0)1(04.000047.0)( 002.017.02.1max

47.0

ZZd eeZ ,

740.0

6)1(35

1

1ˆ

14.0074.0

Zdd

e

,

6.16.016.3)log5(1.0 Zem ,

we obtain

031.0)740.0(04.0825.0 6.1740.01 ed .

Then we compute Re , sB (Eq. (1.10)) and the friction factor fc (Eqs. (3.4) and

(1.4)):

4.262Re X ,

2.9]1[5.8)5.5Reln5.2(55.255.2 )Re(ln0594.0)Re(ln0705.0

eeBs ,

2.332.91000/2.02

1.16368.0ln

4.0

1

2368.0ln

1

sf B

D

hc

.

Finally, we determine the resistance factor c from Eq. (3.11) (with d and

d ):

2.161.16

6.96031.0

2

1

2.33

1

2

112/1

2

2

2/1

2

2

hcc d

d

f

.

c) The value of Manning’s n follows from the interrelation between n and c

derived in Problem 1.4:

031.081.9

1.16

2.16

11 6/16/1

g

h

cn .


Problem 3.6

Consider the Mississippi River (at Natchez) at its bankfull stage (Winkley 1977,

Schumm 1977; see also Yalin and Scheuerlein 1988): mmD 4.0 , mh 63.14 ,

00007.0S , mB 1771 .

a) Determine the resistance factor c and the flow rate Q using:

1. the method of Yalin and da Silva (present monograph, 2.5.2);

2. the method of White, Paris and Bettess (1979);

3. the method of van Rijn (1984) (knowing that smu /39.1 ).

b) Compare your estimates of the flow rate Q with the measured value

smQ /35963 3 .

Solution:

a) 1. Method in present monograph (Sub-section 3.1.2)

i) Determination of the type of bed forms. First we determine v , and crY :

smgShv /1.063.1400007.081.9 ,

1.10101000

)1000/40.0(5.161863/1

12

33/1

2

3

Ds ,

034.0]1[045.013.0 068.0015.0392.0 2

eeYcr .

Next we determine X , and Z :

4010

)1000/40.0(1.06

DvX ,

3.45034.0

1

)1000/40.0(5.16186

1.010001 22

crscr YD

v

Y

Y

,

36575D

hZ .

Then we plot the point )3.45;40( XP on the graph in Figure 2.29. P lies in

the D -region whose upper boundary (which corresponds to 36575Z and which

is not lower than the level 140 ) is well above P. Finally we plot the point

)121/;36575/( hBDhP on the graph in Figure 2.30, and observe that P lies

in the A-region. Thus the bed forms are dunes superimposed on alternate bars.

However, since the steepness of alternate bars is invariably much smaller than that

of dunes (see Sub-section 3.1.2, ‘Derivation’), the influence of alternate bars on the

value of c will be ignored in this solution.


35

ii) Determination of the dune length. We compute d from Eq. (2.17), viz

m

d eZ

ZZh

)400)(40(01.016 .

Here

1.124004.036575055.004.0055.0 XZm ,

and thus

med 0.8836575

)40036575)(4036575(01.0163.146 1.12

.

iii) Determination of the dune steepness. We determine d from Eq. (2.22), viz

mddd

de )()(1

max

.

Substituting in this relation (see Eqs. (2.20), (2.23), (2.21), (2.24))

04.0)1(04.000047.0)( 002.017.02.1max

47.0

ZZd eeZ ,

54.1

6)1(35

1

1ˆ

14.0074.0

Zdd

e

,

6.16.016.3)log5(1.0 Zem ,

we obtain

034.0)54.1(04.0 6.154.11 ed .

iv) Determination of the resistance factor. First we determine 802Re X

(fully rough flow) and 5.8sB (see Figure 1.5). Then we evaluate the friction

factor from Eq. (3.4) (with Dks 2 (Eq. (1.4)):

5.305.81000/4.02

63.14368.0ln

4.0

1

20368.0ln

1

sf B

D

hc

.

Finally we evaluate the resistance factor from Eq. (3.11) (with d and

d ):

8.1463.14

0.88034.0

2

1

5.30

1

2

112/1

2

2

2/1

2

2

hcc d

d

f

.


v) Determination of the flow rate. We use Eq. (3.34):

)(/383461.08.1463.141771 13 QsmBhcvQ .

2. Method of White, Paris and Bettess (1979) i) Determine the values of , Y and Z (see part (1) of this solution):

1.10 , 54.1Y , 36575Z .

ii) Determine )( , n and A from Eqs. (3.25), (3.26) and (3.27):

47.0)1(70.01)(65.2)(log4.1 e ,

438.01.10log56.01log56.01 n ,

212.014.01.10

23.014.0

23.0

A .

iii) Determine grF from Eq. (3.24):

696.0212.0)212.054.1(47.0))(( AAYFgr .

iv) Determine c from Eq. (3.23):

)10log(32

)1/(1

ZY

Fc

ngr

2.11)3657510log(3254.1

696.0)1438.0/(1

.

v) Determine Q (from Eq. (3.34)): )(/290191.02.1163.141771 2

3 QsmBhcvQ .

3. Method of van Rijn (1984) In this method, u ( sm /39.1 ) is to be used instead of S ( 0007.0 ). i) Determination of T. First we compute crv :

smYD

v crs

cr /015.0034.01000

)1000/40.0(5.161862/12/1

.


37

Then, using the given smu /39.1 and also 5.30fc (determined earlier), we

compute T from Eq. (3.32):

2.8015.0

015.0)5.30/39.1()/(

2

22

2

22

cr

crf

v

vcuT .

ii) Determination of the dune height and steepness. We compute d and d from

Eqs. (3.30) and (3.31), respectively:

mTehZ Td 14.1)25)(1(11.0 5.03.0 ,

011.0)25)(1(015.0 5.03.0 TeZ Td .

iii) Determination of the resistance factor c . First we determine sK (from Eq.

(2.72)):

meeDK dds 303.0)1(14.11.10004.03)1(1.13 011.02525

.

Then we compute c from Eq. (3.28), with hR :

7.15303.0

63.1411ln5.211ln5.2

sK

hc .

iv) Determination of the flow rate. This part of the question is not applicable in this

case, for smu /39.1 has not been arrived at independently. Rather, it was

computed on the basis of the measured mQ ( sm /35963 3 ; see part (b)) as

BhQu m / .

b) The ratios of the computed iQ ’s to the measured flow rate )( mQ are:

1. 07.135963/38346/1 mQQ (Present method; Sub-section 3.1.2);

2. 81.035963/29019/2 mQQ (White, Paris and Bettess).


Problem 3.7

Consider the following conditions observed in the Zaire River at Ntua Nkulu

(Peters 1978; see also Yalin and Scheuerlein 1988), where mmD 75.0 :

mB 500 , 000058.0S , smQ /12300 3 .

Estimate the flow depth h.

Solution:

From the resistance equation (3.34), we obtain

3/1

22

2

gScB

Qh . (1)

For the present Q , B and S , Eq. (1) gives

3/2102 ch , (2) where ),,( ShDc c .

The following computation procedure can be adopted to determine h from Eq. (2).

Adopt ih

For i = 1, 2, ..., until satisfied, do:

1. Determine ),,( ShDc ici using the computer program RFACTOR.

2. Knowing ic from step 1, determine 1ih from Eq. (2): 3/21 102 ii ch .

3. If ii hh 1 , then the problem is solved: ihh ;

If ,1 ii hh then 1 ii , and repeat steps 1 to 3.

The results of computations conducted as described above are shown below.

It follows that mh 4.18 .

Problem 3.8

In the 0.76m-wide, 21m-long sediment transport flume of the Queen’s University –

Graduate Research Laboratory, six experimental runs were conducted to investigate

the formation of bars: the bed material was mmD 1.1 sand. All runs were carried

i )(mhi ic

1

2

3

4

5

10.0

15.9

18.0

18.4

18.4

16.3

13.5

13.1

13.1


39

out starting from a flat initial bed. The initials flows (at 0t ) were uniform; the

values of Q, h and S were as indicated in the Table below.

a) Plot these initial flows as data-points on the )/;/( DhhB -plane in Figure 2.30.

b) What was the configuration in plan view of the bars (one-row or multiple-row)

observed in each run?

c) Determine the bar length for the conditions of Run CS-5 using both the

equations by Boraey and da Silva (2014) and Ikeda (1984).

Solution:

a, b) The values of hB / and Dh / for each run are shown below.

On the )/;/( DhhB -plane of Figure 2.30, the points representing the runs CS-3 to

CS-6 are situated between the lines 1,0L and 1,2L (determined by the relations

(2.42)-(2.46)). In contrast to this, the points representing the runs CS-1 and CS-2

are lying above the line 1,2L . Thus, we have multiple-row bars in runs CS-1 and

CS-2, and one-row bars (alternate bars) in runs CS-3 to CS-6.

c) 1. Method by Boraey and da Silva (2014)

First we determine v and Re :

smgShv / 045.0025.00081.081.9 ,

9010

)1000/3.02(045.0Re

6

skv (fully rough flow).

Run )/( slQ )( cmh 3

10S

CS-1 2.11 1.0 8.1

CS-2 2.86 1.2 8.1

CS-3 4.15 1.5 8.1

CS-4 6.12 2.2 8.1

CS-5 9.73 2.5 8.1

CS-6 11.00 4.1 9.0

Run hB / Dh /

CS-1 76.0 9.1

CS-2 63.3 10.9

CS-3 50.7 13.6

CS-4 34.5 20.0

CS-5 30.4 22.7

CS-6 18.5 37.3


Since the flow is fully rough, a is given by Eq. (2.50). First we determine DB /

and the values of )/(1 DB , )/(2 DB and )/(3 DB :

760)1000/0.1/(76.0/ DB ,

426.5393.1)/ln(028.1)/(1 DBDB ,

0987.0)/(5765.0)/( 266.02 DBDB ,

524.005.2

11)/(

]75.3)/(001.0[3

DBe

DB .

Next we determine a from Eq. (2.50). This yields:

mDBeDBBDhDB

a 02.5)]/()/([6 3)]/()/([

12

.

2. Method by Boraey and da Silva (2014)

First we determine u and Fr:

smBh

Qu /512.0

025.076.0

)1000/73.9(

,

07.1024.081.9

512.0 22

gh

uFr

Since 64.0Fr , we determine a from Eq. (2.57). This yields:

mhBDBBa 19.64.3076076.03.5)/()/(3.5 45.045.0

Note:

Since DB / is relatively small, both equations give results that can be viewed as

comparable (see comments at the end of Sub-section 2.4.2).

Problem 3.9

Explain why the flume-data, corresponding to a given B )( const and D

)( const , determined for various h and/or S form on the )/;/( DhhB -plane the

1/1-declining straight lines (see the data in Figure 2.30; see also Problem 3.8, part

(a)).


41

Solution:

The ratio hB / can be developed as

Dh

const

Dh

DB

h

D

D

B

h

B

//

]/[ ,

which, in the )/;/( DhhB -plane, is a 1/1-declining straight line.

Problem 3.10

In a wide tilting flume, whose bed is covered by a layer of sand having mmD 5.1 ,

a flow having smQ /44.0 3 , 600/1S , and 5.3 is introduced. Will this

flow generate dunes, ripples, alternate bars or multiple bars?

Solution:

First we determine , crY (Eq. (1.34)) and Y :

9.37101000

)1000/5.1(5.161863/1

12

33/1

2

3

Ds ,

042.0]1[045.013.0 068.0015.0392.0 2

eeYcr ,

147.05.3042.0 crYY .

From ,/)/()/(2 DhSDvY ss we obtain

mh 22.0600/1

147.0)1000/5.1(65.1

.

Hence

smgShv /060.022.0)600/1(81.9 ,

9010

)1000/5.1(060.06

*

DvX ,

147)1000/5.1(

22.0

D

hZ .


Then we plot the point )5.3;90( XP on the graph in Figure 2.29, and

observe that P lies in the D -region whose upper boundary corresponds to 147Z .

Thus, for the present flow conditions, dunes will form.

Next, we will investigate the possibility of occurrence of bars, and for this purpose

we need to determine the flow width B. The determination of B can be carried out

as follows.

First, we determine the dune length. Since the present 90X is larger than 42.66,

we have (see Figure 2.20)

mhd 32.122.066 .

Then we compute the dune steepness from Eq. (2.22), viz

m

dddde )()(

1max

.

Here (see Eqs. (2.20), (2.23), (2.21), (2.24))

042.0)1(04.000047.0)( 002.017.02.1max

47.0

ZZd eeZ ,

287.0

6)1(35

1

1ˆ

14.0074.0

Zdd

e

,

01.16.016.3)log5(1.0 Zem .

Consequently,

025.0)287.0(042.0 01.1287.01 ed .

Now we compute fc and c (ignoring the (negligible) contribution of bars – if they

are also present). Eq. (3.4) (with Dks 2 ) and Eq. (3.11) (with d and

d ) give

7.165.81000/5.12

22.0368.0ln

4.0

1

2368.0ln

1

sf B

D

hc

,

5.1322.0

32.1025.0

2

1

7.16

1

2

112/1

2

2

2/1

2

2

hcc d

d

f

.

Finally, using this value of c, we determine B:

mhcv

QB 47.2

060.05.1322.0

44.0

.


43

Hence we have

2.1122.0/47.2/ hB and 147/ ZDh . Since the point representing these values on the )/;/( DhhB -plane of Figure 2.30

is below the line 1,0L , bars will not form and we have dunes only.


Regime channels and their computation 45

CHAPTER 4

Regime channels and their computation

Problem 4.1

Consider the North Saskatchewan River: smQ /4386 3 (bankfull flow rate),

mmD 31 (gravel).

a) Determine the regime characteristics RB , Rh and RS . Use the equations in

Sub-section 4.5.2.

b) Compare the values of RB , Rh and RS obtained in a) with the corresponding

values in Table 4.1 (which were obtained with the computer program BHS-

STABLE). Explain the differences. (Hint: Determine RB , Rh and RS with the

method in Sub-section 4.5.2, but replace the power approximation (4.60) by

)/11ln()/1( skhc .

Solution:

a) First, using 045.0crY (gravel), we determine crv :

smYD

v crs

cr /150.0045.01000

)1000/31(5.161862/12/1

.

Then we compute RB , Rh and RS from Eqs. (4.62), (4.64) and (4.65),

respectively:

mv

QB

crR 8.242

150.0

438642.142.1

,

mv

QDh

crR 13.7

150.0

4386

0.7

)1000/31(

0.7

7/37/17/37/1

,

00032.04386

150.0

)1000/31(

0.7

81.9

150.0)0.7(7/3

7/1

27/3

7/1

2

Q

v

Dg

vS crcr

R .


46

b)

Let us mark the results computed in part (a) by the subscript a, those in Table 4.1

(and obtained with the computer program BHS-STABLE), by p. The Table above

indicates that pRaR hh )()( , while pRaR SS )()( , but aRaR Sh )()(

pRpR Sh )()( )0023.0//)(( 22 gvgv crR . Here the discrepancy is analyzed

in terms of Rh only.

The difference between aRh )( and pRh )( is due to the different expressions of Rc .

In the computer program, pRc )( was expressed by the logarithmic law; in the part

(a), the power approximation of that law was used for the expression of aRc )( (see

(4.60)). If the resistance equation (4.63) is evaluated by the same expressions of

RB and RS as in part (a) (viz by Eqs. (4.62) and (4.59)), but pRc )( is expressed by

the logarithmic law

s

pRpR

k

hc

)(11ln5.2)( ,

then we determine

cr

pRpR

v

Q

D

hh

55.3

1

2

)(11ln)( .

Substituting the present Q , D and crv in this relation, we obtain

17.48))(4.177ln()( pRpR hh ,

and thus mh pR 79.6)( , which, with the aid of Eq. (4.59), gives

00034.079.681.9

150.0 22

R

crR

gh

vS .

Problem 4.2

A gravel channel is in regime. Knowing that the bankfull flow rate is

smQ /3000 3 and the regime width is mBR 7.211 , determine the grain size D,

the regime flow depth Rh , and the regime slope RS .

Part (a) Table 4.1

)(mBR 242.8 242.6

)(mhR 7.13 6.79

RS 0.00032 0.00034


Solution:

For a gravel channel, the characteristics RB , Rh and RS are given by Eqs. (4.62),

(4.64) and (4.59), viz

crR

v

QB

42.1 , (1)

7/37/1

0.7

crR

v

QDh , (2)

R

crR

gh

vS

2 . (3)

From Eq. (1) we obtain

smB

Qv

R

cr /135.07.211

300042.142.12

2

2

2

, (4)

which, after the substitution in Eq. (2) , yields

7/37/1

135.0

3000

0.7

DhR i.e. 7/142.10 DhR . (5)

Eq. (3) gives, with the aid of Eqs. (4) and (5),

7/1

2

42.1081.9

135.0

DSR

i.e. 7/141078.1 DSR . (6)

For a gravel regime channel, we have 1/)( crRR YY , and also 045.0crY .

Hence

045.065.1

1 )(

D

ShY RR

R and thus RRShD 47.13 . (7)

Solving three unknowns ( D , Rh , and RS ) from three equations (5), (6), (7), we

determine

mD 025.0 ( cm5.2 ), mhR 15.6 , and 4100.3 RS .

Problem 4.3

A straight laboratory flume has rigid (plexiglass) walls; the flume width is mB 2 ,

the bed is formed by a layer of sand having mmD 5.1 . Determine the regime


48

values 1Rh and

1RS of the regime channel 1R if smQ /1 3 and

smQs /100.5 35 .

Solution:

The sand is coarse. Hence we assume that the flow is fully rough ( 5.8sB ,

5.0 ) and that the sediment is transported as bed-load only. We will use, for the

sake of simplicity, Bagnold’s original bed-load formula (1.43); for the value of

we have

9.37101000

)1000/5.1(5.161863/1

12

33/1

2

3

Ds .

Substituting in the resistance equation (3.34), viz

3/1

22

2

gScB

Qh , (1)

the present values smQ /1 3 and mB 2 , we obtain

3/1

2

025.0

Sch . (2)

On the other hand, we have smBQq ss /105.22/100.5/ 255 , and thus

(see Eq. (1.27))

107.0)1000/5.1(5.16186

105.210002/32/1

52/1

2/32/1

2/1

D

q

s

s

. (3)

Modifying the dimensionless Bagnold’s formula (first equation in (1.43)) so as to

correspond to an undulated bed (see Section 3.2), and using Eq. (3), we determine

107.0)()( 22/1 crccs YYYB , (4)

where fc cc / . Considering that 25.45.05.8 sB and 042.0crY (see

Eq. (1.34)), we obtain from Eq. (4)

2

2

2

2

24 042.0103.6

Y

c

c

c

cY

f

f, (5)

which, with the aid of )/)(65.1/1( DhSY , yields

Regime channels and their computation 49 2

2

2

2

2

042.065.1

1001.0

D

hS

c

c

c

c

h

DS

f

f. (6)

Since ),,( ShDcfcf and ),,( ShDc c , the two unknowns

1Rh and 1RS can

be determined by solving simultaneously Eqs. (2) and (6). The following

computational procedure is adopted here to solve these equations:

Adopt 1h and 1S

For i = 1, 2, …, until satisfied, do:

1. Knowing D, ih and iS , determine fic and ic with the aid of the program

RFACTOR.

2. Knowing ic and iS , determine 1ih from Eq. (2).

3. Knowing fic , ic , iS and 1ih , determine 1iS from Eq. (6).

4. If ii hh 1 and ii SS 1 , then the problem is solved: iR hh 1

, iR SS 1

;

Otherwise, 1 ii , and repeat steps 1 to 4.

The results of calculations are shown below. (Note: In order to ensure convergence,

2S was determined using 1h (rather than 2h )).

Hence mhR 50.01 and 0025.0

1RS .

Problem 4.4

Determine RB , Rh and RS for a river whose bankfull flow rate is smQ /1500 3

and the average grain size is .7.0 mmD

Solution:

We compute the regime characteristics corresponding to smQ /1500 3 and

mmD 7.0 with the computer program BHS-STABLE, which yields

mBR 2.225 , mhR 86.5 , and 00013.0RS .

i )(mh S fc c

1 0.750 0.00100 19.8 10.8

2 0.598 0.00292 19.2 8.9

3 0.476 0.00239 18.7 9.2

4 0.497 0.00226 18.8 9.2

5 0.507 0.00262 18.8 9.0

6 0.490 0.00225 18.7 9.2

7 0.508 0.00240 18.8 9.1

8 0.500 0.00250


50

Problem 4.5

Consider the Peace River ( mmD 31.0 ). The measured bankfull characteristics

are smQ /9905 3 , mB 2.619 , mh 33.9 , and 000084.0S (Ref. [12d] in

Appendix D). Is this river in regime? Solution: We compute the regime characteristics corresponding to smQ /9905 3 and


mBR 6.576 , mhR 8.17 , and 000028.0RS . Comparing these regime values with the measured characteristics stated in the

question, we obtain

107.1/ RBB , 152.0/ Rhh , 10.3/ RSS . The last two ratios are noticeably different from unity. Hence the river is not in

regime.

Problem 4.6

Consider a straight region of a stream flowing in a cohesionless alluvium:

mmD 1 . It is assumed that the stream is in regime. The total volumetric sediment

transport rate (past the undulated bed) at the regime stage is

smQ Rs /100.3)( 33 . Determine the regime characteristics RB , Rh , RS , and

the value of the bankfull flow rate bfQ . (Assume that the flow is fully rough and

that the sediment is transported as bed-load only; take 7.0/ fc cc ).

Solution:

The purpose of this problem is to reveal the regime characteristics, the precision in

the value of the transport rate being of secondary importance. Hence we will use

the Bagnold’s bed-load formula (1.43), modified for the case of an undulated bed

(see Section 3.2):

)()( 22/1crccs YYYB .

Here

25.45.05.8 sB (fully rough flow),

hShS

D

hSY

s

606)1000/1(65.1

1

,

and (for the present 3.25)/( 3/123 Ds )

037.0]1[045.013.0 068.0015.0392.0 2

eeYcr .


Consequently,

)037.09.296(2.73 2/12/1 hSSh . (1) The total volumetric sediment transport rate is given by

BD

BqQ ssbs

2/1

2/32/1

. (2)

Combining Eqs. (1) and (2), using the subscript R to indicate the regime state, and

considering that smQ Rs /100.3)( 33 , we obtain

32/12/1 100.3)037.09.296(0093.0 RRRRR ShShB , (3)

which, taking into account that QuhB RRR , yields

1

2/1

2/1

)037.09.296(323.0 RR

R

RR ShS

huQ . (4)

The following computational procedure is adopted here to determine bfQQ from

Eq. (4): Adopt 1Q


1. Knowing iQ and D, determine the regime characteristics iRB )( , iRh )( and

iRS )( with the computer program BHS-STABLE.

2. Compute ))()/(()( iRiRiiR hBQu .

3. Using iRh )( and iRS )( determined in step 1 and iRu )( determined in step

2, compute 1iQ from Eq. (4).

4. If ii QQ 1 , then the problem is solved: ibf QQQ ;

Otherwise, 1 ii , and repeat steps 1 to 4. The calculations are summarized in the Table below.

Hence smQbf /74 3 .

i )/( 3 smQ )(mBR )(mhR RS )/( smuR

1 500 105.8 3.60 0.000329 1.31

2 141 51.9 2.34 0.000479 1.16

3 89 40.3 2.01 0.000530 1.10

4 78 37.3 1.93 0.000549 1.08

5 75 36.9 1.90 0.000550 1.07

6 74 36.5 1.89 0.000550 1.07

7 74


52

Problem 4.7

Eq. (4.7) gives the regime flow depth Rh in terms of the regime values of the

Froude number. Show that in terms of the regime values of avu and , Rh is

given as

RavB

crR

u

vQh

)(

2/12/1

and

2/12/1

2/1

)( crRRB

Rvc

Qh

,

respectively.

Solution:

Consider Eq. (4.31), viz

3/13/122 )]([

g

QvSch cr

RRBR , (1)

and the resistance equation (4.30), which can be written as

R

RavRR

gh

uSc

22 )(

. (2)

Combining Eqs. (1) and (2), we obtain

3/13/12

2 )(

g

Qv

gh

uh cr

R

RavBR ,

i.e.

RavB

crR

u

vQh

)(

2/1*

2/1

, (3)

which is the first relation sought.

Substituting RRRav vcu *)( in Eq. (3), we obtain

2/1**

*2/1

*

2/1*

2/1111

crR

cr

RBRR

cr

BR

vv

v

c

Q

vc

vQh

,

i.e.

2/12/1

2/1

)( crRRB

Rvc

Qh

, (4)

which is the second relation sought.


Problem 4.8

Consider a constN curve on the );( *Fr -plane corresponding to a given value

of , and let P be the point of this curve where c (which also varies along this

curve when * varies) has its minimum value. What is the inclination of the

tangent (implying */ Fr ) at the point P?

Solution:

Using the relation (3.43), we can write

)()( 2/3*

32/3*

2

cconstcN

Fr ,

which yields

2/1

*32/3

**

2

* 2

33

c

d

dccconst

d

dFr.

For mincc , we have 0/ * ddc , and therefore

2/1

*3min

* 2

3

cconst

d

dFr ,

which indicates that the inclination of the tangent (in the log-log );( *Fr -plane) at

the point P is 1/2.


54

Formation of regime channels; meandering and braiding 55

CHAPTER 5

Formation of regime channels; meandering and braiding

Problem 5.1

Consider the inclined (by 1/3) part of the boundary 1,2L between the one-row and

multiple-row bars in Figure 5.2a. The ordinate hB of this part of the boundary

varies as a function of Dh . Show that this ordinate can be expressed also as a

function of (only) the dimensionless combination

5

2

gSD

QΜ ,

and determine this function.

(Treat the flow as two-dimensional; use 6/1)2/(66.7 Dhc f (Eq. (1.13)).

Solution:

The expression of the inclined (by 1/3) part of the boundary 1,2L between the one-

row and multiple-row bars in Figure 5.2a is (see Eq. (2.42))

3/1

25

D

h

h

B. (1)

Consider the friction equation (1.16), which can be expressed as

gShBhcQ f . (2)

Substituting here the value of fc given by Eq. (1.13), viz

6/16/1

82.62

66.7

D

h

D

hc f , (3)

we obtain

gShD

hhBQ

6/22222 )82.6(

, (4)

and thus

56 Fluvial processes: 2nd edition − Solutions manual

3/162

2

5

2

)82.6(

D

h

h

B

gSD

Q )( M . (5)

Eliminating )/( Dh between Eqs. (1) and (5), we determine

18

16

2

25

)82.6(

h

BΜ i.e. 18/11.14 Μ

h

B, (6)

which is the function sought.

Problem 5.2

Determine whether the rivers below are meandering or braiding (the characteristics

given, which are derived from Ref. [7e] in Appendix E, correspond to bankfull flow

rate):

a) Mississippi River:

,/0.42450 3 smQ ,5.0 mmD ,0.1382 mB ,13.20 mh ;000047.0S

b) Yamuna River:

,/0.5.2122 3 smQ ,15.0 mmD ,9.205 mB ,40.6 mh ;000328.0S

c) Savannah River:

,/0.849 3 smQ ,8.0 mmD ,7.106 mB ,18.5 mh ;00011.0S

d) Brahmaputra River:

,/5.24762 3 smQ ,3.0 mmD ,0.9455 mB ,52.1 mh 000252.0S .

Solution:

Consider Figure 5.2a. The upper boundary of meandering is the line M,BL

)( 1,2L , which conveys the following relations (see Eqs. (2.42) and (2.43)): When :200)/( Dh

The stream is meandering if 3/1)/(25)/( DhhB ;

Otherwise, it is braiding. When :200)/( Dh

The stream is meandering if 150)/( hB ;

Otherwise, it is braiding.

The lower boundary of meandering, viz the line ML , is ignored throughout, for the

hB / -values of large natural streams are invariably larger than their counterparts

implied by ML .

The above stated conditions lead to the following conclusions.


a) Mississippi River:

20040260 D

h and 1506.68

h

B; the stream is meandering.

b) Yamuna River:

20042700 D

h and 1502.32

h


c) Savannah River:

2006475 D

h and 1506.20

h


d) Brahmaputra River:

2005067 D

h and 1506220

h

B; the stream is braiding.

Note:

As can be inferred from Eq. (7.108), viz SS /0 , the prominence of meandering

is determined by the ratio of the initial or “valley” slope 0S to the river slope

(along the stream length) S . No prediction with regard to the prominence of

meandering can be made if 0S is not specified. And if 0SSR , the stream remains

“straight” ( 1 ).

Problem 5.3

The initial alluvial channel which has a flat bed and which can be treated as two-

dimensional, is determined by mmD 1.1 , 5102/ QQs , 100/ 00 hB .

Determine the range of S-values for which the flow developing in this initial

channel will certainly tend to (and perhaps will) exhibit braiding – not meandering.

Use Bagnold’s formula; take 3.0 .

Solution:

Consider Bagnold’s formula (1.45), where Sh 0 (see Eq. (1.7)):

cr

ss

cr

ssb Y

h

DShuShuq

)( 0 .

From this expression it follows that

cr

s

sbsb Yh

DS

q

q

Q

Q

,


which yields

cr

sbs Yh

D

Q

QS

1. (1)

Since 100/ 00 hB , the stream will tend to exhibit braiding provided that (see

Figure 5.2a and Eq. (2.42))

3/1

0

0

0 25

D

h

h

B i.e.

3

0

03

0

25

B

h

h

D. (2)

Using Eq. (2) in Eq. (1), where S and h are to be viewed as 0S and 0h (initial

channel), we obtain

cr

sbs YB

h

Q

QS

3

0

030 25

1

. (3)

For 3.0 , 5102/ QQsb , 100/1/ 00 Bh and 038.0crY (which

corresponds to 8.27 ), the inequality (3) gives

0011.0038.0100

125

3.0

10265.1

33

5

0

S ,

which indicates that the stream will tend to exhibit braiding as long as 0011.00 S .

Problem 5.4

The regime development of a gravel channel takes place by meandering:

,/230 3 smQ .20mmD Determine the interrelation between its B and S near

the end of the regime development.

Solution:

Near the end of the regime development of a gravel channel, we have 0~ ss qQ

and thus 1 . Hence

165.1

1*

crcr DY

hS

Y

Y ,

and thus

S

Y

D

h cr65.1 . (1)


Since the regime development is by meandering, the channel is to be represented in

Figure 5.2a by a point below the boundary-line 1,2L , at any stage of its

development.

Using the computer program BHS-STABLE we reveal that the regime value of the

flow depth of the present stream is mhR 96.1 , and therefore 98/ DhR . Note

from Figure 5.2a that the 1/3-inclined part of the boundary-line 1,2L is valid for all

Dh / which are smaller than 200 . Hence it is valid for the present case of

98/ DhR ( 200 ). Clearly the stream under study may not be in regime. Yet all

what has been said above with reference to the regime flow depth Rh is equally

valid for the developing flow depth h , for h is smaller than Rh throughout the

development interval RTtT 0ˆ (see Figure 4.5).

Considering the aforementioned, and bearing in mind that the 1/3-inclined part of

the line 1,2L is given by Eq. (2.42), we can write

3/1

25

D

h

h

B, (2)

which, in conjunction with Eq. (1), gives

3/1

65.125

S

Y

h

B cr . (3)

For gravel channels, the resistance equation is of the following form

6/1

6/106/1

82.62

)66.7(D

gSBhgSh

D

hBhQ

. (4)

Eliminating h between Eqs. (3) and (4) we determine

3/1

3/13/1

10/310/610/310/6

10/110/6

)65.1(1

82.625

S

Y

SBg

DQB cr ,

i.e.

3/110/110/630/1910/16 71.4 crYDQSB . (5)

For smQ /230 3 , mmD 20 and 045.0crY , Eq. (5) gives

6.2930/1910/16 SB , which is the interrelation sought.


Problem 5.5

If Q, D and c are treated as constants, then various constant values of S imply, in

the )/;/( DhhB -plane, the straight lines inclined by –2.5 (see the ScFr 2

const lines in Figure 5.8). What would be the inclination of these lines if we had

a (flat) gravel-bed and only Q and D were treated as constants?

Solution:

Consider the resistance equation (3.34), viz

gShBhcQ . (1)

Multiplying and dividing the right-hand side of Eq. (1) by 2/5hD and solving for

hB / , we obtain

2/5

2/5

1

D

h

ScDg

Q

h

B, (2)

which confirms that for given Q, D, c, various constant values of S imply (in the

log-log )/;/( DhhB -plane) the straight lines inclined by –2.5.

Consider now the c-expression for a (flat) gravel-bed channel (see Eqs. (1.13) and

(1.4)):

6/16/1

82.62

66.7

D

h

D

hc )( const . (3)

Substituting Eq. (3) in Eq. (2), we determine

6/16

2/5

1

82.6

D

h

SDg

Q

h

B, (4)

which indicates that, if only Q and D are constants, then the inclination of the lines

corresponding to various constant values of S is 6/16 ( 667.2 ).

Problem 5.6

Consider the development of a regime channel in the interval RTt 0 : the

(constant) flow rate is smQ /500 3 ; the average grain size of the cohesionless

alluvium is mmD 9.0 .

a) The initial channel (at 0t ), which has a flat bed, has mB 700 and

500/10 S . Determine its flow depth 0h , and plot the corresponding point

0P on the )/;/( DhhB -plane in Figure 5.2a.


b) The adjusted initial channel (at 0Tt ) has mB 105ˆ0 . Determine its flow

depth 0h , and plot the corresponding point 0P on the )/;/( DhhB -plane in

Figure 5.2a. (Take 00ˆ SS ).

c) Determine the regime characteristics RB and Rh , and plot the corresponding

point RP on the )/;/( DhhB -plane in Figure 5.2a.

d) What is the inclination of the straight line connecting the points 0P and 0P ?

What is the inclination of the straight line connecting the points 0P and RP ?

Solution:

a) Consider the resistance equation (3.34), which, for the initial channel, can be

expressed as follows:

00000 hgSchBQ . (1)

Since the bed is flat, 0c can be evaluated, on the basis of Eqs. (1.13) and (1.4), as

6/10

02

66.7

D

hc . (2)

Using Eq. (2) in Eq. (1) and solving 0h , we obtain

10/3

020

2

3/12

082.6

SgB

DQh . (3)

For smQ /500 3 , mmD 9.0 , mB 700 and 500/10 S , Eq. (3) gives

mh 66.1)500/1(7081.982.6

)1000/9.0(50010/3

22

3/12

0

,

and we have )1844/;2.42/( 0000 DhhBP .

b) In this case, the resistance equation (3.34) implies

00000ˆˆˆˆˆ hSgchBQ , (4)

which, for the present smQ /500 3 , mB 105ˆ0 and 500/10 S , gives

3/2

0

3/1

20

2

2

0 )ˆ(49.10)500/1(81.9ˆ105

500ˆ

c

ch . (5)

Since ),,ˆ(ˆ 00ˆ0 0DShc c , Eq. (5) is transcendental, and the value of 0h can be

determined with the aid of the following procedure.


Adopt 10 )ˆ(h


1. Knowing D, ih )ˆ( 0 and 0S , determine ic )ˆ( 0 with the aid of the computer

program RFACTOR.

2. Knowing ic )ˆ( 0 from step 1, determine 10 )ˆ( ih from Eq. (5).

3. If ii hh )ˆ()ˆ( 010 , then the problem is solved: 100 )ˆ(ˆ ihh .

Otherwise, 1 ii , and repeat steps 1 to 3.

The results of computations carried out with the aid of this procedure are

summarized below.

Hence mh 52.1ˆ0 , and we have )1689/ˆ;1.69ˆ/ˆ(ˆ

0000 DhhBP .

c) We determine the regime characteristics RB , Rh (and RS ), with the aid of the

computer program BHS-STABLE, which gives

mBR 1.111 , mhR 64.3 (and 00028.0RS ).

Consequently, we have )4044/;5.30/( DhhBP RRRR .

d) The values of hB / and Dh / of the initial channel, the adjusted initial channel

and the regime channel determined in parts (a), (b) and (c) are summarized below.

hB / Dh /

Initial channel ( 0P ) 42.2 1844

Adjusted initial channel ( 0P ) 69.1 1689

Regime channel ( RP ) 30.5 4044

If 00 PP I and

RPP 0ˆI are the inclinations of the straight lines connecting the points

00ˆ; PP and RPP ;ˆ

0 , respectively, then

i )()ˆ( 0 mh i ic )ˆ( 0

1 2.00 21.8

2 1.34 15.7

3 1.67 19.3

4 1.46 17.1

5 1.58 18.4

6 1.51 17.6

7 1.55 18.1

8 1.52 17.7


1/662.5)1844/1689log(

)2.42/1.69log(

)/log()/ˆlog(

)/log()ˆ/ˆlog(

00

0000ˆ00

DhDh

hBhBPP

I ,

1/194.0)1689/4044log(

)1.69/5.30log(

)/ˆlog()/log(

)ˆ/ˆlog()/log(

0

00ˆ0

DhDh

hBhB

R

RRPP R

I .

Problem 5.7

The adjusted initial alluvial stream has smQ /140 3 , 001.0ˆ0 S , ,5.0 mmD

mh 30.1ˆ0 . Will such a stream be meandering? (Hint: determine 0c first).

Solution:

Knowing mmD 5.0 , 001.0ˆ0 S and mh 30.1ˆ

0 , we determine first 0c with the

computer program RFACTOR:

9.15ˆ0 c . Then we evaluate 0B from the resistance equation:

mhSgch

QB 60

30.1001.081.99.1530.1

140

ˆˆˆˆˆ

0000

0

.

Hence

2.4630.1

60

ˆ

ˆ

0

0 h

B and 2600

1000/5.0

30.1ˆ0

D

h.

Since 200/ˆ0 Dh and 150ˆ/ˆ

00 hB , the corresponding point 0P (on the

)/;/( DhhB -plane) is between the horizontal parts of the boundary-lines L and

ML . Hence the stream will be meandering (refer to the solution of Problem 4.2).

Problem 5.8

Prove that the paths l in Figure 5.5 must merge (tangentially) to the respective

RFr)( -curves when the points gR PP and sR PP are approached (the paths l

cannot intersect a RFr)( -curve at the points mentioned with a finite angle).

Solution:

Consider the approach of the “moving point” m to the regime point RP at the later

stages of development (Figure 1).


Figure 1

Let 1 and 2 be the horizontal and vertical distances of m to RP , 3 being the

distance of m to “its” RFr)( -curve. The coordinates and Fr are dependent on h

and S (but not on B), and so are 1 and 2 . In contrast to this, 3 varies mainly

with B (for B appears in the expression of crBDvQN / which specifies a Fr -

curve).

Since the regime development of B is much faster than that of h and S (Figure 4.5),

the moving point m must tend to approach the point RP so as to have (in the later

stages), 13 ( and/or )2 . But this means that the trajectory of m must merge

(tangentially) to the RFr)( -curve when the point RP is approached.

Problem 5.9

How many times is the magnitude of the relative time-variation of S, viz

,/)/( SS larger than that of ?/)/( hh How many times is it larger than 22 /)/( cc ? Treat Q and B as constants; assume that the bed is flat.

Solution:

Consider the resistance equation (3.34), i.e.

gShchBQ 2222 , (1) where (since the bed is assumed to be flat)

3/1

22

266.7

D

hc . (2)

Hence from Eq. (1) it follows that


ShgB

DQ 3/10

2

3/12

6.46

. (3)

Differentiating this expression with respect to , and taking into account that the

left-hand side is to be treated as a constant, we obtain

03

10 3/103/7

ShS

hh , (4)

i.e.

011

3

10

S

S

h

h, (5)

which yields

3

10

|/)/(|

|/)/(|

hh

SS. (6)

Differentiating Eq. (2), viz

3/13/1

3/1

23/122 )(

)2(

66.7

266.7 hconsth

DD

hc

, (7)

we determine

h

h

c

c 3

11 2

2, (8)

which, in conjunction with Eq. (6), gives

10|/)/(|

|/)/(|22

cc

SS. (9)


Geometry and mechanics of meandering streams

67

CHAPTER 6


Problem 6.1

The ratio of the sinuosities of two sine-generated meandering channels (channels 1

and 2) is 202.2/ 21 ; the ratio of their deflection angles is 0.2)/()( 2010 .

What are the values of 10 )( and 20 )( ?

Note: The following polynomial approximation is available to compute )( 00 J :

60

40

2000 )3/(3163866.0)3/(2656208.1)3/(2499997.21)( J

120

100

80 )3/(0002100.0)3/(0039444.0)3/(0444479.0 .

Solution:

Since )(/1 00 J (Eq. (6.10)), we have

202.2))((

))((

100

200

2

1

J

J, (1)

which, for the present 10 )( ( 20 )(0.2 ), gives

))(0.2(202.2))(( 200200 JJ . (2) Using the polynomial approximation above and denoting 3/)( 20 by x , we obtain

from Eq. (2)

642 271.44325.43568.17202.1 xxx

0894.1890.8011.25 12108 xxx . (3)

Solving Eq. (3), we determine 291.0x , i.e. rad873.0)( 20 . Hence

50)( 20 and 100)( 10 .

Problem 6.2

For which values of 0 do sine-generated meandering streams have aRB / (i.e.

RB / at the apex-section) equal to 0.37?


68

Solution:

We have (see Eq. (6.13))

37.0)( 000 JR

B

a

. (1)

Using the polynomial approximation in Problem 6.1 and denoting 3/0 by x , we

obtain from Eq. (1)

9753 133.0949.0797.3750.63 xxxxx

037.0001.0012.0 1311 xx , (2)

which, for ]803.0;0[ radx , i.e. for ]138;0[0 , yields

radx 128.0 and radx 693.0 . Hence, we have 37.0/ aRB for 220 and 1190 .

Problem 6.3

Write a general computer program to draw the centreline of a sine-generated

meandering stream from iO to 1iO (i.e. between two consecutive crossovers).

The deflection angle 0 and the flow width B are given. Take BM 2 ; use

);( yx cartesian coordinates with the origin at the crossover iO .

Solution:

The following procedure is suggested for the determination of the cartesian

coordinates of the centreline-points (see Figure 2).

Figure 2


69

i) Initialize variables:

Input the values of 0 and B ;

Compute )( 00 J (see Problem 6.1);

Compute )(/22 00 JBBL ;

Set 0/ Ll ; 0x ; 0y ;

Set 001.0)/( Ll .

ii) Compute the x- and y- coordinates of the centreline-points:

10 Compute ))/(2cos(0 Ll

Compute cos)/()/( LlLx and sin)/()/( LlLy ;

Compute )/( LxLxx and )/( LyLyy ;

Compute )/(// LlLlLl ;

If 2/1/ Ll then

go to 10

else

Stop

Problem 6.4

Use the computer program of Problem 6.3 to draw the centreline of a sine-

generated meandering stream having 1100 . The flow width is mB 40.0 .

Solution:

Figure 3


70

Problem 6.5

A river is “endeavouring” to achieve its regime state by meandering. The valley

slope is 500/10 S , the regime slope being 1350/1RS . The centreline of the

river closely follows a sine-generated curve.

a) At a certain stage of its regime development, the slope of the river is

670/1S . What is the value of the deflection angle 0 at that stage?

b) What is the maximum 0 this river can possess?

Solution:

a) First we determine the sinuosity of the stream at the stage ( 670/1S ):

34.1670/1

500/10 S

S . (1)

Since )(/1 00 J (Eq. (6.10)), Eq. (1) implies

0746.0)( 00 J . (2) Approximating )( 00 J by the polynomial in Problem 6.1, and denoting 3/0 by

x , we obtain from Eq. (2)

8642 044.0316.0266.1250.2254.0 xxxx

00002.0004.0 1210 xx , (3)

which yields radx 348.0 . Hence 600 .

b) The maximum 0 occurs at the regime stage, when

7.21350/1

500/10 RS

S . (4)

But )(/1 00 J , and therefore Eq. (4) implies

0370.0)( 00 J , (5) i.e.

8642 044.0316.0266.1250.2630.0 xxxx

00002.0004.0 1210 xx (6)

(where 3/0x ). Solving Eq. (6), we determine radx 583.0 . Hence 1000 .

Meandering-related computations

71

CHAPTER 7


Problem 7.1

Let b be the “small” distance between two vertically averaged streamlines s at a

point P of a meandering flow (in plan view): the deviation angle at P is , the

variation of the flow depth h is negligible. Prove that

sns

b

b

1 and

sns

U

U

1,

where s and sn are the natural coordinates.

Solution:

On the basis of Figure 4, where l and s are the coordinate lines and streamlines,

respectively, we can write

sbn

sbbbs

)( . (1)

In the limit, Eq. (1) gives

snb

s

b

, (2)

which is the first relation sought.

For the continuity of fluid motion, we have

0)(

s

bU

s

Ub

s

bU, (3)

which, with the aid of Eq. (2), gives

0)(

snbU

s

Ub

s

bU ,

i.e.


72

Figure 4

sns

U

U

1,

which is the second relation sought.

Problem 7.2

Prove that at a point P of the flow plan, the curvature r of the coordinate line l

and the curvature sr of the (vertically averaged) streamline s are interrelated as

follows

cos11

rsrs

.

Solution:

Let s be the deflection angle of the streamline s passing through the point P ,

being the deflection angle of the coordinate line passing through the same point

(Figure 5). Clearly

s and thus s . But (see Figure 5)

r

l while

ss

r

s ,

where the negative signs are because the increment of l and s causes the

decrement of and s ( r and sr are treated as positive). Hence

r

l

r

s

s

.


73

Figure 5

Dividing this relation by s , considering that cos/ sl , and approaching the

limit (replacing ’s by ’s), we obtain

cos11

rsrs

,

which is the relation sought.

Problem 7.3

Prove (starting from Eqs. (7.1)-(7.3)) that the equations of motion and continuity of

a vertically-averaged flow in a curved channel can be expressed in natural

coordinates );( srs as

2

22 )(

M

sc

UghJ

s

hU

sr

s

gJr

U

2

0)(

s

hU,

where sJ and sr

J are the free surface slopes along s and sr .

Solution:

We replace in Eqs. (7.1)-(7.3) u by U , )( nR by sr , l by s (and thus l

clRn )/1( by s ). Moreover, we use 0v and


74

)( hzs

J bs

; )( hz

rJ b

srs

.

This gives

2

2

2

22

)()(

Μ

s

Μ

bc

UghJ

c

Uhz

sgh

s

hU

(1)

srbss

ghJhzr

ghr

hU

)(

2

(2)

0)(

s

hU, (3)

which are the expressions stated in the question.

Problem 7.4

Consider an infinitely long circular open channel (as in Figure 5.23b in Yalin

1992): the flow is uniform and stationary ( constR , 0/ clh , 0/ clu ).

a) Determine the reduced forms of the equations of motion and continuity (7.4)-

(7.6) which correspond to this special case.

b) Determine the expressions of u and h as functions of Rn / .

Solution:

a) Substituting 0/ clh , 0/ clu and 0 in Eqs. (7.4)-(7.6), viz

2

22

Μcc

b

hc

u

l

h

l

z

nR

Rg

nu

(1)

2

222

Μ

b

c hc

u

n

h

n

zg

nR

u

lnR

Ru

(2)

0

nR

hu

n

hu

l

hu

nR

R

c

, (3)

we obtain

02

2

Μc

b

hc

u

l

z

nR

Rg (4)

n

h

n

zg

nR

u b2

(5)

00 . (6)


75

b) For the sake of simplicity, let us assume that the channel bed is flat:

0 nzb ; ccb Slz . Using these relations in Eqs. (4) and (5), we

determine

Rn

Sc

gh

u cΜ

/1

22

(7)

R

n

n

h

gR

u1

2

. (8)

Note that if R (straight channel) then Eq. (7) reduces to the familiar wide

channel resistance formula where h remains constant along the flow width (see Eq.

(8)).

Problem 7.5

Prove on the basis of the equations of motion and continuity that if Mc is constant,

or if it varies only as a function of sKh / , then, in the case of a flat bed, these

equations can yield only an ingoing flow – no matter how large the value of 0

might be (see Figure 7.1).

Solution:

Consider the longitudinal equation of motion (7.4) which can be expressed as

2

11

Mcl

h

FrFr

S

rh

, (1)

where ghuFr /2 is the local Froude number. We focus the attention on the flow

sections that are in the neighbourhood of the apex ia . Since the superelevation (i.e.

the most intense r -variation of h ) occurs in the flow sections mentioned, we can

identify lh with zero:

2

1

McFr

S

rh

. (2)

i) If 0 is “small”, then the flow is ingoing. In this case (as can be inferred from

Figure 6.16a) 0 (and thus 0/ r ) in the apex region and Eq. (2) reduces

into

ScFr M2 ,

which is the resistance equation of the individual stream filaments (i.e. flow can be

viewed as a basically straight flow which is somewhat deformed by its wavy

boundaries).


76

ii) If 0 is “large” (Figure 6.16b), then the flow is outgoing. In this case the largest

positive occur at the apex region ia . Figure 6 shows the nature of the

r -variation

Figure 6

of and r / . Note that r / changes its sign along B . Now at the apex-

section S decreases with r , h increases with r , and u remains approximately

constant along r (with the exception of the neighbourhood of the banks). Hence 2// ugShFrS may monotonously vary with r along B . However, FrS /

certainly maintains its (negative) sign along B (Figure 7). But this means that 2/1 Mc (and thus Mc ) must necessarily vary with r along B . And its variation

should, in fact, be of the type shown in Figure 7.

Figure 7

Problem 7.6

Adopt the following notation max)/( ccy ; max)/()( cacay . a) How does the ratio ay vary with 0 ?

b) Explain why the curve max)( c in Figure 6.19 tends to become

indistinguishable from the straight line 0max)( c , when 00 .


77

Solution:

a) The ratio y varies with c and 0 as shown by the curve-family in Figure 6.18,

where each 0

C -curve corresponds to a given 0 . This curve-family can be

represented by (see Eq. (6.46))

)](2sin[ 0ccy , (1)

in which 0c is a function of 0 (given by Eq. (6.47)). If 4/1c (apex), then Eq.

(6.46) reduces into

)]4/1(2sin[ 0cay . (2)

The graph of Eq. (2), with 0c calculated from Eq. (6.47), is shown in Figure 8.

Figure 8

As follows from this figure, the graph of ay versus 0 takes the form of a S-like

curve.

c) If 0 is “small”, then the meandering flow takes place in a basically straight

channel with the small-amplitude wavy banks (infinitesimal / ). The central

part of the flow does not “feel” the infinitesimal waviness of the banks, and the

streamline cl is practically straight. But this means that max)( c (at iO ) is the same

as 0 (Figure 9), i.e. that

000

)(lim0

.


78

Figure 9

Problem 7.7

Consider the stream determined by smQ /1500 3 and mmD 7.0 ( 65.1/ s , 33 /10 mkg , sm /10 26 ). The regime development of S is totally by

meandering (no degradation). Use for the total friction factor 0.13c at all stages.

a) Determine the regime values RB , Rh , RS and R)( 0 , assuming that the slope

0S of the initial channel is four times larger than RS )4( 0 RSS .

b) Consider now the development stage when S is twice larger than RS

).2( RSS

1. What are the values of the angles 0 and ac )( at this stage?

2. What is the transport rate avsq )( at this stage? (Take RBB 95.0 ).

3. What is the channel expansion velocity aW at this stage? (Take

5.6WW ).

Solution:

a) We compute the regime characteristics corresponding to smQ /1500 3 and


mBR 225 , mhR 86.5 , and 00013.0RS . The relation (7.108) gives for the sinuosity R at the regime state

))((

14

00

0

RRR

JS

S

, (1)

i.e.

025.0))(( 00 RJ . (2) Using the polynomial in Problem 6.1, and denoting 3/)( 0 R by Rx , we obtain

from Eq. (2)

8642 044.0316.0266.1250.2750.0 RRRR xxxx

00002.0004.0 1210 RR xx ,


79

which yields radxR 652.0 . Hence 112)( 0 R .

b) 1. We have

22

40 R

R

S

S

S

S , (3)

which, in conjunction with )(/1 00 J (Eq. (6.10)), yields

05.0)( 00 J . (4) Using the polynomial in Problem 6.1, and denoting 3/0 by x , we obtain from

Eq. (4)

8642 044.0316.0266.1250.2500.0 xxxx

00002.0004.0 1210 xx . Solving this equation, we determine radx 507.0 , and thus 870 .

The value of ac )( can be estimated from Eq. (6.46), with 4/1c and 0c and

max)( c determined from Eqs. (6.47) and (6.48), respectively.

For rad521.10 ( 87 ), Eqs. (6.47) and (6.48) yield

083.025.011.00198.00033.0 20

30

40 c

and

084.0115.0)(3.4

0120.00max

ec .

Substituting these values, together with 4/1c , into Eq. (6.46), one obtains

073.0)]084.025.0(2sin[084.0)( ac .

2. First we determine the flow depth h from the resistance equation (3.34), i.e.

3/1

22

2

gScB

Qh .

Using in this equation smQ /1500 3 , mBB R 21422595.095.0 , 0.13c

and 00026.000013.022 RSS , we obtain

mh 85.4 . Then we determine v , Re and :


80

smghSv /111.000026.085.481.9 ,

4.15510

)1000/7.02(111.0Re

6

skv

(fully rough turbulent flow),

7.17101000

)1000/7.0(5.161863/1

12

33/1

2

3

Ds .

Finally we evaluate avsq )( from Eq. (3.49), which can be expressed as

)()( 2crcfcavs YYuDq .

Here (see Eqs. (1.34), (1.24) and (1.46))

032.0]1[045.013.0 068.0015.0392.0 2

eeYcr ,

09.1)1000/7.0(5.16186

111.01000 22

D

vY

s

,

411.0

2368.0ln

11

2

1ln

2

11

D

h

B

DB

s

s


9.12 (see Eq. (1.63)).

Moreover

smk

hvu

sf /93.2

1000/7.02

85.411ln

4.0

111.011ln

,

4.26111.0

93.2

v

uc

ff ,

5.04.26

0.13

fc

c

c .

Consequently,

smq avs /10)032.009.15.0(93.25.0)1000/7.0(411.0)( 242 .

3. The value of aW is given by Eq. (7.109), viz


81

WWM

avsaca J

qW )(

)()( 00

.

Substituting here the computed 084.0)( ac , smq avs /10)( 24 , the given

5.6WW , and taking into account that mBBM 12846ˆ6 0 , while

5.0)( 00 J (see Eq. (4)), we determine yearmsmWa /67.0/1013.2 8 .

Problem 7.8

Prove that the expansion velocity of the angle 0 , viz dtd /0 , is related to the

apex-expansion velocity aW as follows

a

a

R

W

dt

d 0

0

.

Solution:

Consider Eqs. (6.9) and Eq. (7.96), viz

LRa

021 and

2

0 L

dt

dWa ,

respectively. Multiplying these equations with each other, and thus eliminating

2/L , we obtain the relation sought.


82

83

Software Instructions - Fortran

The source codes (files Bhsstabl.for and Rfactor.for) of the programs BHS-

STABLE and RFACTOR, as well as their compiled counterparts (application files

bhsstabl.exe and rfactor.exe) and examples of their results’ files can be downloaded

from the book page on the CRC Press website (download zip files entitled:

BHSSTABL_Fortran_zip and RFACTOR_Fortran.zip): https://www.crcpress.com/9781138001381

or https://www.crcpress.com/Fluvial-Processes-2nd-Edition/Silva-Yalin/p/book/9781138001381

To run BHS-STABLE:

1. In your computer make a new directory entitled ‘Fluvial Processes’ and save

all files extracted from the corresponding zip file into this new directory.

2. In the new directory ‘Fluvial Processes’, click twice on the application file

Bhsstabl.exe. This will direct you to a DOS Window. Answer the questions

asked. e.g.

Name results file: RESULTS1

Flow rate (m3/s)= 1669.7

Grain size (mm)= 0.18

Note: The name of the results file should not exceed eight characters.

3. When the word ‘Finished’ appears on the top left corner of the DOS Window,

that means that the program has finished the calculations.

4. Close the DOS Window.

5. Find the ASCII results file ‘RESULTS1’ in the directory ‘Fluvial Processes’.

Open this file with e.g. NOTEPAD or WORD, and find the values of BR, hR and

SR, as well as some other pertinent information. (Refer to ‘Example 1’ in Table

4.1, p. 121 of the monograph, where the computed regime characteristics for the

example above are shown in detail).

To run RFACTOR:

Follow the procedure above, but replace Bhsstabl by Rfactor.

Note that Rfactor asks four questions. Answer the questions asked. e.g.

Name results file: RESULTS2

Grain size (mm)= 0.33

Flow depth (m)= 1.25

Slope= 0.0002

(Refer to the ‘Numerical example’ in pages 80 and 82-83 of the book, where the

computed results for this example are shown in detail).

https://www.crcpress.com/Fluvial-Processes-2nd-Edition/Silva-Yalin/p/book/9781138001381

84

Software Instructions - MATLAB

The source codes (suite of files ending with .m) of the programs BHS-STABLE

and RFACTOR can be downloaded from the book page on the CRC Press website

(download zip files entitled: BHSSTABL_MATLAB_zip and

RFACTOR_MATLAB.zip).

Book page: https://www.crcpress.com/9781138001381

or https://www.crcpress.com/Fluvial-Processes-2nd-Edition/Silva-Yalin/p/book/9781138001381

To run BHS-STABLE:

1. In your computer make a new directory entitled ‘BHS-Stable’ and save all files

extracted from the corresponding zip file into this new directory.

2. Open MATLAB and find the icon ‘Browse for Folder’.

3. Select the folder ‘BHS_Stable’.

4. In the folder ‘BHS-Stable’ find the main program ‘BHS-STABLE’ and double

click to open it.

5. Enter the values of Q (in m3/s) and D (in mm) in the data input section.

6. Select a name for the results file and enter it in the ‘output’ variable (just below

the input data section), noticing that the results file can only be a ‘~.txt’ file (i.e.

its name must end with .txt).

7. Click the ‘Run’ button in the editor section of the MATLAB toolbar.

8. Find the results file ‘~.txt’ in the folder ‘BHS-Stable’.

9. Open the results file with e.g. NOTEPAD or WORD, and find the values of BR,

hR and SR, as well as some other pertinent information as in Table 4.1, p. 121 of

the monograph.

To run RFACTOR:

1. In your computer make a new directory entitled ‘RFactor’ and save all files

extracted from the corresponding zip file into this new directory.

2. Open MATLAB and find the icon ‘Browse for Folder’.

3. Select the folder ‘RFactor’.

4. In the folder ‘RFactor’ find the main program ‘RFACTOR’ and double click to

open it.

5. Enter the values of D (in mm), h (in m) and S in the data input section.

6. Select a name for the results file and enter it in the ‘output’ variable (just below

the input data section), noticing that the results file can only be a ‘~.txt’ file (i.e.

its name must end with .txt).

7. Click the ‘Run’ button in the editor section of the MATLAB toolbar.

8. Find the results file ‘~.txt’ in the folder (BHS-Stable).

9. Open the results file with e.g. NOTEPAD or WORD.

https://www.crcpress.com/Fluvial-Processes-2nd-Edition/Silva-Yalin/p/book/9781138001381

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