Chapter 4: Flowing Fluids andChapter 4: Flowing Fluids and Pressure Variation

By

Dr Ali Jawarneh

1

Outline• In this lecture we will become familiar with:

– Different approaches for viewing velocity.– Flow patterns and streamlines.

Different flow classifications– Different flow classifications.• We will also discuss the concept of

convective and local accelerationconvective and local acceleration• Euler’s Equation• Bernoulli EquationBernoulli Equation• Rotation and Vorticity• Separation

2

Separation

4.1: Velocity & Descriptions of Flow

• There are two ways of expressing the equations for fluids in motion:

– The Lagrangian approachThe Eulerian approach– The Eulerian approach

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4

Lagrangian Approach

The Lagrangian approach is based on recording the motion of a specific fluid particle at all time

g g pp

the motion of a specific fluid particle at all time.• The position vector for a particle can be given as:

Diff ti ti ddd

kji zyxtr ++=)(

• Differentiating: kjidtdz

dtdy

dtdxtV ++=)(

kjiV )( kji wvutV ++=)(

To describe the motion of the flow field , the motion f ll fl id ti l t b id d

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of all fluid particles must be considered simultaneously.

Eulerian Approachpp• This approach focuses on a certain point in

space and describes the motion of fluid particles passing through this point.Th l it f fl id ti l ill b• The velocity of fluid particles will be described depending on the location of the point in passing through it in space and time:point in passing through it in space and time:

),,,(1 tzyxfu =

),,,(),,,(

3

2

tzyxfwtzyxfv

==

6

Eulerian Approachpp• This approach is favoured on the Lagrangian

approach in most engineering problems in fluid mechanics.Th i th f i th• There is another way of expressing the velocity using this approach as a function of position along a streamline and time:position along a streamline and time:

),( tsVV =

7

Streamlines• They are lines drawn through the

flow field so that the velocity vector yat each and every point on the streamline is tangent to the streamline at that instant.

• A group of streamlines construct what is known as a flow pattern.

• Streamlines are very effective in yillustrating the geometry of the fluid flow or even the speed of the flow. As the speed is inversely proportional to the spacing between the streamlines.

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At each point, the derivative  is the slope of a line that is tangent to the curve. 

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p p gThe line is always tangent to the blue curve; its slope is the derivative. Note derivative is positive where green, negative where red, and zero where black.

Streamlines• A dividing streamline

is a streamline that follows the flow divisiondivision.

• The point of division is called theis called the stagnation point, where the velocity is yequal to zero.

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Flow ClassificationUniform Flow: Non-uniform Flow:• The velocity does not

change from point to i t l f th

• The velocity changes along the streamlines

ith i di tipoint along any of the streamlines in the flow field

either in direction or magnitude.Th t liflow field.

• The streamlines are straight and parallel

• The streamlines may not be straight and/or parallelstraight and parallel. parallel.

V 0.0∂=

V 0.0∂≠

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0.0s∂

0.0s≠

∂

Flow ClassificationUniform Flow:

Non-Uniform Flow:The magnitude of the velocity does not change along thefluid path but the

The magnitude of fluid path, but the

direction does, so the flow is nonuniform.

magnitude of the velocity increases as the duct

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converges, so the flowis nonuniform

Flow Classification• In addition to uniform and nonuniform, the flow can be

l ifi d tclassified to:– Steady Flow

V 0.0t

∂=

∂∂– Unsteady Flow

• In steady flow the velocity at any given point does not

V 0.0t

∂≠

∂• In steady flow, the velocity at any given point does not

vary in magnitude or direction with time.• One can see that the flow can be steady and uniform,

steady and nonuniform, unsteady and uniform, or unsteady and nonuniform.

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Laminar and Turbulent Flow• Turbulent flow is mainly characterised by the

mixing action throughout the flow field, caused by eddies of varying size.L i fl th th h d i th• Laminar flow on the other hand is very smooth-looking.Th fl i j d d if it i l i t b l t• The flow is judged if it is laminar or turbulent depending on Reynolds number (Re = ρVD/μ):

Re<2100: Laminar Flow– Re<2100: Laminar Flow– Re>3000: Turbulent Flow

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Laminar and Turbulent Flow

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4.2: Methods for Developing Flow Patterns

Analytical methods-Analytical methods

-Computational Methods, CFD

-Experimental Methods

In Experimental Methods, two ways to visualize the flow patterns:patterns:

a- Pathline: is a line drawn through the flow field in such a way that it defines the path that a given particle of fluida way that it defines the path that a given particle of fluid has taken. Ex.: PIV

b- Streakline: is to inject dye or smoke in the flow fieldb Streakline: is to inject dye or smoke in the flow field and to observe the dye or smoke trace as it travels downstream.

16Note: Streamline, streakline and pathline are identical and the same for steady flow.

Figure 4.7Predicted streamline pattern over theVolvo ECC prototype (Courtesy of J. Michael Summa, Analytical Methods Inc.).

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Figure 4.8gPathlines of floating particles.

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Figure 4.9Figure 4.9Smoke traces about an airfoil with a large angle of attack. (Courtesy of Education Development Center, Inc., Newton, MA.).

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4.3: Acceleration• Using the Cartesian coordinate system,

and applying the Eulerian approach:

kjiV wvu ++• Where:

kjiV wvu ++=

)( tzyxfu

)(),,,(),,,(

2

1

tzyxfwtzyxfvtzyxfu

==

),,,(3 tzyxfw =

In the Lagrangian fromulation, the velocity is a function of time only

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time only. In the Eulerian formulation, the velocity at a point in the flow field is a function of both spatial coordinates and time.

Acceleration

• The acceleration of a fluid particle in the x-direction is given by:

dua =

• Applying the chain rule for differentiating:

dtax =

• Applying the chain rule for differentiating:u

ddzu

ddyu

ddxuax ∂

∂+

∂∂

+∂∂

+∂∂

=tdtzdtydtxx ∂∂∂∂

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Acceleration

• Or: uuwuvuua ∂+

∂+

∂+

∂=

• Similarly:

tzw

yv

xuax ∂

+∂

+∂

+∂

=

Similarly:

tv

zvw

yvv

xvuay ∂

∂+

∂∂

+∂∂

+∂∂

=tzyx ∂∂∂∂

wwwwvwua ∂+

∂+

∂+

∂=

tzw

yv

xuaz ∂

+∂

+∂

+∂

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Acceleration• In general:

wvua ∂+

∂+

∂+

∂=

)()()()(

• It can be seen that there are two parts forming

tzw

yv

xua

∂+

∂+

∂+

∂

the acceleration: – Derivatives with respect to position (convective

acceleration)acceleration).– Derivative with respect to time (local acceleration).

• Local acceleration is zero in steady flow while• Local acceleration is zero in steady flow, while convective acceleration is zero in uniform flow.

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• Example: Gi 2+= ytxuGiven:

Wh t i th l ti t i t 1 20

22

=−=

+=

wtytxvytxu

What is the acceleration at a point x=1 m, y=2 m, and at a time t=3 s?

S l ti ∂∂∂∂Solution:tu

zuw

yuv

xuuax ∂

∂+

∂∂

+∂∂

+∂∂

=

x))(ytxt()t)(yxt(a ++−++= 022 2 x))(ytxt()t)(yxt(ax ++++ 022

228321 s/matand,y,x@ x =→===

vvvv ∂∂∂∂tv

zvw

yvv

xvuay ∂

∂+

∂∂

+∂∂

+∂∂

=

)yxt()t)(ytxt()t)(yxt(a y −++−−++= 202 22

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)y())(y())(y(y

258321 s/matand,y,x@ y =→=== 25828 s/ma ji +=

Acceleration by Applying the Lagrangian approach

tetsVV ),(=

)()( deVdVdV t

Use the chain rule

)()(dt

Vedtdt

a tt +==

Use the chain rule

tV

dtds

sV

dtdV

∂∂

+∂∂

= ))((tdtsdt ∂∂

tV

sVV

dtdV

∂∂

+∂∂

= )(2tsdt ∂∂

t eVde=

nt er

VetV

sVVa )()(

2

+∂∂

+∂∂

=

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nerdt r: is the local radius of curvature of the pathline.

Example:pThe velocity of water flow in the nozzle shown is given by the following equation: V=2t / (1- 0.5 x /L)2 , where L=4 ft. When x=0.5 L and t= 3 sec, what is the local acceleration along th t li ? Wh t i th tithe centerline? What is the convective acceleration? Assuming one-dimensional flow prevailsprevails.

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Solution: eVeVVVa )()(2

+∂

+∂

=Solution: nt er

ets

Va )()( +∂

+∂

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4.4: Euler’s Equationq• There are three causes for pressure

variation in flowing fluid:– Weight effectsg– Acceleration– Viscous resistance

• Pressure variation is needed to overcome viscous resistance that acts in an oppositeviscous resistance that acts in an opposite direction to the motion of the fluid.

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Pressure Variation• Also, if a fluid mass accelerates in a certain

direction, this means that there is a net force in the direction of acceleration, and consequently, pressure must decrease in the direction ofpressure must decrease in the direction of acceleration.

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Pressure Variation Due to Acceleration

• Considering the element which is being accelerated in the l direction:the l direction:

∑ = ll MaF∑ ll

lMaWAppAp =Δ−ΔΔ+−Δ αsin)( lppp )(

laAlAlgAppAp ΔΔ=ΔΔ−ΔΔ+−Δ ραρ sin)(

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Pressure Variation Due to Acceleration

• Simplifying:

laglp ραρ =−ΔΔ

− sin

lalz

lp ργ =

ΔΔ

−ΔΔ

−

• Taking the limit as Δl→0Δl→0

lalz

lp ργ =

∂∂

−∂∂

−

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Pressure Variation Due to Acceleration

• Rearranging:

lazpl

ργ =+∂∂

− )(Inviscid Flow

• This is basically Euler’s equation of motion for a fluid.

• The pressure must decrease in the direction of flow

• Note that if acceleration is zero, the equation reduces to: Czp =+ γ

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equation reduces to: Czp + γ

Acceleration of a tank of liquidq• Considering the open

tank of liquid which is accelerated in the x-directiondirection

• The force on the left is greater than that onis greater than that on the right of the tank and so the water surface readjusts its surface.

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Acceleration of a tank of liquidq• Applying Euler’s

equation on the surface A′B′:

lazpl

ργ =+∂∂

− )(

αργ cos)( xazdld

−=

ga

dldz x αcos

−=

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Acceleration of a tank of liquidq

a αcosig

ax αα cossin =

ax=αtan

• If we apply Euler’s

g=αtan

pp yequation at the bottom of the tank:

xadxdp ρ−=

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Example:Example:

A liquid with a specific weight of 100 Ibf/ft3 is in th d it Thi i i l ki d f li id th tthe conduit. This is a special kind of liquid that has zero viscosity. The pressures at points A and B are 170 psf and 100 psf respectivelyand B are 170 psf and 100 psf, respectively. Find the acceleration.

36

Solution:

37

Example:p

If the velocity varies linearly with distancethrough this water nozzle, what will be thepressure gradient , dp/dx, halfway through the

l ? A t d d i i id flnozzle? Assume steady and inviscid flow

38

Solution: uuwuvuua ∂+

∂+

∂+

∂=Solution:

tzw

yv

xuax ∂

+∂

+∂

+∂

nt er

VetV

sVVa )()(

2

+∂∂

+∂∂

=rts ∂∂

30300@21 +=

VxccV

55,2/1,503050,80,1@30,30,0@

2

1

==+=======

VxatxVcVxcVx

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• Example: The closed tank shown, which is full f /of liquid, is accelerated downward at 2/3 g and to

the right at one g. Here L=2 m, H=3 m, and the liquid has a specific gravity of 1 3liquid has a specific gravity of 1.3. Determine pC-pA and pB-pA.

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Apply Euler Eq. at A-B: Solution:za)zp(

zργ =+

∂∂

−

pp ∂∂ 3/4251))()(3/1()3/2( mNSzpg

zp

water −=−=∂∂

⇒−−=+∂∂ γργ

kpappppp BA 75124251⇒−∂(1)

Apply Euler Eq. at C-B:

kpappz AB 75.124251

3=−⇒−==

∂(1)

3/12753)( mNSgaxpazp

x waterxx −=−=−=−=−=∂∂

⇒=+∂∂

− γγρρργ

∂

From (1) and (2):

kpappmNppxp

BCCB 506.25/12753

23 =−⇒−=

−=

∂∂

kpapp AC 26.38=−

(2)

( ) ( )

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ppp AC

Discussion: if static, p A+γh=pB, so pB - p A= 38 259 pa

Discussion: pC>pB due to acceleration

4.5: Bernoulli Equationq• Starting with the Euler’s equation

derived in the previous lecture along streamline:

tazp ργ =+∂

− )(

• The acceleration along the pathline i i

tps

ργ∂

)(

is given as:

F t d fl l l l ti itV

sVVat ∂

∂+

∂∂

=

• For steady flow, local acceleration is equal to zero:

sVVat ∂∂

=

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s∂

Bernoulli Equationq• Substituting in the equation:

sVVzp

s ∂∂

=+∂∂

− ργ )(

• Rearranging:0)

2(

2

=++Vzp

dd ργ

• Integrating the equation:

)2

( pds

ργ

g g q

CVzp =++2

2

ργ

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Bernoulli Equationq• This is what is known as the Bernoulli equation

which states that the sum of the piezometric pressure and the kinetic pressure is constant.Th ti l b itt i th f ll i• The equation can also be written in the following form:

Vp 2

O

Cg

Vzp=++

2γ

• Or:C

gVh =+2

2

44

Bernoulli Equationq• Bernoulli’s equation can be applied along

streamline if:– The flow is steady

Th fl i i ibl– The flow is incompressible– The flow is inviscid (viscous effects negligible)

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• Example: The velocity in the outlet pipe from this reservoir is 6 m/s and h=15 m Because of the roundedreservoir is 6 m/s and h=15 m. Because of the rounded entrance to the pipe, the flow is assumed to be irrotational. Under these conditions, what is the pressure , pat A?

-150 0 0VA=6

• Solution:gVz

γp

gVz

γp A

AA

22

221

11 ++=++ kpa.pA 15129=

0 0

46

9810

Application of the Bernoulli Equation- Stagnation Tube- Stagnation Tube

• The stagnation tube is a simple device that can be used for measuring the velocitymeasuring the velocity of the flow with the aid of Bernoulli equation.of Bernoulli equation.

• Applying Bernoulli equation between qpoints 1 and 2:(equal z)

2 21 1 2 2p V p V+ = +

472g 2g

+ = +γ γ

Stagnation Tubeg• Point 2 is a stagnation

point (V=zero):

)(22 ppV =

• But:

)( 121 ppV −=ρ

)(2

1

dlpdp

+==γγ

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Stagnation Tubeg• Substituting:

))((221 ddlV γγ

ρ−+=

• Consequently:

glV 21 =

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- Pitot Tube• The Pitot tube is another

device that can be used for measuring the velocity of the flowthe flow.

• It becomes extremely useful in pressurised pipes and for gases.g

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Pitot Tube• Applying Bernoulli

equation between points 1 and 2:

2

22

21

21

1 22zVpzVp γργρ ++=++

• But V1=zero, rearranging:

21

22

11

2 )()(2 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡+−+= zpzpgV

γγ

51

Pitot Tube• Or (in terms of

i t i h d)peizometric head):

)(2 21 hhgV −=

• For the special case of gases, the effect of

l ti ( i ht)elevation (weight) becomes negligible and so the equationso the equation reduces to:

ρ/2 pV Δ=

52

ρ/2 pV Δ=

• Example: Find the velocity at the pipe center.

• Solution:- Apply Bernoulli (1&2)

0.0 (stag.)

M t th h d t ti E

2

222

1

211

22z

gV

γpz

gV

γp

ww

++=++

( g )

)(gVpp w 12

21

12 γ=−

- Manometer or the hydrostatic Eq.:

21 121121 p)y/()/(yp M =+−++ γγγ

62.4 Ibf/ft313.6 x 62.4

ρw =1.94 slug/ft3

From Eq (1) & Eq (2)

21 121121 p)y/()/(yp wMw +++ γγγ

)(Ib/ft.pp 25265 212 =−

g)( 2

32.2 ft/s2

- From Eq.(1) & Eq.(2):53

ft/s.γg)()ppVw

282121 =−=

• Example: A pitot-static probe connected to a water manometer is used to measure the velocity ofwater manometer is used to measure the velocity of air. If the deflection (the vertical distance between the fluid levels in the two arms) is 7.3 cm. Determine )the air velocity. Take the density of air to be 1.25 kg/m3 .

54

• Solution:- Apply Bernoulli between point 1& 2:

222

211 zVpzVp

++++

0.0 (stag.)

)()pp(VV 12 21 −==2

221

11

22z

gγz

gγ airair

++=++ )(VVair

12 ρ==

- Apply the hydrostatic equation between point 1& 2:

21 0730 p).(γp w =− pa.pp 1371621 =−

- From Eq.(1) and Eq.(2):

m/s...

ρ)p(pVV

air

833251

1371622 212 =

×=

−==

55

56

• Example: The maximum velocity of the flow past i l li d i i h h l ia circular cylinder is twice the approach velocity.

What is Δp between the point of highest pressure and point of lowest pressure in a 40 m/s wind?and point of lowest pressure in a 40 m/s wind? Assume irrotational flow and the air density is 1.2 kg/m3.g

• Solution:22 VpVp

0.0 (stag.)

222

111

22z

gV

γpz

gV

γp

++=++

kpa.)/(.VρgVγpp 84328021

222

22

22

21 ====−

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4.6: Rotation & Vorticity

58

59

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Rotation & Vorticityy• The rate of rotation of flow about the z-axis is given

as: 1 v u( )∂ ∂Ω =

• Similarly:

z ( )2 x y

Ω = −∂ ∂

x1 w v( )2 y z

∂ ∂Ω = −

∂ ∂ y1 u w( )2 z x

∂ ∂Ω = −

∂ ∂2 y z∂ ∂ 2 z x∂ ∂

1 w v 1 u w 1 v u( ) ( ) ( )2 y z 2 z x 2 x y

∂ ∂ ∂ ∂ ∂ ∂Ω = − + − + −

∂ ∂ ∂ ∂ ∂ ∂i j k

• Vorticity is twice the rate of rotation, consequently, the vorticity vector can be given as:

2 y z 2 z x 2 x y∂ ∂ ∂ ∂ ∂ ∂

the vorticity vector can be given as:

w v u w v u2 ( ) ( ) ( )y z z x x y

∂ ∂ ∂ ∂ ∂ ∂ω = Ω = − + − + −

∂ ∂ ∂ ∂ ∂ ∂i j k

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y y

• Example: Is the following equation irrotational?

• Solution: ji )xy()yx(V 22 +−=

yu

xv

∂∂

=∂∂ ?

22 xy −≠

The flow is rotational.

62

4.7: Pressure Distribution in Rotating FlowsFluid rotates as a rigid bodyu d otates as a g d body

- Rotation of tank of liquid

C id i• Considering a cylindrical tank of liquid rotating at a constantrotating at a constant speed ω, and applying Euler’s equation in the qradial direction:

razpdrd ργ =+− )(

V 2

63rVar

2

−= rV ω= For a liquid rotating as a rigid body

Rotation of tank of liquidq• Substituting for the

acceleration:2( )d p z r

drγ ρ ω− + = −

• Integrating along the r-direction:

Crzp =−+2

22ωργ2

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• Example: A U-tube is rotated about one leg, f fbefore being rotated the liquid in the tube fills

0.25 m of each leg. The length of the base of the U tube is 0 5 m and each leg is 0 5 m long WhatU-tube is 0.5 m, and each leg is 0.5 m long. What would be the maximum rotation rate (in rad/sec) to ensure that no liquid is expelled from the outerto ensure that no liquid is expelled from the outer leg?

65

• Solution:2

22

222

21

11 22ωrργzpωrργzp −+=−+

0.0 0.0

z1=0.0, r1=0.0, z2=0.5 m, r2=0.5 m

rad/sgzω 26462 2 == rad/s.r

ω 264622

==

66

• Think!

21 VV =

672211

21

hAhAVV=

Example:A tank of liquid (S=0.8) that is I ft in diameter and 1ft high (h=1 ft) is rigidly fixed (as shown) to arotating arm having a 2 ft radius. The arm rotatessuch that the speed at point A is 20 ft/s. If thepressure at A is 25 psf what is the pressure at B?pressure at A is 25 psf, what is the pressure at B?

68

Solution:Solution:

69

Example:A closed tank of liquid (S=1.2) is rotatedabout a vertical axis, and at the same timethe entire tank is accelerated upward at 4m/s2. If the rate of rotation is 10 rad/s,what is the difference in pressure betweenppoints A and B (PB-PA)? Points B is at thebottom of the tank at a radius of 0.5 mfrom the axis of rotation and point A is atfrom the axis of rotation, and point A is atthe top on the axis of rotation.

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Solution:Solution:

71

4.8: The Bernoulli Equation in Irrotational FlowIrrotational Flow

• Starting with the Euler’s equation normal to the streamline: 2V∂streamline:

nV(p z) a ( )

r r∂

− + γ = ρ = ρ −∂

Vp 2

Cg

Vzp=++

2γBernoulli’s equation can be freely appliedBernoulli s equation can be freely applied anywhere in the flow field (along streamline or cross streamlines, given that:

The flow is steadyThe flow is steadyThe flow is incompressibleThe flow is irrotational Th fl i i i id ( i ff t

72

The flow is inviscid (viscous effects negligible)

Rotation in Flows with Concentric StreamlinesIt is interesting to realize that a flow field rotating with circular streamlines canIt is interesting to realize that a flow field rotating with circular streamlines can be irrotational; that is, the fluid elements do not rotate. Consider the two-dimensional flow field shown in Fig. 4.19. The circumferential velocityon the circular streamline is V, and its radius is r. The z-axis is perpendicular to the page. As before, the rotation of the element is quantified by the rotation of the bisector, Eq. (4.24), which is

73

as expected. This type of circular motion is called a “forced” vortex.

74

h C i t t I thi th i f ti l l it i i lwhere C is a constant. In this case, the circumferential velocity varies inversely with r, so the velocity decreases with increasing radius. This flow field is known as a “free” vortex. The fluid elements go around in circles, but do not rotate.

75

The difference between element rotation and deformation for flowswith circular streamlines is shown in Fig. 4.20. For the rotational flowshown in Fig. 4.20a the fluid elements rotate but they do not deform.I th i t ti l fl h i Fi 4 20b th l tIn the irrotational flow shown in Fig. 4.20b, the elementscontinuously deform but do not rotate. In other words, the elementsdeform to maintain a constant orientation (no rotation).

76

Example:A free vortex in air rotates in a horizontal planeA free vortex in air rotates in a horizontal planeand has a velocity of 40 m/s at a radius of 4 kmfrom the vortex center. Find the velocity at 10km from the center and the pressure differencebetween the two locations. The air density is 1.2kg/m3kg/m3.

77

Si fl i i t ti l th B lli ti f diff

SOLUTION:Since flow is irrotational, use the Bernoulli equation, for pressure difference.

78

Pressure V i ti iVariation in a Cyclonic Storm

79

80

A cyclonic storm is characterized by rotating winds with a low-pressure region in the center. Tornadoes and hurricanes are examples of cyclonic storms. A simple

d l f th fl fi ld i l i t i f d t t th tmodel for the flow field in a cyclonic storm is a forced vortex at the center surrounded by a free vortex, as shown in Fig. 4.22. This model is used in several applications of vortex flows. In practice, however, there will be no discontinuity in the slope of the velocity distribution as shown in Fig. 4.22, but rather a smooth p y gtransition between the inner forced vortex and the outer free vortex. Still, the model can be used to make reasonable predictions of the pressure field.

The model for the cyclonic storm is an illustration of where the Bernoulli equationThe model for the cyclonic storm is an illustration of where the Bernoulli equation can and cannot be used across streamlines. The Bernoulli equation cannot be used across streamlines in the vortex at the center because the flow is rotational. The pressure distribution in the forced vortex is given by Eq. (4.13b). The Bernoulli equation can be used across streamlines in the free vortex since the flow is irrotational.

Take point 1 as the center of the forced vortex and point 2 at the junction of theTake point 1 as the center of the forced vortex and point 2 at the junction of the forced and free vortices, where the velocity is maximum. Let point 3 be at the extremity of the free vortex, where the velocity is essentially zero and the pressure is atmospheric pressure p0. Applying the Bernoulli equation, Eq. (4.42),

f

81

between any arbitrary point in the free vortex and point 3, one can write

82

83

Pressure Coefficient, Cp, p

h h V 2op 2

oo

h h VC 1 ( )VV / 2g

−= = −For liquid:

F G VFor Gas: 2op

2 oo

p p VC 1 ( )1 VV2

−= = −

ρ2

Where: h=piezometric head=p/γ +z

V p = reference velocity pressure84

Vo,po= reference velocity,pressure

Pressure Distribution around a circular Cylinder Ideal Fluidcircular Cylinder-Ideal Fluid• Consider the ideal

flow across the shown cylinder:V l it t i t B i• Velocity at point B is zero and hence the flow decelerates fromflow decelerates from A to B, implying that the static pressure pwould increase.

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• The opposite happens from B to C.

• Despite the fact that ththe pressure decreases from C to D the momentum ofD, the momentum of the flow is sufficient to make it travel to point pD.

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• This is further• This is further illustrated in the figure showing theshowing the distribution of Cp:– Positive Cp is drawn

inward.– Negative Cp is drawn

outwardoutward.

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4.9: SeparationPhenomenon occurs when the flow separate from the boundary and a

i l ti tt irecirculation pattern is generated in the region.

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fDownstream of the midsection, the pressuregradient is adverse and the fluid particles in theboundary layer, slowed by viscous effects, cany y , y ,only go so far and then are forced to detouraway from the surface. This is called theseparation point A recirculatory flow called aseparation point. A recirculatory flow called awake develops behind the cylinder. The flow inthe wake region is called separated flow. The

di t ib ti th li d f ipressure distribution on the cylinder surface inthe wake region is nearly constant, as shownin Fig. 4.25b. The reduced pressure in theg pwake leads to increased drag.

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Distribution around a Cylinder-Realy• What happens in reality is

that after the flow passesthat after the flow passes mid-section (maximum velocity point), viscous resistance leads to slow theresistance leads to slow the velocity, therefore, the fluid near the boundary can proceed only a very shortproceed only a very short distance against the adverse pressure gradient before stopping completelybefore stopping completely causing the flow to be separated from the boundary

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boundary.

Distribution around a Cylinder-Realy

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Fl t l t ( ) Id l fl tFlow past a plate. (a) Ideal flow past a plate. (b) Real flow past a plate.

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