L13-RVF Energy I

8/3/2019 L13-RVF Energy I

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Rapidly Varied Flow --- the Energy Principle

The Energy Equation

Application of the Energy Equation

Flow over a smooth step

Specific Energy

Subcritical, Critical and Supercritical Flow

The Froude Number

Venturi Flume


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Common phenomenon! -- Develop mainly at hydraulic structures, e.g. flowmeasuring structures like weirs and flumes, regulation structures like gatesand valves, and discharge structures like spillways.

Important! -- Hydraulic structures form part of major water engineeringschemes, for irrigation, water supply, drainage, sewage treatment,hydropower, etc.

Difficult! -- Water surface is highly curved and the assumptions of

hydrostatic pressure distribution and parallel streamlines that areappropriate for uniform and gradually varied flows do not apply.

Solvable! -- Good approximation can be obtained by using the energy andmomentum principles (usually sufficiently accurate for engineering

purpose).

Introduction

Rapid changes in stage and velocity occur whenever there is a

sudden change in cross-section, a very steep bed-slope or someobstruction in the channel Rapidly varied flow.

Page 79


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Bernoulli equation for uniform flow in horizontal channel:

Energy Equation for Open Channel Flow

HZg

V

g

p

2

2

Page 79


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Assuming hydrostatic pressuredistribution, at point A:

Energy Equation

AA gyp AA ygp

Bernoulli equation:

HZ

g

Vy

2

2

y--- water depth; Z--- elevation of the interested point above datum

or

HZ

g

V

g

p

2

2

* Valid for most of the engineering purposes because practical channels havevery small bed slopes (e.g. less than 1:100, corresponding to less than 0.57o).

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Question: In an steady uniform flow, the upstream

depth and discharge are known. Calculate the flowconditions (e.g. y2) after being interrupted by araised bed level (Z) when neglecting energy losses.

Energy equation + continuity equation

Bernoulli equation to 1 and 2:

Zg

Vyg

Vy 22

2

22

2

11

Continuity equation:

QAVAV 2211 qb

QyVyV 2211

Zgy

qy

gy

qy

2

2

2

22

1

2

122

What principles should be used?

Page 80


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Cubic equation with the only unknown y2.

Three solutions to the equation only one is correct for this situation.

0222 22

1

2

1

2

2

3

2

qy

qgyZgygy

Which solution is the correct one for y2?

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Specific Energy

Definition (Bakhmeteff 1912):Specific energy, Es--- energy of the flow referred to channel bed as datum:

For steady flow, the specific energy equation can be written as:

Concept of specific energy--- provide the extra information for solving theabove problem!

g

VyEs 2

2

g

AQ

yEs 2

)(2

In a rectangular channel of width b, Q/A = bq/by= q/y:

2

2

2gy

qyEs

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Specific Energy

--- Cubic equation in yfor a given Es(three solutions but only two will be positive)

qis constant:

constant2

)(

2

2

g

qyyEs or 2

constant)(

yyEs

Considering only positive solutions, the equation gives a curve with twoasymptotes:

y 0, Es

y , Es y

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Resolve the Problem

Uniform flow interrupted by a smooth hump:

Bernoulli equation:

Zg

Vy

g

Vy

22

2

22

2

11

ZEE ss 21

Point A on the curve corresponds to the specific energy at point 1 in the channel.

Point 2 in the channel must be at either point B or point B on the specific energy

curve.

The depth represented by B and B are alternate depths, i.e. the depth with the

same specific energy.

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Resolve the Problem

All points between 1 and 2 must lie on the specificenergy curve between A and B or B.

To reach point B, there are chances that

(for those points locate at the curves between B andB) which is physically impossible.

Flow depth at section 2 must correspond to point B

on the specific energy curve.

ZEE ss 21

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Maximum Step Size

The maximum step for a given

discharge and a given approachingspecific energy can be determined asthe difference between theapproaching specific energy and theminimum specific energy.

If the bottom step is bigger than themaximum step for the givenapproaching specific energy anddischarge, either the discharge or the

approaching specific energy has to bemodified.

yc

y2

y1

y

Es

Zmax

Z>Zmax

Es1

Es1

new

y1new

In the case when the discharge is kept constant:the approaching specific energy has to rise in order to provide a

larger limit for the maximum step.

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Maximum Step Size

when the approaching specific energy remains uncharged,

the discharge has to fall in order to obtain a new minimum specificenergy and thus to produce a bigger maximum step for the givenapproaching specific energy.

yc

y1

y

Es

Zmax

Z>Zmax

Es1

ycnew

q1

qnew

< q1

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Flow under a sluice gate

Neglecting head losses, the depth before and after the gate are alternatedepths as the specific energy is the same at both locations.

y1

y2

Energy level

y2

y1

y

Es

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Example

a) Determine the depth of flow downstream of a section in which the bedrises by 0.2m over a distance of 1m.

b) Evaluate the maximum step that the bed can rise for the givenapproaching specific energy and discharge.

Data:

Q= 10m3/s; b= 5m;

y1= 1.25m; Z= 0.2m

Find y2and Zmax

The discharge in a rectangular channel is 10m3/s, where the channel width

is 5m and the maximum depth is 2m. The normal flow depth is 1.25m.

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Solution (a)

Data:

Q= 10m3/s; b= 5m;

y1= 1.25m; Z= 0.2m

Find y2

Neglect frictional losses:

ZEE ss 21 38.12

525.1

10

25.12

2

21

11

gg

VyEs

where

2

2

2

2

2

2

2

222

2039.0

2

5

10

2 y

y

g

yy

g

VyEs

2.0

2039.038.1

2

2

2 y

y

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Solution (a)

Solution corresponding to:

Specific Energy

0

0.5

1

1.5

2

2.5

3

3.5

0.00 1.00 2.00 3.00 4.00 5.00 6.00

E (m)

h(m)

m18.12.038.112 ZEE ss

y2 0.9m

Z

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Solution (a)

y2 0.9 m isused as the initial estimate in a trial-and-error procedure for:

y2 (m) Es2 (m)0.9 1.15

1.0 1.2

0.96 1.18

* The depth of the raised section is 0.96m or the water level (stage) is1.16m. There is a drop of 9cm when the bed has raised 20cm.

2

2

2

2039.018.1

yy

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Solution (b)

Since , the maximum step can be sought when Es2 has the

minimum value, i.e. when:

Hence:

ZEE ss 21

02

2 dy

dEs

2

2

22

2039.0

yyEs

0)2(2039.01

2039.0

3

2

2

2

2

2

2

2

ydy

yyd

dydEs

3

24078.01 y m74.014078.03

2 y

m11.174.0

2039.074.0

2039.022

2

2min2 y

yEs

m27.011.138.121max ss EEZ

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Example

A rectangular channel expands smoothly from a width of 1.5m to 3.0m.Upstream of the expansion, the depth of the flow is 1.5m and the velocityof the flow is 2.0m/s. Estimate the depth of the flow after the expansion.

Data:

b1 = 1.5m; b2 = 3m; y1 = 1.5m; V1 = 2m/s

Find y2

Solution:

Since there is no change in the elevation of the channel bed, theupstream specific energy Es1 is equal to the downstream one Es2, or

21 ss EE

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Example

where

The velocity at the downstream station is:

The two positive solutions are: y2 = 1.6m, and y2 = 0.28m.

Right after an expansion of the channel, only a depth that is biggerthan the upstream depth is physically possible, and hence y2 = 1.6m.

m7.181.92

0.25.1

2

22

1

11

g

VyE

s and g

VyE

s 2

2

2

22

222

11

2

2

5.1

3

5.15.12

yyA

AV

A

QV

And therefore

m7.111.0

81.92

5.1

2 2222

2

2

2

2

222

yy

yy

g

VyEs 011.07.1

2

2

3

2 yyor

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Summaries

Know how to derive the energy equation for open channelflow

Be able to apply the energy equation to solve the problem

of uniform flow interrupted by a smooth hump

Understand the concept of specific energy and be able touse it to calculate and explain some open channel flowphenomena, e.g. uniform flow interrupted by a smoothhump

Documents

L13-RVF Energy I