3 Fluids Second Part

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Fluids (suite)

Content of the LectureSee the contents of the first part ofFluids

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3.2 hydrodynamicshydrodynamics deal with the motion of fluids.An ideal fluid (= incompressible and non- viscous) is assumed.Non-compressibility is accepted for most liquids.

Non-compressibility is accepted for gases ONLY when pressuredifference is not great.Non-viscous flow approximation is accepted since gravitationalforces (and forces arising from pressure difference) are greaterthan sheering forces acting on adjacent sliding laminae of fluid.

3.2.1 steady flow- line of flow = path followed by any element of a moving fluid.- streamline = curve whose tangent, at any point, is flow

direction.- when flow starts, it is non-steady, but after certain period of

time, it becomes steady.- steady implies that:

1.velocity at any given point remains constant, i.e. no

velocity change takes place with time in the same point, butthere could be velocity change from a site to another (highervelocity at narrow cross-sections, and where velocityincreases, pressure will be decreased). For example, above awing of aircraft (in level flight), pressure is lower thanbeneath it. This pressure difference is the raison of airlift (andwhen changing the attack angel, the pressure above the wings

increases, and the airplane stalls).2.fluids in different flow tubes do not mix.3.streamlines coincide with lines of flow.

- homogeneous flow = straight parallel flow tubes with similarvelocities.

- average velocity: uniform velocity over the entire cross section(of a pipe) resulting in the same discharge.

3.2.2 turbulent flow- It occurs when velocity exceeds the critical one, Vc , and

vertices develop (this could take place behind an obstacle).

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= RD

VC

where R Reynolds number (dimensionless), D tube diameter(m), density (kg m-3), viscosity coefficient (N s m-2 = kg s-1m-1):

R = =s m

DV kg m .m.ms

kg

C

3

1 1

1

experimentation showed that in a tube with diameter 0.01 mfor R< 2000 flow is laminar for R 2000-3000 flow is unstablefor R> 3000 flow is turbulent

Example:Both in the case of water and air flow in a tube whose diameteris 10-2 m, calculate the velocity that keeps the flow laminar, andthat which makes the flow turbulent.

Solution:viscosity coefficient of water at 20oC = 0.001 N s m-2

density of water at 20oC = 1000 kg m-3

= 2000V0.001 Nsm

1000 kg m 0.01 m

2

3

= 0.20 m s

-1

= 0.72 km hr -1

= 17.280 km day-1

= 3000V0.001 Nsm

1000 kg m 0.01 m

2

3

= 0.30 m s-1

= 1.08 km hr -1

= 25.92 km day-1

viscosity coefficient of air at 20oC = 181 x 10-7 N s m-2

density of air at 20oC = 1.3 kg m-3

= 2000

= 2.78 mS -1

V181 N sm

1. kg m 0.01 m

*2

3

10

3

7

= 2.78 * 60 * 60 * 10-3km hr-1

= 10 km hr -1= 240 km day-1

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= 3000

= 4.18 mS -1

V181 N sm

1. kg m 0.01 m

*2

3

10

3

7

= 4.18 * 60 * 60* 10

-3

km hr

-1

= 15 km hr -1

= 360 km day-1

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3.2.3continuity equation(Note: this item is already given earlier, it is referred to here as it related to next subject)

Due to mass conservation, under steady flow through differentcross-sections, constant mass travels during time increment:

= mm1 2

volume density = volume density* *1 1 2 2area distance = area distance* * * *1 1 1 2 2 2

Now, divide by time to obtain the mass passing during 1 second,

areadistance

time= area

distance

time* * * *1 21 1 2 2 A V = A V* * * *1 21 1 2 2 Since the liquid is

incompressible, its density is canceled,A V = A V* *

1 21 2

V =A

A1

2

1

* V2

area *distance

time= area *

distance

time

1 21 2

v o l u m e

t i m e=

v o l u m

t i m e

1 2 d is ch a rg e = d isch a rg1 2Q Q

Meaning: to increase velocity, we reduce the cross section.Example: in order to have velocity at point 1 double that at point2, the cross section at point 1 must be halfthat at point 2.Practical observation: 1. cross-section reduction at balloonorifice, 2. pressing the end of garden water tube.

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3.2.4 Bernoulli equationDue to energy conservation principle, when steady flow takesplace through different cross-sections of a tube, the total energyat any section is constant (= total energy at any other section).

At any point, there are 3 components of total energy:- potential energy m g h (work done =force * distance)

- kinetic energy1

2mV2 (here V = velocity)

- pressure energy PV (here V= volume)h (in meters) elevation above reference, P pressure (Pascal).(Note that each term has the same dimensions, try to verify!)

mgh1+ 12 mV12+P1V1 = mgh2+12 mV22+P2V2

to obtain two other forms of the equation:first method:

- divide by the mass m(to obtain energy corresponding to unit mass):

This is the most IMPORTANT FORM

gh1+ 12 V12+ P1/1 = gh2+ 12 V22+ P2/1and a special case: in a horizontal tube, since h1 = h2,

1

2 V12+ P1/1 =

1

2 V22+ P2/1

or, second method:

- divide by the weight mg(to obtain energy corresponding to unit weight):

h1+1

2 V12/g + P1/1g = h2+

1

2 V22/g + P2/1g

and a special case: in a horizontal tube, since h1 = h2,1

2 V12/g + P1/1g =

1

2 V22/g + P2/1g

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applications:

1- Torricelli reservoir theorem

in a large reservoir with a small lower-end discharge outletat top h = h1 and V1 = very small velocity (negligible = zero).at lower-end h2 = zero (it is at the reference datum level).P1 = P2 = atmospheric pressure (both ends open to air)so, in the unit mass formula of Bernoulli Equation:

gh1 +1

2 V12 + P1/1 = gh2 +

1

2 V22 + P2/1

when we introduce these values, the formula readsgh1 + zero + Po /1 = zero +

1

2 V22 + P0 /1

gh1 =1

2 V22

V = 2 g h

if the reservoir contains a liquid above which there is air underpressure, and we have a gauge measurement for this pressurizedair, we already know that the absolute pressure Pis the sum

P Po Pg= +

so,

gh1 + (Pg + Po)/1 =1

2 V22 + Po/1

gh1 + (Pg + Po_ Po)/1 =

1

2 V22

gh1 + Pg/1 =1

2 V22

V = 2 g h +P

1

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2- constricted section in ahorizontal tube

for a horizontal tube with a constricted section, we use the unitmass formula of Bernoulli equation:

gh1 +1

2 V12 + P1/1 = gh2 +

1

2 V22 + P2/1

but h1 = h2, so

1

2 V12+ P1/1 =1

2 V22+ P2/1

1

2V

2V

22 1

1 2P P

=

we note that:

velocity pressure

at the constricted section high low

at the non-constricted section low high

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3- Venturei velocity meterusing constricted tube and manometer

Problem:

to obtain a formula to estimate flow velocity using pressuredifference, cross-sections and density.

Solution:

first, write the formula of the constricted section in a horizontal

tube as follows:1

2V

2V

22 1

1 2P P

=

second, use the continuity equation to expressV2 (at theconstricted section) in function of V1 (at the non-constrictednormal cross-section) and the ratio of the two cross-sections, takethe square of both sides, substitute and arrange the terms:

V =

A

A2

1

2

* V1

V =A

A2

1

2

22

2

21

* V

1

2*V V

21

1 2P PA

A

1

2

2

22

1

=

1

2

2V

P P1

1 2A

A

1

2

2

21

=

1

2

2

V

P P1

1 2A A

A

1 2

2

2 2

2

=

VP P

11 22 =

22

2 2

A

A A

2

1 2

VP P

11 2

=

22

2 2

A

A A

2

1 2

and to calculate discharge (Q), multiply the velocity by the crosssection area (as usual).

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Note that the value ofV2 could be obtained by changing thesubscripts (change 1 by 2, and change 2 by 1) in the last formula.

In fact, pressure difference

P1-P2can be measured in a

manometer fitted to the constricted and the non-constricted cross

sections, where the density Hg and the height h are those formercury in the manometer:

P1-P2 = Hg g h

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Venturi meter

Pitot tube

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4- Pitot tube velocity meterwithout a manometer, we can use the head difference betweenwater level in two fine tubes perpendicularly inserted into ahorizontal main flow tube, to measure flow velocity.

The idea: tube 1 (inserted close to flow source) has an inlet endparallel to flow direction while the inlet end oftube 2 (far fromsource) is perpendicular to flow direction. So, in tube 2, flowvelocity at its inlet end is practically zero (obstacle to flow), andliquid level is higher (higher pressure).First: Apply the unit mass formula of Bernoulli equation for themain horizontal tube, and insert zero for velocity at tube (# 2):

1

2 V12+ P1/1 =1

2 V22+ P2/11

2 V12+ P1/1 = zero + P2/1

1

2 V12 = P2- P1/1

VP P

1 =

22 1

Second: the pressure difference P2-P1 (head difference in the

two perpendicular tubes) is equivalent to the term g h,P2-P1 = g hso

V1= 2 g h

(this is the same as Torricelli formula, do you remember ?). Butif flow is turbulent, we use a Pitot tube constant k:

V1=K 2 gh

and discharge = VelocityQ Area*

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5- vacuum water-pump

Problem:We need to partially evacuate a glass vessel (for example, to

accelerate water infiltration through a thick saturated paste on aporcelain filter disk - since water does not drain freely from suchpaste due to retention by the fine soil particles).

Solution: A normal velocity inflow (introduced by water-tap) isdrained-out with much higher velocity due to passage through asmall-diameter outlet of the outer compartment (this implies that

water pressure is lower at the outlet than at the inlet).Consequently, air molecules above the outer water-outlet tubegradually become forced to move outside the system and expulsedthrough the outlet of the inner compartment. Progressively, apermanent airflow is created and partial vacuum is generated (air pressure is lower than atmospheric in the inner compartment).Before starting, this inner compartment must be tightly connectedto the vessel that we like to evacuate partially.

6- Atomizer (spray can)

Problem: we would like to force the liquid inside a bottle to get-out in the form of fine spray.

Solution: an external air jet (air current with pressure higher thanthat of the internal air pressure in the container confining theliquid) is admitted, the injected air passes through a finer tube, soits velocity increases (accelerated) and a region of low pressure isformed beneath it. This low-pressure will force the liquid to rise-up in the fine tube immersed in it and will disperse out as finespray mist.

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3 Fluids Second Part