energy equation.pdf

8/18/2019 energy equation.pdf

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Lecture 4Lecture 4Energy Equation for An IdealEnergy Equation for An Ideal

FluidFluid

Basic Hydraulic Principles CourseBasic Hydraulic Principles CourseRobert R. Holmes, Jr., PhD, P.E.Robert R. Holmes, Jr., PhD, P.E.


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WelcomeWelcome

Please put your phone onPlease put your phone on mutemutePleasePlease do not place us on holddo not place us on holdas sometimes youras sometimes your ““hold musichold music ””plays while you are on another callplays while you are on another call


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Please Ask Questions ThroughoutPlease Ask Questions Throughout

LectureLectureRaise your hand on the WEBEXRaise your hand on the WEBEX

I will recognize you and ask you to takeI will recognize you and ask you to takeyour phone off mute to askyour phone off mute to ask


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Main Points for this LectureMain Points for this Lecture

Continuity EquationContinuity Equation

Energy EquationEnergy Equation

C Z Y g

V Z Y

gV

B B B

A A A =++=++

22

22

B B A A V AV A

AV Q=

=


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Understanding the EnergyUnderstanding the Energy

Equation and its use isEquation and its use is CRUCIALCRUCIALto success in this courseto success in this course


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Overview of LectureOverview of Lecture

Flow characterization by time dependencyFlow characterization by time dependency

StreamlinesStreamlinesContinuity equationContinuity equation

Energy equationEnergy equationWork a problem exampleWork a problem exampleQuestions on concepts or homeworkQuestions on concepts or homeworkproblems in chapter 4problems in chapter 4Other questionsOther questions


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Flow CharacterizationFlow Characterization

Fluid flow may be either steady orFluid flow may be either steady or

unsteady.unsteady.Steady flow exists when none of theSteady flow exists when none of the

variables in the flow problem change withvariables in the flow problem change withtime.time.If any of the variables change with time,If any of the variables change with time,

the condition of unsteady flow exists.the condition of unsteady flow exists.This discussion and most of this courseThis discussion and most of this coursedeals with steadydeals with steady --flow only.flow only.


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Steady Flow Implies:•Stage t=noon on Feb 24 = Stage t= midnight on Feb 24•Q t=noon on Feb 24 = Q t= midnight on Feb 24•Vt=noon on Feb 24 = V t= midnight on Feb 24

Is the above steady flow?


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To apply energy concepts, we oftenTo apply energy concepts, we often

have to make some steady flowhave to make some steady flowapproximationsapproximations

Example: We assume a steady flow at theExample: We assume a steady flow at the

peak of a flow in order to use indirectpeak of a flow in order to use indirectcomputation methods.computation methods.


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StreamlinesStreamlinesPath linesPath lines ——trace made by a single particletrace made by a single particleover a period of timeover a period of time

StreamlinesStreamlines ——Curve that is tangent to theCurve that is tangent to thedirection of velocity at every point on thedirection of velocity at every point on thecurvecurveStreamtubesStreamtubes ——In 2In 2 --dimensional space,dimensional space,this is the area between two streamlines.this is the area between two streamlines.

It resembles a tube or passageway.It resembles a tube or passageway.For Steady Flow, a path line and a streamline are identical

Let’s put aside the concept of streamlines for a bit while we developsome other concepts. We will come back to it.


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Continuity EquationContinuity EquationMatter can neither be created nor destroyedMatter can neither be created nor destroyed

is the principle ofis the principle of conservation of massconservation of mass Application of the conservation of

mass to steady flow in the streamtuberesults in the Equation of Continuity


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•In a small interval of time, Δ t, fluid at the beginning of the pipemoves Δs 1, which Δs 1 =v1 Δt, where v 1 is velocity in the area ofpipe that has a cross sectional area = A 1

Note: At all points

in this pipe, theparticles of thefluid movetangential to thestreamline, whichwould beexpected insteady flow

Δs 1

Δs 2

•If A1 is the cross sectional area of the beginning of the pipe.

The mass contained in the maroon area is M1=ρ 1 A1 ΔS 1= ρ 1 A1 v1 Δ t where ρ is the density of the fluid

•Similarly, the fluid moving through the enlarged section of pipe in Δt time has amass equal to M2=ρ 2 A2 ΔS 2= ρ 2 A2 v2 Δt.

•Because mass is conserved and the flow is steady, the mass that crosses A 1 in Δ t isthe same as the mass that crosses A 2 in Δ t or M 1=M2. Or:

ρ 1 A1 v1 Δt= ρ 2 A2 v2 Δt

For a steady flow in an incompressible fluid, ρ is constant. A 1 v1 = A 2 v2 =Q Q=Flow rate

Equation of Continuity


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Energy EquationEnergy Equation

Apply the conservation of energy Apply the conservation of energy

Assume an ideal fluid Assume an ideal fluid there is nothere is noshearingshearing stressstress therethere is no energy lossis no energy loss

from frictionfrom frictionConsider Consider ………… ....


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Streamline A

Streamline

Streamline

•Potential Energy —Weight times the distance above the datum. In ourdiscussion in this class, we will use units for energy as energy per pound offlowing water. Therefore, for section A in the diagram, the potential energy ofthe one pound parcel of water at A is Z A foot-pound per pound. We typically

allow the units of weight to cancel and express the units only in length terms.In English units, that will be feet. In hydraulics, we call the potential energyhead . So, we would say that at point A, we have Z A feet of head .


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Streamline A

Streamline

Streamline

•Pressure Potential Energy — A parcel of water, which is neutrallybuoyant, could rise to the surface without expenditure of energy. Its effectivepotential energy per pound of fluid is Z A + Y A. Y A is the pressure potentialenergy and is equal to the pressure at point A divided by the unit weight (p/ ρg).

Y A is known as the pressure head .

•Notice that the effective potential energy for any parcel of fluidat section A is the same and equal to the sum of the water depthplus the elevation of the bed above the datum.

Also note….the sum of the head and the pressure head, which above wecall the effective potential energy, is also known as the piezometric head


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Streamline A

Streamline

Streamline

Kinetic Energy: From Physics, we know that the kinetic energy ofa particle of mass m and a speed of v is defined as KE=1/2 mv 2. Fora pound of water, it has a mass m=1/g (from Newton’s 2 nd lawwhere weight is a Force (F) and F=ma, where in this case a is theacceleration of gravity (g)) Therefore, Kinetic energy of a onepound parcel of water KE=1/2(1/g)v 2=v 2/2g

Also known as velocity head


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Streamline

The sum of all 3 energies is called the total head or total energy

AStreamline

Streamline

For an IDEAL FLUID (no frictional resistance in the fluid), the totalenergy along a streamline is constant

C Z Y g

V Z Y

gV

B B B

A A A =++=++

22

22

Numerous engineering problems can be solved by this simplified situation,however, if friction losses are large, the results will be poor.

Example


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ExampleExample

Write down what you know and your unknowns:Write down the equations at your disposal to help you solve things:

A A v A = A B vB =Q

C Z Y g

V Z Y

gV

B B B

A A A =++=++

22

22

Continuity

Energy Equation

Example

V A, VB, D B

Example


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ExampleExample

We can solve for V A just by using the continuity equation.

A A v A =QContinuity

p

(5*8) v A =280

v A

=280/(5*8)

v A =7 ft/sec

Example


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ExampleExample

ft Z Y g

V A A A 011.825.25761.025.25)2.32(2

7

2

22

=++=++=++

Energy Equation

p

Utilizing the Energy Equation, I can solve for the total energy at Section A.

C Z Y g

V Z Y g

V B B

B A A

A =++=++

22

22

Example


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ExampleExample

1)2.32(22

011.822

++=++= B

B B B

B DV Z Y g

V ft

p

Still Utilizing the Energy Equation, I can solve items at section Bbecause since this is an IDEAL fluid, the total energy at section A isequal to the total energy at section B

C Z Y g

V Z Y g

V B B

B A A

A =++=++

2222

Example


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ExampleExample

1)2.32(2

8.2011.8 2

2

++= B

B

D D

ft

We don’t know V B or D B yet…..but we can once again use the continuity equation

A B vB=(D B*10)*VB =Q v B=Q/(D B*10)=280/(D B*10)=2.8/D B

B B

D D

ft += 2174.12

011.7

This is a cubic equation which can be solved by trial and error


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See Text Book on Page 31See Text Book on Page 31Different values of DB are assumed until the right hand side ofDifferent values of DB are assumed until the right hand side of thethe

equation is equal to 7.011equation is equal to 7.011

B

B

D D

ft += 2174.12

011.7

Using the computed valueof D B of 1.284, computethe velocity in section B


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Homework Problems Chapter 4Homework Problems Chapter 4


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Problem 4.1Problem 4.1

Part A. Use Continuity Equation

Remember:1. Write down all you

know and what youdon’t know2. Write down equations

B 1=20

B 2=20

V2=6 ft /s


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You know that Velocity Head

g

V

2

2Part B.


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Part C.Total Head

Piezometric Head


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Write the energy equation from Section 1 to 2

Part D.

22

22

11

21

22 Z Y

gV

Z Y g

V ++=++


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Part B


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Problem 4.3Problem 4.33.Compute the discharge in the 20-foot wide rectangular channel shownbelow. Draw and label the total head line and the water surface near the gate.

22

22

11

21

22 Z Y

gV

Z Y g

V ++=++

A 1 v1 = A 2 v2 =Q

Equations to Use


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Problem 4.3Problem 4.33.Compute the discharge in the 20-foot wide rectangular channel shownbelow. Draw and label the total head line and the water surface near the gate.


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Problem 4.4Problem 4.44.Compute the discharge and depth in the contracted section for the indicated

rectangular channel.

We have enough data to computevelocity head at section 1

Apply Continuity to getvelocity head at Section 2

Problem 4 4Problem 4.4


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Problem 4.4Problem 4.44.Compute the discharge and depth in the contracted section for the indicated

rectangular channel.

Apply the Energy Equation

Assume various values of D2 andsolve for Left Hand Side (LHS)

energy equation.pdf

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