PHYS222 – LSSU – BazlurSlide 1 Chapter - 11 Fluids.

PHYS222 – LSSU – Bazlur Slide 1

Chapter - 11

Fluids


Fluids 11.1 Mass Density11.2 Pressure11.3 Pressure and Depth in a Static Fluid11.4 Pressure Guages11.5 Pascal’e Principle11.6 Archimedes’ Principle11.7 Fluid in Motion11.8 The Equation of Continuity11.9 Bernoulli’s Equation11.10 Applications of Bernoulli’s Equation11.11 Viscous Flow


11.1 Mass Density

Definition of Mass Density

The mass density of a substance is the mass of a substance per unit volume:

V

m

SI Unit of Mass Density: kg / m3


11.1 Mass Density


11.1 Mass Density

Example-1: Blood as a Fraction of Body Weight

The body of a man whose weight is 690 N contains about5.2x10-3 m3 of blood.

(a) Find the blood’s weight and (b) express it as a percentage of the body weight.

kg 5.5mkg1060m102.5 333 Vm

V

m


11.1 Mass Density

N 54sm80.9kg 5.5 2 mgW(a)

(b) %8.7%100N 690

N 54Percentage


11.1 Mass Density

Definition of Weight Density

The Weight density of a substance is the Weight of a substance per unit volume:

V

mg

SI Unit of Weight Density: N / m3


Specific Gravity• Is relative density.• It is a ratio of the densities of two materials. • Dimensionless.

• Using water as a reference can be convenient, since density of water is (approximately) 1000 kg/m³ or 1 g/cm³ at 4C.

• Specific gravity of Mercury is 13.6, means it is 13.6 times massive or heavier than water.

Reference ofDensity

Substance ofDensity Gravity Specific


11.2 Pressure

A

FP

SI Unit of Pressure: 1 N/m2 = 1Pa

Pascal


Pressure• Pressure is force per unit area.

P = F/A• Pressure is not a vector quantity. • To calculate the pressure we only consider the

magnitude of the force.• The force generated by the pressure of the static fluid

is always perpendicular to the surface of the contact.

Pa = N/m2 = 1.450 x 10-4 lb/in2 (very small quantity)bar = 105 Pa = 100 kPaAtmospheric pressure is 101.3 kPa

= 1.013 bar = 1.450 lb/in2 = 760 torrlb/in2 or psi = 6.895 x 103 Pa


Pressure causes Perpendicular ForceThe forces of a liquid pressing against a surface add up

to a net force that is perpendicular to the surface.

Components parallel to the surface cancels out

Components vertical to the surface adds up


11.2 Pressure

Example-2: The Force on a Swimmer

Suppose the pressure acting on the backof a swimmer’s hand is 1.2x105 Pa. Thesurface area of the back of the hand is 8.4x10-3 m2.

(a) Determine the magnitude of the forcethat acts on it.

(b) Discuss the direction of the force.


11.2 Pressure

A

FP

N 100.1

m104.8 mN102.13

2325

PAF

Since the water pushes perpendicularly against the back of the hand, the forceis directed downward in the drawing.


Shape of the Dam• Would you prefer Dam-A or Dam-B?• Justify your answer.

Dam-A Dam-B


11.2 Pressure

Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere

Weight of the column of the air of 1m2 = 1.013x105 N

Mass of the column of the air of 1m2 = 1.013x105 N/g = 10326.2 kg


11.3 Pressure and Depth in a Static Fluid

012 mgAPAPFy

mgAPAP 12

Vm

Relation between Pressure and Depth

Since the column is in equilibrium,The sum of the vertical forces equal to zero



VgAPAP 12

AhV

AhgAPAP 12

hgPP 12

The pressure increment = gh = weight density x heightPressure at a depth, h = P1 + gh = Atmospheric Pressure + Liquid Pressure

So, Liquid Pressure at a depth, h = gh = weight density x depth


PressureWater pressure acts perpendicular to the sides of a

container, and increases with increasing depth.



Conceptual Example-3: The Hoover Dam

Lake Mead is the largest wholly artificial reservoir in the United States. The waterin the reservoir backs up behind the damfor a considerable distance (120 miles).

Suppose that all the water in Lake Meadwere removed except a relatively narrowvertical column.

Would the Hoover Dam still be neededto contain the water, or could a much lessmassive structure do the job?



Example-4: The Swimming Hole

Points A and B are located a distance of 5.50 m beneath the surface of the water. Find the pressure at each of these two locations.



Pa 1055.1

m 50.5sm80.9mkg1000.1Pa 1001.15

233

pressure catmospheri

52

P

ghPP 12


Where to put the pump?• Pump-X pushed the water up.• Pump-Y removes the air from the

pipe and air pressure outside pushes the water up.

• Pump-Y has a limit, what is that?

ghPatm

Pump-X Pump-Y

mm 10306m 306.10

sm80.9 mkg101.0

Pa 1001.1233

5

g

Ph atm

ghPatm


Pressure Gauges• One of the simplest pressure

gauges is the mercury barometer used for measuring atmospheric pressure.

• 760 mm of mercury


11.4 Pressure Gauges

ghPatm

mm 760m 760.0

sm80.9mkg1013.6

Pa 1001.1233

5

g

Ph atm

Point A and B are at the same depth, so,

PB = PA

Patm = P1 + gh

= 0 + gh


Open-tube manometer• The phrase “open-tube” refers to

the fact that one side of the U-tube is open to atmospheric pressure.

• The tube contains mercury• Its other side is connected to the

container whose pressure P2 is to be measured.

PB = PA

P2 = Patm + gh

P2 - Patm = gh

• P2 – Patm is called the guage pressure

• P2 is called the absolute pressure



AB PPP 2

ghPPA 1

ghPP atm pressure gauge

2

absolute pressure


sphygmomanometer

blood-pressure instrument: an instrument used to measure blood pressure in an artery that consists of a pressure gauge, an inflatable cuff placed around the upper arm, and an inflator bulb or pressure pump.

Systolic pressure is 120 mm of mercury

Diastolic pressure is 80 mm of mercury




11.5 Pascal’s Principle

Pascal’s Principle

Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls.



m 012 gPP

1

1

2

2

A

F

A

F

1

212 A

AFF



Example-7: A Car Lift

The input piston has a radius of 0.0120 mand the output plunger has a radius of 0.150 m.

The combined weight of the car and the plunger is 20500 N. Suppose that the inputpiston has a negligible weight and the bottomsurfaces of the piston and plunger are atthe same level. What is the required inputforce?



N 131m 150.0

m 0120.0N 20500 2

2

2

F

1

212 A

AFF


11.6 Archimedes’ Principle

APPAPAPFB 1212

ghPP 12

ghAFB

hAV

gVFB

fluiddisplaced

of mass



Archimedes’ Principle

Any fluid applies a buoyant force to an object that is partiallyor completely immersed in it; the magnitude of the buoyantforce equals the weight of the fluid that the object displaces:

fluid displaced

ofWeight

fluid

forcebuoyant of Magnitude

WFB



If the object is floating then the magnitude of the buoyant forceis equal to the magnitude of itsweight.



Example-9: A Swimming Raft

The raft is made of solid squarepinewood. Determine whetherthe raft floats in water and ifso, how much of the raft is beneaththe surface.



N 47000

sm80.9 m 8.4 mkg1000 233

max

gVVgF waterwaterB

m 8.4m 30.0m 0.4m 0.4 raftV



N 47000

N 26000

sm80.9 m 8.4 mkg550 233

gVgmW raftpineraftraft

The raft floats!



gVwaterwaterN 26000

Braft FW

If the raft is floating:

23 sm80.9 m 0.4 m 0.4 mkg1000N 26000 h

m 17.0sm80.9 m 0.4 m 0.4 mkg1000

N 2600023

h



Conceptual Example-10: How Much Water is Neededto Float a Ship?

A ship floating in the ocean is a familiar sight. But is allthat water really necessary? Can an ocean vessel floatin the amount of water than a swimming pool contains?




11.7 Fluids in Motion

In steady flow the velocity of the fluid particles at any point is constant as time passes.

Unsteady flow exists whenever the velocity of the fluid particles at a point changes as time passes.

Turbulent flow is an extreme kind of unsteady flow in which the velocity of the fluid particles at a point change erratically in both magnitude and direction.



Fluid flow can be compressible or incompressible. Most liquids are nearly incompressible.

Fluid flow can be viscous or nonviscous.

An incompressible, nonviscous fluid is called an ideal fluid.



When the flow is steady, streamlines are often used to representthe trajectories of the fluid particles.



Making streamlines with dyeand smoke.


11.8 The Equation of Continuity

The mass of fluid per second that flows through a tube is calledthe mass flow rate.



2222 vAt

m

111

1 vAt

m

distance

tvAVm



222111 vAvA

Equation Of Continuity

The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow.

SI Unit of Mass Flow Rate: kg/s



Incompressible fluid: 2211 vAvA

Volume flow rate Q: AvQ



Example-12: A Garden Hose

A garden hose has an unobstructed openingwith a cross sectional area of 2.85x10-4m2. It fills a bucket with a volume of 8.00x10-3m3

in 30 seconds.

Find the speed of the water that leaves the hosethrough (a) the unobstructed opening and (b) an obstructedopening with half as much area.



AvQ

sm936.0

m102.85

s 30.0m1000.824-

33

A

Qv

(a)

(b) 2211 vAvA

sm87.1sm936.0212

12 v

A

Av


Bernoulli’s EquationFor steady flow,

The speed,

Pressure, and

Elevation

of an incompressible and nonviscous fluid are related by Bernoulli’s equation.

2222

121

212

11 gyvPgyvP


11.9 Bernoulli’s Equation

The fluid accelerates toward the lower pressure regions.

According to the pressure-depthrelationship, the pressure is lowerat higher levels, provided the areaof the pipe does not change.


Bernoulli’s Equation• Bernoulli’s Equation from Work-energy theorem:• Work done on an object by external non-conservative forces is

equal to the change in total mechanical energy.• The pressure within a fluid is caused by collisional forces, which

are non-conservative.• Therefore, when a fluid is accelerated because of a difference

in pressures, work is being done by non-conservative forces.

2222

11

212

1

21

0fnc

mgymvmgymv

EE

EEW



2222

11

212

1nc mgymvmgymvW

VPPAsPsFsFW 12


Bernoulli’s Equation• Whenever a fluid is flowing in a horizontal pipe and encounters

a region of reduced cross-sectional area, the pressure of the fluid drops.

• Can be explained by the Newton’s 2nd law.• When moving from the wider region 2 to the narrower region 1,

the fluid speeds up according to the equation of continuity.• According to the 2nd law, th eaccelerating fluid must be

subjected to an unbalanced force.• There can be an unbalanced force only if the pressure in region

2 is higher than the pressure in region 1.

2222

11

212

1

21

0fnc

mgymvmgymv

EE

EEW


Bernoulli’s Equation• Explanation:• In figure-b on the top surface of the blue fluid element, the

surrounding fluid exerts a pressure P.• This pressure give rise to a force, F = P/A• On the botom of the blue fluid element, the pressure is slightly

higher, P+P.• As a result, the force on the bottom surface has a magnitude of

F+ F = (P+P)/A.• The net force pushing the fluid element up the pipe is F=P/A


Bernoulli’s Equation• When the fluid element moves through its own length s, the

work done is the

W = F s = P A s = P V

The total work done on the fluid element in moving it from region 2 to region 1is the sum of the small increments of work P V done as the element moves along the pipe.

This sum amounts to Wnc = (P2 – P1)V

Wnc = (P2 – P1)V = E1 – E2



2222

11

212

112 mgymvmgymvVPP

2222

11

212

112 gyvgyvPP

Bernoulli’s Equation

In steady flow of a nonviscous, incompressible fluid, the pressure, the fluid speed, and the elevation at two points are related by:

2222

121

212

11 gyvPgyvP


Bernoulli’s EquationThe term

has a constant value at all positions in the flow.

constant221 gyvP


Bernoulli’s EquationBernoulli’s Equation

reduces to the result for static fluids (v1=v2), as it is when the cross-sectional area remains constant.

2222

121

212

11 gyvPgyvP

ghPP

yygPP

gygyPP

gyPgyP

12

2112

2112

2211

)(


11.10 Applications of Bernoulli’s Equation

Conceptual Example-14: Tarpaulins and Bernoulli’s Equation

When the truck is stationary, the tarpaulin lies flat, but it bulges outwardwhen the truck is speeding downthe highway.

Account for this behavior.


Bernoulli’s EquationBernoulli’s Equation

reduces to

When all parts have the same elevation (y1 = y2).

The term P+ ½ remains constant throughout a horizontal pipe.

If increases, P decreases and vice versa.

2222

121

212

11 gyvPgyvP

222

12

212

11 vPvP

constant221 vP









Example-16: Efflux Speed

The tank is open to the atmosphere atthe top. Find and expression for the speed of the liquid leaving the pipe atthe bottom.



2222

121

212

11 gyvPgyvP

atmPPP 2102 v

hyy 12

ghv 212

1

ghv 21


11.11 Viscous Flow

Flow of an ideal fluid.

Flow of a viscous fluid.


11.11 Viscous Flow

Force Needed to Move a Layer of Viscous Fluid withConstant Velocity

The magnitude of the tangential force required to move a fluid layer at a constant speed is given by:

y

AvF

coefficient of viscosity

SI Unit of Viscosity: Pa·s

Common Unit of Viscosity: poise (P)

1 poise (P) = 0.1 Pa·s


11.11 Viscous Flow

Poiseuille’s Law

The volume flow rate is given by: L

PPRQ

812

4


11.11 Viscous Flow

Example-17: Giving and Injection

A syringe is filled with a solution whose viscosity is 1.5x10-3 Pa·s. The internalradius of the needle is 4.0x10-4m.

The gauge pressure in the vein is 1900 Pa.What force must be applied to the plunger,so that 1.0x10-6m3 of fluid can be injected in 3.0 s?


11.11 Viscous Flow

Pa 1200

m104.0

s 0.3m100.1m 025.0sPa105.18

8

44-

363

42

R

LQPP


11.11 Viscous Flow

Pa 120012 PP

Pa 19001 P

Pa 31002 P

N25.0m100.8Pa 3100 252 APF


ForceA


ForceA


ForceA


ForceA

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PHYS222 – LSSU – BazlurSlide 1 Chapter - 11 Fluids.

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