BASIC FLUID MECHANICS (ECW 211)

JULIANA BINTI MARTIN

BKBA 3.13

013-9809070

EXT: 2574

CONTENT

PROGRAM OUTCOMES

COURSE SYLLABUS

COURSE OUTCOMES

LESSON PLAN

COURSE ASSESSMENT

PROGRAME OUTCOMES PO1 - Ability to acquire and apply basic knowledge of science,

mathematics and engineering.

PO2 - Ability to communicate effectively with technical personal

and the public.

PO3 - Ability to identify, formulate and solve engineering

problems.

PO4 - Ability to function on multi-disciplinary teams.

PO5 - Ability to function effectively as an individual and in a

group with leadership, managerial and entrepreneurial

capabilities.

PO6 - Understanding of the social, cultural, global and

environmental responsibilities and ethics for sustainable

development.

PO7 - Recognising the need to undertake lifelong learning and

possessing/acquiring the capacity to do so.

COURSE OUTCOMES

CO1 - Apply basic knowledge on various fluid properties

and problems related to fluid mechanics.

CO2 - Apply concept of hydrostatic pressure in determining

forces exerted by fluids on plane surfaces under static

condition.

CO3 - Apply concept of up thrust, buoyancy of objects

immersed in fluids in determining the stability of

floating bodies.

CO4 - Apply concepts and application of the continuity, energy

and momentum equations and flow measurement in fluid

mechanics.

COURSE ASSESMENT

GRADING %

TEST 1 30%

SOFT SKILLS 10%

FINAL EXAMINATION 60%

TOTAL 100%

COURSE ASSESMENT

ASSESSMENT CHAPTER GRADING ROOM

QUIZ

Quiz 1 Quiz 2

Chapter 1 & 2 Chapter 3

2% 2%

DURING LECTURE

CLASS (ANYTIME)

ASSIGNMENT Chapter 4 & 5 6%

DURING LECTURE

CLASS-LAST WEEK

REMARKS NO RE-QUIZ, ZERO MARK WILL BE GIVEN FOR STUDENT NOT ATTEND THE CLASS WITHOUT MC/LETTER. NEED TO SUBMIT AT THE END OF THE CLASS. ASSIGNMENT IS COMPULSARY-WILL BE GIVEN AT THE LAST WEEK (1 hr LECTURE) BEFORE STUDY WEEK.

COURSE ASSESMENT

QUESTION CHAPTER

5 Questions

Question 1 Question 2 Question 3 Question 4 Question 5

Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5

REMARKS DONT WRITE USING PENCIL, PLEASE WRITE USING PEN PLEASE WRITE USING BALL POINT PEN RATHER THAN GEL PEN BRINGS YOUR CALCULATOR DURING FINAL EXAMINATION SESSION STANDARD MARKING ( BEWARE) WRITE ALL UNITS/FORMULA.

CHAPTER ONE

1.1 FLUID AS CONTINUUM

1.2 UNITS AND DIMENSION USED IN

ENGINEERING FLUIDS

At the end of this topic student should:

Be able to explain the continuum concept of fluid. (CO1-PO1) Be able to identify the units and dimension used in engineering fluids.(CO1-PO3)

WHAT IS HYDRAULICS

WHAT IS

FLUID MECHANICS

?

Mechanics of fluids Its that branch of engineering science which deals with the

behaviour of fluid under the

conditions of rest & motion

Greek word HUDAR , means WATER Its that branch of engineering science deals with water ( at rest or in motion) Or its that branch of engineering science which is based on

experimental observation of

water flow.

INTRODUCTION

FLUID MECHANICS

FLUID MECHANICS is a study of the behavior

of liquids and gases either at rest (fluid

statics) or in motion (fluid dynamics).

The analysis is relate continuity of

mass and energy with force and

momentum.

FLUID is a substance which deforms continuously under the action of shearing

force (however small it is may be)

IMPORTANT OF FLUID

MECHANICS TO

ENGINEER

To determine the

stability of floating and

submerged objects

pontoons, ships

To determine the

hydrostatic forces

dams

To determine flow and

energy losses in pipe

To design fluid

machines pumps

and turbines

To determine flow rate,

energy dissipation from

spillway and flow in

open channels such as

rivers

IMPORTANT OF FLUID MECHANICS

DIFFERENCE BETWEEN SOLID AND FLUID

Have preferred shape

Hard & not easily deformed

Cannot deformed continuously under shear force

SOLID Does not have any preferred shape

Soft & easily deformed

Deformed continuously under shear force

FLUID

3 CONDITIONS OF FLUIDS

The study of incompressible fluid under static conditions (hydrostatics)

That dealing with the compressible static gases- aerostatics

STATICS

Deals with the velocities, accelerations and pattern of flow only

Force and energy causing velocities and accelerations are not deal under this head.

KINEMATICS

Deal with the relationship between velocities and accelerations of fluid with the FORCES @ ENERGY causing them.

DYNAMICS

CONCEPT OF FLUID

In FLUID:

-The molecules can move freely but are constrained through a traction force called cohesion.

-This force is interchangeable from one molecule to another.

For GASES:

-It is very weak which enables the gas to disintegrate and move away from its container.

-A gas is a fluid that is easily compressed and expands to fill its container.

-It fills any vessel in which it is contained. There is thus no free surface.

For LIQUIDS:

-It is stronger which is sufficient enough to hold the molecule together and can withstand high compression, which is suitable for application as hydraulic fluid such as oil.

-On the surface, the cohesion forms a resultant force directed into the liquid region and the combination of cohesion forces between adjacent molecules from a tensioned membrane known as free surface.

1.1 FLUID AS CONTINUUM

Continuum mechanics and its concept

It is a branch of mechanics that deals with the analysis of the kinematics and mechanical behaviour of materials modelled as a continuum. (eg. solids and fluids), (eg. liquids and gases)

A continuum concept assumes that the substance of the body is distributed uniformly throughout, and completely fills the space it occupies.

Fluid properties is depends on their molecular structure.However, engineering applications hardly analyses fluids at molecular level.

It is the fluids bulk behavior of main concern in engineering applications.

CONTINUUM CONCEPTS

Atoms are widely spaced in the

gas phase.

However, we can disregard the

atomic nature of a substance.

View it as a continuous,

homogeneous matter with no

holes, that is, a continuum.

This allows us to treat properties

as smoothly varying quantities.

Continuum is valid as long as size

of the system is large in

comparison to distance between

molecules.

Fluid as a continuum

A continuous substance where quantities such as velocity and pressure can be taken as constant at any section irrespective of the individual fluid particle velocity.

PRESSURE

Pressure acts perpendicular to the

surface and increases

at greater depth.

area

forcepressure

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

A measure of the amount of force exerted on a surface area

1.2 UNITS AND DIMENSION USED IN

ENGINEERING FLUIDS

WHAT IS

UNITS?

WHAT IS

DIMENSION

?

Standardized system of measurements used to

describe the magnitude of

the dimension

A properties that can be measured

Measurable properties used to describe a body/system

The standard element, in terms of which these dimensions can be

described quantitatively & assigned

numerical values.

VARIOUS SYSTEM OF UNIT

Parameter SI UNITS c.g.s system of unit

Imperial units ( British

Gravitational system; English

Units)

Length Meters (m) Centimeters (cm) Foot (ft)

Mass kilogram(kg) Gramme (g) Pound ( Ib)

Time Seconds (s) Seconds (s) Seconds (s)

Temperatur

e

Degree Celcius

(oC)

Degree Fahrenheit ( oF)

The primary quantities which are also referred to as basic dimensions, such as L for length, T for time, M for mass and F for force.

Student also expected to be familiar with the various systems of units used in engineering. These systems include :

As any quantity can be expressed in whatever way you like it is sometimes easy to become

confused as to what exactly or how much is being referred to. This is particularly true in the

field of fluid mechanics.

DERIVED UNIT

1.3 DENSITY, RELATIVE DENSITY

SPECIFIC WEIGHT, SPECIFIC GRAVITY,

SPECIFIC VOLUME AND VISCOSITY

At the end of this topic student should: Be able to apply basic knowledge of various fluid properties.(CO1-PO1) Be able to acquire various fluid properties in identify and solving problems related to fluid engineering problem.(CO1-PO3) Be able to formulate the relationship between shear, stress and velocity gradient from the Newtons law of viscosity. (CO1-PO3)

1. DENSITY

Regardless of form (solid, liquid,

gas) we can define how much mass

is squeezed into a particular space

Density of a material is defined by

the amount of matter per unit

volume.

Density of material may be referred

to in many ways.

1.1 MASS DENSITY,

Definition

Density of a fluid, , is defined as the mass per unit volume

It is denoted by the Greek symbol, .

== V m3 kgm-3

kg

m

water= 1000 kgm-3

air =1.23 kgm-3

1.2 SPECIFIC WEIGHT,

Definition Specific weight of a fluid, , is defined as the weight of the fluid per unit volume . Force exerted by gravity, g, upon unit volume of substance

= w V

= g

Units: N/m3

= the density of the material (kgm-3)

g = acceleration due to gravity (ms-2)

Water = 9.81 X 103 N/m3

1.3 RELATIVE DENSITY

@ SPECIFIC GRAVITY, SG

Definition A ratio of the mass density of a substance to the mass density of water at standard temperature (4 C) and atmospheric pressure.

Units: dimensionless

Cw

s

Cw

sSG

4@4@

Unit is none, since ratio is a pure number. SG is a dimensionless quantity

2. SPECIFIC VOLUME, V

Definition The reciprocal of the mass density i.e. the volume per unit mass or the inverse of density

Units: m3/kg

v = 1/ = V/m

Dynamic

Kinematic

3. VISCOSITY

3.1 DYNAMIC VISCOSITY,

Definition Dynamic viscosity, , is defined as the Shear force per unit area (shear stress, ) needed to drag a layer of fluid with a unit velocity past

another layer at a unit distance away from it in the fluid

Measure of internal friction of fluid particles Molecular cohesiveness Resistance fluid has to shear (or flow)

Units:

Water:

Air:

3.2 KINEMATIC VISCOSITY,

Definition It defined as the ratio of dynamic viscosity to mass density

v

= dynamic viscosity = mass density

Will be found to be important in cases in which significant viscous and

gravitational forces exist.

Typical values:

Water = 1.14x10-6 m2/s;

Air = 1.46x10-5 m2/s;

Units: m2/s or stokes (10,000 St = 1m2s-1)

NEWTON LAW OF VISCOSITY

When fluid moves, it generates shearing stress

If no movement between the moving fluid particles no shear stresses developed

Fluid particles which in contact with solid boundaries will adhere to these boundaries will have same velocities as the solid boundaries

Movement of a fluid over solid boundary can be visualized as layers of a fluid moving one above the other.

The velocity of fluid layers increases as the distance from the solid boundary increases

y

v

Flowing passing over a solid boundary

TEMPERATURE VS VISCOSITY

(LIQUID AND GASES)

Liquids

Gases

Viscosity

Temperature

Viscosity is caused by the cohesive forces between the molecules in liquids

and by the molecular collisions in

gases, ant it varies greatly with

temperature.

The viscosity of liquid decreases with temperature, whereas the viscosity of

gases increases with temperature.

This is because in a liquid the molecules possess more energy at

higher temperature and they can

oppose the large cohesive

intermolecular forces more strongly.

As a result, the energized liquid molecules can move more freely.

In gases, the intermolecular activities are negligible and the gas molecules at

high temperature move randomly at

higher velocity.

Viscosity in gases

Due to intermolecular collision between randomly moving particles

For gas, temperature , amount of intermolecular collision , viscosity

Viscosity in liquid

Due to intermolecular collision between liquid particles

For liquid, temperature , intermolecular collision is weakened, viscosity

VISCOSITY IN GASES & LIQUIDS

NEWTON LAW OF VISCOSITY

It is important to evaluate the magnitude of the shear stress generated by the moving fluid

Newtons Law of viscosity:

= shear stress

= viscosity of fluid

du/dy = shear rate, rate of strain

or velocity gradient

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 .

NEWTONIAN &

NON NEWTONIAN FLUID

Fluid Newtons law of viscosity

Newtonian fluids obey refer

Example: Air, Water, Oil, Gasoline, Alcohol, Kerosene, Benzene, Glycerine

Fluid Newtons law of viscosity

Non Newtonian fluids not obey refer

NON NEWTONIAN FLUID

*The slope of a curve at a point is the apparent viscosity of the fluid at that point

EXAMPLE 1

1. The lower plate as shown below is fixed while the upper

one is free to move under the action of a mass of 50g.

Castor oil with absolute viscosity 650 x 10-3 Ns/m2

occupies the space between these two plates. The area

of contact of the upper plate with the oil is 0.7m2, find the

velocity of the upper plate when the distance separating

the plates is 0.5cm.

Hint:

Answer: du = 5.4mm/s

A

F

dy

du = 650 x 10-3 Ns/m2 y =

0.5cm

m=50g

pulley

Stationary

2. A vertical gap 25mm wide of infinite extent contains oil

of relative density 0.95 and viscosity 2.4Pa.s. A metal

plate 1.5m x 1.5m x 1.6mm, weighing 55N is to be lifted

through the gap at a constant speed of 0.06 m/s.

Determine the force required.

Hint:

Answer: F = 110.4 N

0.06m/s

dy dy

F

W

25 mm

A

F

dy

du

EXAMPLE 2

EXAMPLE 3

3. Crude oil at 20 C fills the space between two concentric

cylinders of diameters 150mm and 156mm respectively.

Both cylinders are 250mm in height. If the inner cylinder is

to be rotated at a constant speed of 12 rev/min while

keeping the outer cylinder stationary, calculate the torque

required. The fluid properties of the crude oil at 20 C are:

i) specific gravity = 0.86

ii)kinematic viscosity = 8.35 x 10-6 m2/s

Hint: Linear velocity,

Answer: T = 0.002Nm

rFT

A

F

dy

du

rv

EXAMPLE 4

4. A vertical cylinder of diameter 180mm rotates

concentrically inside another cylinder of diameter

181.2mm. Both the cylinders are 300mm high. The

space between the cylinders is filled with a liquid whose

viscosity is unknown. Determine the viscosity of the fluid

if torque of 20 Nm is required to rotate the inner cylinder

at 120 rpm.

Answer : =0.696 Ns/m2

5. 145 mm radius inner cylinder is placed in stationary of

outer cylinder with 150mm radius. Both cylinders are

250mm long. The inner rotates at an angular velocity of

1 revolution per second (rps). Torque of 0.75Nm is

required to maintain this velocity. Determine the

viscosity of the liquid that fills the space between the

cylinder.

Answer: = 0.120 Ns/m2

EXAMPLE 5

1.4 COMPRESSIBILITY AND BULK MODULUS,

VAPOUR PRESSURE, SURFACE TENSION,

AND CAPILLARITY

At the end of this topic student should: Be able to define the fluid parameters.(CO1-PO1) Be able to apply bulk modulus, surface tension and capillarity in solving fluid engineering problem.(CO1-PO1) Be able to use the Newtons law of viscosity which are the relationship of shear stress and velocity gradient in solving fluid engineering problems (CO1-PO3)

4. SURFACE TENSION,

Surface tension

defined as the force acting a unit length of a line drawn in the liquid surface

Surface tension

Surface tension tend to reduce the surface area of a body of liquid

The internal pressure within the droplet, p and the surface tension forces, must be in equilibrium.

p

Surface tension

Taking vertical equilibrium of the forces acting on the droplet

The magnitude of surface tension forces are very small compared to other forces

Normally are neglected

22 rpr

rp

2

2

prUnits : N/m

5. VAPOR PRESSURE, Pv

Vapor pressure

defined as the pressure at which a liquid turns to vapour

the pressure exerted by its vapor in phase equilibrium with its liquid at a given temperature

The molecules which moves above the surface of the liquid exert pressure in the confined surface

Vapor pressure

Pvapour = P saturation

Units: N/m2 or Pascal

6. CAPILLARITY

When a liquid comes into contact with a solid surface:

- Adhesion forces: forces between solid and liquid

- Cohesion forces: forces within liquid

If cohesive forces > adhesive forces, the meniscus in a glass tube will take a shape as in figure (a) and (b).

Figure (a) and (b)

Capillary effect is

the rise or fall of a

liquid in a small-

diameter tube

gdh

cos4

grh

cos2

dh

cos4@ @

Units= m @ mm

where h = height of capillary rise (or depression)

= surface tension

= wetting (contact) angle

= specific weight of liquid

r = radius of tube

7. COMPRESSIBILITY &

BULK MODULUS Definition The change of pressure corresponding to frictional change in volume of fluid where temperature remains constant

Gases are much more compressible compared to liquids

Liquids are considered incompressible

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.

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

strainvolumetric

pressureinchangeK

/

pK

VdV

pK

/

Units: N/m2

6. 1 When the pressure exerted on a liquid is

increased from 550 kN/m2 to 1000 kN/m2, the

volume is decreased by 1%. Determine the bulk

modulus of the liquid.

Answer: K = 45x106 N/m2

EXAMPLE 6

6.2 Water at 20 C has a bulk modulus of 21.8 x

108 N/m2. Find the increase in pressure that is

required to decrease its volume 1%.

Answer: dP = of 21.8 x 106 N/m2

50

6.3 Determine the bulk modulus of a liquid if it

undergoes a 0.1% decrease in volume when

subjected to a pressure change from 100kPa to

6.5Mpa.

Answer: 6.4GPa

51

7. The pressure at a depth of 4.5km in the ocean is

50 MN/m2. The density of sea water at the surface

is 1040 kg/m3 and its average bulk modulus is 2.4

x 103 MN/m2. Calculate the:

7.1 Change in specific volume

Answer: -20.03x10-6 m3/kg

7.2 Specific volume at 4.5km depth

Answer: 941.51x10-6 m3/kg

7.3 Specific weight at 4.5 km depth

Answer: 10.4 kN/m3

EXAMPLE 7

for your attention