notes on open channel flow - unibs.it · open channel flow: passage over a sill (hump, bump:...

NOTES ON OPEN CHANNEL FLOW

Profili di moto

permanente in un

canale e in una serie

di due canali -

Boudine, 1861

Prof. Marco Pilotti

Facoltà di Ingegneria, Università

degli Studi di Brescia

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

M. Pilotti - lectures of Environmental Hydraulics

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 (Tyrolean) 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.



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

By placing a sluice gate one can raise the water level and greatly increase the amount of discharge released

by the weir.



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

Two different problems:

(1) Q0 ,L and c are given; find out , i.e., Q(L) : FUNCTIONAL VERIFICATION problem

If Fr < 1, start downstream (station A) with a temptative value of Q(L) and a corresponding hi(Q(L) ) and compute the

corresponding profile in a backward fashion. Change Q(L) until Q0 is found. If Fr > 1, start upstream (B) knowing hi and Q0

and integrate the equation moving downward. In the former case the procedure is iterative, not in the latter.

(2) Q0, c and are given; find out L: DESIGN problem

If Fr < 1, start downstream (station A) with the known values of [Q(L),hi ] and compute profile in a backward fashion. When

Q(s)= Q0, then L = s. If Fr > 1, start upstream (B) with the known value Q(s),hi and compute the profile until

Q(s)= Q0 - . then L = s.

( )hEg

hqB

Q −==α2

∫− qdsQ0

∫ qds

∫ qds

Structural variables: L, c (usually constrained)

Hydraulic variables: Q(L), η = (Q0 -Q(L))/ Q0 (efficiency)

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 stretch where

flow is entering, 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

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