Module 6
Lecture 5: Energy equation and numerical
problems
Energy in Gradually Varied
Open channel flow
 In a closed conduit there can be a pressure gradient that drives the flow.
 An open channel has atmospheric pressure at the surface.
 The HGL (Hydraulic Gradient Line) is thus the same as the fluid surface.
Module 6
Energy equation applied to open channel
v12
p2
v22
+ α1
+ z1 =
+ α2
+ z 2 + hL
2g
2g
γ
γ
p1
p1 and p2 : pressure forces per unit width at sections 1 & 2 respectively
v1 and v2 : velocity of flow at sections 1 & 2 respectively
α 1 and α 2 : energy coefficients
z1 and z2: elevations of channel bottom at sections 1 & 2 respectively w.r.t any
datum
hL: energy head loss of flow through the channel
Module 6
Energy equation applied to open channel
Contd…
Simplifications made to the Energy Equation:
1.
Assume turbulent flow (α = 1).
2.
Assume the slope is zero locally, so that z1 = z2
3.
Write pressure in terms of depth (y = p / γ).
4.
Assume friction is negligible (hL = 0).
i.e.
y1 +
2
1
v
2g
= y2 +
v
2
2
2g
E1 = E2
Module 6
Specific Energy Equation
Energy at a particular point in the channel = Potential Energy + Kinetic Energy
2
v
E = y+
2g
or
2
Q
E = y+
2 gA 2
where y is the depth of flow, v is the velocity, Q is the discharge, A is the crosssectional flow area and E is the specific energy i.e energy w.r.t channel bottom.
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Total energy
 When energy is measured with respect to another fixed datum , it’s
called Total Energy
2
v
E = y+z+
2g
where z is the height of the channel bottom from the datum
 Pressure head (y) is the ratio of pressure and the specific weight of
water
 Elevation head or the datum head (z) is the height of the section
under consideration above a datum
 Velocity head (v2/2g) is due to the average velocity of flow in that
vertical section
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Example Problem
Channel width (rectangular) = 2m, Depth = 1m, Q = 3.0 m3/s, Height above
datum = 2m. Compute specific and total energy
Ans: A = b*y = 2.0*1.0 = 2 m2
Specific energy =
Total energy =
2
Q
E = y+
2
3
E =1+
2gA2
2 * 9.81 * 2 2
Datum height + specific energy = 2.0 + 1.20 = 3.20 m
Module 6
Specific Energy Diagram
The specific energy can be plotted graphically as a function of depth of
flow :
E = Es + Ek
2
Q
where
E = y+
2gA2
Es = y ( Static energy)
2
Q
Ek =
2 gA 2
( Kinetic energy)
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Specific Energy Diagram
Contd…
Module 6
Specific Energy Diagram
Contd…
As the depth of flow increases,
the static energy increases and
the kinetic energy decreases,
y
∞
Es ⇒ y
∞ 1 Ek
The total energy curve
approaches the static energy
curve for high depths and the
kinetic energy curve for small
depths
As discharge (Q) increases, the specific energy curves move to the upper right
portion of the graph
Module 6
Thus, for flat slope (+ other assumptions…) we can graph y against E:
(Recall for given flow, E1 = E2 )
Curve for
different, higher Q.
For given Q and E, usually have 2 allowed depths:
Subcritical and supercritical flow.
Module 6
The specific energy is minimum (Emin) for a particular  critical depth –
Depth
Froude’s number = 1.0. & velocity = Vc.
Emin  only energy value with a singular depth!
Depths < critical depths – supercritical flow (Calm, tranquil flow)
Froude Number > 1.0. V > Vc.
Depths > critical depths – subcritical flow (Rapid flow, “whitewater”)
Froude Number < 1.0. V < Vc.
Example: Flow past a sluice gate
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Critical Depth and Froude Number
At the turning point (the left-most point of
the blue curve), there is just one value of
y(E).
This point can be found from
It can easily be shown that at
,
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Critical Depth and Froude Number
Contd…
The Froude number can be defined as:
(Recall that the Reynolds number is the ratio of acceleration to viscous forces).
The Froude number is the ratio of acceleration to gravity
Perhaps more illustrative is the fact that surface (gravity) waves move at a speed
of
Flows with Fr < 1 move slower than gravity waves.
Flows with Fr > 1 move faster than gravity waves.
Flows with Fr = 1 move at the same speed as gravity waves.
Module 6
Critical Depth and Froude Number
Contd…
Flows sometimes switch from supercritical to subcritical:
(The switch depends on upstream and downstream velocities)
Gravity waves: If you throw a rock into the water, the entire circular wave will
travel downstream in supercritical flow.
In subcritical flow, the part of the wave trying to travel upstream will in fact move
upstream (against the flow of the current).
Module 6
Example Problem
Will the flow over a bump be supercritical or subcritical?
Left
Right
Left
Right
or
As it turns out:
Left = subcritical
Right = supercritical
Using the Bernoulli equation for frictionless, steady, incompressible flow along a
streamline:
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Example Problem
Contd…
Apply Bernoulli equation along free surface streamline (p=0):
For a channel of rectangular cross-section,
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Example Problem
Contd…
Substitute Q = V z b into Bernoulli equation:
To find the shape of the free surface, take the x-derivative:
Solve for dz / dx:
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Example Problem
Contd…
Since
Subcritical flow with dh / dx > 1
Supercritical flow with dh / dx > 1
if flow is subcritical
subcritical: Fr < 1
supercritical: Fr > 1
dz / dx < 1
dz / dx > 1
if flow is supercritical
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Hydraulic Jump
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Hydraulic Jump
Contd…
There is a lot of viscous dissipation ( = head loss ) within the hydraulic jump.
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Hydraulic Jump
Contd…
Apply the momentum equation:
Momentum equation is used here as there is an unknown loss of energy (where
mechanical energy is converted to heat).
But as long as there is no friction along the base of the flow, there is no loss of
momentum involved.
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Hydraulic Jump
Contd…
Momentum balance:
The forces are hydrostatic forces on each end:
(where
and
are the pressures at centroids of A1 and A2 )
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Hydraulic Jump
Contd…
If y1 and Q are given, then for rectangular channel
is the pressure at mid-depth.
Here, entire left-hand side is known, and we also know the first term on the
right-hand side. So we can find V2.
Module 6
Exercises
1) A rectangular channel 4m wide has a flow discharge of 10.0 m3/s and depth of
flow as 2.5 m. Draw specific energy diagram and find critical and alternate
depth.
2) A triangular channel with side slopes having ratio of 1:1.5 has a discharge
capacity of 0.02 m3/s. Calculate:
a. critical depth
b. Emin
c. Plot specific energy curve
d. Determine energy for 0.25 ft and alternate depth
e. Velocity of flow and Froude number
f. Calculate required slopes if depths from d are to be normal
depths for given flow.
Module 6
Highlights in the Module
 Reynolds' transport theorem (Leibniz-Reynolds' transport theorem) is a 3D generalization of the Leibniz integral rule.
 Control volume is a definite volume specified in space. Matter in a control
volume can change with time as matter enters and leaves its control surface.
 Reynolds Transport Theorem states that the total rate of change of any
extensive property B of a system occupying a control volume C.V. at time ‘t’ is
equal to the sum of:
a) the temporal rate of change of B within the C.V.
b) the net flux of B through the control surface C.S. that surrounds
the C.V.
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Highlights in the Module
Contd…
 St.Venant equations are derived from Navier-Stokes Equations for shallow
water flow conditions.
 The solution of the St. Venant equations is known as dynamic routing,
which is generally the standard to which other methods are measured or
compared.
 Forces acting on the C.V. in an open channel flow are gravity force,
friction force , contraction/expansion force, wind shear force and unbalanced
pressure forces.
 Solutions to St. Venant equations :
 Method of characteristics
 Finite Difference methods : Explicit, Implicit
 Finite Element Methods
Module 6