Chapter. 1 IntroductionFundamental Fluid and Flow Properties

이동근ME Bldg. (517)

Tel. 051-510-2365Download at http://npt.pusan.ac.kr

Reference:1. Fluid Mechanics, Frank MWhite, 6th Edition, 2008, McGraw Hill2. Munson et al., Fundamentals of Fluid Mechanics, 8th Edition, 2006, John Wiley &Sons, Inc

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Contents

Chap. 1 Introduction

- Dimensions, Dimensional Homogeneity, and Units

- Concept of Continuum

- Measures of Density, Specific Weight & Gravity of Fluids

- Compressibility of Fluids: Bulk modulus, Compression of gas, Mach #

- Viscosity

- Surface tension of liquid

Chap. 2 Fluid Statics

Chap. 3 Elementary Fluid Dynamics : the Bernoulli equation

Chap. 4 Fluid Kinetics

Chap. 5 Finite Control Volume Analysis

Chap. 6 Differential Analysis of Fluid Flow

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What is a fluid?

• Three most common phases of matter are solid, liquid and gas.

• Liquids and gases are together called fluids.

• Later we’ll give another definition for ‘‘fluid’’, based on its behavior under shear forces.

Exercise : From a technical point of view, what is the difference between gases andvapors?

Exercise : What are the fourth phase of matter?

Intermolecular Attraction Forces

Molecules Volume and Space

Solid Strong Relative positions arerather fixed

Definite volume

Definite shape

Liquid Medium Free to change their relative positions

Definite volume

Indefinite shape

Gas Weak Practically unrestricted Indefinite volume

Indefinite shape

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What is Mechanics?

• Mechanics studies the motion and deformation ofmaterial bodies under applied loads.

• It involves force, energy, motion, deformation, material properties, etc.

• When the material is in solid phase it is called solid mechanics.

• When the material is in liquid or gas phases it is called fluid mechanics.

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Statics (bodies at rest)

Dynamics (bodies in motion)

Design of a wind turbine requires an understanding of both solid and fluid mechanics.

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Branches of Fluid Mechanics

• Fluid mechanics deals with the behavior of fluids at rest (fluid statics) and in motion (fluid dynamics).

• Hydrodynamics studies liquids (incompressible flow) in motion.

• Hydraulics studies liquids flowing in pipes, ducts and open channels.

• Gas dynamics studies compressible flow of gases with high density changes.

• Aerodynamics is similar to gas dynamics, but also covers low speed flows. It focuseson air flow.

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A Day Full of Fluid Mechanics

1. You wake up in the morning and the room is comfortably warm. Free convection driven by the radiator warm up the room all night.

2. You wash your face. The water first passes through a flow meter outside your house and travels inside pipes before coming to the faucet.

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A Day Full of Fluid Mechanics

4. You decide to do some cleaning before heading to school. Fluid mechanics is involved in proper suction of air and filtering dust.

3. You have your breakfast. The coolantmoving inside the tiny pipes at the back of your refrigerator, and the air circulating insideit are both fluids.

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A Day Full of Fluid Mechanics

5. You watch the morning news on the TV.

Fluid mechanics is involved in countlessmilitary/defense applications.

www.nasa.gov

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A Day Full of Fluid Mechanics

6. Weather forecast says it will be a sunny day.

We were lucky being far away from the Typhoon Jabi sweeping the main land of Japan.

Today computational fluid dynamics is used for weather

forecast

7. You get on your car. When you turn the engine on, the fuel is first pumped from the tank, mixed with air, and combustion gases do work inside the engine.

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A Day Full of Fluid Mechanics

8. You can also use the metro, which needs to be designed considering the air flow related comfort of passengerswaiting at the stations.

9. Before your class you may have to print your homework. Inkjet printing and electronics cooling also involve fluid mechanics.

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A Day Full of Fluid Mechanics

10. Before sleeping you watch some sports.

Both experimental and computational tools are used to improve performance in various sports.

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A Day Full of Fluid Mechanics

11. Some recent application fields: Metallic powder production for Metal 3D printing

CFD through healthy lung airway

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Before we start - Dimensions and Units

• In the MLT system basic dimensions are

Mass [M], Length [L], Time [T]

Exercise: Express the dimensions of the following quantities in terms of M, L and T.

Acceleration [???]

Energy [???]

Pressure [???]

Angular velocity [???]

Torque [???]

• For some problems, Temperature [𝜃] also serves as a basic dimension.

• Equations should be dimensionally homogeneous, i.e. dimensions of the left and right hand sides of an equation should be the same.

e.g., Speed of a uniformly accelerated body is given by

[m/s] [m/s2]

[s]

V = V0 + at

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Dimensions and Units (cont'd)

Exercise: A website claims that the distance travelled by a freelyfalling body is given by the following formula. Is it dimensionallyhomogeneous?

Exercise: The force F, due to the air blowing against a car in a wind tunnel is given by

where V is the air speed, is the density of air, 𝐴 is the projected cross sectional area of the car facing the air, and Cd is a constant called the drag coefficient. Determine the dimensions of Cd.

d = 4.90 t2

𝐹 =𝐶𝑑𝑉2𝐴

2

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Dimensions and Units (cont'd)

Exercise: one ME student provided the following wrong answer for the shownmidterm question. The instructor put a comment on the student’s paper saying‘‘You should be able to notice that this answer cannot be correct’’. What does theinstructor expect from the student?

Torque =2𝐻𝑅3

c+

𝑅3

2c=𝑅3

2𝐻 +1

2

Question: A concentric cylinder viscometer is formed by rotating the inner cylinder of a pair or closely fitting cylinders. The outer cylinder is kept fixed and the gap between the cylinders is filled with a fluid of viscosity . Using the parameters given in the figure, determine the torque that needs to be applied on the inner cylinder to rotate it at a constant speed of .

Student′s wrong answer:

𝐻

Fluidviscosity:

𝑅

c

c

c

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Before we start - Significant Figures

Exercise: A different version of the previous question was asked. This time numerical values are given for all the parameters and the required torque is asked. One student provided the following answer and the instructor put a comment on the student’s paper saying ‘‘Are you serious?’’.What does the instructor expect from the student?

Student’s answer: Torque = 15.908633 Nm

• Number of significant figures that needs to be used in an answer depends on the significant figures of the numbers that you used to calculate it.

• For measured quantities, number of significant figures depends on the minimum reading scale of the measuring device.

• The shown ruler CANNOT measure a length as 2.345 cm.

Exercise: We used the above ruler to measure the diameter and the height of acylinder as 10.0 cm and 25.5 cm, respectively. Calculate the volume of thecylinder.

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Significant Figures (cont'd)

• Often the significant figures of the numerical values given in a problem are notknown.

• An engineer needs to use his/her engineering intuition to judge the correct number of significant figures to use in a final calculated value and do a proper rounding.

• Some textbooks use a general rule such as ‘‘All final answers will be rounded to 3significant figures’’.

Exercise: a) A student calculated the value of the water flow rate coming out of a faucet as 20.4560 liters per minute. How should it be reported?

b) A student calculated the thrust generated by a jet engine as 485643.271 N. How should it be reported?

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Before we start - Greek Letters

• The incomplete list is

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Before we start - 1st order Taylor series Expansion

• Consider a function f(x). If it is known at a certain point x0 as f(x0), we cancalculate it at a nearby point x1 = x0 + x as follows

• This is the mathematical tool we use to perform differential analysis.

Exercise: Consider a differential fluid volume of size x × y × z. If the pressure at its centroid is p what are the pressures at its faces?

xy

z Pressure at the centroid is p.

x

y

z

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Concept of Continuum

• At the microscopic scale, fluids are composed of molecules.

has about 1025 molecules has about 1025 molecules

• Question : Is it possible to keep track of all these molecules ?

A glass of water

1 m3 air

• Answer : Practically impossible and not necessary for most engineering problems.

• Rather, we study most engineering problems at the macroscopic scale.

• That is we treat fluids as continuum and do not concern with the behavior of individual molecules.

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Concept of Continuum (cont'd)

Microscopic level: Each fluid molecule shown below moves at a different speed in a different direction.

Macroscopic level: The speed at point A is 90 km/h. The direction of air flow at point A is as shown below.

• 90 km/h is the average speed of molecules in the very small volume surrounding point A.

• We can say that the fluid particle (not the fluid molecule) located at point A is

A

Streamline showing theflow direction

moving with a speed of 90 km/h.

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Concept of Continuum (cont'd)

• Continuum assumes that fluid and flow properties, such as density, velocity, pressure, temperature, etc. vary continuously throughout the fluid.

• It does not pay attention to individual fluid molecules or worries about the gaps between them.

• In continuum, the smallest element of a fluid is NOT a fluid molecule, but rather a fluid particle, which contains enough number of molecules to makemeaningful statistical averages.

• Question: Is continuum a reasonable assumption?

• Practical answer: Yes, in many engineering problems.

• Technical answer: Depends on the Knudsen number.

Kn = = Average distance traveled by molecules between collisions (mean free path)

D Characteristic dimension of the flow field

• If Kn < 0.1 → continuum is valid. In this course we will treat fluids as continuum.

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Concept of Continuum (cont'd)

Exercise : Estimate Kn for the foregoing automobile example which can be scaled as 1 m. Air at standard atmospheric conditions has a mean free path of8 x 10 – 8 m.

Exercise : What will be the limiting characteristic length that will break the validity of the continuum assumption for an application using standard atmospheric air? What happens at micro and nano scales?

Exercise : How much the mean free path of air should be increased so that wecan start questioning the validity of continuum for flow around a missile witha characteristic length of 10 m? Is it possible to have such high mean free path values at the outermost regions of the atmosphere?

Exercise : What about in a liquid?

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Fundamental Flow and Fluid Properties

It is common to use p and T to define a thermodynamic state. Then otherproperties can be expressed as a function of these two:

= (p,T) , h = h(p,T) , = (p,T) , etc

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Density

Greek letter ‘‘rho’’

• ( ) [ kg/m3 ]

•

•

Mass contained in a unit volume of a fluid. = m / ∀

Density determines the inertia of a unit volume of fluid and hence its acceleration when subjected to a given force. Gases are easier to accelerate than liquids.

• Density also determines the amount of gravitational force (weight) acting on afluid body. Weight of gases is neglected more often than that of liquids.

• Fluids have a very wide range of densities.

(at the standard condition)

H2 CH4 Air Water Mercury

Density [kg/m3]

0.084 0.67 1.2 998 13600

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Density (Cont'd)

Add heat and allow piston to go up to keep pconstant

T : increases : decreases p: increases

: increases

• Density in general is a function of p and T, i.e. = (p,T)

• If a fluid’s density is a function of pressure only (not temperature) it is called abarotropic fluid, a simplification mostly used in meteorology (기상학).

• Following examples demonstrate how density of an ideal gas (p = 𝑅T) can change with temperature and pressure.

Remove heat to keep T constant

= (p,T) → d =

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Density (Cont'd) vs Compressiblity

• Changing por T may result in a change in .

pT

dp+

Tp

dT

Total change of Change of due to the change in p

Change of due to the change in T

dpE =

d/or using m = ∀= constant

d pE = −

d∀/∀

Used to define coefficient of thermalexpansion,

Used to define bulkmodulus of elasticity, E

Amount of differential pressure change (d p ) necessary to create a differential change in volume (d ∀), of a fluid of volume ∀.

Change of pChange ofT

p∀

p + d p ∀ + d ∀ + d

Push down

m m

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Compressibility of liquid

• Large E indicates higher resistance to compression.

• Incompressible fluid (E → ∞) is an idealization used mostly for liquids that have

constant density over a given range of conditions of a flow problem.

• In this course we’ll consider all liquids to be incompressible (constant density) unless otherwise mentioned.

• We’ll take water = 1000 kg/m3 (constant) unless otherwise mentioned.

P = 21.5 MPa

∀= 0.99 unit

Exercise: At 1 atm (~100 kPa) pressure and 15 oC temperature, we need to increase pressureto 21.5 MPa to compress a unit volume of water by 1 %. Calculate E of water under these conditions. Compare the value to the modulus of elasticity of steel.

P = 1 atm∀ = 1 unit

water water

compress

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Compressibility of gas

• Gases are much more compressible than liquids.

• E of a gas that compresses (or expands) depends on the process.

For example an ideal gas ( p = 𝑅T ) can go through

Isentropic process

(𝜌 𝑘

= constant )p

Isothermal process( T = constant )

(𝜌

= constant )p

p

Isothermal

Isentropic

Exercise: Calculate the isentropic bulk modulus of air (𝑘 = 1.4) at standardconditions. Compare the compressibility of air and water.

𝑘: specific heat ratio

𝐸𝑣 = 𝑑𝑝

𝑑= 𝑝 ∝ 𝐸𝑣 =

𝑑𝑝

𝑑= 𝑘𝑝 ∝ 𝑘

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Compressibility of gas (cont'd)

• Speed of sound (c) is a critical parameter in compressibility.

• Mach number (Ma) is the non-dimensional parameter that can be used to check the

importance of compressibility in gas flows.

Ma =V Characteristic speed in a flow problem

c Speed of sound=

• As a rule of thumb, flows with Ma < 0.3 can be approximated to be incompressible.

• For an incompressible fluid, which is an idealization, c → ∞ and Ma → 0.

• Although gases are much more compressible than liquids, they can also be treated as incompressible (constant density) in many engineering applications unless Ma > 0.3.

• Exercise : A car is traveling in still air at a speed of 90 km/h. Calculate Mach number of the flow and decide if compressibility effects are negligible or not.

• In this course we’ll consider air to be incompressible with air = 1.2 kg/m3, unless otherwise mentioned.

𝑐 = 𝑑𝑝/𝑑 = 𝐸𝑣/

Greek letter‘‘gamma’’

• Specific weight (비중량):

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Specific gravity & weight

• Specific gravity (Relative density): ( s ) [ unitless ]

Ratio of density of a substance to the reference density of water at 4 oC.

= = gVolumeWeight of unit volume of a substance.

( ) [ N/m3 ] Weight

1000 kg/m3

s =𝜌

𝜌w ater (at 4 ℃)

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Viscosity

• ( ) [Pa⋅s] or [centiPoise, cP = 10-3 Pa⋅s]

•

•

Measure of a fluid’s resistance to shear (or angular) deformation.

It is about the “fluidity” of a fluid. It shows a fluid’s resistance to change shape.

• Experiment: Consider a solid block firmly attached to two parallel plates.

B’

A

• The block deforms elastically if a force F is applied to the upper plate.

Shear stress ∝ Angular deformation

• If we gradually increase the force, the block will break apart at some point.

Greek letter ‘‘mu’’

B

A

F

33

Viscosity (cont'd)

• Think of a similar experiment with a layer of fluid between the parallel plates.

• Vertical fluid element AB will deform continuously as long as the shear force isapplied by moving the top plate.

F

t0

A

B

• First observation: Top plate will reach a speed of U0 and the velocity profile within the fluid will be linear. No-slip condition will dictate the fluid speeds at y = 0 and y = ℎ. Couette flow

t1

B’t2

B’’Shear deformation

u = U0

u = 0 xy

ℎ

U0

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Viscosity (cont'd)

x

y

ℎ

• Angular deformation rate of line AB (how fast the angle d changes in time) is

Exercise : Show that this deformation rate is equal to which is also equal to

• Second obervation: Shear stress acting on a surface parallel to the flow (such as the surface of the top plate) will be proportional to the angular deformation rate

Shear stress ∝ Rate of angular deformation or

A

B

t0 d

B’ u =U0

u = 0

U0

𝑢 =𝑈0

ℎ𝑦at 𝑡0+dt

𝑑

𝑑𝑡

𝑑𝑢

𝑑𝑦

𝑈0

ℎ

𝜏 =𝐹

𝐴∝

𝑑

𝑑𝑡=

𝑈0

ℎ

𝑈0dt

35

Viscosity (cont'd)

y

𝜏

𝜏x

• Newton’s Law of Viscosity: Shear stress on a surface tangent to the flow direction is proportional to the

• rate of shear strain (rate of angular deformation)

• or to the velocity gradient on the surface (change of velocity in the direction normal to the surface).

• For Newtonian fluids (air, water, gasoline, oils, etc.) the above proportionality islinear and the proportionality constant is known as viscosity.

Coefficient of viscosity,

Absolute viscosity,

Dynamic viscosity,

Viscosity

Shear force acts in the x which the force acts is in y direction.

𝜏 ∝𝑑

𝑑𝑡

𝑑𝑑𝑡

= 𝑈0

ℎ=

𝑑𝑢

𝑑𝑦

𝜏𝑥𝑦 = 𝜇𝑑

𝑑𝑡= 𝜇

𝑑𝑢

𝑑𝑦U0

36

Viscosity (cont'd)

Exercise : a) What is the sign convention for stress? b) Are the stresses shown in the previous figure negative or positive?

Exercise : For the problem shown on the right, determine the sign of the shear stress in thelower and upper fluids. Show the direction ofshear forces acting by the fluid on the plates.

yx

U0

𝜏𝑥𝑦+ = 𝜇 𝑑𝑢

𝑑𝑦 ℎ+< 0

𝜏𝑥𝑦− = 𝜇 𝑑𝑢

𝑑𝑦 ℎ−> 0

𝜏𝑥𝑦+< 0

𝜏𝑥𝑦−> 0

• For problems in the cylindrical coordinate system we need to be careful to write Newton’s law of viscosity in correct form. Herewe see two cylinders, inner one is fixed andouter one is rotating.

𝜃 r

𝜏𝑟𝜃 = 𝜇𝑑𝑉𝜃

𝑑𝑟

37

Viscosity (cont'd)

Exercise : Velocity distribution of the pressure driven flow of a Newtonian fluid betweentwo fixed, wide parallel plates is as shown. Umax is the maximum centerline speed. The Newtonian fluid has a viscosity of 2 Pa∙s. Umax = 0.9 m/s and h = 5 mm. Determine a) shearstress at the bottom wall, b) shear stress at the top wall, c) shear stress on the centerlineplane.

2ℎy

x 𝑢 = 𝑈𝑚𝑎𝑥 1 −𝑦

ℎ

2

𝜏−ℎ = 𝜇 𝑑𝑢

𝑑𝑦 −ℎ= −𝜏+ℎ > 0 ; 𝜏0 = 0

Exercise : Fully developed, pressure driven flow inside a constant diameter pipe has theshown parabolic velocity profile. For a Newtonian fluid with viscosity and centerlinevelocity of Vmax , calculate the force exerted by the fluid on the pipe wall over a pipesection of length L.

2𝑅𝑉𝑧 = 𝑉𝑚𝑎𝑥 1 −𝑟

𝑅

2

L

𝜏𝑤 = const

38

Viscosity (cont'd)

𝐻

Exercise : A concentric cylinder viscometer may be formed by rotating the inner cylinder of a pair or closely fitting cylinders. A torque of 0.15 Nm is required to turn the inner cylinder at 20 rad/s while keeping the outer cylinder fixed.Determine the viscosity of the Newtonian fluid in the clearance gap of theviscometer. Assume that the velocity distribution inside the gaps is linear.

c = 0.5 mm𝐻 = 0.2 mD = 0.1 m = 20 rad/sT = 0.15 Nm

= ?

c

c

D

𝜏𝑟𝜃 = 𝜇𝑑𝑉𝜃

𝑑𝑟

= 𝜇𝐷/2

𝑐= const

∴ 𝑇 = 𝜋𝐷𝐻𝜏𝑟𝜃

A missing term?resolution?

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Viscosity (cont'd)

• Definition of fluid: A fluid deforms continuously under the application of a shear (tangential) force, no matter how small the force is.

• It is more difficult to deform highly viscous fluids.

• Viscosity can be measured using

• capillary tube viscometer

• falling sphere viscometer

• concentric cylinder viscometer

• Saybolt viscometer

• Kinematic viscosity: ( ) [m2/s] Greek letter ‘‘nu’’

Air Water SAE 30 motor oil Honey Ketchup

𝑎𝑖r 50 𝑎𝑖r 25,000 𝑎𝑖r 500,000 𝑎𝑖r 2,500,000 𝑎𝑖r

= /

숙제: 다양한점도계에대해 Wikipedia에서조사하여 2 page로요약해서 1주후수업시간에제출

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Viscosity Behavior of FluidsFluids

Inviscid (ideal)

Viscous

Newtonian Non-Newtonian

Time independent Time dependent

Shear thinning (Pseudoplatic)

Bingham plastic

Shear thickening (Dialatant)

Thixotropic Rheopectic

Bingham plastic

Shear thinning

Shear thickening

Inviscid fluid

Elas

tic

solid

Newtonian

1

𝜏

𝑑𝑑𝑡

41

Viscosity Behavior of Fluids (cont'd)

• Newtonian behavior simple (it is linear). Common fluids such as water, air, oils

behave as Newtonian.

• Inviscid (ideal) fluids have = 0 and they do not exist in real world. It is an

idealization.

• Bingham plastics do not flow below a certain amount of shear stress. (toothpaste,

mayonnaise).

• Shear thinning fluid become thinner under increased shear stress. (wall paint,

blood).

• Shear thickening fluids become thicker under increased shear stress. (printing ink,

corn starch-water mixture, quicksand).

• For thixotropic fluids viscosity decreases with time (the longer the shear force is

applied) (lipstick).

• For rheopectic fluids viscosity increases with time (solidifying concrete).

42

Viscosity (cont'd)

• Effect of pressure on viscosity is small and often neglected.

• Viscosity of liquids decrease with increasing temperature.

• Viscosity of gases increase with increasing temperature.

• General (T ) relations

with A, B, C, D being fluid dependent constants

Exercise : As mentioned above, viscosities of liquids and gases change with temperaturein different ways due to two different mechanisms that affect their viscosities. What are these mechanisms? Hint: Think about temperature’s effect on the level of molecularactivity and on the magnitude of intermolecular attraction separately.

air

water

Andrade equation

Sutherland equation

T

= 𝐴𝑒𝐵𝑇

=𝐶𝑇1.5

𝑇 + 𝐷

43

Surface Tension

This non-surface molecule isbeing attracted equally bythe molecules all around it.

Greek letter ‘‘sigma’’

• ( 𝜎 ) [N/m]

• Surface tension is due to the asymmetric cohesive forces acting on the molecules at a free surface (interface between a liquid and a gas).

• This asymmetry will result in a hypothetical skin (membrane) all around the surface with a surface tension force acting tangential to it.

For this molecule on the surface, cohesive forces are not symmetric. It is being pulled inward.

44

Surface Tension

Example: Surface tension is known to create a pressure difference across a curved interface of two fluids. Consider a spherical liquid droplet suspended in a gas, such as a water droplet in air.

• The pressure difference Δp = p𝑖 − po can be related to the surface tension 𝜎 on droplet’s surface, and the radius of the droplet 𝑅 as follows.

• Forces acting on one-half of the droplet are

Surface tension force acts all along the surface

Net pressure force is due to the Δpdifference

Forces should be in balance for static equilibrium

𝜎 2𝑅 = Δp 𝑅2

Gas, po

Liquid droplet

p𝑖 Δp =?

Perimeter along which 𝜎acts

2𝜎→ Δp =

𝑅

Force per length

Projectedcircular areaover which Δp acts.

45

Surface Tension (cont'd)

Example: Capillary rise is observed when an open ended capillary (small diameter) tube is immersed into a liquid. This isrelated to surface tension.

ℎ

Curved meniscus

𝜎

• A curved meniscus will be formed inside the tube, with a Δp across.

• Surface tension force acting along the perimeter of the meniscus will support theforce due to the pressure difference across it.

Vertical force balance

𝜎

D𝜎

𝜎

Capillaryglass tube

ℎ =?

ℎ =4 𝜎 cos(𝜃)

gD

pup = patm

pdown = patm - gh

𝜌𝑔ℎ(𝜋𝐷2

4) = σ(𝜋𝐷)cos(𝜃)

46

Surface Tension (cont'd)

Example: Surface tension affects the shape of a liquid droplet put on a surface.

• The contact angle depends on the properties of the liquid,gas and the solid surface.

• Some surfaces repel the liquid (hydrophobic surfaces) and some surfaces attract the liquid (hydrophilic surfaces).

• Contact angle quantifies the wettability of a solid surface.

• A raincoat needs to be designed as hydrofobic, i.e. low wettability with high contact angle.

Gas

Liquid

Solid

Horizontal force balance gives

σ𝑙,𝑠 σ𝑔,𝑠

σ𝑔,𝑙

𝜎𝑙,𝑠 = 𝜎𝑔,𝑠 + 𝜎𝑔,𝑙 cos(-)