lecture 5: energy equation and numerical problems

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

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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 ,

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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,

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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: 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

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Example Problem Contd…

Hydraulic Jump

Module 6

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.

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 )

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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

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