OPEN CHANNEL FLOW: quantitative profiles in complex channels

M. Pilotti - lectures of Environmental Hydraulics

Initially, in order to understand what is the actual applicability scope of uniform motion we asked how much

far away is far ? When an M1 profile is considered a first guess can be provided by the following table, that is

valid for infinitely wide rectangular channel, according to Bresse’s solution



Direct step method

(distance calculated from depth)

fb SSdx

dE −=

y1 at the position x1 along the channel is known as a boundary condition. From the qualitative discussion of

the profile, one knows what is the asymptotic depth (e.g., if M1, it will tend to h0). Accordingly one selects in an

adaptive way a depth value y1 < yi+1 < h0 for the section at the unknown station xi+1, computing the

corresponding values Ei+1 and (Sf)i+1 (that depend only on yi+1 in prismatic channels). The unknown station xi+1

is obtained by discretization of the dynamic equation. If Fr < 1 then x grows from downstream to upstream

[ ]1

11

)()(21

+

++

+−

−−=ififb

iiii

SSS

EExx⇒[ ] )()()(

2

1)( 1111 iiififiibii xxSSxxSEE −+−−=+− ++++

Let us first consider a method which is very convenient but is valid only in prismatic channel, where,

indeipendently from x, one knows A(x) and Sb(x)

If Fr > 1 then x grows upstream to

downstream and the equation is

[ ]1

11

)()(21

+

++

+−

−+=ififb

iiii

SSS

EExx



Standard step method

(depth calculated from distance)

Let us now consider a general method which can be as convenient as the direct step if solved explicitly or just

a bit more complex if solved implicitly. Its scope is not limited to prismatic channels.

We know the boundary condition yi at the position xi along the channel. By a first order approximation of the

energy balance equation, we obtain :

The unknown value yi+1 at the known position xi+1 is such that F=0

This equation is non linear and must be solved by, e.g., a Newton Raphson method.

As a first guess for the iteration, a first estimate y*i+1 of yi+1 can be obtained as

that can also be used as an explicit approximation of the energy balance equation

fSdx

dH −= [ ] )()()(21

111 iiififii xxSSHH −+−=− +++⇒

)()( 1*

1*

1*

1 ++++ ′−= iiii yFyFyy

2/)]()()[()( where,0)( 11111 iiififiiii xxSSHHyFyF −++−== +++++

)()()(2 12

1

2

11 iiifiii

ii xxSHygA

Qzy −−+−−= +

+++ α

IMPLICIT

EXPLICIT



Implicit


If the channel is a prismatic channels (see classwork 3) where we know the boundary condition yi at the

position xi along the channel, the standard step method can be given as:

That is an implicit equation that provides the unknown value yi+1 at the known position xi+1.

Alternatively, the method can be written in a fully explicit way as:

Which works equally well provided the space increment xi+1 - xi are adequately small.

As shown before for the direct step method, if Fr > 1 then the integration moves from the boundary

condition, given at a point x0 upstream, to downstream, so that x grows in the same direction of the main

flow. If Fr < 1 then the boundary condition x0 is given downstream and the integration moves upstream. In

this case if x grows from the position of the boundary condition to upstream,

[ ][ ]

+−

+−

−+= +

+

++ 1

122

11 )()(

2

1

)()(21

1

)(ififb

ii

iiii SSS

FrFr

xxyy⇒21 Fr

SS

dx

dy fb

−−

=

[ ]ifbi

iiii SS

Fr

xxyy )(

)(1

)(2

11 −

−−+= +

+

Explicit


[ ][ ]

+−

+−

−−= +

+

++ 1

122

11 )()(

2

1

)()(21

1

)(ififb

ii

iiii SSS

FrFr

xxyy [ ]ifb

i

iiii SS

Fr

xxyy )(

)(1

)(2

11 −

−−−= +

+

OPEN CHANNEL FLOW: inlet of a mild/steep channel


Let us consider the outlet of a lake in a channel whose geometry is known. How can we find out the discharge Q that

flows out from the lake ? To solve the problem, as a first step we have to make an educated guess on Sb: is it mild or

steep ?

≡=

=

=

0

32

2

06/1

0

hhhgB

Qh

ShBhhkQ

c

c

boS

To provide an answer, let us refine our sensibility on this point. Let us consider an infinitely wide channel with critical

slope. It must be

From this system, we may obtain Sb either

as a function of (q=Q/B, ks): 9/22

9/10

_ qk

gS

Scriticalb =

or as a function of (h, ks): 3/12_ hk

gS

Scriticalb =

Conclusions:

1. The rougher the channel, the higher the critical slope Sb_critical : e.g., a mountain creek could have a mild slope if

it is very rough.

2. The higher the Q (or h), the smaller the critical slope Sb_critical : e.g., a channel could go from mild to steep if Q

increases

OPEN CHANNEL FLOW: inlet of a mild/steep channel


OPEN CHANNEL FLOW: inlet of a short and wide channel


Real case can be a bit more complicate but can be easily solved if one understands the rules of this iterative game.

Let us consider this case where the geometry of the channel, the water depth in the lake and the elevation of the gate

are known

As a first step, we have to make an educated guess on Sb: is it mild or steep ?

OPEN CHANNEL FLOW: inlet of a short and wide channel - mild slope case


1. According to our guess, Sb is mild

2. Solve Energy balance at 1 to find both h0 and Q under the assumption (true or false ?) of infinitely long

channel.

3. Make an energy balance between 2 and 3 to find h3 .

4. Compute the M1 and verify if the specific energy on the sill (with possible entry loss) complies with the water

stage in the lake. If not satisfied, decrease Q and go back to point 3)

5. When Q has been found, compute the M3 profile from station 2 and the M2 from station 4. Compute also the

specific force of these profiles and locate station 5.

6. Is the sluice gate submerged (i.e., the M3 profile is not present) ? If no, end of the game; otherwise go back to

3) and go on playing until convergence of Q.

OPEN CHANNEL FLOW: inlet of a short and wide channel - steep slope case


If our guess is wrong and the channel is steep…

1. Compute Q under the assumption (true or false ?) of infinitely long channel.

2. Make an energy balance between 2 and 3 to find h3 .

3. Compute the S1 and S2 profiles. Compute also the specific force of these profiles and locate the station 4 of

the hydraulic jump. If the S2 profiles exist, than go on to the following step, otherwise the outlet of the lake is

submerged by the S1 profile, the channel is not infinitely long and the discharge must be decreased, going

back to point 2)

4. Compute the S3 profile from 2 downstream to station 5.

OPEN CHANNEL FLOW: inlet of a an infinitely long wide channel


Let us consider a wide channel (Y/B << 1) which is originated from a reservoir where water is motionless

Let us suppose that the channel is infinitely long so that we can disregard the influence of boundary conditions.

In general term we can write an energy balance between the reservoir and the flow at the inlet of the channel, also

considering the presence of a local dissipation that is proportional to the kinetic energy. Q is unknown

If the slope Sb of the channel is mild, then we should have normal depth up to the channel inlet, so that we have to

solve

=

+=−

boo

oo

o

SRAQ

gA

Qh

gA

QE

χ

ξ2

2

2

2

22

Accordingly, by increasing Sb Q increases as well until

the critical condition is obtained at the inlet

If the channel is steep, then the channel inlet is a

transition through the critical depth between mild

and steep channel. The system is solved for

the critical depth and Q, that is indipendent from Sb

=

+=−

dh

dA

gA

QgA

Qh

gA

QE

c

c

cc

3

2

2

2

2

2

1

22ξ

Which is solved for

the normal depth and

Q

OPEN CHANNEL FLOW: inlet of a an infinitely long wide channel


In both cases, provided that the water stage E in the lake is known, one has to solve a non linear equation to find h

and then Q

( )g

SRkhE bo

os

21

3/42

ξ++=

( )dh

dAg

gAhE c

c

21 ξ++=

In rectangular cross section the second equation greatly simplifies, in the form

22

3 chE

+= ξ

OPEN CHANNEL FLOW: inlet of a short and wide channel - mild slope case


If the channel has a length l that is not infinitely long, then we may have a backwater (rigurgito) or drawdown

(chiamata) effect caused by the boundary condition located at the channel outlet. In this case the discharge must be

computed through an iterative procedure.

Let us suppose that the channel outlet is into another reservoir, that constrains the level of water at the channel

end, he . If the channel is mild, the discharge Q computed from the equation seen before is only an initial guess

1) Using Q, compute the critical depth hc

2) If he < hc compute a M2 profile starting from hc, otherwise either a M2 profile (hc <he < ho ) or a M1 one (he > ho )

3) Compare the computed water depth at the channel sill (inlet) with the normal depth. If it is higher, than Q must be

decreased; otherwise it must be increased.

If he = E + Sb l, then Q = 0;

If he > E + Sb l, the flow is reversed from downstream to upstream. The mild slope channel turns into an adverse slope one

OPEN CHANNEL FLOW: inlet of a short and wide channel - steep slope case


If the channel is steep, there might be a backwater effect caused by the boundary condition at the channel outlet, that

could cause an hydraulic jump within the channel stretch. If the specific force of the S1 profile is larger than the

specific force of the accelerated supercritical S2 profile, the hydraulic jump moves backward locating closer and

closer to the channel inlet where eventually there might be a submerged hydraulic jump.

This happens when he is close to the upstream energy level E

If he = E + Sb l, then Q = 0 (as before, l is the channel length).

If he > E + Sb l, the flow is reversed from downstream to upstream. The original channel turns into an adverse slope

one

Now you know what “Infinitely Long” means

OPEN CHANNEL FLOW: passage over a sill (hump, bump: soglia)


When the flow passes over an hump, several situations may happen, depending on the Froude number and on

Energy content. Locally there is a sudden curvature of the flow, the channel is not prismatic and the theory on

water surface profiles is of no use. However an energy balance can be accomplished to study this transition.

Let us first suppose that

1. no head loss is present

2. the sill height a is small with respect to the energy upstream.

3. The channel is infinitely long downstream and upstream, so that the depth of the flow approaching the sill and

downstream of it is the normal depth

If the slope is mild, water depth on the sill lowers more than the sill height. If the slope is steep, the effect of rise of

the sill bed prevails

020101 ; EEEaEHH ==+=



Sometimes the height of the sill is such that the specific energy of the normal flow of the approaching current is not

sufficient to pass over it. In such a case the flow upstream must gain energy and we have to distinguish between

mild and steep channel

AN IMPORTANT EXAMPLE: Broad crested, round nose, horizontal crest weir

• Upstream corner well rounded to prevent separation• Geometrical requirements as in figure above and in the specific publications


WEIRS: Broad crested horizontal crest weir




But an head loss is almost

inevitable so that

(0: normal flow

1: on the sill;

2: downstream;

0m: upstream)

Making an energy balance

starting downstream (2), one

sees that in a mild channel

the level upstream is

higher (M1). Depending on

The length of the channel, this

could affect Q

And in a steep channel,

Starting upstream, one

sees that the rise on the

hump is stronger and

The level downstream

Is greater than the

Normal depth (S2)

2100

210010121202 ;;;

HHEE

HHHHHHHHHHHH

m

mm

∆+∆+=∆+∆+=∆−=∆−=≡

2102

210221210100 ;;;

HHEE

HHHHHHHHHHHH mm

∆−∆−=∆−∆−=∆−=∆−=≡

OPEN CHANNEL FLOW: passage through a contraction (1)


The same situation occurring when a flow passes over an hump can be observed in the passage through a

contraction. Usually a contraction can be caused by the piers or abutments of a bridge

If no localized losses are

Present, then the specific

energy is constant



Sometimes the Energy

upstream isn’t enough…



Although one can suppose that no

head loss is present,this is not

generally true.

Accordingly, the flow must gain

energy to compensate for the

localized head loss. This happens

upstream if Fr < 1 (M1) and

downstream if Fr > 1 (S2)

The process is similar to the one

considered for the passage over a

bump

21

11

00

0

0

SHHFrif

MHHFrif

HHH

m

V

Vm

→≡>→≡<

=∆−

OPEN CHANNEL FLOW: Transitions


As a first approximation one can disregard the energy losses implied in a transition. In such a case the following situations arise for a sudden rise/fall of the bed or contraction/expansion

OPEN CHANNEL FLOW: Transitions in subcritical flow with head loss


Let us consider an abrupt drop in the channel bed. If we have an head loss we cannot directly use an energy balance and we have to revert to a momentum balance, under the same assumptions usually used to derive Borda’s head loss in a pipe.

22

)( 22

2

221

1

2

22

2

11

2

bh

gbh

Qahb

gbh

Q

gA

Q

gA

Q

γγγγ

γβγβ

+=++

Π+=Π+

HgA

Qha

gA

QhaHHEaE

HHH

∆++=++<∆∆+=+

∆+=

22

2

221

2

121

21

22;

If we now consider an energy balance

we get under reasonable assumptions

( )g

VVH

2

221 −=∆

0021 EaHEaHEE <−∆+≡−∆+=

Accordingly, provided that is

the drawdown effect is diminished by the localized loss

OPEN CHANNEL FLOW: Transitions in subcritical flow


==

∆++=++

+=++

bhVbhVQ

Hg

Vha

g

Vh

bhQV

ahbQV

2211

22

2

21

1

22

2

21

1

22

22

)( γργρ

Let us solve for h2, neglecting the meaningless negative root

)()(1

)()(

)(1

0)()(2

2)(2

12

2122

12

222122

2122

21

2212

222

222

22

21212

ahVVVg

ahg

VVVVVV

gh

ahVVVg

hh

gbhbhVahgbbhVV

++−≅++−+−=

=+−−−

+=++

( ) ( ) ( )2211

22121

22

2121

22

21 2

1)()(

1

2

1

2

1VV

gahVVV

gahVV

ghahVV

gH −=+−−−++−=−++−=∆

That is a reasonable approximation

OPEN CHANNEL FLOW: Variable discharge due to lateral inflow/outflow


Main hypothesis:

•Steady motion in a rectangular channel (base is B) with a small and constant slope; gradually varied flow

•Negligible weight component in the direction of motion and of shear along the wall; α and β = 1

Let us consider the equation of momentum balance

and its component along the main flow direction

Where we suppose that the outflow velocity is V. The LHS varies with s because both h and Q are a function of s

Let us now consider the mass balance equation

∫∫∫∫ +=⋅−∂∂

S

n

WSW

dSdWgdSnVVdWVt

σρρρ rrrrrr)()(

)()2

(

)()(

)()(

*

22

*

*

VQVQBh

QBh

ds

d

VQVQdsdsMds

d

dsVQdssdssMdsVQM

oi

oi

oi

−=+

−=Π+

++Π++=+Π+

ρργ

ρ

ρρ

)(2

)( *2

2

VQVQds

dQ

Bh

Q

Bh

QhB

ds

dhoi −=+− ρρργ

)(

)()(

oi

oi

QQds

dQ

dsQdssQdsQsQ

−=

++=+

OPEN CHANNEL FLOW: lateral outflow - Q decreasing along the flow direction


Case A: Qi=0; discharge decreasing along the flow direction

Which can be combined to obtain

If we now consider the flow specific energy E

It varies with s as a function of h and Q

o

o

Qds

dQ

VQds

dQ

Bh

Q

Bh

QhB

ds

dh

−=

−=+− ρρργ 2)(

2

2

0)()2

()( 2

2

2

2

=+−=−+−ds

dQ

Bh

Q

Bh

QhB

ds

dhV

Bh

Q

ds

dQ

Bh

QhB

ds

dh ρργρρργ

22

2

2 hgB

QhE +=

22

32

2

1

hgB

Q

Q

E

hgB

Q

h

E

ds

dQ

Q

E

ds

dh

h

E

ds

dE

=∂∂

−=∂∂

∂∂+

∂∂=

Lateral outflow (on the left) and

inflow (on the right)

Drop Intake of a small hydropower

plant



If one consider that

The momentum balance equation can be written as

or, more simply

And alternatively

Both equations require an additional equation for water overflowing out of the channel. Usually it is in the form

Although an analytical solution is possible if µ is constant, a numerical solution provides a more general approach

Bh

Q

Q

EhB

Bh

QhB

h

EhB

ργ

ργγ

=∂∂

−=∂∂

2

2

0=∂∂+

∂∂

ds

dQ

Q

EhB

ds

dh

h

EhB γγ

0=ds

dE

ds

dQ

EhgB

hEg

ds

dQ

h

Q

Q

Bghds

dh

)23(

)(2

)(

122 −

−−=

−−=

water overflow from the channel happens without decreasing the energy per unit

weight of the water flowing in the channel. Its value will be determined on the

basis of the boundary condition

( ) 2/32 chgQds

dQo −−=−= µ

in an alternative way, this equation provides the

differential equation that governs the water surface

profile. It can be integrated numerically.



E constant and Q decreasing along the flow: use of the Specific discharge curve

Two different classes of problem:

(1) L and c are given; find out , i.e., Q(L) : FUNCTIONAL VERIFICATION problem

If Fr < 1, starts downstream (station A) with a temptative Q(L) and a corresponding hi(Q(L) ) and compute profile in a

backward fashion. Change Q(L) until Q(0) is found. If Fr > 1, starts upstream (B) knowing hi and Q(0) and integrate the

eqaution moving downward.

(2) Q(0) and are given; find out L with c being usually constrained : DESIGN problem

If Fr < 1, starts downstream (station A) with the known value Q(L),hi and compute profile in a backward fashion. When

Q(s)= Q(0), then L = s. If Fr > 1, starts upstream (B) with the known value Q(s),hi and compute profile until

Q(s)=Q(0) - . then L = s.

( )hEg

hqB

Q −==α2

∫− qdsQ )0(

∫ qds

∫ qds



The efficiency of the lateral weir can be increased by operating downstream on the boundary condition.

OPEN CHANNEL FLOW: lateral inflow - Q increasing along the flow direction


Case B: discharge increasing along the flow direction

Here we need the velocity component V* of the entering discharge along the flow directon. Often this quantity

can be set = 0, so that

Which is an equation stating the conservation of the specific force (SF)

Accordingly, the SF is constant whilst E is not. The constant value of the

Specific Force, S, must be determined on the basis of the boundary condition.

The SF equation must be considered along with the mass conservation equation

where the entering discharge Qi is a known function.

*2

2 2)( VQ

ds

dQ

Bh

Q

Bh

QhB

ds

dhiρρργ =+−

ds

dQ

Bh

QhB

Bh

Q

ds

dh

)(

2

2

2ργ

ρ

−−=

0)( =Π+Mds

d

iQds

dQ =



In order to investigate the possible profiles, we consider

whose maximum is the critical depth. As one can see, whilst Q increases with s, in a subcritical flow the depth

decreases. the contrary happens in a supercritical flow. In both cases the section where the critical depth occurs

can only be located downstream. In both cases, E decreases moving from upstream to downstream, due to the

entering discharge that has no momentum in the average flow direction

−=+=

2;

2

222 BhS

BhQ

Bh

Bh

QS

γρ

γρ

4/3

3/1

3/11

3

2

=

g

BSQ

ρ



In this case, only an S2 profile is possible.

Actually E, which is a specific quantity,

keeps decreasing along the flow entrance

flow stretch, because dq enters with 0

momentum in the flow direction.

Accordingly, at the end the flow must gain

energy to attain a final downstream

normal flow that is more energetic the the

one upstream

If Fr> 1, it might happen that the overall

inflow cannot be supported by the specific

force of the normal flow upstream. In

such a case this situation may occur.

Being a mild profile, one must start

downstream from the critical depth and

compute the profile moving upstream

OPEN CHANNEL FLOW: Bridge and culvert


When flows interact with the invert of a bridge, a sudden

reduction of the hydraulic radius happens, so that also the

stage-discharge relationship of the bridge is modified.

The upstream propagating M1 profile is strongly conditioned

by the boundary condition exerted by the bridge

Firenze, 1966, Ponte Vecchio

OPEN CHANNEL FLOW: Culvert (tombino o botte a sifone)


Often a small channel is use to convey water from one side to the other of a levee

(often a road). The hydraulic behaviour can be quite complex and, apart from the

geometry, depends on the level upstream (hm) and downstream (hv) and on the

culvert length (L).

a) Initially, when both hm and hv are small: open channel flow through a contraction

b) Then, when hm grows but both L and hv are small: orifice flow

c) Eventually, pressure flow

The transition between 1 and 3 implies a reduction of RH.

Accordingly, a strongly backwater effect may occur

OPEN CHANNEL FLOW: Bridge


Bridges are the most common obstruction and they can strongly condition the upstream water surface profile. On the other

hand, a wide variety of situations is possible and this must be treated on a case by case basis

In general terms, passage through a bridge usually implies a

contraction, due to piers and abutments.

OPEN CHANNEL FLOW: Specific Discharge


Specific discharge for E constant( )

( )hEg

hqB

Q

khh

hEdh

dQ

hEg

hAQ

−==

≡→+=→=

−=

α

α

2

20

2)(

In a rectangular cross section

NOTES ON OPEN CHANNEL FLOW - UniBs Water...

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