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Water Engineering Reference Notes

Open Channel Flow

D G Bertram

[email protected]

Ext 3251



REFERENCE TEXTS:

TITLE AUTHOR PUBLISHER

Hydraulics in Civil and

Environmental Engineering

A. Chadwick &

J. Morfett

E. & F. N. Spon ***

Civil Engineering Hydraulics Featherstone & Nalluri BSP Professional Books

(Blackwell)

**

Fluid Mechanics for Civil Engineers N. B. Webber Chapman and Hall **

Open Channel Hydraulics R. H. French McGraw Hill Book

Company

**

Open Channel Hydraulics Ven Te Chow McGraw Hill Book

Company

*

Fluid Mechanics V. L. Steeter

E. B. Wylie

McGraw Hill Book

Company

*

Fluid Mechanics & Hydraulics

[Schaum's Outline series S.I. Edition]

R. B. Giles McGraw Hill Book

Company

*

Open Channel Flow F. M. Henderson Macmillan *

An Introduction to Engineering Fluid

Mechanics

J. A. Fox Macmillan

Mechanics of Fluids W. J. Duncan

A. S. Thom

A. D.Young

Edward Arnold

Mechanics of Fluids B. S. Massey Van Nostrand

Reinhold

Fluid Mechanics for Engineering

Students

J. R. D. Francis Edward Arnold



CONTENT

1. Introduction to open channel flow - Chezy, Manning, Darcy Weisbach.

2. Determination of normal depth of flow.

3. The energy principle in open channel flows.

4. Theory of gradually varied flows (dy/dx).

5. Direct Step Method for gradually varied flows.

6. Classification of open channel flows.

7. Rapidly varied flows - the energy method.

8. Force momentum principle in open channel flows.

9. Hydraulic jump equations and energy dissipators.

10. Location of hydraulic jumps

Tutorial 2

Tutorial 1

Tutorial 3



INTRODUCTION TO OPEN CHANNEL FLOW:

Open channel flow refers to any flow in which the flowing stream is not completely enclosed by solid boundaries

but has a free surface exposed to atmospheric pressure.

Examples of open channel flow are: Rivers, Canals, artificial channels, Spillways, etc.

Rectangular Channel

Free

Surface

Natural River

Pressure Velocity

V

P=ρgy



CLASSIFICATION OF OPEN CHANNEL FLOW:

Depending on the variation of depth and mean velocity with space and time, open channel flow may be classified

into different types.

• Steady Flow: The depth (y), discharge (Q), and mean velocity (V) at any section do not change with time

(t). In mathematical terms, for steady flow:

0=dt

dQ, 0=

dt

dy, 0=

dt

dV

Example of this kind of flow: Constant flow through a long length of an irrigation canal.

• Unsteady Flow: The depth (y), discharge (Q), and mean velocity (V) vary with time (t). In mathematical

terms, for unsteady flow:

0≠dt

dQ, 0≠

dt

dy, 0≠

dt

dV

Examples: flow in a river during flood, tides in estuaries, surge waves, etc. We intend to cover this area in year 4.

Open Channel flow

Steady Flow Unsteady Flow

Uniform Flow Non-uniform Flow

Gradually Varied Flow

(GVF)

Rapidly Varied Flow

(RVF)



• Uniform Flow: The depth (y), discharge (Q), and mean velocity (V) do not change along the length of the

channel (x) at any given instant. In mathematical terms, for uniform flow:

0=dx

dQ, 0=

dx

dy, 0=

dx

dV

• Non-uniform or varied Flow: The depth (y), discharge (Q), and mean velocity (V) change along the length

of the channel (x). In mathematical terms, for non-uniform flow:

0≠dx

dQ, 0≠

dx

dy, 0≠

dx

dV

In varied flow there is an imbalance between gravity and friction forces and the net result could accelerate or

decelerate the flow. Varied flow can be accelerated dV/dx>0, or decelerated dV/dx<0.

Non-uniform or varied Flow may be further subdivided into:

• Gradually varied (GVF): the flow depth (y), as well as the other parameters vary slowly from one section

to another.

• Rapidly varied flow (RVF): the flow depth (y), as well as the other parameters change abruptly over a

short distance.

Uniform Varied Flow Uniform Uniform Varied Flow

Weir Hydraulic

jump Drop

GVF RVF GVF RVF RVF

y



UNIFORM FLOW:

Consider a sloped prismatic channel in the following. The liquid in motion provokes a friction force at solid

boundary along the wetted perimeter:

Friction forces = τo Pdx

Longitudinal component of

the gravity force =ρg Adx Sinθ

In uniform flow:

Friction force = Gravity force

τo Pdx =ρg Adx Sinθ

τo =(ρg Adx Sinθ)/ Pdx

τo =ρg (A/P) Sinθ

τo =ρg R Sinθ for θ<5o

Sinθ = tanθ=S

ττττo =ρρρρg R S where τo is the boundary shear stress at the channel bed and walls and R is the

Hydraulic radius of the channel.

CHEZY EQUATION:

This equation was proposed by the French engineer, Antoine Chezy in 1769.

Chezy assumed that the force resisting the flow per unit area of the stream bed is proportional to the square of

the velocity:

τo ∝ V2

θ

dx

Area=A ρg Adx Sinθ τo Pdx

Wetted

Perimeter=P



τo = K V2 = ρg R S K is a constant of proportionality

V = RSK

gρ

V=C RS Chezy equation or flow rate Q=AC RS

Where V is the mean velocity of the flow and C is the Chezy coefficient and has the dimensions of TL . C

may be expected to depend on the Reynolds number 4 νVR in smooth channels and on the relative roughness,

ks/R in rough channels. It varies widely over a typical range of 20 (rough channels) to 70 (smooth channels).

Several empirical methods for the prediction of C are available. The most commonly used method is due to the

Irish Engineer, Robert Manning.

MANNING EQUATION:

Robert Manning in 1889 found that:

n

RC

6

1

=

SRn

V 3

21

= Manning equation (SI Units)

or flow rate 2

1

3

2

SRn

AQ =

where n is a roughness factor. Ven T. Chow's book gives examples of values for n.



Typical values of Manning's roughness n.

Type of surface Description n

Concre

te

Perspex open channel flume 0.01

Culvert, with bends, connections and some debris 0.013

Cast on steel forms 0.013

Cast on smooth wood forms 0.014

Unfinished, rough wood form 0.017

Excavate

d o

r

dre

dg

ed

channe

ls

Earth, after weathering, straight and uniform 0.022

Gravel, uniform section, clean 0.025

Earth bottom, rubble sides 0.030

Rivers 0.03-0.05

Floodplains 0.03-0.1

DARCY-WEISBACH EQUATION:

The headloss due to friction in steady uniform flow is given by:

For pipes:

g

V

D

Lλh f

2

2

=

For non-circular sections:



Dπ

Dπ

P

AR

4

2

== D =4R

gR

Vλ

L

h f

8

2

= gR

VλS f

8

2

= RSg8

Vλ

=

Darcy-Weisbach Equation

It can be noted that when comparing Chezy Eq. with Manning, and Darcy Weisbach Eqs., we obtain: C =

n

R 6

1

=λ

g8

The values of λ can be obtained using the Moody Diagram, see below.

• Laminar flow: Flow is considered to be laminar, if Reynolds number Re= 20004

≤ν

VR. In

laminar flow the relative roughness has no influence on the friction factor, therefore Re

64=λ

• Smooth turbulent zone in which the friction factor is a function of Reynolds number only.

51.2

Relog2

1 λ

λ=

• Transitional turbulent zone in which λ is a function of relative roughness k/4R, and Re.

• Rough turbulent zone in which λ is a function of relative roughness k/4R, only.

k

R

λ

8.14log2

1=

Colebrook-White equation for smooth and rough turbulent zone:

]4

51.2

8.14[log2

1

λVR

ν

R

k

λ+−=



Moody Diagram



THE CONCEPT OF "NORMAL DEPTH":

In open channel flow when the body weight force is balanced by bed friction, then we have uniform flow. During

uniform flow, the depth of flow is known as "normal depth".

Examples of Normal Depth:

If channel slope is constant, flow will reach normal depth, yn, sooner or later.

THE CALCULATION OF NORMAL DEPTH:

The value of normal depth can be determined from Manning's equation 2

1

3

2

SRn

AQ = (or Chezy or Darcy-

Weisbach equation).

yn Body weight

Friction

yn yn



The following methods are available for calculating normal depth:

1. Direct calculation (only for wide channels).

2. From a graph.

3. Using Trial and error.

4. Using Newton-Raphson iterative technique.

1. NORMAL DEPTH (SPECIAL CASE; VERY WIDE RECTANGULAR CHANNEL):

In a wide rectangular channel (B>12y):

yyB

By

P

AR ≈

+==

2

2

1

3

2

SRn

AQ = 2

1

3

2

Syn

ByQ n

n= 2

1

3

5

Syn

BQ n=

6.0)(

SB

Qnyn =

or 6.0

n )S

qn(y = where q =dischrge per unit width.

Alternatively one can calculate Q, if given yn, S, n.

2. THE CALCULATION OF NORMAL DEPTH USING GRAPH:

If B is known, use Fig. 6.1 from the Ven T. Chow book (shown below).

• Calculate

SB

Qn

3

8

• From Manning equation we know:

3

8

3

2

3

8

B

AR

SB

Qn=

B

y



• Using Fig. 6.1 and

3

8

3

2

B

AR, one can find

B

yn , hence yn.

3. THE CALCULATION OF NORMAL DEPTH USING TRIAL AND ERROR:

2

1

3

2

SRn

AQ = 3

2

)(P

AA

S

Qn=

3

2

3

5

P

A

S

Qn=

3

2

3

5

3

5

)2( n

n

yB

yB

S

Qn

+

=

3

2

3

5

3

5

)2( nn yBS

QnyB +=

If B is known, try different values of yn until LHS=RHS (Trial & Error Method). This method is the most suitable

for use in exams.

4. THE CALCULATION OF NORMAL DEPTH USING NEWTON-RAPHSON METHOD:

The most suitable method for solution by computer or spreadsheet is the Newton-Raphson technique, an

Iterative Method

We have seen:

3

2

3

5

3

5

)2( nn yBS

QnyB +=

2

3

↑ )2()( 2

3

2

5

2

5

nn yBS

QnyB +=

)2

1()( 2

3

2

5

B

y

SB

Qny n

n +=



0)()2

()( 5.15.15.2 =−−SB

Qn

B

y

SB

Qny n

n

Assume:

5.1

1 )(SB

QnK =

BSB

QnK

2)(

5.1

2 =

then:

012

5.2 =−− KyKy nn 12

5.2)( KyKyyf −−= 2

5.15.2)( Kyyf −=′

Newton-Raphson:

1st Iteration:

)(

)(

1

112

yf

yfyy

′−= Continue till LHS=RHS or y=yn

(or till error is very small)



TUTORIAL 1

1. A rectangular channel is 3m wide and has a slope of 1/500.

(a) If the discharge is 10 m3/sec and the normal depth of the flow is 1 m determine the roughness coefficient

'n' and state whether the flow is subcritical or supercritical.

(b) If the normal depth of the flow is 0.5 m determine the discharge. Is flow subcritical or supercritical?

(c) If the discharge is 25 m3/sec determine the normal depth of flow and state whether this is subcritical or

supercritical.

(d) At approximately what normal depth of flow would critical conditions occur? Assume Q=25 m3/s and the

bed slope is changed. (n = 0.00952).

Ans: (a) n=0.00952, flow supercritical

(b) Q=3.67 cumecs, flow supercritical

(c) Yn =1.975 m, flow subcritical

(d) Yn = 1.92 m, S = 1 in 465

2. A canal has a trapezoidal cross-section with side-slopes 1:1 a base width of 2.5m and a uniform bed slope of 1/400. If the

Chezy coefficient is 100, determine:

(a) The discharge if the depth of uniform flow is 2 m.

(b) The equivalent value of Mannings 'n'.

(c) If 'n' remains constant the normal depth of flow for discharge of 15 m3/sec.

Ans: (a) Q = 47.3 m3/sec

(b) n = 0.01

(c) yn = 1.072 m



3. An open channel of trapezoidal section 2.5 m wide at the base and having sides at 60o to the horizontal has a bed slope of

1/500. It is found that when the flow 1.24 m3/sec the depth of water in the channel is 0.35m. Assuming the validity

of Mannings equation calculate the flow rate when the depth is 0.5m.

Ans: Q = 2.215 (m3/sec)

3. A rectangular channel bed ABC is 3 m wide and consists of a portion AB of slope 1/100 and BC of 1/900 slope. The

depth of flow on AB is 0.8 m. If Mannings 'n' = 0.02 discuss the nature of the flow on AB and find the normal depth

on BC.

Ans: YnBC = 1.78 m.



THE ENERGY CONCEPT IN OPEN CHANNEL

FLOW:

TOTAL ENERGY& SPECIFIC ENERGY

Total energy relative to a fixed datum is:

g

V

gρ

PZ

2

2

++ as in Bernoulli's Equation

For hydrostatic pressure this becomes: g

VyZ

2

2

++

In open channel hydraulics we often make the datum along the channel bed and the energy is termed "SPECIFIC

ENERGY". Specific Energy, E, is defined as the energy of the flow per unit weight of fluid referred to the channel

bed as datum.

g

VyE

2

2

+=

Datum

Z

Bed y

gρ

P=

Water Surface

Energy Line

V2/2g



One can apply the energy principle between any two cross-sections in the flow:

f

2

222

2

111 h

g2

VyZ

g2

VyZ +++=++

hf is the total energy loss between sections 1 & 2 mainly composed of bed friction and is equal to dxSf for

shallow slopes. Z1-Z2 is the change in bed slope and is equal to dxSo for shallow slopes.

Hence dxSEdxSE f2o1 +=+

fo SS

Edx

−∆

=

where fS is the average slope of the energy line between any two sections in the flow.

Energy Line

Datum

Z2

Bed 2y

Water Surface

V22/2g

y1

Z1

V12/2g

fS

dxSf

dx

1 2



CONCEPT OF MINIMUM SPECIFIC ENERGY:

If we consider a channel with a constant discharge Q then depending on the channel slope (hence depth y) there

is a depth where the energy is minimum.

Note: The point of minimum energy is the critical depth.

For Minimum energy 0dy

dE=

g

VyE

2

2

+= 2

2

gA2

QyE +=

dy

dA

gA

Q1

dy

dE3

2

−=

3

2

gA

BQ10 −= 1

gA

BQ3

2

= 1gA

BV2

=

In a Rectangular Channel:

A=By

1gA

BV2

= 1gy

V 2

= gyV =

E = y + V2/2g

Specific Energy

y

V=Q/A dA/dy=B

In terms of

B

y Area=By

Q=Constant

Minimum

Energy

y y

Q=constant

45o



The point of minimum energy is the critical depth crcr gyV =

CRITICAL DEPTH (RECTANGULAR CHANNEL):

1gA

BQ3

2

= 1ygB

BQ33

2

= 1ygB

Q32

2

=

g

qy

23

cr = 3

2

crg

qy =

Examples:

Note: What is the difference between normal and critical depth of flow? See sketch below.

A=By q = Q/B

q=Discharge/Unit Width

ycr

ycr

Subcritical Supercritical

Subcritical Supercritical



Minimum Energy (Rectangular Channel):

g2

VyE

2

cr

crcr += 2

yyE cr

crcr += crcr y5.1E =

when

V>Vcr ; (y<ycr) Flow is supercritical, rapid 1gy

VFr >=

V<Vcr ; (y>ycr) Flow is subcritical, tranquil 1gy

VFr <=

V=Vcr ; (y=ycr) Flow is critical 1gy

VFr ==

Fr is know as the Froude Number.

Depths y1 and y2 have equal energy and are conjugate depths but with different Froude Numbers.

crcr gyV =

Minimum

Energy

45o

Ecr

g2

VyE

2

+=

Subcritical

y

y2

ycr

y1



GRADUALLY VARIED FLOW

THE GENERAL EQUATION FOR GRADUALLY VARIED FLOW IN A RECTANGULAR

CHANNEL:

Total energy (H) relative to a fixed datum is:

g2

VyZH

2

++=

∴ )g2

V(

dx

d

dx

dy

dx

dZ

dx

dH 2

++=

where dx

dH is slope of energy line -Sf

dx

dZ is the slope of bed -So

∴

So - Sf = )gA2

Q(

dx

d

dx

dy2

2

+ )gA2

Q(

dA

d

dx

dA)

gA2

Q(

dx

d2

2

2

2

=

3

2

3

2

gA

Q

dx

dy

dy

dA

gA

Q

dx

dA−=−=

3

2

gA

BQ

dx

dy−=

Datum Z

Bed

ygρ

P= Water Surface

Energy Line

V2/2g



∴

So - Sf 3

2

gA

BQ

dx

dy

dx

dy−=

3

2

1gA

BQ

SS

dx

dy fo

−

−=

This is the General Equation for gradually varying flows predicting the change in flow depth in the downstream

direction.

A few points about the equation:

(1) What happens when So = Sf ?

The flow would be uniform because 0dx

dy= .

(2) What happens when 0gA

BQ1

3

2

=− ?

The flow depth would be critical ycr; but ∞=dx

dy at that point. This is a point of discontinuity in the equation and

therefore it can not be used at critical depth.

y

x

yc

Shallow Slope Steeper Slope

y x

Sf

So



RULE:

Start near ycr and work upstream towards subcritical flow.

Start near ycr and work downstream towards supercritical flow.

∴ This is therefore two separate calculations.

(3) The term Q2B/gA

3 often approximated for a rectangular section by:

A=By ; Q=q B

∴

3

2

3

32

3

2

gy

q

gA

Bq

gA

BQ===

From Lecture 3: 3

2

crg

qy =

∴

3cr

3

2

3

2

)y

y(

gy

q

gA

BQ===

and the general equation becomes:

3cr

fo

)y

y(1

SS

dx

dy

−

−= For rectangular channels only

B

y



(4) There is a further approximation used when the river or estuary is considered very wide. For B/y>12, the

hydraulic radius R ≅ y (flow depth).

From Manning equation

o3

2

SRn

AQ =

For uniform flow A = Byn ; R = yn

∴ 3

10

n

2

22

o

yB

QnS =

and for non-uniform gradually varying flow:

f3

2

SRn

AQ =

3

10

2

22

f

yB

QnS =

∴ )S

S1(SSS

o

fofo −=−

])y

y(1[S 3

10

n

o −=

and finally the general equation approximates to:



]

)y

y(1

)y

y(1

[Sdx

dy

3cr

3

10

n

o

−

−= For wide rectangular channels only

This equation could be applied between any two cross-sections:

−

−=

∆∆

3cr

3

10

n

o

)y

y(1

)y

y(1

Sx

y

A simple example of flow passing under a sluice gate on a concrete spillway:

The spillway slope is 1/1000 ∴ So= 0.001

Mannings 'n' for the spillway n = 0.015

The spillway considered very wide and q = 1.71m3/s/m

2y

Water Surface y1

∆x

Sluice

0.3m 0.2m

∆x



Step 1: Calculate yn from Manning's equation.

Step 2: Calculate ycr from criterion for critical depth.

m67.0g

qy 3

2

cr ==

Step 3: From y=0.2 y=0.3 ∴ ∆y=0.1m,

m25.0y =

−

−=

∆ 3

3

10

o

)25.0

67.0(1

)25.0

88.0(1

Sx

1.0

∆x=27.7m

m88.0S

nqy

6.0

o

n =

=



CALCULATION OF GRADUALLY VARIED FLOW:

The methods used in practice are:

(1) Standard Step Method (not covered)

(2) Direct Step Method

DIRECT STEP METHOD:

2

SSS 21 ff

f

+= ;

2

SSS 21 oo

o

+=

Energy principle is:

dxSg2

Vy

g2

VydxS f

2

2

2

2

1

1o ++=++

2f1o EdxSEdxS +=+

fo SS

Edx

−∆

=

Energy Line

Bed 2y

Water Surface

V22/2g

y1

V12/2g

fS

dxSf

dx dxSo



TABULAR SOLUTION FORM

Y A P R R4/3

V V2/2g E ∆E Sf Sf So-Sf ∆x Σx

BY B+2Y A/P Q/A Y+V2/2g 0

∆E Sf ∆x

BY B+2Y A/P Q/A Y+V2/2g 0+∆x

Note: When doing direct step method pay no attention to sign. It is always positive.

Known

Point

Next

value

of y

3

4

22

R

Vn

3

4

22

R

Vn



CLASSIFICATION OF GRADUALLY VARIED FLOW PROFILES:

Classification by slope: (Five Slopes)

Classification By Region (Three Regions)

• Region 1 y>yn and ycr

• Region 2 y between yn and ycr

• Region 3 y less than yn and ycr

yn

ycr

Mild (M)

ycr

yn

Steep (S)

yn =∞

ycr

yn negative! can not exist

Adverse (A)

Horizontal (H)

ycr

ycr yn =

Critical (C)



ALL POSIBILITIES:

H1 H2 H3

M1 M2 M3

C1 C2 C3

S1 S2 S3

A1 A2 A3

Can not Exist

Does not exist

Uniform Flow

M3

Steep Slope

ycr

yn

So>Scr

S3

S2

S1

Asymptotes

To Horizontal

Horizontal Slope

ycr

(N.B. No H1 as yn=∞) So = 0

H2

H3

Asymptotes

To Horizontal

yn = y cr

(N.B. C2 equivalent to uniform flow)

C3

C1

So = Scr

Mild Slope

yn

ycr

So < Scr

M2

M1

Asymptotes

To Horizontal

Asymptotes

To yn line



Some Examples in Civil Engineering:

ycr

(N.B. A1 does not exist)

Adverse Slope

Horizontal

Asymptote

A2

A3

Mild Slope (M)

ycr

yn M3

M1

Mild

Steep yn

Hydraulic Jump

Sluice Gate yn

ycr

M1

Mild

M3

Hydraulic Jump



ycr

yn M2

ycr

S2

Sluice Gate

yn

ycr S1

Steep

S3

yn

ycr

Hydraulic Jump



GVF TUTORIAL

1. A large tank in a fish farm is supplied with fresh water by a 1m wide rectangular channel leading to a

control sluice gate and then into the tank, as shown in Fig. 1. Assuming the water level just upstream of

the sluice gate is lm deep, and the supply channel is laid horizontally with Manning's 'n' value 0.012, use

the Direct Step Method in 0.1m depth Increments to determine the water level at A, 946m upstream of

the sluice gate. The discharge along the channel is 1m3/s.

2. A long rectangular channel with a mild slope of 1 in 900 suddenly changes to a steep slope of 1 in 4.,

The channel is 5 m wide throughout, has a Manning's n value of 0.02 and is designed to carry a

maximum discharge of 155 m3/s. Assuming critical conditions exist at the junction of the two slopes,

calculate the water surface profile for approximately 150 m upstream of the Junction using depth

increments of 0.5 m throughout.

3. A wide rectangular river slopes at 1 in 1600 carrying a discharge per unit width of 2 m2/s with a

Manning's "n" value of 0.025. An underflow sluice gate is placed in the river across its entire width, the

bottom of the sluice gate being 0.3 m above the channel bed. Calculate the water surface profile from the

sluice gate to the point upstream where normal depth occurs and assume no energy losses at the sluice

gate.

4. A rectangular channel 3m wide, slopes at 1 in 40 and carries a total discharge of 5 m3/s when Manning's

"n" value is 0.02. A sluice gate is placed across the channel width with the bottom of the gate 0.2m

above the channel bed. Use the Direct Step Method to calculate the water surface profile downstream of

the sluice gate. Your depth increments should be smaller than 0.1m.

Ans : Q1 – 1.4m; Q2 – 6.3m approx; Q3 – 4.3km; Q4 – 85.4m

Sluice gate

Large Tank 946m

A

Q=1m3/s

1m



RAPIDLY VARYING FLOW

In rapidly varied flows if we are considering short reaches and small energy losses then the energy method is

used. For large or unknown energy losses, force-momentum principle is used e.g. hydraulic jump.

RAPIDLY VARIED FLOW- ENERGY METHOD:

Rapidly varied flow occurs whenever there is a sudden change in the geometry of the channel over a short

length. Typical examples are flow over broad-crested weirs and flow through venturi flumes.

Flow over broad-crested weir is considered, with two possible flow regimes.

Ecr is dominant energy: E d/s is dominant energy:

Ecr = h + 3/2ycr > E d/s= 2

n

2

ngy2

qy + E d/s= 2

n

2

ngy2

qy + > Ecr = h + 3/2ycr

yn is the u/s and d/s of weir and the depth on the top of

the weir is given by:

Ecr = 2

2

ngy2

qy + Ed/s = 2

weir

2

weirgy2

qyh ++

yd/s

Ecr E d/s

yn y u/s

ycr

h

E d/s

Ecr

yn

y n yweir

h



There are three roots to above equation:

• 1 negative ×

• 1 positive large = y u/s

• 1 positive small = y d/s

WORKED EXAMPLE OF RAPIDLY VARIED FLOW USING ENERGY METHOD:

A long rectangular channel with no friction loss has a normal flow depth of 1m when the discharge is 7m3/s. The

channel is 3m wide and there are constrictions inside the channel as shown below. Calculate the water level and

the energy line in the channel.

A represents a venturi flume contracting to 1.5 m in width.

B is a venturi flume of width 2.5m with a small weir at the channel bed 0.15m high

C is a weir 0.5m high

There are three roots for yweir:

• negative ×

• positive large = y weir

• positive small ×

3m 2.5m 1.5m 3m

0.15m 0.5m

1m

A B C

SECTION

PLAN



Energy Levels:

1. Downstream energy: Ed/s= 2

n

2

ngy2

qy + =

( )2

2

)1(g2

3/71+ =1.276m

2. 0.5m weir 3

2

crg

qy = = 3

2

g

)3/7(=0.826 Ecr=3/2(0.826)+0.5=1.739m

3. 0.15 weir 3

2

crg

qy = = 3

2

g

)5.2/7(=0.906 Ecr=3/2(0.906)+0.15=1.51m

4. 3

2

crg

qy = = 3

2

g

)5.1/7(=1.308 Ecr=3/2(1.308)=1.962m

Ecr 0.5weir > E d/s ∴ 1.739m dominates control section at C

Ecr 0.15weir < Ecr 0.5weir ∴ 1.739m dominates flow at B

Ecr venturi > E cr weirs 1.962> 1.839 ∴ control section at A

Determination of water levels:

1. Critical flow over 0.5m weir yc=0.826m WL=0.826+0.5

2. U/S of 0.5m weir Ecr = 2

2

gy2

qy + , 1.739=

2

2

gy2

)3/7(y +

y sub.=1.63m ; y sup.=0.466m

3. Over 0.15m weir

Ed/s = 2

weir

2

weirgy2

qyh ++ ; 1.739=0.15+

2

weir

2

weirgy2

)5.2/7(y + yweir=1.37m



U/S Weir =1.63m as D/S

4. At venturi ycr=1.308m

Ecr = 2

2

gy2

qy + ; 1.962=

2

2

gy2

)3/7(y + y

sub.=1.89m ; y sup.=0.44m

0.15m

0.466m

1.37m

0.5m 0.44m

1.308m

1.63m

1m

Ed/s=1.276m

Energy Line Eu/s

=1.962m Ecr=1.739m

1.89m 0.83m 1.63m



THE CONCEPT OF FORCE-MOMENTUM IN OPEN CHANNEL FLOW:

According to Issac Newton law of motion: Net Force=Rate of change of momentum

= Vdt

dm∆ =ρQ(∆V)

P P mgSin pdx Q V Vo1 2 2 1− + − = −θ τ ρ ( )

For a horizontal reach mg Sinθ 0

For a short distance τopdx is small

∴ P1-P2=ρQ(V2-V1)

P1=hydrostatic force=1

21

2

1 1 1ρ ρgy B gA Z= where Z1 is distance to the centroid of area

P2=1

22

2

2 2 2ρ ρgy B gA Z=

∴ ρg(A1 Z1- A2 Z2)= ρQ ( )Q

A

Q

A2 1

− = ρ ρQ

A

Q

A

2

2

2

1

−

dx

θ

mg Sinθ

τo Pdx P2

P1

V1 V2

+



or Q

gAA Z

Q

gAA Z

2

1

1 1

2

2

2 2+ = + = Specific Force

Specific Force 1 = Specific Force 2

A jump forms when SF1 = SF2

For rectangular channels, often expressed as per unit width:

q B

gAB y

y q B

gAB y

y2 2

1

1 1

1

2 2

2

2 2

2

2 2+ = +

Specific force per unit width:

q

gy

y q

gy

y2

1

1

2 2

2

2

2

2 2+ = +

Applied to a Hydraulic Jump in Rectangular Channel:

q

gy

y q

gy

y2

1

1

2 2

2

2

2

2 2+ = +

Specific Force 1 = Specific Force 2

y1

y2



If q and y1 are known, then y2 can be calculated. i.e. K2

y

gy

q2

1

1

2

=+

Kq

gy

y= +

2

2

2

2

2 Ky

q

g

y2

2

2

3

2= +

0 5 02

3

2

2

. y Kyq

g− + = Solve by trial and error or Newton-Raphson

RAPIDLY VARIED FLOW- FORCE MOMENTUM METHOD:

Significant energy losses over short distances. Apply Newton 2nd

law i.e. Net force = rate of change of

momentum.

PER UNIT CHANNEL WIDTH

y

Specific

Energy (E)

y

Specific

Force (SF)

∆E y2

y1



(i) Forces 1

2

1

22

2

1

2ρ ρ τ θgy gy pdx mgSin− + −

consider slope horizontal mgSinθ = 0

Bed friction small τ = 0

(ii) Rate of change of momentum (per unit channel width)

mdV

dtq V V= −ρ ( )1 2

(iii) 1

2

1 12

2

1

2

1 2

2

1 2

ρ ρ ρg y y q V V qy y

( ) ( ) ( )− = − = −

gy y y y q y y y y

22 1 2 1

2

2 1 1 2( )( ) ( ) /− + = −

2

2

1

2

211221

2

yyyy)yy)(yy(g

q2+=+=

(iv) Assume q and y1 are known, then solve for y2

y y y yq

g1 2

2

1

2

2

220+ + =

ay by c2

2

2 0+ + =

y1

y2

F1

F2=1/2ρg y22

ρg y1 ρg y2

Body

Weight

τ

dx



∴ yb b ac

a

y y y q g

y2

21

2

1

4

1

2

1

4

2

8

2=

− ± −=

− + + /

(v) yy y q gy

y

y y q gy2

1

2

1

2 2

1

3

1

1 1

2

1

31 8

2

1 8

2=

− + +=

− + +/ /

or ( )y

y

V

gyFr2

1

1

2

1

1

2

1

21

21

21

81

1

21 8 1= +

−

= + −



HYDRAULIC JUMP EQUATIONS AND

ENERGY DISSIPATORS:

Practical applications of the hydraulic jump are many; it is used

(1) to dissipate energy in water flowing over dams, weirs, and other hydraulic structures and thus prevent

scouring downstream from the structures

(2) to recover head or raise the water level on the downstream side of a measuring flume and thus maintain

high water level in the channel for irrigation or other water-distribution purposes;

(3) to increase weight on an apron and thus reduce uplift pressure under a masonry structure by raising the

water depth on the apron;

(4) to increase the discharge of a sluice by holding back tailwater, since the effective head will be reduced if

the tail water is allowed to drown the jump;

(5) to remove air pockets from water-supply lines and thus prevent air locking

JUMP IN HORIZONTAL RECTANGULAR CHANNELS:

For supercritical flow in a horizontal rectangular channel, the energy of flow is dissipated through frictional

resistance along the channel resulting in a decrease in velocity and an increase in depth in the direction of flow.

A hydraulic jump will form in the channel if the Froude number Fr1, of the flow, the flow depth y1, and a

downstream depth y2 satisfy the equation:

( )[ ]1Fr812

1

y

y 2

1

1

2 −+=

This is known as the Belanger

equation.



This equation may be represented by the curve in Fig. 1. This curve has been verified satisfactorily with many

experimental data and will be found very useful in the analysis and design for hydraulic jumps.

Fig 1. Relation between F1 and y2/y1 for a hydraulic jump in a horizontal rectangular channel.

TYPES OF JUMP

Hydraulic jumps on horizontal floor are of several distinct types. According to the studies of the U.S. Bureau of

Reclamation, these types can be conveniently classified according to the Froude number Fr1 of the incoming flow

(Fig. 2), as follows:



For Fr1= 1, the flow is critical, and hence no jump can form.

For Fr1= 1 to 1.7, the water surface shows undulations, and the jump is called an undular jump.

For Fr1= 1.7 to 2.5, a series of small rollers develop on the surface of the jump, but the downstream water surface

remains smooth. The velocity throughout is fairly uniform, and the energy loss is low. This jump may be called a

weak jump.

For Fr1= 2.5 to 4.5, there is an oscillating jet entering the jump bottom to surface and back again with no

periodicity. Each oscillation produces a large wave of irregular period which, very commonly in canals, can travel

for miles doing unlimited damage to earth banks and ripraps. This jump may be called an oscillating jump.

For Fr1= 4.5 to 9.0, the downstream extremity of the surface roller and the point at which the high-velocity jet

tends to leave the flow occur at practically the same vertical section. The action and position of this jump are

least sensitive to variation in tailwater depth. The jump is well-balanced and the performance is at its best. The

energy dissipation ranges from 45 to 70%. This jump may be called a steady jump.

For Fr1= 9.0 and larger, the high-velocity jet grabs intermittent slugs of water rolling down the front face of the

jump, generating waves downstream, and a rough surface can prevail. The jump action is rough but effective

since the energy dissipation may reach 85%. This jump may be called a strong jump.

It should be noted that the ranges of the Froude number given above for the various types of jump are not clear-

cut but overlap to a certain extent depending on local conditions.



Fig 2. Various types of hydraulic jump.

BASIC CHARACTERISTICS OF THE JUMP:

Several basic characteristics of the hydraulic jump in horizontal rectangular channels are to be discussed below:

Energy Loss

The loss of energy in the jump is equal to the difference in specific energies before and after the jump. It can be

shown that the loss is



21

3

12

yy4

)yy(2E1EE

−=−=∆

The ratio ∆E/E1 is known as the relative loss.

Efficiency

The ratio of the specific energy after the jump to that before the jump is defined as the efficiency of the jump. It

can be shown that the efficiency is

)Fr2(Fr8

1Fr4)1Fr8(

E

E2

1

2

1

2

12

3

2

1

1

2

+

+−+=

This equation indicates that the efficiency of a jump is a dimensionless function, depending only on the Froude

number of the approaching flow.

The relative loss is equal to (1 - E2/E1); this also is a dimensionless function of Fr1.

Height of Jump

The difference between the depths after and before the jump is the height of the jump, or hj = y2 - y1. Expressing

each term as a ratio with respect to the initial specific energy,

1

1

1

2

1

j

E

y

E

y

E

h−=

where

1

j

E

h, is the relative height,

1

1

E

y is the relative initial depth, and

1

2

E

y, is the, relative sequent depth. All

these ratios can be shown, to be dimensionless functions of Fr1. For example:



2Fr

3Fr81

E

h

2

1

2

1

1

j

+

−+=

Fig 3. Characteristic curves of hydraulic jumps in horizontal rectangular channels.

Since the relative, loss, efficiency, relative height, and relative initial and sequent depths of a hydraulic jump in a

horizontal rectangular channel are functions of Fr1, they can be plotted against Fr1 resulting in a set of

characteristic curves (Fig. 3). With reference to these curves, the following interesting features may be noted:

• The maximum relative height hj/E1 is 0.507, which occurs at Fr1 = 2.77.

• 2. The maximum relative depth y2/E, is 0.8, which occurs at y1/E1 = 0.4 and Fr1= 1.73.

• Experiments have shown that the transition from an undular jump to a direct jump takes place approximately at this point Fr1 =1.73.

• When Fr1= 1, the flow is critical and y1 = y2 = 2/3E1.



• When Fr1 increases, the changes in all characteristic ratios become gradual.

The characteristic curves will provide the designer with a general idea about the range of conditions under which

the structure is to be operated. For instance, in the design of a sluice gate involving a jump below the gate, such

curves will show clearly the formation of the jump for different gate openings under a given head. The above

discussion applies to horizontal rectangular channels. For horizontal nonrectangular channels, similar curves may

also be prepared.

Length of Jump

The length of a jump may be defined as the distance measured from the front face of the jump to a point on the

surface immediately downstream from the roller. This length cannot be determined easily by theory, but it has

been investigated experimentally by many hydraulicians.

The experimental data on length of jump can be plotted conveniently with the Froude number Fr1, against a

dimensionless ratio L/(y2 -y1), L/y1, or L/y2. The plot of Fr1, vs. L/y1, is probably the best, for the resulting curve can

be best defined by the data. For practical purposes, however, the plot of Fr1, vs. L/y2 is desirable, because the

resulting curve shows regularity or a fairly flat portion for the range of well established jumps.

A curve of Fr1, vs. L/y2, (Fig. 4) based on the experimental data of six test flumes has been prepared by the

Bureau of Reclamation. The curve shown in Fig. 4 was developed primarily for jumps occurring in rectangular

channels. In the absence of adequate data, this curve may also be applied approximately to jumps formed in

trapezoidal channels.



Fig 4. Length

in terms of sequent depth y2 of jumps in horizontal channels.

(Based on data and recommendations of U.S. Bureau of Reclamation).



LOCATING THE HYDRAULIC JUMP

Occurs when open channel flow goes from supercritical to subcritical flow through critical depth.

• Cannot use energy principle.

• Need to use force momentum on a control volume.

• Ignoring body weight and bed friction, including hydrostatic forces only.

• Per unit width

)VV(qgy2

1gy

2

121

2

1

2

2 −=− ρρρ

)y

q

y

q(

g

q)

2

y

2

y(

21

2

1

2

2 −=−

q

gy

y q

gy

y2

1

1

2 2

2

2

2

2 2+ = + =Specific force/unit width

• Jump forms when equal specific forces exist either side of jump.

(6) Simpler case when y2 (jump) equals normal depth yn downstream e.g. flow under a sluice gate.

• yn can be calculated using Manning's Equation

• use q

gy

y q

gy

y2

1

1

2 2

2

2

2

2 2+ = + and find y1 by trial& error

• Use Direct Step Method, GVF profiles yg to y1 to find ∆x.

(7) More complex case where y2 >yn

• Calculate both GVF profiles separately

dx θ

mg Sinθ

τo Pdx P1

P2

V1 V2

y2=yn

Sluice

yg y1

∆x



• Calculate specific forces for both profiles + Plot

on graph

(8) Consider jump drowning back SF2>SF1 Jump moves upstream

SF1>SF2 Jump moves downstream

Example:

There is a hydraulic jump downstream of a sluice gate (see below). The flow discharge is 12m3/s. The channel is

rectangular, 7m wide, and has a horizontal slope. The water surface profiles for H3, and H2 are given below. Find

length and position of the jump.

Sluice

yg

H2

∆x

ycr

H3

So=0

Sluice

0.2m

H2

∆x

ycr

H3

So=0

Profile H3

Dist. From Weir (m) 0 1 2 2.5

y(m) 0.2 0.3 0.44 0.56

Profile H2

Dist. From Weir (m) 1 2 3 4

y(m) 0.98 0.95 0.92 0.88



Calculate specific forces for both profiles:

Specfic force/Unit width=2

y

gy

q 22

+ e.g. for y=0.3 SF=2

3.0

)3.0)(81.9(

)7/12( 22

+ =1.044

The depth in which the specific forces are equal is y1=0.44m. This depth, y1, occurs at 2m from weir.

Froude Number, 87.144.081.9

1

744.0

12

gy

A/QF

1

1 =××

==

From plot of F1 versus L/y2:

L/y2=4.2 L=4.2×0.95 ≈ 4m (Length of the jump)

Profile H3

Dist. From Weir (m) 0 1 2 2.5

y(m) 0.2 0.3 0.44 0.56

Profile H2

Dist. From Weir (m) 1 2 3 4 6 7.5

y(m) 0.98 0.95 0.92 0.88 0.78 0.68



HYDRAULIC JUMP TUTORIAL

1. A very wide rectangular channel with Manning’s n value of 0.0159 has a steep slope of 1/40 followed by

a horizontal reach 133 m long, terminating in a vertical drop. If the critical depth at this drop is 1 m,

determine whether the jump will form upstream or downstream of the change in slope. Use depth

increments of 0.2 m in any numerical integration.

Ans. Depth at end of H2 profile = 1.4 m, Jump forms downstream.

2. Write down an expression for the rate of change of depth of steady non-uniform flow in a long prismatic

channel and deduce what simplifications follow from the assumption of (a) a rectangular cross-section

and (b) a very wide cross-section.

A very wide rectangular channel carries a discharge per unit breadth of 2.5m3/sec/m with a normal depth

of 1.5 m. The slope of the bed is 1 in 1000. An underflow sluice gate maintains an upstream head of 5.5

m and the breadth of the piers may be ignored. Show that a hydraulic jump will occur, and determine its

position assuming the value of Manning’s ‘n’ to remain constant. (Use depth increments of approx. 0.1 m

in any numerical integration, Neglect losses through the gate).

Ans. Jump forms 21 m downstream.

3. A channel 3 m wide with vertical sides has a uniform slope of 1 in 1600 and a Manning’s n value of

0.013. If the depth of flow in a long uniform reach is 1 m calculate the discharge. Streamlined screens

with an effective area of 60% are to be introduced into the channel about midway along its length.

Determine whether or not the screens form a control section and if so find the depth of flow 640 m

upstream. (Use depth increments of 0.1 in). Indicate what form the surface profile is likely to take

downstream of the screens.

Ans. Screens form control, depth = 1.04 m.

4. A broad rectangular channel carries 1 m3 /sec/m width at a normal depth of 1.5 m. A pipe 0.5 m diameter

spans across the channel resting on the channel bed. Assuming all the losses to be concentrated at the

downstream side of the pipe and a hydrostatic pressure distribution throughout, find the average velocity

at the pipe and the depth upstream of the pipe.

Ans. Depth at pipe = 1.475 m. Vel. = 1.025 m/s U/s depth = 1.50 m.

Water Eng OCF Notes

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