1

流體力學 (Fluid Mechanics)(Spring 2009)

教科書 (Text Book)

Munson, B. R., Young, D. F., and Okiishi, T. H., Fundamentals of Fluid Mechanics, 5th edition, John Wiley & Sons (Asia) Pte Ltd, 2006.

Pte Ltd = Private Limited

Chapter 1

Basic Properties of Fluids

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Chapter 1 Basic Properties of Fluids

Fluid mechanics( 流体力學 ) 是什麼 ?

－屬於 applied mechanics( 應用力學 ) 之領域 －研究氣体及液体在靜止及運動之行為 (Behavior). －所研究的流體種類包括，由血液至原油，其所流經 的管徑大小，由幾個 microns(10-6m) 至 4 ft 直徑之油 管 (800 miles長 ) 。 －研究問題範圍 為何飛機的表面需要平滑 ? 而高爾夫球的表面須要粗糙 ( 或 dimpled) 以增加其飛行效率 ?

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有些很有趣的問題可以利用很簡單流體力學之觀念去回答，例如：

－火箭為何能在外太空中沒有空氣裡可以飛行？－何以超音速飛機，經過你之後才聽到聲音？－何以在河床平坦之處的流水仍然流動很大？－水龍頭流出之水，何以有時候平滑，有時候為不平滑 之流動？－何以能夠利用飛機模型去研究飛機建造之設計？－何以流線型的汽車或卡車設計，能夠增加汽油使用的 里程數？

流體力學是一門非常重要而實際的課題。

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§1.1 Some characteristics of fluids

－ What is a fluid?

－ What is the difference between a solid and a fluid?

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What is the difference between a solid and a fluid?

Ans: Descriptive answer: Solid － "hard" ; not easily deformed Fluid － "soft" ; easily deformed

From molecular structure of materials: Solid － densely spaced molecules with large intermolecular cohesive forces to maintain its shape, and to not be easily deformed.

Fluid － the molecular are spaced further apart, the intermolecular forces are smaller than for solid, and the molecules have more freedom of movement.

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流體 (Fluid) 包括液体 (Liquid) 及氣體 (Gases) ，其區別如下：

Liquids － easily deformed, not easily compressed, － free surface.

Gases － greater molecular spacing and freedom of motion with negligible cohesive intermolecular forces. － no free surface

8

Definition of fluid and how fluids deform :

A fluid is defined as a substance that deforms continuously flow when acted on by a shearingstress of any magnitude.

A shearing stress (F/A) is created whenever a tangential force acts on a surface.

When common solids such as steel or other metals are acted on by a shearing stress, they will initially deform (usually a very small deformation),but they will not continuously deform(flow).

What is a fluid?

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Rheology:

Some materials, such as slurries, tar, putty, toothpaste, and so on, are not easily classified since they will behave as a solid if the applied shearing stress is small.

But if the stress exceeds some critical value, the substance will flow. The study of such material flow is called rheology.

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Fluid continuum

Although the molecular structure of fluids is important in distinguishing one fluid from another, it is not possible to study the behavior of individual molecules when trying to describe the behavior of fluid at rest or in motion.

Rather, we characterize the behavior by considering the average, or macroscopic, value of the quantity of interest, where the average is evaluated over a small volume containing a large number of molecules.

The spacing for gases at normal P&T is on the order of 10-6 mmThe spacing for liquid at normal P&T is on the order of 10-7 mmThe number of molecules/mm3 is on the order of 1018 for gases. 1021 for liquids.

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Thus, we assume that all the fluid characteristics we are interested in (pressure, velocity, etc) vary continuously throughout the fluid － that is, we treat the fluid as a continuum.

One area of fluid mechanics, for which the continuum concept breaks down, is in the study of rarefied gases such as would be encountered at very high altitudes.

Spacing between air molecules can become large

The continuum concept is no longer acceptable

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§1.2 Dimensions, Dimensional Homogeneity, and Units

It is necessary to develop a system for describing fluid characteristics both qualitatively and quantitatively.

The qualitative aspect serves to identify the nature, or type of the characteristics -- such as length, time, stress, and velocity.

The quantitative aspect provides a numerical measure of the characteristics. The quantitative description required both a standard (or a unit) and a number.

A standard (or unit) for a length might be a meter or foot for time might be an hour or second for mass might be a slug or kilogram

14

Basic dimensions (or primary quantity):

The qualitative description is conveniently given in terms of certain primary quantities, such as

Length (L), Time (T), Mass (M), and Temperature(θ) MLT θ systemor Length (L), Time (T), Force (F), and Temperature(θ) FLT θ systemor Length (L), Time (T), Mass (M), Force (F), and Temperature(θ) FMLT θ system

15

Secondary dimensions(or secondary quantity):

Secondary quantities are based on primary quantities (or basic dimensions) , such as area [=] L2, velocity [=] LT-1, density [=] ML-3 and so on v ≐ LT-1 or v[=]LT-1

"The dimensions of a velocity equal length divided by time"

16

Secondary quantities

v (velocity) = L/T LT≐ -1

(stress) = F/A [=] FL-2 or (stress) = F/A [=] ML-1T-2

Table 1.1 (p.4)

a list ofdimensions

17

Table 1.1 (p. 4)Dimensions Associated with Common Physical Quantities (continued on next slide)

18

Table 1.1 (p. 4)continued

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Dimensional Homogeneity:

All theoretically derived equations are dimensionally homogeneous.

Dimensional homogeneity unit [=] unit

Same

unit [=] unit + unit

Same Same

20

For example:

v = v0 + at -----(1.1) dimensional homogeneity LT-1 [=] LT-1 + LT-2 T LT-1 [=] LT-1 + LT-1

d = ------(1.2) dimensional homogeneity where g = 32.2 ft/s2

d = 16.1 t2 ------------------(1.3) 16.1 is a constant with dimensions of LT-2

Eq(1.1) & Eq(1.1) are general homogeneous equation Eq(1.3) is restricted homogeneous equation

Dimensional analysis (in Chapter 7)

2

gt 2

21

Example 1.1 Investigate the dimensional homogeneity of the following formula:

Q = 0.61A

Solution:

On next page

gh2

22

Q = 0.61A

Left Side [=] L3T-1

Right Side [=] 0.61(L2)

Right Side [=] Left Side = pure constant

Thus Q = 0.61A is a general dimensional homogeneity

(valid for any constant system of units)

)L)(LT(2 2

2/12/122 )()(261.0][ LLTL 13261.0][ TL

261.0

gh2

gh2 Q = volume rate of flow(volume/time) [=] L3T-1

A = area [=] L2 g = acceleration of gravity [=] LT-2

h = height [=] L

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Alternately, Q = 0.61A = 0.61A where g = 32.2 ft/s2

Q = 4.90A

Left Side [=] L3T-1

Right Side [=] 4.90L2L1/2

[=] 4.90L5/2

for left side = right side L3T-1 [=] 4.90L5/2

4.90 [=] L3T-1L-5/2 [=] L1/2T-1

it is not a constant, instead it has a dimension of L1/2T-1

(or ft1/2/s)

Thus Q = 4.9A is a restricted homogeneous equation (valid for a specific system of units)

gh2h2.322

h

h

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§1.2.1 System of Units

For example, if we measure the width of a black board and say 10 units wide --------no meaning 10 meters wide ------meaningful “meter”--- standard length

A unit must be established for each of the remaining basic quantities( length, force, mass, time, and temperature).There are several systems of units in use and we shall consider three systems that are commonly used in engineering.

British Gravitational (BG) system International System (SI) English Engineering (EE) System

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*British Gravitational (BG) System -- BG units (FLT )

Basic Length (L) --- foot (ft)Dimensions Time (T) --- second (s) Force (F) --- pound (lbf or lb) Mass (M) --- slug (slug) Mass is defined from Newton's Second law

force = mass * acceleration or 1 lbf = 1 slug * 1 ft/s2 or 1 slug = 1 lbf / 1 ft/s2

Temperature() ---- Fahrenheit(0F) or Rankine(0R) 0R = 0F + 459.67

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*International System (SI) ---- SI units (MLT )

In 1960 the Eleventh General Conference on Weights and Measures, the international organization responsible for maintaining precise uniform standards of measurements, formally adopted the international system of Units as the international standard

Length (L) ---- meter(m) Time (T) ----- second (s) Mass (M) ---- kilogram (kg)

Temperature() ----- kelvin (k) ºk wrong k = 0C+273.15

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Force(F) ------ newton (N) F = m a 1 N = (1kg) (1m/s2)

weight W = m g W(N) = m(kg) g(m/s2) where g = 9.807 m/s2 = 9.81m/s2

work W = F S W(Joule, J) = F(N) S(m)

power P = F V P(watt) = F(N) V(m/s)

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*English Engineering (EE) System --- EE Units (FMLT )

Length(L) --- foot(ft)Time(T) --- second(s)Mass(M) --- pound mass(lbm)Temperature(θ) --- degree Rankine(0R)Force(F) --- pound(lbf)

Newton's Second law

weight W=mg/gc )//(

)/()()(

2

2

fmc

mf lbsftlbg

sftglbmlbW

cg

amF

)lb/s/ftlb(g

)s/ft(a)lb(m)lb(F

f2

mc

2m

f

where gc=32.174 lbm ft / s2 / lbf

29

Compare

In BG unit In EE unit

1slug=32.174 lbm

Units in this textbook

BG Units ***Using a consistant system of units SI Units throughout a given solution.EE Units ---sparingly *Conversion tables: Table 1.3;1.4; 1.5~1.8 (inside of back & front covers)

)s/ft(a)slug(m)lb(F 2f

)lb/s/ftlb(174.32

)s/ft(a)lb(m)lb(F

f2

m

2m

f

amF

30

System of Units BG(FLT )

SI(MLT )

EE(FMLT )

British Gravitational

System

International System

English Engineering

System

Length(L) ft(foot) m(meter) ft(foot)

Time(T) S(second) S(second) S(second)

Mass(M) slug(slug) kg(kilogram) lbm(pound mass)

Temperature() Raukine(0R)0R=0F+459.67

Kelvin(k)k=0c+273.15

Raukine(0R) or Fahrenhert(0F)

   

Lbf (Pound force)Force Lbf (Pound force) N (Newton)

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cg

amF

f

2m

c lb

s/ftlb174.32g

fkgForce ][

c

2

f g

s/m8.9kg][kg

f

2

c kg

s/kgm8.9g

  1lbf=1slug1ft/s2

1slug = 1 ft/s2 / 1 lbf

1Newton=1kg1m/s2  

 

  gc=1 gc=1

     

     

     

cg

amF

cg

amF

BG(FLT )

SI(MLT )

EE(FMLT )

1 pound-force = 0.453 kilogram-force

1 pound-force = 4.448 newton

1 newton = 0.224 pound-force

1 newton = 0.101 kilogram-force

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Example 1.2

Given:

tank Ff=?

kgmoH

362

2s/ft7a

w

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Solution: Newton's second law of motion to the water body

amF

mawFf

mamgFf

Nlb

Nlb

lb

lblb

sftkg

slugkgsft

kg

slugkg

mamgF

ff

f

ff

f

4301

448.472.96

72.96

27.1745.79

/759.14

136/2.32

59.14

136 22

table 1.3

table 1.3

35

Table 1.3 (back cover)Conversion Factors from BG and EE Units to SI Units

36

Table 1.4 (back cover)Prefixes for SI Units

37

§1.3 Analysis of Fluid Behavior

Fundamental laws involved in the study of fluid mechanics(same as you have encountered in physics and other mechanics courses)

---Newton's laws of motion (the 1st, 2nd, and 3rd laws)

---Conservation of mass

---the first and second laws of thermodynamics

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Fluid statics ------- the fluid is at restFluid mechanics Fluid dynamics --- the fluid is moving

Fluid properties Fluid behavior

For example

Compressible fluid Incompressible fluid

Viscous fluid Inviscous fluid

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Fluid properties:

Heaviness –density() –specific weight( γ ) –specific gravity(SG)

Fluidity –viscosity

Compressibility Vapor pressure Surface tension

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§1.4 Measures of Fluid Mass and Weight

§1.4.1 Density()

= m/v [=] slug/ft3 ( in BG units ) [=] kg/m3 ( in SI units )

= f ( different fluids ) Table 1.5 & 1.6 Δ liquid = f (Δ Pressure, Δ temp. ) = small Table B.1 p. 761

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= 1.94 slug/ft3

= 999.9 kg/m3

( or 1000 kg/m3 ) Table B.1 & B.2 p. 761

This property is not commonly

The specific volume, υ = 1/ used in fluid mechanics but

is used in thermodynamics.

oH2

gas

at 600F

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§1.4.2 Specific weight (γ )

γ = weight/volume = mg/ = g

γ = g [=] lbf/ft3 in BG Units Table 1.5 [=] N/m3 in SI Units 1.6

§1.4.3 Specific Gravity ( SG )

SG = at some specified temperature

SG = independent of the system of units used

C4at OH2

WaterofDensity

fluid theofDensity

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§1.5 Ideal Gas Law

Ideal (or perfect) gas law, or Equation of state

p = RT ------------------------ (1.8)

where P = absolute pressure = Normal force on a plan surface/area [=] lbf/ft2 abs ( psfa ) or lbf/in2 abs( psia ) in BG Units [=] N/m2 ( Pascal, Pa ) in SI Units

absolute pressure --- relative to absolute 0 pressure gauge pressure --- relative to the local atmospheric pressure

absolute pressure (psia) = gauge pressure (psig) + local atmospheric pressure

44

Standard Sea – level atmospheric pressure = 14.696 psia or 101.33 kpa ( abs. ) or 14.7 psia or 101 kpa ( abs. )

= density T = absolute temperature R = gas constant in BG Units R [=] in SI Units R[=] table 1.7 & 1.8 (front page)

)Rslug/(lbft 0f

)kkg/(J

45

46

47

Table 1.7 (front cover)Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (BG Units)

48

Table 1.8 (front cover)Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (SI Units)

49

Example 1.3 Given: Compressed air tank = 0.84 ft3 ; p = 50 psig, T = 700F, Patm = 14.7 psia Question: = ? w = ?

Solution: = p/RT

R = 1716 ft-lb/slug0R T = 700F = 70+459.6 = 529.60R

3ftslug/ 0.01025

6.5291716

8.9316

psia7.64psi7.14psi50psig50p

flb 0.277

84.02.3201025.0

gw

2f2

2

2f ft/lb8.9316

ft1

in144

in

lb7.64

50

Table 1.3 (back cover)Conversion Factors from BG and EE Units to SI Units

51

Table 1.4 (back cover)Prefixes for SI Units

52

§1.6 Viscosity

Density Specific weight Viscosity ------ "fluidity" of a fluid

measures of the "heaviness" of a fluid

free to move

P(force)

Fixed plateShearing strain

p

shearing stressδA

B B'

A

δa

53

Figure 1.2 (p. 13)(a) Deformation of material placed between two parallel plates. (b) Forces acting on upper plate.

54

Figure 1.3 (p. 14)Behavior of a fluid placed between two parallel plates.

55

Velocity=Ua

P (force)

yu B B'

A water

Fixed plate

no slip condition

u = u(y) For t 0 0

ABB' tan = Since a = Ut

= Rate of shearing strain =

b

Uyu

Velocitygradient

b

U

dy

du

b

a

b

U

t

b

tU

56

b

U

tlim

0δt

dy

du

b

U

From experiment p α τ & τ α

or τ α

dy

du

dy

du

Shearingstress

Rate of shearing strainor velocity gradient

For common fluids, such as water, oil, gasoline, and air.

τ = ----------------(1.9)where μ = absolute viscosity. or dynamic viscosity or viscosity

dy

du

57

τ =

τ vs. linear (Newtonian fluids) slope = viscosity, μ μ = constant

Newtonia fluid

dy

du

dy

du

58

Non-Newtonian Fluids

τ vs. μ ≠ constant non-linear non-newtonian fluids slope = apparent viscosity, μ ap

Non-Newtonian Fluids Bingham plastic fluid --- Toothpaste, mayonnaise Shear thinning fluid --- latex paint Newtonian fluid ---water, oil, air Shear thickening ---

water - corn starch mixture water - sand mixture (quick sand)

Shearing

Stress, τ

Bingham Plastic

Shear thinning

Newtonian

Shear thickeing1

μ ap

Rate of shearing strain, dy

du

dy

du

59

Figure 1.5 (p. 16)Variation of shearing stress with rate of shearing strain for several types of fluids, including common non-Newtonian fluids.

60

Dimensions of Viscosity τ = μ = τ [=] μ [=] FTL-2 [=] ML-1T-1

[=] lbf s / ft2 in BG units Table 1.7 (P.14) [=] N s / m2 in SI units Table 1.8 (P.14)

in CGS ( cm-gm-s) system

poise = dyne s / cm2

= g / (cm s)

1 cp = 10-2 poise = 0.001 kg/(m s) = 2.09 10-5 slug/(ft s)

dy

du

du

dy12 LT

L

L

F

61

μ = absolute viscosity = Kinematic viscosity or dynamic viscosity [=] L2/T or viscosity [=] ft2/s BG units [=] FT/L2 Table 1.5-1.8 [=] m2/s SI units (inside of front cover) Table 1.5-1.8 [=] lbf s/ft2 BG units [=] N s/m2 SI units [=] dyne s/cm2 CGS units [=] cm2/s CGS units (poise) ( stoke, st )

ρ

μν

62

63

64

65

66

67

68

elTf

smallpf

arg)(

)(

For liquids, μ dec. as T inc. For gases, μ inc. as T inc.

For liquid, T inc. Cohesive forces are reduced μ dec.For gas, T inc. Exchange of momentum of gas molecules between adjacent layers μ inc

Empirical Correlations. (μ =f(T) )----- for gases, the Sutherland equation -----------(1.10) C & S are constants T = absolute temperature----- for liquid, Andrade's equation ------------(1.11) D & B are constants

T is absolute temperature

Because of molecular structure

ST

CT 2/3

TBDe /

69

Figure 1.6 (p. 17)Dynamic (absolute) viscosity of some common fluids as a function of temperature.

70

Example 1.4 Given: Reynolds number (Re) =

μ = 0.38 N s/m2 = 0.38 kg/(m s) S.G. = 0.91 (specific gravity) D = 25 mm diameter pipe v = 2.6 m/s

Determine: Re = ? in (a). SI units (b). BG units

μ

ρνD

= fluid densityv = mean fluid velocityD = pipe diameterμ = viscosity

71

Solution: SG = = 0.91 × 1000 kg/m3 = 910 kg/m3

(a). In SI units Re =

=

(b). In BG units Re =

= =156

OH2

C4at OH 02

SG

2

33

m/SN38.0mm10

m1)mm(25)s/m(6.2)m/kg(910

D

156S/mkg1

N1

NS

mkg

1562

2

2

2f

22

3-3

33

3

m/SN1

ft/Slb10089.2S/m0.38N

1m

3.281ftm1025

1m/s

3.281ft/s2.6(m/s)

1kg/mft

slug101.94

)910(kg/mD

2f

f2

S/ftslug1

lb1lb/)S/ftslug(156

μ = 0.38 N s/m2 = 0.38 kg/(m s)S.G. = 0.91 (specific gravity)D = 25 mm diameter pipev = 2.6 m/s

72

Example 1.5 Given: u =

where v = 2 ft/s (mean velocity) = 0.04 lbf S/ft2

h = 0.2 in Determine (a) τ (Shearing stress) on the bottom wall (b) τ (Shearing stress) on mid-plane ( Plane parallel to the walls and passing through the centerline )

h

y

x

hu

])h

y(1[

2

v3 2

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Solution: τ = u =

(a) y =

(b) y = 0

dy

du

])h

y(1[

2

v3 2

2h

vy3]

h

1

h

y2[

2

v3

dy

du

)/1(21600)

12

12.0(

)()/(23

2sy

in

ftin

ftysft

ft01667.0in12

ft1in2.0

)flow ofdirection in (/4.14

)]/1()01667.0(21600)[/(04.0

2

2

ftlb

sftslbdy

du

f

fbottom

0)021600(dy

dumidplane

=

74

§1.7 Compressibility of Fluids

§1.7.1 Bulk Modulus

How easily can the volume (and thus the density) of a given mass of the fluid be changed when there is a pressure?

How compressible is the fluid?

A property that is commonly used to characterize compressibility is the bulk modulus, Ev, defined as

------------------------------(1.12) " － " means p

/d

dpE v

↓↑

75

m = ρ or = m/ρ ---------(1.13)

Ev = Bulk modulus or bulk modulus of elasticity [=] F/L2 [=] lbf/ft2 (psf) in BG units [=] N/m2 (pa) in SI units

Large values for Ev relatively incompressibleFor water (600F & atmosphere), dp = 3120 psi compress a unit volume of water 1% Ev = 3.12e5 incompressible

Compressibility (c) = 1/Ev [=] Small value c incompressibility

/)(

/1)/1(

/)/( 2 d

dp

d

dpd

dp

mmddp

Ev

2/

1

LF

76

§1.7.2 Compression and Expansion of Gases

From the ideal or Perfect gas law (or the equation of state)

p =ρRT ---------------------------(1.8) ---------------------------(1.14) If T = const. (Isothermal process)

1. Cconst

p

(1.16)---------)/(/

1

1

1 ppd

dC

d

dpE

cv

RTp

77

If the compression or expansion is frictionless and no heat is exchanged with the surroundings (isentropic process), then p / (k) = const.--------------------------------(1.15)

where k = Cp/Cv (Specific heat ratio) － tables 1.7 & 1.8 (p.14) R = Cp-Cv tables B.3 & B.4 (p.832 , 833) p = absolute pressure

-------------------(1.17)

For air k = 1.40 at P = 14.7 psia Ev = 20.6 psia (isentropic bulk modulus)For water

Ev = 312,000 psia

kp/d

dpE v

78

Example 1.6A cubic foot of helium at an absolute pressure of 14.7 psi is compressed isentropically to 1/2 ft3. What is the final pressure?

Solution: Since k = 1.66 (p.16)

m = v P1 = Pf

psia=6,696 psfa

constk

p

k

p

k

p

f

f

1

1

k

f

fk

1

1

m

p

m

p

5.46)5.0

1(7.14pp 66.1

kf

k1

1f

kfk

1

79

§1.7.3 Speed of Sound─ Disturbances introduced at some point in the fluid propagates at finite velocity consequence of the compressibility of fluids.For example

─Similarly

propagate fluid

acoustic velocity,

or speed of sound ,C

Flowing fluidShut down downstream suddenly disturbance propagates to upstream

Vibrates of loud speaker diaphragm

Localized disturbance

80

)19.1....(..........E

C or

)18.1....(..........d

dpC

V

for fluid (gas and liquid)

Since the disturbance is small, there is negligible heat transfer and the process is assumed to be isentropicFor gases, EV = kp (eq 1.17)

RTP

)20.1.........(kRTC

kpC

For air at 60℉ k = 1.4 & R = 1716 (ft-lbf/slug.oR) C = 1117 ft/s (Appendix B

tables B.3 and B.4)For water at 20 ℃ Ev=2.19x109N/m2 & ρ=998.2kg/m3

and P=ρRT

for gas only

C = 1481 m/s (or 4860 ft/s)

81

Example 1.7

A jet aircraft fly speed(v) = 550 mphat attitude(h)=35,000 ft, temperature (T)= -66 ,℉K = Cp/Cv = 1.40(assumed)

Determine: the ratio of the speed of the aircraft, V, to that of the speed of sound, C 。

82

Solution: The speed of sound in air at h=35,000ft and T=-66 , ℉ k=1.40 C = = The aircraft speed is V =

Note: .Mach number = Ma = v/cIf Ma < 1.0 the aircraft is flying at subsonic speedIf Ma > 1.0 the aircraft is flying at supersonic speed

For compressible fluid flow withMa < 0.3 incompressible fluid

kRTsftRRsluglbfft /973)6.45966)(./.1716)(44.1(

sftmile

ft

s

hr

hr

Miles/807

1

5280

60

min1

min60

1550

8185.0/986

/807

sft

sft

C

V

83

§1.8 Vapor pressureEvaporation ─ because some liquid molecules at the surface have sufficient momentum to overcome the intermolecular cohesive forces and escape into the atmosphere evaporation Liquid

Liquid Liquid Liquid

Vapor pressure = f (temperature) Appendix B (tables B.1&B.2) or tables 1.5 and 1.6

t = 0vaccum

t = t1 > 0molecules

t = t2 > t1

moleudes

pressure

equilibrium

Pressure exert on the liquid surface = vapor pressure

evaporation

Vapor is saturated

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Boiling ─ the formation of vapor bubbles within a fluid mass; it is initiated when the absolute pressure in the fluid reaches the vapor pressure, Water at atmospheric pressure = 14.7 psia Boiling point = 212 (100 )℉ ℃

vapor pressure( of water ) = 14.7 psia at 212℉ or boiling point = 212 ℉

at atmopheric presure=14.7

Important of vapor pressure ─ flowing fluid → fluid velocity↑→ pressure↓→ boling will occur→ Cavitation

`

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§1.9 Surface Tension

─At the interface between a liquid and a gas, or between two immiscible liquids, forces developed in the liquid surface which cause the surface to behave as if it were a "skin" or "membrane" stretched over the fluid mass.

Although such a skin is not actually present, this conceptual analogy allows us to explain several phenomena

Molecules along the surface are subjected to a net force toward the interior.

Steel needle floated because of surface tension (force)

HgCohesive forces in the surface

H2O

Waxed surfaceH2O

surface

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The apparent physical consequence of this unbalanced force along the surface is to create the hypothetical skin or membrane.

A tensile force may be considered to be acting in the plane

of the surface along any line in the surface.

The intensity of the molecular attraction per unit length

along any line in the surface is called surface tension(σ)

σ= f ( temperature)

[=] F/L

[=] lbf/ft BG unit tables 1.5& 1.6

[=] N/m SI unit Appendix B (table B.1 and B.2)

σ ↘ as T ↗

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2πRσ= PπR△ 2

where P = P△ i - Pe

Pi = internal pressure

Pe = external pressure

R

2P

one-half of a

liquid drop

R

2PPP 1

R

2PPP 2

R

4PPP 21

?21 PPP

airP1

Bubblefilm

Pe airP2

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The rise (or fall) of a liquid in a capillary tube --- having a liquid – gas – solid interface

  g(πR2h) = 2πRσcos θ

)22.1(cos2

r

cos2

gRRh

where

θ= contact angle = f (liquid , surface)

γ = g

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Example1.8

Solution:

σ= 0.0728 N/m for water at T = 20 ℃

ρg = 9.789 kN/m3 (table B.2 , P.857)

h < 1mm = 1X10-3m θ 0

gRh

cos2

Find R=?

For water T=20℃ & h<1mm

tube

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101)(

cos2101

cos2

gRm

gR

m0148.01010789.9

10728.02R

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