Ch15 testing

7/28/2019 Ch15 testing

http://slidepdf.com/reader/full/ch15-testing 1/37

Chapter 15: Fluid Motion

Fluids Characteristics of fluids

• Microscopically molecules of a fluid do not have long-range

order. But liquids do have short-range order unlike gases

• Fluids can flow and conform to the boundaries of a container

• Fluids cannot sustain a shearing stress



Fluids (cont’d)

Some useful quantities

Density for

V

m

V

m

for uniform solid or liquid

unit kg/m3 (at STP for a gas)

333

kg/m2.1;kg/m100.1 air water

3333 kg/m1035.11;kg/m101.0 lead styrofoam

Pressure for area

m V mass ,volume

A and normal force F

dA

dF

A

F p

A

0

unit pascal2m1/ N1Pa in SI unit

in SI unit

other useful units:

25 in/lb7.14torr 760Pa1001.1atm1

height of a column of Hg corresponding to this pressure (mm)



Pressure

Pressure of a liquid at rest (uniform density)

A

A

2 y

1 y

fluid level

imaginary

box

2 F

1 F

mg

A p F 11

A p F 22

0)( 1221 gA y y F F

weight of the imaginary box

hg

g y y p p

)( 1212

mg

h

hg p p 12



Pressure (cont’d)

Pressure of a fluid at rest : gauge pressure

A

A

2 y

1 y

fluid level

imagi

nary

box

2 F

1 F

mg

h

hg p p 12

0 p

gauge pressure:

gauge

gauge

p p p

p p p

202

101

atmospheric pressure

atmospheric

pressure



A simple model for atmospheric pressure

Pressure of a gas at rest ( )kp

gdy y y g p pdp )( 2112

When ,0, 1212 y y p p

kpgdy gdydp

kpg dydp /

)(

1212 y ykg e p p



Barometer Pressure measurement using a liquid:

(measures absolute pressure)

h

0 p

0 p

atmospheric

pressure

From )( 2112 y y g p p

as 0122 ,,0 p ph y p

gh p 0

If the liquid is mercury, for 1 atm :

m/s83.9,kg/m106.13

,Pa103.101atm1

33

3

0

g

p

Hg

mm760m758.0/0 g ph



Manometer

Pressure measurement using a liquid :

(measures gauge pressure)

g p h

liquidtank

manometer

0 p

level 1

level 2

)( 2112 y y g p p

0,

,,

101

22

y p p

h y p p

gh p p p g 0

gauge pressure



Pascal’s law

Pressure applied to an enclosed fluid is transmitted

undiminished to every portion of the fluid and walls

of the containing vessel

piston

weight

liquid

P

h

ext p

p

(area A)

w

gh p p pext

0

pressure at P

pressure due

to weight w:

w/A

atmospheric

pressure

pressure due to

liquid above P

incompressible

ext p p



Pascal’s law (cont’d)

Hydraulic lever

1 x

2 x1 F

2 F

1 A2 A

2211// A F A F p

1122 )/( F A A F

1212 1/ F F A A

12122211 )/( x A A x A x A x

111

121112222 )/()/(

W x F

x A A F A A x F W



Buoyancy

Origin of buoyancy

• Consider a submerged massless object filled with the same fluid

as the fluid that surrounds the object

1 P

2 P

1 F

2 F

0)( 12 g m F F F f net

buoyancymass of fluid in the object

The object is at rest

• Now fill the object with another material

g m F f b

buoyant force

0)( 12 g m F F F objnet

g m F f b

opposite dir.

mf g



Buoyancy

Origin of buoyancy (cont’d)

• Consider a portion of fluid at rest in a container surrounded by animaginary boundary represented in dashed line.

• Since the portion of the fluid defined by the surface in dashed line

is at rest, the net force on this portion due to pressure must be

equal to that of the weight of the fluid inside the surface, and

opposite in direction.

dF

dF

dF

dF

dF

dF

dF

dF

• The same argument can be applied when the

imaginary portion of the fluid is replaced by an

object that occupies the same space.

X

0fluid bfluidnet g V F g m F d F

Fnet mg

Vg g m F fluidfluid b buoyant force



Buoyancy

Archimedes’s principle

• The buoyant force on a partly or completely submerged objectis equal to the weight of the displaced fluid:

Vg

g m g m g m F F

obj f

obj f objbnet

)(

• If

0, net obj f F

apparent weight g mVg W objobj f

)(

0, net obj f F

• If

The object will rise until a part of it comes out above the fluid

Surface when the average density increases to f

Object sinks

Object floats



What fraction of an iceberg is submerged in the sea water?

Let’s assume that the total volume of the iceberg is Vi.

Then the weight of the iceberg Fgi is

Since the whole system is at its

static equilibrium, we obtain

g V F ii gi

Let’s then assume that the volume of the iceberg

submerged in the sea water is Vw. The buoyant force B

caused by the displaced water becomes

g V B ww

g V g V wwii

Therefore the fraction of the

volume of the iceberg

submerged under the surface of the sea water is

890.0/1030

/9173

3

mkg

mkg

V

V

w

i

i

w

About 90% of the entire iceberg is submerged in the water!!!

Example



• A fake or pure gold crown?

Is the crown made of puregold?

Tair =7.84 N

Twater =6.86 N0mg T air

0 Bmg T water

0 BT T air water

N980.0

water water water air gV T T B

34 m1000.1 crownwater V V

kg800.0/ g T m air

33 kg/m1000.8/ crowncrown V m

gold=19.3x103 kg/m3

Example



• Floating down the river

What depth h is the bottom of the raft submerged?

A=5.70 m2 wood=6.00x102 kg/m3

g m B g m B raft raft 0

g Ah g V g m B water water water water )()(

g V g m raft raft raft )(

g V g Ah raft raft water )()(

m0632.0 A

V h

water

raft raft

Example



Density and Pressure

Example

• Oil and water

=0.700 g/cm3

=1025 kg/m3

h1=8.00 m

h2=5.00 mPa101.56

m)00.8(

)m/s80.9)(kg/m1000.7(

Pa1001.1

5

232

5

101

gh P P

Pa1006.2 5

21

gh P P bot



Ideal fluid flow

Ideal fluids in motion

• Incompressible ( density is constant at any position)• No internal friction (no viscosity)

• Steady (non-turbulent) flow- the velocity at a point is

constant in time.

flow tube

flow lines

flow line : The path of an individual particle in a

moving fluid

steady flow: A flow whose pattern does not change

with time. Every element passing

through a given point follows the same

flow line

streamline : A curve whose tangent at any point isin the direction of the fluid velocity at

that point

flow tube : The flow lines passing through the edge

of an imaginary area such as A A



Continuity equation

Continuity equation I (incompressible fluid)

1 A

2 A

dt v1

dt v2

1v

2v

Avdt dV /

The volume of the fluid that passes through

area A during a small time interval dt :

Avdt dV

dt v AdV dt v AdV 222111 ;

•In an ideal fluid the density is constant.

•In a time interval dt the mass that flowsinto Area 1 is the same as the mass that

flows out of Area 2.

22112211

v Av Adt v Adt v A

The mass of a moving fluid does not change as it flows.



Continuity equation

Volume flow rate

1 A

2 A

dt v1

dt v2

1v

2v

Avdt dV

222111222111 v Av Adt v Adt v A

Avdt dV / volume flow rate

Continuity equation II

(compressible fluid)



Bernoulli’s equation

Work done by pressure

1 A

2 A

dt vds 11

dt vds 22

1v

2v222 A p F

111 A p F

1 y2 y

dV

dV

dV p p

ds A pds A pdW

)( 21

222111

Change in kinetic energy

2211 ds Ads AdV

)()2/1( 2

1

2

2 vvdV dK

)( 12 y ydVg dU

Change in potential energy



Bernoulli’s equation

dU dK dW

Energy conservation

dV p pdW )( 21

)()2/1( 2

1

2

2vvdV dK

)( 12 y ydVg dU

dV p p )( 21 )()()2/1( 12

2

1

2

2 y ydVg vvdV

)()()2/1( 12

2

1

2

221 y y g vv p p

const.)2/1()2/1(2

222

2

111 v gy pv gy p



Example

Venturi meter

2

22

2

112

1

2

1v pv p

)1(2

12

2

2

12

121 A

Av p p

1

2

12 v

A

Av

h

gh p p 21

1)/(

22

21

1

A A

ghv



Example Torricelli’s theorem

The velocity of the fluid coming out of a hole in a tank as

shown in the figure can be calculated using Bernoulli’s

equation.

At the top surface the velocity

of the fluid is zero. The pressures

at the top surface and at the hole

are the same, namely, the

atmospheric pressure.

22 )2/1()2/1( holeholeholetoptoptop v gy pv gy p

ghvv y y g holeholeholetop

2)2/1()( 2



Example Siphon

Suppose a U-shaped piece of pipe is completely submerged in

water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of

water, the water does not run out of the pipe. WHY?



Example Siphon

Suppose a U-shaped piece of pipe is completely submerged in

water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of

water, the water does not run out of the pipe. WHY?

Air cannot enter the pipe. As the water starts running out of the pipe,

a near vacuum is created in the topmost region of the inverted U. The

pressure here drops to near zero. The atmospheric pressure on the

surface of the water in the bucket pushes the water into the U-shaped

pipe.



Example Siphon

If a U-shaped hose or pipe connects a liquid-filled container at a

higher altitude to a container at a lower altitude over a barrier, theliquid can be siphoned into the container at the lower altitude.

Atmospheric pressure helps to push the liquid over the barrier.



Example Siphon

When P1>P2, the fluid can be siphoned from the left to the

right bucket.



• Water garden

A water hose 2.50 cm in diameter is used by a gardener to fill a

30.0-liter bucket. The gardener notices that it takes 1.00 min to fillthe bucket. A nozzle with an opening of cross-sectional area 0.500

cm2 is then attached to the hose. The nozzle is held so that water

is projected horizontally from a point 1.00 m above the ground. Over

what horizontal distance can the water be projected?

/sm1000.5s0.60

min00.1

cm100.0

m00.1

L00.1

cm1000.1

min1.00

L0.30rateflowvolume 33

333

velocity)initialtheof component-xthe:( 0022211 x x vv Av Av A

m/s0.10m10500.0

/m1000.524

34

2

110

s

A

v Av x

m52.4s;452.0

m/s80.9

m)00.1(22

2

102

2

0

t v x

g

t t gt t v y x y

Example



Consider a water tank with a hole.

(a)Find the speed of the water

leaving through the hole.

• Example : A water tank h =0.500 m

y1 =3.00 m

201

2

10

2

1 gy P gyv P

m/s13.3

2)(2 121

gh y y g v

(b) Find where the stream hits the ground.

s782.02

10 0

2

1 t t v gt y y y

m45.210 t vt v x x

x

y



Find the speed at Point 1.

• Example : Fluid flow in a pipe

A2=1.00 m2 A1=0.500 m2

h =5.00 m

2

2

201

2

102

1

2

1 gyv P gyv P

2211 v Av A

1

2

12 v

A

Av

2

2

1

2

101

2

102

1

2

1 gyv A

A P gyv P

m/s4.11

)/(1

22)(21

2

21

112

2

2

12

1

A A

ghv gh y y g

A

Av



Viscosity and turbulence

Viscosity

Viscosity is internal friction in a fluid, and viscous forces opposethe motion of one portion of a fluid relative to another.




Drag

If a fluid in laminar flow flows around an obstacle, it exerts a viscousdrag on obstacle. Frictional forces accelerate the fluid backward

against the direction of flow and the obstacle forward in the direction

of flow.

laminar flow

adjacent layers of fluid slide

smoothly past each other andflow is steady




Turbulence

When the speed of a flowing

fluid exceeds a certain critical

value the flow is no longer laminar.

The flow patter becomes extremely

irregular and complex, and it changes

continuously in time. There is nosteady flow pattern. This chaotic flow

Is called turbulence.



Problems

Problem 1

2.00 mThe upper edge of a gate in a dam runsthe water surface. The gate is 2.00 m high

and 4.00 m wide and is hinged along the

horizontal line through its center. Calculate

the torque about the hinge arising from

the force due to the water.

Solution

Denote the width and depth at the bottom of the gate by w and H.

The force on a strip of vertical thickness dh at a depth h is:

and the torque about the hinge is After integrating from h=0 to h=H, you get the torque:

)(wdh ghdF

.)2/( dh H h gwhd

.1061.212/43

m N gwH



Problem 2

Solution

(a) From Archimedes’s principle so

An object with height h, mass M, and a uniform cross-sectional area A

floats upright in a liquid with density .

(a)Calculate the vertical distance from the surface of the liquid to thebottom of the floating object in equilibrium.

(b) A downward force with magnitude F is applied to the top of the object.

At the new equilibrium position, how much farther below the surface of

the liquid is the bottom of the object than it was in part (a)?

(c) Calculate the period of the oscillation when the force F is suddenlyremoved.

, Mg gLA )./( A M L

(b) The buoyant force is: With the result of part (a) solvingfor x gives:.)( F Mg x L gA

)./( gA F x

(c) The force is always in the direction toward the equilibrium, namely, a

restoring force Therefore the “spring constant” is

and the period of the oscillation is

. gAx Mg F F buoyres gAk .)/(2/2 gA M k M T



Problem 3

Solution

You cast some metal of density m in a mold, but you are worried that there

might be cavities within the casting. You measure the weight of the casting

to be w, and the buoyant force when it is completely surrounded by water

to be B. (a) Show that is the total volume

of any enclosed cavities. (b) If your metal is copper, the casting’s weight

is 156 N, and the buoyant force is 20 N, what is the total volume of any

enclosed cavities in your casting? What fraction is this of the total volume

of the casting?

(a) Denote the total volume V. If the density of air is neglected, the buoyant

force in terms of the weight is:

Therefore

]/)/[( 0V g w g gV B mwater water

)/()/(0 g w g BV mwater

)./()/(0 g w g BV mwater

(b) .1052.2)/()/( 34 m g w g B Cuwater

The total volume of the casting is

),/( g B water The cavities are 12.4% of the total volume.

P bl 4



Problem 4

Solutions

A U-shaped tube with a horizontal portion of

length contains a liquid. What is the difference

in height between the liquid columns in the

vertical arms (a) if the tube has an accelerationtoward the right? (b) if the tube is mounted on

a horizontal turntable rotating with an angular

speed with one of the vertical arms on the

axis of rotation?

a

(a)Consider the fluid in the horizontal part of the tube. This fluid with mass

, is subject to a net force due to the pressure difference between

the ends of the tube, which is the difference between the gauge pressures

at the bottoms of the ends of the tubes. Now this difference is

and the net force on the horizontal part of the fluid is

or

(b) Similarly to (a) consider the fluid in the horizontal part of the tube. As in (a)

the fluid is accelerating. The center of mass has a radial acceleration of

magnitude so the difference in heights between the columns

is

),( R L y y g

A

,)( a A A y y g R L .)/()( g a y y R L

,2/2 rad a

).2/()/)(2/( 222 g g

A A

Ch15 testing

Documents

Transcript of Ch15 testing