(Part ii)- open channels

Open Channel Hydraulics

1

Hydraulics

Dr. Mohsin Siddique

Assistant Professor

Steady Flow in Open Channels

� Specific Energy and Critical Depth

� Surface Profiles and Backwater Curves in Channels of Uniform sections

� Hydraulics jump and its practical applications.

� Flow over Humps and through Constrictions

� Broad Crested Weirs and Venturi Flumes

2

Specific Energy and Critical Depth

Basic Definitions

� Head

� Energy per unit weight

� Energy Line

� Line joining the total head at different positions

� Hydraulics Grade Line

� Line joining the pressure head at different positions

3


Basic Definitions

� Open Channel Flow

1Z

g

V

2

2

1

Datum

So

1y

2Z

g

V

2

2

2

2y

HGL

EL

Water Level

lhg

VyZ

g

VyZ +++=++

22

2

222

2

111

4

� Slopes in Open Channel Flow

� So= Slope of Channel Bed = (Z1-Z2)/(Δx)= -ΔZ/Δx

� Sw= Slope of Water Surface= [(Z1+y1)-(Z2+y2)]/Δx

� S= Slope of Energy Line= [(Z1+y1+V12/2g)-(Z2+y2+V2

2/2g)]/Δx

= hl/ΔL

1Z

g

V

2

2

1

Datum

So

1y

2Z

g

V

2

2

2

2y

HGL

EL

Water

Level

Sw

S

∆L

∆x


Basic Definitions

5


Basic Definitions

� Slopes in Open Channel Flow

1Z

g

V

2

2

1

Datum

So

1y

2Z

g

V

2

2

2

2y

HGL

EL

Water

Level

Sw

S

∆L

∆xFor Uniform Flow

y1=y2 and V12/2g=V2

2/2gHence the line indicating the bed of the channel, water surface profile and energy line are parallel to each other.For θ being very small (say less than 5 degree) i.e ∆x=∆L

So=Sw=S6


Basic Definitions

� Froude’s Number (FN)

� It is the ratio of inertial forces to gravitational forces.

� For a rectangular channel it may be written as

� FN= 1 Critical Flow

> 1 Super-Critical Flow

< 1 Sub-Critical Flow

gy

VNF =

William Froude (1810-79)

Born in England and engaged

in shipbuilding. In his sixties

started the study of ship

resistance, building a boat

testing pool (approximately 75

m long) near his home. After his

death, this study was continued

by his son, Robert Edmund

Froude (1846-1924). For

similarity under conditions of

inertial and gravitational forces,

the non-dimensional number

used carries his name.

7


(Rectangular Channels)

� Specific Energy � Specific Energy at a section in an open channel is the energy

with reference to the bed of the channel.

� Mathematically;

Specific Energy = E = y+V2/2g

For a rectangular Channel

q = Discharge per unit width m3/s per m

B

( )Vy

B

VBy

B

AV

B

Qqwhere

yE

yE

yg

q

gV

====

+=

+=

2

2

2

2

2 y

8

Datum


� As it is clear from E~y diagramdrawn for constant discharge forany given value of E, there wouldbe two possible depths, say y1 andy2. These two depths are calledAlternate depths.

� However for point Ccorresponding to minimumspecific energy Emin, there wouldbe only one possible depth yc. Thedepth yc is know as critical depth.

� The critical depth may be definedas depth corresponding to minimumspecific energy discharge remainingConstant.

� E~y Diagram or E-Diagram

Static Head

Line

BQqwhereyEyg

q/2

2

2=+=

9


� For y>yc , V<Vc Deep Channel

� Sub-Critical Flow, Tranquil Flow, Slow Flow.

� For y<yc , V>Vc Shallow Channel

� Super-Critical Flow, Shooting Flow, Rapid Flow, Fast Flow.

10

Specific Energy and Critical Depth Relationship Between Critical Depth and Specific Energy

for rectangular channels

22

2cc y

g

V=

Substituting in eq. (1)

( )

( )( )

( )g

yVc

g

yVc

ccg

yV

c

g

q

c

cg

q

c

gy

q

dydE

gy

q

dydE

gy

q

gV

cc

cc

cc

y

y

yVqy

y

yyy

yyE

2

3

3

3/1

3/1

23

3

2

32

22

2

2

2

22

22

22

2

2

2

2

)2(

)(

01

1

)1(

=

=

==

=

==

=−=

−=

+=+=

Q

Q

)5(23

2min

cc

y

cc

yE

yEE c

=

+==

11

)4(1

)3(22

2

=

=

c

c

c

gy

V

y

gVc

Froude # =1 !!

Example

12

Problem

13

( ) 3/12

g

q

cy =

( ) 2/1

cc gyV =

2/13/2486.1oSAR

nQ =

Problem 11.38

� Water is released from a sluice gate in a rectangular channel 1.5m wide such that depth is 0.6 m and velocity is 4.5 m/s. Find

(a). Critical Depth for this specific energy

(b). Critical Depth for this rate of Discharge

(c). The type of flow and alternate depth by either direct solution or the discharge Curve.

� Solution

B=1.5 m

y=0.6 m

V= 4.5 m/sec

(a)

mEy

mg

VyEnergySpecific

c 088.13

2

632.12

2

==

=+=

sluice gate

14

Critical flow

Problem 11.38

(b)

(c)

CriticalSuperisFlowTherefore

ymg

qy

mpermvyq

mByVAVQ

c

−

>=

=

==

===

906.0

sec/7.2

sec/05.4

3/12

3

CriticalSuperisFlow

gy

VFN

−

>== 1855.1

2

2

2

2

3

)81.9(2

7.2632.1

2

sec/05.4

yy

gy

qyE

mByVAVQ

+=

+=

===

my

my

yy

46.1

6.0

037156.0632.1 23

=

=

=+−

15


(Non Rectangular Channels)

� Hydraulic Depth

� The hydraulic depth, yh for non rectangular channel is the depth of arectangular channel having flow area and base width the same as the flowarea and top width respectively as for non rectangular channel.

dy

T

A y

dy

T

A yh

16

Specific Energy and Critical DepthRelationship Between Critical Depth and Specific Energy

� Froude’s number may be numerically calculate as

3

2

2

22

gA

TQF

T

Ay

gA

QF

gy

VF

N

h

TA

h

N

N

=

==

=

ycyg

Q

T

A

Therefore

dy

dEflowCriticalfor

TgA

Q

dy

dE

TdydASince

=

=

=

−=

=

23

3

2

0

1

17

dy

dA

gA

Q

dy

dE

gA

QyEEq

3

2

2

2

1

2)1.(

−=

+=⇒

Problems11.45

� A Trapezoidal canal with side slopes 1:2 has a bottom width of 3m and carries a flow of 20 m3/s.

a). Find the Critical Depth and Critical velocity.

b). If the canal is lined with Brick (n=0.015), find the critical slope for the same rate of discharge.

� Solution

B=3m

T

A= (B+xy)y

P= B+2y(1+x2)1/2

T= B+2xy

1

x

Q=20 m3/s

x= 2

18

Problem 11.45

�Q2/g = A3/T

� (b)

Q2/g y A T A3/T

40.775 1 5 7 17.85

2 14 11 249.45

1.2 6.48 7.8 34.88

1.25 6.883 8.004 40.74

1.2512 6.885 8.0048 40.77

224433.0

2/1

2/3

=

=

cS

SP

A

n

AQ

19

Problem

� The 50o triangular channel in Fig. E10.6 has a flow rate Q 16 m3/s. Compute (a) yc, (b) Vc, and (c) Sc if n 0.018.

� Answer: (a). 2.37m, (b). 3.41m, (c ) 0.00542

2020

Surface Profiles and

Backwater Curves in Channels of

Uniform sections

21

Steady Flow in Open Channels

� Specific Energy and Critical Depth

� Surface Profiles and Backwater Curves in Channels of Uniform Sections

� Hydraulics jump and its practical applications.

� Flow over Humps and through Constrictions

� Broad Crested Weirs and Venturi Flumes

22

Types of Bed Slopes

� Mild Slope (M)yo > yc

So < Sc

� Critical Slope (C)yo = yc

So = Sc

� Steep Slope (S)yo < yc

So > Sc

So1<Sc

So2>Sc

yo1

yo2

yc

Break

23

yo= normal depth of flowyc= critical depthSo= channel bed slopeSc=critical channel bed slope

Occurrence of Critical Depth

� Change in Bed Slope

� Sub-critical to Super-Critical

� Control Section

� Super-Critical to Sub-Critical

� Hydraulics Jump

Control Section

So1<Sc

So2>Sc

yo1

yo2

yc

Break where

Slope changes

Dropdown Curve

So1>Sc

So2<Sc

yo1

yo2

yc

Hydraulic Jump

24

Occurrence of Critical Depth

� Change in Bed Slope

� Free outfall

� Mild Slope

� Free Outfall

� Steep Slope

yb~ 0.72 yc

So<Sc

yo yc

3~10 ycBrink

yc

So>Sc

25

Non Uniform Flow or Varied Flow.

� For uniform flow through openchannel, dy/dl is equal to zero.However for non-uniform flow thegravity force and frictionalresistance are not in balance. Thusdy/dl is not equal to zero whichresults in non-uniform flow.

� There are two types of nonuniform flows.

� In one the changing conditionextends over a long distance andthis is called gradually varied flow.

� In the other the change may occurover very abruptly and the transitionis thus confined to a short distance.This may be designated as a localnon-uniform flow phenomenon orrapidly varied flow.

So1<Sc

So2>Sc

yo1

yo2

yc

Break

26

Energy Equation for Gradually Varied Flow.

1Z

g

V

2

2

1

Datum

So

1y

2Z

g

V

2

2

2

2y

HGL

EL

Water Level

lhg

VyZ

g

VyZ +++=++

22

2

222

2

111

Theoretical EL

Sw

hL

∆X

∆ L

S

27Remember: Both sections are subject to atmospheric pressure


( )

( ) ( )

profilesurfacewateroflengthLWhere

SS

EEL

LSLSEE

Now

forL

ZZ

X

ZZS

L

hS

hZZg

Vy

g

Vy

o

o

o

oL

L

=∆

−

−=∆

∆+∆−=

<∆

−≈

∆

−=

∆=

+−−+=+

)1(

6,

22

21

21

2121

21

2

22

2

11

θ

An approximate analysis of gradually varied, non uniform flow can be achieved by considering a length of stream consisting of a number of successive reaches, in each of which uniform occurs. Greater accuracy results from smaller depth variation in each reach.

28


3/4

22

2/13/21

m

m

mm

R

nVS

SRn

V

=

=

The Manning's formula (or Chezy’s formula) is applied to average conditions in each reach to provide an estimate of the value of S for that reach as follows;

2

2

21

21

RRR

VVV

m

m

+=

+=

In practical, depth range of the interest is divided into small increments, usually equal, which define the reaches whose lengths can be found by equation (1)

29

Problem 11.59� A rectangular flume of planer

timber (n=0.012) is 1.5 m wideand carries 1.7m3/s of water. Thebed slope is 0.0006, and at acertain section the depth is 0.9m.Find the distance (in one reach) tothe section where depth is 0.75m.Is the distance upstream ordownstream ?

3

1

2

Rectangular Channel

0.012

1.5

1.7 / sec

0.0006

0.9

0.75

o

n

B m

Q m

S

y m

y

=

=

=

=

=

=

B

y

30

Problem 11.59

Solution

1 2

2 2

4 / 3

2

1

2

2

1

2

1 1 1

2 2 2

1 1

2 2

&

1.5 0.9 1.35

1.5 0.75 1.125

1.5 2 0.9 3.3

1.5 2 0.75 3

/ 0.41

/ 0.375

/ 1.26 / sec

/ 1.51 / sec

o

m

m

Since

E EL

S S

V nS

R

A x m

A x m

P x m

P x m

R A P

R A P

V Q A m

V Q A m

−∆ =

−

=

= =

= =

= + =

= + =

= =

= =

= =

= =

2 2

4 /3

1 2

2 2

1 21 2

0.3925

1.385

& 0.000961

2 2

317.73

m

m

m

m

o

o

R m

V m

V nS

R

E ENow L

S S

V Vy y

g gL

S S

m Downstream

=

=

= =

−∆ =

−

+ − +

∆ =

−

=

31

upstream

Problem

32

5ft

yo

1ft

L

Problem

33

Problem

34

Problem 11.66

� The slope of a stream of a rectangularcross section is So=0.0002, the widthis 50m, and the value of Chezy C is43.2 m1/2/sec. Find the depth foruniform flow of 8.25 m3/sec/m of thestream. If a dam raises the water levelso that at a certain distance upstreamthe increase is 1.5m, how far from thislatter section will the increase be only30cm? Use reaches with 30cm depth.

� Given That

1.5m

yo

0.3m

1/ 2

3

0.0002

50

43.2 / sec

8.25 / sec/

508.25 43.2 0.0002

50 2

6.1

o

oo o

o

oo

o

o

S

B m

C m

q m m

Aq y C S

P

yy

y

y m

=

=

=

=

=

=+

=

L

35

Problem 11.66

y A P R V E E1-E2 Vm Rm S S-So ∆L Σ∆L

By B+2y A/P q/y y+v2/g (v1+v2)/2 Vm2/(RmC2)

(E1-E2) /(S-So)

m m2 m m m/s m m m/s m m/m m/m m m

7.6 380 65.2 5.82 1.09 7.66

0.295 1.11 5.74 0.000115 -0.000085 -3454.33 -3454.33

7.3 365 64.6 5.65 1.13 7.365

7.0

6.7

6.4

V=C(RS)1/2

36

Water Surface Profiles in Gradually Varied

Flow.

1Z

g

V

2

2

1

Datum

So

1y

2Z

g

V

2

2

2

2y

HGL

EL

Water Level

g

VyZHeadTotal

2

2

++=

Theoretical EL

Sw

hL

∆X

∆L

37


Flow.

( )

01

0

)2(1

.

1

1

2

.tan..

22

2

2

3

22

3

2

2

2

2

22

1

=−

−∴

=

−

−=

−

=−+−=−

−+=

−

++=

++=++=

N

o

N

NNo

F

SS

dx

dyflowuniformFor

F

SSo

dx

dyor

gsindecreaisflow

ofdirectionalongheadtotalthatshowssignve

gy

qFF

dx

dySS

gy

q

dx

dy

dx

dZ

dx

dH

gulartanrecastionseccrossgConsiderin

gy

q

dx

d

dx

dy

dx

dZ

dx

dH

xdirectionhorizontalincedistrwHheadtotaltheatingDifferenti

gy

qyZ

g

vyZH

Q

Equation (2) is dynamic Equation for

gradually varied flow for constant value

of q and n

If dy/dx is +ve the depth of flowincreases in the direction of flow andvice versa

38

Important assumption !!


Flow.

3/10

3/10

22

3/10

22

2/13/5

2/13/2

1

1

=∴

=

=

=

=

≈

y

y

S

S

y

qnS

channel

gulartanrecinflowuniformFor

y

qnS

orSyn

q

orSyn

V

yR

channelgulartanrecwideaFor

o

o

o

o

21 F

SS

dx

dy o

−

−=

� Consequently, for constant q and n,when y>yo, S<So, and the numeratoris +ve.

� Conversely, when y<yo, S>So, andthe numerator is –ve.

� To investigate the denominator weobserve that,

� if F=1, dy/dx=infinity;

� if F>1, the denominator is -ve; and

� if F<1, the denominator is +ve.

39

Classification of Surface Profiles

� Mild Slope (M)

yo > yc

So < Sc

� Critical Slope (C)

yo = yc

So = Sc

� Steep Slope (S)

yo < yc

So > Sc

� Horizontal (H)

So = 0

� Adverse (A)

So = -ve

� Type 1: If the streamsurface lies above both thenormal and critical depth offlow (M1, S1)

� Type 2: If the streamsurface lies between normaland critical depth of flow(M2, S2)

� Type 3: If the streamsurface lies below both thenormal and critical depth offlow. (M3, S3)

40

Water Surface ProfilesMild Slope (M)

1

2

3

1:1

2 :1

3:1

oo c

N

oo c

N

o

o c

N

S Sdy Vey y y Ve M

dx F Ve

S Sdy Vey y y Ve M

dx F Ve

S Sdy Vey y y Ve M

dx F Ve

− +> > = = = + ⇒

− +

− −> > = = = − ⇒

− +

− −> > = = = + ⇒

− −

yc

Note:

For Sign of Numerator compare

yo & y

For sign of denominator compare

yc & y

If y>yo then S<So and Vice Versa

41

Water Surface ProfilesSteep Slope (S)

1

2

3

1:1

2 :1

3:1

oc o

N

oc o

N

oc o

N

S Sdy Vey y y Ve S

dx F Ve

S Sdy Vey y y Ve S

dx F Ve

S Sdy Vey y y Ve S

dx F Ve

− +> > = = = + ⇒

− +

− +> > = = = − ⇒

− −

− −> > = = = + ⇒

− −

Note:


yo & y


yc & y


yc

42

Water Surface ProfilesCritical (C)

1

3

1:1

2 :1

oo c

N

oo c

N

S Sdy Vey y y Ve C

dx F Ve

S Sdy Vey y y Ve C

dx F Ve

− +> = = = = + ⇒

− +

− −= > = = = + ⇒

− −

Note:


yo & y


yc & y


yo=yc

C2 is not possible

43

Water Surface ProfilesHorizontal (H)

( ) 2

( ) 3

1:1

2 :1

oo c

N

oo c

N

S Sdy Vey y y Ve H

dx F Ve

S Sdy Vey y y Ve H

dx F Ve

∞

∞

− −> > = = = − ⇒

− +

− −> > = = = + ⇒

− −

Note:


yo & y


yc & y


yc

H1 is not possible bcz water has to lower down

44

Water Surface ProfilesAdverse (A)

( ) 2

( ) 3

1:1

2 :1

oo c

N

oo c

N

S Sdy Vey y y Ve A

dx F Ve

S Sdy Vey y y Ve A

dx F Ve

∞

∞

− −> > = = = − ⇒

− +

− −> > = = = + ⇒

− −

Note:


yo & y


yc & y


yc

A1 is not possible bcz water has to lower down

45

Problem 1

46

� Classify the water surface profile for given data,

� Channel type: rectangular

� n=0.013

� B=1.6

� So=0.0005, Q=1.7m

� y= varies from 0.85-1.0m

� Solution:

� Calculate normal depth of flow yo using Manning’s equation

21 F

SS

dx

dy o

−

−=

( ) 2/1

0

3/2

2/1

0

3/2

2

11S

yB

ByBy

nSAR

nQ

o

oo

+==

y< yo

Problem 1

47

� Determine Critical depth

� Thus yo > yc � slope is Mild and profile type will be one of M type

� Since y lies between both yo and yc i.e. yo >y> yc, the profile type is M2

( )m

g

qyc 486.0

81.9

6.1/7.13/1

23/12

=

=

=

Problem 2

48

� Classify the water surface profile for given data,

� Channel type: rectangular

� n=0.013

� B=1.6

� So=-0.0004 < 0 adverse slope

� Q=1.7m

� y= varies from 0.85-1.0m

� Solution:

� Calculate normal depth of flow yo:

21 F

SS

dx

dy o

−

−=

yyo >∞=

Problem 2

49

� Determine Critical depth

� Thus yo > yc � Since slope is Adverse and profile type will be one of A type

� Since y lies between both yo and yc i.e. yo >y> yc, the profile type is A2

( )m

g

qyc 486.0

81.9

6.1/7.13/1

23/12

=

=

=

Problem 3

50

� B=15m, rectangular channel, Q=1400cfs, yo=6ft. � y=2.8m� Determine y to yo depth increase of decrease towards

downstream. � Solution: Determine yc

� yo< yc mean So>Sc, Steep slope (S)

� Now comparing all depths

� yc > yo >y

Hence the slope type is S3

+=+

+=

−

−=

21 F

SS

dx

dy o

Water depth will increase

Practice problems

51

Thank you

� Questions….

� Feel free to contact:

52

Engineering

(Part ii)- open channels