flow in open channel i sem_13_14 [compatibility mode]

10/12/2013

1

CE F312 Flow in Open Channel

by Dr. Ajit Pratap Singh

Flow in Open Channel

By

Dr. Ajit Pratap Singh

Civil Engineering Department



CE F312: Flow in Open Channel


Brays Bayou

Concrete Channel

Uniform Open Channel FlowUniform Open Channel FlowUniform Open Channel FlowUniform Open Channel Flow Open Channel FlowOpen Channel FlowOpen Channel FlowOpen Channel Flow1. Uniform flow - Manning’s Eqn in a prismatic

channel - Q, v, y, A, P, B, S and roughness are all

constant as discussed in last classes

2. Critical flow - Specific Energy Eqn (Froude No.)

3. Non-uniform flow - gradually varied flow (steady

flow) - determination of floodplains

4. Unsteady and Non-uniform flow - flood waves





Open Channel Flow• Liquid (water) flow with a ____ ________

(interface between water and air)

• relevant for

– natural channels: rivers, streams

– engineered channels: canals, sewer lines or culverts (partially full), storm drains

• of interest to hydraulic engineers

– location of free surface

– velocity distribution

– discharge - stage (depth) relationships

– optimal channel design

free surface



Properties of Open Channels

• Free water surface

– Position of water surface can change in space and time

• Many different types

– River, stream or creek; canal, flume, or ditch; culverts

• Many different cross-sectional shapes

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TYPES OF CHANNELS

• Prismatic and Non-prismatic ChannelsA channel in which the cross-sectional shape and sizeand also the bottom slope are constant is termed as aprismatic channel. Most of the man-made (artificial)channels are prismatic channels over long stretches.The rectangle, trapezoid, triangle and circle are some ofthe commonly-used shapes in man-made channels. Allnatural channels generally have varying cross-sectionsand consequently are non-prismatic.

• Rigid and Mobile Boundary ChannelsOn the basis of the nature of the boundary openchannels can be broadly classified into two types: (i) rigidchannels and (ii) mobile boundary channels.

Definition of Geometric Elements

R = hydraulic radius = A/P

D = hydraulic depth = A/T

y = depth of flow

d = depth of flow section

T = top width

P = wetted perimeter

A = flow area


by Dr. Ajit Pratap Singh Common Geometric Properties Cot α = z/1

α

z

1

Classification of Flows• Steady and Unsteady

– Steady: velocity at a given point does not change with time

• Uniform, Gradually Varied, and Rapidly Varied

– Uniform: velocity at a given time does not change within a given length of a channel

– Gradually varied: gradual changes in velocity with distance

• Laminar and Turbulent

– Laminar: flow appears to be as a movement of thin layers on top of each other (if Re < 500)

– Turbulent: packets of liquid move in irregular paths ((if Re >1000)

(Temporal)

(Spatial)



• Based on the relative importance of inertia and gravity forces, flow may be critical, sub-critical and super-critical flow:

– Fr = 1 for critical flow

– Fr < 1 for subcritical flow

– Fr > 1 for supercritical flow



Flow Classification

• Uniform (normal) flow: Depth is constant at every section along length of channel

• Nonuniform (varied) flow: Depth changes along channel

– Rapidly-varied flow: Depth changes suddenly

– Gradually-varied flow: Depth changes gradually

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CLASSIFICATION OF FLOWS

• Steady and Unsteady Flows

A steady flows occurs when the flow properties, such asthe depth or discharge at a section do not change withtime. As a corollary, if the depth or discharge changeswith time the flow is termed unsteady.

• Flood flows in rivers and rapidly-varying surges in canalsare some example of unsteady flows. Unsteady flowsare considerably more difficult to analysis than steadyflows.



• Uniform and non-uniform Flows

• If the flow properties, say the depth of flow, in an openchannel remain constant along the length of channel, theflow is said to be uniform. As a corollary of this, a flow inwhich the flow properties vary along the channel istermed as non-uniform flow or varied flow.

• A prismatic channel carrying a certain discharge with aconstant velocity is an example of uniform flow (Fig. (a)).



• Flow in a non-prismatic channel and flow with varyingvelocities in a prismatic channel are examples of variedflow. Varied flow can be either steady or unsteady.



• Gradually-varied and Rapidly –varied Flows

• If the change of depth in a varied flow is gradual so thatthe curvature of streamlines is not excessive, such a flowis said to be a gradually –varied flow (GVF). Thepassage of a flood wave in a flood wave in a river is acase of unsteady GVF (Fig. (b)).



• A hydraulic jump occurring below a spillway or a sluicegate is an example of steady RVF. A surge , moving upa canal (Fig. (c)) and a bore traveling up a river areexamples of unsteady RVF.



• Spatially-varied flow

• Varied flow classified as GVF and RVF assumes that noflow is externally added to or taken out of the canalsystem. The volume of water in a known time interval isconserved in the channel system. In steady-varied flowthe discharge is constant at all sections. However, ifsome flow is added to or abstracted from the system theresulting varied flow is known as a spatially varied flow(SVF).

• SVF can be steady or unsteady. In the steady SVF thedischarge while being steady-varies along the channellength. The flow over a bottom rack is an example ofsteady SVF (Fig (d)).

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• The production of surface runoff due to rainfall, known as overland flows, is a typical example unsteady SVF.



• Classification Thus open channel flows are classified for purposes of identification and analysis.

• Fig. 1.1(a) through (d) shows some typical examples of the above types of flows

21

Classification of Channel Flow: An Example



22



State of Flow

• Flow in open channels is affected by viscous and gravitational effects

• Viscous effects described by Reynolds number, Re = VR/ν

• Gravitational effects described by Froude number, Fr = V/(gD)1/2



Viscous Effects

• For Re < 500, viscous forces dominate and flow is laminar

• For Re > 2000, viscous forces are weak and flow is turbulent

• For Re between 500 and 2000, there is a transition between laminar and turbulent flow

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Gravitational Effects

• Critical flow is the point where velocity is equal to the speed of a wave in the water

• For F = 1, flow is critical

• For F < 1, flow is subcritical

– Wave can move upstream

• For F > 1, flow is supercritical

– Wave cannot move upstream



Equations of Motion

• There are three general principles used in solving problems of flow in open channels:

– Continuity (conservation of mass)

– Energy

– Momentum

• For problems involving steady uniform flow, continuity and energy principles are sufficient



Conservation of Mass

• Since water is essentially incompressible, conservation of mass (continuity) reduces to the following: discharge in = discharge out

• Stated in terms of velocity and area:

Q = V1A1 = V2A2



Velocity ProfilesVelocity Profiles

Water SurfaceWater Surface

Maximum VelocityMaximum Velocity

Maximum VelocityMaximum Velocity



Open Channel Flow: Uniform Flow

• Uniform Flow

– Discharge-Depth relationships

• Channel transitions

– Control structures (sluice gates, weirs…)

– Rapid changes in bottom elevation or cross section

• Critical, Subcritical and Supercritical Flow

• Flow Mesurements

normal depth



Normal depth is function of flow rate, and

geometry and slope. One usually solves for normal

depth or width given flow rate and slope information

B

b

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Normal depth implies that flow rate, velocity, depth,

bottom slope, area, top width, and roughness remain

constant within a prismatic channel as shown below

Q = C

V = C

y = C

S0 = C

A = C

B = C

n = C

UNIFORM FLOW



• Given a long channel of constant slope and cross section find the relationship between discharge and depth

• Assume

– Steady Uniform Flow - ___ _____________

– prismatic channel (no change in _________ with distance)

• Use Energy, Momentum, Empirical or Dimensional Analysis?

• What controls depth given a discharge?

• Why doesn’t the flow accelerate?

Open Channel Flow: Discharge/Depth Relationship

P

no acceleration geometry

Force balance

A

l

dhl

40

γτ −=



Open Conduits:Dimensional Analysis

• Geometric parameters

• Does Fr affect shear?

P

AR =Hydraulic radius (Rh)

Channel length (l)

Roughness (k)

No!gy

VFr =



• If the suitable form of the Reynolds no. can be found, theresults obtained in pipe analysis can be transformed forChannel Analysis.

• Characteristic length dimension in pipe case : Diameter

• Equivalent Characteristic length dimension for channels:Hydraulic radius

• Re for channel = ρ V R/µ

• For a pipe flowing full, R = D/4

• Re for channel = Re for pipe/4

• Therefore for laminar channel flow Re < 500 and forturbulent channel flow Re > 1000

• Note that upper limit of Re is not so well defined forchannels and is normally taken to be 2000.



• For pipe case, Darcy-Weisbach formula forpipe friction wasintroduced and Moody’sdiagram is developedaccordingly.

• A similar diagram forchannels may beobtained

(I) V

8gRSf

4R2g

fVS

L

h

4R2g

fLVh

pipefor 2gD

fLVh

2

0

2

0f

2

f

2

f

=⇒

×==⇒

×=

=



Uniform Flow

• Equations are developed for steady-state conditions

– Depth, discharge, area, velocity all constant along channel length

• Rarely occurs in natural channels (even for constant geometry) since it implies a perfect balance of all forces

• Two general equations in use: Chezy and Manning formulas

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Reach of an Open-channel flow



37

Steady-Uniform Flow: Force Balance

Shear force =________

0sin =∆−∆ xPxwA oτθ

sin θ P

Awτ o =

R = P

A

θθ

θsin

cos

sin≅=S

θ

W

θ

W sin θ

∆x

a

b

c

d

Shear force

Energy grade line

Hydraulic grade line

W cos θ

g

V

2

2

Wetted perimeter = __

Gravitational force = ________

Hydraulic radius

Relationship between shear and velocity? ___________

τoP ∆ x

P

ρgA ∆x sinθ

Turbulence

(II) wRSτo =



f

8g

fρ

8w C where,

RSCV

RSfρ

8wV

wRSρV8

fτ

ρV8

fτ

2

o

2

o

=×

=

=⇒

×=⇒

==⇒

=



• Let a state of rough turbulent flow

(IV) t coefficien sChezy' is C where

RSCV RSf

ρgV

(III) and (II) From

×=⇒×=

το ∝ V2 or τo = KV2 (III)

11/2T L ofdimension with the

f

8gC Thus

(IV) and (I) From

−

=



Chezy Equation (Summary)• Introduced by the French engineer Antoine Chezy in

1768 while designing a canal for the water-supply system of Paris

• Balances force due to weight of water in direction of flow with opposing shear force

• Note: V is mean velocity, R is hydraulic radius(area/wetted perimeter), S is the slope of energygradeline, and C is the Chezy coefficient

• Chezy’s coefficient C is a depends on Reynolds no.and boundary roughness

V = C RS



Manning Equation

• Similarly, the Manning equation which is anempirical relationship can be expressed as:

• Note: V is mean velocity (m/sec), R ishydraulic radius (m), S is the slope ofthe energy gradeline (m/m), and n is theManning roughness coefficient

2/13/21SR

nV =

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Manning Equation (1891)

• Similarly, Manning’s equation in SI (metric) units

n Manning' isn where

SRn

1V 1/22/3=

1/6R

n

1C and =

Ganguillet-Kutter Formula

• Where n is roughness coefficient known as Kutter’s ‘n’. This is different from Manning’s n



R

n

S

0.00155231

n

1

S

0.0015523

C

++

++

=



Manning’s Roughness (n)

• Roughness coefficient (n) is a function of:

– Channel material

– Surface irregularities

– Variation in shape

– Vegetation

– Flow conditions

– Channel obstructions

– Degree of meandering

Manning’s n

Lined Canals n

Cement plaster 0.011

Untreated gunite 0.016

Wood, planed 0.012

Wood, unplaned 0.013

Concrete, trowled 0.012

Concrete, wood forms, unfinished 0.015

Rubble in cement 0.020

Asphalt, smooth 0.013

Asphalt, rough 0.016

Natural Channels

Gravel beds, straight 0.025

Gravel beds plus large boulders 0.040

Earth, straight, with some grass 0.026

Earth, winding, no vegetation 0.030

Earth , winding with vegetation 0.050

n = f(surface roughness, channel irregularity, stage...)

d = median size of bed material

min is d if 038.0

ftin is d if 0.031dn

6/1

1/6

dn =

=



+−=

−

fRe

0.6275

14.8R

k2log

f

1

channelsfor formula equivalent WhiteColebrook

s

+−=

=

0

s0

2

0

8gRSRe

0.6275V

14.8R

klog8gRS2V

equations above Combining

V

8gRS f But,



• f-Re diagram channels may be derived using above equation. However,

• The application of these equations to a particular channel more complex than the pipe case due to extra variables involved say R changes with depth and channel shape.

• Validity is questionable because of the presence of free surface has considerable effect on the velocity distributions.

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Discharge using Manning’s Equation

• Using continuity equation (Q=VA), Manning’s equation for English units can be written as

• And for metric units

Q =1.486

nAR

2 / 3S

1/ 2

Q =1

nAR

2 / 3S

1/ 2



Conveyance

• For uniform flow, A, R and n are constant thus

Q = KS1 / 2

3/21AR

nK =

• The term K is conveyance, given as



Normal Section Factor

• Based on Manning’s equation, define a normal section factor for the portion dependent on geometry, namely

Z n = AR2 / 3

• For simple geometries, Zn can be written as a function of yn

• For complex geometries, use tabulated values



Computing Normal Flow

Three cases to consider:

�Find Q for known values of yn and S

�Find S for known values of Q and yn

�Find yn for known values of Q and S

• From Manning’s equation, see that

2/11SZ

nQ n=



Computing Normal Flow

2/11SZ

nQ n=

• For known values of yn and S, can use Manning’s equation directly

• Based on relationship between yn and Zn, determine the value of Zn

• Compute discharge from Manning’s equation, namely



Computing Normal Slope

2

1

=

n

Z

QS

n

• For known values of yn and Q, can use Manning’s equation directly

• Based on relationship between yn and Zn, determine the value of Zn

• Rearrange Manning’s equation, and compute slope as

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Computing Normal Depth

• For known values of Q and S, can compute Zn as

2/1S

nQZn =

• Normal depth (yn) is then determined from the relationship between yn and Zn

Note

• For each regular channel section, we can construct the curves R versus y and AR2/3

versus y:

(a) For example Circular with diameter of 2.0 m



AR2/3 versus y Curve

• We can construct the curves R versus y and AR2/3

versus y for Trapezoidal, with bottom width = 3.0m and side slopes H:V 2.5





Channel Design

• Previously, we have assumed that channel dimensions and slope are known

• In design situations, usually discharge is and need to determine geometry and slope

• In an open channel, usually have allowable velocity range

– Minimum of 2-3 ft/s to move grit

– Maximum velocity to prevent scouring



Channel Design

• For circular pipes, can determine minimum slope needed to achieve a minimum velocity of 2 ft/s



Channel Design

• From basic principles, we know that:

– For a given slope and roughness, Q increases with increase of section factor (Zn=AR2/3)

– For a given area, Zn is maximum for a minimum wetted perimeter (P)

• Based on these principles, we can develop guidelines for efficient hydraulic sections for given channel shapes

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Design Procedure

• For a specified Q, select S and estimate n Compute section factor from Q, n and S

• Select a shape, and express section factor in terms of y; solve for y

• Try different values of S, n and shape and compute cost for each

• Verify that velocity exceeds 2-3 ft/s

• Add freeboard if open section

Rectangular Channel

CE F312 Flow in Open Channel by Dr. Ajit Pratap Singh

2y Bor 2

202y

AdP

0dy

dP condition, P minimumFor

area, sectional cross-given x aFor

22

y

APB2y PBut

y

ABBy A

2

2

2

==⇒

==⇒=+−=⇒

=

⇒

+=+=⇒+=

=⇒=

By

ByyAdy

y

yAy

Rectangular Channel…

• Rectangular channel section will be mosteconomical when either the depth of flow isequal to half of bottom width of the channel orhydraulic radius al to the half of the depth offlow.



24

22

2/

2

2

2

R Radius, Hydraulic

Similarly,

2

yB

B

B

B

BB

yy

A

By

P

A==

×+

×

=

+

==



Trapezoidal Channel

• Derive P = f(y) and A = f(y) for a trapezoidal channel

• How would you obtain y = f(Q)?

z1

B

y( )

zyy

Az12yP

Bz12y PBut

zyy

AByzyBA

y2

2zyBB A

2

2

−++=⇒

++=

−=⇒+=

++=

2/13/21

oh SAR

n

Q =



A being constant, differentiating P with respect to y

( )( )

( )

2

y

P

A R and

z12y 2zyB

z12zy

y zyBy zyBABut

z12zy

A

0zy

Az12

dy

dP

2

2

2

2

2

2

2

==

+=+

+=++

⇒+=

+=+

=−−+=

Problem

• A channel of trapezoidal section has to convey a discharge of 7 m3/sec of water at a velocity of 1.75 m/sec. The sides of the channel have a slope of 1 vertical to 2 horizontal. Find the sectional dimensions of the most economical channel. Also find the slope of the channel.

• For conveying the same discharge if a rectangular channel 1.25 m wide and 3.6m deep is provided what would be the saving in power in km length of the channel? Take C = 60.



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Flow in Round Conduits

• Circular sewers are very easily manufactured• A circular section gives the maximum area for a given

perimeter and thus gives greatest H.M.D. when runningfull or half full.

• Most efficient section at these flow conditions• The circular section has uniform curvature all round and

hence it offers less opportunities for deposits• Minimum grades are enough so long as circular sewer

flow more than half full• Both the wetted area as well as the wetted perimeter

varies with the depth of flow. Thus the condition of areaof flow section being constant can not applied all thetime.



(i) Condition for maximum discharge

( )sinθ-θ2

r

2

2

θrcos

2

θ2rsin

θ2π

πr

OAB ∆ of AreaOADBO arc of AreaA Area Wetted

2

2

=

×

−=

−=

rθ PPerimeter Wetted =

rθ PPerimeter Wetted =

r

C

y

T

A θ

O

D

B

0P

A

dθ

d

S and Cgiven for P

A when maximum be willDischarge

SP

ACRSACAVQ

3

0

3

0

3

0

=

⇒

===



( ) ( )( )

o

22

2

32

308θ

0sinθ3.cosθ-2θ

0rsinθθ2

rcosθ1

2

r3rθ

0dθ

dPA

dθ

dAP.3

P

dθ

dPA

dθ

dAP.3A

=

=+

=−−

−

=

−

=

−

⇒

( )

( )

0.5733r0.29DR

308θ and sinθθ2θ

r

P

AR

,Similarily

diameter is d whereD 0.95 y or

1.8988rcos261ry

2

308-360rcosry

o

==⇒

=−==

=

=+=⇒

+=



Circular Sections

• For partially full circular sections (with diameter do), the geometric elements are a function of depth (y)

• Values have been tabulated in non-dimensional form using do as a scaling parameter

• Note that maximum flow occurs at a depth of 0.94do



Geometric Elements for Circular Pipes



Efficient Hydraulic Sections

Section Most efficient

Trapezoidal Base < depth

Rectangular Width = 2 x depth

Triangular No specific relationship

Circular Semicircle if open Circle if closed

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by Dr. Ajit Pratap Singh Optimal Channels - Max R and Min P CE F312 Flow in Open Channel


• For other most economical sections, refer class notes and text book.



Design Procedure

• For a specified Q, select S and estimate n Compute section factor from Q, n and S

• Select a shape, and express section factor in terms of y; solve for y

• Try different values of S, n and shape and compute cost for each

• Verify that velocity exceeds 2-3 ft/s

• Add freeboard if open section



Computation of uniform flow

section. channel theofcapacity

carrying of measure is andsection channel

theof conveyance is and RACK where

SRSACAVQ

=

=== K

2/3

1/22/3

ARn

1K where

SARn

1AVQ

OR

=

==

S

QnAR OR

ARK Conveyance

2/3

2/3

=

=⇒=∴ KnS

Q



• L.H.S. depends only on the geometry of the wetted area

• Note that for a given value of n, Q and S, there is only one possible depth of flow at which the uniform flow will be maintained in any channel section. This depth is known as normal depth (yn)

• Similarly, for a given value of n, S and the depth of flow, there is only one possible discharge for maintaining a uniform flow through any channel section. This discharge is known as normal discharge (Qn)



Momentum and Energy Equations

• Conservation of Energy

– “losses” due to conversion of turbulence to heat

– useful when energy losses are known or small

• ____________

– Must account for losses if applied over long distances

• _____________________________

• Conservation of Momentum

– “losses” due to shear at the boundaries

– useful when energy losses are unknown

• ____________

Contractions

Expansion

We need an equation for losses

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Open Channel Flow: Energy Relations

2g

V2

11α

2g

V2

22

α

xSo∆

2y

1y

x∆

L fh S x= D

______

grade line

_______

grade line

velocity head

Bottom slope (So) not necessarily equal to EGL slope (Sf)

hydraulic

energy



Control Volume for Open Channels

• Conservation of energy applied to control volume results in the following:

L

2

2222

2

1111 h

2g

Vαyz

2g

Vαyz +++=++

where z1,z2 are elevations of the bed,

y1, y2 are depths of flow,

V1, V2 are velocities,

α1, α2 are kinetic energy corrections, and

hL is the frictional loss.

Energy Relationships: Conservation of Energy



Energy Relationships: Conservation of Energy

2 2

1 1 2 21 1 2 2

2 2L

p V p Vz z h

g ga a

g g+ + = + + +

2 2

1 21 2

2 2o f

V Vy S x y S x

g g+ D + = + + D

Turbulent flow (α ≅ 1)

z - measured from

horizontal datum

y - depth of flow

Pipe flow

Energy Equation for Open Channel Flow2 2

1 21 2

2 2o f

V Vy S x y S x

g g+ + D = + + D

From diagram on previous slide...



Energy Coefficient

• The term associated with each velocityhead (α) is the energy coefficient

• This term is needed because we are usingthe average velocity over the depth tocompute the total kinetic energy

• Integrating the cubed incrementalvelocities is not equal to the cube of theintegrated incremental velocities



Specific Energy

• Specific energy is defined as the energy in a channel section measured w.r.t. channel bed

• If slope of channel is relatively flat, we can let z1 = z2 = 0

• For energy coefficient (α) = 1, specific energy (E) is thus

g

VyE

2

2

+=

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Specific Energy… Note

The syllabus of Tutorial III will be up to this Portion. Next Classes, I’ll discuss about critical flow equations (for prismatic channels) and applications of specific energy diagrams and discharge diagrams.





Significance of Specific Energy…

• The sum of the depth of flow and the velocity head is the specific energy:

g

VyE

2

2

+=

If channel bottom is horizontal and no head loss

21EE =

y - _______ energy

g

V

2

2

- _______ energy

For a change in bottom elevation

1 2E y E- D =

xSExSE o ∆+=∆+ f21

y

potential

kinetic

+ pressure

P

A

Critical Flow

T

dy

y

T=surface width

Find critical depth, yc

2

2

2gA

QyE +=

0=dy

dE

dA =0dE

dy= =

3

2

1

c

c

gA

TQ=

Arbitrary cross-section

A=f(y)

2

3

2

FrgA

TQ=

22

FrgA

TV=

dA

AD

T= Hydraulic Depth

2

31

Q dA

gA dy-

0

1

2

3

4

0 1 2 3 4

E

y

yc

Tdy

More general definition of Fr



( )

1Fr1gD

V1

gD

V

T

AD D

g

V

T

A

g

VA

T

A

g

Q

T

ADDepth hydraulc and AVQ

2

2

3232

=⇒=⇒=∴

==∴

=⇒=

==

Q

T

A

g

Q

satisfy must surfaceat water T width topandA area sec- xthe

Q), dischargegiven a(for flow of state criticalin is flow theif Thus

32

=



Critical Depth

• For higher values of specific energy there are two values that give the same specific energy

• If Q is fixed, for the smaller depth there will be a higher velocity (supercritical) and for the larger depth there will be a lower velocity (subcritical)

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Critical depth is used to characterize channel flows --

based on addressing specific energy E = y + v2/2g :

E = y + QE = y + QE = y + QE = y + Q2222/2gA/2gA/2gA/2gA2 2 2 2 where where where where Q/A = q/y and q = Q/bQ/A = q/y and q = Q/bQ/A = q/y and q = Q/bQ/A = q/y and q = Q/b

Take Take Take Take dE/dy = (1 dE/dy = (1 dE/dy = (1 dE/dy = (1 –––– qqqq2222/gy/gy/gy/gy3333) and set ) and set ) and set ) and set = 0= 0= 0= 0.... qqqq = = = = constconstconstconst

E = y + qE = y + qE = y + qE = y + q2222/2gy/2gy/2gy/2gy2222

y

E

Min E Condition, q = C



Solving Solving Solving Solving dE/dy = (1 dE/dy = (1 dE/dy = (1 dE/dy = (1 –––– qqqq2222/gy/gy/gy/gy3333) and set ) and set ) and set ) and set = 0= 0= 0= 0....

For a rectangular channel bottom width b,

1.1.1.1. EEEEminminminmin = 3/2Y= 3/2Y= 3/2Y= 3/2Ycccc for critical depth for critical depth for critical depth for critical depth y = yy = yy = yy = ycccc

2.2.2.2. yyyycccc/2 = V/2 = V/2 = V/2 = Vcccc2222/2g/2g/2g/2g

3. yc = (Q2/gb2)1/3

Froude No. = v/(gy)1/2

We use the Froude No. to characterize critical flows



Y vs E E = y + qE = y + qE = y + qE = y + q2222/2gy/2gy/2gy/2gy2222

q = constq = constq = constq = const





In general for any channel shape, B = top width

(Q(Q(Q(Q2222/g) = (A/g) = (A/g) = (A/g) = (A3333/B) /B) /B) /B) at y = yat y = yat y = yat y = ycccc

Finally Fr = v/(gy)1/2 = Froude No.

Fr = 1 for critical flowFr = 1 for critical flowFr = 1 for critical flowFr = 1 for critical flowFr < 1 for subcritical flowFr < 1 for subcritical flowFr < 1 for subcritical flowFr < 1 for subcritical flowFr > 1 for supercritical flowFr > 1 for supercritical flowFr > 1 for supercritical flowFr > 1 for supercritical flow

Critical Flow in Open Channels



Critical Flow & its Computation• When the depth of flow of water

over a certain reach of a givenchannel is equal to the criticaldepth yc, the flow is described ascritical flow

• Zc is a section factor for thechannel.

• For a prismatic channel sectionfactor is a function of depth of flow

• When Discharge is given, fromabove equation, Zc can beobtained. Z is f(y) and a curvebetween Z and y can be obtainedand with the help of this curve,corresponding to the known valueof Z=Zc, the value of depth of flowcan be obtained which will becritical depth yc

c

c

Zg

Q

T

AA ==

When the depth of flow isgiven, from above equation,corresponding value of thesection factor Z can beobtained from the same andfrom this value of Z, thecritical discharge Qc can beobtained Qc =Z g

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Critical Flow…

• From necessary condition for minimum specific energy, can define section factor for critical flow (Zc):

g

QDAZc ==

where Zc is the critical section factor,

A is the area of flow,

D is the hydraulic mean depth, and

Q is the discharge



Computing Critical Flow

• If yc is known, can compute Zc for specific geometry. Then critical discharge is given as

gZQ cc =



Computing Critical Depth

• If Q is known, can compute Zc as

• For simple geometries, Zc can be written as a function of yc For complex geometries, use tabulated values

g

QZc =



Characteristics of Critical State of Flow

• Specific energy is a minimum for givendischarge

• Discharge is a maximum for a given specificenergy

• Velocity head is equal to the half of the hydraulichead in a channel of small slope

• Froude no. is equal to unity



Critical Flow: Rectangular channel

yc

T

Ac

(I) gA

TQ1

3

c

c

2

=

3

cgyqor =

Only for rectangular channels!

cTTB ==

Given the depth we can find the flow!

Let q be the discharge per unit width of the channel section i.e. Q = Bq

Corresponding to a critical depth of flow yc, the area of rectangular channel section i.e. Ac = byc

Substituting the value of Q and A in eq. (I)

T

yBBq3

c

322

=g

1/32

c

3

c

2

g

q yOr

yg

q

=

=⇒



( )( )

( ) (IV) yy-E2gqOr

(III) 2gy

qyE

By2g

qBy

2gA

QyE

2

2

2

2

2

2

2

=

+=⇒

+=+=

For a given specific energy E, the condition for discharge per unit

width q to be maximum may be obtained by

( )( )

(V) y 2

3E

yyE2

3y2ye2g0

dy

dq

2

2

=⇒

−

−==

Substituting the value of E for maximum discharge in eq. (IV)

g

q2

3

32

y

gyyy-y2

32gq

=⇒

=

=

3 roots (one is negative)

How many possible depths given a

specific energy? _____2

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18

Critical Flow Relationships:Rectangular Channels

3/12

=

g

qyc cc yVq =

=

g

yVy

cc

c

22

3

g

Vy

c

c

2

=

1=gy

V

c

cFroude number

velocity head =

because

g

Vy cc

22

2

=

2

c

c

yyE += Eyc

3

2=

forcegravity

forceinertial

0.5 (depth)

g

VyE

2

2

+=

Kinetic energy

Potential energy



Applications

• Use of sp. Energy and discharge diagrams areused to analyze channel section in the case oftransition

• A portion of channel with varying x-section whichconnects one uniform channel to another whichmay or may not have same x-sectional area.These transitions may be either sudden orgradual transition

• Variation of channel may be caused either byreducing or increasing the width or by raising orlowering the bottom of the channel.



Applications …

• The transitions are being used for metering flow.The main devices used here are Venturi flume,Parshall flume, standing wave flume etc.

• The transitions are being used for dissipation ofenergy by providing a sudden drop in the channelbottom or by providing gradual expansion orcontraction with well rounded corners.

• The transitions are being used for reduction invelocities in irrigation channels in order to preventscouring velocities or increase in velocities in orderto prevent shoaling for navigation channels.

• Often channel x-section is reduced, if a bridge hasto be constructed across it.



Reduction in x-section: An Example

• By decrease in width

• By decrease in depth or raised bottom

• Or by combination of both



Case I: Decrease in width

• (a) If width at u/s is B1 and at transition it is B2 which isbasically reduced from B1 to B2 but it is more than Bc

• (b) If width at u/s is B1 and at transition it is B2 which isbasically reduced from B1 to B2 but it is equal to Bc

• (c) If width at u/s is B1 and at transition it is B2 which isbasically reduced from B1 to B2 but it is less than Bc

• Note that analysis of above cases depends upon thetype of flow whether it is sub-critical, critical orsupercritical

• For the analysis in these cases, mainly dischargediagram is used.



Case I (a)

(1) For given data, calculate specificenergy at the u/s

( )222

2

221

2

yB2g

QyEE

equation usingerror & by trial y Calculate (2)

+==

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Case I (b)

(1) For given data, calculate specific energy

at the u/s

( )3/21

c

c

Eg

Q84.1B

equation using B width critical Calculate (2)

=



Case II: Decrease in depth

• (a) If depth at u/s is y1 and at transition it is y2 which is basically reduced from y1 to y2 but it is more than yc

• (b) If width at u/s is y1 and at transition it is y2 which is basically reduced from y1 to y2 which is equal to yc

• (c) If width at u/s is y1 and at transition it is y2 which is basically reduced from y1 to y2 but it is less than yc

• Note that analysis of above cases depends upon thetype of flow whether it is subcritical, critical orsupercritical.

• In the analysis mainly specific energy diagram is used.



Principle

2

2

1

2

22

++=

++

g

Vyz

g

Vyz

2

22

22 gy

qy

g

VyHE +=+==

zHH ∆=− 21



(1) In a rectangular channel 3.5 m wide,laid at a slope of 0.0036, uniform flowoccurs at a depth of 2.0 m. Find howhigh can a hump be raised on thechannel bed without causing a changein the upstream depth. If the upstreamdepth is to be raised to 2.4 m whatshould be the height of the hump?Assume Manning’s n = 0.015.



Problems

(2) Water flows in a 4.0 m wide rectangularchannel at a depth of 2.50m and a velocityof 2.25 m/sec. If the width of the channel isreduced to 2.50 m and bed of channel israised by 0.20 m at a section downstreamside, will the water surface in the channelupstream be disturbed? Why? Alsodetermine the depth of flow over the raisedbed if it is disturbed and show how will thelevel of water surface in the channel beaffected?



(3) Water flows in a rectangular channel 3 m wide at avelocity of 3 m/s at a depth of 3 m. There is anupward step of 0.61 m. What expansion in widthmust take place simultaneously for this critical flowto be possible?

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Problems

(4) Water flows in a 4.0 m wide rectangularchannel at a depth of 2.0m and avelocity of 1.5 m/sec. Determine (i) thewidth at contraction which just causescritical flow without a change in the u/sdepth (ii) the depth in the contractionwhen the width at the throat is 50%more than the above value.



Problems..

(5) Water flows in a 4.0 m wide rectangularchannel at a depth of 2.0m and avelocity of 1.5 m/sec. Determine (i) theheight of hump required to producecritical flow without affecting the u/sdepth (ii) the depth over the humpwhen the height of the hump is half theabove value.



METERING FLUMES

• Critical depth of flow may be obtained at certain sections

in an open channel where the channel bottom is raised by

the construction of low hump or the channel is

constructed by reducing its width.

• Since at critical state of flow the relationship between the

depth of flow and discharge is definite and is independent

of the channel roughness and other uncontrollable

factors, provides a theoretical basis for the measurement

of discharge in open channels. As such various devices

which have been developed for the flow measurement are

based on principle of critical flow.



FLUMES

Flume is a term applied to the devices in which theflow is locally accelerated due to a

• a stream lined lateral contraction in the channel

sides

• the combination of lateral contraction togetherwith a stream lined hump in the invert channel bed.

Flumes are usually designed to achieve criticallyflow in the narrowest section together with smallafflux. And are applicable where deposition ofsolids must be avoided.



VENTURI FLUME:

• The venturi flume is a structure in a channel which

has a contracted section called throat,

downstream of which follows a flared transition

section designed to restore the stream to its

original width.

• It is an open channel counterpart of a Venturi-

meter, which is for measuring discharge in open

channels.



Plan of venturi flume:

Elevation of venturi flume:

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• The velocity at the throat section is always

less than the critical velocity. Therefore the

discharge passing through it will be function

of the deference between the depth of the

flow upstream of the entrance section and at

the throat.

• At the throat section there will be a drop in

the water surface and this drop in the water

surface can be related to the discharge.



• Further since the velocity of the flow at the throat is

less than the critical velocity, standing wave or

Hydraulic jump will not be formed at any section in

the venturi flume.

• Let B, H, V be normal breadth, depth of the flow

and velocity at the entrance of the flume.

• Let b, h, v be the breadth, depth of the flow and

velocity at the throat.

• So, at the throat the velocity v>V



Hence there will be drop in water level at the throat as

the total energy remains constant

Due to continuity of flow

Q = BHV = bhv

Let A = BH , a = bh => V = (a/A)v

By Bernoulli's equation

H + V2/2g = h + v2/2g

v2 – (a2/A2)V2 = 2g(H – h)

by rearranging we get

v2[(A2 – a2)/A2] = 2g(H – h)

Discharge Q = avCE F312 Flow in Open Channel


In the actual case the discharge obtained is slightly less than the above value

due to the losses in the flume. So in order to get the actual discharge multiply

the above equation with coefficient of discharge Cd .

Although the float bed of venturi flume is simpler to construct , it is sometimes

necessary to raise the invert in the throat to attain critical conditions.

( )hH2g

aA

Aa Q Discharge Theretical

22−

−

=∴

( )

discharge oft coefficien theis C where

hH2g

aA

AaC Q Discharge Actual

d

22d −

−

=∴



STANDING WAVE FLUME OR CRITICAL DEPTH FLUME



• It is a structure in a channel which has narrowed throat having a

raised floor (or hump) at the bottom as shown which act as broad

crest weir.

• The downstream of the throat section is followed by the flared

transition section to restore the stream to its original width.

• For any discharge flowing in the channel the velocity of the flow at

the throat of the flume is greater than critical velocity.

• As such a standing wave or hydraulic jump is formed at or near

the down stream end of raised floor.

• Since the velocity at the throat is greater than the Critical velocity.

The depth of flow at a section on upstream of the entrance of the

flume remains unaffected by variations in the down stream depth

until the downstream depth of submergence becomes greater

than about 0.7 of the up-stream.

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• Therefore as long as the downstream depth

of the flow is kept below the limiting value,

the discharge passing through a standing

wave flume will be a function of only the

depth of the flow ‘H’ (above the raised floor)

at a section well upstream of the entrance

section.

• This is similar to the Venturi-flume, except

that in this case a standing wave is formed.



Considering an upstream section and a section at the throat, if

the depths of flow at these sections are ‘H’ and ‘h’ respectively

( )hH2gv

2

vh

2

VH

1

1

22

−=∴

=+=+∴ Hgg

( )

( ) ( )1/2321

321

1

hhH2gbhhH2gbQ

hH2gbhbhv Q Discharge Theretical

−=−=

−==∴

For discharge to be maximum the quantity (H1h2 – h3)

should be maximum



Thus we find that for the condition of maximum discharge

the depth of flow at the throat should be 2/3 the total

energy head. For this condition

( )

1

21

321

H3

2h

03hh2H

0hhHdh

d

=⇒

=−⇒

=−

( )

ghhh2

32gv

hH2gv 1

=

−=⇒

−=∴



Which means for the condition of maximum

discharge the depth of flow at throat is equal to

critical depth.

substituting h = (2/3) H1 in discharge expression

Q = 1.705 bH11.5

=> Q = 1.705b(H + V2/2g)1.5



Problem

• A rectangular channel is 1.20 m wide and itswidth is narrowed to 0.60 m to form a throatregion of venturiflume. The depths of flow in thetwo sections being 0.6m and 0.55m. Neglectinglosses, determine the discharge through thechannel assuming its bed to be horizontal.

If a hump of 0.18 m height is provided at thethroat region so as to produce a standing waveon the downstream side, find the depth of flowon the upstream side for the same discharge.

Problem 2

• A 1.0 m wide rectangular section channel is considerded to a depth of 0.5m, the depth of water ahead of the flume being 0.4. If a hump of 0.15m height is place at the throat so that standing wave flume occurs on d/s, determine the discharge through the channel assuming its bed to be horizontal and cd=0.95. apply correction for velocity of approach.



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Metering of Stream flow





Flumes• Flumes include various

specially shaped and stabilized channel sections that are used to measure flow.

• Use of flumes is similar to use of weirs in that flow is related to flow depths at specific points along the flume.

Parshall FlumeCE F312 Flow in Open Channel


Measuring flow using orifice plates



Weirs



Weir Shapes and FormulasWeir Shapes and Formulas

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Open Ditches Flumes

Flume Upstream Measure



RCB Flume with

Pressure Transducer



RCB Flume in Drain Ditch

Notice Drop



Furrow Irrigation

flow in open channel i sem_13_14 [compatibility mode]

Engineering

ajit pratap singh flow

example of uniform flow

varied flow gvf

depth of flow sectiont

ajit pratap singhce

subcritical flow fr

supercritical flowce

uniform flow mannings