What is Fluid?????

• A fluid may be liquid, vapour or gas. It has no permanent shape but takes up the shape of a containing vessel or channel or is shaped by external forces (eg the atmosphere).

• A fluid consists of atoms/molecules in random motion (translation) and in continual collision with the surroundings.

• Fluids are readily deformable, and flow.• Solids have ‘frozen’ molecules that vibrate and do not

translate. Solids resist change of shape.

What is Fluid?????

FT

Attached plates

Solid(Rectangular

Block) B B

FT

For a solid, application of a shear stress causes a deformation which, if modest, is not permanent and solid regains original position.

What is Fluid?????

Fluid at rest

Sliding (shearing occur between fluid layer)

FT

FT

For a fluid, continuous deformation takes place with an infinite number of layers sliding over each other. Deformation continues until the force is removed.

What is Fluid?????

FT

FT FT

FT

to t1 t2to 0<t<a

a) Solid a) Fluid

to<t1<t2

A fluid is a substance for which a shear stress tends to produce unlimited deformation.

Dimensions and Unit

1. Primary Dimensions

Primary dimension SI Unit BG Unit Conversion Factor

Mass (M) Kilogram(kg) Slug 1slug = 14.5939kg

Length (L) Meter(m) Foot (ft) 1 ft = 0.3048 m

Time (T) Second (s) Second (s) 1 s = 1 s

Temperature (Q) Kelvin (K) Rankine ® 1 K = 1.8 R

Dimensions and Unit

Secondary Dimension SI Unit .In primary dimen

sion

velocity m/s ms-1 LT-1

acceleration m/s2 ms-2 LT-2

forceN

kg m/s2 kg ms-2 M LT-2

energy (or work)Joule J N m,

kg m2/s2

kg m2s-2 ML2T-2

powerWatt WN m/s

kg m2/s3

Nms-1

kg m2s-3 ML2T-3

Dimensions and Unit

Secondary Dimension SI Unit .In primary dime

nsion

pressure ( or stress)Pascal P,

N/m2,kg/m/s2

Nm-2

kg m-1s-

2

ML-1T-2

density kg/m3 kg m-3 ML-3

specific weightN/m3

kg/m2/s2

kg m-2s-

2 ML-2T-2

relative densitya ratio

no units.

1no dimension

Viscosity

A fluid offers resistance of motion due to its viscosity or internal friction. Viscosity arises from movement of molecules from one layer to another moving at a different velocity. Slower layer tend to retard faster layers hence resistance

Viscosity

Shear stress A

F y

xShear strain

Viscosity

Rate of Shear strain t

yt

x

ty

x 1

y

u

Viscosity

For most fluids used in engineering it is found that the shear stress is directly proportional to rate of shear when straight and parallel flow is involved

y

utcons tan

y

u

Viscosity

where y

uis velocity change in y direction

So, at any point is the true velocity gradientdy

du

dy

du Newton's law of viscosity

The constant of proportionality; is called the dynamic viscosity or just viscosity

EXAMPLE 0

1) Determine the unit of dynamic viscosity;

2) Kinematic viscosity; is defined as the ration of dynamic viscosity to fluid density; Determine the unit of Kinematic viscosity

EXAMPLE 1Circular plate slides over the larger flat surface on a thin film of liquid that has a thickness of 0.02 mm. The plate has an diameter of d=6.3cm and mass of 20 g. If the plate is given an initial velocity of 5m/s, calculate the force required to move the plate at a steady velocity.Assume dynamic viscosity of liquid =1.805 kg/m-s

u=5 m/s

y=0.02 mm=1.805kg/m-s

Surface

EXAMPLE 1.1

A board 1m by 1m that weight 25 N slides down an inclined ramp (slope = 20O) with a velocity of 2.0 cm/s. The board is separated from the ramp by a thin film of oil with a velocity of 0.05 N.s/m2. Neglect the edge effect, calculate the spacing between the board and the ramp

EXAMPLE 1

A board 1m by 1m that weight 25 N slides down an inclined ramp (slope = 20O) with a velocity of 2.0 cm/s. The board is separated from the ramp by a thin film of oil with a velocity of 0.05 N.s/m2. Neglect the edge effect, calculate the spacing between the board and the ramp

EXAMPLE 1.2

A piston moves inside a cylinder at a velocity of 5 m/s, as shown in figure. The 150 mm diameter piston is centrally located within the 150.2 mm inside diameter cylinder. The film of oil separating the piston from the cylinder has an an absolute viscosity of 0.40 N.s/m2. Assuming a linear velocity profile, find the

a) Shear stress in the oil

b) Force F required to maintain the given motion

c) Force by which the required force would change if the velocity increased by a factor of 2

EXAMPLE 1.2

Viscosity Fluids which do not obey the Newton's law of viscosity are called as non-Newtonian fluids. Generally non-Newtonian fluids are complex mixtures: slurries, pastes, gels, polymer solutions etc.,

Viscosity

Variation of Viscosity with Temperature

For gasesn

oo T

T

ST

STTT

oo

o

)(

2/3

Power law

Sutherland law

Viscosity Variation of Viscosity with Temperature

For Liquid bTae 2

00

0

ln

T

Tc

T

Tba

For water To = 273.16K, mo = 0.001792 kg/(m.s)

and a = -1.94, b = -4.80 , c = 6.74

CONTIMUUMFrom a microscopic point of a view a fluid is not continuous and homogenous substance but consists of atom or molecules in random motion and with relatively large space between them. Under such circumstances it has no meaning refer to the velocity at a point in a fluid because that point may be empty space at particular instant. When we refer to the velocity of a fluid, we usually imply a quantity of fluid consisting of an enormous number of atoms of molecules-fluid velocity is a macroscopic concept. microscopic

macroscopic

DENSITY

Fluid

XTotal Mass = m1

Total Volume = V1

i

i

v

mAverage Density =

DENSITY

Density at point =

V

mv 0lim

m/V

V

Average Density

Specific Weight

The weight per unit volume of a fluid is called its specific volume and equal to g, the product of its density and the acceleration of gravity

gair = (1.204 kg/m3) (9.807m/s2) = 11.8N/m3 or = 0.0752lbf/ft3

gwater = (998kg/m3) (9.807m/s2) = 9790N/m3 or = 62.4lbf/ft3

Specific Gravity

Specific gravity (SG) which is the ratio of density to the standard density of some reference fluid at 20oC and 1 atm

3/204.1 mkgSG gas

air

gasair

3/998 mkgSG liquid

water

liquidwater

Compressibility and the Bulk modulus

V

P

VV

PP

Bulk modulus (K) = (change in pressure) / (volumetric strain)

Compressibility and the Bulk modulus Bulk modulus (K) = (change in pressure) / (volumetric strain)

Where the volumetric strain is the ratio of the change in volume to the initial volume

KPVV //

or dVVdPK /

Compressibility and the Bulk modulus From the mass conservation;

0 dVVddm

or d

VdV )(

d

VVdP

K

So that

d

dPK or

Compressibility and the Bulk modulus

Typical values of Bulk Modulus:

K = 2.05x109 N/m2 for water

K = 1.62 x 109 N/m2 for oil.