physics 111physics.valpo.edu/courses/p111/lectures/lecture21notes.pdf · earth’s gravitational...

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Physics 111

Title page

Thursday,November 11, 2011

Physics 111 Lecture 21

• Ch 15: FluidsPascal’s PrincipleArchimede’s PrincipleFluid FlowsContinuity EquationBernoulli’s EquationToricelli’s Theorem

• Wednesday, 8 - 9 pm in NSC 118/119• Sunday, 6:30 - 8 pm in CCLIR 468

Help sessions

AnnouncementsThursNov

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This week’s lab will be another physicsworkshop - on fluids this time. No quizthis week.

labs

AnnouncementsThursNov

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P2A = P

1A+ mg

P2A = P

1A+ ρ( AΔh)g

P2= P

1+ ρg(Δh) Δ ΔP g h= ρ ( )

equilibrium

Ch 15: Fluid MechanicsThursNov

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If our system is in equilibrium,the net force must be 0. So...

NOTE: our derivation here assumes a uniformdensity of molecules at a given layer in theatmosphere. In the real atmosphere, densityNovreases with altitude. Nevertheless, ourpressure and force balance diagram appliesso long as our layer is sufficiently thin so thatwithin it, the density is approximately constant.

note


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If the atmosphere is in equilibrium (which wouldimply a uniform temperature and no windsblowing), the pressure at a given height abovethe surface would be the same around the Earth.

The same arguments can be made forpressure under water. All other things beingequal, the pressure at a given depth belowthe surface is the same.

to Pascal


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2

A scuba diver explores a reef 10 m below thesurface. The density of water is 1 g/cc. What isthe external water pressure on the diver?

? Scuba diver (W1)


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Worksheet #1In solving the last problem, we applied aprinciple that we haven’t even defined yet, butthat probably made good sense to us.

We said that the surface pressure at thebottom of the atmosphere equaled thepressure in the surface layer of water.

If this weren’t true, the water would fly out of theoceans or sink rapidly toward the ocean floor!

Pascal’s Principle


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In fact, Pascal’s Principle guarantees this willbe true. It states:

The pressure applied to an enclosed liquid istransmitted undiminished to every point inthe fluid and to the walls of the container.

Which means, that the pressure below thesurface of the water is equal to the surfacepressure + the pressure due to the columnof water above a given level.

con’t


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Worksheet #2

CQ2 Pascal’s Principle


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What happens to a cork when we try tosubmerge it in water?

It shoots right back up to the surface.

What’s responsible for the motion of the cork?

There must be a force acting upward onthe cork greater in magnitude than gravity.

Archimede’s Principle


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But what happens to thecork when it gets to thesurface?

It floats!

So what must be the net force on thecork as it’s floating on the surface?

ZERO! What’s changed?

con’t


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3

What have we noticed about our strangeunderwater force on the cork?

• It’s greater than gravity when the corkis completely submerged.

• It’s equal to gravity when the cork floatson the surface, only partially submerged.

• Our new force relates to the volume ofthe cork that’s underwater!

con’t


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Archimedes had this whole process figuredout some 2000 years ago! He said,

A body wholly or partially submerged ina fluid is buoyed up by a force equal to theweight of the displaced fluid.

So, the cork naturally float with just theright portion of its volume under the water’ssurface so that the buoyant force upwardfrom the water equals the gravitational force.

con’t


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If we have a cork with density of0.8 g/cc, what fraction of its volumewill be below the surface in a pool ofwater when it reaches equilibrium?

? Floating cork (W3)

Worksheet #3


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CQ4: Buoyant Force

Worksheet #4


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We’re now going to examine the behavior ofliquids as they flow or move through pipes, theatmosphere, the ocean,...

Let’s trace out the motion of a given piece orparcel of water as if flows through a channel.

These lines, which tell uswhere a parcel has been andin which direction it is going,are called trajectories.

Fluid Flows: trajectories


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If the flow is in a conditionknown as steady state (notvarying) then the trajectoriesare the same as thestreamlines.

The streamlines tell us the instantaneousdirection of motion of a parcel in a flow,whereas the trajectories trace out exactlywhere the parcel has actually been.

streamlines


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4

turbulent regionReal flows are also often viscous.Viscosity describes the internal“friction” of a fluid, or how well onelayer of fluid slips past another.

viscosity


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Real flows often result in turbulence -- acondition in which the flow becomes irregular.

To simplify our problems, we’re going tostudy the behavior of a class of fluids knownas “ideal” fluids.

1) The fluid is nonviscous (no internal friction)

2) The fluid is incompressible (constant density)

3) The fluid motion is steady (velocity, densityand pressure at each point remain constant)

4) The fluid moves without turbulence.

Ideal Fluids


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This is really just a conservation of massargument. It says that if I put in 5 g of watereach second at the left end of my hose, thenunder steady-state flow conditions, I mustget out 5 g of water each second at the rightend of the hose.

Δmin

Δt=Δm

out

Δt5 g/s 5 g/s

Continuity Equation


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For ideal fluids in steady-state (unchanging)flows, this must be true regardless of theshape of the hose. For instance, I could havea hose that’s narrower at the left end wherethe fluid enters the hose than it is at the rightend where fluid leaves.

5 g/s 5 g/s

Nevertheless, the mass entering at the left eachsecond must equal the mass exiting at the right.

con’t


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v1 v2A1 A2

The mass / time entering at the left side is

Δmin

Δt= ρ

ΔV1

Δt= ρ

A1Δx

1

Δt= ρA

1v

1

And similarly, the mass / time leaving at right is

Δmout

Δt= ρ

ΔV2

Δt= ρ

A2Δx

2

Δt= ρA

2v

2

con’t


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These two quantities must be equal, leavingus with the relationship

ρA1v

1= ρA

2v

2 A v A v1 1 2 2=

con’t


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v1 v2A1 A2

5

An ideal fluid flows through a pipeof cross-sectional area A. Suddenly,the pipe narrows to half it’s originalwidth. What is the ratio of the finalto the initial speed of the fluid flow?

? Continuity (W5)

1) 4:12) 2:13) 1:14) 1:25) 1:4

2Ar∝Worksheet #5


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In examining flows through pipes in theEarth’s gravitational field, Bernoulli founda relationship between the pressure in thefluid, the speed of the fluid, and the heightoff the ground of the fluid.

The sum of the pressure (P), the kinetic energyper unit volume (0.5ρv2), and the potentialenergy per unit volume (ρgy) has the samevalue at all points along a streamline.

Bernoulli’s Equation


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P + 1

2mV

v2 + mV

gy = constant

P v gy+ + =1

22ρ ρ constant

We can derive Bernoulli’s Equation usinga conservation of energy argument.

con’t


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Skip

ΔK = 1

2mv

22 − 1

2mv

12

The change in kineticenergy of the fluidbetween the two endsmust equal the network done on the fluid.

derivation

gh2A1

A2

v1

v2

h1

P1

P2

Δx1

Δx2

The hatched regions havethe same mass of fluid.

And the work done bygravity on the fluid isgiven by

Wg= −ΔU

= −(mgh2− mgh

1)

= mg(h1− h

2)


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W1= P

1A

1Δx

1= P

1ΔV

W2= −P

2A

2Δx

2= −P

2ΔV

Next, we consider thework done by thepressure forces at eachend of the pipe:

derivation con’t

The hatched regions havethe same mass of fluid.

Giving a network onthe fluid of

Wnet

=W1+W

2+W

g

= P1ΔV − P

2ΔV + mg(h

1− h

2)


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gh2A1

A2

v1

v2

h1

P1

P2

Δx1

Δx2

Wnet= ΔK

Now, putting it alltogether, we have

derivation con’t

gh2A1

A2

v1

v2

h1

P1

P2

Δx1

Δx2

P

1ΔV − P

2ΔV + mg(h

1− h

2) = 1

2m(v

22 − v

12 )

P

1ΔV + mgh

1+ 1

2mv

12 = P

2ΔV + mgh

2+ 1

2mv

22

Rearranging we get…


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gh2A1

A2

v1

v2

h1

P1

P2

Δx1

Δx2

6

Dividing by ΔV…

derivation the end

P

1+

mgh1

ΔV+

mv12

2ΔV= P

2+

mgh2

ΔV+

mv22

2ΔV

Finally, identifying density…

P gh v P gh v1 112 1

22 2

12 2

2+ + = + +ρ ρ ρ ρ


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gh2A1

A2

v1

v2

h1

P1

P2

Δx1

Δx2Airplane wings

Airplane wing

Bernoulli’s Principle: P

1+ 1

2ρv

12 = P

2+ 1

2ρv

22

If v1 > v2, then P1 < P2. That creates a pressuregradient force known as “lift.”

Figures copyrighted by ALLSTAR: www.allstar.fiu.edu/aero


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Curveballs

Curveball

Bernoulli’s Principle:

P

1+ 1

2ρv

12 = P

2+ 1

2ρv

22

If v1 > v2, then P1 < P2. That creates a pressuregradient force that causes the “curveball.”

Figure copyrighted by Cislunar Aerospace:muttley.ucdavis.edu/Book/Sports/instructor/curveball-01.html

v1

v2

P1

P2


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A large water tower is drained by a pipe ofcross section A through a valve a distance15 m below the surface of the water in thetower. If the velocity of the fluid in the bottompipe is 16 m/s and the pressure at the surfaceof the water is 1 atm, what is the pressure ofthe fluid in the pipe at the bottom? Assumethat the cross-sectional area of the tank ismuch bigger than that of the drain pipe.

? Water tower (W6)

Worksheet #6


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The pressure in both theseplaces is approximately theatmospheric pressure Pa

Torricelli’s Theorem

H

v = 0 m/s

vexit = ? And since the cross-sectional area of thecontainer is muchgreater than that ofthe hole, we canapproximate thespeed of the fluiddescent in thecontainer as v = 0.

Bernoulli’s Equationnow reduces to arather simple form…

P + 1

2ρv2 + ρgy = C


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P

a+ 1

2ρv2 + ρgH = P

a+ 1

2ρv

exit2

H

v = 0 m/s

con’t

vexit = ? Bernoulli’s Equationnow reduces to arather simple form…

v gHexit = 2

The pressure in both theseplaces is approximately theatmospheric pressure Pa


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7

How a mercury barometer works!

Mercury Barometer

Evacuated chamber (i.e., P = 0)

H

Pa

Bernoulli’s Principle now says

P

a+ 1

2ρv

02 + ρgy

0= P + 1

2ρv 2 + ρg(H + y

0)

aPHgρ=

H =

Pa

ρg

The height of themercury column isusually reported inthe nightly weather!


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How tall will the mercury column bewhen the air pressure is at thestandard value of 101.3 kPa? Thedensity of mercury is 13.6 g/cc.

? Barometer (W7)

Worksheet #7

H

Pa


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