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Lecture 5: Energy equation and numerical problems
Module 6
Energy in Gradually VariedOpen channel flow
Module 6
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.
Energy equation applied to open channel
p1 and p2 : pressure forces per unit width at sections 1 & 2 respectivelyv1 and v2 : velocity of flow at sections 1 & 2 respectively
and : energy coefficientsz1 and z2: elevations of channel bottom at sections 1 & 2 respectively w.r.t any datumhL: energy head loss of flow through the channel
1α 2α
Lhzg
vpzg
vp+++=++ 2
22
22
1
21
11
22α
γα
γ
Module 6
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.
gvyg
vy 2222
2
21
1 +=+
21 EE =
Module 6
Energy equation applied to open channel Contd…
Specific Energy Equation
Module 6
Energy at a particular point in the channel = Potential Energy + Kinetic Energy
where y is the depth of flow, v is the velocity, Q is the discharge, A is the cross-
sectional flow area and E is the specific energy i.e energy w.r.t channel bottom.
2
2
2
2
2
gAQyE
org
vyE
+=
+=
Total energy
When energy is measured with respect to another fixed datum , it’s
called Total Energy
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 sectionModule 6
gvzyE 2
2++=
Example Problem
Module 6
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 =
Datum height + specific energy = 2.0 + 1.20 = 3.20 m
2
2
2gAQyE +=
2
2
2*81.9*231+=E
Specific Energy Diagram
Module 6
The specific energy can be plotted graphically as a function of depth of flow : E = Es + Ekwhere
)(2 2
2
energyKineticgAQEk =
)( energyStaticyEs =
2
2
2gAQyE +=
Specific Energy Diagram Contd…
Module 6
As the depth of flow increases,
the static energy increases and
the kinetic energy decreases,
The total energy curve
approaches the static energy
curve for high depths and the
kinetic energy curve for small
depths
ks EyEy 1∞⇒∞
Specific Energy Diagram Contd…
Module 6
As discharge (Q) increases, the specific energy curves move to the upper right
portion of the graph
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
Module 6
Critical Depth and Froude Number
It can easily be shown that at ,
Module 6
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
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 gravityPerhaps 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
Critical Depth and Froude Number Contd…
Example Problem
Will the flow over a bump be supercritical or subcritical?
As it turns out:Left = subcriticalRight = supercritical
Using the Bernoulli equation for frictionless, steady, incompressible flow along a streamline:
or
Left
Left
Right
Right
Module 6
Apply Bernoulli equation along free surface streamline (p=0):
For a channel of rectangular cross-section,
Module 6
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:
Module 6
Example Problem Contd…
Since subcritical: Fr < 1supercritical: Fr > 1
Subcritical flow with dh / dx > 1 dz / dx < 1Supercritical flow with dh / dx > 1 dz / dx > 1
if flow is subcritical if flow is supercritical
Module 6
Example Problem Contd…
Hydraulic Jump
Module 6
There is a lot of viscous dissipation ( = head loss ) within the hydraulic jump.
Module 6
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.
Module 6
Hydraulic Jump Contd…
Momentum balance:
The forces are hydrostatic forces on each end:
(where and are the pressures at centroids of A1 and A2 )
Module 6
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
Hydraulic Jump Contd…
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.
Exercises
Module 6
Highlights in the Module
Reynolds' transport theorem (Leibniz-Reynolds' transport theorem) is a 3-D 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.
Module 6
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