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BFC 10403 FLUID MECHANICS

BFC 10403 – NOOR ALIZA AHMAD

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CHAPTER 1.0: Principles of Fluid

1.1 Introduction to Fluid Mechanics

1.2 Thermodynamic Properties of a Fluid:

Density, specific weight, specific gravity ,

viscocity (kelikatan)berat tentu,

compressibility (kemampatan), Bulk

modulus (modulus pukal), dynamic &

kinematic viscosity (kelikatan dinamik dan

kinematik) ,surface tension

(ketegangan permukaan) and capillarity

(kererambutan).


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CHAPTER 1.0: Principles of Fluid


Mechanics is the oldest physical science that deals with

both stationary and moving bodies under the influence of

forces. The branch of mechanics that deals with bodies at

rest is called statics, while the branch that deals with

bodies in motion is called dynamics.

The subcategory fluid mechanics is

defined as the science that deals with the behavior of

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

and the interaction of fluids with solids

or other fluids at the boundaries.


4 BFC 10403 – NOOR ALIZA AHMAD


- Fluid engineering applications is enormous:

breathing, blood flow, swimming, pumps, fans,

turbines, airplanes, ships, rivers, windmills, pipes,

icebergs, engines, filters, jets, and sprinklers, to

name a few…..

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1.1 Introduction to Fluid Mechanics (Cont’d)

- From the point of view of fluid mechanics, all matter consists of only two states, fluid and solid. -Distinction between a solid and a fluid is made on the basis of the substance’s ability to resist an applied shear (or tangential) stress that tends to change its shape. - Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid. The fluid moves and deforms continuously as long as the shear stress is applied.


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Figure 1 illustrates a solid

block resting on a rigid

plane and stressed by its

own weight. The solid

sags into a static

deflection, shown as a

highly exaggerated

dashed line, resisting

shear without flow. Figure 1



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Figure 2

- The liquid and

gas at rest in

Figure 2 require

the supporting

walls to eliminate

shear stress.



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- The liquid retains its volume and forms a free

surface in the container.

- If the walls are removed, shear develops in the

liquid and a big splash results.

- If the container is tilted, shear again develops,

waves form, and the free surface seeks a horizontal

configuration, pouring out over the lip if necessary.



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1.2 Thermodynamic Properties of a Fluid:

a) Density, (ketumpatan)

- Density is highly variable in gases and

increases nearly proportionally to the pressure

level.

)/( 3mkgV

m


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1.2 Thermodynamic Properties of a Fluid: (Cont’d)

- Density in liquids is nearly constant; the density

of water (about 1000 kg/m3) increases only 1

percent if the pressure is increased by a factor of

220. Thus most liquid flows are treated

analytically as nearly “incompressible.”

- Compare their densities at 20°C and 1 atm:

Mercury: = 13,580 kg/m3

Hydrogen: = 0.0838 kg/m3


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1.2 Thermodynamic Properties of a Fluid( Cont’d):

The density of liquids and solids

depends more strongly on

temperature than it does on

pressure. At 1 atm, for example,

the density of water changes

from 998 kg/m3 at 20°C to 975

kg/m3 at 75°C, a change of 2.3

percent, which can still be

neglected in many engineering

analyses.

0C Density (kg/m3)

0

10

20

30

40

50

100

999.8

999.2

998.2

995.7

992.3

988.0

958.4

Nota : 1000 kg/m3 = 1.94 slugs/ft3


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b) Specific volume, (Isipadu tentu)

1

m

V

c) Relative density (ketumpatan relatif)

and is defined as the ratio of the density of a

substance to the density ofsome standard substance

at a specified temperature (usually water at 4°C,

for which H2O 1000 kg/m3).



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d) Specific gravity, SG (graviti tentu)

the density of a substance is

given relative to the density of a

well-known substance.

COatH

SG0

2 4

Substances SG

Water

Blood

Seawater

Gasoline

Mercury

Wood

Gold

Ice

Air

1

1.05

1.025

0.7

13.6

0.3-0.9

19.2

0.92

0.0013

e) specific weight, s (berat tentu)

s = g ( N/m3)



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Example 1:

Given Specific Gravity of Mercury is 13.55 (20 0C). Calculate mercury’s density.

Solution:

3/100055.13

mkg

merkuri

33 /106.13 mkgxmerkuri



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Example 2:

Calculate the gasoline’s density at 200C. The mass

and volume are 60 kg dan 0.5 m3 respectively

Solution:

)/( 3mkgV

m

3/1205.0

60mkg



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f) Compressibility, (kebolehmampatan) refers to the

change in volume (V) of a substance that is subjected

to a change in pressure on it.

Added

Volume

Bar

0

0.5

1.0

1.5

0

25

50

75

-the usual quantity used to measure this phenomenon is the bulk modulus of elasticity or simply bulk modulus, E



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g) Bulk Modulus, E (Modulas Pukal)

Pressure changes needed for changing the volume

Initial Volume

Volume changes Showing the pressure increment with volume reduction

E , harder to compress



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h) Viscosity, (Kelikatan)

is a property that represents the internal resistance

of a fluid to motion or the “fluidity,” and that property

is the viscosity.

The force a flowing fluid

exerts on a body in the

flow direction is called the

drag force, and the

magnitude of this force

depends, in part, on

viscosity



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To obtain a relation for viscosity, consider a fluid layer

between two very large parallel plates (or equivalently, two

parallel plates immersed in a large body of a fluid)

separated by a distance l .

Now a constant parallel force F

is applied to the upper plate

while the lower plate is held

fixed. After the initial transients, it

is observed that the upper plate

moves continuously under the

influence of this force at a

constant velocity V.



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The fluid in contact with the upper plate sticks to

the plate surface and moves with it at the same

velocity, and the shear stress acting on this fluid

layer is ( = F/A) where A is the contact area

between the plate and the fluid. Note that the fluid

layer deforms continuously under the influence of

shear stress.

Details : See Cengel( 2005). Fluid Mechanics. Mc Graw Hill



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Fluids for which the rate of deformation is proportional to

the shear stress are called Newtonian fluids

Water, air, gasoline, and oils (Newtonian fluids)

Blood and liquid plastics (non-Newtonian fluids)

In one-dimensional shear flow of Newtonian fluids, shear

stress can be expressed by the linear relationship

Shear stress:



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Example 3:

SAE 30 Oil at 200C of 0.005 in is placed in between two

layer. The bottom layer is fixed while upper layer moves

with acceleration 13 ft/s. Calculate shear stress for the

oil.

Solution:

= (9.20 x 10-3)[ 13/(0.005/2)]

= Ib/ ft2



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Example 4:

Benzene at 200C has a coefficient of viscocity,

0.000651 Pa.s. Calculate the shear stress to deform

this fluid at velocity gradient of 4900s-1?

Solution:

= 0.000651 x 4900 = 3.19Pa



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i) Dynamic Viscosity, (Kelikatan dinamik) defined as shear force per unit area

Units: Ns/m2, kgm-1s-1,Poise P

Typically

Water =1.14 kgm-1s-1, Air =1.78 kgm-1s-1,

Mercury =1.552 kgm-1s-1, Paraffin Oil=1.9 kgm-1s-1,



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Dynamic viscocities of

some fluids at 1 atm

and 200C (unless

otherwise stated)



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j) Kinematic Viscosity, Kelikatan

kinematik, the ratio of dynamic

viscosity to density

Two common units of kinematic

viscosity are m2/s and stoke

(1 stoke 1 cm2/s 0.0001 m2/s).

Dynamic viscosity, in general, does

not depend on pressure, but kinematic

viscosity does.



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Typically

Water =1.14 x 10-6 , m2/s Air =1.46 x 10-5 m2/s ,

Mercury =1.145 x 10-4 m2/s , Paraffin Oil =2.375 x 10-3 m2/s



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a drop of blood forms a hump on a

horizontal glass!

a drop of mercury forms a near-perfect

sphere and can be rolled just like a

steel ball over a smooth surface!

water droplets from rain or dew hang

from branches or leaves of trees!

k) Surface Tension,s (Ketegangan permukaan)



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b

Fs

2

In these and other observances, liquid

droplets behave like small spherical

balloons filled with the liquid, and the

surface of the liquid acts like a

stretched elastic membrane under

tension. The pulling force that causes

this tension acts parallel to the surface

and is due to the attractive forces

Between the molecules of the liquid. The

magnitude of this force per unit

length is called surface tension s and

is usually expressed in the unit N/m



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How surface tension arises??

Let see at microscopic view in

Figure . By considering two

liquid molecules, one at the

surface and one deep within the

liquid body. The attractive forces

applied on the interior

molecule by the surrounding

molecules balance each other

because of symmetry.



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But the attractive forces acting on the surface molecule

are not symmetric,and the attractive forces applied by

the gas molecules above are usually

very small.

Therefore, there is a net attractive force acting on the

molecule at the surface of the liquid, which tends to pull

the molecules on the surface toward the interior of the

liquid.

This force is balanced by the repulsive forces from the

molecules below the surface that are being

compressed.



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The resulting compression effect causes the liquid to

minimize its surface area. This is the reason for the

tendency of the liquid droplets to attain a spherical

shape, which has the minimum surface area for a

given volume.



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That why with amusement,

that some insects can land on

water or even walk on water

and that small steel needles

can float on water. These

phenomena are again made

possible by surface tension

that balances the weights of

these objects.



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l) Capillary effect (Kererambutan)

Another interesting consequence of

surface tension is the capillary

effect,which is the rise or fall of a liquid in

a small-diameter tube inserted into the

liquid.

This effect is usually expressed by saying that

water wets the glass (by sticking to it) while

mercury does not



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The contact angle for wetting and nonwetting fluids.

A liquid is said to

wet the surface

when < 90° and

not to wet the

surface when >

90°.



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The capillary rise of water and

the capillary fall of mercury in a

small-diameter glass tube.

The phenomenon of capillary effect

can be explained microscopically

by considering cohesive forces (the

forces between like molecules,

such as water and water) and

adhesive forces (the forces

between unlike molecules, such as

water and glass). The liquid

molecules at the solid–liquid

interface are subjected to both

cohesive forces by other liquid

molecules and adhesive forces by

the molecules of the solid.



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The magnitude of the capillary

rise in a circular tube can be

determined from a force

balance on the cylindrical liquid

column of height h in the tube

(see figure)

The bottom of the liquid column

is at the same level as the free

surface of the reservoir, and

thus the pressure there must be

atmospheric pressure.



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This balances the atmospheric pressure acting at

the top surface, and thus these two effects cancel

each other. The weight of the liquid column

is approximately

Equating the vertical component of the surface

tension force to the weight gives



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cos

2

gRh s

Solving for h gives the capillary rise to be

R = constant

Example 5:

A 0.6-mm-diameter glass tube is inserted into

water at 20°C in a cup. Determine the capillary

rise of water in the tube



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Solution:

With assumption,

1 There are no impurities

in the water and no

contamination on the

surfaces of the glass tube.

2 The experiment is

conducted in atmospheric

air.



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h = 2(0.073 N/m)

1000 kg/m3( 9.81m/s2)(0.3 x 10-3)

1 kg . m/s2

1 N

= 0.050 m = 5.0cm

(cos 00)



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