Chapter 1 Fundamentals · 2018. 4. 20. · • state: solid. liquid increasing spacing increasing...

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Chapter 1

Fundamentals

No-slip of real fluid

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Chapter 1 Fundamentals

Contents1.1 Scope of Fluid Mechanics

1.2 Historical Perspective

1.3 Physical Characteristics of the Fluid State

1.4 Units and Density

1.5 Compressibility, Elasticity

1.6 Viscosity

1.7 Surface Tension, Capillarity

1.8 Vapor Pressure

Objectives

- Present concept of fluidity

- Introduce fundamental properties of fluid

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1.1 Scope of Fluid Mechanics

• Problems of water supply

flood prevention

navigation need to know fluid phenomena

water power

irrigation

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• d'Alembert (1744)

"The theory of fluids must necessarily be based upon experiment"

• d'Alembert paradox theory - ideal, inviscid fluid

practice - real fluid (viscous)

• Two schools theoretical group → hydrodynamics

practical group → hydraulics

• Navier and Stokes

→ general equations for viscous fluid → equation of motion

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[Re] Navier-Stokes equation

Claude-Louis Navier (1785-1836, French engineer) and George Gabriel

Stokes (1819-1903, UK mathematician & physicist)

- one continuity equation + three momentum equations

- model the weather, ocean currents, water flow in a pipe, the air's flow

around a wing, and motion of stars inside a galaxy

- design of aircraft and cars, the study of blood flow, the design of power

stations, the analysis of pollution,

- exact solution - one of the seven most important open problems in

mathematics → “Gifted (2017)”

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( )2 2 2

2 2 2

13x x

p u u ug aqx xx y z

ρ µ µ ρ ∂ ∂∂ ∂ ∂

− + + =+ + ∇ ⋅ ∂ ∂∂ ∂ ∂

( )2 2 2

2 2 2

13y y

p v v vg aqy yx y z

ρ µ µ ρ ∂ ∂∂ ∂ ∂

− + + =+ + ∇ ⋅ ∂ ∂∂ ∂ ∂

( )2 2 2

2 2 2

13z z

p w w wg aqz zx y z

ρ µ µ ρ ∂ ∂∂ ∂ ∂

− + + =+ + ∇ ⋅ ∂ ∂∂ ∂ ∂

( ) ( ) ( ) 0u v wt x y zρ ρ ρ ρ∂ ∂ ∂ ∂+ + + =

∂ ∂ ∂ ∂

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• New problems in modern times

- dispersion of man's wastes in lakes, rivers, and oceans

→ Environmental Fluid Mechanics (Hydraulics)

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• state: solid

liquid increasing spacing increasing inter

fluid gaseous and latitude of molecular cohesive

plasma particle motion force

• fluid – continuum (연속체) → no voids or holes


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stressstrain

solid fluid

tension

elastic deformation

→ permanent distortion

unable to support tension

(surface tension)

compressionelastic deformation

(compressible fluid)

shear (tangential forces)

permanent distortion or flow

(change shape) to infinitesimal

shear stress

(연속적 변형 또는 흐르는 물질)

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stress

real fluid (viscous fluid; 실제유체)

ideal fluid

(non-viscous

Fluid; 이상유체)

in motion at restat rest and

in motion

compression

(pressure)○ ○ ○

shear ○ × ×

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incompressible fluid (비압축성 유체) compressible fluid (압축성 유체)

① Compressibility is of small important. ① Compressibility is predominant.

② Liquids and gases may be treated

similarly.

② Behavior of liquids and gases is quite

dissimilar.

③ Fluid problems may be solved with

the principles of mechanics.

③ Thermodynamics and heat transfer

concepts must be used as well as

principles of mechanics.

17/961.4 Units and Density

물리량 SI unit English Unit (FSS) Korean Unit

Length (L) metre (m) feet (ft) 자=0.3m, 리=392m

Mass (M) kilogram (kg) slug (-) 근=600g

Time (t) second (s) second (s) 경=2시간, 나절=6시간

Temp. (T) kelvin (K)degree Rankine (°

R)도

Basic units (기본단위)

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￭ International System of Units; SI system – metric system (국제표준단위계)

[Cf] English Unit System (영국계 단위계)

f• Frequency ( ): hertz (HZ = s-1) F

F ma=

2/ /a v t L t= = 2Lt− /v L t= 1Lt−

21kg m/s 1N( )F Newton∴ → ⋅ =

• Force,

→ introduce Newton's 2nd law of motion

Force = mass × acceleration

(m/s2)

(m/s)

￭ 기본단위 vs 유도단위

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E2 2kg m /s ( )E FL J Joule= → ⋅ =

P2 3/ / kg m sP E t J s= → = ⋅

p ,σ τ2 2/ N/m Pa (pascal) kg/m sp F A= → = = ⋅

T

• Energy, (work)

• Power,

• Pressure, ; Stress,

• Temperature, : degree Celsius (°C)

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ρ

3kg/mMV

ρ = →

• Density (밀도),

= mass per unit volume → heaviness

~ depends on the number of molecules per unit of volume

~ decreases with increasing temperature

γ

3 2 2N/m kg/m sWV

γ = → = ⋅

• Specific weight (weight density; 단위중량),

= weight (force) per unit volume

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(1.1)

W Mg=g

gγ ρ∴ =

[Re](Newton’s 2nd law of motion)

= acceleration due to gravity

1/ ρ• Specific volume=volume per unit mass=

• Specific gravity, s.g. , ~ r. d. (relative density; 비중)

= ratio of density of a substance to the density of water at a specified

temperature and pressure

. . f fw w

s gρ γρ γ

= =

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[Re] s.g. of sea water = 1.03

s.g. of soil = 2.65

s.g. of mercury = 13.6

SI unit English unit

1,000 kg/m3 1.94 slugs/ft3

9,806 N/m3 62.4 lb/ft3

9.81 m/s2 32.2 ft/s2

ρ

γ

g

Water at 5 °C

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Density measurement

Pycnometer (specific gravity bottle; 비중병)– Equipment to weigh accurately a known volu

me of liquid.– The specific weight of the liquid,

– W2 is total weight of the pycnometer,– W1 is the weight of empty pycnometer– γt is specific weight of the liquid

2 1t

W WV

γ−

=

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Density measurement

Hydrometer (비중계)– Float with different immersions

in liquids of different densities

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α

β,γ Γ

,δ ∆εζη

,θ Θ

ι

• Greek Alphabet

Alpha angle

Beta [beitə] angle

Gamma specific weight, circulation

Delta thickness of boundary layer

Epsilon eddy viscosity, height of surface roughness

Zeta

Eta

Theta

Iota [aioutə]

Kappa [kæpə]κ

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,λ Λ

µ

νξοπ

ρ

,σ ∑

τ

Lambda

Mu [mju:] dynamic viscosity

Nu kinematic viscosity

Xi [gzai, ksai] vorticity

Omicron

Pi [pai]

Rho mass density

Sigma Sigma Xi, Scientific Research Society, 1886

honor society for scientists & engineers

Tau shear

Upsilon,υ ϒ

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,ϕ Φ

χ

,ψ Ψ

,ω Ω

Phi [fai] Phi Beta Kappa

Chi [kai]

Psi [psai, sai] stream function

Omega angular velocity

• Prefixes

E exa 1018

P peta 1015

T tera 1012

G giga 109

M mega 106

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µ

M mega 106

k kilo 103

h hecto 102

da deca 101

d deci 10-1

c centi 10-2

m milli 10-3

micro 10-6

n nano 10-9

p pico 10-12

f femto 10-15

a atto 10-18

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유체역학 계산 (문제풀이)• 동차성의 원리

- 모든 항은 동일한 차원을 가져야 함. 또한 동일한 단위계로 계산하여야 함

• 정확도

- 계산과정에서 유효숫자가 4자리 이상으로 표시되더라도,해답은 3자리로 반올림함.

이유는 유체의 물성데이터나 실험 결과의 정확도가 유효숫자 3자리를 넘지 않기 때문임

• 문제풀이

- 주어진 데이터를 나열하고, 문제 풀이에 필요한 그림(개념도, 그래프)을 그림

- 유체 원리에 해당하는 수학적 식을 쓴다.

- 식에 대입하는 수치들의 단위를 포함하고 동차성의 원리를 따르는지 점검

- 식을 풀고 해답은 유효숫자 3자리로 표기함

- 구한 답이 공학적인 상식에 부합하는 지 판단함. 또한 의미를 찾아서 자신만의 토론

을 붙임

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• Elastic behavior to compression

• Compressibility (압축성) ≡ change in volume due to change in pressure

solid - modulus of elasticity, E (N/m2)fluid - bulk modulus

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1

dVdpV

∝1

dVdp EV

→ = −

1

1

( , )dp dpE V const fn p TdV dVV

= − = − ≠ = → p E↑ → ↑

1

1 1dVCE V dp

= = −

• Stress-strain curve (E↑, difficult to compress)

= modulus of compressibility (m2/N)

Minus means that increase in

pressure causes decrease in

volume

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, 1E C= ∞ ρ

[Re] large E or small C → less compressible• incompressible fluid (inelastic):

→ constant density =const.

~ water is incompressible fluid.

• compressible fluid

→ changes in density → variable density

~ gas

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Pressure

106 N/m2

Temperature, °C

0° 20° 50° 100° 150°

0.1 1950 2130 2210 2050

10.0 2000 2200 2280 2130 1650

30.0 2110 2320 2410 2250 1800

100.0 2530 2730 2840 2700 2330

Bulk modulus of water, E (106 N/m2)


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Compressibility Modulus of Elasticity, E (kPa)

steel 1/80 of water water 2,170,500

mercury 1/12.5 of water sea water 2,300,000

nitric acid 6 of water mercury 26,201,000

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E

E

- increases as pressure increases.

- is maximum at about 50 °C.

→ The water has minimum compressibility at about 50 °C.

1dpE VdV

= −

1

V pV E∆ ∆

≈ −

2 1 2 1

1

V V p pV E− −

≈ −

• For the case of a fixed mass of liquid at constant temperature

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62,200 10 PaE = ×6

2 1 7 10 Pap p= + ×

2 1 2 1

1

0.0032V V p pV E− −

∴ ≈ − = −

2 1(1 0.0032)V V∴ = −

0.3%V∆ ≈

[Ex] For water; @ 20˚C

decrease

→ water is incompressible

63

7 10 Pa 70101.3 10 Pap×∆ = ≈

×기압

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1.6 Viscosity

[Re] From Wikipedia

Viscosity is a measure of the resistance of a fluid which is being

deformed by either shear stress or tensile stress. (상대적인 운동에 저항

하는 성질)

Viscosity ~ "thickness" or "internal friction"

▪ water ~ "thin (묽다)", having a lower viscosity

▪ honey ~ "thick (진하다)", having a higher viscosity

The less viscous the fluid is, the greater its ease of movement (fluidity).

Viscosity describes a fluid's internal resistance to flow and may be

thought of as a measure of fluid friction.

dvdy

τ µ=

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1.6 Viscosity

For example, high-viscosity felsic (규장질) magma will create a tall, steep

stratovolcano, because it cannot flow far before it cools, while low-viscosity

mafic (철질암) lava will create a wide, shallow-sloped shield volcano.

All real fluids (except superfluids) have some resistance to stress and

therefore are viscous, but a fluid which has no resistance to shear stress is

known as an ideal fluid or inviscid fluid.

[Re] super fluid – a fluid having frictionless

flow, and other unusual properties

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1.6 Viscosity

• Two types of fluid motion (real fluid)

1) laminar flow:

- viscosity plays a dominant role

- fluid elements or particles slide over each other in layers (laminar)

- molecular diffusion

[Ex] flow in a very small tube, a very thin flow over the pavement, flow in

the laminar flow table

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1.6 Viscosity

Re Vdν

=

where V = flow velocity; d = characteristic length; ν = kinematic viscosity

diameter of pipe,depth of stream

2) turbulent flow:

- random or chaotic motion, eddies of various sizes are seen

- common in nature (streams, rivers, pipes)

- large scale mixing between the layers

• Reynolds number

[Ex] flows in the water supply pipe, flows in the storm sewer pipe, flows in

the canals and streams

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1.6 Viscosity

• Reynolds experiments

laminar flow: Re < 2,100

transition: 2,100 4,000

The same fluid withdifferent velocity

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1.6 Viscosity

cubestreamline

source

sink

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1.6 Viscosity

Smoothsurface

Eddyingmotion

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1.6 Viscosity

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1.6 Viscosity

Symmetric shape,No separation

Re < 1

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1.6 Viscosity

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1.6 Viscosity

Separation, eddyformation

Growthof eddy

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1.6 Viscosity

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1.6 Viscosity

후방난류

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1.6 Viscosity

Spiralsecondaryflow

Secondaryflow

대규모와류

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1.6 Viscosity

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1.6 Viscosity

• Deformation of solid between two plates

Shear stress total angular displacement

yxdGdyζτ =

∝

yxddyζτ ∝

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1.6 Viscosity

• Laminar flow between two plates

2 1d d dvdt dv dtdy dy dy−

= = =

2 2 1 1

2 1 2 1

;( )

d v dt d v dtd d v v dt

= =− = −

strain = angular displacement (전단변형도)

[Re]

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1.6 Viscosity

τ• Experiment has shown that, in many fluids, shearing (frictional) stress

per unit of contact area, is proportional to the time rate of relative

strain (Newton).

/dv dvdt dtdy dy

τ∴ ∝ =

dvdy

τ µ= →

(velocity gradient)

Newton’s equation of viscosity (1.2)

µwhere = coefficient of viscosity

= dynamic (absolute) viscosity (점성계수)

• Viscosity ~ measure of fluid's resistance to shear or angular deformation

~ internal resistance of a fluid to motion (fluidity)

Large → sticky, difficult to flowµ

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1.6 Viscosity

du duu y t u t y tdy dy

= + ∆ ∆ − ∆ = ∆ ∆

/du duy t y tdy dy

= ∆ ∆ ∆ = ∆

[Re] rate of angular deformation (전단변형도의 시간변화율)

(i) displacement of AB relative to CD

(ii) angular displacement of AC

(iii) time rate of angular deformation

/du dut tdy dy

= ∆ ∆ =

dvdy

τ µ=

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1.6 Viscosity

µ• dynamic viscosity,

/F Aτ =

[ ] 2 2 1 2 2/ kg/(m s ) PaMLt L ML tτ − − − = = ⋅ = 1

1dv Lt tdy L

−− = =

[ ]1 2

1 1 21/ kg/m s N s / m Pa s

dv ML t ML tdy t

µ τ− −

− −−

∴ = = = = ⋅ = ⋅ = ⋅

11 ( ) 10 Pa spoises Poiseuille −⇒ = ⋅

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1.6 Viscosity

ν• kinematic viscosity (동점성계수),µνρ

=

[ ]1 1

2 1 23 m /s

ML t L tML

ν− −

−−

= = =

(1.3)

1 m2/s = 104 stokes =106 centistokes

,τ µ

τ τ

• Remarks on Eq. (1.2)

① are independent of pressure.

[Cf] friction between two moving solids

② Shear stress (even smallest ) will cause flow (velocity gradient).

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1.6 Viscosity

0 0dvdy

τ= → = µregardless of

③ Shearing stress in viscous fluids at rest will be zero.

dvdy

≠ ∞ τ→ ≠ ∞④ At solid boundary, ( (no infinite shear))→ Infinite shearing stress between fluid and solid is not possible.

⑤ Eq. 1.2 is limited to laminar (non-turbulent) fluid motion in which

viscous action is predominant.

[Cf] For turbulent flow

ε µ( )dvdy

τ µ ε= +

where ε = eddy viscosity

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1.6 Viscosity

⑥ Velocity at a solid boundary is zero.

→ No-slip condition (continuum assumption; 점착조건, 비활조건)

• Newtonian and non-Newtonian fluids

i) Newtonian fluid ~ water

ii) Non-Newtonian fluid ~

plastic, blood, suspensions, paints,

polymer solutions → rheology

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1.6 Viscosity

1dvdy

τ τ µ− =

ndvKdy

τ

=

1τ

1n >

1n <

• Non-Newtonian fluid

1) plastic, = threshold

2) Shear-thickening fluid

Shear-thinning fluid

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1.6 Viscosity

• Couette flow: laminar flow in which the shear stress is constant

- thin fluid film between two large flat plates

- thin fluid film between the surfaces of coaxial cylinders

dv Vdy h

=

Vh

τ µ∴ =

~ linear velocity gradient

~ constant

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1.6 Viscosity

[IP 1.1] Couette flow

Upper plate is moving on the oil (µ = 0.0652 N·s/m2) with the constant velocity

0.2 m/s. Find force F. The thickness of oil is 0.1 mm, and the contact area of the upper plate is 75 cm2.

[Solution]

2 23

2 2 2

(0.2 / )(0.0652 / ) 130.4 /(0.1 10 )

(130.4 / )(75 10 ) 97.8 ~10

F Adv Vdy h

m sN s m N mm

F N m m N kg

τ

τ µ µ

τ −

−

=

= =

= ⋅ =×

= × = 중

64/961.6 Viscosity

gas liquid

main cause of

viscosity

exchange of molecule's momentum →

interchange of molecules between

the fluid layers of different velocities

intermolecular cohesion

effect of

temperature

variation

temp↑ → molecular activity↑

→ viscosity↑ → shearing stress↑

temp↑ → cohesion↓

→ viscosity↓ → shear stress↓

[Re] Viscosities of gas and liquid

Friction forces result from

- cohesion for liquid

- momentum interchange between molecules for gas

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1.6 Viscosity

Two layers tendto stick togetheras if there issome viscositybetween two.

[Re] Exchange of momentum

fast-speed layer (FSL)

molecules from FSL speed up molecules in LSL

molecules from LSL slow down molecules in FSL

low-speed layer (LSL)

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1.6 Viscosity

[Re]

- Exchange of momentum (운동량 교환): exchange momentum in either

direction from high to low or from low to high momentum due to random

motion of molecules ~ molecular scale

- Transport of momentum (온동량 전달): transport of momentum from

layers of high momentum (high velocity) to layers of low momentum ~

continuum scale

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Viscosity measurement

Rotational Viscometers

Outer cylinder is rotating at constant angular velocity (각속도), ωMeasure torque (회전모멘트) on the inner cylinder

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Rotational Viscometers- Measure torque, T on the inner tank

2 322

R h V R VTR h

π µ π µ= +

∆ ∆

T F r Ardvdy

τ

τ µ

= =

=

1

1

1

2

1

2

2

2

2

3

2

2

2

2

2

i

i

i

i

i

i

T Fr Ar

dV V

dy R

A R h

r R

R hVT

RV

h

A R

Rr

R VT

h

τ

τ µ µ

π

πµ

τ µ

π

πµ

= =

= =∆

=

=

=∆

=∆

=

=

=∆

10

Viscosity measurement

V Rω=

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• Surface tension

- occur when the liquid surfaces are in contact with another fluid (air) or

solid

- Forces develop in the liquid surface that cause the surface to behave as

if it were a “skin” or “membrane” stretched over the liquid mass

• Engineering problems related to surface tension

- capillary rise of liquids in narrow spaces

- formation of liquid drops

- small models of larger prototype

→ dam, river model

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σ /F L• surface tension, ( , N/m)- force per unit length

- force attracting molecules away from liquid

- function of (relative sizes of intermolecular cohesive and adhesive

forces to another body)

- as temp↑ → cohesion↓ → σ↓ ☞ Table A2.4b, p. 694

72/961.7 Surface Tension, Capillarity

Consider the small element dx dy of a surface of double curvature with radii R1and R2

0F∑ =

0( ) 2 sin 2 sinip p dxdy dy dxσ α σ β− = +

where pi = pressure inside the curvature; po = pressure inside the

curvature

1

sin ,2dxR

α ≈2

sin ,2dyR

β ≈ 12( sin )dx R α=

01 2

1 1ip p R R

σ

∴ − = +

(1.4)

→ for the force components normal to the element

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• Cylindrical capillary tube

- due to both cohesion (응집력) and adhesion (부착력)

cohesion < adhesion → rise (water)

cohesion > adhesion → depression (mercury)

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1 2R R R= = ≈

0p hγ= −

0ip =

For a small tube, given conditions are as follows

(liquid surface section of sphere)

(hydrostatic pressure)

(atmospheric)

01 2

1 1ip p R R

σ

− = + 2h

Rγ σ∴ =

Substitute above conditions into Eq. 1.4: (1.5)

cosr R θ=2 2 cos

/ cosh

r rσ θγ σ

θ∴ = =

2 coshr

σ θγ

=

By the way,

(1.6)

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h r hθr ≤

in which = capillary rise → ↑ → ↓

= angle of contact

= radius of tube 2.5 mm for spherical form

12r > h

[Ex] water and mercury

If mm, is negligible for water.

• Pressure measurement using tubes in hydraulic experiments

→ Ch.2 manometer

~ capillarity problems can be avoided entirely by providing

tubes large enough to render the capillarity correction negligible.

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• Formation of curved surface, droplet

- At free liquid surface contacting the air, cohesive forces

at the outer layer are not balanced by a layer above.

→ The surface molecules are pulled tightly to the lower layer.

→ Free surface is curved.

[Ex] Surface tension force supports small loads (water strider).

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0 1.0ip p− =

0.0728σ =At 20°C, N/m ← App. 2

[IP 1.10] For a droplet of water (20 °C), find diameter of droplet

Given: kPa

01 2

1 1 2ip p R R R

σσ

− = + =

∴1R⋅

0.000146m 0.146mmR∴ = = 0.292d→ =

[Sol]

(1.4)

1×103 N/m2 = 2(0.0728)

mm

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[IP 1.11] Find height of capillary rise in a clean glass tube of 1 mm

diameter if the water temperature is 10°C or 90°C.

[Sol]

From App. 2 Table A 2.4b;

σ γ

σ γ@ 10°C =0.0742 N/m, = 9.804 kN/m3

@ 90°C =0.0608 N/m, = 9.466 kN/m3

80/96


Use Eq. 1.16

For water, 0θ =

2 coshr

σ θγ

=

1̀02(0.0742)(1) 0.030m=30mm

9804(0.0005)h = =

902(0.0608)(1) 0.026m=26mm

9466(0.0005)h = =

(1.5)

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1.8 Vapor Pressure

• Vapor pressure is the partial pressure exerted by ejected

molecules of liquid.

• 증기압 – 분자의 운동에너지에 기인

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1.8 Vapor Pressure

• Liquids tend to vaporize or evaporate due to molecular thermal

vibrations (molecular activity).

- Evaporation takes place because some liquid molecules at the

surface have sufficient momentum to overcome the intermolecular

cohesive forces and escape into the atmosphere.

- If the container is closed with a small air space left above the surface,

a pressure will develop in the space as a result of the vapor that is

formed by the escaping molecules.

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1.8 Vapor Pressure

55.2 kPavp =2.34 kPavp =

0.00017 kPavp =

• volatile liquids (휘발성 액체):

~ easy to vaporize → high vapor pressure

gasoline: at 20 °C

water: at 20 °C

mercury: at 15.6 °C

• mercury: low vapor pressure and high density → difficult to vaporize

→ suitable for pressure-measuring devices

• Vapor pressure → Table A2.1 and A2.4b

temperature↑ → molecular activity↑ → vaporization↑ → vapor pressure↑

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1.8 Vapor Pressure

In the interior and/or boundariesof a liquid system

• Cavitation (공동현상): App. 7 (p. 672)

In a flow fluid wherever the local pressure falls to the vapor pressure of

the liquid, local vaporization occurs.

→ Cavities (공동) are formed in the low pressure regions.

→ The cavity contains a swirling mass of droplets and vapor.

→ Cavities are swept downstream into a region of high pressure.

→ Then, cavities are collapses suddenly.

→ it causes erosion (pitting) of solid boundary surfaces in machines, and

vibration

→ boundary wall receives a blow as from a tiny hammer

[Cf] 지하동공 (싱크홀) -> 지반함몰

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1.8 Vapor Pressure

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1.8 Vapor Pressure

• Prevention of cavitation

~ cavitation is of great importance in the design

of high-speed hydraulic machinery such as

turbines, pumps, in the overflow and underflow

structures of high dams, and in high-speed

motion of underwater bodies (submarines,

hydrofoils).

→ design improved forms of boundary surfaces

→ predict and control the exact nature of

cavitation → set limits

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1.8 Vapor Pressure

atm vp p≤

• Boiling:

= rapid rate of vaporization caused by an increase in temperature

= formation of vapor bubbles throughout the fluid mass

~ occur (whatever the temperature) when the external absolute pressure

imposed on the liquid is equal to or less than the vapor pressure of the

liquid

~ boiling point = fn (imposed pressure, temp.)→ boiling occurs

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1.8 Vapor Pressure

altitude

(El. m)

(kPa),

absolute

(kPa),

absolute

boiling point

(°C)remark

m.s.l. 101.3 101.3 at 100°C 100 well cooked

12,000 19.4 19.9 at 60°C 60 undercooked

vpatmp

• Boiling point of water at high altitude

- 해발이 낮은 곳에서 대기압은 101.3 kPa 이나,- 높은 산에 올라가면 공기가 희박하여 기압이 낮아짐- 해발 12,000 m에서는 대기압이 19.4 kPa로 떨어짐 (Table A2.5b)- 한편 물의 증기압은 온도의 함수로서 기온이 60 °C 일 때 19.9 kPa 임 (Table A2.4b)- 따라서 해발 12,000 m에서는 물이 60 °C 에서 끓기 시작함.

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1.8 Vapor Pressure

• Evaporation (증발): When the space surrounding the liquid is too large,

the liquid continues to vaporize until the liquid is gone and only vapor

remains at a pressure less than or equal to pv.

[IP 1.12] For a vertical cylinder of diameter 300

mm, find minimum force that will cause the water

boil. (patm = 100 kPa)

patm

F

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1.8 Vapor Pressure

'vp p≤

' 100 31.16FpA

∴ = − =

2(0.3)(100 31.16) 4.87 kN4

F π∴ = − =

[Sol] From Table A2.4b: At 70 °C, Pv = 31.16 kPa; patm = 100 kPa

For water to boil;

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Homework Assignment # 1

Due: 1 week from today

1. (Prob. 1.10)

A rocket payload with a weight on earth of 8,900 N is landed on the moon

where the acceleration due to the moon’s gravity gm ≈ gn /6. Find the mass of

the payload on the earth and the moon and the payload’s moon weight.

2. (Prob. 1.27)

The flux of momentum F of a flow through a pipe is F = QγV / gn . Given

that Q is the flowrate with dimensions of ( L3 / t ), what are the dimensions of

F ?


92/96

3. (Prob. 1.46)

Calculate velocity gradients for y = 0, 0.2, 0.4, and 0.6 m from the

boundary, if the velocity profile is a quarter-circle having its center 0.6 m

from the boundary.


93/96

4. (Prob. 1.49)

The velocity profile for laminar flow in a pipe is

where r is measured from the centerline where υ = υcand R is the pipe radius. Find τ(r) and dυ / dr . Plot both versus r.

( ) 2 2 1 /( )– c rr Rυ υ=


94/96

5. (Prob. 1.60)

Crude oil at 20°C fills the space between two concentric cylinders 250

mm high and with diameters of 150 mm and 156 mm. What torque is

required to rotate the inner cylinder at 12 rpm, with the outer cylinder

remaining stationary? Ignore end effects.


95/96

6. (Prob. 1.69)

The weight falls at a constant velocity of 50 mm/s in the cylinder.

Calculate the approximate viscosity of the oil.


96/96

7. (Prob. 1.72)

Calculate and plot the maximum capillary rise of water (20℃) to be

expected in a vertical glass tube as a function of tube diameter (for

diameters from 0.5 to 2.5 mm).

8. (Prob. 1.82)

At what temperature will water boil at an altitude of 6,100 m? See

Appendix 2.


Chapter 1Chapter 1 Fundamentals1.1 Scope of Fluid Mechanics��슬라이드 번호 4슬라이드 번호 5슬라이드 번호 6슬라이드 번호 71.2 Historical Perspective��1.2 Historical Perspective��1.2 Historical Perspective��1.2 Historical Perspective��1.3 Physical Characteristics of the Fluid State��1.3 Physical Characteristics of the Fluid State��1.3 Physical Characteristics of the Fluid State��1.3 Physical Characteristics of the Fluid State��1.3 Physical Characteristics of the Fluid State��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��Density measurementDensity measurement 1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.4 Units and Density��1.5 Compressibility, Elasticity��1.5 Compressibility, Elasticity��1.5 Compressibility, Elasticity��1.5 Compressibility, Elasticity��1.5 Compressibility, Elasticity��1.5 Compressibility, Elasticity��1.5 Compressibility, Elasticity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��1.6 Viscosity��Viscosity measurement슬라이드 번호 681.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��Consider the small element dx dy of a surface of double curvature with radii R1 and R21.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.7 Surface Tension, Capillarity��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��1.8 Vapor Pressure��슬라이드 번호 91Homework Assignment # 1��Homework Assignment # 1��Homework Assignment # 1��Homework Assignment # 1��Homework Assignment # 1��

Chapter 1 Fundamentals · 2018. 4. 20. · • state: solid. liquid increasing spacing increasing...

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Transcript of Chapter 1 Fundamentals · 2018. 4. 20. · • state: solid. liquid increasing spacing increasing...