fap0015 ch10 fluids

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Ch 10: FLUIDS1. Density, specific gravity

Pressure Pascal’s Principle

2. luid Static :

Buoyancy & Archimedes’ Principle

. u a : uContinuity Equation

Bernoulli’s Equations scos ty

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Lesson OutcomesAt the end of the lesson, students should be able to:

1. define density, specific gravity and pressure .

2. state Pascal’s Principle and apply the principle to solve related problems .

3. define Buoyancy and state Archimedes’ Principle .

4. apply Archimedes’ Principle to solve related problems.5. apply continuity equation to solve related problems.

6. define viscosity .

7. state Bernoulli’s Equation and apply the equation to solverelated problems .

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• Fluids refer to both the gaseous and liquid state of matter.

• Out of the three states of matter, only solids have uniqueshapes.

• A liquid flows from place to place and takes the shape of the container and its volume can be chan ed si nificantlonly by the application of large forces.

• A as has neither a fixed sha e nor a fixed volume .

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Density and specific gravity• Density is the ratio of the mass to the volume .

)(kg/m 3mmass == ρ

• Densit of the substance is the same re ardless of thetotal amount we have in a system.

•given volume.

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S ecific Gravit(relative density)

• pec c grav y o a su s ance s e ne as e ra o othe density of that substance to the density of water

O. .

• The density of the water at 4.0o

C equals to3, .

• Specific gravity is a numerical value based on relative

.

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Pressure in Fluids• Pressure is defined as force per unit area , where theorce an t e area must e at r g t ang es to one anot er.

F P pressure ==

F

ressure un n s m = pasca a

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Atmospheric and Gauge Pressure

• Atmospheric pressure, P atm , is a direct result of the weight.

• P atm varies with weather changes and with elevation .

• It has the following value in SI units

kPa101Pa10011 N/m10011 525 =×=×= ..Patm

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Atmospheric and Gauge Pressure

• The gauge pressure, P g , is defined as the pressuredetermined by the tire gauge .

• Absolute pressure, P = P + P atm .

• A flat tire does not have zero pressure in it, it is at P atm , to,than P atm by this amount,

atmg −= = a + atm

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Fluid Pressure•

• In order to derive our desired equation, we make use of

containing liquid of density ρ .

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.which one column ofthe fluid is outlined.

e u s a res .

b. The free bodyagram, s ow ng

the vertical forcesactin on thecolumn.

The column bottom’s area is A at a depth h from the top.

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The force exerted equals to the weight of the liquid column, mg .

mg AP APF y =−−= 12 0

m AP AP +=gAh AP AP ρ += 12

=−=ghPP ρ += 12

The pressure at equal depths within a uniform liquid is the same .

12

The equation derived is valid for fluid whose density is constantand does not change with depth.

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Reasoning and Discussion

,

decreases. This allows the air in the bubble to expand andoccu a lar er volume.

nswer:

(a) The diameter of the bubble increases.

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Measurement of Pressure

• Many devices have been invented to measure pressure.

• e s mp est s t e open-tu e manometer.

ghPP atm ρ +=2

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• The top surface of the fluid is open to atmosphere with a pressure

P . If the cross-sectional area of the tube is A, the downward forceexerted by P 0 is APF top 0=At the bottom of the hei ht h, the downward force is F lus the

weight of the liquid, recalling that m = ρ V , and that V = hA for acylinder with height h and area A.ghAVgmgW ρ ρ ===

( )ghA APW F F topbottom ρ +=+= 0

This weight,

• We have

hPghA APF

P bottom ρ +=+== 0 )(• The pressure as the bottom is

A Ao om

If the depth increase by amount h , the pressure increase by an amount ρ gh ,

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Barometer• Fill one glass tube -opened in one end and

closed at the other-with fluid of density ρ.

• Inverted the tube and placed it open end onthe bowl filled with the same fluid.

•The highest difference is directly related to theatmospheric pressure that pushes down on the

.

•The measurement of the depth gives theatmos heric ressure at the bottom.

ghPatm ρ =

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Conceptual QuestionCould you use a straw to sip a drink on the moon? Explain.

REASONING AND SOLUTION • When you drink through a straw, you draw the air out of the

straw, and the external air pressure leads to the unbalanced .requires the presence of an atmosphere. The moon has no

the moon.

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Example• The Titanic was found in 1985 lying on the bottom of the

North Atlantic at a depth of 2.5 miles. What is the pressure

at this depth? ρseaw =1025 kg/m3

•Applying this equation

0 += gPPbottom ρ

( ) m16095.28.910251001.1 5 ×××+× mi=

Pa101.4 7×=

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Pascal’s PrinciplePascal’s Principle states that: the pressure applied to anenclosed fluid is transmitted undiminished to every portion

of the fluid and the wall of the containing vessel.

inout PP = F F

⎥⎢=⎥⎢ inout

only on the depth.

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A hydraulic Lift, A force F in exerted on small piston causes aout .

F in

out in A A ⎥

⎦

⎢

⎣

=⎥⎦

⎢

⎣

F out

out

out

in

inout in A A

PP === inin

out out F

AF =hus

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• Let us assume that A out is 100 times greater than A in. Then by

ushin down on iston 1 with the force F , we ush u ward on piston 2 with a force of 100 F in. Our force has been magnified100 times.

• As in the Fig. Piston 1 being pushed down via a distance d 1,

Thisdisplace a volume of fluid A ind 1.

• Same volume flows into cylinder 2, and caused piston 2 to rise adistance d 2. Equating the volumes:

A1221 T us

A out

nout in ==

•For previous example d = d /100

•Our force at piston 2 has been magnified 100 times, but thedistance it moves has been reduced 100 times.

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Example• To inspect a 14500-N car, it is raised with a hydraulic lift. If the

radius of the small piston, as in slide 23, is 4.0 cm, and thera us o e arge p s on s cm, n e orce a mus eexerted on the small piston to lift the car. What is the pressurea lied to the fluid in the lift?

• Solution: Using Pascal principle

Aout

out in A

=

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’

they do when outside the fluid..

• The weight acts downwards, but when an object ismmerse n wa er, an up- rus or uoyan orce sexerted by the water.

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Archimedes:

Born: 287 BC in Syracuse, SicilyDied: 212 BC in Syracuse, Sicily

• Archimedes principle states that a body wholly or artiall immersed in a fluid is buo ed u b a force

equal to the weight of the fluid it displaces .• It can be applied to determine a person’ body–fat

percentage, floatation, ships boat etc.

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Archimedes’ PrincipleThe net upwards force, F F F B 12 −=

1212 −=−=12F −=

=mV ==

= the weight of the displaced fluid

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CONCEPTUAL CHECKPOINT A flask of water rests on a scale. If you dip your finger into the water,without touching the flask, does the reading on the scale(a) increase, (b) decrease, or (c) stay the same?

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Reasoning and Discussion

Your finger experiences an upward buoyant force when it is dipped intothe water. By Newton’s third law, the water experiences an equal and

.

transmitted to the scale, which in turn gives a higher reading.

into the water, its depth increases. This results in a greater pressure at

the bottom of the flask, and hence a greater downward force on the flask.e sca e rea s s ncrease ownwar orce.

Answer

(a) The reading on the scale increases.

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Conceptual questions• Two objects have different densities and but same

volumes. Will they experience different or sameuoyan orces

• The buoyant force depends only on volume anddensity of immersed objects. Therefore, both will

.

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• -

above the water.•

• (b) If the river carries the log into the ocean, does,

decrease, or stay the same? Explain.

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Solution F b

It increases because the

W (b)

b =

buoyant force is proportional to the

a

density of thedisplaced fluid and

e ens y o sawater is greater

fresh water.

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Fluids in Motion• The flow of a fluid is said to be steady or laminar or

.

• Steady flows , the velocity of the fluid particles at any.

• Unsteady flow exists whenever the velocity at a point in.

• The fluid flows become irregular and turbulent or non-.

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Fluids in Motion• In describing the motion of a fluid, it is helpful to make

simplifying assumptions:

• Consider it as an ideal fluid , which is simpler to handlemathematically and yet provide useful results.

• The density of the fluid is assumed to remain constant , so

that the fluid is incompressible .• The fluid is supposed to be non-viscous , the viscosity is a

measure of how resistive the fluid is to flow.

• The fluid is in steady flow .

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Equation of ContinuityHave you ever used your thumb to control the water flowing fromthe end of the hose?

The water flow increase when you reduced the cross-sectional area

This kind of fluid behavior is described by the equation of continuity.

The mass of fluid per second that flows through a tube is called the mass flow rate , or volume flow rate (Q ).

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• The useful equation that can be derived is the equation of

=, .Upstream

Downstream

2mΔ 1mΔ

The mass that enters the tube at one end must equal the massthat leaves at the other, provided there are no leaks .

Mass s conserved ne t er create nor estroye as u ows.

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Equation of Continuity

2mΔ 1mΔ

At the same time Δt

AA , 22211121 Δ=ΔΔ=Δ t vt vmm ρ ρ

v 1∝V ===

cons an us 2211 == vvv

At It’s the volume per second that passes through the tube.

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Example

4.0 m. The depth of the river before the rapids is 2.7 m; thedepth in the rapids is 0.85 m. Find the speed of water flowingin the rapids, given that its speed before the rapids is 2.2 m/s.

Assume the river has a rectangular cross section.

Solution

1 1 2 2

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Bernoulli’s effect• Bernoulli’s principle states that where the velocity of a fluid is

high , the pressure is low and where the velocity is low , the.

• It depends on an application of work-energy theorem to fluids,usin the relation between the ressure , P , of the fluid, itsspeed , v and its height , h .

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The work done on the left sideositive work : 111111 x AP xF W Δ=Δ=

The work done on the right side:(negative work): 222222 x AP xF W Δ−=Δ−=Work done by gravity:(negative work):

)( 123 y ymgW −−=

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According to the work-energy theorem , the net work done is

The net work done: W = W 1 + W 2 + W 3 = K 2 − K 1 = ΔK .

( ) 1222211121

222 mgymgy x AP x APvvm +−Δ−Δ=−

m 1 = m 2 = ρ V = ρ A1 x1 = ρ A 2 2

1221122gy

V gy

V V P

V Pvv

V +−−=−

( ) 12212

1222

gygyPPvv ρ ρ ρ +−−=−

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11

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12212

122

11gygyPPvv ρ ρ ρ ρ +−−=−

22221

211 2

121 gyvPgyvP ρ ρ ρ ρ ++=++

The equation just derived is Bernoulli’s equation and themore general form of it is as given below:

1 2

const vP =++ 2-

of the fluid with its speed and height .

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Bernoulli’s effect,

n n p y ry n r• If the water flow in the same level,( y 2 - y 1 = 0) and the

.11 22 vPvP +=+ameter s c ange , n t s t e equat on s

1 2 = .2

As the fluid s eeds u its ressure decreases .

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Fluid Pressure in a Pipe of Varying Elevation

• The fluid speed is constant and the change in the kineticenergy is ( ΔK =0).

+=+ PPconstant=+ gyP ρ

As the height within a fluid increases, the pressure decreases.

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Exam le•

110 kPa flowing with a speed of 1.4 m/s. Whenthe i e narrows to one-half its ori inal diameter what is

•• (b) the pressure of the water?

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v = v( ) P +1 2 = P + 1 2(b)

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1v1 = 2 v2(a) P1 +2 ρ v1

2 = P2 + 2 ρ v2

2(b)

2 21

4v1 = 2

4v2

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Test our understandin• It is not surprising that a wind blowing directly on an open

.

• Use Bernoulli’s principle to explain how a wind blowing-

make the door close. (Assume that the door open inwards.)

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Answer • From Bernoulli’s e uation an increase in the

flow speed v corresponds to a decrease in theair ressure P . The reduced air ressure onthe “outdoor” side of the door makes the door

swin toward that side closin it.

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a) A Force F is applied to the top plate, whichis in contact with av scous u .

b) Because of the force F ,

the to late and its adjacent layer of fluidmove with constant

L

velocity v.Laminar flow

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Viscosity (cont )

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Viscosity (cont.)• It is found that the force F required to maintain uniform

relative motion of the plate is directly proportional to thearea o e er p a e an o e spee v o e mov ng p a e.

• The force is inversely proportional to the separation L of .

•

⎟ ⎠

⎜⎝

= L

AF η

more viscous the fluid, the larger the value of coefficient of viscosity, .

• Its unit is poise (P), 1 poise= 0.1 Pa.svA=η

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C l Q i

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Conce tual Question

• When a person’s blood pressure is taken, it is

measured on the arm, at approximately thesame level as the heart. How would theresults differ if the measurement were to be

made on the patient’s leg?

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A

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Answer

• Assumin the le is below heart level, as in a

standing person, the blood pressure will be

that the pressure in a fluid increases with

.

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Conceptual Question

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Conceptual Question• Since metal is more dense than water, how is it

ossible for a metal boat to float?

• A metal boat can float if it dis laces a volume of water whose weight is equal to the weight of the boat. This can

be accomplished by giving the boat a bowl-like shape.

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Summary

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Summary.• Pressure is defined as force per unit area.

• The ressure at de ths of static is which is the same for samedepth.

• Pascal’s Principle, the pressure applied to an enclosed fluid isghPP atm ρ +=2

ransm e un m n s e o every por on o e u an ewall of the containing vessel.

• Archimedes rinci le states that a bod wholl or artiallimmersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid.

• e mass a en ers e u e a one en mus equa e mass aleaves at the other, provided there are no leaks.

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Summary

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Summary• Bernoulli’s principle states that where the velocity of a fluid is

high, the pressure is low and where the velocity is low, the.

22221

211 11 gyvPgyvP ρ ρ ρ ρ ++=++

• The value of η depends on the fluid between the plates, the

tangent force, separation distance and inversely proportional tothe velocity and area of the plate.vAFL=η

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fap0015 ch10 fluids

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