1 introduction to fluid mechanics. chapter 1. phys 404 dr nazir mustapha

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Introduction to Fluid Mechanics.Chapter 1.

Phys 404

Dr Nazir Mustapha

Al Imam University College of Sciences

Dr Nazir Mustapha – 323-223-C-1

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Sunday Phys 404 (291)

Phys 404 (291)

Phys 404

(291)

Phys 404

(291)

Monday

Tuesday

Wednesday Phys 404 (291)

Phys 404 (291)

Phys 404 (291)

Phys 404 (291)

Thursday Phys 404 (291)

Phys 404 (291)

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Exams- Midterm 1: week: 2- Midterm 2: week: 4

- Final: week: 6 – or - 7

Grading: - Midterm1: 20

- Midterm 2: 20- Home works, participation &

quizzes: 20-Final Exam: 40

-Please note that a quiz will be given at the end of each chapter.

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PHYISCS 404 - Chapter 1-2Physics 404 –Fluid Mechanics

Dr. Nazir Mustapha Office: SR 104, Third Floor

Phone: 2582167

e-mail: [email protected]

[email protected]

OFFICE HOURSSun:10:00 – 11:20am, LECTURE HOURS

Sun & Wed: 1,2,3 &4 and Thurs: 1, 2

SCHEDULE

Lectures are on: Sunday, Wednesday and Thursday.

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Textbooks:1. Yunus A. Çengel and John M. Cimbala. Fluid Mechanics: Fundamentals and Applications.Second Edition, McGraw Hill, 2010. 2.Fox, McDonald & Pritchard. Introduction to Fluid Mechanics. 5th Edition, Wiley, 2004

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Mechanical properties of Matter and fluids.

Contents:• Definition of a fluid• Density,• Pressure, • Fluid flow,• Pascal Law• Boyle’s Law• Bernoulli's equation and its application,

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Introduction• Fluid mechanics is a study of the behavior of

fluids, either at rest (fluid statics) or in motion (fluid dynamics).

• The analysis is based on the fundamental laws of mechanics, which relate continuity of mass and energy with force and momentum.

• An understanding of the properties and behavior of fluids at rest and in motion is of great importance in Physics.

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1.1 Definition of FluidA fluid is a substance, which deforms

continuously, or flows, when subjected to shearing force.

In fact if a shear stress is acting on a fluid it will flow and if a fluid is at rest there is no shear stress acting on it.

Fluid Flow Shear stress – Yes

Fluid Rest Shear stress – No

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Definition of a Fluid

A fluid is a substance that flows under the action of shearing forces القص If a fluid is at rest, we know that the . قوىforces on it are in balance.

A gas is a fluid that is easily compressed. It fills any vessel in which it is contained.

A liquid is a fluid which is hard to compress. A given mass of liquid will occupy a fixed volume, irrespective of the size of the container.

A free surface is formed as a boundary between a liquid and a gas above it.

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Definition of a Fluid

• “a fluid, such as water or air, deforms continuously when acted on by shearing stresses of any magnitude.” - Munson, Young, Okiishi

WaterOilAirWhy isn’t steel a fluid?

WaterOilAirWhy isn’t steel a fluid?

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Fluid Deformation between Parallel Plates

Side viewSide view

Force F causes the top plate to have velocity U.Force F causes the top plate to have velocity U.What other parameters control how much force is What other parameters control how much force is required to get a desired velocity?required to get a desired velocity?

Distance between plates (b)Distance between plates (b)

Area of plates (A)Area of plates (A)

F

b

U

Viscosity!Viscosity!

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Shear stress in moving fluid

• If fluid is in motion, shear stress are developed if the particles of the fluid move relative to each other. Adjacent particles have different velocities, causing the shape of the fluid to become distorted

• On the other hand, the velocity of the fluid is the same at every point, no shear stress will be produced, the fluid particles are at rest relative to each other.

Moving plate Shear force

Fluid particles New particle position

Fixed surface

Shearing Forces

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Fluid Mechanics

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•Liquids and gases have the ability to flow.

•They are called fluids.

•There are a variety of “LAWS” that fluids obey.

•Need some definitions.

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Differences between liquid and gases.

Liquid Gases

Difficult to compress and often regarded as incompressible.

Easily to compress – changes of volume is large, cannot normally be neglected and are related to temperature.

Occupies a fixed volume and will take the shape of the container.

No fixed volume, it changes volume to expand to fill the containing vessels.

A free surface is formed if the volume of container is greater than the liquid.

Completely fill the vessel so that no free surface is formed.

Pressure in a Fluid

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•The pressure is just the weight of all the fluid above you.

•Atmospheric pressure is just the weight of all the air above an area on the surface of the earth.

•In a swimming pool the pressure on your body surface is just the weight of the water above you (plus the air pressure above the water).

Pressure in a Fluid

17

•So, the only thing that counts in fluid pressure is the gravitational force acting on the mass ABOVE you.

•The deeper you go, the more weight above you and the more pressure.

•Go to a mountaintop and the air pressure is lower.

Pressure in a Fluid

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Pressure acts perpendicular to the surface and increases at greater depth.

Pressure in a Fluid

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Buoyancy

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Net upward force is called the buoyant force!!!

Easier to lift a rock in water!!

Displacement of Water

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The amount of water displaced is equal to the volume of the rock.

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Archimedes’ Principle

• An immersed body is buoyed up by a force equal to the weight of the fluid it displaces.

• If the buoyant force on an object is greater than the force of gravity acting on the object, the object will float.

• The apparent weight of an object in a liquid is gravitational force (weight) minus the buoyant force.

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Flotation

• A floating object displaces a weight of fluid equal to its own weight.

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Flotation

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Gases

• The primary difference between a liquid and a gas is the distance between the molecules.

• In a gas, the molecules are so widely separated, that there is little interaction between the individual molecules.

• IDEAL GAS.

• Independent of what the molecules are.

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Boyle’s Law

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Boyle’s Law

• Pressure depends on density of the gas.

• Pressure is just the force per unit area exerted by the molecules as they collide with the walls of the container.

• Double the density, double the number of collisions with the wall and this doubles the pressure.

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Boyle’s Law

Density is mass divided by volume.

Halve the volume and you double the density and thus the pressure.

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Boyle’s Law

• At a given temperature for a given quantity of gas, the product of the pressure and the volume is a constant

P1V1 P2V2

Elements that exist as gases at 250C and 1 atmosphere

Figure 1.7: The Pressure-Volume Relationship:Boyle’s Law

The Gas LawsThe Gas Laws

The Pressure-Volume Relationship: Boyle’s Law• Mathematically:

• A sample of gas contained in a flask with a volume of 1.53 L and kept at a pressure of 5.6×103 Pa. If the pressure is changed to 1.5×104 Pa at constant temperature, what will be the new volume? (in-class example).

The Gas LawsThe Gas Laws

PV

1constant constantPV

2211 VPVP

P 1/V

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

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Atmospheric Pressure

• Just the weight of the air above you.

• Unlike water, the density of the air decreases with altitude since air is compressible and liquids are only very slightly compressible.

• Air pressure at sea level is about 105 newtons/meter2.

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Barometers

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Buoyancy in a Gas

• An object surrounded by air is buoyed up by a force equal to the weight of the air displace.

• Exactly the same concept as buoyancy in water. Just substitute air for water in the statement.

• If the buoyant force is greater than the weight of the object, it will rise in the air.

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Buoyancy in a Gas

Since air gets less dense with altitude, the buoyant force decreases with altitude. So helium balloons don’t rise forever!!!

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Bernoulli’s Principle

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Bernoulli’s Principle.

• Flow is faster when the pipe is narrower.

• Put your thumb over the end of a garden hose.

• Energy conservation requires that the pressure be

lower in a gas that is moving faster.

• Has to do with the work necessary to compress a

gas (PV is energy, more later).

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• When the speed of a fluid increases, internal pressure in the fluid decreases.

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Why the streamlines are compressed is quite complicated and relates to the air boundary layer, friction and turbulence.

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Bernoulli’s Equation


A fluid flowing through a constricted pipe with streamline flow. The fluid in the section of length Δx1 moves to the section of length Δx2 . The volumes of fluid in the two sections are equal.

The speed of water spraying from the end of a hose increases as the size of the opening is decreased with thumb.

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• The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume has the same value at all points along a streamline.

• This result is summarized in Bernoulli’s equation:

• P + 1/2 ρ v2 + ρ gy = constant

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The Venturi tube

(a) Pressure P1 is greater Than the pressure P2 since v1 < v2. This device can be used to measure the speed of fluid flow.

Solution: Because the pipe is horizontal,

y1 = y2, and applying Bernoulli’s equation:

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The horizontal constricted pipe illustrated in figure known as a Venturi tube, can be used to measure the flow speed of an incompressible fluid. Let us determine the flow speed at point 2 if the pressure difference P1 ̶ P2 is known.

Figure 15: Venturi Tube

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For steady flow, the speed, pressure, and elevation of an incompressible and nonviscous fluid are related by an equation discovered by Daniel Bernoulli (1700–1782).

Figure 16

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In the steady flow of a nonviscous, incompressible fluid of density ρ, the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) are related by:

Figure 17:

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A fluid flowing through a constricted pipe with streamline flow. The fluid in the section of length Δx1 moves to the section of length Δx2 . The volumes of fluid in the two sections are equal.

The speed of water spraying from the end of a hose increases as the size of the opening is decreased with thumb.

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A fluid can also change its height. By looking at the work done as it moves, we find:

This is Bernoulli’s equation. One thing it tells us is that as the speed goes up, the pressure goes down.

Fluids at rest

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Stress on an object submerged in a static fluid

• The force exerted by a static fluid on an object is always perpendicular to the surfaces of the object.

• The force exerted by the fluid on the walls of the container is perpendicular to the walls at all points.

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Density

The density of a fluid is defined as its mass per unit volume. It is denoted by the Greek symbol, .

=V m3kgm-3

If the density is constant (most liquids), the flow is incompressible.

If the density varies significantly (e.g. some gas flows), the flow is compressible.

(Although gases are easy to compress, the flow may be treated as incompressible if there are no large pressure fluctuations)

water= 998 kgm-3

air =1.2kgm-3

kgm

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Pressure

Pressure is the force per unit area, where the force is perpendicular to the area.

p=A m2

Nm-2

(Pa)

NF

This is the Absolute pressure, the pressure compared to a vacuum.

pa= 105 Nm-2

1psi =6895Pa

The pressure measured in your tyres is the gauge pressure, p-pa. 1 Pa = 1 N/m2, in the SI system.

Atmosphere Pressure and the Barometer

Figure 1.2: PressureFigure 1.2: Pressure

A

FP

•Standard atmospheric pressure = 760 mm of Hg

•1 atm = 760 mmHg = 760 torr = 101.325 kPa.

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Barometer

Pressure = ForceArea

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm

Absolute Pressure

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Example: (a) What are the total force and the absolute pressure on the bottom of a swimming pool 22.0 m by 8.5 m whose uniform depth is 2.0 m? (b) What will be the pressure against the side of the pool near the bottom?

(a)The absolute pressure is given by P=P0+ρgh, and the total force is the absolute pressure times the area of the bottom of the pool.

5 2 3 3 2

0

5 2

5 2 7

1.013 10 N m 1.00 10 kg m 9.80 m s 2.0 m

1.21 10 N m

1.21 10 N m 22.0 m 8.5 m 2.3 10 N

P P gh

F PA

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(b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the bottom,

5 21.21 10 N mP

Uniform Pressure

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Uniform Pressure: If the pressure is the same at all points on a surface.

Properties of Fluids

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Density: SI Units: Kg/m3

Properties of fluids

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Blood as a fraction of body weight

• The body of a man whose weight is 690 N contains about 5.2 × 10 ̶̶̶̶ 3 m3 of blood.

• (a) Find the blood’s weight and?• (b) Express it as percentage of the body weight?

( in class example).

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A simple device for measuring the pressure exerted by a fluid

• The device consists of an evacuated cylinder that encloses a light piston connected to a spring.

• As the device is submerged in a fluid, the fluid presses on the top of the piston and compresses the spring until the inward force exerted by the fluid is balanced by the outward force exerted by the spring.

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Force and Pressure

• dF =P dA • Where P is the pressure

at the location of the area dA.

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Example1: The water bed

• The mattress of a water bed is 2m long by 2m wide and 30 cm deep.

• (a) Find the volume of the mattress?

• V = (2m)(2m)(0.3m) = 1.2 m3

• (b) Find the weight of the water in the mattress?

• M = V= (1000 kg/m3)(1.2m3) = 1.2 x 103 kg

• and its weight is: Mg = (1.2 x 103 kg)( 9.80m/s2) = 1.18x 104N

• (c) Find the pressure exerted by the water on the floor when the bed rests in its normal position. Assume that the entire lower surface of the bed makes contact with the floor?

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The water Bed

• When the bed is in its normal position, the area in contact with the floor is: 2 x 2 = 4 m2 ; thus we find that:

• (d) What if the water bed is replaced by a 50 kg ordinary bed that is supported by four legs? Each leg has a circular cross section of radius 2 cm. what pressure does this bed exert on the floor? HW

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Variation of Pressure with depth

Pressure in a fluid acts equally in all directions

Pressure in a static liquid increases linearly with depth

p=increase in depth (m)

pressure increase

g h

The pressure at a given depth in a continuous, static body of liquid is constant.

p1 p2

p3 p1 = p2 = p3

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Problem:

In a movie, Tarzan evades his captors by hiding underwater for many minutes while breathing through a long, thin reed. Assuming the maximum pressure difference his lungs can manage and still breathe is calculate the deepest he could have been.

Solution: The pressure difference on the lungs is the pressure change from the depth of water

2

3 3 2

133 N m85 mm-Hg

1 mm-Hg 1.154 m 1.2 m

1.00 10 kg m 9.80 m s

PP g h h

g

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A parcel of fluid (darker region) in a larger volume of fluid is

singled out. The net force exerted on the parcel of fluid must be zero because it is in equilibrium.

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Exercises: A Pain in your Ear• The difference between the total

pressure at the bottom of the pool and atmospheric pressure:

• Pbot – P0 = ρ gh

= (1000 kg / m3)(9.8 m/s2 )(5m)

= 4.9 × 104 Pa

The force on the eardrum:

F =( Pbot – P0 ) A = 5N

• Estimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of a pool is 5 m deep.

• g = 9.8 m/s2 (SI)

• ρwater = 1000 kg / m3

• Solution: We estimate the surface area of the eardrum to be approximately:

• 1cm2 = 1x10-4 m2. • The air inside the middle ear is

normally at atmospheric pressure P0.

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Example: A pain in the Ear.

Because a force on the eardrum of this magnitude is extremely uncomfortable, swimmers “pop their ears” while under water, an action that pushes air from the lungs into the middle ear. Using this technique equalizes the pressure on the two sides of the eardrum and relieves the discomfort.

F = ( Pbot – P0) A =( 4.9 × 10 4 pa) (1×10 ̶ 4 cm2) = 5 N

b) We estimate the surface area of the eardrum to be approximately:

1 cm2 = 1 × 10 ̶ 4 m2, This means that the force on it is:

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Ideal Gas Law• Gases are highly compressible in comparison to

liquids, with changes in gas density directly related to changes in pressure and temperature through the equation:– P = RT

• where– P is the absolute pressure, is the gas density– R is the gas constant– Ru is the universal gas constant, – T is the absolute temperature

• SI unit of the temperature is the Kelvin scale.

http://en.wikipedia.org/wiki/Pressure

http://en.wikipedia.org/wiki/Universal_gas_constant

http://en.wikipedia.org/wiki/Absolute_temperature

Density Form of Ideal Gas Law

P= RT

Ideal gas Equation

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T(K) = T(0C) + 273.15The gas constant R is different for each gas.

Many gases can be treated as ideal gases :•Air – nitrogen, oxygen, hydrogen, helium, argon, neon, krypton and carbon dioxide.

•Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators should not be treated as ideal gases.

Density, Specific Gravity, and Mass of Air in a room

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Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m × 5 m × 6 m at 100 kPa and 25 0C.Solution: The density, specific gravity, and mass of the air in a room are to be determined.Assumptions At specific conditions, air can be treated as an ideal gas.Properties The gas constant of air is R = 0.287 kPa. m3/ kg. K.Analysis The density of the air is determined from the ideal-gas relation: P= RT


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3

3

/17.1

)15.27325)(./.287.0(

100

mkg

KKkgmkPa

kPa

RT

P


84

00117.0/1000

/17.13

3

mkg

mkgSG

water

air

Then the specific gravity of the air becomes:


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Finally, the volume and the mass of the air in the room are:

Volume: V = (4 m) (5 m)(6 m) = 120 m3

mass: m = ρ V = (1.17 kg/m3)(120 m3) = 140 kg.

Discussion: Note that that we converted the temperature to the unit K from 0C before using it in the ideal-gas relation.

• Highly compressible.• Occupy the full volume of their containers.• When gas is subjected to pressure, its volume decreases.• Gases always form homogeneous mixtures with other

gases.• Gases only occupy about 0.1 % of the volume of their

containers.

General Characteristics of GasesGeneral Characteristics of Gases

Four Physical Quantities for GasesFour Physical Quantities for Gases

Phys. Qty.

Symbol SI unitOther common units

pressure PPascal (Pa)

atm, mm Hg, torr, psi

volume V m3 dm3, L, mL, cm3

temp. T K °C, °F

moles n mol

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Because the increase in pressure is the same on the two sides, a small force F1 at the left produces a much greater force F2 at the right.

Diagram of a hydraulic press

A vehicle undergoing repair is supported by a hydraulic lift in a garage.

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Pascal Law

Pascal ‘s Law: a change in the pressure applied to a fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

French Scientist Blaise Pascal ( 1623 – 1662)

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hydraulic press

• In the hydraulic press illustrated before:

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Hydrostatic Application: Transmission of Fluid Pressure

• Mechanical advantage can be gained with equality of pressures

• A small force applied at the small piston is used to develop a large force at the large piston.

• This is the principle between hydraulic jacks, lifts, presses, and hydraulic controls

11

22 F

A

AF

22

11 F

A

AF

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Example: the car lift

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Measuring pressure (1)Manometers

h

p1 p2=pa

liquiddensity

x y

z

p1 = px

px = py

pz= p2 = pa

(negligible pressure change in a gas)

(since they are at the same height)

py - pz = gh

p1 - pa = gh

So a manometer measures gauge pressure.

• Closed Systems => manometers

Figure 1.3: PressureFigure 1.3: Pressure

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Manometer, how it works?( السائل ( أو الغاز ضغط مقياس

• Manometer is an instrument used to measure the pressure of a gas or vapor. There are several types of manometers. The simplest kind consists of a U-shaped tube with both ends open. The tube contains a liquid, often mercury, which fills the bottom of the U and rises in each of the arms. The person using this type connects one of the arms to the gas whose pressure is to be measured. The other arm remains open to the atmosphere. In this way, the liquid is exposed to the pressure of the gas in one arm and atmospheric pressure in the other.

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Measuring Pressure (2)Barometers

A barometer is used to measure the pressure of the atmosphere. The simplest type of barometer consists of a column of fluid.

p1 = 0vacuum

h

p2 = pa

p2 - p1 = gh

pa = gh

examples

water: h = pa/g =105/(103*9.8) ~10m

mercury: h = pa/g =105/(13.4*103*9.8) ~800mm

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Measurement of Pressure: Barometers

The first mercury barometer was constructed in 1643-1644 by Torricelli.

He showed that the height of mercury in a column was 1/14 that of a water barometer, due to the fact that mercury is 14 times more dense that water.

He also noticed that level of mercury varied from day to day due to weather changes, and that at the top of the column there is a vacuum.

Evangelista Torricelli (1608-1647)

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Measuring Pressure• Two devices for measuring

pressure:

• a) an open-tube manometer and

• b) a mercury barometer.

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• An immersed body is buoyed up by a force equal to the weight of the fluid it displaces.

• If the buoyant force on an object is greater than the force of gravity acting on the object, the object will float.

• The apparent weight of an object in a liquid is gravitational force (weight) minus the buoyant force.

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Buoyancy

• Net upward force is called the buoyant force!!!

• Easier to lift a rock in water!!

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Displacement of Water

• The amount of water displaced is equal to the volume of the rock.

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Flotation

• A floating object displaces a weight of fluid equal to its own weight.

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Flotation

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Buoyant Forces and Archimedes’s Principle

• The upward force exerted by water on any immersed object is called the Buoyant force.

• We can determine the magnitude of a buoyant force by applying some logic and Newton’s second Law.

• Archimedes’s Principle States that the magnitude of the Buoyant Force always equals the weight of the Fluid displaced by the object.

• The buoyant force acting on the steel cube is the same as the buoyant force acting on a cube of liquid of the same dimensions.


The external forces acting on the cube of liquid are:

1- The force of gravity Fg and

2-The buoyant force FB.

Under equilibrium conditions:

FB = Fg

h: height of the cube


Buoyant Force:• FB = mg

• Or FB = Vρ g

• FB = mg = Vρ g

• ΔP = FB/A

• FB = (ΔP)A = (ρ g h)A

• FB = ρ gV

• V is the volume of the cube.

• mg is the weight of the fluid in the cube.

• The mass of the fluid in the cube is m= ρ V.

• The pressure difference ΔP between the bottom and top faces of the cube is equal to the buoyant force per unit area of those faces.

Example of a buoyant force.

•Hot-air balloons. Because hot air is less dense than cold air.Then:

•A net upward force acts on the balloons.

• (a) A totally submerged object that is less dense than the fluid in which it is submerged experiences a net upward force.

• (b) A totally submerged object that is denser than the fluid sinks.



Quiz: • A glass of Water contains a

single floating ice cube. When the ice melts, does the water level go up, go down, or remain the same?

• When a person in a rowboat in a small pond throws an anchor overboard, does the water level of the pond go up, go down, or remain the same?

Exercises: A Pain in your Ear• The difference between the total

pressure at the bottom of the pool and atmospheric pressure:

• Pbot – P0 = ρ gh

= (1000 kg / m3)(9.8 m/s2 )(5m)

= 4.9 x 104 Pa

The force on the eardrum:

F =( Pbot – P0 ) A = 5N

• Estimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of a pool is 5 m deep.

• g = 9.8 m/s2 (SI)

• ρwater = 1000 kg / m3

• Solution: We estimate the surface area of the eardrum to be approximately:

• 1cm2 = 1×10 ̶ 4 m2. • The air inside the middle ear is

normally at atmospheric pressure P0.

113

Example:AirWaterOilGasolineAlcoholKeroseneBenzeneGlycerine

Fluid Newton’s lawof viscosity

Newtonian fluids obey refer

Newton’s’ law of viscosity is given by;

dy

du (1.1)

• The viscosity is a function only of the condition of the fluid, particularly its temperature.

• The magnitude of the velocity gradient (du/dy) has no effect on the magnitude of .

= shear stress = viscosity of fluiddu/dy = shear rate, rate of strain or velocity gradient

Newtonian and Non-Newtonian Fluid

Velocity gradient

114

115

So:

Viscosity

117

119

Fluid Newton’s lawof viscosity

Non- Newtonianfluids

Do not obey

•The viscosity of the non-Newtonian fluid is dependent on the velocity gradient as well as the condition of the fluid.

Newtonian Fluids a linear relationship between shear stress and the velocity gradient

(rate of shear), the slope is constant the viscosity is constant

non-Newtonian fluids slope of the curves for non-Newtonian fluids varies

Newtonian and Non-Newtonian Fluid

98

Figure 1.1 Shear stress vs. velocity gradient

Bingham plastic : resist a small shear stress but flow easily under large shear stresses, e.g. sewage sludge, toothpaste, and jellies.Pseudo plastic : most non-Newtonian fluids fall under this group. Viscosity

decreases with increasing velocity gradient, e.g. colloidal substances like clay, milk, and cement.

Dilatants : viscosity decreases with increasing velocity gradient, e.g. quicksand.

121

ViscosityViscosity

Viscosity is the resistance that a fluid offers to flow when subject to a shear stress.

Isaac Newton: “the resistance which arises from the lack of slipperiness of the parts of a fluid, other things being equal, is proportional to the velocity with which the parts of the liquid are separated from each other”

122

Primary Units:

In fluid mechanics we are generally only interested in the top four units from this table.

SI. Units

Quantity SI Unit

Length Metre, m

Mass Kilogram, kg

Time Seconds, s

Temperature Kelvin, K

Current Ampere, A

Luminosity Candela

123

Derived Units

Quantity SI Unitvelocity m/s -

acceleration m/s2 -

force Newton (N) N = kg.m/s2

energy (or work) Joule (J) J = N.m = kg.m2/s2

power Watt (W) W = N.m/s = kg.m2/s3

pressure (or stress) Pascal (P) P = N/m2 = kg/m/s2

density kg/m3 -

specific weight N/m3 = kg/m2/s2 N/m3 = kg/m2/s2

relative density a ratio (no units) dimensionless

viscosity N.s/m2 N.s/m2 = kg/m/s

surface tension N/m N/m = kg/s2

124

Fluid Properties (Continue)

Specific weight:Specific weight of a fluid, • Definition: weight of the fluid per unit volume • Arising from the existence of a gravitational force • The relationship and g can be found using the

following:

Since = m/Vtherefore = g (1.3)

Units: N/m3

Typical values:Water = 9814 N/m3; Air = 12.07 N/m3

125

Specific gravityThe specific gravity (or relative density) can be defined in two ways:

Definition 1: A ratio of the density of a substance to the density of water at standard temperature (4C) and atmospheric pressure, or

Definition 2: A ratio of the specific weight of a substance to the specific weight of water at standard temperature (4C) and atmospheric pressure.

(1.4)

Unit: dimensionless.

Cw

s

Cw

sSG

44 @@


126

Viscosity:

• Viscosity, , is the property of a fluid, due to cohesion and interaction between molecules, which offers resistance to shear deformation.

• Different fluids deform at different rates under the same shear stress. The ease with which a fluid pours is an indication of its viscosity. Fluid with a high viscosity such as syrup deforms more slowly than fluid with a low viscosity such as water. The viscosity is also known as dynamic viscosity.

Units: N.s/m2 or kg/m/s

Typical values:Water = 1.14 ×10 ̶ 3 kg/m/s; Air = 1.78 ×10 ̶ 5 kg/m/s


127

Kinematic viscosity, Definition: is the ratio of the viscosity to the density;

• will be found to be important in cases in which significant viscous and gravitational forces exist.

Units: m2/s

Typical values: Water = 1.14x10-6 m2/s; Air = 1.46x10-5 m2/s;

In general, viscosity of liquids with temperature, whereas

viscosity of gases with in temperature.

/


128

Bulk Modulus All fluids are compressible under the application of an

external force and when the force is removed they expand back to their original volume.

The compressibility of a fluid is expressed by its bulk modulus of elasticity, K, which describes the variation of volume with change of pressure, i.e.

Thus, if the pressure intensity of a volume of fluid, , is increased by Δp and the volume is changed by Δ, then

Typical values:Water = 2.05x109 N/m2; Oil = 1.62x109 N/m2

strainvolumetric

pressureinchangeK

/

pK

p

K

129

Vapor Pressure A liquid in a closed container is subjected to a

partial vapor pressure in the space above the liquid due to the escaping molecules from the surface;

It reaches a stage of equilibrium when this pressure reaches saturated vapor pressure.

Since this depends upon molecular activity, which is a function of temperature, the vapor pressure of a fluid also depends on its temperature and increases with it.

If the pressure above a liquid reaches the vapor pressure of the liquid, boiling occurs; for example if the pressure is reduced sufficiently boiling may occur at room temperature.

130

•A reservoir of oil has a mass of 825 kg. The reservoir has a volume of 0.917 m3. Compute the density, specific weight, and specific gravity of the oil.

•Solution:

3/900917.0

825mkg

m

volume

massoil

3oil m/N882981.9x900g

mg

volume

weight

9.01000

900SG

C4@w

oiloil

Example 1.2

131

End of Chapter 1.

Chapter 2: Forces in Static FluidsThis section will study the forces acting on or generated by fluids at rest.Objectives:• Introduce the concept of pressure.• Prove it has a unique value at any particular elevation.• Show how it varies with depth according to the hydrostatic equation and• Show how pressure can be expressed in terms of head of fluid.This understanding of pressure will then be used to demonstrate methods of pressure measurement that will be useful later with fluid in motion and also to analyse the forces on submerges surface/structures.

Chapter 2: Forces in Static Fluids

133

•Define and apply the concepts of density and fluid pressure to solve physical problems. •Define and apply concepts of absolute, gauge, and atmospheric pressures.•State Pascal’s law and apply for input and output pressures.•State and apply Archimedes’ Principle to solve physical problems.

Objectives:

134

Mass Density

2 kg, 4000 cm3

Wood

177 cm3

45.2 kg

; mass m

Densityvolume V

Lead: 11,300 kg/mLead: 11,300 kg/m33

Wood: 500 kg/mWood: 500 kg/m33

4000 cm3

Lead

Same volume

2 kgLead

Same mass

135

Example 1: The density of steel is 7800 kg/m3. What is the volume of a 4-kg block of

steel?

4 kg3

4 kg;

7800 kg/m

m mV

V

V = 5.13 × 10 ̶ 4 m3V = 5.13 × 10 ̶ 4 m3

What is the mass if the volume is 0.046 m3?3 3(7800 kg/m )(0.046 m );m V

m = 359 kgm = 359 kg

136

Relative Density

The relative density rof a material is the ratio of its density to the density of water (1000 kg/m3).

Steel (7800 kg/m3) r = 7.80

Brass (8700 kg/m3) r = 8.70

Wood (500 kg/m3) r = 0.500

Steel (7800 kg/m3) r = 7.80

Brass (8700 kg/m3) r = 8.70

Wood (500 kg/m3) r = 0.500

Examples:Examples:

31000 kg/mx

r

137

Pressure

Pressure is the ratio of a force F to the area A over which it is applied:

Pressure ; Force F

PArea A

Pressure ; Force F

PArea A

A = 2 cm2

1.5 kg

2

-4 2

(1.5 kg)(9.8 m/s )

2 x 10 m

FP

A

P = 73,500 N/m2P = 73,500 N/m2

138

Pressure

Pressure is the ratio of a force F to the area A over which it is applied:

Pressure ; Force F

PArea A

Pressure ; Force F

PArea A

A = 2 cm2

1.5 kg

2

-4 2

(1.5 kg)(9.8 m/s )

2 x 10 m

FP

A

P = 73,500 N/m2P = 73,500 N/m2

139

Fluid PressureA liquid or gas cannot sustain a shearing stress - it is only restrained by a boundary. Thus, it will exert a force against and perpendicular to that boundary.

• The force F exerted by a fluid on the walls of its container always acts perpendicular to the walls. Water flow

shows F

140

Fluid PressureFluid exerts forces in many directions. Try to submerse a rubber ball in water to see that an upward force acts on the float.

• Fluids exert pressure in all directions.

F

141

Pressure vs. Depth in Fluid

Pressure = force/area

; ; mg

P m V V AhA

Vg AhgP

A A

h

mgArea

• Pressure at any point in a fluid is directly proportional to the density of the fluid and to the depth in the fluid.

P = gh

Fluid Pressure:

142

Independence of Shape and Area.

Water seeks its own level, indicating that fluid pressure is independent of area and shape of its container.

• At any depth h below the surface of the water in any column, the pressure P is the same. The shape and area are not factors.

• At any depth h below the surface of the water in any column, the pressure P is the same. The shape and area are not factors.

143

Properties of Fluid Pressure

• The forces exerted by a fluid on the walls of its container are always perpendicular.

• The fluid pressure is directly proportional to the depth of the fluid and to its density.

• At any particular depth, the fluid pressure is the same in all directions.

• Fluid pressure is independent of the shape or area of its container.

• The forces exerted by a fluid on the walls of its container are always perpendicular.

• The fluid pressure is directly proportional to the depth of the fluid and to its density.

• At any particular depth, the fluid pressure is the same in all directions.

• Fluid pressure is independent of the shape or area of its container.

144

Example 2. A diver is located 20 m below the surface of a lake (ρ = 1000 kg/m3). What is the pressure due to the water?

h = 1000 kg/m3

P = gh

The difference in pressure from the top of the lake to the diver is:

h = 20 m; g = 9.8 m/s2

3 2(1000 kg/m )(9.8 m/s )(20 m)P

P = 196000 Pa = 196 kPaP = 196000 Pa = 196 kPa

145

Atmospheric Pressure

atm atm h

Mercury

P = 0One way to measure atmospheric pressure is to fill a test tube with mercury, then invert it into a bowl of mercury.

Density of Hg = 13,600 kg/m3

Patm = gh h = 0.760 m

Patm = (13,600 kg/m3)(9.8 m/s2)(0.760 m)

Patm = 101,300 Pa = 1.013 × 105 PaPatm = 101,300 Pa = 1.013 × 105 Pa

146

Absolute Pressure

Absolute Pressure:Absolute Pressure: The sum of the pressure due to a fluid and the pressure due to atmosphere.

Gauge Pressure:Gauge Pressure: The difference between the absolute pressure and the pressure due to the atmosphere:

Absolute Pressure = Gauge Pressure + 1 atmAbsolute Pressure = Gauge Pressure + 1 atm

hP = 196 kPa

1 atm = 101.3 kPa

P = 196 kPa

1 atm = 101.3 kPa

Pabs = 196 kPa + 101.3 kPa

Pabs = 297 kPaPabs = 297 kPa

147

Pascal’s Law

Pascal’s Law: An external pressure applied to an enclosed fluid is transmitted uniformly throughout the volume of the liquid.

FoutFin AoutAinPressure in = Pressure outPressure in = Pressure out

in out

in out

F F

A A

148

Example 3. The smaller and larger pistons of a hydraulic press have diameters of 4 cm and 12 cm. What input force is required to lift a 4000 N weight with the output piston?

FoutFin AouttAin

; in out out inin

in out out

F F F AF

A A A

2

2

(4000 N)( )(2 cm)

(6 cm)inF

2; 2

DR Area R

F = 444 NF = 444 N

Rin= 2 cm; R = 6 cm

149


• An object that is completely or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.

2 kg

2 kg

The buoyant force is due to the displaced fluid.

The block material doesn’t matter.

150

Calculating Buoyant Force

FB = f gVf

Buoyant Force:

h1

mg

Area

h2

FB

The buoyant force FB is due to the difference of pressure P between the top and bottom surfaces of the submerged block.

2 1 2 1; ( )BB

FP P P F A P P

A

2 1 2 1( ) ( )B f fF A P P A gh gh

2 1 2 1( ) ( ); ( )B f fF g A h h V A h h

Vf is volume of fluid displaced.

151

Example 4: A 2-kg brass block is attached to a string and submerged underwater. Find the buoyant force FB and the tension in the rope T.

All forces are balanced:

mg

FB = gVT

Force diagram

FB + T = mg FB = wgVw

3

2 kg;

8700 kg/mb b

b bb b

m mV

V

Vb = Vw = 2.30 x 10-4 m3

Fb = (1000 kg/m3)(9.8 m/s2)(2.3 x 10-4 m3)

FB = 2.25 NFB = 2.25 N

152

Example 4 (Cont.): A 2-kg brass block is attached to a string and submerged underwater. Now find

the tension in the rope. (T=?)

mg

FB = gVT

Force diagram

FB + T = mg T = mg - FB

FB = 2.25 NFB = 2.25 N

T = (2 kg)(9.8 m/s2) - 2.25 N

T = 19.6 N - 2.25 N

T = 17.3 NT = 17.3 N

This force is sometimes referred to as the apparent weight.

153

Floating objects:

When an object floats, partially submerged, the buoyant force exactly balances the weight of the object.

FB

mg

FB = f gVf mx g = xVxg

f gVf = xVxg

f Vf = xVxf Vf = xVxFloating Objects:

If Vf is volume of displaced water Vwd, the relative density of an object x is given by:

Relative Density:x wd

rw x

V

V

154

Example 5: A student floats in a salt lake with one-third of his body above the surface. If the density of his body is 970 kg/m3, what is the density of the lake water?

1/3

2/3

Assume the student’s volume is 3 m3.

Vs = 3 m3; Vwd = 2 m3; s = 970 kg/m3

w Vwd = sVsw Vwd = sVs

3

w3

32 m;

3 m 2s wd s

w s

V

V

3

w

3 3(970 kg/m )

2 2s

w = 1460 kg/m3w = 1460 kg/m3

155

Problem Solving Strategy

1. Draw a figure. Identify givens and what is to be 1. Draw a figure. Identify givens and what is to be found. Use consistent units for P, V, A, and found. Use consistent units for P, V, A, and ..

2.2. Use absolute pressure PUse absolute pressure Pabsabs unless problem unless problem

involves a difference of pressure involves a difference of pressure P.P.

3.3. The difference in pressure The difference in pressure P is determined by P is determined by the density and depth of the fluid:the density and depth of the fluid:

2 1

m F; = ; P =

V AP P gh

156

Problem Strategy (Cont.)

4.4. Archimedes’ Principle: A submerged or floating Archimedes’ Principle: A submerged or floating object experiences an object experiences an buoyant force buoyant force equal to the equal to the weight of the displaced fluid: weight of the displaced fluid:

4.4. Archimedes’ Principle: A submerged or floating Archimedes’ Principle: A submerged or floating object experiences an object experiences an buoyant force buoyant force equal to the equal to the weight of the displaced fluid: weight of the displaced fluid:

B f f fF m g gV

5. Remember: m, ρ and V refer to the displaced fluid. The buoyant force has nothing to do with the mass or density of the object in the fluid. (If the object is completely submerged, then its volume is equal to that of the fluid displaced.)

157

Problem Strategy (Cont.)

6.6. For a floating object, FFor a floating object, FBB is is

equal to the weight of that equal to the weight of that object; i.e., the weight of the object; i.e., the weight of the object is equal to the weight object is equal to the weight of the displaced fluid: of the displaced fluid:

6.6. For a floating object, FFor a floating object, FBB is is

equal to the weight of that equal to the weight of that object; i.e., the weight of the object; i.e., the weight of the object is equal to the weight object is equal to the weight of the displaced fluid: of the displaced fluid:

x or x f x f fm g m g V V

FB

mg

158

Summary

; mass m

Densityvolume V

31000 kg/mx

r

Pressure ; Force F

PArea A

Pressure ; Force F

PArea A

21 Pa = 1 N/mPascal:

P = ghFluid Pressure:

159

Summary (Cont.)

FB = f gVf

Buoyant Force:Buoyant Force:Archimedes’ Archimedes’ Principle:Principle:

in out

in out

F F

A APascal’s Pascal’s

Law:Law:

2.1 Fluids Statics

The general rules of statics ( as applied in solid mechanics) apply to fluids at rest . From earlier we know that:• a static fluid can have no shearing force acting on it, and that• any force between the fluid and the boundary must be acting at right angles to the boundary.

Force normal to the boundary

Pressure Force normal to the boundary•This statement is also true for curved surfaces, in this case the force acting at any point is normal to the surface at that point.

•This statement is also true for any imaginary plane in a static fluid

•We use this fact in our analysis by considering elements of fluid bounded by imaginaryPlanes.

We also know that:

•For an element of fluid at rest, the element will be in equilibrium – the sum of the components of forces in any direction will be zero.• The sum of the moments of forces on the element about any point must also be zero.

2.2. Pressure

• If the force exerted on each unit area of a boundary is the same, the pressure is said to be uniform.

Example

Pressure and manometersexercise 1.1

Exercise 1.3

Exercise 1.4

Exercise 1.5

172

End of Chapter 2

1 introduction to fluid mechanics. chapter 1. phys 404 dr nazir mustapha

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