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 ==+=
OPEN CHANNEL FLOW: passage over a sill (hump, bump: soglia)
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
OPEN CHANNEL FLOW: passage over a sill (hump, bump: soglia)
M. Pilotti - lectures of Environmental Hydraulics
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)
M. Pilotti - lectures of Environmental Hydraulics
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
OPEN CHANNEL FLOW: passage through a contraction (2)
M. Pilotti - lectures of Environmental Hydraulics
Sometimes the Energy
upstream isn’t enough…
OPEN CHANNEL FLOW: passage through a contraction (1)
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
==
∆++=++
+=++
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
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
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
OPEN CHANNEL FLOW: lateral outflow - Q decreasing along the flow direction
M. Pilotti - lectures of Environmental Hydraulics
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.
OPEN CHANNEL FLOW: lateral outflow - Q decreasing along the flow direction
M. Pilotti - lectures of Environmental Hydraulics
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.
OPEN CHANNEL FLOW: lateral outflow - Q decreasing along the flow direction
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
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 =
OPEN CHANNEL FLOW: lateral inflow - Q increasing along the flow direction
M. Pilotti - lectures of Environmental Hydraulics
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
OPEN CHANNEL FLOW: lateral inflow - Q increasing along the flow direction
M. Pilotti - lectures of Environmental Hydraulics
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
M. Pilotti - lectures of Environmental Hydraulics
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)
M. Pilotti - lectures of Environmental Hydraulics
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