Fluid Mechanics-61341

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Fluid Mechanics-61341

An-Najah National UniversityCollege of Engineering

Chapter [5]

Fluid Mechanics-2nd Semester 2010- [5] Flow of An Incompressible Fluid Dr. Sameer Shadeed1

Dr. Sameer Shadeed

Chapter [5]

Flow of An Incompressible Fluid

Euler’s Equation

Dr. Sameer Shadeed2 Fluid Mechanics-2nd Semester 2010- [5] Flow of An Incompressible Fluid

Euler’s Equation

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Euler’s Equation

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Bernoulli’s Equation

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Bernoulli’s Equation

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Constanthead

Elevation

head

Velocity

head

Pressure

The Energy Line (EL) and the Hydraulic Grade Line (HGL)

Each term in the Bernoulli’s equation is a type ofhead

P/g = Pressure Head

V2/2gn = Velocity Head

Z = Elevation head

EL is the sum of these three heads

HGL is the sum of the elevation and the pressureheads

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The Energy Line (EL) and the Hydraulic Grade Line (HGL)

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V2/2gEL

V2/2g

HGL

Understanding the graphical approachof EL and HGL is key to understandingwhat forces are supplying the energy thatwater holds

Point 1: Majorityof energy stored inthe water is in thePressure Head

Point 2: Majorityof energy stored in

The Energy Line (EL) and the Hydraulic Grade Line (HGL)

Q

Z

P/ g

P/g

Z

1

2HGL

of energy stored inthe water is in theelevation head

If the tube wassymmetrical, thenthe velocity wouldbe constant, andthe HGL would belevel

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Bernoulli’s Equation (Uniform Cross Section)

For uniform cross sections streamtubes, the velocity across the entire section is uniform as a result Bernoulli’sequation becomes:

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Example 1

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Example 1 (Solution)

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Application of Bernoulli’s Equation

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Torricelli’s theorem

Torricelli’s Theorem

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An ideal fluid is one that isincompressible and has noresistance to shear stress. Idealfluids do non actually exist, butsometimes it is useful toconsider what happen to anideal fluid in a particular fluidflow problem in order to simplifythe problem

Taking the datum at the center of the nozzle andchoosing the center streamline give h = z + p/gg in thereservoir where velocities are negligible

Writing Bernoulli’s equation for a streamline between thereservoir and the tip of the nozzle shown as in Fig. 5.4

Torricelli’s Theorem

reservoir and the tip of the nozzle shown as in Fig. 5.4

hgVg

Vh

g

Vphz

p

nn

n

22

0pifresultsequations'Torricelli,2

2

2

22

11

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For freely falling body

Torricelli’s Theorem

hgV

asuV

20

22

22

hgVg

Vh

hgV

nn

n

22

202

2

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Torricelli’s Theorem (Free Jets)

The velocity of a jet of water is clearly related to thedepth of water above the hole

The greater the depth, the higher the velocity

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Example 2

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Example 2 (Solution)

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Example 2 (Solution)

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Example 2 (Solution)

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Example 2 (Solution)

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Example 3

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Example 3 (Solution)

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Example 3 (Solution)

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Application of Bernoulli’s Equation

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Application of Bernoulli’s Equation

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stagnation pointstagnation point

Stagnation Points On any body in a flowing fluid, there is a stagnationstagnation pointpoint. Somefluid flows over and some under the body. The dividing line (thestagnation streamline) terminates at the stagnation point.

The velocity decreases as the fluid approaches the stagnationpoint. The pressure at the stagnation point is the pressure obtainedwhen a flowing fluid is decelerated to zero speed

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stagnation pointstagnation point

Example 4

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Example 4 (Solution)

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Example 5

Determine the difference in pressure between points1 and 2. Hint: Point 1 is called a stagnation point,because the air particle along that streamline, when ithits the biker’s face, has a zero velocity

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Assume a coordinate system fixed to the bike (from thissystem, the bike is stationary, and the world moves past it).Therefore, the air is moving at the speed of the bike. Thus, V2

= Velocity of the Biker

Apply Bernoulli’s equation from 1 to 2

Example 5 (Solution)

Point 1 = Point 2

P1/gair + V12/2g + z1 = P2/gair + V2

2/2g + z2

Knowing the z1 = z2 and that V1= 0, we can simplify the equation

P1/gair = P2/gair + V22/2g

P1 – P2 = ( V22/2g ) gair

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Example 5 (Solution)

If the Biker is traveling at 5 m/s, what pressure does he feel

on his face if the gair = 12.01 N/m3 ?

We can assume P2 = 0, because it is only atmospheric pressurepressure

P1 = ( V22/2g )(gair)

P1 = ((5)2/(2(9.81)) x 12.01

P1 = 15.3 N/m2 (gage pressure)

If the biker’s face has a surface area of 300 cm2

He feels a force of 15.3 x 300x10-4 = 0.46 N

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Application of Bernoulli’s Equation

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Application of Bernoulli’s Equation

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Example 6

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Example 6 (Solution)

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Example 6 (Solution)

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Example 6 (Solution)

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Application of Bernoulli’s Equation

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Application of Bernoulli’s Equation

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Example 7

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Example 7 (Solution)

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Example 7 (Solution)

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Example 7 (Solution)

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Example 7 (Solution)

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Example 8

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Example 8 (Solution)

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Example 8 (Solution)

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The Work Energy Equation

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The Work Energy Equation

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The Work Energy Equation

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Example 9

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Example 9 (Solution)

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Example 9 (Solution)

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Example 10

Calculate the power output of this turbine

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Example 10 (Solution)

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Example 10 (Solution)

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Example 11

Water is pumped from a large lake into an irrigationcanal of rectangular cross section 3 m wide, producing theflow situation shown in the figure. Calculate the requiredpump power assuming ideal flow.

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Example 11 (Solution)

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Example 11 (Solution)

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