The nature of river water The energy approach to fluid mechanics

The nature of river water The energy approach to fluid mechanics
Edward J. Hickin: River Hydraulics and Channel Form
Chapter 2
The energy equation for open-channel flow
The nature of river water
The energy approach to fluid mechanics
Some definitions
The energy equation
Specific energy and alternative flow regimes
Critical flow
Critical flow, wave celerity, and the Froude number
Subcritical and supercritical flow: transitions and controls
Flow transitions in three dimensions
Accounting for energy losses
Some Concluding Remarks
References
Because most natural rivers are capable of deforming their channels by eroding and depositing
sediment, at first it might seem a bit odd that we should concern ourselves with the behaviour of
water flowing through channels with rigid boundaries. But there are two good reasons for taking
this approach.
First, some channels do have rigid boundaries and they deserve our attention. In many cases the
first few orders of channels in a river network may be flowing on bedrock. No matter how large
the discharge carried through such bedrock channels, the boundary deforms so slowly by erosion
that, for all practical short-term purposes, it can be regarded as rigid or non-alluvial in character.
The second and more important reason is that rivers are such exceedingly complex physical
systems that we cannot hope to understand them without first simplifying reality in order to
grasp the character of the general forces at work in channels. When we understand the workings
of the simple case, we can then consider increasingly more complex ones which more closely
match the behaviour of real rivers. The most important complexity in this regard is the ability of
alluvial rivers to mould their channels to accommodate the forces in flowing water, a matter we
Chapter 2: The energy equation for open-channel flow
will take up again in Chapter 4. Meanwhile we need to back up and have a close look at how
flowing water responds to the forces acting on it.
Some of the models of flow developed in this chapter do not have much real-world application
because they are just too much of a simplification of reality. Nevertheless, they do serve the
useful purpose of revealing the quality of important forces at work in the flow even if they do not
allow us to quantify them precisely. Other quite simple models turn out to be remarkably good
at predicting real river behaviour. Part of our job here is to learn to tell the difference between
them.
Much of the discussion to follow assumes that you are familiar with the nature of basic
dimensions used in mechanics and with the requirement of dimensional homogeneity in
equations describing physical systems. If these notions are not familiar you may find it very
useful before proceeding to review these matters in the previous chapter.
The nature of river water
The water in rivers is actually a complex mixture of water, dissolved matter of both organic and
inorganic origin, and suspended particles ranging in size from clay to sands and in some cases
even gravel. Although this fluid mixture varies from one river to another, the properties of pure
water so dominate its character that, for purposes of our model building, we can regard rivers
simply as moving bodies of pure water. We will need to relax this assumption in certain
circumstances, but it holds reasonably well in general.
Because water is so abundant and its simple two-element formula so familiar, there is perhaps a
tendency to think of it as a very simple compound. It turns out, however, that water is rather
complicated stuff. It consists of the molecule H2O in which two small hydrogen atoms are
covalently bonded to the same hemisphere of the relatively large oxygen atom. Although
electrically balanced overall, the asymmetry of this covalent bond means that there is a relative
abundance of electrons and an excess negative local charge on one end of the H2O structure
counterbalanced by an excess local positive charge at the other. In consequence the water
molecule behaves rather like a weak dipole magnet. Some but not all of these molecules are in
2.2
Chapter 2: The energy equation for open-channel flow
turn joined to form tetrahedral clusters in which positively charged hydrogen ions are ionically
bonded to negatively charged oxygen ions. These clusters of molecules are separated one from
the other by unbonded water molecules that move freely and serve to lubricate the bonded
substructures, allowing flowage to occur. Furthermore, the whole structure of bonded and
unbonded molecules is remarkably dynamic with molecules exchanging rapidly between clusters
and flow layers such that a given intermolecular hydrogen bond breaks and reforms some 1012
times each second!
It is the strength of the bonding of hydrogen to oxygen ions in particular in water - a sort of
molecular glue - that is expressed in the physical property of viscosity. Viscosity is a measure of
the ease with which a fluid will deform when subjected to some stress. In order for water to
'flow', the electrical force bonding the water molecules one to the other must be overcome. The
rate at which water flows (deforms or changes shape) reflects the balance between the stress
acting on it (such as pipe pressure or gravity) and the internal molecular forces resisting the
deformation. Obviously these molecular or viscous forces must be very great in fluids like oil or
molasses which flow only sluggishly, and much smaller in others such as water and alcohol
which under the same stress flow quite readily.
Some physical properties of pure water, including viscosity, are listed in Figure 2.1. Clearly,
these common properties of water (and of most other liquids) are quite temperature dependent.
Temperature
oC
0
5
10
15
20
25
30
40
50
60
70
80
90
100
Specific weight
Density
γ
ρ
Nm
-3
9 805
9 807
9 804
9 798
9 789
9 777
9 764
9 730
9 689
9 642
9 589
9 530
9 466
9 399
Dynamic viscosity
µ x 10-3
-3
Nsm
999.8
1000.0
999.7
999.1
998.2
997.0
995.7
992.2
988.0
983.2
977.8
971.8
965.3
958.4
1.781
1.518
1.307
1.139
1.002
0.890
0.798
0.653
0.547
0.466
0.404
0.345
0.315
0.282
kgm
-2
2.1: Some physical properties of water (SI units)
2.3
Kinematic viscosity
ν x10-6
m2s-1
1.785
1.519
1.306
1.139
1.003
0.893
0.800
0.658
0.553
0.474
0.413
0.364
0.326
0.294
Chapter 2: The energy equation for open-channel flow
At about 4oC thermal agitation of the water molecules is minimal and the number of hydrogen
bonds, and thus molecule cluster size, are at a maximum, corresponding to peak density for the
liquid phase. As water temperature increases, more of the hydrogen bonds are broken so that
molecular cluster size declines along with density and specific weight. Internal strength (internal
resistance to deformation, or viscosity) also declines; for example, Figure 2.1 indicates that, at
20oC, the absolute or dynamic viscosity of water is little more than half that at 0oC, and about
twice that at 50oC.
Two different coefficients of viscosity are listed in Figure 2.1. The first is the absolute or
dynamic viscosity (µ) and has the dimensions Nsm-2 while the second is the kinematic viscosity
(ν = µ/ρ) with the dimensions m2s-1. Unless otherwise specified, discussions involving the
viscosity of water usually refer to the absolute or dynamic viscosity.
The temperature of water is quite variable among rivers (near freezing in high alpine or arctic
streams and tepid in the tropics) and may vary significantly on a seasonal basis in a given river.
It follows that the related physical properties of water listed in Figure 2.1 will vary in nature as
well.
The energy approach to fluid mechanics
The physics of flow can be studied in various ways. Hydrodynamics involves the study of the
behaviour of ideal fluids deduced mathematically from various assumptions about the physics
involved. As such it is the theoretical branch of the science concerned with fluid flow.
Engineers faced with solving real rather than theoretical problems in relation to flow have
adapted certain ideas in hydrodynamics but more often have replaced them altogether with
alternative engineering solutions because the theory of flow is incomplete and therefore not very
practical in application. These practical engineering solutions form the applied field of
hydraulics. Most river science is concerned with both theory and practice and useful notions of
both sorts are combined in the study of the fluid mechanics of open channel flow. We are
careful to specify open-channel flow because much of fluid mechanics was developed to
2.4
Chapter 2: The energy equation for open-channel flow
understand the behaviour of flow in closed conduits (pipes). Indeed, many of our models of
open-channel flow have been directly adapted from earlier thinking about the behaviour of flow
in pipes.
Some Definitions
It is sometimes useful to visualize the pattern of flow direction in moving water in terms of
streamlines. Streamlines are imaginary lines across which there is no flow component, so that
the velocity vector at any instant is tangential to every point on it (Figure 2.2a). A streamtube or
flow filament may be thought of as an imaginary tube in which the walls are made up of
contiguous streamline bundles across which there can be no flow (Figure 2.2b). A common
application of this concept is to imagine that the entire flow is a streamtube so that the conduit
boundary and the streamtube walls coincide.
Flow through a streamtube may be steady or unsteady. Unsteady flow varies with time so that an
observer standing on the bank of a river, for example, would see temporal fluctuations in the
velocity and the depth of flow. Such fluctuations imply that both positive and negative
acceleration characterize such flow. On the other hand, if the flow were steady in the river, our
observer would note only a constant velocity and constant depth over time. Only in a steady
flow does a streamline coincide exactly with the actual pathline followed by an individual water
particle; in unsteady flow streamlines and pathlines are distinctly different. We should note that
2.2: The streamline and streamtube
2.5
Chapter 2: The energy equation for open-channel flow
it is quite possible for certain flow phenomena to be regarded as either steady or unsteady,
depending on the observer's point of view. For example, a wave moving through the water
surface will pass our riverbank observer as unsteady flow while another observer moving
downstream with the same wave in a canoe would see the wave as part of a steady flow. Indeed,
changing our frame of reference can greatly simplify (or complicate!) the solutions to certain
flow problems.
Flow through a circular streamtube of constant diameter (Figure 2.2) is said to be spatially
constant or uniform. Thus uniform flow may be steady or unsteady but it must remain constant
with respect to distance along the streamtube. Uniform flow through a river channel means that
velocity and depth do not vary with distance down the channel and that all streamlines therefore
are parallel to both the bed of the channel and to the free water-surface (Figure 2.3a). In such
uniform flows the pressure everywhere in the flow simply is hydrostatic and thus depends only
on the flow depth beneath the free surface. Conversely, in flows where streamlines converge or
diverge the fluid is respectively accelerating and decelerating and the flow is nonuniform. In
cases of curvilinear flow, additional centrifugal forces act on the flow and pressures are no
longer simply hydrostatic. In the vertical longitudinal section (Figure 2.3b) concave streamlines
imply a downward acting centrifugal force which augments gravity and gives rise to pressures in
excess of hydrostatic while the converse relationship holds for convex streamlines.
Even though flow may not be strictly steady and uniform, practically it can be conveniently
considered to be so over short distances if changes in direction and spacing of the streamlines are
gradual enough. In such cases we speak of gradually varied flow, in contrast with rapidly varied
flow in which the assumption of uniformity is no longer tenable even as an approximation.
2.6
Chapter 2: The energy equation for open-channel flow
water surface
streambed
(a) Uniform flow
(b) Nonuniform flow
2.3: Longitudinal streamline profiles illustrating uniform and nonuniform flow in an open channel.
A fundamental concept in steady uniform flow mechanics is the continuity principle. It states
that the discharge (Q) must remain constant along a streamtube and at all points is defined as
Q = A V ........................................................................ (2.1)
where A is cross-sectional area of the flow and V is mean flow velocity. At successive cross
sections 1, 2, 3, 4,......n, the continuity principle requires that (Figure 2.2):
Q = A1 V 1 = A2 V 2 = A3 V 3 = A4 V 4 = ...........An V n ................................ (2.2)
We will often find it useful to consider the particular case of a rectangular channel, for which the
appropriate form of equation (2.1) is
Q = WD V ....................................….............................. (2.3)
where D is the depth of flow.
At other times we will find it useful to look, not at the three-dimensional flow picture described
by equations (2.1) - (2.3), but rather at longitudinal slices through the flow.
Such two-
dimensional flow has no width and the two-dimensional discharge q = Q/W in the case of our
rectangular channel is obtained by dividing both sides of equation (2.3) by W:
2.7
Chapter 2: The energy equation for open-channel flow
q = d V ............................................................................ (2.4)
Thus we can think of q as the discharge per unit width of a rectangular channel.
Water moves through a channel as a result of net applied forces of one sort or another. As we
noted in Chapter 1, if we are to concern ourselves with forces we must turn to the study of
dynamics and therefore to Newton's laws of motion. One of the most important in this regard, is
Newton's second law which states that:
F = ma……………………………………………....... (1.7)
or that the force F necessary to accelerate a mass m at a certain rate a is equal to the product ma.
When a force acts over some distance (s) we say that work (w = Fs) has been done. The
relationship between work and energy becomes apparent if both sides of equation (1.7) are
multiplied by a length s in the direction of the force:
Fs = w = mas………………………………………….. (2.5)
It follows from the principles of uniformly accelerated motion discussed in Chapter 1 that, since
v22 = v12 + 2as, by rearrangement,
a=
v22 - v12
……………………………………………. (2.6)
2s
Substituting this expression for a in equation (2.5) yields
v22 - v12
Fs = w = m  2s  s =
1
2
m(v22 - v12)…………………………... (2.7a)
We have in effect integrated both sides of equation (2.5) with respect to length s so that in
general,
2.8
Chapter 2: The energy equation for open-channel flow
s2
s2
∫ Fds=
m
s1
∫ ads =
1
2
m(v22 - v12) ………………………… (2.7b)
s1
This is the energy equation which states that the work done on a body as it moves from s1 to s2 is
€
€
1
equal to the kinetic energy (KE = 2 mv2) acquired by the body. This is a quite fundamental
equation in dynamics and has many applications to problems in fluid mechanics as we shall soon
see.
The force F in equation (2.7) and its derivatives is a resultant or net force. That is, it is the
difference between the total impressed force and any force such as friction or viscous drag. For
the moment, however, you are asked to set aside the reality that water exhibits various properties
including internal resistance to motion, and assume that it is an ideal fluid, incompressible and
inviscid (without viscosity). Such imaginary Newtonian fluids, as they also are known, move in
such a way that their energy is completely conserved (without loss in time or space). Although
these assumptions may seem to be outrageous excursions from the real world you may be
surprised at just how much we can learn about river flows by exploring them.
Because equations (2.6) and (2.7) were developed, and will be familiar to you, in terms of solidbody motion it is easier to deal with their application to fluids if we conceptualize them as
applying to a fluid element such as that shown in Figure 2.4.
This element is Δn high, Δs long, and has an implied unit width (normal to the page); its motion
vertical distance, z
is referred to a streamline along which it
Δs
p
moves distance s. There are only two forces
w sin θ
Δn
acting on such an element: that resulting from
∂p
p+
Δs
∂s
w= γ Δs Δ n
horizontal datum
2.4: Definition diagram for the
forces acting on a fluid element.
the pressure gradient in the direction of
motion, –(∂p/∂s)ΔsΔn, and that due to its
θ
weight resolved in the direction of motion, –
s
(γΔsΔn)sin q. Alternatively, this weight
component of force may be expressed with
respect to the vertical height z above the
datum, or -(γΔsΔn) ∂z/∂s.
2.9
Chapter 2: The energy equation for open-channel flow
Thus equation (2.7) in the case of this element can be written:
-(∂p/∂s)ΔsΔn - (γΔsΔn) ∂z/∂s = ma ……………………….……… (2.8)
The partial derivatives are necessary here because the two component forces may be changing
with time as well as with distance s.
If ρ is fluid density so that m = ρΔsΔn, and as is its acceleration in the s direction, equation (2.8)
becomes
-(∂p/∂s)ΔsΔn - (γΔsΔn) ∂z/∂s = ρΔsΔnas …………….……….. (2.9)
which simplifies to
∂/∂s (p+γz) + ρas = 0……………………….……… (2.10)
Equation (2.10) is the Euler equation, named for Leonhard Euler, a famous 18th century Swiss
mathematician. Although the Euler equation is not as useful in direct application as its integrated
form, the Bernoulli equation, it is introduced here because it will serve to clearly remind us of
the physics involved in the flow. Two important properties of this relationship need to be
appreciated.
First, the term (p + γz) in equation (2.10), known as the piezometric pressure, is constant
throughout a body of still water (as = 0) so that ∂/∂s(p+γz) = 0, regardless of the direction s. In
other words, in the case of still water where there is a free water surface a distance y above the
datum for z (i.e., the water depth), pressure p at any height z is hydrostatic so that
p = γ(y - z).................................................................... (2.11)
and therefore,
(p + zγ) = a constant = γy...................................................... (2.12)
Should the fluid element begin to move, however, the acceleration term ρas becomes non-zero
so that the piezometric pressure clearly no longer will remain constant throughout the fluid.
2.10
Chapter 2: The energy equation for open-channel flow
Second, the acceleration term as in equation (2.10) is a result of velocity variation in both space
and time (non-uniform and unsteady flow). Thus we can say that
dv
= as
dt
ds ∂v
∂v
= dt ∂s + ∂t
∂v
∂v
= v ∂s + ∂t ………………… (2.13)
ds
That is, to an observer moving in the direction of flow (s) at velocity dt = v with the water
€
element shown in Figure 2.4, the rate of change in the resultant velocity with respect to time (i.e.,
acceleration, as) is the sum of the velocity change attributable to the change in position (i.e,
∂v
∂v
distance s), v ∂s , and that change in v specifically attributable to the change in time, ∂t . The
velocity change related to distance is termed the advective acceleration while that related to time
is termed the local acceleration.
∂v
In steady flows, by definition there is no local acceleration and for this special case, as = v ∂s ,
which upon substitution in equation (2.10) yields:
∂/∂s(p+γz) + ρv ∂/∂s = 0............................................. (2.14)
Equation (2.14) can be integrated directly with respect to distance s to yield the Bernoulli
equation (named for Daniel Bernoulli, the Swiss physicist and mathematician who developed the
kinetic theory of gases in the 18th century):
p + γz + 12 ρv2 = constant .................…...................... (2.15a)
or
p/γ+ z + v2/2g = constant H..............…...................... (2.15b)
Since equation (2.15) was obtained by integrating a force equation with respect to distance, it is
an energy equation and the constant H is termed the total energy or total head. The derivation
employed here is quite general but it usefully highlights the dependence of the Bernoulli
equation on steady flow conditions; otherwise, of course, there would be an additional term from
the integration of the expression for the local acceleration, a mathematical operation involving
considerable difficulties and one best avoided if possible.
2.11
Chapter 2: The energy equation for open-channel flow
Because equation (2.15) is an energy equation, it could have been derived directly from equation
(2.7). Indeed, this derivation is a useful exercise for us to undertake because it helps to clarify
the meaning of the terms in the Bernoulli equation and their relation to certain measurable
quantities routinely used to describe the flow and the geometry of the conduit. It also will be
instructive if we initially consider this direct derivation in terms of the pipe flows for which the
ideas were first developed. Once we have accomplished this task we can set about adapting the
Bernoulli concepts to the free-surface flows we encounter in rivers.
Figure 2.5 shows a run of circular piping
A
pipe
crosssectional
area A1
B
(although any cross-sectional shape could be
L2
employed) in which water flows from a largep2
L1
v1
v2
p1
Flow
direction
z1
z2
pipe
crosssectional
area A 2
diameter section to a smaller diameter section.
Our analysis will focus on a control volume of
the fluid as it moves through the transition. We
assume that no energy is lost through this pipe
transition (it is an ideal fluid moving through a
horizontal datum
frictionless pipe transition) and that the sum of
potential and kinetic energy in section A (PEA +
2.5: A definitional long-section through a
run of pipe of variable diameter for deriving
the energy equation.
KEA) plus the additional kinetic energy acquired
as a result of the work (W) done in moving the
water through the pipe transition, must equal the
sum of kinetic and potential energy in section B (PEB + KEB).
Thus we can write an equation to describe the circumstances in Figure 2.5 as follows:
PEA + KEA + W = PEB + KEB...................................... (2.16)
or more specifically,
1
1
mgz1 + 2 mv12 + W = mgz2 + 2 mv22 ..................…............ (2.17)
Work (W = Fs) here is the net work done by the pressure forces (the opposing pressure acting
over the respective cross-sectional areas) and the distances moved by the control volume are
respectively L1 and L2. So the net work involved is
2.12
Chapter 2: The energy equation for open-channel flow
W
= F1L1 - F2L2
= p1A1L1 - p2A2L2
Thus we can substitute this expression for W in equation (2.17) yielding:
1
1
mgz1 + 2 mv12 + [p1A1L1 - p2A2L2] = mgz2 + 2 mv22…………….. (2.18)
Dividing equation (2.18) throughout by the fluid volume (Vo = A1L1 = A2L2), and noting that
m/Vo = r, results in:
1
1
ρgz1 + 2 ρv12 + [p1 - p2] = ρgz2 + 2 ρv22…………………….. (2.19)
Noting further that ρg = γ, we can make this substitution and regroup the terms in equation
(2.19) to yield:
p1 v12
p
v2
+
+ z1 = 2 + 2 + z2 ……………………………….. (2.20)
γ 2g
γ 2g
which is simply a particular version of the general form of the Bernoulli equation:
€
p
γ
€
pressure
head
+
v2
2g
+ z
velocity
head
= H …………………………….. (2.15b)
total
head
The equation (2.15b) components each have the dimension of length and these lengths may be
related to the observed physical properties of pipe flow as shown in Figure 2.6. If a small open
tube is connected to such a pipe the static pressure will force the fluid up the tube until the height
of the free water-surface reaches p/γ m above the centre of the pipe. Early experiments by the
French engineer H. Pitot in 1732 showed that the sum of the velocity head and the pressure head
could be measured by placing a second small open tube (now known as a pitot tube) in the flow
with its open end facing into the flow. The difference between the height of the water column in
the static tube and that in the pitot tube is the velocity head. The velocity head commonly is
measured in this way in laboratory experiments so that the pressure difference becomes the basis
for computing flow velocity.
2.13
Chapter 2: The energy equation for open-channel flow
Since energy in this system is conserved, the locus of the free water-surface elevations in a series
of pitot tubes along the pipe describes the horizontal (invariant) energy line. The locus of the
free water-surface in a corresponding series of static tubes describes the hydraulic grade line
which will rise and fall in response to velocity changes along the pipe. In the limit, as velocity
goes to zero, the elevation of the free water-surface in the static and pitot tubes converges.
The application of these ideas to open channel flow is direct and straightforward, although not
made without some important assumptions; the application of equation (2.20) to the longitudinal
section of a rectangular open channel is illustrated in Figure 2.7. The velocity head, v2/2g, has
direct correspondence to the pipe flow component provided we assume that the velocity is
constant over the entire cross-section, an assumption never strictly met in real channels but
closely enough approximated that we do not introduce severe errors by making it. Provided the
downstream bed and water-surface slopes are not too great (say, less than 5o), the pressure
head, p/γ, at any point in the flow can be taken as the hydrostatic head and therefore is simply
equal to the depth below the water surface (y - z); see equation (2.11). Thus the sum (p/γ + z) for
any point in the flow must represent the height of the water surface above datum and if z is taken
as the bed elevation, p/γ must equal the flow depth, y.
energy line
v2 /2g
hydraulic grade line
H
ρ/γ
v
p
z
horizontal datum
2.6: Definition diagram for the Bernoulli energy equation terms as they apply to flow through a pipe.
2.14
Chapter 2: The energy equation for open-channel flow
We can now rewrite equation (2.15) in these terms:
v2
y + 2g
+ z
= H ........................................... (2.21)
The requirement of a low water-surface slope is necessary because the pressure head is strictly
the vertical distance below the water surface, h, and not y (the flow depth measured normal to
the water surface). The approximation is close for small water-surface inclines (see Figure 2.7),
however, and y/h = cos q (for q = 5o, y/h = 0.996 and for q = 10o, y/h = 0.985).
 ∂

(p + γz) + ρa s = 0 serves as an important reminder that the assumption of
 ∂s

The Euler equation 
hydrostatic pressure distribution can only be true in the vertical direction s if the acceleration
term, as, is zero. In other words, there must be no vertical acceleration associated with strongly
curved streamlines such as those depicted in Figure 2.3.
Fortunately, the conditions of relatively low water-surface slope and hydrostatic pressure
distribution are very often closely approximated in rivers so that equation (2.21) enjoys quite
wide application. Nevertheless, we must remember these limitations and carefully assess each
application to assure ourselves that the assumptions underlying equation (2.21) are sensibly met.
θ
v12 /2g
θ
p1 /γ
ΔH
energy
line
y1
hydraulic
grade line
water
surface
2
2
v /2g
v
y2
z1
p /γ
2
H
channel
boundary
z2
horizontal datum
2.7: A longitudinal profile of a two-dimensional open-channel flow illustrating the application of the terms of
the Bernoulli equation.
2.15
Chapter 2: The energy equation for open-channel flow
Another assumption that deserves our careful assessment here is the fundamental notion of
energy conservation. You will note that Figure 2.7 now includes a head-loss term, ΔH, but that it
is ignored in equation (2.21). Head loss is that component of the energy which is consumed
overcoming the frictional resistance to flow. Thus, it represents a 'leakage' of energy from our
assumed energy-conserving system and obviously can only be ignored if it is insignificant in
relation to the total energy of the flow. It turns out, fortunately, that this is not an unreasonable
assumption provided we are dealing with short reaches of channel only. For problems involving
long reaches of channel, however, the total energy at the end of the reach must be discounted by
the frictional head loss in order for the Bernoulli equation to apply without error. Later we will
return to examine much more closely this whole business of flow resistance.
Equation (2.21) can be used to solve a variety of channel transition problems where we want to
know how the flow will change in response to certain downstream changes in the boundary
conditions. For example, Sample Problem 2.1 shows a solution to a simple problem where both
upstream and downstream flow depths are known.
Sample Problem 2.1
v12 /2g
Total energy line
v 2 /2g
flow direction
y1 = 2.0 m
cross-section
of fallen log
2
y 2 = 0.5 m
Problem: A tree falls across a 5 m-wide rectangular stream channel and causes the water to back up on the upstream
side to a depth of 2 m while water discharges under the tree trunk in a 0.5 m deep flow and continues downstream.
For this short channel transition the channel bed can be regarded as horizontal. What is the discharge of the stream?
Solution: Initially, we treat this as a two-dimensional problem. Since the bed is horizontal (z1 = z2) from equation
(2.21) we get
v 12
2.0 + 2g
v 22
= 0.5 + 2g
and from two-dimensional continuity
2.0v1 = 0.5v2 and therefore v2 = 4.0v1
It follows from substitution for v2 that
v 12
2.0 + 2g
= 0.5 +
2.16
(4.0v1)2
2g
v 12
= 0.5 +16.0 2g
Chapter 2: The energy equation for open-channel flow
Sample Problem 2.1 (cont)
and
v 12
1.5 = 16 2g
Noting that g = 9.806 ms-2, it follows that
1.5
v12 = 15.0 (2g) = 1.961;
Therefore, q = (1.4)(2.0) = 2.8 m2s-1
v 12
v 12
v 12
- 2g = 2g (16.0 -1.0) = 15.0 2g
v1 =
1.961
= 1.4 ms-1
Q = (2.8)(5.0) = 14.0 m3s-1
and
A more commonly encountered flow transition problem is outlined in Sample Problem 2.2. Here
we are interested in knowing how the velocity and depth of flow will change as the river flows
Sample Problem 2.2
?
v1
1.5 m
?
20 cm
v
2
y
2
Problem: Water flows through a 5 m-wide rectangular channel and over a 20 cm step up in the bed. If the discharge
is 15 m3s-1, and the initial depth upstream of the step is 1.5 m, what will be the depth and velocity of the flow
downstream of the transition?
Solution: Initially, we treat this as a two-dimensional problem.
Noting that q = Q/w = 15/5 = 3 m2s-1 and V1
=q/y1 = 3/1.5 = 2 ms-1, we set up the Bernoulli equation for this transition as follows:
y1 +
v12
v22
+
z
=
y
+
1
2
2g
2g + z2
4.00
1.500 + 2g
v 22
+ 0 = y2 + 2g
+ 0.200
You will note that bed elevation upstream of the transition is the datum so that z2 = 20 cm = 0.200 m (implicit in the
3-digit lengths is a measurement accuracy to the nearest millimetre). For g = 9.806 ms-2, this Bernoulli equation
simplifies to
v 22
1.504 = y2 + 19.612
From continuity, y2v2 = 3, so that v2 = 3/y2. Substituting for v2 above and rearranging yields
y23 - 1.504y22 + 0.459 = 0
which by iteration (see Appendix 3.1) has three solutions: y2 = 1.167 m, 0.818 m, and -0.481 m.
The negative depth clearly has no physical meaning and of the positive solutions, y2 = 1.167 m is taken as correct
because it is the closest to the pre-transition condition (where y1 = 1.5 m). The solution y2 = 1.167 m implies that
the water surface must drop by 0.133 m over the step [(1.167 + 0.20) -1.50) = -0.133 m]. From continuity it
also follows that v2 = 3/1.167 m = 2.571 ms-1.
2.17
Chapter 2: The energy equation for open-channel flow
over an upward step in the bed. Once again we set the Bernoulli terms for each side of the
transition equal to each other and use continuity to close the set of equations. In this case,
however, we are left with a cubic equation for which there are three solutions. Setting aside the
mathematical difficulty of solving a cubic equation for the moment, deciding which of our
solutions to Sample Problem 2.2 is correct and physically possible presents a problem. Clearly
the negative solution is not a physically real solution but it is not readily apparent, however,
which of the two positive solutions is correct. It turns out that y2, closest to the initial y1, is the
appropriate solution (i.e., y2 = 1.167 m) but in order to understand why this is so we must
approach this particular type of problem from the perspective of the specific energy,
an
extremely useful concept introduced to fluid mechanics by B.A. Bakhmeteff in 1912.
approach this particular type of problem from the perspective of the specific energy, an
extremely useful concept introduced to fluid mechanics by B.A. Bakhmeteff in 1912.
Specific energy and alternative flow regimes
Specific energy, E, is defined as the energy of flow in relation to the bed (rather than to an
external datum), and thus is described by the expression:
v2
E = y + 2g ............................................................. (2.22)
In a sense equation (2.22) recognizes that, for short reaches of channel, changes in specific
energy related to the downstream decline in bed elevation or water-surface slope (both measured
with respect to an external horizontal datum) are negligible compared to that related to local
changes in depth and velocity. Thus we can adopt the bed itself as a datum, greatly simplifying
the energy equation and allowing us to explore the relation between the velocity and depth heads.
Since the simplest case allows the clearest development of this concept of specific energy, for
now we will consider the two dimensional version of flow in a rectangular channel of fixed
width in which q = Q/w, the discharge per unit width of channel. Thus equation (2.22) can be
rewritten in the form:
2.18
Chapter 2: The energy equation for open-channel flow
E=y+
q2
........................................................... (2.23)
2gy2
or for the case where discharge is constant along the channel,
q2
(E - y)y2 = 2g = a constant .......................................... (2.24)
The graph of the relationship between E and y described by equation (2.22) as it applies to flow
over an upward step in the bed (see Sample Problem 2.2) is shown in Figure 2.8.
2
2g
y
v 22
v1
total energy grade line
E1
E2
2g
E=y
Δz
A
y1
v1
y1
E1
y2 v 2
E2
y= 2
E
3
Δy
B
increasing q
y2
C
Δz
45o
0
yc
B'
A'
y=0
E min
E
2.8: The specific energy curve and its application to the channel transition problem
(after Figure 2-3 in F.M. Henderson, 1966, Open Channel Flow, Macmillan Publishing Company, New York).
In the physically meaningful (positive y) domain the graph of the cubic equation (2.24) is
bounded by the 45o angle formed by the asymptotes (E-y) = 0 and y = 0 in the first quadrant.
Prior to the step, the water possesses specific energy E1 and flows at a depth y1, conditions
corresponding to point A on the specific energy curve. It is clear from this curve, however, that
a second alternative combination of depth and velocity also is possible at E1, corresponding to
point A'. Although the specific head is the same at this alternative condition, depths are much
lower and velocity must therefore be higher than at point A (since q is constant). As the flow
moves over the step the head declines by the length of the step height so that E1 - E2 = Δz. As
the flow accelerates over the step, energy is transferred from the depth head to the velocity
2.19
Chapter 2: The energy equation for open-channel flow
head and we move from points A to B on the specific energy curve in Figure 2.8. Because the
total energy is conserved, these changes must be accompanied by a drop in the elevation of the
water surface over the step, a perhaps surprising result because most people find it to be rather
counter-intuitive.
Once again, we find that an alternative depth and velocity occurs, this time at
point B' (the second positive and smaller depth solution noted in Sample Problem 2.2).
These circumstances beg an obvious question: why does the flow adjust to condition B rather
than condition B'? Both alternative points B and B' represent physically real equilibrium flow
conditions. The answer to our question lies in the accessibility of the alternative depths and
velocities to the precursor flow. First we should note that the cubic equation graphed as a heavy
line in Figure 2.8 represents our specified condition of a constant discharge. Other curves can be
drawn for higher (or lower) discharges, as shown by the faint-line curves, but all specific energy
changes in our example must follow the heavy-line curve. For example, it is not possible to
jump from point A to point B' (nor to A') simply because such a trajectory would require an
increase in the two-dimensional discharge. Such a jump could be achieved, however, if the flow
also encountered a local channel contraction at the step. A constriction of just the right degree
would increase local q and provide direct access to the alternative flow conditions at B'.
Nevertheless, if discharge is constant and the channel has a fixed width, all specific energy
changes must follow a given specific energy curve.
The second possibility that might at first appear plausible is that point B' is reached by flow
adjustments which simply follow the curve around the apex to settle on these alternative
conditions. But such an adjustment is not physically possible because it requires that the specific
energy drop below E2 and return to it again. Such an effect could be achieved only by a local
rise in the upstream bed just high enough above the step to achieve a depth yc (at point C) but it
is not a possibility for a simple upward-step transition of the kind shown in Figure 2.8.
So we may conclude that, if width remains unchanged through a simple upward-step transition,
the point B' in Figure 2.8 simply is not accessible from an upstream flow represented by point A.
Similar physical reasoning leads us to conclude that B is not accessible from A'. From all of this
follows a useful rule: of the two solutions to the energy equation applied to step transitions, the
2.20
Chapter 2: The energy equation for open-channel flow
appropriate depth/velocity condition in the transition will be that nearest the initial conditions
upstream.
We might also expect that a 'negative step' or downward step in the bed will yield results
consistent with the processes described above. Indeed, Sample Problem 2.3 shows that where
approaching flow is represented conceptually by a point on the upper limb of the energy
Sample Problem 2.3
?
v1
?
1.0 m
v2
y
2
20 cm
Problem: Water flows through a 10 m-wide rectangular channel and over a 20 cm step down in the bed. If the
discharge is 25 m3s-1, and the initial depth upstream of the step is 1.0 m, what will be the depth and velocity of the
flow downstream of the transition?
Solution: Once again, we initially treat this as a two-dimensional problem. Noting that q = Q/w = 25/10 = 2.5
m2s-1 and v1 =q/y1 = 2.5/1.0 = 2.5 ms-1, we set up the Bernoulli equation for this transition as follows:
y1 +
v12
v22
+
z
=
y
+
1
2
2g
2g + z2
6.25
1.000 + 2g
v 22
+ 0 = y2 + 2g
- 0.200
You will note that bed elevation upstream of the transition is taken as the datum so that z2 = -20 cm = 0.200 m (implicit in the 3-digit lengths is a measurement accuracy to the nearest millimetre). Of course we could
just as easily set the lower bed as datum and avoided a negative z2. For g = 9.806 ms-2, this Bernoulli equation
simplifies to
v 22
1.519 = y2 + 19.612
From continuity, y2v2 = 2.5, so that v2 = 2.5/y2. Substituting for v2 above and rearranging yields
y3 - 1.519y22 + 0.319 = 0
which by iteration (see Appendix 1) yields the positive solutions: y = 1.342 m and 0.584 m
Only the solution y2 = 1.342 m is accessible from y = 1.0 implying that the water surface must rise by 0.142 m
over the downward step [1.342 - (1.0 +.0.2) = +0.142 m]. From continuity it also follows that v2 = 2.5/1.342 m =
1.863 ms-1.
2.21
Chapter 2: The energy equation for open-channel flow
equation, a negative step results in a reduction in the velocity head, an increase in the depth head,
and a rise in the water surface over the transition. Thus we can see that, for these conditions
described by the upper limb of the energy equation, the water surface is out of phase with the
bed, rising over pools and falling over shoals.
But what about the case where the upstream flow approaching a transition has a depth and
velocity combination corresponding to the lower-limb condition A' in Figure 2.8?
Problems 2.4 and 2.5 illustrate the flow response in just such a case.
Sample
Here the nearest
Sample Problem 2.4
?
-1
v1 = 5.0 ms
?
20 cm
1.0 m
v2
y
2
Problem: Water flows through a 5 m-wide rectangular channel and over a 20 cm step up in the bed. If the discharge
is 25 m3s-1, and the initial depth upstream of the step is 1.0 m, what will be the depth and velocity of the flow
downstream of the transition?
Solution: Noting that q = Q/w = 25/5 = 5 m2s-1 and V1 =q/y1 = 5/1 = 5 ms-1, we set up the Bernoulli equation for
this transition as follows:
25.00
1.000 + 2g
Simplifying,
v 22
+ 0 = y2 + 2g
+ 0.200
v 22
2.075 = y2 + 19.612
From continuity, y2v2 = 5, so that v2 = 5/y2. Substituting for v2 above and rearranging yields
y23 - 2.075y22 + 1.275 = 0
which by iteration has two real solutions: y2 = 1.531 m and 1.224 m.
The solution y2 = 1.224 m is taken as correct because it is the closest to the pre-transition condition (where y1 = 1.0
m). In this case the solution y2 = 1.224 m implies that the water surface must rise by 0.424 m over the step [(1.224
+ 0.20) - 1.0 = - 0.424 m]. From continuity it also follows that v2 = 5/1.224 m = 4.085 ms-1.
2.22
Chapter 2: The energy equation for open-channel flow
solution to the initial conditions also is on the lower limb of the specific energy curve. In this
case the energy equation indicates that the water surface rises over the upward-step transition and
falls over a negative step, the reverse of the case on the upper limb.
Clearly, quite different
responses result, depending on whether the upper or lower limb of the energy curve is
accessed by the flow. In this latter case we note that the water surface and bed geometry are inphase with the water surface, rising over shoals and falling over pools.
We should also note here that the particular problem treated in Sample Problem 2.5 involves a
Sample Problem 2.5
?
v1 = 4.5 ms-1
?
0.75 m
v2
y
2
20 cm
Problem: Water flows through a 100 m-wide rectangular channel and over a 20 cm step down in the bed. If the
discharge is 338 m3s-1, and the initial depth upstream of the step is 0.75 m, what will be the depth and velocity of
the flow downstream of the transition?
Solution: Noting that q = Q/w = 338/100 = 3.38 m2s-1 and v1 =q/y1 = 3.38/0.75 = 4.507 ms-1, we set up the
Bernoulli equation for this transition as follows:
y1 +
0.750 +
which simplifies to
v12
v22
+
z
=
y
+
1
2
2g
2g + z2
20.313
2g
v 22
+ 0.20 = y2 + 2g
+0
v 22
1.986 = y2 + 19.612
From continuity, y2v2 = 3.38, so that v2 = 3.38/y2. Substituting for v2 above and rearranging yields
y23 - 1.986y22 + 0.583 = 0
which by iteration yields y = 1.807 m and 0.664 m
In this case only the solution y2 = 0.664 m is accessible from y1 = 0.75 m implying that the water surface must
fall by 0.286 m over the downward step [0.664 - (0.75 + 0.2 ) = - 0.286 m]. From continuity it also follows that
v2 = 3.38/0.664 m = 5.090 ms-1.
2.23
Chapter 2: The energy equation for open-channel flow
transition over which the streamlines almost certainly are strongly curvilinear and associated
with some vertical acceleration. Since we can no longer assume that the acceleration term as in
the Euler equation is zero, it would be prudent to treat the solutions obtained here with some
caution. Experience suggests that, although the indicated direction of change in the variables is
correct, the precise degree of predicted change is less accurate. Thus, Sample Problem 2.5 is
nudging the application limit of the Bernoulli equation and as we shall see later, such problems
sometimes are more appropriately solved in terms of the momentum exchanges involved rather
than by approximating the energy balance.
Setting aside these potential errors for the moment, these examples illustrate that flow behaviour
on the upper limb of the specific energy curve in Figure 2.8 is quite fundamentally different from
that associated with the lower limb. For this reason the entire domain of relatively low velocities
and large depths (y>yc) is known as lower regime flow (or subcritical flow) and the
corresponding domain of alternative high velocities and small depths y<yc) is known as upper
regime flow (or supercritical flow).
Clearly we need to know more about the critical condition C in Figure 2.8, corresponding to yc,
and which separates these two domains, one from the other.
Critical Flow
In order to derive the equations for critical flow, first we need to note from Figure 2.8 that
critical flow occurs at point C where the specific energy, E, is a minimum. Thus, the defining
equation for critical flow can be obtained by differentiating equation (2.23) with respect to depth
and setting the differential expression equal to zero (the subscript c indicates critical flow
conditions):
Equation (2.23) states that E = y +
Differentiating,
q2
q 2 -2
=
y
+
y .......................……....................... (2.23)
2gy 2
2g
q 2 -3
q2
dE
=1y =1=0
€
g
gy 3
dy €
2.24
€
€
€
from which it follows that,
Chapter 2: The energy equation for open-channel flow
q2 = gyc3......................................................…............. (2.25)
Dividing both sides by yc2,
vc2 = gyc ...................................................................... (2.26)
Another useful form of equation (2.25) is obtained by rearrangement:
yc =
3
q2
.................................................................. (3.27)
g
The relationship between yc and E can be derived from equations (2.23) and (2.25).
€
For critical flow,
or alternatively
Ec = yc +
Ec = yc +
vc2
g = yc
€
From equation (2.26)
vc 2 yc 2
2gy c 2
q2
2gyc2
= yc +
vc 2
2g
.............................................. (2.28)
vc2
yc
=
2g
2
and therefore
€
Making the appropriate substitution in equation (2.28) gives
yc
Ec = yc + 2
or
Ec =
3
2
yc
2
and by rearrangement,
yc = 3 Ec.......................................................... (2.29)
These equations relating E and y in critical flow have been developed for the case of a fixed
discharge, q. It is also of considerable practical
y
interest to consider how q varies with y for a given
specific energy. In terms of the specific energy
curve in Figure 2.8, if we fix E = E1, for example,
Eo
y =
c
0
q
max
2
3
Eo
q
we can focus on the changes in q that will occur as
we move vertically from the upper subcritical
asymptote at E = y to the lower limiting
supercritical asymptote at y = 0. At the two
2.9: The discharge-depth curve for a fixed specific
energy, Eo.
2.25
Chapter 2: The energy equation for open-channel flow
asymptotes, q = 0, at the lower limit because y = 0, and at the upper limit because the entire
specific head is in the form of the depth head so that once again y = 0 . Between these two
limits, however, a vertical line through A and A' in Figure 2.8 passes through isolines of
increasing discharge to reach a maximum discharge qmax, beyond which q declines to zero once
again. The general form of the q/y relationship for some fixed specific energy Eo, therefore,
must appear as depicted in Figure 2.9
The maximum discharge is shown in Figure 2.9 as occurring at the critical depth, yc. The proof
of this correspondence is shown readily by differentiating an appropriate form of equation (2.24)
and solving
dq
= 0, thus:
dy
q2 = 2gEoy2 - 2gy3
Rearranging equation (2.24),
€
and differentiating,
2q
dq
= 4gEoy - 6gy2
dy
When 4gEoy - 6gy2 = 0
€
4Eo = 6y
and
2
y = 3 Eo ......................................................... (2.30)
Equation (2.30) essentially is equivalent to equation (2.29) describing the conditions in critical
flow. Thus we can conclude that, not only does critical flow occur at the minimum specific
energy for a given discharge, but it also corresponds with the maximum discharge for a given
specific energy. In Figure 2.9 the locus of critical depth across a range of specific energies and
2
discharges (generalizing to all points C) plots as the straight line y = 3 E.
2.26
Chapter 2: The energy equation for open-channel flow
Critical flow, wave celerity, and the Froude number
As we have seen, the condition of critical flow is associated with a critical depth and also with a
corresponding critical velocity, vc. Where specific energy Eo is fixed and discharge can vary we
might also think in terms of a critical discharge (qc = qmax).
The relationship between the critical velocity and the physical process of depth adjustment in
open channels is fundamental to understanding the nature of the flow in rivers. In most rivers the
water surface is smooth in the sense that changes in the water-surface elevation are spatially
gradual. Most of us simply take this circumstance as normal even though changes in the
boundary (such as upward and downward steps, contractions and expansions) may be quite
abrupt. But why is this so?
The answer to this question lies in the fact that, although boundary irregularities generate
disturbances at a point in the flow, the resulting water-surface elevation changes are quickly and
continually dispersed throughout the water surface. Small elevation changes are propagated
radially outwards from the point of disturbance in the form of small gravity waves. In a sense
we might think of these waves as the physical mechanism by which information about the
boundary is transmitted to the rest of the flow.
We will not explore wave theory in this account and simply take as our starting point the general
expression for the velocity of an oscillatory wave in the free water-surface that you will find
developed in most texts on the subject:
c2 =
gL
2πy
tanh
.................................……................. (2.31)
2π
L
Here c and L are respectively wave celerity (velocity in standing water) and wavelength in water
€ by disturbances in open channels are long waves of
of depth y. Typically the waves €
propagated
low amplitude (or 'shallow-water waves') in which 2πy/L is small so that tanh
these circumstances equation (2.31) reduces to
2.27
€
€
2πy 2πy
=
. In
L
L
Chapter 2: The energy equation for open-channel flow
c2 =
and therefore
gL 2πy
= gy......................................…............. (2.32a)
2π L
c = gy .............................................……................. (2.32b)
€
Clearly, equations (2.26) and (2.32) are identical, indicating yet another important property of
critical flow. We conclude that critical velocity equals exactly the velocity with which
disturbances in the flow are propagated through the free-water surface. The relationship between
vc and c is expressed in an important dimensionless ratio called the Froude number, F, as
F=
v
gy
................................................................... (2.33)
Thus, the definitional property that we should now recognize is that, in critical flow, F = 1.0,
v = vc, and y = yc. It also follows that, in subcritical flow, F<1.0, v<vc, and y>yc, while in
supercritical flow, F>1.0, v>vc, and y<yc. Thus, to return to specific energy diagram in Figure
2.8, the lower regime alternative depths at A and B occur in the subcritical flow domain where
Froude numbers are less than unity and the upper regime alternative depths A' and B' occur in the
supercritical flow domain where Froude numbers are greater than unity.
The importance of the Froude number in specifying the state of flow is implicit in the specific
energy equation and we might note here that, from equation (2.23) we get
v2
E = y + 2g
and
=y+
q2
2gy2
y
= y + 2 F2 ............................... (3.34)
dE
2
dy = 1 - F .................................................... (2.35)
We can recognize critical flow in rivers by the presence of standing waves. We have all seen
how disturbances to the surface of standing water spread out radially. In subcritical flow this
radial pattern of wave propagation persists but is distorted because of the influence of the mean
downstream velocity of the river. To an observer standing on the river bank, waves moving in
2.28
Chapter 2: The energy equation for open-channel flow
the downstream direction will travel relatively fast with an enhanced resultant downstream
velocity = gy + V . Waves moving in the upstream direction, however, will appear to our
observer to be moving much more slowly at the reduced upstream resultant velocity = gy - V .
In critical flow, on the other hand, water-surface disturbances will move downstream at
velocity 2 gyc (since c = vc = gyc ) but in the upstream direction they will stand still
with respect to an observer on the bank because their upstream resultant velocity is zero
( gyc - gyc ).
It is quite common, for example, to see in steep gravel-bed rivers, trains of standing (or
stationary) waves spread out for some distance downstream from a site where a large boulder on
the bed produces local acceleration to critical flow. Of course, in fully developed supercritical
flow in which F>1.0, all surface disturbances will be swept downstream because v > gyc .
Subcritical and supercritical flow: transitions and controls
The introduction of the Froude number allows a general confirmation of the water-surface/bed
phases we already have noted in the sample problems. Recalling the Bernoulli equation (2.21)
applied to a rectangular channel we can write:
v2
H = y + z + 2g
= E + z = a constant, H........................... (2.36)
which can be differentiated with respect to distance x along the channel, giving
dE dz
dx + dx = 0
which might also be written
dy dE
dz
=
dx dy
dx
dE
Expressing dy in terms of the Froude number (from equation (2.35)) yields
dy
dz
2) = (1
F
dx
dx .................................................... (2.37)
Once again we see that, if there is a downward step in the bed (i.e., dz/dx is negative), then the
left-hand side of equation (2.37) must be positive. It follows that, when the flow is subcritical
(F<1.0), dy/dx must be positive, indicating a rise in the water surface over the step. Similarly,
2.29
Chapter 2: The energy equation for open-channel flow
for supercritical flow (F>1.0), dy/dx must be negative, implying a drop in the water-surface over
the step. Similar reasoning for the case of an upward step (positive dz/dx) confirms that
subcritical flow is out of phase with the bed and drops over the step while supercritical flow is in
phase and rises over the step.
This relationship between the propagation rate for disturbances and the mean flow-velocity
highlights a fundamental difference between subcritical and supercritical flow. Subcritical flow
can be influenced by downstream conditions because the related effects are transmitted upstream
at a faster rate than the river flows downstream. Consequently, an obstruction or waterfall might
produce a respective upstream backwater or drawdown effect in subcritical flow in a way that
simply is not possible in supercritical flow. Subcritical flow is able to adjust to a channel
transition somewhat before it actually arrives at the source of disturbance so that the complete
process of adjustment often is made through gradually varied flow. In summary, we say that
subcritical flow is subject to downstream control.
Although this and the following explanation may seem a little anthropomorphic, it captures the
physics of the phenomenon to say that, because supercritical flow is moving faster downstream
than the upstream propagation of disturbances ahead of it, it arrives unexpectedly at sources of
disturbance downstream so that necessary flow adjustments must be made abruptly at relatively
severe transitions. Because supercritical flow must be induced by some upstream condition
which raises the Froude number above unity, we say that supercritical flow is subject to
upstream control. These are not intuitively comfortable notions to most people and you might
find it useful to remember an old engineering saying that goes, “unlike subcritical flow,
supercritical flow doesn't know what it’s doing 'til it gets there”; or a variant that says,
“supercritical flow gets to where it's going before it knows that it's there!”
In this regard it may be helpful to consider the control exerted on the flow by a simple sluice
gate. Such gates commonly are used to control flows in laboratory flumes, irrigation canals and
reservoir outfalls; Figure 2.10 illustrates the case where such a sluice gate is slowly lowered into
an open-channel flow. With the gate at position a above the water surface, the depth and
velocity accord with the specific energy level Ea at point A in the subcritical flow domain
(F<1.0). When the gate is lowered to position b, the depth and velocity here are thus forced to
2.30
Chapter 2: The energy equation for open-channel flow
conform to those consistent with the lower specific energy level at point B . Once the flow
passes under the gate, however, depth increases as the flow adjusts back along the specific
energy line from B to A. Only the subcritical alternative depth is accessible from A because
access to the supercritical domain would require further flow acceleration so that specific energy
could first fall to critical at yc and there is no physical mechanism to produce such acceleration
of the flow.
sluice
gate
water surface at
gate position d
a
water surface at
gate positions b
a, b and c
water surface at gate
level a, b, c, and d.
flow
depth
ye
E
ya
c
y
b
y
c
d
yd'
ye'
B
C
D'
ya'
subcritical
flow domain
A
F<1.0
A'
E'
supercritical
flow domain
F>1.0
0
E min
Ed
Ea
specific energy, E
2.10: Control characteristics of a sluice gate set at various heights above the bed of a river.
If the sluice gate is lowered further to position c, forcing on the flow the critical depth yc, as the
depressed water surface moves downstream of the gate it will either again rise back to A through
B to reestablish the flow at the original depth ya, or else it will adjust to the alternative depth y'a
corresponding to A' on the specific energy curve. Both alternative depths are accessible from C
and which one is accessed depends on the downstream conditions. Generally, if there is no
downstream control (such as another sluice gate), the flow will access the supercritical
alternative depth. Once the gate has been closed to the critical depth it is not possible to close it
further without affecting the flow conditions upstream.
Remember, at critical depth yc the
discharge is at a maximum for the given specific energy and the specific energy is at the
minimum necessary to maintain that discharge. Consequently, by lowering the sluice gate to
position d we make an impossible demand on the flow. Although we now have reduced the depth
head, there is not a sufficient corresponding increase in the velocity head necessary to maintain
2.31
Chapter 2: The energy equation for open-channel flow
the discharge required by continuity. In effect, we have slipped off the specific energy curve in
Figure 2.10 and onto one corresponding to a lower discharge. The physical consequences of this
circumstance is that the flow backs up behind the sluice gate until the upstream water-level rises
to the depth ye where the new specific energy Ed and the required discharge are once again
equilibrated.
In summary, several general observations can be made about the behaviour of the flow under a
sluice gate. First, although advective acceleration of the flow under the sluice gate lowers the
specific energy there, as long as conditions remain in the subcritical flow domain, the flow
immediately downstream of the gate simply will return to the initial flow conditions. Second, if
flow depth downstream of a sluice gate remains at or less than the sluice-gate opening (i.e., it
does not return to the upstream subcritical flow conditions), then flow under and beyond the
gate must be supercritical. Third, the lowest possible setting of a sluice gate that does not
interfere with the upstream flow occurs at critical depth. Finally, we can see from this example,
that a sluice gate can exert a downstream control on subcritical flow but that it exerts an
upstream control in the case of supercritical flow.
water surface
a
q max
H = E o = constant
LAKE
sluice gate
discharge/depth
curve
b
y
c
c
c
yc
o
yb y a
q
OUTFLOW
critical outflow depth
b
a
c
b
weir
2.11: Discharge-depth relations in a lake or reservoir outflow across
a broad-crested weir controlled by a sluice gate.
It also is quite instructive to consider the case of a sluice gate which dams the outflow of such a
large body of water that the lake level remains sensibly unchanged when the gate is opened for
only a short period of time. As before, we assume that no energy is lost through the transition
and that the outflow has a rectangular cross-section. In this case, shown in Figure 2.11, the lake
2.32
Chapter 2: The energy equation for open-channel flow
outflow is zero when the gate is closed and flow occurs across a broad-crested weir as the
gate is opened.
We specify a broad-crested weir so that the pressure distribution throughout
can be regarded as hydrostatic. Flow over a short or sharp-crested weir would be influenced by
the drawdown of the free-falling outflow and would violate the assumptions of our simple
model.
When the sluice gate shown in Figure 2.11 is closed at a, qa = 0 and the total head is the depth of
water H above the weir. When the gate is opened to position b water begins to flow out under
the gate and the outflow equilibrates when the depth of water over the weir drops from ya to yb;
the total head H remains constant although specific energy over the weir must be declining. As
the sluice gate is gradually opened further, the depth of flow over the weir will continue to
decrease as discharge increases to the maximum flow for the given head. Clearly, the discharge
/depth curve applicable here is simply the subcritical flow domain of the curve for a fixed
specific energy, Eo, shown in Figure 2.9. Thus we know that, since position c on the
depth/discharge plot in Figure 2.12 corresponds to q = qmax, the depth of flow over the weir
must be critical at y = yc. Furthermore, if we raise the sluice gate further, the depth of flow over
the weir will remain unchanged, and the flow will continue to discharge at the maximum rate
fixed by the available head.
A general conclusion we can derive from this example is that all lake or reservoir outflows
across uncontrolled broad-crested weirs will discharge at critical depth (and velocity). Indeed,
if we know the magnitude of the head H in Figure 2.12, we can use these relationships to solve
a variety of flow problems such as those considered in Sample Problems 2.6 and 2.7. We must
not forget, however, that we have assumed that there is no resistance to flow in our model. In
considering a real outflow we might have to modify our predictions made here in order to
account for the fact that H may decline through the outflow as a result of friction. In general
such modifications will not alter the general conclusion (that flow will be critical) but it might
mean that the critical flow is restricted to a smaller portion of the weir than implied by our
uniform flow model.
Implicit in the sluice gate examples considered above is the notion that it is not always possible
to simultaneously satisfy both the energy equation and continuity. Just as some settings of a
2.33
Chapter 2: The energy equation for open-channel flow
sluice gate will obstruct an orderly transition of the flow, causing a backwater, so it is that
channel width contractions or bed steps above a certain magnitude will cause the same problem.
When the flow is thus obstructed by too severe a transition, it is said to be choked. The
Sample Problem 2.6
Problem: If the total head H = 1.0 m for the outflow shown in Figure 2.11, what is the two-dimensional discharge at
the outflow when the sluice gate is fully raised?
Solution: Since the outflow depth must be critical, from equation (2.30) we know that yc = 2/3Eo = 2/3H =2/3 x 1.0
= 0.667 m. We also know from equation (2.33) that, in critical flow vc =
gyc so that in this case vc =
-1
9.806 x 0.667 = 2.557 ms . Thus the two-dimensional discharge is q = vcyc = 1.706 m2s-1.
We might note here that, for all problems of this type, the two-dimensional discharge is specified by the
general relationship:
q = 2/3H 2/3gH ............................................................(3.38)
Sample Problem 2.7
Problem: If the water-surface of the outflow shown in Figure 2.11 is 45 cm below the level of the lake when the
sluice gate is fully raised, what is the discharge at the outflow?
Solution:
From the energy equation we know that H = y + v2/2g and therefore that H - y = 0.45 = v2/2g. Thus, v
= 0.45 x 2g = 2.971 ms-1. But since the flow across the weir must be critical, we also know that
v = vc = 2.971 ms-1 and that 2.971 =
gyc . It follows that 8.825 = gyc and that yc = 0.900 m. Hence the two-
dimensional discharge q = 2.971 x 0.900 = 2.674 m2s-1.
Sample Problem 2.8
Problem: What is the maximum step height possible before a choke forms in the flow transition described in
Sample Problem 2.2 (a 5m-wide rectangular channel carrying 15 m3s-1 discharge, 1.5 m deep upstream of the step)?
Solution:
The maximum step height Dy is the difference between the upstream specific energy, E1 and the
minimum possible specific energy at critical flow, Ec (i.e., Dy = E1 - Ec) . Noting that q = Q/w = 15/5 = 3 m2s-1
and V1 =q/y1 = 3/1.5 = 2 ms-1, upstream of the step, specific energy E1 is given by
v 12
4.00
E1 = y1 + 2g = 1.500 + 2 x 9.806 = 1.704 m
v c2
The minimum specific energy at critical flow is Ec = yc + 2g where yc can be determined from equation (2.27) as
3
yc =
q2
g
3
=
3.02
vc = 3/0.972 = 3.086 ms-1.
9.806 = 0.972 m; and from continuity,
3.0862
Thus Ec = 0.972 + 2 x 9.806 = 1.458 m.
Therefore the maximum permissible step height Dy = E1 - Ec = 1.704 - 1.458 = 0.246 m
2.34
Chapter 2: The energy equation for open-channel flow
transition limits beyond which choking will occur are readily determined.
Recalling the discussion in relation to Figure 2.8, in the case of a positive step the maximum step
height is simply the difference between the upstream specific energy and the minimum possible
specific energy (when the flow is critical); an example of such a determination is worked in
Sample Problem 2.7.
Sample Problem 2.9 poses a problem which is insoluble. We know from Sample Problem 2.8
that, under the specified flow conditions the maximum possible height of the step is 0.246 m, so
clearly a step of 0.355 m cannot be negotiated by the flow because the specific energy over the
step must fall below that necessary to maintain the constant discharge. Of course, this physically
impossible situation does not prevent us from deriving an equation and herein lies a warning: we
must remain alert to the fact that generating a Bernoulli-based cubic expression does not
necessarily mean that we have solved the flow transition problem at hand nor does a failure to
reach convergence in our mathematical iteration imply that our computations are in error!
Sample Problem 2.9
?
v1
1.5 m
?
35 cm
v
2
y
2
Problem: Water flows through a 5 m-wide rectangular channel and over a 35 cm step up in the bed. If the discharge
is 15 m3s-1, and the initial depth upstream of the step is 1.5 m, what will be the depth and velocity of the flow
downstream of the transition?
Solution: Noting that q = Q/w = 15/5 = 3 m2s-1 and V1 =q/y1 = 3/1.5 = 2 ms-1, we set up the Bernoulli equation for
this transition as follows:
v12
v22
y1 + 2g + z1 = y2 + 2g + z2
4.00
1.500 + 2g
which simplifies to
v 22
+ 0 = y2 + 2g
+ 0.350
v 22
1.354 = y2 + 19.612
From continuity, y2v2 = 3, so that v2 = 3/y2. Substituting for v2 above and rearranging yields
y23 - 1.354y22 + 0.459 = 0
It is left to the reader to verify that this equation has no meaningful solution (there is a physically meaningless
negative solution at y2 = - 0.498 m).
2.35
Chapter 2: The energy equation for open-channel flow
Flow transitions in three-dimensions
In the discussion to this point we have kept our analysis simple by assuming a rectangular
channel of constant width because such a regular cross-section readily lends itself to a twodimensional approach to the problems. The flow principles established for this simple case,
however, are readily extended to three-dimensional flow and to other forms of channel crosssection (trapezoidal, semi-circular, parabolic, for example). We will not fully develop these
alternative models here although they are readily available elsewhere (an excellent source is
Henderson, 1966). It will be useful for our purposes, however, to briefly consider the adaption
of the energy equation to transitions in rectangular channels of variable width.
The key to dealing with three-dimensional flow transition problems in rectangular channels is to
recognize that equations (2.21) and (2.22), respectively for the total and specific energy, can also
be expressed in terms of discharge, Q, as follows:
Q
(Q/wy)2
+
+ z = H ....................................................... (2.39)
wv
2g
Q
(Q/wy)2
+
= E ............................................................. (2.40)
wv
2g
It turns out that, in terms of the specific energy/depth relationships, a channel expansion has
exactly the same effect on the flow as an increase in depth (a negative step) and a channel
contraction is analogous to a reduction in depth (a positive step). Furthermore, the previously
discussed contrasted directions of change associated with subcritical flow on the one hand, and
with supercritical flow on the other, also apply in the case of responses to changes in channel
width. Thus, an accompanying width change can enhance a response to a bed-elevation change
or it can counter it.
Similarly, a flow can be induced to go critical by reducing the channel width just as it can be
choked by making it encounter too severe a contraction. These relationships are best explored
through some examples such as those developed in Sample Problems 2.10 to 2.14 to follow.
2.36
Chapter 2: The energy equation for open-channel flow
Sample Problem 2.10
y1 = 2.00 m
65.00 m
85.00 m
Q = 100 m 3 s-1
y2 = ?
v2 = ?
Plan view of
a channel
contraction
Problem: Water flows through an 85 m-wide rectangular channel at a rate of 100 m3s-1. If the flow is contracted
to 65 m width, what is the depth and velocity in the contracted section of the channel if the approach depth is 2.0 m?
Determine the extent and direction of change in the water-surface elevation through the channel transition.
Q
100.0
Solution: From continuity we get v1 = w y = 85.00 x 2.000 = 0.5882 ms-1 and from equation (2.39)
1 1
(Q/w2y2)2
0.58822
we can state that y1 + 2g
+ z1 = y2 +
+ z2
2g
so that
2.000 +
and
0.58822
+ 0 = y2 +
2g
2.0176
= y2 +
[100.0/(65.00 x y2)]2
+0
2g
0.1207
y 22
y23 - 2.0176y22 + 0.1207 = 0
or
Solving in the usual way, the two positive solutions are y2 = 0.263 m and 1.987 m. Since the flow
upstream of the contraction is in the subcritical flow domain (F1 = 0.588/ 9.806 x 2.00 = 0.13), the correct
alternative depth is y2 = 1.987 m. From continuity it follows that v2 = 100/(65.00 x 1.987) = 0.774 ms-1.
Since y2 =1.987 m, it also follows that the water-surface must fall by 2.000 - 1.987 = 0.013 m through the
contraction. Thus we can see that a contraction in channel width has the same qualitative effect on the flow as an
upward step in the bed (a boundary contraction in the vertical plane).
Sample Problem 2.11
Problem: To what width would the channel in Sample Problem 2.10 have to be contracted to achieve critical flow?
How far would the water surface drop under these conditions? What degree of contraction will choke the flow?
Solution: We know from equation (2.30) that, in this case where E1 = 2.0176 m,
yc = 2/3 Eo = 2/3(2.0176) = 1.345 m
From equation (2.26) it follows that vc = gyc = 9.806 x 1.345 = 3.632 ms-1.
100
From continuity,
w = 1.345 x 3.632 = 20.471 m
So in order to achieve critical flow the channel would have to be contracted from 85 m down to about 20
m. The drop in the water surface, Δh, is given by
Δh = y1 - y2
= 2.000 - 1.345 = 0.655 m.
Since a channel width of 20.471 m represents the maximum discharge for the available specific energy, it is also the
limiting condition for choking; i.e., any further contraction will here choke the flow.
Sample Problem 2.12
2.37
Chapter 2: The energy equation for open-channel flow
Q = 180 m 3 s-1
Δh = 5.0 cm
w2 = ?
60 m
Plan view of
a channel
expansion
y2 = ?
v2 = ?
y1 = 2.25 m
Problem: Water flows through a rectangular and symmetrical, but otherwise unspecified, channel expansion as
shown above. Given Q =180.0 m3s-1 and a 2.25 m-deep flow in the 60 m-wide approach channel, determine the
degree of expansion if the water surface rises by 5.0 cm. Calculate the mean depth and velocity in the expansion.
Solution: At the outset we might note from physical reasoning that, since the water surface drops in a contraction
(see Sample Problem 2.10), the rise in the water surface here implies that the channel width must expand to some
degree in this transition.
We know that, since the bed is sensibly horizontal, the increase in the water surface, Δh, must be equal to
the difference in flow depth through the transition, or
Δh = y2 - y1 and therefore y2 = y1 + Δh = 2.25 + 0.05 = 2.30 m
Q
180.0
From continuity we get v1 = w y = 60.00 x 2.250 = 1.333 ms-1 and, noting that z1 = z2, from equation (2.39)
1 1
 180.0 2
2
2.300w2
2
(Q/w
y
)
1.333
2 2
we can state that
2.250 + 2g
= 2.300 +
=
2.300
+
2g
19.612
which simplifies to
2.341 = 2.300 +
312.297
w 22
and further to
0.041w22 = 312.297; w2 = 87.275 m
Thus the 5 cm rise in the water-surface implies a channel expansion from 60.000 m to 87.275 m.
180.00
Continuity dictates that v2 = 87.275 x 2.300
=
0.897 ms-1.
Sample Problem 2.13
Δ h = 5 cm
Problem: Determine the height of the upward step at the expansion
in Sample Problem 2.12 necessary to counteract the 5 cm rise in the
water surface.
2.25 m
Δz
2.25 m - Δ z
Solution: In order for the water surface to neither rise nor fall through the expansion it is necessary that
180.0
180.0
y2 = 2.25 - Δz. Furthermore, we know from continuity that y2v2 = w
= 87.275 = 2.062 m2s-1. Thus,
2
2.0622
 y2 
2
1.333
0.2168
we can state that
E1 = 2.250 + 2g
= 2.342 = y2 + 19.612 + Δz = y2 +
+ Dz
y 22
Since Δz = 2.25 - y2,
2.342 = y2 +
0.2168
y 22
+ 2.25 - y2 =
0.2168
y 22
+ 2.25
0.2168
So, y22 = 0.092 = 2.36 and y2 = 1.535 m. Therefore, the step height to counteract the water-surface rise
associated with the flow expansion must be Δz = 2.25 - y2 = 2.250 - 1.535 = 0.715 m.
2.38
Chapter 2: The energy equation for open-channel flow
Sample Problem 2.14
100.0 m
Problem: Shown on the right are the dimensions of a
rectangular channel transition that involves both a
negative step and an expansion. If the channel is
discharging 500 m3s-1 of water and the approaching
flow has a mean depth of 3.00 m, calculate the mean
Froude number in the channel expansion. Although
the expansion is shown to occur abruptly, we will
assume, as in previous examples, that frictional head
loss is negligible so that energy in the transition is
conserved.
3.000 m
Water surface
0.500 m
Y2
120.0 m
Solution: The Bernoulli equation for this transition is
y1 +
3.000 +
v12
v22
+
z
=
y
+
1
2
2g
2g + z2
 500.0 2
100.0 x 3.0
19.612
which simplifies to
+ 0 = y2 +
3.142 = y2 +
Flow direction
 500.0 2
 120.0 x y2
19.612
0.885
- 0.5
y 22
- 0.5
or
y23 -3.642y22 + 0.885 = 0
Solving by iteration yields two positive solutions at y2 = 0.534 m and 3.573 m. Since the approach flow is
1.67
subcritical (F1 =
= 0.307), the correct solution is 3.573 m. From continuity we get v2 =
9.806 x 3.000
500.0
1.166
-1
120.0 x 3.573 = 1.166 ms & in the channel expansion, F2 = 9.806 x 3.573 = 0.197.
On the other hand, if the discharge through the channel is 2 000 m3s-1, the Bernoulli equation becomes
 2000.0 2
 2000.0 2
100.0 x 3.0
 120.0 x y2
3.000 +
+
0
=
y
+
- 0.5
2
19.612
19.612
which simplifies to
5.766 = y2 +
14.164
y 22
or
y23 - 5.766y22 + 14.164 = 0
Once again, solving by iteration yields two positive solutions at y2 = 1.919 m and 5.253 m. But here the
6.667
approach flow is supercritical (F1 =
= 1.229) and y2 = 1.919 m is the correct alternative depth.
9.806 x 3.000
2000
8.685
From continuity, v2 = 120.0 x 1.919 = 8.685 ms-1 and F2 =
= 2.002.
9.806 x 1.919
For Q = 500 m3s-1 the subcritical flow through the transition involves an increase in the water-surface
elevation of 0.073 m [3.573 - (3.000 + 0.500)] but in the supercritical flow at Q = 2000 m3s-1 the water surface
drops by an impressive 1.581 m [1.919 - (3.000 + 0.500)] as the flow enters the expansion. Of course, a discharge
of 2000 m3s-1 would not be contained by this channel if the flow were not supercritical because the equivalent
subcritical flow depth in the expansion (5.253 m) would significantly exceed the bank height.
2.39
Chapter 2: The energy equation for open-channel flow
Accounting for energy losses
In all of these flow-transition problems employing the energy equation, the solution follows
from the assumption of energy conservation in which E1=E2. In real rivers this can never be true
because energy is always “lost” in overcoming frictional drag at the boundary and is drained
further to drive turbulence and secondary flow (lateral rather than downstream flow). In channel
transitions where the boundary is smooth and the boundary transition is such that it causes just
gradually varied flow, the loss in energy or head may be very small or even negligible. Flow
over rough boundaries through abrupt channel transitions, however, may involve considerable
head loss and the energy equation will only describe flow accurately if there is an accounting for
this loss.
In problems where there is significant energy loss (ΔE) we modify the energy balance so that
E1 = E2 + ΔE ................................................ (2.42)
thereby recognizing that the initial energy E1 is partitioned into that available for allocation to
the depth and velocity heads in the flow after the transition (E2) and a component used in
overcoming frictional drag and turbulence.
In most flow-transition problems we know the amount of initial energy (E1) available and we
need to compute the flow parameters for E2. Rearranging equation (2.42) yields E1-ΔE = E2 ,
reminding us that, whenever energy is lost in a channel transition, E1<E2 by the amount of the
energy loss. For example, one of our first two-dimensional step problems (Sample Problem 2.2)
is reworked below (Sample Problem 2.15) to allow for a two per cent head loss as the flow
adjusts to an upward step in the bed (ie, E1 is discounted to 98%).
The two per cent head-loss in Sample Problem 2.15 only amounts to about 3.4 cm but the effect
on the energy-based solutions for depth and velocity over the step are significant (see Figure
2.12).
Although the directions in which changes occur remain unchanged from the fully energyconserved case, the reduction in total head forces the flow to adopt a higher velocity and
reduced flow depth in order to maintain continuity of flow through the transition. In other
words, the flow state has moved much closer to critical flow. Indeed, if we were to impose a
2.40
Chapter 2: The energy equation for open-channel flow
Flow parameter
Sample Problem 2.2
(no head loss)
Sample Problem 2.15
(2% head loss)
Available head
Flow depth, y2
Flow velocity, v2
Water-surface elev.
Froude number, F2
1.704 m
1.167 m
2.571 m/s
-0.133 m
0.76
1.670 m
1.066 m
2.814 m/s
-0.234 m
0.87
Relative
Change
-2.0%
-8.7%
+9.5%
+14.5%
2.12: Comparison of flow parameter changes in Sample Problems 2.2 and 2.15 illustrating the effects of
a 2% head loss through the transition.
five per cent head loss in this particular case we would find that there is no longer enough energy
to maintain the discharge and the flow would choke.
Sample Problem 2.15
?
v1
1.5 m
?
20 cm
v
2
y
2
Problem: Water flows through a 5 m-wide rectangular channel and over a 20 cm step up in the bed. If the discharge
is 15 m3s-1, and the initial depth upstream of the step is 1.5 m, what will be the depth and velocity of the flow
downstream of the transition if there is a 2 per cent head loss due to frictional drag?
Solution: Initially, we treat this as a two-dimensional problem.
Noting that q = Q/w = 15/5 = 3 m2s-1 and V1
=q/y1 = 3/1.5 = 2 ms-1, we set up the Bernoulli equation for this transition as follows: 1.70395676


v2
v 2
0.98 y1 + 1 + z1 = y 2 + 2 + z 2
2g
2g




4.00
v 2
0.98 1.500+
+ 0 = y2 + 2 + 0.200
2g
2g


This equation simplifies to
v 22
0.98(1.704) -0.200 = 1.452 = y2 + 19.612
From continuity, y2v2 = 3, so that v2 = 3/y2. Substituting for v2 above and rearranging yields
y23 - 1.470y22 + 0.459 = 0
which by iteration has an appropriate solution at y2 = 1.066 m. This depth change implies that the water surface
must drop by 0.133 m over the step [(1.066 + 0.20) -1.50) = -0.234 m]. From continuity it also follows that v2 =
3/1.066 m = 2.814 ms-1.
Consider the three-dimensional flow transition problem treated in Sample Problem 2.14. If we
now account for the previously ignored energy loss resulting from the rapid expansion of the
2.41
Chapter 2: The energy equation for open-channel flow
flow, the problem can be reworked as in Sample Problem 2.16. Here we allow for a rather
extreme ten per cent loss of initial energy to friction and turbulence (ie, E1 is discounted to
90%). The comparative flow parameters for each case are summarized in Figure 2.13.
Sample Problem: 2.16
Problem: Shown on the right are the dimensions of
a rectangular channel transition which involves both
a negative step and an expansion. If the channel is
discharging 500 m3s-1 of water and the approaching
flow has a mean depth of 3.000 m, calculate the
mean Froude number in the channel expansion.
Because the channel expansion occurs abruptly, we
will assume that there is a 10% frictional head loss
incurred as the flow moves through the transition.
100.0 m
3.000 m
Water surface
0.500 m
Y2
120.0 m
Solution: The Bernoulli equation for this transition
is
y1 +
v12
v22
+
z
=
y
+
1
2
2g
2g + z2
3.000 +
 500.0 2
100.0 x 3.0
19.612
so that
Flow direction
+ 0 = y2 +
 500.0 2
 120.0 x y2
3.142 = y2 +
19.612
- 0.5
0.885
- 0.5
y 22
Here E1 = 3.142 m so discounting for the energy loss (0.90 x 3.142 = 2.828 m) yields
2.828 = y2 +
0.885
- 0.5
y 22
which simplifies to
y23 -3.328y22 + 0.885 = 0
Solving by iteration gives a (subcritical) solution for y2 = 3.244 m. From continuity we get
1.284
500.0
v2 =
= 1.284 ms-1 & in the channel expansion, F2 =
= 0.228.
120.0x3.244
9.806x3.244
Flow parameter
Sample Problem 2.14
(no head loss)
Sample Problem 2.16
(10% head loss)
Relative
Change
Available head
Flow depth, y2
Flow velocity, v2
Water-surface elev.
Froude number, F2
3.142 m
3.573 m
1.166 m/s
+0.073 m
0.20
2.828 m
3.224 m
1.284 m/s
-0.276 m
0.23
-10.0 %
-9.8 %
+10.1%
+15.0%
2.13: Comparison of flow parameter changes in Sample Problems 2.14 and 2.16 illustrating the effects
of a 10% head loss through the transition.
2.42
Chapter 2: The energy equation for open-channel flow
Here again, the reduced available head has resulted in a lower depth of flow and an increased
velocity in order to maintain flow continuity through the transition. In this low-Froude-number
flow the relative changes in depth and velocity are about the same as the reduction in the
available head. Note that, because less energy is available to allocate to the depth head, the
water-surface elevation now actually drops through the expansion, contrary to our expectations
in the energy-conserved case.
As we noted at the outset, the degree of energy loss through short gradual transitions in river
channels is negligible and usually can be ignored. If it is suspected that energy loss is
significant then an allowance should be made for it. Losses of a few per cent of the initial head
are common and a ten per cent loss is rather extreme. As in many aspects of river science,
experience is the ultimate guide to the appropriate discounting factor.
Some concluding remarks
There are many circumstances where energy loss in a river will always be significant and
difficult to evaluate. Because there is a sensitive relationship between changes in the head and
the responses in the flow depth and velocity, the energy approach is not always the most robust
basis for describing the flow. It is particularly suspect at high Froude numbers where energyexchange responses are at their most sensitive. But we have some other tools available for such
circumstances and one of these, the momentum approach, is the subject of the next chapter.
References
Henderson, F.M., 1966, Open Channel Flow, Macmillan, New York.
2.43
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