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1. Fluid as a continuum2. Measures of fluid mass and weight3. Ideal gas law4. Velocity field5. Flow classifications6. Flow visualization methods7. Stress field8. Viscosity9. Types of fluids10. Surface tension11. Classification of fluid motion12. Bulk modulus13. Compression and expansion of gases14. Speed of sound15. Vapour Pressure16. Classification of Fluid Motions

Chapter 2 Fundamental Concepts

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1. Continuum assumption

• Continuum assumption:

– Valid if the mean free path of the molecules is (10-9 mfor gases) much less than the smallest significant dimension of the problem (<10-6 m)

– Both air and water under normal conditions satisfy this condition

• Under the continuum assumption

– fluid properties are continuous functions of position and time

• E.g. ρ=ρ(x,y,z,t)

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2. Measures of fluid mass and weight

∀= /mρ1. Averaged density: mass per unit volume

(In SI, kg/m3) (2.1)

2. Specific volume: volume per unit mass. It is the reciprocal of the density

1/υ ρ= (2.2)

gργ =3. Specific weight γ: Weight per unit volume

(In SI, kg/m3·m/s2 = N/m3) (2.3)

4. Specific gravity SG: a non-dimensional quantity

Fluids: ρ depends slightly on T and P under normal conditions.Gases: depend significantly on T and P

2 @4/

H O CSG ρ ρ= o (2.4)

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3. Ideal gas law

Ideal gas law: P RTρ= (2.5)

P: absolute pressure (N/m2), measured relative zero pressure,

p: gage pressure (Pa), measured by a pressure gage,

p0: Standard sea-level atmospheric pressure, 101.33 kPa.

T: absolute temperature (K); K = °C + 273.15 (K: Kelvin)

R: gas constant (air is R = 286.9 (Joule/(kg K)), depending on

gas types and is related to molecular weight

0P p p= +

An ideal gas or perfect gas: hypothetical gas consisting ofidentical particles of zero volume, with no intermolecular forces. The constituent molecules undergo elastic collisions with the walls of the container.

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Example 2.1A compressed air tank has a volume of 0.05 m3. When the tank is filled with air at a gage pressure of 10 MPa, determine the density of the air and the mass of air in the tank. Assume the temperature is 15 oC and the atmospheric pressure is 101 kPa.

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4. Velocity field• Velocity

– One of the most fundamental quantities in fluid mechanics,

– Velocity at a point: the instantaneous velocity of the fluid particle passing through the point

• Velocity is a vector:

– Expressed in terms of scalar components

( , , , )V V x y z t=r r

kwjviuVrrrr

++= (2.6)

ux

z y

v wVv

Magnitude: 222 wvuV ++=

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An example of velocity vector

u

vV

22 vuV +=θ

u

v=θtan

Resultant velocity

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5. Type of flows-various classifications

(1). Does the flow properties depend on time?

� No → Steady: flow properties do not change with time:

where η represents any fluid property

0t

η∂ =∂

0 or ( , , )x y zt

ρ ρ ρ∂ = =∂

0 or V ( , , )V

V x y zt

∂ = =∂

vv v

�Yes → Unsteady: flow properties change with time:

0≠∂∂

t

η

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(2). One, two-, and three-dimensional flows• Based on the number of space coordinates

required to specify the velocity field

• Most of real flows are three-dimensional• Some flows can be simplified into 1 or 2-D

−=2

max 1R

ruu

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(3). Is velocity constant over a cross-section?

� Yes → Uniform flow

• Uniform flow at a cross section: velocity is constant at this cross-section

• Uniform flow field: a flow with constant velocity U

x

Uniform flow at a section

�No → non-uniform flow

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6. Flow visualization methods• Pathlines

The path traced out by a moving particle

tFor a specific fluid particle

• StreaklineA line joining particles passing through a fixed location in space

tAt a fixed location of the flow field

y

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• Streamlines– A line that is tangent to the direction of velocity at every point

in the flow at a given instant

uvdxdy // = (2.7)dxdy/

Vr

v

uα

y

xstreamline

Slope:

Streamline equation:

tan /v uα =

Question: Can a fluid particle pass across a streamline? Why?

•TimelinesA set of adjacent fluid particles thatwere marked at the instant in time

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An example of streamlines

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Summary of Last Lecture

• Ideal gas law: (how to use it)

• Velocity vector concept

• Flow visualization techniques

– In particular: streamlines

• streamline equation:

P RTρ=

uvdxdy // =

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Example 2.1a

A velocity field is given by

Find:1. Determine the velocity of a particle at point (2, 8)

2. Equations of streamlines in the x-y plane

3. Plot the streamline passing through the point (2, 8)

4. If the particle passing through the point (x0, y0) is marked at time t = 0, determine the location of the particle at t = 6 s?

5. Show that the equation of the particle path is the same as the equation of the streamline.

)/(3.03.0 smjyixVrrr

−=

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7. Stress field• Type of forces on fluid particles

– Body forces: e.g. gravity, electromagnetic forces

– Surface forces: e.g. pressure and friction

• Surface forces ⇒ stresses (forces per unit area)

Normal stress:

Shear stress:

nnA

n AFn

δδσδ

/lim0→

=

ntA

n AFn

δδτδ

/lim0→

=

(2.8)

(2.9)

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Stress components in Cartesian coordinate system:

xxA

xx AFx

δδσδ

/lim0→

=xy

Axy AF

x

δδτδ

/lim0→

=

xzA

xz AFx

δδτδ

/lim0→

=

(2.10)(2.11)

(2.12)

•First subscript: direction normal to the surface•Second subscript: direction of shear stress

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Stress at a point has nine components:

zzzyzx

yzyyyx

xzxyxx

στττστττσ

(2.13)

x

z

y

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8. Viscosity

Rate of deformation

y

tu

y

l

δδδ

δδδα ==

y

u

t δδ

δδα =⇒

Taking limit:dy

du

dt

d =αRate of shear strain

xyx

x

F

A

δτδ

=

Shear stress:

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Shear stress is proportional to rateof deformation − Newtonian fluids

dy

duyx ∝τ

Or: dy

duyx µτ =

where µ is dynamic viscosity

(2.14)

Kinematic viscosity ν:

ρµν /= (2.15)

Question:for a fluid at rest, is there any stress? why?

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Variation of shear stress with rate of shear strain for several types of fluids

23Fluid viscosity depends on the type of fluids and temperature

Dynamic viscosity of some fluids as a function of temperature

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9. Types of fluids

• Newtonian fluids

– Shear stress is proportional to deformation rate

τ = µdu/dy

– Most of common fluids, e. g. water, air, gasoline

• Non-Newtonian fluids

– Shear stress is not proportional to deformation rate

τyx = k(du/dy)n

– Bingham fluids, shear thinning and shear thickening fluids, …

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Summary of Last Lecture

• Stresses: force per unit area (N/m2)

– 3 normal stress and 6 shear stresses

– Symbols and sign conventions

• Viscosity: a measure for “fluidity” of a fluid

– Definition for Newtonian fluids (τyx = µdu/dy);

– Dynamic viscosityµ has a unit of kg/(m·s)or Pa·s , or N·s/m2;

– The kinematic viscosity (ν =µ/ρ, m2/s)

– Viscosity depends on type of fluid, pressure and temperature

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(a)The kinematic viscosity of the liquid (SI units)(b)The shear stress and its direction on the upper plate(c)The shear stress and its direction on the lower plate

Example 2.2An infinite plate is moved over a second plate. Assume a linear velocity distribution in fluid. The dynamic viscosity of the liquid µ=6.5×10-4 kg/(m⋅s) and SG=0.88. Determine:

y

x

U=0.3 m/s

d=0.3 mmfluid

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10. Surface tension

• Develops at interfaces with other liquids or solids due to the attraction between molecules as shown in the figures below

• Surface tension: force per unit length (N/m)

• The interface acts like a stretched elastic membrane (with a contact angle θ), creating surface tension σ

• θ and σ depend on the type of liquid and contact surface

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10.1. Pressure inside a drop of fluid:

Free-body diagram

σ: surface tension (N/m)

∆p = pi − pe

pi: internal pressure

pe: external pressure

Force around the edge: 2πRσ

Force due to the pressure difference: ∆pπR2.

Using force balance:

2πRσ = ∆pπR2

⇒ ∆p = 2 σ/R (2.15a)

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– Insignificant for large scale of motions– Important for some of small scale motions: e.g.

capillary rise (or depression)

– θ: angle of contact

10.2. Engineering applications

Vertical force balance:

Due to surface tension:2πRσ cosθ

Due to weight:γπR2h

⇒

⇒

2 cosh

R

σ θγ

=

1/h R∝Wet surface Free body Nonweting surface

(2.15b)

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Example 2.3: Analysis of capillary effect in a tubeCreate a graph for capillary rise or fall of a column of water or mercury as a function of tube diameter D. Find the minimum diameter of each column required so that the height magnitude will be less than I mm.Solution:

Assumptions: (1)∆h measures to the middle of meniscus(2)Neglect the volume in the meniscus region

Water:σ = 72.8×10-3N/m; θ = 0; ρ = 1000 kg/m3

Mercury:σ = 484×10-3N/m; θ = 140o; ρ = 13550 kg/m3

∆h z

To let ∆h = 1 mm:Dmin (water) = mm

Dmin (mercury) = mm

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11. Compressibility of fluids

1. Incompressible fluids

– density change negligible:

– Liquids and gases with small Mach number (M =

V/C<0.3, C is the speed of sound and V is flow velocity)

2. Compressible fluids

– density change is not negligible

– e. g. flow of gases (M>0.3); some flows of liquids under extremely high pressure ( water hammer);

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12. Bulk modulus

Bulk compressibility modulus (or modulus elasticity)

– Ratio of pressure change to relative density change (compressibility)

– Has the unit of pressure (Pascal:N/m2)

– Check Appendix A of Fox et al. Book for the values for common fluids

)/( ρρd

dpEv = (2.16)

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13. Compression and expansion of gases

When gases are compressed (or expanded) the relationship between pressure and density depends on the nature of the process;

• Under constant temperature conditions (isothermal):

constantp

ρ= (2.17)

(2.18)pEv =

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• Under isentropic conditions (frictionless compression and no heat exchange with surroundings):

constantk

p

ρ=

K: ratio of the specific heat at constant pressure, cp, to the specific heat at constant volume cv (R = cp – cv).

Note: Pressure in the above equations has to be the absolute pressure.

kpEv =

(2.19)

(2.20)

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Example 2.4 One cubic meter of helium at an absolute pressure of 101.33 kPa is compressed isentropically to 0.5 m3. What is the final pressure. The value for k is 1.66.

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Summary of Last Lecture

• Concept of surface tension and capillary effect

• Compressibility of Fluids

– Compressible and incompressible fluids

– Bulk modulus of fluids: a measure of compressibility

• Compression and expansion of gases

– Isothermal conditions

– Isentropic conditions

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14. Speed of sound

• Disturbances in fluids travel at speed of sound;• Speed of sound in fluids:

ρρvE

d

dpc == (2.21)

• For gases undergoing isentropic process Ev = kp

/c kP ρ= (2.22)

kRTc = (2.23)

• Making use of the ideal gas law, it follows that

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•Air at 20 oC with k = 1.4 and R = 286.9

smkRTc /14.343)2015.273(9.2864.1 =+××==

• Water at 20 oC, Ev = 2.19 GN/m2 and ρ = 998.2 kg/m3

smE

c v /12.14812.998

1019.2 9

=×==ρ

For example:

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Example 2.5 A jet aircraft flies at a speed of 1000 km/h at an altitude of 12,000 m, where the temperature is -50 oC. Determine the ratio of the speed of the aircraft, V, to that of the speed of sound, c, at the specified altitude. Assume k = 1.40.

Solution:

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15. Vapour Pressure

Vapour pressure: a pressure developed in a vacuum space left above the liquid in a closed container.

•It depends on temperature. Temperature ↑, vapour pressure ↑•Boiling: when the absolute pressure reaches the pv,

•Importance in engineering applications:

For fluid passing valve and pumps, when p< pv ⇒vaporbubble (low p) ⇒ swept along high p ⇒ bubble collapse ⇒may damage the structures − cavitation (picture above).

•Values for water can be found in Table A.8

water

evaporatepv

water

To vacuum

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16.1. Viscous and inviscid flows

• All real flows are viscous

• Flows are assumed inviscid when viscosity is not important for problems of concern

– Total drag on an airfoil

• Analysis of inviscid flows is much simpler than viscous flows

– Laplace equation (for irrotational flows)

– Navier-stokes equations (for viscous flows)

16. Classification of Fluid Motions

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Example of Viscous and inviscid flows

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16.2 Laminar and turbulent flows

• Laminar flow– Fluid particles move in smooth layers

• Turbulent flow

– Fluid particles mix rapidly with random three-dimensional velocity fluctuations

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16.3. Internal and external flows

• Internal flows

– Flows completely bounded by solid surfaces

e. g. flow in pipes

• External flows

– Flows over bodies immersed in unbounded fluid, e. g. flow around a submarine

• Both internal and external flows may be laminar or turbulent, compressible or incompressible

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Summary of Last Lecture• Classification of fluid motion

• Concept of vapour pressure & its importance in engineering

Continuum Fluid Mechanics

Inviscid flowµ=0

Viscous

Laminar Turbulent

Internal ExternalCompressible Incompressible