ch5 energy in steady flow

7/27/2019 Ch5 Energy in Steady Flow

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Ch 5 Energy in Steady Flow
http://e/%ED%8C%91%E7%AF%8B%EE%81%A6%D1%A7/%CD%A2%CE%84/LECTURE%20NOTE/Clips/V4_5.mov


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Outline

The fundamental equations of fluid dynamics

The continuity equation (principle ofconservation of mass)

The energy equation (Bernoulli equation)

Application of the energy equation


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Derive the Bernoulli (energy) equation

Use of the Bernoulli Equation

Introduce the momentum equation for a fluid

Demonstrate how the momentum equation andprinciple of conservation of momentum is used to

predict forces induced by flowing fluids

Objectives


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5.1 Energies of a Flowing Fluid

kinetic energy

22

2

1

2

1

, VmVKEenergyKinetic ==

g

V

g

VmV

Weight

KE

2

22

212

21

=

=

=

=== === gWweight volumemmass

Kinetic energy per unit weight


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5.1 Energies of a Flowing Fluid (Con)

Potential energy

Due to gravity

hgh

gh

W

EP

Weight

EP=

==

...

ghmghEP == .


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5.1 Energies of a Flowing Fluid (Con)

Pressure head

Fluids have energy in the form of pressure.

Weight

Energyessure

lengthr

p

hhpFrom

Pr

===


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5.2 Equation for Steady Motion of An Ideal

Fluid Along A Streamline

gdAdzds

dzgdAdsagdAdsadG === coscos

Pressure forces:

upstream end +pdA

downstream end -(p+dp)dA

Gravity force:

dt

dudAdsgdAdzdAdpppdA =+ )(

(Streamline dir)

maF=

Deriviation of Bernoull i equation


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)2

(2u

ds

d

ds

duu

dt

ds

ds

du

dt

du===

dt

dudAdsgdAdzdAdpppdA =+ )(

For steady flow ,velocity (u) only varies

with distance(s)

Substituting for du/dt

0)2

(2

=++g

u

g

pz

ds

d

C

g

u

g

pz =++

2

2

Bernouillis equation

(Along the streamline)


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Assumption ()

)(0 fluidInviscid=

)(. fluidibleIncompressconst=

Along the streamline (1-D)

Steady flow


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Bernoulli Equation

=g

p

=+ zg

p

The Bernoulli Equation is a

statement of the conservation

of ____________________Mechanical Energy

Pressure head

z =Elevation head

Velocity head

Piezometric head

Total head

Energy Head Line

Hydraulic Grade Line

Cg

uzg

p =++2

2

=g

u2

2

=++g

uzg

p2

2

Bernoullis eqn. is a useful relationship between p, V and z


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5.3 Energy line (EL) and

Hydraulic Grade line(HGL)

It is often convenient toplot mechanical energy

graphically using heights.

Hydraulic Grade Line

Energy Line (or totalenergy)

PH G L z

g= +


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Fig Energy Grade Line (EGL) and Hydraulic

Grade Line (HGL) for an one-dimensional flow

Energy Grade Line


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)2

(2

g

uz

g

p ++

Energy Grade Line (EGL)= Total head line

Hydraulic grade line (HGL) = Piezometric head line

(Piezometer)

+ z

p

()


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For the two cross- sections

g

u

g

pz

g

u

g

pz

22

2

222

2

111 ++=++

Eliminate the constant in the Bernoulli equation?

Apply at two points along a streamline.

Bernoulli equation does not include

Mechanical energy to thermal energy

Heat transfer


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Solving Steps with Bernoulli Equation

Three selections of steps for solving problems

and one solving-method .

1.Selecting the datum plane.

2.Selecting computation cross sections: It should be

the cross-section in uniform or gradually varied flowswith variables already known as many as possible.

3.Selecting the computed point : For the tube flow, it is

usually on the tube axis; while for the open channelflow, it is usually on the free surface.

4.Listing the energy equation and solving the problem.

Bernoulli Equation: Simple Case


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Bernoulli Equation: Simple Case

(V = 0)

Reservoir (V = 0)Put one point on the

surface, one pointanywhere else

g

uz

g

p=++

2

2

z

Elevation datum

Pressure datum

1

2

Same as we found using statics

We didnt cross any streamlinesso this analysis is okay!

22

11 z

g

pz

g

p+=+

g

pzz

221 =

Hydraulic and Energy Grade Lines


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Mechanical Energy

Conserved

Hydraulic and Energy Grade Lines

(neglecting losses for now)

The 2 cm diameter jet is

5 m lower than the

surface of the reservoir.

What is the flow rate(Q)?

z

p

g

z

2

2

V

g

Elevation datum

Pressure datum? __________________Atmospheric pressure

z

2

2V

g

2

2

p Vz C

gg+ + =

How do we compensate


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How do we compensate

for energy losses?

We add an energy loss term!

On which side?


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Energy Equation in Steady Total Flow

Real fluid steady total flow energy equation ( per unit

weight fluid )

In which,

zspecific elevation energy elevation head

specific pressure energy pressure head,

piezometric weight )

specific kinetic energy velocity head)

specific potential energy piezometric head)

total specific energy total head

gp

g

v

2

2

g

pz

+

g

v

g

pzH

2

2

++=

wh average specific energy loss (head loss).


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Kinetic Energy Correction Factor

We have assumed in the derivation of Bernoulli

equation that the velocity at the end sections (1)and (2) is uniform. But in a practical situation this

may not be the case and the velocity can very

across the cross section. A remedy is to use a

correction factor for the kinetic energy term in the

equation.

= AVdAu33 /


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0 01

2

z1

hw

1

2

z2z

p1

p2

1v12

2g2v22

2g

Piezometric head

line

Total headline

p

v 2

2g


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v21

2

2

1Water surface

v1

1v12

2g

2v22

2g

z1

z2

hwTotal head line

Piezometric head line


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Summary

By integrating F=ma along a streamline we found That energy can be converted between pressure, elevation, and

velocity

That we can understand many simple flows by applying the

Bernoulli equation

However, the Bernoulli equation can not be applied toflows where viscosity is large or where mechanical energy

is converted into thermal energy.


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Question

A diffuse pipe locates horizontally, as is shown in

the figure. If the head loss is neglected, then the

character of the pressures exerting on the cross-section's center is:

A. p1>p2

B. p1=p2

C. p1


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Applications of Bernoulli Equation

Stagnation tube Pitot tube

Free Jets Orifice Venturi Sluice gate Sharp-crested weir

Applicable to contracting

streamlines (accelerating flow).

Pi T b


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Pitot Tubes

Can connect a differential pressure transducer todirectly measure V2/2g Can be used to measure the flow of water in

pipelinesPoint measurement!

Pit t T b


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Pitot Tubes

g

p

g

p

g

ABB

=+

2

2

Total energy at A = Total energy at B

Static pressure tube

0gHpB =)( 0 hHgpA +=

ghpp BAB 2)(2

==

H. De Pitot

(1675-1771)

Th V t i M t


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The Venturi Meter


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Example Venturi Tube

Given: Water 20oC, V1=2 m/s, p1=150 kPa,

D=6 cm, d=3 cm

Find: p2 and p3

Solution: Continuity Eq.

Bernoulli Eq.

2

12

112

2211

==

=

d

D

VA

A

VV

AVAV

D Dd

1

2

3

( )

( )

kPap

Pa

VdDp

VVpp

g

Vz

p

g

Vz

p

120

2]3/61[2

1000000,150

]/1[2

)(2

22

2

24

21

41

22

2112

22

22

21

11

=

+=

+=

+=

++=++

Similarly for 2 3, or 1 3

Pressure drop is fully recovered, since we

assumed no frictional losses

kPap 1503

=

Nozzle: velocity

increases, pressure

decreases

Diffuser: velocity

decreases, pressure

increases

( ) ]/1[

)(24

212

Dd

ppV

=

Knowing the pressure drop 1 2 and

d/D, we can calculate the velocity and

flow rate


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Example

Given: Velocity in outletpipe from reservoir is 6

m/s and h = 15 m. Find: Pressure at A.

Solution: Bernoulli

equation

kPap

g

Vhp

gVp

gh

g

Vz

p

g

Vz

p

A

AA

AA

AA

A

0.129

)81.9

1815(9800)

2(

20

200

22

2

2

22

11

1

=

==

++=++

++=++

Point 1

Point A


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ch5 energy in steady flow

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