Fluid Engineering Mechanics - جامعة نزوى · Chapter 2 Fluid Properties: Density, specific...

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Chapter 2

Fluid Properties: Density, specific volume, specific weight,

specific gravity, compressibility, viscosity, measurement of

viscosity, Newton's equation of viscosity, Surface tension,

capillarity and pressure

Dr. Muhammad Ashraf Javid

Assistant Professor

Department of Civil and Environmental Engineering

1

Fluid Engineering Mechanics

Physical Properties of Fluids

2

Density

Specific Volume

Specific Weight

Specific Gravity

Compressibility

Viscosity

Surface Tension

Pressure

Buoyancy

Density

3

It is also termed as specific mass or mass density.

It is the mass of substance per unit volume .i.e., mass of fluid per

unit volume.

It is designated with symbol of ρ (rho)

ρ =mass/volume

=M/L3

Fundamental Units=kg/m3, slug/m3, g/cm3

4

2

3

2

L

FT

L

M

L

FT

a

FMMaF

Note: Density of water at 4oc=1000kg/m3, 1.938slug/ft3, 1g/cm3

Specific Volume

4

It is defined as volume of substance per unit mass.

It is designated with υ.

MLmassvolume // 3

MaF

2

4

FT

L

Fundamental Units=m3/kg, m3/slug, cm3/g

Relationship Between Density and Specific Volume

5

3

3

//

//

LMvolumemass

MLmassvolume

/1

/1

Specific Weight

6

It is the weight of substance per unit volume or say it is

the weight of fluid per unit volume.

It is designated by γ (gamma).

3L

W

volume

weight

3L

Mg MgW

2223 TL

M

TL

ML 2T

Lg

Note: Specific weight of water at 4oc=9810N/m3, 62.4lb/ft3, 981dyne/cm3

Relation Between and

7

volume

weight

volume

Mass &

g

MgW

Effect of Temperature and Pressure on

Specific Weight

8

As the equation of state for a

perfect gas is given by

Where

P =absolute pressure

υ =specific volume

T =absolute temperature

R =gas constant

For perfect gases

mR=8312 N-m/(kg-k)

Where

m=molecular weight of gas

RTP

1RT

P

RT

g

P

/ g

RT

gP

T

Pconstant)(

R

gconstant,

TP

1&

RTP

Effect of Temperature and Pressure on

Specific Weight

9

Since

Assuming constant

pressure

Assuming constant

temperature

V

W

volume

weight

nT

1

nP

1,0n

1,0n

Specific Gravity (Relative Density)

10

It is the ratio of density of a substance and density of water at 4oC.

It is the ratio of specific weight of substance and specific weight of

water at 4oC.

It is the ratio of weight of substance and weight of an equal volume

of water at 4oC.

water

fluid

water

fluid

water

fluid

W

WS

Remember: T

P1

&

CTatW

CTatW

CTat

CTat

CTat

CTatS

o

water

o

fluid

o

water

o

fluid

o

water

o

fluid

Note: Specific gravity of liquid is measured w.r.t. water while for

cases of gases it is measured w.r.t. standard gas (i.e. air)

Class Problems

11

12

Compressibility

Compressible fluids

Incompressible fluids

In fluid mechanics we deal with both compressible and

incompressible fluids of either variable or constant density.

Although there is no such thing in reality as incompressible fluid, we

use this terms where the change in density with pressure is so small

as to be negligible. This is usually the case with liquids.

Ordinarily, we consider the liquids as incompressible.

We may consider the gases to be incompressible when the pressure

variation is small compared with absolute pressure.

Compressibility

14

Compressible fluids

Fluids which can be compressed.

Fluid in which there is a change in volume with change in pressure

P1

P2

v1

v2

12 PP

12 vv

As a result of change in volume, density and specific weight of fluid also changes. Hence, for compressible fluids,

21

21

21

vv

Compressibility

15

Incompressible fluids

Fluids which can not be compressed.

Fluid in which there is no change in volume with change in pressure

P1 P2

v1

v2

12 PP

12 vv

As a result, no-change in volume, density and specific weight of fluid. Hence, for incompressible fluids,

21

21

21

vv

v2

Compressibility (Volumetric Strain)

16

Volumetric Strain is the ratio of change in volume and original volume.

P1

P2

v1

v2

12 PP

12 vv

11

21

1

21

21

1

21

/

//

v

dv

v

vv

Mv

MvMv

Volumetric strain=change in specific volume/original specific volume.

1

21

1

21

d

Compressibility

17

Bulk Modulus or Volume Modulus of Elasticity (Ev):

It is defined as ratio of volumetric stress to volumetric strain

Ev= volumetric stress/volumetric strain

Ev=change in pressure/compressibility

1v

dv

dpEv

1

d

dpEv

Viscosity

18

The viscosity of a fluid is a measure of its resistance to shear or angular

deformation.

It is the property of a fluid by mixture of which it offers resistance

to deformation under the influence of shear forces. It depends

upon the cohesion and molecular momentum exchange between

fluid layers.

It can also be defined as internal resistance offered by fluid to flow.

It is denoted by μ.

It is also termed as coefficient of viscosity or absolute viscosity or

dynamic viscosity or molecular viscosity.

Factor affecting viscosity

19

1. Cohesion

2. Molecular momentum

1. Cohesion: It is the attraction between molecules of fluid. More

the molecular attraction (cohesion) more is the viscosity (resistance

to flow) of fluid.

It is dominant in liquids.

2. Molecular momentum: Molecules in any fluid change their

position with time and is known as molecular activity. More the

molecular activity more will be viscosity of the fluid.

It is dominant in gases

A B

Effect of temperature on viscosity

20

For Liquids:

In case of liquids, cohesion (molecular

attraction is dominant). Therefore, if the

temperature of liquid is increased, its

cohesion and hence viscosity will

decrease.

For Gases:

In gases momentum exchange is

dominant. Therefore, if the temperature

of gases is increases, its momentum

exchange will increase and hence

viscosity will increase.

T

1

T

Kinematic Viscosity

21

It is ratio of absolute viscosity and density of fluid.

It is denoted by (nu)

Newton’s Equation of Viscosity

22

Consider two parallel plates, in which lower plate is fixed and upper is

moving with uniform velocity ‘U’ under the influence of force ‘F’. Space

between the plates is filled with a fluid having viscosity, μ.

F= Applied force (shearing force)

A= Contact area of plate(resisting area)

Y=gap/space between plates

U= Velocity of plate

As the upper plate moves, fluid also moves in the direction of applied force

due to adhesion.

Y

u

dy

du

U Moving plate

Fixed plate

Force, F

Newton’s Equation of Viscosity

23

Factors affecting Force, F

Hence,

Where, μ is coefficient of viscosity

Assuming linear velocity profile (as shown in figure)

YFiiiUFiiAFi

1)(;)(;)(

Y

AUF

Y

AUF

dy

du

Y

U

A

F

Newton’s Equation of Viscosity

24

At boundaries the particles of fluid adhere to wall and so their velocities are zero relative to wall. This so called non-slip condition occurs in viscous fluids

Newton’s Equation of Viscosity

25

The above equation is called as Newton’s equation of viscosity.

The equation shows that the shearing stress is directly proportional

to the velocity gradient.

In the above equation

du/dy= velocity gradient or rate of change of deformation or shear

rate

μ = absolute viscosity

τ=shear stress

dy

du

Dimensional Analysis of Viscosity

26

Viscosity

This expression is used to write

fundamental unit of viscosity

Kinematic Viscosity

2

2 )/(

L

FT

TLL

FL

U

Y

A

F

T

L

L

MLT

M2

3

LT

M

MLTFL

TMLT

2

2

2

Unit of Viscosity

27

Viscosity

Widely used unit is Poise =0.1N.s/m2

Kinematic Viscosity

Widely used unit is Stoke=10-4m2/s

LTM /

TL /2

SI BG CGS

N-s/m2 Lb-s/ft2 Dyne-s/cm2

(Poise, P)

Kg/(m-s) Slug/(ft-s) g/(cm-s)

SI BG CGS

m2/s ft2/s cm2/s

(stoke)

Problem

28

smxx

/1088.5850

105 263

Problem

29

A flat plate 200mm x 750mm slide on oil (μ =0.85N.s/m2) over a

large surface as shown in fig. What force, F, is required to drag the

plate at a velocity u of 1.2m/s if the thickness of the separating oil

film is 0.6mm?

NF

AY

UF

AY

UA

dy

duAF

dy

du

Y

U

A

F

255

7.02.01000/6.0

2.185.0

Here, t = Y

Problem

30

A space 16mm wide between two large plane surfaces is filled with

SAE 30 Western lubricating oil at 35oC(Fig). What force is required

to drag a very thin plate of 0.4m2 area between the surfaces at a

speed u=0.25m/s (a) if this plate is equally spaced between the two

surfaces? (b) if t=5mm?

Solution:

Y=16mm A=0.4m2 u=0.25m/s

T=35oC μ =0.18N.s/m2 (from figure A.1)

F=? If Y=8mm

(a)

8mm dy

du

Y

U

A

F

2121 AAFFF

Solution to Problem

31

8mm

N

Ay

uA

y

uF

AAFFF

5.4

4.01000/8

25.018.04.0

1000/8

25.018.0

21

21

2121

Solution to problem

32

(b): F=? If t =5mm

y1=11, y2=5mm

N

Ay

uA

y

uF

AAFFF

24.54.01000/11

25.018.04.0

1000/5

25.018.0

21

21

2121

5mm

33

Shear Stress ~ Velocity gradient curve

34

Ideal fluid

Newtownian Fluid

Non-Newtownian fluid

Ideal plastic

Real solid

Ideal solid/elastic solid

Real solid

Shear Stress ~ Velocity gradient curve

35

Ideal Fluid: The fluid which does not offer resistance to flow

Newtownian Fluid: Fluid which obey Newtown’s law of viscosity

slope of curve ( )is constant

Non-Newtonian fluid: Fluid which does not obey Newtown’s Law

of viscosity

slope of curve ( )changing continuously

dy

du

00

dy

du

dydu /~

dydu /~

Shear Stress ~ Velocity gradient curve

36

Ideal Solid: solid which can never be deformed under the action of

force

Real solid: solid which can be deformed under action of forces

Ideal Plastic: These are substances which offer resistance to shear

forces without deformation upon a certain extent but if the load is

further increased then they deform

Real Plastic: These are substances in which there is deformation

with the application of force and it increases with increase in applied

load.

0dy

du

Problem

37

38

Exercise Problems

39

Measurement of Viscosity

40

The following devices are used for the measurement of viscosity

1. Tube type viscometer

2. Rotational type viscometer

3. Falling sphere type viscometer

Falling Sphere Viscometer

41

It consists of a tall transparent

tube or cylinder and a sphere of

known diameter.

The sphere is dropped inside the

tube containing liquid and time of

fall of sphere between two points

(say A and B) is recorded to

estimate the fall velocity (s/t)of

sphere inside liquid.

Where ‘s’= distance between

point A and B and ‘t’ is the time

of travel.

From this velocity of fall, viscosity

is estimated from the expression

of fall sphere type viscometer.

s

A

B

Fig. Falling Tube type viscometer

W

FB

FD

Dt

D

Falling Sphere Viscometer

42

D=Ds=Diameter of sphere

Dt=Diameter of tube or cylinder

Vt=velocity of sphere in tube (s/t)

s=Distance between points A and B

t=time taken by sphere to cover

distance (s)

W=weight of sphere=γ*(Vol)

= γs(πD3/6)

FB=Force of Buoyancy

= γL(πD3/6)

FD=Drag force

= (3πμVD)

Note: V is not equal to Vt

s

A

B

Fig. Falling Tube type viscometer

W

FB

FD

Dt

D

Stoke’s Law

Falling Sphere Viscometer

43

Buoyancy: It is the resultant upward thrust exerted by the fluid on a sphere. It is the tendency of fluid to lift the body and it is equal to weight of volume of fluid displaced by the body (Archimedes Principal).

Drag Force: It is a resisting force generated by the liquid on the moving object which is acting in the opposite direction of movement .

Vt=velocity of sphere in tube with wall effect

V=velocity of sphere in tube without wall effect

V>Vt

...4

9

4

91

2

ttt D

D

D

D

V

V

3

1

tD

Dif

63

6

;0;0

33 DVD

D

WFFF

SL

DBy

Falling Sphere Viscometer

44

The above equation is governing equation for falling sphere type

viscometer.

For a particular temperature, D, γs and γL are constant. So we can write ;

Thus, velocity of fall is inversely proportional to viscosity and is indicative of viscosity in falling sphere type viscometer.

Note: This method can only be used for transparent liquids

LS

LSLS

V

D

DDV

DDVD

18

663

663

2

2233

V

1

Problem: 11.1.10

45

Problem: 11.1.10

46

Tube Type Viscometer

47

In tube type viscometer, liquid is

placed in a container to a certain

level. Valve in the bottom is opened

to fill the flask of known volume.

Time taken to fill the flask is

recorded which gives measure of

viscosity of liquid

H’=Average imposed head causing

flow=H+L-h/2

VL=volume of flask

D=Diameter of tube

L=length of tube

h=fall of liquid level in container to

fill liquid in flask

H’

h H

V

Tube Type Viscometer

48

Lets consider two points 1 and 2

and apply energy equation.

Where, are potential,

pressure, and velocity head and

HL=head loss

The head loss in tube type

viscometer is due to friction loss in

tube and is represented by

H’

h H

V

1

2 Datum

LHg

VPz

g

VPz

22

2

222

2

111

LHg

VH

20000'

2

2

g

VPz

2&,

2

VD

LHH FL 2

32

(ii)

(i)

Tube Type Viscometer

49

Eq (ii) is a Hagen Poiseulli law for laminar flows in tube.

Moreover, for laminar flow in tube viscometer V<< 1 therefore V2 can

be neglected. Hence Eq (i) becomes as

VD

LH

VD

LH

2

2

32'

3200000'

VD

LHH FL 2

32

L

HDQ

A

Q

D

LH

128

'

32'

4

2

2

4DA

AVQ

QL

HD

128

'4

Equation for tube

type viscometer

A=Cross-sectional area

of tube and V is average

velocity

Tube Type Viscometer

50

Where

kt = constant of tube type

viscometer

QL

HD

128

'4

t

V

time

VolumeQ L

tLV

HgD

L

128

'4

hdVL

2

4

t

tKt

Note: Equation of tube type viscometer is applicable for laminar

flows. For flows in pipes, flow will be laminar if Re≤2000 and flow

will be turbulent if Re>4000

2000

VDReno Reynolds

Re=2000-4000 Transition flow

Problem; 11.1.7

51

Solution:

D=0.0420in= .0420/12 ft

L=3.1in=3.1/12ft

VL=60mL=0.00212 ft3

H’=(10+9.5)/2=9.75 in

t=128.7sec

tLV

HgD

L

128

'4

sft /0000227.0/ 2

=H’

Problem

52

Rotational Type Viscometer

53

It consists of two concentric

cylinders. Small cylinder is

placed inside the bigger cylinder.

The gap (space) between

cylinders is filled with liquid up

to a height, h,.

Then either inner or outer

cylinder is fixed and other is

rotated by applying a constant

torque.

Revolution per minute (RPM) is

measured which is indicative of

viscosity.

h

r1

r2

Δr

Rotational Type Viscometer

54

r1=outer radius of inner cylinder

r2=inner radius of outer cylinder

r=mean radius=(r1+r2)/2

Δr=gap (space) between cylinders

h=height of liquid

F=shearing force

T=Applied torque (mean torque)=F x r

h

r1

r2

Δr

ArT

AF

Ardy

duT

dy

du

r

uAr

r

uT

AF

(1)

Rotational Type Viscometer

h

r1

r2

Δr

Where

N=RPM(Revolution per minute)

ω=angular frequency

hrA 2area Resisting

60

2

VelocityAngular

Nru

ru

radian/min 60/2/minrevolutionN

radian/s 2/srevolutionN

radian/s 2/srevolution 1

radian 2revolution 1

N

N

Rotational Type Viscometer

56

Now substituting the values of

A and u in Eq. (1)

Where

Kr=Rotational viscometer

constant

NKT

Nr

hrT

rrhr

rNT

r

15

260

2

32

r

hrKr

15

32

By re-arranging the formula, the

absolute viscosity using

rotational type viscometer can

be obtained as

For a particular viscometer,

both T and Kr are constant and

therefore

Nk

T

r

1

N

1

Problem: 11.1.6

57

Solution:

Rotation type viscometer

h=300mm=0.3m

OD of inner cylinder=100mm=0.1m

ID of outer cylinder=102mm=0.102m

Δr=(102-100)/2=1mm=0.001m

Torque=T= 8 N-m

N=1/4 rev/s=60/4 RPM

Neglect mechanical friction

h=300mm

r1=50mm

r2=51mm Δr

Problem: 11.1.6

58

r

hrK

NK

TNKT

r

r

r

15

32

2/4.20 mNs

Surface Tension

59

The tension force created at the imaginary

thin surface due to unbalanced-molecular

attraction is termed as surface tension.

v2 A

B

Molecule A in figure above is situated at a certain depth below the

surface. It is acted upon by equal force from all sides whereas

molecule B (situated at the surface) is acted upon by unbalanced

forces from below.

Thus a tight skin/film/surface is formed at the surface due to inward

molecular pull.

Types of molecular attraction

60

Cohesion: It is the attraction force between the molecules of same

material

Adhesion: It is the attraction force between the molecules of

different materials

Surface tension depends upon the relative magnitude of cohesion

and adhesion but primarily it depend upon the cohesion.

With the increase in temperature cohesion reduces and hence

surface tension also reduces.

Concept of surface tension is used in capillarity action

Capillarity

61

It is the rise or fall of a liquid in a small diameter (< 0.5”) tube due

to surface tension and adhesion between liquid and solid.

For capillary action diameter of tube is less than 0.5inch while for

large diameter tubes this phenomenon become negligible.

The curved surface that develops in tube is called meniscus

v2

h

σ θ

D

v2

h

σ

θ

D

Water Mercury θ<90 θ>90

Capillarity

62

D= diameter of tube

γ=specific weight of liquid

h=capillary rise/fall

θ=angle of contact or contact angle

σ=force of surface tension per unit length v2

h

σ θ

D

Derivation of expression for capillary rise/fall

Let’s take

Weight of column of liquid acting downward=

Vertical component of force of surface tension=

0Fy

hDvol 2

4

cosD

Capillarity

63

Equating both equations

The above equation is used to compute capillary rise/fall.

Note: Fall has –ve sign

hDD 2

4cos

Dhor

hD

hDD

cos4

cos4

4cos 2

Problem. 2.29

64

Solution:

At 50oF

With θ=0o

True static height=6.78-1.174in=5.61in

inft

Dh

174.10979.0

12/04.041.62

00509.04cos4

ftlbftlb /00509.0,/41.62 3

Surface Tension of water

65

Temperature

- t -

(oF)

Surface Tension of Water in

contact with Air

- σ -

(10-3 lb/ft)

32 5.18

40 5.13

50 5.09

60 5.03

70 4.97

80 4.91

90 4.86

100 4.79

120 4.67

140 4.53

160 4.40

180 4.26

200 4.12

212 4.04

Vapor Pressure of liquids

66

Vapor Pressure: It is the pressure at which liquid transforms into

vapors or it is the pressure exerted by vapors of liquid.

All the liquids have tendency to

release their molecules in the space

above their surface.

If the liquid in container have limited

space above it, then the surface is filled

with the vapors.

These vapors when released from liquid exert pressure known as

vapor pressure.

It is the function of temperature. More the temperature more will

be vapor pressure.

Vapor Pressure of liquids

67

Saturated vapor pressure: It is the vapor pressure that

corresponds to the dynamic equilibrium conditions (saturation) i.e.,

when rate of evaporation becomes equal to rate of condensation.

Boiling vapor pressure: The pressure at which vapor pressure

becomes equal to atmospheric pressure.

Effect of Temperature and Pressure on Ev

of Water

68

Pressure, dp Temp, T

Ev Ev

50oC

Ev(max)

Temperature is constant Pressure is constant

Relation Between Ev and Compressibility

69

As

1v

dv

dpEv

1

1

v

dvEv

ilitycompressibEv

1

Problem

70

1

12

1

dp

d

dpEv

/1

/1

g

Solution

71

Sample MCQs

72

1.Specific gravity of a liquid is equal to

(a). Ratio of mass density of water to mass density of liquid

(b). inverse of mass density

(c). Ratio of specific weight of liquid to specific weight of water

(d). None of all

2. What happens to the viscosity of a liquid when its temperature is raised?

(a). The viscosity of the liquid increases

(b). The viscosity of the liquid stays the same

(c). The viscosity of the liquid decreases

(d).The temperature of a liquid does not rise

3. What happens to the specific weight of a liquid when its temperature is raised?

(a). It increases

(b). It stays the same

(c). It decreases

(d). The temperature of a liquid does not rise

Sample MCQs

73

4. In a falling sphere viscometer, according to force balance we have

(a). Weight=Drag+Buoyancy

(b). Drag=Weight - Buoyancy

(c). Buoyancy=Weight +Drag

(d). None of all

5. The kinematic viscosity is

(a). Multiplication of dynamic viscosity and density

(b). division of dynamic viscosity by density

(c). Multiplication of dynamic viscosity and pressure

(d). None of the above

6. As a result of capillary action (a) liquid rise in capillary tube

(b) liquid falls down in capillary tube

© Both (a) and (b)

(d) None of all