Download - Properties of Fluids-A

7/31/2019 Properties of Fluids-A

1/41

NATURE OF ENGINEERING MECHANICS

Interaction

INTRODUCTION

Bodies nRest/MotioForces:

Tension

Compression

Shear

Solids:

Rigid

Deformable

Fluids:Liquids

Gases

- Velocity

- Acceleration

- Time

SIMILARITY IN THE BRANCHES OF MECHANICS

a. Basic Principles and Fundamental Concepts

b. Newtons Laws of Motion

c. Principle of Work and Energy

d. Principle of Impulse and Momentum


2/41

UNITS OF MEASUREMENTS

System Mass ( M ) Length ( L ) Time ( T ) Force ( F )

SI Kilogram ( kg ) Meter ( m ) Second ( s ) Newton ( N ) *

English Slug * Foot ( ft ) Second ( s ) Pound ( lb )

Metric Kilogram Meter ( m ) Second ( s ) Kilogram (kgF)*

*Derived Units from Newtons Second Law of Motion

a. Standard Acceleration of Gravity ( g ):

SI: g = 9.81 m/sec2 English: g = 32.2 ft/sec2

b. Mass (M) and Weight (W): W = Mg

c. Conversion Factors:

1 kgF = 2.2 lb 1 kgF = 9.81 N 1 lb = 4.448 N

MaF 8.9/MakgF


3/41

d. Common Prefixes of SI Units:

SI Prefix Abbreviation MultipleGiga G 109

Mega M 106

Kilo K 103

Centi c 10-2

Milli m 10-3


4/41

CLASSIFICATION OF FLUIDS

a. Compressibility:

- incompressible

- compressible

b. Viscosity:

- Non-viscous

- Viscous

- Newtonian

- Non-Newtonian

c. Ideal Fluid:

- Incompressible and non-viscous


5/41

FUNDAMENTAL CONCEPTS

Mechanics of Fluids and Hydraulics, these aresciences that deals with the laws of liquids and gases at restor in motion and of their practical applications. Fluidmechanics is based on the same fundamental principlesthat are employed in mechanics of solids. Hydraulics

utilized experimental techniques to developed empiricalformulas which supplied answers to practical problems.

DEFINITION OF A FLUID

Deforms continuously when subjected to shear stress


6/41

Three Branches of Fluid Mechanics

1. Fluid Statics, is the study of the mechanics offluids at rest.

2. Kinematics, deals with velocities and streamlineswithout considering forces or energy involved.

3. Fluid Dynamics, is concerned with the relationsbetween velocities and accelerations and theforces exerted by or upon fluids in motion.


7/41

1.1 DISTINCTIONS BETWEEN A GAS AND A LIQUID

GAS

1. Molecules are very far apart.

2. Very compressible.3. Expand indefinitely when external pressure is removed.

4. Can be held in equilibrium only when it is completely enclosed.

5. No free surface

LIQUID

1. Molecules are little closer.

2. Relatively incompressible.

3. Cohesion between molecules holds them together except of its

own vapor and does not expand indefinitely.

4. Can be held in equilibrium both in an open and close space.

5. May have a free surface (i.e a surface from which all pressure is

removed, except that of its own vapor) .


8/41

A. FLUID STATICS

PART 1: PROPERTIES OF FLUIDS

1. Specific weight, y: the force exerted by gravity on a unit volume of fluid, or

Units:English System:

Metric System:

S.I. :

Note: This relationship is true for liquids where varies slightlyregardless of temperature and pressure. For gases, is

computed from Charles and Boyles laws which give

English:

R- gas constant

Tabsolute temperature, ( 460 + F)

V

W

3ft

lb

3cm

gr3

m

tons

3m

N

RT

p

or

where p - absolute pressure

R)lbf/lbmol(1545ft


9/41

S.I.

T=K, K =273 + C

Alternative unit of R: J/kg-K; Btu/lb-R; ft-lbf/lb-R

2. Specific Volume,

The reciprocal of the density, or

English:

RT

p

2m

Np

KkgmNR

:

1

slugs

ft3

gr

cm3

kg

m3:Metric :.IS

K)m/kmol(8314N

R)lbmol(1.986Btu/


10/41

3. Mass Density,

The mass per unit volume of the fluid, or

Units: English,

Metric,

S.I.,

:)(rho

VM

gVW

g

3

ft

slugs

3cm

gr

3

m

kg

The density of water at 4 C (39.2 F ) is 1.00 g/cc ( or 1.00 g/mL ), equivalent

to 1000 kg/cu. m.


11/41

4. Specific Gravity; s: the dimensionless ratio of the fluid specific weight (or density) to

the standard reference fluid, water (for liquids),and air (for gases):

Note: At 39.2 F (4 C) the specific weight of water is approximately

or

5. Dynamic (Absolute) Viscosity, (mu): the property of a fluid which determines the

amount of its resistance to a shearing stress.

Note: The shearing stress, T (Tau), and the viscosity, (Mu) are related by,

Units of :

English System: S.I:

;

water

liquid

water

liquid

liquids

34.62ft

lbw ;9810 3

m

N

dx

dv

,,

22

mm

N

ft

lb

,dx

dv

2

sec

ft

lb 2m

sN

2)(

cm

sdynePpoise

:Metric

;

air

gas

air

gas

gass

3205.1 m

kgair

ml

g

cm

g

m

kgw 0.10.11000 33

s

mkg

is the velocity gradient

2

1.0m

sN At 20 C, = 1 cP

= 0.01P =mN-s/m2


12/41

Newtons Law of Viscosity

Fluid between parallel plates:

a

F

y

b b

Velocity Distribution:

Assumptions/Observations:

a. Closely spaced plates, y is very small.

b. Lower plate is fixed.

c. Force F exerted on upper plate with area A imparting a velocity u to the plate.

d. No slip at the boundaries, velocity of fluid same as boundary.

e. Line ab moves to the line ab after time interval.

f. Velocity distribution is linear.

g. Based on experimental observations: The applied force is proportional to thearea and velocity gradient. is the constant of proportionality.

h. Differential form ( Newtons Law of Viscosity):

y

AuF

y

u or

dy

du


13/41

Temp, C Specificweight

, N/m3

Density ,kg/ m

Viscosity ,

kg / (m-s)

10-3 =

KinematicViscosity, m2/s10-6 v =

SurfaceTension

, N / m

100 =

VaporPressure

head

p / , m

Bulk modulusof elasticity

K, N..m2

10-2 K =

0 9805 999.9 1.792 1.792 7.62 0.06 204

5 9806 1000.0 1.519 1.519 7.54 0.09 206

10 9803 999.7 1.308 1.308 7.48 0.12 211

15 9798 999.1 1.140 1.141 7.41 0.17 214

20 9789 998.2 1.005 1.007 7.36 0.25 220

25 9779 997.1 0.894 0.897 7.26 0.33 222

30 9767 995.7 0.801 0.804 7.18 0.44 223

35 9752 994.1 0.723 0.727 7.10 0.58 22440 9737 992.2 0.656 0.661 7.01 0.76 227

45 9720 990.2 0.599 0.605 6.92 0.98 229

,50 9697 988.1 0.549 0.556 6.82 1.26 230

'55 9679 985.7 0.506 0.513 6.74 1.61 231

60 9658 983.2 0.469 0.477 6.68 2.03 228

65 9635 980.6 0.436 0.444 6.58 2.56 226

70 9600 977.8 0.406 0.415 6.50 3.20 225

75 9589 974.9 0.380 0.390 6.40 3.96 223

80 9557 971.8 0.357 0.367 6.30 4.86 221

85 9529 968.6 0.336 0.347 6.20 5.93 217

90 9499 965.3 0.317 0.328 6.12 7.18 216

95 9469 961.9 0.299 0.311 6.02 8.62 211

100 9438 958.4 0.284 0.290 5.94 10.33 207

PHYSICAL PROPERTIES TABLE OF WATER IN SI UNITS


14/41

Illustrative Examples:

a) The specific weight of water at ordinary pressure and temperature is

. The specific gravity of mercury is 13.55. Compute the

density of water and the specific weight and density of mercury.

Solution.

./81.93

mkN

gw

w

2

3

81.9

81.9

s

mm

kN

31000m

kg

Note : kgm

sN11

2

3313381.955.13

m

kN

m

kNs watermercurymercury

33550,13100055.13

m

kg

m

kgs watermercutymercury

2

3

81.9

100081.9

s

mm

Nx


15/41

b) A gallon of water weighs 4.08 lb. Compute the following:

1) Mass in slugs

2) Mass in kg

3) Volume in cu. ft.

Solution.

1) Mass in slugs:

2) Mass in kg:

3) Volume in cu. ft. :

g

WM

22.32

08.4

s

ft

lbM )(1267.0

2

slugsft

slb

281.9

448.408.4

s

mlb

Nlb

M

)(85.1

2

kgm

sN

gal

ftgalxV

48.7

11

3

3134.0 ft

A i i h 16 N/ i d


16/41

c) A certain gas weighs 16 N/cu. m. at a certain temperature and

pressure. What are the values of its density, specific volume, and

specific gravity relative to air weighing 12 N/cu. m. ?

Solution.

The density of the gas is,

The specific volume is,

The specific gravity is,

g

2

3

81.9

16

s

mm

N

3

631.1m

kg

1

3631.1

1

m

kg

kg

m3

613.0

a

s

3

3

12

16

m

Nm

N

333.1

)


17/41

d) The specific gravity of glycerin is 1.26. Compute its density and

specific weight.

Solution.

The density is,

The specific weight is,

water

glycerins

33 26.100.126.1 cm

g

cm

gglycerin

or 33 1260100026.1 m

kg

m

kg

gglyceringlycerin

2381.91260

s

m

m

kg

3360,12

m

N or

3360.12

m

kN

( ) h f h d


18/41

6. Kinematic Viscosity, v (nu): the ratio of the dynamic viscosity to its

mass density, or

Units of :

English System: S.I:

Metric System:7. Adhesion: the property of a liquid that enables it to stick to another

body.

8. Cohesion: the property of a liquid that resists tensile stress.

9. Surface Tension, (sigma): the effect of cohesion between particlesof the liquid at its surface.

Unit: English System, S.I,

sec

2ft)(

sec

2

ststokecm

sec

2cm

stoke

ft

lb

m

N

10 C ill i hi i d h i d dh i If dh i h10 C ill i hi i d h i d dh i If dh i h


19/41

10. Capillarity: this is due to cohesion and adhesion. If adhesion has a

greater effect than cohesion, the liquid will rise at the point where it

comes in contact with another body. If cohesion is prevalent, the

liquid will depressed.

10. Capillarity: this is due to cohesion and adhesion. If adhesion has a

greater effect than cohesion, the liquid will rise at the point where it

comes in contact with another body. If cohesion is prevalent, the

liquid will depressed. The capillary rise in a tube can be expressed,

where

grrh

cos2cos2

= surface tension in units offorce per unit length = wetting angle = specific weight of liquid

r = radius of tubeh = capillary rise = density of liquidg = gravitational acceleration

hr

1111 V P h d b h


20/41

11.11. Vapor Pressure, : the pressure exerted by the vapor as

evaporation of the liquid takes place within a closed space.

Units: English System,

S.I.,

Metric System,

vp

2ft

lb

or 2in

lb

2m

N

2cmgr


21/41

Property Definition Dimension

(SI Units)

Common Values

For Water ( 4 C) *

, Mass Density mass per unit volume ML-3

(kg/m3)

1,000 kg/m3

, Specific Weight weight per unit

volume

FL-3

(N/m3)

9,800 N/m3

S, Specific gravity /water (for liquid) 0 1

p, Pressure Normal force per unit

area

FL-2

(N/m2)

, Absolute Viscosity Newtons law of

viscosity

FL-2T

(N s/m

2

)

1.52 x 10-3 N s/m2

, Kinematic Viscosity / L2T-1

(m2/s)

1.52 x 10-6 m2/s

* Note: The values of fluid properties vary with temperature

FLUID PROPERTIES MOST COMMONLY CONSIDERED

CCOMPRESSIBILITY OF LIQUID


22/41

CCOMPRESSIBILITY OF LIQUID

The compressibility ( change in volume due to change in pressure ) of

a liquid is inversely proportional to its volume modulus of elasticity,

also known as the bulk modulus. This is defined as

where

Rearranging the definition of as an approximation, we may used

for the case of a fixed mass of liquid at constant temperature,

or

dpdv

v

dv

dpvE

v

v = specific volumep = pressure

,vE

vEp

vE

pp12

1

12

Coefficient of Compressibility

= coefficient of compressibility

vE

1

E l 1 R d 1 ti i t th E li h t If thi i th


23/41

Example 1. Reduce 1 centipoise to the English system. If this is the

dynamic viscosity of water at 68 F (20C), what is the kinematic

viscosity in the English system?

Solution.

1 centipoise

1 lb = 444,800 dynes

a) The dynamic viscosity in English system

b) The Kinematic viscosity in the English system

scmg

cmsdynepoise

01.001.001.0

2

poise01.02

21254.2

800,444101.0

ftin

incm

dyneslb

cmsdynes

2

510088.2

ft

slbx

g where )(32.62 3 waterftlb at F68

sec

10079.1

32.62

2.3210088.22

5

5ft

xx

E l 2 Th ki ti i it it f 1 ft2/ i i l t t h


24/41

Example 2. The kinematic viscosity unit of 1 ft2/sec is equivalent to how

many stokes?

Solution:

Example 3.A volume of 15.5 cu. ft of a certain fluid weighs 782 Ib.

Compute the mass density.Solution:

22

1254.2sec

1

ftin

incmft

sec929

2

cm

V

M

gV

W

32

5.15

sec

2.32

782

ftft

lb

3567.1

ft

slugs

or stokes929

Example 4 A barangay of 8 000 population has an average water


25/41

Example 4. A barangay of 8,000 population has an average water

consumption per person per day of 110 gallons. Compute the average

total rate of consumption in cubic feet per sec. and in gallons per

minute. (1 cu. ft. = 7.48 gallons)

Solution:

a) Average Total Consumption: 8,000 persons x 110

b) Average Total Consumption:

dayperson

gallons

day

galx 41088

hrday

hr

ft

gal

day

galx

ATCsec3600

2448.7

1088

3

4

sec362.1

3ft

hrday

hr

day

galx

ATC

1

min6024

10884

mingal

min111.611

gal

Example 5 Compute the unit weight of dry air at 15 C and an absolute


26/41

Example 5. Compute the unit weight of dry air at 15 C and an absolute

pressure of 14.7 psi (pounds per square inch). Also solve for the

kinematic viscosity.

Solution.

As determined by Bearden, the dynamic viscosity of air at 20 C is

0.0001819 poise and changes at that temperature at a rate of

0.0000005 poise /C

a)

where:

RT

p

FR

ft

ft

in

in

lb

594603.53

1447.14

2

2

2

30765.0

ft

lb

R

ftR

3.53 air

FC 5915


27/41

b)

The Kinematic viscosity is,

at 0000005.050001819.015 C

)(0001794.0 Ppoise or2

sec

cm

dyne

22 54.212

800,444

sec0001794.0

in

cm

ft

in

lb

dynescm

dyne

27 sec1074.3

ftlbx

g

3

22

7

0765.0

sec2.32

sec1074.3

ft

lb

ft

ft

lbx

sec1058.1

24 ftx

Example 6 At a depth of 8 km in the ocean the pressure is 81 8 MPa


28/41

Example 6. At a depth of 8 km in the ocean the pressure is 81.8 MPa.

Assume specific weight at the surface is 10.05 kN/cu.m and that the

average volume modulus is for the pressure

range.(a) What will be the change in specific volume between that at the

surface and at that depth?

(b) What will be the specific volume at that depth?

(c ) What will be the specific weight at that depth?

Solution.

(a) Change in specific volume,

2

91034.2

m

Nx

vEp

vv

vEpv

2

9

2

6

3

3

2

1034.2

108.81

1005.10

81.9

m

N

x

m

Nx

m

N

x

s

m

v

kg

mxv

35

10412.3

( b ) Specific volume at that depth of 8 km


29/41

( b ) Specific volume at that depth of 8 km

kg

mxvvv

35

12 10412.3

kgmxv

3

5

1

210412.31

kg

m

x

g

v

35

12

10412.3

kg

mx

kg

Nx

s

m

v3

5

3

2

2 101412.3

1005.10

81.9

kg

mxv

34

2 10420.9


30/41

(c) Specific weight at that depth 8 km

g22

gv

2

2

1

23

4

281.9

10420.9

1

sm

kg

mx

3414,10

m

N

Example 7 The radius of the tube as shown in the figure is 1 mm


31/41

Example 7. The radius of the tube as shown in the figure is 1 mm.

The surface tension of water at 20 C is equal to 0.0728 N/m . For a

waterglass = 0

Solution.

( a) Capillary rise in the tube in mm

h

2r

grh

cos2

001.081.91000

0cos0728.02 .0148.0 m mm8.14

( b ) Total force due to surface tension


32/41

( b ) Total force due to surface tension

( c ) Weight of water

))(cos( dF

002.0)0cos0728.0( F

NxF4

1057.4

VW mmmN 0148.0001.098102

3

Nx 41057.4


33/41

Example 8. Calculate the density, specific weight and specific volume ofchlorine gas at 25 C and under a pressure of 600 kPa absolute. The gasconstant R for chlorine is 117 N m/ kg - K.

Solution:

1. The density of chlorine gas is

2. The specific weight of chlorine is

3. The specific volume of chlorine gas is

RT

p

27325117

106003

x

3

209.17m

kg

g 81.9209.173

820.168

m

N

1

209.17

1

kg

m3

058.0

E l 9 A f l il h i i i f 0 297 N / 2 fl h h


34/41

Example 9. A fuel oil having a viscosity of 0.297 N s/m2 flows through acircular pipe 15 cm in diameter. At the center of the pipe the velocity is 1.20m/s and decreases to a minimum value at the pipe wall. The value of thevelocity at any point in the cross section a distance x from the center is

and r is the radius of the pipe. Compute the shear stress at a point midwaybetween the center and the wall ( x = 3.75 cm).

Solution:

22

4.63xr

V

1.2 m/s

15 cm V m/s

x (m)

V = f(x)

Substituting r = 7.5 cm or 0.075 mand = 0.296 Ns/m2 to V, weget

297.0

075.04.63

22x

V

2468.213201.1 xV

xdx

dV936.426

Note: The negative sign denotes that V is decreasing as x increases.

x (m)


35/41

15 cm

x (m)

3.75 cm = 0.0375 m

The shearing stress at x = 0.0375 m is

dx

dV

0375.0936.426297.0

xdx

dV936.426Note :

2755.4

m

N

PROPERTY CHANGES IN IDEAL GAS


36/41

PROPERTY CHANGES IN IDEAL GAS

For any ideal gas experiencing any process or changes, the equationof state is given by:

(Universal Gas Law) (1)

When temperature is held constant, equation 1 reduces to2

22

1

11

T

Vp

T

Vp

2211 VpVp (Boyles Law) (2)

When the volume of a confined gas is constant, the pressure isproportional to the absolute temperature, or

2

2

1

1

T

p

T

p (3)

If the pressure of a confined gas is unchanged, the volume is directlyproportional to the absolute temperature, or

2

2

1

1

T

V

T

V (4)

Equations (3) and (4) are called Charless laws.


37/41

From equation (1)

k

T

pV R

W

m

T

pV nR

T

pV

nRTpV (General Gas Law)

where: n = number of molesm = mass of the gas in gramsW = atomic or molecular mass of the gas in grams/moleR = universal gas constant


38/41

If a gas is compressed or expand without loss of heat through the wallsof its container, the change in volume is said to be adiabatic or isentropic. Thenthe special pressure volume relation is given as

tconsvpvpkk

tan2211

where: p = absolute pressurev = specific volume

k = adiabatic exponent, the ratio of specific heat at constantpressure to the specific heat at constant volume.

The value of k depends on the molecular structure of thegask = 1.4 for air, hydrogen, oxygen and nitrogen.


39/41

SPECIFIC WEIGHT AND DENSITY OF AIR

TemperatureC

Density ( kg/m3)

Specific Weight (N/m3)

0 1.293 12.68

10 1.248 12.24

20 1.205 11.82

30 1.165 11.43

40 1.128 11.06

60 1.060 10.40

80 1.000 9.81

100 0.946 9.28


40/41

DYNAMIC VISCOSITY OF AIR

Temperature

C

Viscosity

Pa s x 105

0 1.71

10 1.76

20 1.81

30 1.86

40 1.90

60 2.00

80 2.09

100 2.18

Holmans equation for finding of air is

23 00000034.000275.0110716.1 TTx where: is in Pa-s and T is the temperature in C


41/41

Problems for exercise

1. If a certain gasoline weighs 7 kN/m3, what are the values of its density,specific volume, and specific gravity relative to water at 15 C?

2. A cubic meter of air at 101.3 kPa and 15 C weighs 12 N. What is itsspecific volume?

3. The density of alcohol is 790 kg/m3. Calculate its specific weight, specificweight, specific gravity and specific volume.

4. A certain gas weighs 16 N/m3 at a certain temperature and pressure.

What are the values of its density, specific volume, and specific gravityrelative to air weighing 12 N/m3?5. Compute the number of watts which are equivalent to one horsepower.

( 1 HP = 550 ft-lb/sec; W = 107 dynes-cm/sec; 1 lb = 444,800 dynes).6. A city of 6000 population has an average total consumption per person

per day of 100 gallons. Compute the daily total consumption of the city incubic meter per second ( 1 ft3 = 7.48 gallons ).