15. Fluid Motion

1. Density & Pressure

2. Hydrostatic Equilibrium

3. Archimedes’ Principle & Buoyancy

4. Fluid Dynamics

5. Applications of Fluid Dynamics

6. Viscosity & Turbulence

Why is only the “tip of the iceberg ” above water?

ice is only slightly less than water.

Fluid = matter that flows under external forces

= liquid & gas.

solid liquid gas

inter-mol forces strongest medium weakest

volume fixed fixed variable

shape fixed variable variable

Examples of fluid motion:

• Tornadoes.

• Airflight: plane supported by pressure on wings.

• Gas from giant star being sucked into a black hole.

• Brake fluid in a car’s braking system.

• Breathing: air into lung & blood stream.

15.1. Density & Pressure

Avogadro’s number NA = 6.022 1023 / mol . 1 mole = amount of substance containing NA basic elements. ( with NA = number of atoms in 12 g of 12C ).

Fluid: average position of molecules not fixed.

Macroscopic viewpoint: deformable continuum.

Density = mass / vol, [ ] = kg / m3 .

31 /water g cm 1 /g cc 31000 /kg m1 /kg liter 1 1000liter cc1000 ml

310air water

Incompressible = density unchanged under pressure

Liquid is nearly incompressible (molecules in contact).

Gas is compressible.

dVfluid pointdV 0

thousands of molecules

Pressure

Pressure = normal force per unit area

Fp

A 2/p N m

pascal Pa

1 101,300atmosphere atm Pa 101.3 kPa

Pressure is a scalar.

The pressure at a point in a fluid is the magnitude

of the radial force per unit area acting on a fluid

point at that position.

14.7 pounds per square inchpsi

A

F n

A n

F

Fluid point

15.2. Hydrostatic Equilibrium

Hydrostatic equilibrium :

Fnet = 0 everywhere in fluid

Fluid is at rest.

Fext 0 gives rise to pressure differences.

netF F x F x x

P x P x x A

PV

x

netF d Pf

V d x

P f

x

Let f be the force density within the fluid :

Force experienced by the fluid element:

( f is the force per unit volume experienced by a small fluid element due to pressure differences )

Hydrostatic Equilibrium with Gravity

Fluid element: area A, thickness dh, mass dm.

Net pressure force on fluid element:

pdF p d p A p A A d p

Gravitational force on fluid element:

gdF g d m g A d h

Hydrostatic Equilibrium : 0p gdF dF

d p g d h d pg

d h

Liquid (~incompressible):

0p p g h

Example 15.1. Ocean Depths

(a) At what water depth is the pressure twice the atmospheric pressure?

(b) What’s the pressure at the bottom of the 11-km-deep Marianas Trench,

the deepest point in the ocean?

Take 1 atm = 100 kPa & water = 1000 kg/m3 .

0p p g h 0p ph

g

3 2

2 3

1 100 10 / /

9.8 / 1000 /

atm Pa atm N m Pa

m s kg m

10 m

(a)

(b) Pressure increases by 1 atm per 10 m depth increment.

3 111 10

10

atmp m

m

1100 atm 110 MPa

Measuring Pressure

Barometer = device for measuring atmospheric pressure

0p p g h

0 0p vacuum inside tube:

313.6 /Hg g cm 3 313.6 10 /kg m

Hgp g h 2 3 39.8 / 13.6 10 /m s kg m h

133.28 /kPa m h

For p = 1 atm = 101.3 kPa :

101.3

133.28 /

kPah

kPa m 0.760 m 760 mm

Cf. h = 10 m for a water barometer

Manometer

Manometer = U-shaped tube filled with liquid to measure pressure differences.

fluid atm Hgp p g h

Gauge pressure = excess pressure above atmospheric.

Used in tires, sport equipments, etc.

E.g., tire gauge pressure = 30 psi tire pressure = 44.7 psi

Pascal’s law:An external pressure applied to a fluid in a closed vessel is uniformly transmitted throughout the fluid.

equal p

Example 15.2. Hydraulic Lift

In a hydraulic lift, a large piston supports a car.

The total mass of car & piston is 3200 kg.

What force must be applied to the smaller piston to support the car?

11

2

m g AF

A

11

1

Fp

A

2p

2

1

2

dm g

d

2

2 153200 9.8 /

120

cmkg m s

cm

490 N

Pascal’s law2

m g

A

15.3. Archimedes’ Principle & Buoyancy

Archimedes’ Principle:

The buoyancy force on an object is equal to the

weight of the fluid it displaces.

Buoyancy force:

Upward force felt by an object in a fluid

Neutral buoyancy :

average density of object is the same as that of fluid.

fluid element in equilibrium

Fb unchanged after replacement

Example 15.3. Working under Water

To set up a raft, you need to move a 60-kg block of concrete on the lake bottom.

What’s the apparent weight of the block as you lift it underwater?

The density of concrete is 2200 kg / m3 .

app g bF F F

b wF m g

g cF m g

w cV g cw

c

mg

1 wapp c

c

F m g

210001 60 9.8 /

2200kg m s

320 N

~ ½ weight on land

Example 15.4. Tip of the Iceberg

Average density of a typical iceberg is 0.86 that of seawater.

What fraction of an iceberg’s volume is submerged?

0g bF F

b subF m g

g iceF m g

water subV g

0.86

ice iceV g

sub ice

ice water

V

V

Conceptual Example 15.1. Shrinking Arctic

Arctic sea ice is melting as a result of global warming.

Does this contribute to rising sea levels?

No.

Volume of the melted ice (which becomes water) is the same as

that displaced by the floating ice.

Making the Connection

The land based Greenland ice cap occupies some 3 million km3 ,while some 15,000 km3 of ice are afloat in the Arctic Ocean.

Compare the approximate rise in the world’s oceans that would result from complete melting of these two ice volumes.

1. Melting sea ice doesn’t change sea level.

2. Melting land-based ice adds water of volume ~ 86% that of ice. Melting Greenland ice cap adds

0.86 3 106 km3 = 2.58 106 km3

of water to ocean.

3. Ocean covers 71% of earth Ocean surface = 0.71 4 (6.37 103 km)2 = 3.62 108 km2 . Increased sea level = 2.58 106 km3 / 3.62 108 km2 = 7.1 m

Center of Buoyancy

Buoyancy force acts at the center of buoyancy (CB),

which coincides with the CM of the displaced water.

CM must be lower than CB to be stable.

15.4. Fluid Dynamics

Moving fluid is described by its flow velocity v( r, t ).

Streamlines = Lines with tangents everywhere parallel to v( r, t ).

Spacing of streamlines is inversely proportional to the flow speed.

Steady flow: , t v r v r

Small particles (e.g., dyes) in

fluid move along streamlines.

e.g., calm river.

Example of unsteady flow: blood in arteries ( pumped by heart ).

Fluid dynamics: Newton’s law + diffusing viscosity Navier-Stokes equations

slow fast

GOT IT? 15.1.

Photo shows smoke particles tracing streamlines in a

test of a car’s aerodynamic properties.

Is the flow speed greater

(a) over the top, or

(b) at the back?

Conservation of Mass: The Continuity Equation

Steady flowFlow tube: small region with sides tangent,

& end faces perpendicular, to streamlines.

flow tubes do not cross streamlines.

Conservation of mass:

1 1 1 1m A v t Mass entering tube:

2 2 2 2m A v t Mass leaving tube:

1 1 1 2 2 2A v A v

A v const v A

Equation of continuity for steady flow:

Mass flow rate =

[ v A ] = kg / s

Volume flow rate = A v constLiquid:

[ v A ] = m3 / s

v A

Liquid & gas with v < vs : flows faster in constricted area.

Gas with v > vs : flows slower in constricted area.

Steady flow

Conservation of mass:

1 1 1 1m A v t Mass entering tube:

2 2 2 2m A v t Mass leaving tube:

1 1 1 2 2 2A v A v

A v const v A

Equation of continuity for steady flow :

Mass flow rate = [ v A ] = kg / s

Volume flow rate = A v constLiquid:

[ v A ] = m3 / s v A

Liquid : flows faster in constricted area.

Gas with v < vs ound: flows faster in constricted area.

Gas with v > vsound : flows slower in constricted area.

Example 15.5. Ausable Chasm

The Ausable river in NY is about 40 m wide.

In summer, it’s usually 2.2 m deep & flows at 4.5 m/s.

Just before it reaches Lake Champlain, it enters Ausable Chasm, a deep gorge only 3.7 m

wide.

If the flow rate in the gorge is 6.0 m/s, how deep is the river at this point??

Assume a rectangular cross section with uniform flow speed.

1 1 1 2 2 2d w v d w vA v const

1 1 12

2 2

d w vd

w v

2.2 40 4.5 /

3.7 6.0 /

m m m s

m m s 18 m

Conservation of Energy: Bernoulli’s Equation

Same fluid element enters & leaves tube:

2 22 1

1

2K m v v

Work done by pressure upon its entering tube:

1 1 1 1W p A x

Work done by pressure upon its leaving tube: 2 2 2 2W p A x

Work done by gravity during the trip: 2 1gW m g y y

W-E theorem: 1 2 gW W W K 2 21 1 2 2 2 1 2 1

1

2p V p V m g y y m v v

1 1p V

2 2p V

Incompressible fluid: 1 2V V V m

V

21

2p v g y const

Bernoulli’s Equation

Viscosity & other works neglected

15.5. Applications of Fluid Dynamics

Strategy

• Identify a flow tube.

• Draw a sketch of the situation, showing the flow tube.

• Determine two points on your sketch.

• Apply the continuity equation and Bernoulli’s equation.

Example 15.6. Draining a Tank

A large open tank is filled to height h with liquid of density .

Find the speed of liquid emerging from a small hole at the base of the tank.

atmp p

21

2 holev g h

y h

At top surface :

0v

21

2p v g y const

At hole:

atmp p 0y holev v

2holev g h

Venturi Flows

Venturi : a tube with constricted central region C.

Eq. of continuity v larger in C.

Bernoulli eq. p lower in C.

Venturi flow meter measures flow speed by measuring pressure drop in C.

Example 15.7. Venturi Flowmeter

Find the flow speed in the unconstricted pipe of a Venturi flowmeter.

1 1 2 2v A v A

2 21 1 2 2

1 1

2 2p v p v

Bernoulli’s eq.

Continuity eq.

12 1

2

Av v

A

2

211 2 1

2

11

2

Av p p p

A

1 2

1

2

2

1

pv

A

A

Bernoulli Effect

A ping-pong ball supported by downward-flowing air.

High-velocity flow is inside the narrow part of the funnel.

Bernoulli Effect: p v

Example: Prairie dog’s hole

Dirt mound forces wind to accelerate over hole

low pressure above hole

natural ventilation

GOT IT? 15.2.

A large tank is filled with liquid to level h1 .

It drains through a small pipe whose diameter varies.

Emerging from each section of pipe are vertical tubes open to the atmosphere.

Although the picture shows the same liquid level in each pipe,

they really won’t be the same.

Rank order the levels h1 through h4.

1 4 2 3h h h h ordered inversely by flow speed

Flight & Lift

Aerodynamic lift

Top view on a curved ball : spin

Blade pushes down on air

Air pushes up (3rd law) Faster flow, lower P : uplift.

Top view on a straight ball : no spin

Application: Wind Energy

A chunk of air, of speed v & density ,

passing thru a turbine of area A in time t,

has kinetic energy

21

2K v v A t

31

2v A t

available power per unit area = 31

2A vP

Better analysis 38

27vP

3381.2 / 10 /

27kg m m sP 2350 /W m

For 10 / 36 /v m s km h

Present tech gives 80% of this.

0.6 A P

15.6. Viscosity & Turbulence

Smooth flow becomes turbulent.

Viscosity: friction due to momentum transfer between

adjacent fluid layers or between fluid & wall.

B.C.: v = 0 at wall

• drag on moving object.

• provide 3rd law force on propellers.

• stabilize flow.

flow with no viscosity

flow with viscosity