FLUID PROPERTIES (LEC 1&2)

FLUID AS A CONTINUUM

A fluid considered to be a continuum in which thereare no holes or voids ⇒ velocity, pressure andtemperature fields are continuous.

Validity criteria: Smallest length scale in a flow >>average spacing between molecules composing thefluid.

DENSITY (ρ)

Mass/ unit volume (kg/m3)

Density decreases normally with increasingtemperature

ρwater = ρ(T,S,p)

i.e., dependent on• Temperature• Salt content (ρ ≈ 1000 + 0.741⋅S, S in per mille;

S = 3.5% in ocean ⇒ ρ = 1026 kg/m3)• Pressure (but only a small variability)

For a perfect gas the state equation is,

ρ = p/RT, R is gas constant

OK in most cases for common gases in the ordinaryengineering range of pressure and temperature

ρair = 1.23 kg/m3 at atmospheric pressure and 15°C

OTHER DEFINITIONS

Weight = mass × gravity acceleration(W = mg, [N = kg⋅m/s2])

Weight density = density × gravity acceleration(γ = ρg, [N/m3 = kg⋅/(m2s2)])

Specific volume = reciprocal of density(ν = 1/ρ, [m3/kg])

Relative density is the density normalized with thedensity of water at a specific temperature andpressure (normally 4°C and atmospheric pressure):

S = R.d. =ρ/ρwater (often = ρ/1000)

COMPRESSIBILITY

All fluids can be compressed by application ofpressure ⇒ elastic energy being stored

Modulus of elasticity describes the compressibilityproperties of the fluid and is defined on the basis ofvolume

Modulus of elasticity:

1/ VdV

dpE −= [Pa]

For liquids, region of engineering interest is whenV/V1 ∼ 1.

Ewater ~ 2⋅109 Pa

E

p

VV ∆

−=∆ [Pa] (Eq. 2.3a in F&F)

IDEAL FLUID

A fluid in which there is no friction

REAL FLUID

A fluid in which shearing forces always existwhenever motion takes place due to the fluid’s innerfriction – viscosity.

VISCOSITY

- Viscosity is a measure of a fluid’s “innerfriction” or resistance to shear stress.

- It arises from the interaction and cohesion offluid molecules.

- All fluids posses viscosity, but to a varyingdegree. For instance, syrup has a considerablyhigher viscosity than water.

DEFINITION OF DYNAMIC VISCOSITY - µ

Illustrating example:.

Shearing of thin fluid film between two plates. Theupper plate has an area A.

Experiments have shown that for a large number offluids:

h

AVF ∝ (if V och h not too large)

Linear velocity profile ⇒ V/h = dv/dy

Introduction of the proportionality constant µ, nameddynamic viscosity, gives Newton’s viscosity law:

dy

dv

h

V

A

Fµµτ === (eq. 2.9)

µ [Pa⋅s]

Implication of viscosity: a fluid cannot sustain ashear stress without deformation

No-slip condition – water particles adjacent to solidboundary has zero velocity (observational fact)

ν = µ/ρ [m2/s] - Kinematic viscosity

Implications of Newton’s law:- τ, µ independent of pressure (in contrast to

solids)- no velocity gradient ⇒ no shear stress

Restriction of Newton’s law:- law only valid if the fluid flow is laminar in

which viscous action is strong

Laminar flow: smooth, orderly motion in which fluidelements appears to slide over each other in layers(little exchange between layers).

Turbulent flow: random or chaotic motion ofindividual fluid particles, and rapid mixing andexchange of these particles through the flow

Turbulent flow is most common in nature.

Newtonian – non-Newtonian fluids

Examples non-Newtonian fluids:Plastics, blood, suspensions, paints, foods

Shear vs. rate of strain relations for non-Newtonianfluids:

idy

dui ττµττ >=− , Bingham plastic

n

dy

du)(µτ = n>1: Shear-thickening fluid,

n<1: Shear-thinning fluid

SURFACE TENSION, CAPILLARITY

Surface tension effects occur at liquid surfaces(interfaces of liquid-liquid, liquid-gas, liquid-solid).

Surface tension effects are often negligible inengineering problems. Exceptions:

• Bubble formation• Capillary rise of liquids in narrow spaces• Break-up of liquid drops• Formation of liquid drops• Investigations using small physical models

Surface tension, σ [N/m], is thought of as the force inthe liquid surface normal to a line of unit lengthdrawn in the surface.

Surface tension decreases with temperature and isdependent on the contact fluid (surface tensionusually quoted in contact with air).

The surface tension force will support small loads ifliquid surface is curved.

Implications of surface tension

1) Capillary rise/drop

The liquid makes a contact angle θ with the glasstube. A vertical force balance between surfacetension force and weight of the lifted liquid column⇒ σ⋅2πR⋅cosθ = ρg⋅h⋅πR2 ⇒

Rgh ρθσ cos2=

(valid if R<2.5 mm)

θ: angle of contact, dependent on the relationbetween cohesive and adhesive forces

Angle of contact

2) For spherical droplet

Balance between internal pressure force and surfacetension force (see fig.) ⇒

p⋅πR2 = σ⋅2πR ⇒

Rp

σ2=

3) For bubble

Balance between internal pressure force and surfacetension force (see fig.) ⇒

p⋅πR2 = 2⋅σ⋅2πR ⇒

Rp

σ4=

Measurement of surface tension

F = 2σπD⇒ σ = F/(2πD

HYDROSTATICS (LEC 3-5)

Hydrostatics: Study of fluids (water) at rest

No motion ⇒ no shear stress ⇒ viscosity non-significant

Only existing stress for a fluid at rest is normal(compression) stress, i.e. pressure

Characteristics of pressure:

1. Measurement unit [Pa]=[N/m2]2. Pressure is transmitted normal to solid

boundaries or arbitrary sections

3. Pressure is transmitted undiminished to all otherpoints in a fluid at rest

4. Pressure has the same magnitude in alldirections at a point in a fluid at rest (scalarquantity)

Relation between pressure and depth in anincompressible liquid

Assuming constant density and no horizontalpressure variation, the liquid column in the fig belowcan be used the pressure as a function of depth.

Vertical forces acting on column (V, A, and y arevolume, area, and height of column, respectively):

Upward pressure force: pA

Weight (downward): γV = γAy

Vertical force balance:

pA = γAy ⇒

p = γy = ρgy

ABSOLUTE AND RELATIVE PRESSURE

Pressures are measured and quoted in two differentsystems, one relative (gage) and one absolute.

The relation between them are:

Pabsolute = Patmospheric + Pgage

Negative gage pressures are often termed vacuumpressures

Often only relative pressures are of interest

EXAMPLE ON GAGE AND ABSOLUTEPRESSURES

A pressure gage registers a vacuum of 310 mm ofmercury when the atmospheric pressure is 100 kPa,absolute. Calculate the corresponding absolutepressure.

Solution:

Patmospheric = 100 kPa

(Pgage/γHg) = -310 mm Hg ⇒ Pgage = -0.31γHg

γHg = 133.0 kN/m2

Pabsolute = Patmospheric + Pgage ⇒

Pabsolute = 100 - 0.31⋅133.0 = 58.8 kPa

FORCES ON SUBMERGED PLANE SURFACES

Example of applications:- Design of dams, ships, gates, and tanks.

Characteristics of pressure in a fluid at rest:• Constant pressure on plane horizontal surface• Linear pressure variation with depth for constant

density liquid• Pressure acts perpendicular to the surface

Pressure prism: “volume of pressure” on the planesurface

Resultant force is equal to the volume of the pressureprism and acts through its centroid

RELEVANT EQUATIONS – FORCES ON PLANESUBMERGED SURFACES

Resultant force, F:

Ac

hgAc

hF ργ ==

Point of action of resultant force:

Ac

y

cI

cy

py +=

A: area of plane surfacehc: vertical distance liquid surface - area centeryp: distance O - pressure centeryc: distance O – area centerIc: second moment of area about area centeraxis

Median line gives lateral position for center ofpressure for regular plane areas

FORCES ON CURVED SUBMERGEDSURFACES

(1) Resolve the force into two components, onevertical and one horizontal

Pressure intensity on a curved surface. F passesthrough the center of curvature.

(2) The horizontal force is obtained by projecting thecurved surface onto a vertical plane. The horizontalforce is equal to the force on this projected area

projA

projch

HF

,γ=

Projection of the curved surface onto a vertical plane

(3) The vertical force is equal to the weight of thevolume of liquid above the curved survace

The vertical force component, FV, caused by theweight of liquid above the surface

(4)The resultant force is given by

22H

FV

FF +=

and the direction of the resultant force by

HFVF

=φtan

The direction of the resultant force, F, which mustalso pass through C

(5) Remember that there is an equal and oppositeforce acting on the other side of the surface.

ARCHIMEDES PRINCIPLE – BUOYANCYFORCE

Law of buoyancy (Archimedes’ principle):

• “The upthrust (buoyancy force) on a bodyimmersed in a fluid is equal to the weight of thefluid displaced”

Law of flotation:

• “A floating body displaces its own weight of theliquid in which it floats”

Proof of Archimedes principle

Vertical forces acting cylinder surface:

“Downwards”:p1A = γyA

“Upwards”:p2A = γ(y+L)A

“Net pressure force (upthrust)”, FB:

FB = γ(y+L)A - γyA= γLA = γV

FUNDAMENTAL EQUATION OF FLUIDSTATICS

The general relation for pressure in a static fluid is:

g

dz

dpργ −=−=

Implication: pressure varies only with depth and isconstant in a horizontal plane

For a fluid with constant density:

hzzpp γγ =−=− )12

(21

or

γ21

pph

−=

Implications:• pressure varies linearly with depth• pressures may be expressed as head of fluid of

weight density γ• pressures are often quoted as heads in mm Hg or

m H2O

• .2

21

1 Constzp

zp

=+=+γγ

, for all points

in a fluid at rest

MEASUREMENT OF PRESSURE

MANOMETRY

Pressures are equal over horizontal planes withincontinuous columns of the same fluid

Conversion of manometer readings to pressure:

(a) p1 = p2

p1 = px + γlp2 = patm + γ1h ⇒ px = patm + γ1h - γl

(absolute)

(b) p4 = p5

p4 = px + γ1l1

p5 = py + γ2l2 + γ3h ⇒ px - py = γ2l2 + γ3h - γ1l1

Practical considerations in manometry

• Relative densities of manometer liquids varywith temperature (may cause errors)

• Capillarity errors• Inaccurate readings caused by poor menisci due

to surface tension effects• Manometer liquid fluctuations ⇒ (inaccuracy in

readings)• Use of optical devices for extremely precise

readings

BASIC CONCEPTS (LEC 6)

CLASSIFICATION OF FLOWS

Flow characterized by two parameters – time anddistance.

Division of flows with respect to time:• Steady flow – time independent• Unsteady flow – time dependent• Quasi-steady flow – slow changes with time

Division of flows with respect to distance:• Uniform flow – constant section area along

flowpath• Non-uniform flow – variable section area

Examples of types of flow:• Steady uniform flow – flowrate (Q) and section

area (A) are constant• Steady non-uniform flow: Q = constant, A =

A(x).

• Unsteady uniform flow: Q = Q(t), A = constant.

• Unsteady non-uniform flow: Q = Q(t), A =A(x).

VISUALIZATION OF FLOW PATTERNS

Streamline: a curve that is drawn in such a way that itis tangential to the velocity vector at any point alongthe curve. A curve that is representing the directionof flow at a given time. No flow across a stream line.

Streamtube: A set of streamlines arranged to form animaginary tube. Ex: The internal surface of apipeline.

Potential flow: Flow that can be represented bystreamlines.

Streakline: path made by injected dye in a flow field.

1-D, 2-D, AND 3-D FLOWS

Most real flows are three-dimensional.

Many times it is possible to simplify a 3-D flow to bea 1-D or 2-D flow - simplifies greatly the analysis.

For instance, pipe flow is taken to be 1-D andaverage fluid properties are used at each section.

The flow along a streamline (however curved) is 1-Dand the distance is measured along the streamline.

Ex. 2-D flow: flow over a wire (“överfall”), and flowaround a wing

TWO WAYS OF DESCRIBING FLUID MOTION

Lagrangian view: the path, density, velocity andother characteristics of each fluid particle in a flow istraced.

Eulerian view: study the flow characteristics(velocity, pressure, density, etc.) and their variationwith time at fixed points in space.

LAMINAR AND TURBULENT FLOW

Laminar flow• Flow along parallel paths• Shear stress proportional to velocity gradient (τ

= µ⋅du/dy)• Disturbances in the flow are rapidly damped by

viscous action

Turbulent flow• Fluid particles moves in a random manner and

not in layers• Length scales >> molecular scales in laminar

flow• Rapid continuous mixing• Inertia forces and viscous forces of importance

Reynolds experiment

Small velocities ⇒ line of dye intact, movement inparallel layers ⇒ laminar flowHigh velocities ⇒ rapid diffusion of dye, mixing ⇒turbulent flowCritical velocity ⇒ line of dye begin to break-up,transition between laminar and turbulent flow

Two thresholds:

Upper critical velocity – transition of laminar flow toturbulent flowLower critical velocity – transition of turbulent flowto laminar flow

Reynold’s number

Reynolds generalized his results by introduction of adimensionless number (Reynolds number):

µρ

νVDVDR ==

Pipe flow is laminar if R < 2000 and turbulent if R >4000.

The critical Reynolds number, Rc, defining thedivision between laminar and turbulent flow, is verydependent on the geometry for the flow.

Parallel walls: Rc ≅ 1000 (using mean velocity V andspacing D)Wide open channel: Rc ≅ 500 (using mean velocity Vand depth D)Flow about sphere: Rc ≅ 1 (using approach velocityV and sphere diameter D)

GOVERNING EQUATIONS(LEC 6-11)

FLUID SYSTEM AND CONTROL VOLUME

Fluid system: Specified mass of fluid within a closedsurface

Control volume: Fix region in space that can’t bemoved or change shape. Its surface is called controlsurface.

RELATION BETWEEN FLUID SYSTEM ANDCONTROL VOLUME

Let X represent the total amount of some fluidproperty (for instance mass or energy) within aspecified boundaries at a specified time. Let S denotefluid system and CV denote control volume. Then wehave (out stands for transfer out from CV and instands for transfer into the CV):

dtdtdtdt

inCV

outCVCVS dXdXdXdX

−+= (eq. 4.9 in F&F)

CONTINUITY EQUATION

Steady flow

ρ1⋅V1⋅A1 = ρ2⋅V2⋅A2 (4.16a)

Incompressible flow

V1⋅A1 = V2⋅A2 or Q1 = Q2 (Q = V⋅A) (4.17)

V: Average velocity at a section (m/s)A: Cross-section area (m2)Q: Flow rate (m3/s)

Flow in a pipe junction

Q1 + Q2 + Q3 = 0 or V1⋅A1 + V2⋅A2 + V3⋅A3 = 0

Channel flow

d(Vol)/dt = Q1– Q2 (Q12 = 0) (4.18)

Vol: Volume of water in channel betweensection 1 and 2

BERNOULLI’S EQUATION

Bernoulli’s equation is the energy equation for anideal fluid (friction and energy losses assumednegligible).

Bernoulli’s equation may, however, be used withsatisfactory accuracy in many engineering problemsand have the advantage of providing valuable insightabout energy conditions in fluid flow.

Derivation of Bernoulli’s equation for 1-D flow

Consider a streamline and choose a small cylindricalelement for analysis.

Use Newton’s 2nd law.

The first result is Euler’s 1-D equation:

0=++ dzgdVVdp

ρ

Assuming an incompressible fluid we can then derivethe Bernoulli equation:

Hzg

Vp=++

2

2

γ = constant (eq. 5.24 in F&F)

Bernoulli’s equation is a useful relationship betweenpressure, p, velocity, V, and geometric height, z,above a reference plane (datum).

H: energy head (m)z: elevation head above datum (m)V: velocity (m/s)g: gravity acceleration (m/s2)p: pressure (Pa)γ: weight density for the flowing fluid (N/m3)

Quantity Name Measure ofH Energy head Total energyP/γ Pressure head “pressure

energy”Z Elevation head Potential energyV2/(2g) Velocity head Kinetic energy

zp

+γ

: piezometric head or

H.G.L = Hydraulic Grade Line

Validity criterias – Bernoulli’s equation:

1) Along a streamline2) For an ideal fluid3) Steady flow4) Incompressible flow

1-D ASSUMPTION FOR STREAMTUBES WITHA FINITE SECTION AREA

Bernoulli’s Equation may also be applied to streamtubes with larger section areas (like pipes andchannels) if the streamlines are parallel.

If the streamlines are parallel the pressuredistribution is hydrostatic ⇒ the piezometric head

(p/γ + z) is constant over the section perpendicular tothe streamlines.

KINETIC ENERGY CORRECTION FACTOR

For a real fluid, friction will cause a non-uniformvelocity distribution ⇒ the velocity head have to becorrected before use of the Bernoulli equation.

The real kinetic energy is obtained by integrationover the section area and is then expressed in termsof the mean velocity, V, and a correction coefficient,α. The corrected velocity head becomes

g

V

2

2α

Some values on α:

α = 2 (laminar pipe flow)α ≈ 1.06 (turbulent pipe flow)α ≈ 1.05 (turbulent flow in wide channel)

WHY THE ENERGY CORRECTIONCOEFFICIENT α (MOMENTUM COEFFICIENTβ) OFTEN MAY BE OMITTED

1) Most engineering pipe flow problems involveturbulent flow in which α is only slightly morethan unity.

2) In laminar flow where α is large, velocity headsare usually negligible when compared to theother Bernoulli terms

3) The velocity heads in most pipe flows areusually so small compared to the other termathat inclusion of α has little effect

4) The effect of α tends to cancel since it appearson both sides of the equation

5) Engineering answers are not usually required toan accuracy which would justify the inclusion ofα in the equation.

PITOT TUBE

A Pitot tube, see Fig. above, is used to measure the

energy head (h = g

Vz

p

2

20

00 ++

γ) in a fluid flow.

The pressure at point S is called stagnation pressureand is the sum of static and dynamic pressures:

2

20

0

Vpsp ⋅+= ρ

VAPOR PRESSURE

All liquids possess a tendency to vaporize, i.e. tochange from liquid to gaseous phase

Ejected molecules (gaseous) exert their own partialpressure, known as vapor pressure of the liquid

Increased temperature ⇒ increased kinetic energy formolecules ⇒ increase of vapor pressure

Boiling (formation of vapor bubbles throughout thefluid mass) will occur whenever the external pressureis less than the vapor pressure.

Boiling point of a liquid is dependent on imposedpressure and temperature

Volatile liquids, which vaporize more easily, possesshigher vapor pressures

CAVITATION

The rapid vaporization (boiling) of a liquid when itpasses an area with low pressure (equal to the vaporpressure) is called cavitation. Cavitation may lead toserious damages on pipe or hydraulic structures likepumps.

The critical pressure conditions for cavitation are

γγvapor

abscrit

pp =)(

)()(γγγvaporatm

relcrit

ppp −−= (eq. 5.36)

EQUATION FOR A JET TRAJECTORY

A jet trajectory may be described by application ofNewton’s 2nd law ⇒

tVx x ⋅= 0

2

2

0t

gtVz z −⋅=

The jet trajectory is then given by,

22

00

0

2x

Vgx

VV

zxx

z

⋅−⋅= (eq. 5.37)

The top point of the jet is obtained by setting dz/dx =0:

gV

zgVV

x zzz

2

20

max00 =⇒⋅=

The kinetic energy at this point is:

g

VKE x

2

20=

Bernoulli’s equation is valid along the wholetrajectory (p=0):

.2

2

konstHg

Vz ==+

ENERGY AND HYDRAULIC GRADE LINES

ENERGY EQUATION

Lhg

Vz

pMh

g

Vz

p+++=+++ )

2

22

22()

2

21

11(

γγ

hM – energy added by machinery work (pumps,turbines) per unit weight of flowing fluid(m)

hL - energy loss per unit weight of flowing fluid(m)

POWER IN FLUID FLOW

P = γ⋅Q⋅h (W)

P - Power developed by, for instance, pump orturbine

γ - weight density (N/m3)Q - flow rate (m3/s)h - energy per unit weight of flowing fluid that

has been added (hP) or withdrawn (hT)

Efficiency, η = (power output)/(power input)

MOMENTUM EQUATION

DerivationConsider the control volume and the fluid systemdefined within the streamtube in Fig.

Control volume fix between section 1 and 2. Controlvolume identical to the fluid system at time t=t. Attime t + δt the fluidsystem has moved to betweensections 1’ and 2’.Use Newton’s 2nd law and relationship between fluidsystem and control volume ⇒ Momentum equation:

),,( INXVOUTXVQXF −=∑ βρ (x-direction)

),,( INYVOUTYVQYF −=∑ βρ (y-direction)

),,( INZVOUTZVQZF −=∑ βρ (z-direction)

Σ F : Sum of all external forces acting on thecontrol volume (like the streamtube).

ρ: Density of fluidQ: FlowrateVOUT: Velocity out of the control volumeVIN: Velocity in to the control volumeβ: Correction coefficient for momentum

The momentum correction coefficient, β

Typical values:

β = 1.33 (laminar pipe flow)β = 1.02 (turbulent pipe flow)β = 1.02 (turbulent flow in wide channel)

In most cases we can set β = 1 without making anysignificant errors.

METHODOLOGY USING THE MOMENTUMEQUATION FOR A FLUID FLOW PROBLEM

• Define an appropriate control volume

• Define coordinate axes

• Determine all forces (magnitude and direction)acting on the control volume

• Determine flowrate, outflow and inflowvelocities to the control volume (if not given,use continuity equation and Energy (Bernoulli)equation)

• Solve momentum equation

Remember sign (“tecken”) conventions:

1. Outflow velocities are positive and inflowvelocities are negative

2. Velocities and forces are positive in positivecoordinate directions

Ex. An outflow velocity in positive coordinatedirection is positive [+(+VOUT) = +VOUT]

Ex2. An inflow velocity in negative coordinatedirection is positive [-(-VIN) = +VIN]

Ex3. An outflow velocity in negative coordinatedirection is negative [-(+VOUT) = -VOUT]