流體力學（一）

〈 Fluid Mechanics I 〉

方富民 教授 〈 Fuh-Min Fang 〉

Department of Civil Engineering

National Chung Hsing University

February 2020

Outline Textbook： Elementary Fluid Mechanics (7th Edition)

– by Robert L. Street, Gary Z. Watters and John K. Vennard

Subjects: 1. FUNDAMENTAL

Systems of Units Physical Properties of Fluids

2. FLUID STATICS Manometers Force on a Two-dimensional Body – Basic Approach Force on a Body – Control-volume Approach Force on a Body – General Approach Stability of a Floating Body Fluid Masses Subjected to Acceleration

3. KINEMATICS OF FLUID MOTION Lagrangian and Eulerian Views Conservation of Mass Continuity Equation Circulation and Vorticity

4. FLOW OF AN INCOMPRESSIBLE IDEAL FLUID Bernoulli’s Equation Momentum Equation of a Two-dimensional Flow

5. THE IMPULSE-MOMENTUM PRINCIPLE Pipe Bend Abrupt Enlargement in a Closed Passage Hydraulic Jump

6. SIMILITUDE AND DIMENSIONAL ANALYSIS Similarity Laws Buckingham π-theorem

1

Chapter 1 Fundamentals (pp.1-34)

Reading assignment：Chapter 1

States of fluids:

molecular spacing： solid < liquid < gas

inter-molecular cohesive force： solid > liquid > gas

Characteristics of fluids：

(1) Continuum

(2) Incapability of sustaining shear forces without deformation. (Fluids at rest cannot sustain shear stresses.)

(3) Isotropy of pressures in fluids at rest (scalar).

(4) Resultant forces obtained by the integration of the differential forces (produced by the action of pressure on differential areas) are vector quantities. (also see pp. 14)

gasliquid

2

System of Units SI Units：System International Unit (MKS)

FFS Units：Foot ─ Slug ─ Second Unit Dimension SI English FSS Length (L) m ft Mass (M) kg slug Time (t) sec sec

Temperature (T) °K (Kevin) °R (Rankine)

Frequency (f) Hz (sec-1) Hz

Force N (Newton) 2secftsluglbf ⋅=

Energy, Work, Heat Joule BTU (778.2 ftlbf − )

Power watt ( secJ ) HP (550 sec

ftlbf − )

Stresses (pressure, shear) Pa ( 2mN ) psi ( 144ft/lbf 2 × )

Temperature (practical) C° F°

∗ Prefixes： mega (m) = 106

kilo (k) = 103

centi (c) = 10-2

milli (m) = 10-3

micro (µ ) =10-6

∗ See Appendix 1 on pp. 685-688 in the textbook.

Base Units

3

∗ Density (ρ )： ]LM[

VolumeMass

3=

∗ Specific weight ( γ )：VolumeWeight

= ]tL

M[ 22

From Newton’s 2nd law gMW = ⇒ gρ=γ

( 22 s/ft2.32s/m81.9g == )

34

2

2

3

mkg

msN

sm

mN

gofDimensionofDimension

≡⋅

==γ

=ρ

Then 2sm1kg1N1 ⋅= (definition of a “Newton”)

In FSS system： 34

2

2

3

ftslug

ftslbf

sftftlbf

≡⋅

==ρ

ft/s1lbf1slug1 2⋅= or 2s/ft1slug1lbf1 ⋅=

* lbm1 is a mass which leads to a weight of lbf)g1(

2sft2.32g = , thus 1 slug = 32.2 lbm

* Specific volume (υ)：Mass

Volume (ρ

=1 ) = ]

ML[

3

* Specific gravity (S.G.)：OH2

ρρ

(dimensionless)

OH2.)G.S( ρ×=ρ OH2

.)G.S( γ×=γ

4 Equation of state：(for gases)

The specific weight of a perfect gas is obtained by the combination of Boyle’s and Charles’s laws as

TRpg

=γ

where g：gravity (gravitational acceleration) p：pressure T：thermodynamic temperature R：engineering gas constant

See Appendix 2, pp.689-696.

Compressibility and Elasticity

∗ Fluids are compressible thus are elastic media.

∗ In general, fluids can barely sustain tensions.

∗ A ”modulus of elasticity” can be used to define the elastic property of fluid. Since the form of fluids is essentially not rigid, a “bulk modulus of elasticity” is always used, which is on the basis of volume.

As shows in the figures, assume the cylinder and the piston are perfectly rigid. A force (F) applies to the piston and changes the fluid volume from 1∀ to∀ . Meanwhile, the pressure increases. Plotting p against 1/∀∀ produces the

stress－strain diagram. The modulus of elasticity of the fluid is defined as the

slope (absolute value) of the curve at that point.

5

)/(dpdE

1∀∀−=

Noted that for 0)(1

<∀∀

∆ , 0p >∆ , E > 0

It shows that small pressure disturbance travel thought the media (fluids) at a finite speed, which depends on the modulus of elasticity of the fluid. The pressure disturbance moves at a speed of sound “ a ” ,

where sound speed ρ

≅ρ

=E

dpda for small-Mach-number flows.

For incompressible fluids, )a(E ∞→∞→ , which in reality does not exist.

∗ Mach Number (M)speed sound

velocityfluid=

Sub-sonic flows ： 1M <

Super-sonic flows： 1M >

For most engineering calculations, the effect of compressibility may be safely neglected if M < 0.3 (with a relative error of about 9%, corresponding to M2).

6

Viscosity → due to molecular motions in fluids

Flow states： (1) Laminar

There is virtually no large scale mixing between the layers.

(2) Turbulent (mostly seen)

Flow contains random motion fluid particles by rapid mixing. Eddies of a wide range of size are seen in the flow.

In general, the shear (frictional) stress (τ ) in laminar motions is observed to be a function

of the strain rate (the velocity gradient , ydvd )

and a constant , µ , as )ydvd,(fun µ=τ

where µ：absolute (or dynamic) viscosity

For Newtonian fluids： ydvd

µ=τ

Dimension of µ： [ M L-1 t-1 ]

⋅=

⋅=

=

τ

=µ

2s

ft

2

as

ma

ftslbf

ft

ftlbf

spm

p

ydvd

“µ” varies with temperature and types of fluids.

Kinematic Viscosity (ν ) ]tL[

2

==s/fts/m

ρμν

2

2

µ = constant

7

The viscosity of gases increases with an increase of temperature.

The viscosity of liquids decreases with an increase of temperature.

8

9

Surface Tension (force/length)

→ An intermolecular force of a liquid that tends to maintain its surface (or interface between this liquid and another fluid) as it is.

The surface tension depends on the fluid in contact with the liquid surface. For instance, the surface tension of water in air is different from that in oil.

For a cylindrical capillary tube, assume that the liquid surface is a section of sphere.

hpp oi γ=−

then )r2)(cos()r)(h( 2 πθσ=πγ

and r

cos2hγ

θσ=

Noted that based on the assumption that the liquid surface is spherical, the above relationship is hold only for small tubes.

Vapor Pressure ( vp )

It is the pressure at which a liquid is vaporized.

The vapor pressure increases with temperature ∗ See Tables A2.1 and A2.2 (Appendix 2, pp.689-690) Homework of Chapter 1：8, 9, 12, 13, 17, 25, 26, 27, 28, 41, 49, 81

σ

10

Chapter 2 Fluid Statics (pp. 35-89)

When there is no fluid motion, no shear stress can exist (τ = µ dv/dy=0). Therefore, viscosity has NO effect in the analysis. Based on Newton’s 1st law

→ 0Fx =∑ , 0dzdypdzdyp CA =− …○1 E

↑ 0Fz =∑ , 0dWdydxpdydxp DB =−− …A○2 E

A where dzdydxgρdW =

For the small element (dx and dz are small), by Taylor’s expansion

+−

∂∂

+−∂∂

+=!2

)2

dx(

xp)

2dx(

xppp

2

2

2

A

+∂∂

+∂∂

+=!2

)2

dx(

xp)

2dx(

xppp

2

2

2

C

+−

∂∂

+−∂∂

+=!2

)2dz(

zp)

2dz(

zppp

2

2

2

B ( At point “O”, p = p(x, y, z) )

+∂∂

+∂∂

+=!2

)2dz(

zp)

2dz(

zppp

2

2

2

D

where p and all the derivatives are referred to the center of element (point O). Taking up to the 1st order terms, one has

A○1 E

A→ 0dzdy)2

dxxpp(dzdy)

2dx

xpp( =

∂∂

+−∂∂

− ⇒ 0xp=

∂∂

A○2 E

A→ 0dzdydxgdydx)2dz

zpp(dydx)

2dz

zpp( =ρ−

∂∂

+−∂∂

−

+

+

o

0

11

γ−=ρ−=∂∂ g

zp

…………………………………………… A○3 E

∗ For fluids without motion, the hydrostatic pressure is the same at a horizontal plane and the pressure is a function of z only.

For fluids with constant densities (also constant γ ), integration of A○3 E

A leads to

)zz(pp 1221 −γ=− or 22

11 zpzp

+γ

=+γ

where subscripts “1”and“2” are referred to two locations in the fluid.

12

Absolute and Relative Pressure

Like temperature, pressure is presented usually in two systems as

pressure )lative(Re Gage

pressure Absolute

Measuring Devices

(1) Mechanical

Bordon gage：gage pressure

Aneroide gage：absolute pressure

(2) Liquid devices

(3) Pressure transducers

DiaphramElectrical

13

Manometer

It is commonly used in the laboratory for more accurate pressure measurements.

(a) p1 = p2 110 hγp −= ( 101 hzz =− )

γ+= x1 pp

h0p 12 γ+= γ−γ=⇒ hp 1x (b) U-tube manometer (measure differential pressure)

54 pp = , hpp 322y11x γ+γ+=γ+

11322yx hpp γ−γ+γ=−⇒

(c) Inclined manometer (more accurate)

αγ=γ= sinhpx where α is known

(d) Micro-manometer (Research used) Sources of Errors：

(1) Manometer liquids Density may change with temperature.

(2) Capillary effect due to surface tension

(3) Fluctuation of liquid surface

(4) Optical error

1

2

po

zo

po

Air

z1

h1

gage pressure

14

Force on Submerged Plane Surfaces

dApdF = (width 1= )

∫= d)(pF (force/width)

∗ There is also pressure force acting on the other side of the surface if the

plate is submerged.

h1

15

Pressure Force on an Inclined Submerged Plate

AB

C

Dd

1h α+ cosdh1

∇

α+ cosdh2

2h

α

α=− sinhh 12

Force on surface A

( α )

2hhF 21

Aγ+γ

=

Force on surface B

(α

)

2cosd2hhF 21

Bαγ+γ+γ

=

Force on surface C

( α

) d2

cosdhhF 11C

αγ+γ+γ=

Force on surface D

( α

) d2

cosdhhF 22D

αγ+γ+γ=

αCF

DF

AF

BF

( +→ ) α−+α−= cos)FF(sin)FF(F CDBAx

α−γ+ααγ−= cosd)hh(sin)cosd( 12

[ ])hh(αsinαcosdγ 12 −+−= ≡ 0

( +↑ ) α−+α−= sin)FF(cos)FF(F CDABZ

α−γ+ααγ= sind)hh(cos)cosd( 12

]sin)hh(cos[d 122 α−+αγ=

]cossin[d 22 α+αγ=

dγ= ⇒ Buoyancy !!

16 If 0d → , then BA FF =

∇

imaginary surface (thickness=0)

Force on top = Force at bottom

A A1F

2F

∇∇

hinge

21 FF = (opposite direction)

Air

Liquid Same liquid

17

Control Volume Approach

∇

HF

1hγ

2hγ

gate

1VF

2VF

d

1h2h

3h

x-component of total force on gate per unit width：(force/width)

321

H h2

hhF ⋅γ+γ

= (trapezoid)

or 312

31 h2

)hh()hh( ⋅−γ

+⋅γ= (rectangle + triangle)

z-component of total force on gate per unit width：(force/width)

2dhdhFF 3

1VV 21γ+γ=+

Total force 2VV

2Htotal )FF(FF

21++=

Question: Where is the location of the resultant force?

Assume the gate is weightless.

totalF x

z

18

Total Force on a Submerged Body (Control-volume Approach)

•RFLF

bF

UF1V

2V bγ

W3V

vertical projected area vA

horizontal projected area hA

∇

fγ

z

x

The projected areas of pressure forces onto the body：

vARightLeft == hALowerUpper ==

Considering the total force on the body,

x-direction：

Since the pressures are hydrostatically distributed, their distributions corresponding to the (same) projected areas are the same.

R)(

L)(

FF←→

= therefore ∑ = 0Fx

z-direction：

for 1fUF ∀γ= (↓) )(F 321fb ∀+∀+∀γ= (↑)

( +↑ ) W)(WFFF 32fUbz −∀+∀γ=−−=∑

<Buoyancy>

where W is the body weight.

1∀

2∀

3∀

19

General Approach (NOT limited to 2-dimensional case)

Force on “1-side” of the plate：

αγ= sindAdF ( α= sinh )

∫αγ=A

dA)A(sinF

where ∫A

dA is the moment of area “A”

about the O-O axis Ac⇒

Thus AsinF cαγ=

Also, since cc hsin =α thus AhF cγ=

The moment of force dF about O-O：

αγ== sindAdFdM 2

∫αγ= dAsinM 2A

OOI − (second moment of area “A”)

Moment arm： A

IAsin

dAsin

FM

force totalmomenttotal

c

OO

c

A 2

p

−=αγ

αγ===

∫

The location of point c.p. (center of pressure force) is then obtained.

∗See Appendix 3 (pp.697-698)

Noted that from the transfer equation AII 2ccOO +=−

cc

c

c

2cc

p AI

AAI

+=+

= A

I

c

ccp

=−∴

Since 0Ic > , 0Ac > , center of pressure is always below the centroid of a submerged body except for a horizontal plate (where centers of pressure and centroid are at the same elevation).

hp hc

h

centroid

center of pressure

c

p

20

Stability of a Floating Body

For submerged bodies, stability requires the center of buoyancy to be above the center of gravity.

Righting moment > Overturning moment ⇒ Stable.

Submerged body

Submerged body

Non-submerged body

21

Fluid Masses Subjected to Acceleration

Considering a 2-D fluid element, based on Newton’s 2nd Law, one has (x-z plane)

dzdx)xp(F)( x ∂

∂−=+→ ∑

xx a)dzdxg

(am γ=⇒

∑ γ−∂∂

−=+↑ dzdx)zp(F)( z zz a)dzdx

g(am γ

=⇒

or xagx

p γ=

∂∂

− )ga(gz

pz +

γ=

∂∂

−

Also, by definition, ( since p = p(x, z) )

dzzpdx

xpdp

∂∂

+∂∂

= (chain rule)

then dz)ga(g

dxag

dp zx +γ

−γ

−=

At a line of constant pressure, 0dp =

On this line ↓

+−=

z

x

aga

dxdz

the slope of constant-pressure lines

22 The slopes of constant pressure lines are determined from external conditions in specific problems.

(A) Constant Linear Acceleration with 0a x = (A parcel of fluid is accelerated vertically.)

0xp=

∂∂ , =p function of z only

)ga1(

gag

zdpd zz +γ−=

+γ−= , where )(a z +↑

< part of pressure contributed by vertical acceleration > (B) Constant Linear Acceleration ( non-zero xa and za )

z

x

aga

xdzd

+−= zd

zpxd

xppd

∂∂

+∂∂

=

)ga(g

ag

hdzd

zp

hdxd

xp

hdpd

zx +γ

−γ

−

∂∂

+∂∂

=

↓↓

then )hdzd

gag

hdxd

ga(

hdpd zx +

+γ−=

From geometric similarity,

ga

hdxd x

′= ,

gga

hdzd z

′+

= where 2z

2x )ga(ag ++=′

gg

hdpd ′

γ−=⇒

g

23 (C) Central acceleration with constant central acceleration about a vertical

axis and 0a x =

0a z = , ra 2r ω−= (ω is the angular velocity)

ga

rp rγ=

∂∂

− γ=∂∂

−zp

At a constant pressure surface ( )z,r(funp = )

0zdzprd

rppd ≡

∂∂

+∂∂

= ⇒ )zdrdga( r +γ−

ga

rdzd r−=∴

gr

rp 2ωγ=

∂∂ γ−=

∂∂

zp

gr

rdzd 2ω= →Integrate cr

g2z 2

2

+ω

=

⇒ Lines of constant pressures are parabolas. Homework of Chapter 2：

1, 3, 8, 12, 21, 27, 30, 35, 38, 43, 49, 51, 53, 55, 57, 58, 77, 78, 87, 92, 94, 97

24

Chapter 3 Kinematics of Fluid Motion (pp. 91-106) Chapter 4 Systems, Control Volumes, .. (pp. 107-124)

Methods of motion description：

(1) Lagrangian View: ( f = f(t only) )

Fluid “particle” is labeled. The path, density, velocity, etc. corresponding to this particle, are traced as time passes.

Path line: trajectory of a fluid particle position of the particle as a function of time only.

(2) Eulerian View: ( f = f(x, y, z, t) )

based on a particular position in space. Description of fluid motion at each point is given as a function of time.

(Thus all variables are represented as functions of space and time.) Values and variations of the fluid variables are determined at various

spatial points. practical for most engineering problems.

Streamlines:

Curves are drawn in a flow field that the tangent at any point is in the direction of the velocity vector at that point.

There is no flow across these lines. Streak lines:

Lines of fluid particles that have passed through a particular point. In steady flows, the path line, streak line and streamline coincide. Streamtube:

The space between two streamlines.

One-Dimensional Flows (1-D flows)

Fluid variables depend on 1 spatial coordinate.

25

fieldflow

∗ The flow in a streamtube (however curved) is 1-D.

Two Dimensional (2-D) ― 2 spatial coordinates

Three Dimensional (3-D) ― 3 spatial coordinates Along a streamline (in a 1-D flow) (vr = 0)

td)t(sdv = ( in s-direction; v = vs )

sdvdv

sd)s(vd

td)t(sd

td)t(sd

td)t(vda 2

2

s ====

rva

2

r −= (centrifugal acceleration)

where r: radius of curvature at location “s”

∗ Velocities and accelerations are vectors (quantities with directions). ∗ For a 1-D flow, along a streamline 0vr = (no cross flow) * If local streamline is straight )r( ∞→ , 0a r =

In the Eulerian view, in a steady, 2-D field (on x-y plane)

)y,x(uu = , )y,x(vv =

s: tangential direction r: normal direction

26

In terms of displacement and time (Lagrangian view)

td)t(xdu = ,

td)t(ydv =

* Velocity at a point is the same in both the Eulerian and Lagrangian views.

By definition, the acceleration td

)t(udax = td

)t(vda y =

Noted that for )y,x(uu = and )y,x(vv = (Eulerian form)

dyyudx

xuud

∂∂

+∂∂

= , dyyvdx

xvvd

∂∂

+∂∂

=

Then tdyd

yu

tdxd

xu

td)t(udax ∂

∂+

∂∂

== , tdyd

yv

tdxd

xv

td)t(vda y ∂

∂+

∂∂

==

thus yuv

xuua x ∂

∂+

∂∂

= , yvv

xvua y ∂

∂+

∂∂

= (see pp. 67A)

* The above formulae are valid for 2-D cases.

In a polar coordinate system,

tdrdvr = ,

tddrv tθ

=

normal tangential

rv

θrvv

rvva

2tr

tr

rr −∂∂

+∂∂

= Centrifugal acceleration

rvv

θrvv

rvva trt

tt

rt −∂∂

+∂∂

=

27

Conservation of Mass Continuity Equation of 1-D Steady Flows

∗ Steady： Flow variables (such as density, velocity, ….) at any point in a

flow field is independent of time. ( 0t=

∂∂ )

Unsteady：The flow is time dependent.

The 1-D flow assumption implies that the velocity profiles at sections 1 and 2 are uniformly distributed (or at any other cross-sections). For mass conservation, dttORtRI )mm()mm( ++=+

In steady flows, since dttRtR )m()m( +=

Thus dttOtI )m()m( +=

222111 dsAdsA ρ=ρ

td

sdAtdsdA 2

221

11 ρ=ρ

1V 2V

Define mass flow rate as VAm ρ=•

]t/M[

then =ρ=ρ=•

222111 VAVAm constant for a steady flow.

∗ QVAm ρ=ρ=•

, where ]tL[VAQ

3= is the volumetric flow rate.

Assumptions in the analysis:

(1) Steady

(2) One-dimensional flow

28 In an incompressible flow with a constant density ( 21 ρ=ρ ), one has

volumetric flow rate === 2211 VAVAQ constant [ t/L3 ]

* For incompressible flows, the larger the cross section the smaller the corresponding velocity, and vice versa.

In a real flow, the velocity distribution through a cross section is actually not uniform (or non-uniform).

In a 2-D case, ]tL

LtL[

bQq

23=⋅=

* In general, ∫= dA)A(vQ [tL3

]

In 2-D cases, ∫= dy)y(vq [tL2

]

∗ 2-D flow rate )s/mm/s/m( 23 = bQq =

∗ mean velocity AQdAv

A1V A =∫=

Typical examples can be found in PROBLEMS 4.12 (pp. 122 in textbook).

Upper plate

Lower plate

29

Continuity Equation for 2-D Steady Flows

For conservation of mass, dttORtRI )mm()mm( ++=+

For steady flows, (independent of time)

dttRtR )m()m( += thus dttOtI )m()m( +=

where dA)cosds()m( out.s.cdttO ∫ θρ=+ ; dtVds =

and dtdA)θcosV(ρ)m(out.s.cdttO ∫=+

Note:：The tangential component of V (or θsinV ) has no contribution to the flow rate.

Define n as a surface unit vector (outward positive and normal to the surface dA ), then V• n θ= cosV

scalar vector

Therefore, ∫ ρ=+ out.s.cdttO dt)m( V•n ∫ ρ=out.s.c

dtdA V•dA

On the other hand, dtdA)cosV(dA)cosds()m( in.s.cin.s.ctI ∫ θρ=∫ θρ=

= { }∫ •ρ−=∫ −•ρ in.s.cin.s.c dAVdtdA)n(Vdt

Since dttOtI )m()m( += , one gets ∫ ρin.s.c

V•dA ∫ ρ+out.s.c

V•dA 0=

Or ∫ ρ.s.c

V•dA 0= → for steady and compressible flows.

V V -n dA

(Sum of inflow and outflow mass flow rates is zero.)

30

For an infinitesimal control volume in the Cartesian coordinate system

Taylor’s Expansion：for a small x∆

+∆

∂∂

+∆∂∂

+=∆+!2

xxFx

xF)x(F)xx(F

2

x2

2

x

∫ ρAB

V•n dx)2

dyyvv()

2dy

y(dA

∂∂

−∂ρ∂

−ρ−≅ (Angle between V and n is 180°)

∫ ρBC

V•n dy)2

dxxuu()

2dx

x(dA

∂∂

+∂ρ∂

+ρ≅ (Angle between V and n is 0°)

∫ ρCD

V•n dx)2

dyyvv()

2dy

y(dA

∂∂

+∂ρ∂

+ρ≅ (Angle between V and n is 0°)

∫ ρDA

V•n dy)2

dxxuu()

2dx

x(dA

∂∂

−∂ρ∂

−ρ−≅ (Angle between V and n is 180°)

For ∫ ρAB

V•n +dA ∫ ρBC

V•n +dA ∫ ρCD

V•n +dA ∫ ρDA

V•n 0dA =

One gets 0x

uxu

yv

yv

=∂ρ∂

+∂∂

ρ+∂ρ∂

+∂∂

ρ or 0)v(y

)u(x

=ρ∂∂

+ρ∂∂

Continuity equation for compressible, steady, 2-D flows

If the flow is incompressible, the continuity equation is 0yv

xu

=∂∂

+∂∂

For 2-D flows in the polar coordinate system, the continuity equations are

< Compressible flows >

0)v(r

)v(r

)v(r1

trr =ρθ∂

∂+ρ

∂∂

+ρ

<Incompressible flows>

0vr1

rv

rv trr =

θ∂∂

+∂∂

+

( ρ = constant )

31

Circulation

Line integral of the tangential component of velocity around a closed curve. scalar vector

∫∫∫ =α=Γ=ΓCCC

d)cosV(d V•d l

For a 2-D element：(steady, incompressible flow) + dy

BCalongvelocityMean

dxABalongvelocityMean

d

+

≅Γ dy

DAalongvelocityMean

dxCDalongvelocityMean

+

+

dy2

dxxvvdx

2dy

yuudy

2dx

xvvdx

2dy

yuu

∂∂

−−

∂∂

+−

∂∂

++

∂∂

−≅

dydxyu

xv

∂∂

−∂∂

= Note: ( dydx ) is the area of the element.

Vorticity：Differential circulation per unit of area enclosed.

yu

xv

dydxd

∂∂

−∂∂

=Γ

=ξ Cartesian coordinate system

θ∂∂

−+∂∂

=ξr

vrv

rv rtt Polar coordinate system

Suppose these are 2 lines drawn on the element parallel to x- and y-axes.

If the fluid element tends to rotate, within a time interval (dt ),

Vertical Line:

ydtd)

2dy

yuu()

2dy

yuu(d v

∂∂

−−∂∂

+−=θ

(The negative sign is due to sign convention. + )

32

The angular velocity (vertical) is

yu

tdd v

v ∂∂

−=θ

=ω

Horizontal Line:

xdtd)

2xd

xvv()

2xd

xvv(d H

∂∂

−−∂∂

++=θ

The angular velocity (horizontal) is

xv

tdd H

H ∂∂

=θ

=ω

The average angular velocity is

)yu

xv(

21)(

21

Hv ∂∂

−∂∂

=ω+ω=ω

∗ Note: ω=ξ 2

For rotational flows： 0)or( ≠ωξ (see example on pp. 67A)

For irrotational flows： 0)or( ≡ωξ Homework of Chapter 3：1, 2, 3, 4, 5, 6, 7, 10, 11, 12

Homework of Chapter 4：1, 2, 3, 11, 12, 22, 23, 27, 28,

33

Chapter 5 Flow of An Incompressible Ideal Fluid (pp. 125-187)

Incompressible： =ρ constant independent of space and time

0zw

yv

xu

=∂∂

+∂∂

+∂∂ (3-D) )w,v,u(u =

0yv

xu

=∂∂

+∂∂ (2-D) )v,u(u =

Ideal fluid：incompressible + inviscid (there is no frictional effect between fluid layers or between these layers and boundary walls; µ=0)

An assumption to simplify the analysis. Used with great care.

One-Dimensional (1-D) Flows

Euler Equation：(Euler, 1950) Friction is neglected for an ideal-fluid flow.

In a cylindrical fluid system, from Newton’s 2nd law along the direction of motion (along the streamline; 1-D analysis)

+↗∑ =tdvdmddF with dAdsdm ρ=

Thus

+↗ θρ−+−=∑ sindAdsgdA)dpp(pdAdF

dsdzdAdsgdAdp ρ−−= (from geometry)

Then one has

tdvddAdsdzdAgdAdp ρ=ρ−−

θ θ

34

If )s,t(vv = acceleration tdsd

s)s,t(v

t)s,t(v

td)t(vdaS ∂

∂+

∂∂

== (chain rule)

In a steady case ( 0t=

∂∂ ), then v = v(s only)

Thus sd

)s(vd)s(vsvv

tdsd

sv

tdvd

≡∂∂

=∂∂

=

Therefore, dsdvvdAdsdzdAgdAdp ρ=ρ−−

One gets 0zdgvdvpd=++

ρ or 0zd)

g2v(dpd 2

=++γ

∗ gρ=γ

Along the streamline, 0)zg2

vp(d2

=++γ

Finally, integration leads to

total head

Hzg2

vp 2

=++γ

corresponding to a streamline

pressure velocity elevation head head head Bernoulli’s constant

Bernoulli’s equation (in terms of heads with a dimension of [L]) Valid for 1-D, steady cases of ideal (incompressible + inviscid) fluids Derived from Newton’s 2nd Law (see also pp. 36) It is valid along the same streamline.

The sum of the velocity (g2

v 2

) and the

pressure (γp ) heads can be measured by a

Pitot tube. (see pp. 37)

piezometric head ⇒ pressure head + elevation head

1-D

It is horizontal if friction is neglected.

35

Streamtubes of Finite Cross Section (1-D analysis)

Assume streamlines are locally straight and parallel, along the direction “normal” to the streamline, for force balance one has

α=− cosWds)pp( 21 (no centrifugal force)

where =W weight of element dshγ=

and h

zzcos 12 −=α (from geometry)

Then ds)zz(ds)pp( 1221 −γ=− 22

11 zpzp

+γ

=+γ

<Although the relationship is similar to Bernoulli’s equation, it has nothing to do with it.>

∗ When streamlines are straight and parallel, the quantity ( zp+

γ), or the

piezometric head, is constant over the flow cross section normal to the

streamlines. That is, the pressure is hydrostatically distributed.

Examples: In pipe and duct (river) flows

∗ If streamlines are straight and parallel, a single hydraulic grade line (HGL) applies to all the streamlines of the flow.

∗ For 1-D, ideal flows, since the velocity distribution over a cross section of flow is uniform, then the Bernoulli’s constant (H) is the same in the entire flow field.

∇

W

α

36

Applications of Bernoulli’s Equation (1) Flow from an orifice of a huge reservoir

For 1-D flow, Bernoulli’s equation yields

0g2

vpzg2

vp 222

1

211 ++

γ=++

γ

(Datum is set at the nozzle center.)

Take 0p2 = (gage pressure) and assume streamlines at section 2 are straight and parallel, then one has

g)dzdA(dAdzdApdA)dpp( ρ−=γ−++−

〈Newton’s 2nd law in the vertical direction〉

Since gρ=γ , it leads to 0dp =

Also, if section “1” is taken as the reservoir surface, then

21 pp = (atmospheric pressure)

and 1v ≒0 (as the reservoir is huge compared to the orifice)

Then )zz()pp(1)vv(g21

21212

12

2 −+−γ

=− (Bernoulli’s equation)

Finally, 2v ≒ hg2 (Terricelli Theorem, 1643)

∗ Benoulli’s equation can be in a second form in terms of pressures as

=γ+ρ

+ zv2

p 2 constant (see also pp. 34)

○1 E

A A○2 E

A A○3 E

A○1 E

A→ static pressure

A○2 E

A→ dynamic pressure

A○3 E

A→ potential pressure

0 h0

dm

1

z

37

(2) Pitot tube (a velocity measurement device)

Bernolli’s equation yields

S2

SSo2

oo zv2

pzv2

p γ+ρ

+=γ+ρ

+

Then ρ−

=)pp(2v oS

o , ( oS pp − ) is to be measured.

∗ Stagnation pressure 2ooS v

2pp ρ

+=

(3) Bernoulli principle applied to open flow problems.

Section is far away from the dam so that the local streamlines at the section are straight and parallel.

Based on the assumptions that the streamlines are straight and parallel and the velocity distributions are uniform at sections 1 and 2, one has

=γ

+pz constant. The HGL (for all the streamlines) is the liquid surface. The

EL (energy line) will be horizontal and located g2

v 2

above the liquid surface

(assume there is no energy loss due to friction).

* Cavitation：(It occurs only when the fluid is a liquid.)

Form Bernoulli equation =++γ

zg2

vp 2

constant

As the velocity becomes large, the pressure (p) can be very small. Once vpp < (in terms of absolute pressure), evaporation (boiling) occurs and could

damage the structure surface in the long run. This formation of vapor cavity of liquids is called “cavitation”.

)zz( So =

0

weir

38

Momentum Equations for 2-D Flows

From Newton’s 2nd Law, on the x-z plane

dzdxxpdFadm xx ∂

∂−== ( dzdxρdm = )

dzdxgρdzdxzpdFadm zz −

∂∂

−==

Recall in 2-D steady cases, kaiaa zx +=

where zuw

xuua x ∂

∂+

∂∂

= zww

xwua z ∂

∂+

∂∂

= (see pp. 26)

Then )zuw

xuu(dzdxdzdx

xp

∂∂

+∂∂

ρ=∂∂

− ……..………… 1∆

)zww

xwu(dzdxdzdxgdzdx

zp

∂∂

+∂∂

ρ=ρ−∂∂

− …..………. 2∆

⇒ xp1

zuw

xuu

∂∂

ρ−=

∂∂

+∂∂ (x-momentum equation) ………… 3∆

gzp1

zww

xwu −

∂∂

ρ−=

∂∂

+∂∂ (z-momentum equation) …….…… 4∆

⇒ 2-D steady, incompressible, ideal fluid (inviscid)

39

Bernoulli’s Equation for Two-Dimensional (2-D) Flows

dzdx 43 ×∆+×∆ yields

dzgdzzwwdz

xwudx

zuwdx

xuu)dz

zpdx

xp(1

+∂∂

+∂∂

+∂∂

+∂∂

=∂∂

+∂∂

ρ−

For )z,x(pp = in a 2-D steady case, dp1LHSρ

−=

dzg)zu

xw()dxwdzu(

)dzzwwdz

zuu()dx

xwwdx

xuu(RHS

+∂∂

−∂∂

−+

∂∂

+∂∂

+∂∂

+∂∂

=

dzg)dxwdzu(2

wud22

+ξ−+

+= (see definition of ξ on pp. 31)

Being divided by “g” yields

dz)dxwdzu(g1

g2)wu(dpd 22

+−ξ++

=γ

−

Or )dxwdzu(g1z

g2wupd

22

−ξ−=

+

++

γ

Integration leads to

∫ −ξ−=++

+γ

)dxwdzu(g1Hz

g2wup 22

integration constant

Set 222 wuV += , )w,u(u = (2-D case)

then ∫ −ξ−=++γ

)dxwdzu(g1Hz

g2Vp 2

〈steady result〉

vorticity

40

(I) For irrotational flows, 0≡ξ . Then Hzg2

Vp 2

=++γ

∗ In irrotational flow fields, the Bernoulli’s sum is the same everywhere. There is only one energy line (associated with all the streamlines) describing the energy situation.

(II) For rotational flows, along “one” streamline, by definition

uw

dxdz

= or 0dxwdzu =−

then the Bernoulli’s sum is the same along this streamline.

∗ In rotational flows, the Bernoulli’s sum remains constant along the same streamline. Along different streamlines, there are different values of Bernoulli’s sums.

41

dm

Applications of 2-D Bernoulli’s Equation (2-D, steady, incompressible, irrotational)

For streamlines which are curved, there are effects due to curvature.

From Newton’s 2nd law, along the r-direction

+ r

VdsdrcosdWdspds)dpp(2

ρ=θ+−+ (r：positive outwards)

where dsdrdW γ= , rdzdcos =θ (from geometry)

rVdsdrdzdsdsdp

2

ρ=γ+

(centrifugal force)

)dsdr

11(γ

× yields rg

V)zp(rd

d 2

=+γ

(valid in direction perpendicular to streamline)

Noted that in an “irrotational” flow field

Hg2

vzp 2

=++γ

→ a constant independent of r

Taking derivative )Hg2

Vzp(rd

d 2

=++γ

yields

vr = 0

vt = V

dr

streamline

42

0rdVd

g2V2)zp(

rdd

=++γ

or rd

VdgV)zp(

rdd

−=+γ

For irrotational flows, to achieve rd

VdgV

rgV2

−= , one must have

0dVrdrV =+ or 0)Vr(d = , then =Vr constant

Accordingly, if the flow is “irrotational”,

r1V ∝

Assume the fluid is inviscid (there is no energy loss due to friction), the energy line (EL) is horizontal. In an “irrotational” flow field, a “decrease” of velocity is to be expected with an “increase” of distance from the center of curvature (or an increase of r).

Another example:

At section 2, assume streamlines are locally straight and parallel,

=+γ

zp constant

Points D and E have the largest velocity.

Since at point D, z is larger. If cavitation occurs, the most possible location will be at point D, where there is the smallest pressure (since zD > zE).

× D

× E

(Vertical plane)

43

Example of a 2-D irrotational open flow ⇒ sharp crested weir flow

A ′′：stagnation point

At a long distance upstream from the weir, the streamlines are essentially straight and parallel. Therefore, at section 1, the pressure is hydrostatically distributed,

ypA γ=′

Streamlines BB and AA are free streamlines. Their precise positions in space are unknown. However, the pressures of them are everywhere constant (= atmospheric pressure).

A ′′ is the stagnation point. The Bernoulli’s relationship on AA ′′′ is

A2

AA pv2

p ′′′′ =ρ

+ )g2

vy(p2

A +γ=∴ ′′

Homework

Chapter 4 1, 2, 3, 11, 12, 22, 23, 27, 28 Chapter 5 4, 5, 6, 8, 12, 15, 16, 28, 37, 41, 50, 65, 66, 67, 68, 72, 73, 80, 88,

89

A

hydrostatic pressure

44

Chapter 6 The Impulse-Momentum Principle (pp. 189-227)

In a fluid system, Newton’s 2nd law yields

Σ F m(td

d= VC )

sum of external forces = change of linear momentum with respect to time

where ∫= sysdmm VC ∫=

sysm1 V dm

In the discussion of the impulse–momentum principle, the fluids are generally considered to be real. That is, viscosity exists. (However, sometimes its effect may be negligible.)

Σ F m(td

d= VC ) ∫=

systdd V dm

t)dm()dm(

limtsysttsys

0t ∆∫−∫

= ∆+

→∆

dm)dm()dm( OttRttsys ∫+∫=∫ ∆+∆+

1∆

dm)dm()dm(ItRtsys ∫∫∫ +=

2∆

In steady flows,

21 ∆=∆

V V

V V V

V V V

45 For the part of “O”, ρ=dm V•n tdA∆ = ρV•dA t∆

∫O V dm ∫∆= OUT.S.Ct V (ρV•dA)

Similarly, for the part of “I”, ∫ I V dm = − ∫∆ IN.S.Ct V (ρV•dA)

Thus, Σ F = ( )

∫ ∫ •ρ+•ρ∆∆

→∆OUT.S.C IN.S.C

0t)AdV(V)AdV(V

tt

lim

∫= .S.C V (ρV•dA) ( ∗∆ )

∫= .S.C V•

md (The direction of F goes with the direction of V)

where =•

md mass flow rate out of the control volume through dA

V =•

md momentum flux through dA (only through penetrable control surfaces)

* The forces acting on the fluid in a control volume include

(1) body forces：gravity, magnetic force (2) surface forces acting on the control surfaces ∗ The advantage of the momentum principle is that the detailed motion

inside the control volume needs not be known.

In 3-D cases, ∗∆ are

∑ ∫= .S.CxF vx (ρV•dA )

∑ ∫= .S.CyF vy (ρV•dA )

∑ ∫= .S.CzF vz (ρV•dA )

(Compare to a similar form on pp. 29)

V (ρV•dA) V (ρV•dA)

46

Pipe bend

Assume frictionless, or the frictional force is negligible. Also, 1-D analysis is used.

When the flow direction changes, the pipe itself experiences a force from the flow. To fix this pipe, a force F has to be applied.

By taking a control volume “ABCD”, from momentum principle

x-direction:

∑ −α−=→+

x2211x FcosApApF ………………..………..○1 E

and ∫ ρ.S.C x (v V•dA)=+ [ ]12xx vcosvQQvQv

INOUT−αρ=ρ−ρ …..A○2 E

Note: ρ=•

md V•dA= 0 at the (non-penetrable) pipe surface.

<There is no flow across the pipe surfaces.>

From the impulse/momentum theorem, A○1 E

A = A○2 E

)vcosv(QFcosApAp 12x2211 −αρ=−α−

z-direction:

∑ −αρ=+α−−=↑+ )0sinv(QFsinApWF 2z22z

where W is the weight of fluid in the control volume.

To be determined

negative for inflows positive for ouflows

v1

v2

47

2-D structure (gate) in open flows (Assume friction is negligible.)

In the x-direction, )vv(qFFFF 12x21x −ρ=−−=∑→+

(Force/unit width)

Assume at sections and streamlines are straight and parallel, then the local pressure is hydrostatically distributed.

Where 2yF

21

1γ

= (force per unit width in y-direction; see also pp. 35)

2yF

22

2γ

=

2211 yvyvq == (volumetric flow rate per unit width in y-direction)

In the z-direction, 0)00(qFWFF zOBz =−ρ=−−∑ =↑+

It is noted that as the shear force on the gate is negligible, the resultant force is nearly normal to the gate.

Thus. θ

≅cos

FF x .

hard to determine

48

Abrupt enlargement in a closed passage (assume 1-D flow and frictionless) Consider control volume “ABCD”,

2211 vAvAQ ==

The rate of change of momentum

)vv(vA)vv(Qmdv 122212ABCD x −ρ=−ρ=∫•

....A○1 E

2221x ApApF −=∑→+

………….……….……A○2 E (∗Also see footnote in the text book on pp. 195.)

A○1 E

A = A○2 E

A leads to g

)vv(vpp 12221 −=

γ− ….A○3 E

∗ After the sudden expansion (after AB), there are 2 zones where “reverse flows” (or vortices) occur. A significant head (energy) loss is expected.

From Bernoulli equation, (neglect friction at the pipe surfaces)

L2

222

1

211 hz

g2vpz

g2vp

+++γ

=++γ

.…A○4 E

A (for real flows)

head loss due to sudden expansion

Based on the 1-D flow analysis, 21 zz = (same elevations of centerlines)

From A○3 E

A and A○4 E

A, one gets L

21

2212221 h

g2vv

g)vv(v

γpp

+−

=−

=−

and g2

)vv()]vv(v2)vv[(g2

1h2

21122

22

21L

−=−+−=

Borda-Carnot head loss

hL

49

Hydraulic Jump In open channel flow, when a liquid at high speed discharge into a zone of low velocity, a rather abrupt rise (a standing wave) occurs in the liquid surface. This results in significant energy loss due to generation of violent turbulence and eddies.

Assume frictionless, streamlines are straight and parallel at sections and

)vv(q)yy(2

FFF 1222

2121x −ρ⇒−

γ=−=∑→

+

……….……….. 1∆

where q and 1y are usually given and 2y is to be solved (or q and 2y are given and 1y is to be solved) in engineering problems.

At sections and , based on 1-D analysis one has

11 y

qv = 2

2 yqv = (mean velocities at the two cross-sections)

Then )y1

y1(q)

yq

yq(q)vv(q

12

2

1212 −ρ=−ρ=−ρ

1∆ becomes )y1

y1(

gq

2y

2y

12

222

21 −=− …….....………….. 2∆

Or 2

yyg

q2

yyg

q 22

2

221

1

2

+=+ ………………………. 3∆

From 2∆ , 21

212

2121 yyyy

gq)yy()yy(

21 −

=−+ or gq2)yy(yy

2

2121 =+

)y1( 31

× yields 31

22

1

2

1

2

ygq2)

yy(

yy

=+ or 0ygq2)

yy()

yy( 3

1

2

1

22

1

2 =−+

Finally,

++−=

++−=

1

21

31

2

1

2

ygv811

21

ygq811

21

yy

The other root (-) is not valid since y2 / y1 > 0.

50

(A) 1yg

v

1

21 = , 1

yy

1

2 = , 21 yy = , no hydraulic jump.

(B) 1yg

v

1

21 > ,

1

2

yy

> 1 , hydraulic jump occurs.

(C) 1yg

v

1

21 < , 1

yy

1

2 < , it produces a “rise” in energy line

⇒ Physically impossible !

Define a Froude number as yg

vFr = ,

then, when Fr > 1 , hydraulic jump is expected to occur.

To estimate the energy loss due to the occurrence of a hydraulic jump (loss due to friction is not included), from the impulse/momentum theorem one has

21

212

12

212

22

21 yy

yyq)y1

y1(q)vv(q)yy(

2−

ρ=−ρ=−ρ=−γ ………. A○1 E

)yy(yy2gq 2121

2 +=∴ ……………………………………… A○2 E

For )y1

y1(

g2q)yy()

g2vy()

g2vy(E 2

22

1

2

21

22

2

21

1 −+−=+−+=∆ , substitution

of A○2 E

A gives

21

312

21

221

2121

221

21

22

21

21

22

212121

yy4)yy(

yy4)yy()yy(

yy4)yy(1)yy(

yyyy)yy(yy)

2g)(

g21()yy(E

−=

−−−=

+−−=

−++−=∆

∗ Hydraulic jump is an excellent energy dissipater.

Skip pp. 204 ~ 216.

Homework of Chapter 6: 1, 2, 5, 8, 14, 21, 22, 28, 33, 47, 72

)yy)(yy( 2121 +−−

51

Chapter 8 Similitude and Dimensional Analysis (pp. 291-322)

Tools for problem solution:

(A) Analytical methods

(B) Numerical methods

(C) Experimental methods

Use of (A) and (B):

Usually get approximate results. Not applicable for all fluid problems.

Use of (C):

Provide results for verifications of (A) and (B). However, mostly when testing “real thing”(or “prototype”) is not feasible, thus a “scaled version” (or “model”) can be made to “simulate” prototype behavior.

The use of model data to extrapolate the behavior of the corresponding prototype has 2 advantages:

(1) Economics

Typically (but not always), a model is smaller than the prototype. The cost is reduced substantially.

(2) Practicality

In a model test, the environment and test conditions can be easily and rigorously controlled. The measurements can be made handily.

52

Similitude

Complete similarity requires all the following three similarity.

(I) Geometric similarity

The flow fields and boundary geometry of model and prototype have the same shape. The ratios between corresponding lengths in model and prototype are the same.

For example,

m

p

m

p

dd

= ( or mp )d()d(

= )

Accordingly, define the length scale

p

m

p

mL d

d

==λ , then 2L

2

p

m2

p

m

p

m )()dd(

AA

λ===

(II) Kinematic similarity

Geometrically similar flow fields have the same ratios of corresponding velocities and accelerations throughout the flow.

∗ Flows with geometrically similar streamlines are kinematically similar.

(III) Dynamic similarity

To maintain geometric and kinematic similarity between flow fields, the forces acting on corresponding “fluid masses” must be related by the same ratios. This is called dynamic similarity.

m4m

p4p

3

3

2

2

1

1

aMaM

FF

FF

FF

m

p

m

p

m

p ===

∗ Geometric Similarity + Dynamic Similarity ⇒ Kinematic Similarity threshold means target

53 Forces remained in a flow fields：

(1) Pressure force 2p L)p(A)p(F ∆=∆=

(2) Inertia force 222

3I LV)

LV(LamF ρ=ρ==

(3) Gravitational force gLgmF 3G ρ==

(4) Viscous force LVLLVA)

ydud(F 2

v µ=µ=µ=

(5) Elastic force 2E LEAEF ==

(6) Surface Tension LFT σ=

V and L are respectively the characteristic velocity and length.

ρ, E, µ and σ are fluid properties; g is gravitational acceleration.

To have dynamic similarity：

(1) mp

Ip

p

I )FF()

FF( = ; m2

22

p2

22

)LpLV()

LpLV(

∆ρ

=∆ρ or m

2

p

2

)p

V()p

V(∆ρ

=∆ρ

Define “Euler number” as Ep2

V∆ρ

= then mp EE =

(2) mv

Ip

v

I )FF()

FF( = ; mp )LV()LV(

µρ

=µ

ρ

Define “Reynold’s number” as Rν

=µ

ρ=

LVLV then mp RR =

(3) mG

Ip

G

I )FF()

FF( = ; m

2

p

2

)Lg

V()Lg

V( =

Define “Froude number” as FLg

V= then mp FF =

(also see pp. 49)

54

(4) mE

Ip

E

I )FF()

FF( = ; m

2

p

2

)EV()

EV( ρ

=ρ

Define “Cauchy number” as CEV2ρ

= then mp CC =

∗ Noted that since sound speed ρ

=Ea (see pp. 5)

Define “Mach number” as MEV

aV 2ρ

== Note: M2 = C

(5) mT

Ip

T

I )FF()

FF( = ; m

2

p

2

)VL()VL(σ

ρ=

σρ

Define “Weber number” as Wσ

ρ=

2LV then mp WW =

Practically, in most engineering problems, satisfaction of all the similarity conditions (to have dynamic similarity) is not necessary (sometimes, impossible in reality).

For an engineering viewpoint, one should select the “numbers” which involved important forces in the problem. Those involved unimportant (negligible) forces should be discarded in the analysis based on “professional judgments”.

Therefore, in each new problem of similitude, good understanding of fluid phenomena is necessary to determine how the problem be satisfactorily simplified by the elimination of the irrelevant, negligible, or compensating forces.

55

Examples of Reynolds Similarity

(A) Airfoil in a low-speed wind field

In the problem,

(i) There are no surface tension phenomena. (W is gone.)

(ii) The compressibility effect is negligible (incompressible) due to low Mach number. (C or M is gone.)

(iii) Gravity is not important. (F is gone, as there is no free surface.)

Note: To have dynamic similarity, satisfaction of similarity in any 4 out of the 5 numbers (E, R, F, M, W) are required.

Therefore, the Reynolds numbers of the prototype and the model have to be the same to ensure dynamic similarity. (In addition to geometric similarity, the prototype and the model are then kinematically similar.)

Accordingly, if mp )LV()LV(ν

=ν

then m22p22 )LV

D()LV

D(ρ

=ρ

The drag force (D) in the prototype m2m

2p

2m

2p

m

pp D)

LL

()VV

()(Dρρ

= …

As the fluids in the model and prototype are air and water respectively, based on Reynolds number similarity, one has

pLair

waterp

m

p

air

waterm V1V

LL

Vλν

ν=

νν

= where Lλ is the length scale.

Finally, substitution of air

water

Lp

m 1VV

νν

λ= into gives

m2

water

air

water

air2

L

2

water

air2L

water

airp D)(1)(D

νν

ρρ

=λν

νλ

ρρ

=

This is how Vm should be used in model experiments.

air water

56

(B) Fluid through a closed passage

Rp = Rm (same reasons as in (A))

or m1

p1 )

dV()

dV(

ν=

ν ………..

needs to be satisfied to ensure dynamic similarity.

Once Reynolds similarity is hold, complete similitude is then reached and

m221

p221 )

Vpp()

Vpp(

ρ−

=ρ− since m22

221

p22

221 ]

dVd)pp([]

dVd)pp([

ρ−

=ρ−

or m2

m

p

m

pp21p )p()

VV

()pp()p( ∆ρρ

=−=∆ …………………..

To get p∆ in prototype, the procedure is

(i) Select model length scale (to get Lλ )

(ii) Select model fluid (to obtain mν ).

(iii) Based on , calculate mV and use it in the model experiments.

(iv) Measure p∆ in the model then calculate p)p(∆ according to .

57

Example of Froude Similarity

In the case of a ship moving on a liquid surface, considerations are

(i) It involves negligible compressibility effect. (C or M is gone.)

(ii) The model is large so that surface tension is not important. (W is gone.)

(iii) Temporarily, assume frictional effect is not important. (R is gone.)

As the Froude Number (gravitational force) is dominant in the open-flow

problem, thus mp )gV()

gV(

= ……..…………………

Then m22p22 )VD()

VD(

ρ=

ρ ……………………..

In fact, the drag force (D) should also be closely related to frictional force. Thus, Reynolds similarity should also be important besides Froude similarity.

Accordingly, similarity also requires mp )V()V(ν

=ν

……….

To achieve kinematic similarity, and have to be satisfied simultaneously. Equating and leads to (with the same “g”)

⇒ m

p

m

p

VV

=

⇒ 23

m

p

m

p21

m

p

m

p

m

p

m

p )()(VV

===νν

…………...…...

In additional to geometric similarity, implies that there are only 2 alternatives to ensure kinematic similarity.

(i) Select an appropriate model fluid so that is hold.

(ii) If the model fluid is the same as that in the prototype, the model has to be as large as the prototype, which is obviously impractical.

58

Dimensional Analysis

Application of the law of similitude A valuable means of checking engineering calculations Provide a good way to construct the forms of physical equations from knowledge of relevant variables and their dimensions.

There are generally 4 basic dimensions directly relevant to fluid problem：

(i) Mass (M)

(ii) Length (L)

(iii) Time (t)

(iv) Thermodynamic Temperature (T)

Accordingly, the dimension of any dependent flow variables can be represented as some products of the 4 dependent basic dimensions.

For example, based on Newton’s 2nd law, )onaccelerati()mass(Force ⋅=

Then, its dimension can be obtained as ]t

LM[]tLM[]F[ 22 =⋅= , with 3

independent fundamental dimensions involved (M, L and t). On the other hand, in a hydraulic turbine problem, supposed that the power (P) of the turbine is known (based on professional judgments) and is related to

Q：volumetric flow rate in the turbine ]tL[

3

γ：specific weight of fluid ]tL

M[]LF[ 223 =

ET：unit mechanical energy (head), given up by every unit weight of fluid as it passes through the turbine [ L ]

That is, the relationship between the dependent variable (P) and the independent variables (Q , γ and ET) is

)E,,Q(fP Tγ=

59

From the principle of dimensional homogeneity, (Fourier, 1882)

cT

ba EQCP γ=

where coefficient C is a dimensionless constant, which is usually obtained from experiments.

Now, let’s examine the dimensions of the relationship

cb22

a3

3

2

)L()tL

M()tL(

tLM

=

[M]: b1=

[L]: cb2a32 +−= 3 equations, 3 unknowns (a, b and c)

[t]: b2a3 −−=−

⇒ 1a = , 1b = , 1c =

Finally, one gets TEQCP γ= ⇒ Rayleigh’s method

60

π - Theorem (Buckingham, 1915) (i) If “n” variables are functions of each other (such as v,,,,D µρ and g),

then “k” equations of their exponents (a, b, c, …) can be written (where “k” is the largest number of variables among the “n” variables which cannot be combined into a dimensionless group).

(ii) Mostly, “k” is equal to the number “m” of independent dimensions (such as M, L, t, T). In general, mk ≤ .

(iii) Application of dimensional analysis allows expression of the functional relationship in terms of (n-k) distinct dimensionless groups.

In the problem of drag on a ship (pp. 57), the relationship among all the dependent and independent variables is

0)g,v,,,,D(f =µρ

with n = 6 ; number of basic dimensions k = 3 (involving M, L, t)

Then, there are 3 (n – k = 6 – 3) dimensionless groups (3 π numbers).

That is, in the problem one has a dimensionless relationship as

0),,(F 321 =πππ

There are a lot of ways to combine these 6 variables into 3 dimensional groups. Experience suggests that ρ, v and (involved in the inertia force) appear in most dimensionless groups.

By selecting ρ, v and as the repeated variables, one has

),v,,D(f11 ρ=π

),v,,(f22 ρµ=π

),v,,g(f33 ρ=π

By using Rayleigh’s method, one gets a221 )

vD(ρ

=π .

Since the power is less significant than the group itself, take 1a = (or any non-zero number) for convenience.

61

Similarity, µ

ρ=π

v2 ⇒ R

and g

v3 =π ⇒ F

Finally, the relationship of the problems can be represented as

,v

D(F 22ρ

R, F 0) =

or 122 Fv

D=

ρ(R, F)

dimensionless dependent variable dimensionless independent variables

Homework of Chapter 8: 5, 13, 20, 28, 31

62 A

Examples (Chapter 1 -- 3)

Unit conversion (10%) ( 1 ft = 0.3048 m ; 1 Kg = 0.0685 slug )

1 lbf⋅sec /m2 = ? N⋅min/ft2

【Answer】

N45.4ft1

m3048.0slug0685.0

Kg1sec

ftslug1)sec/ftslug(1lbf1 22 =××

⋅=⋅=

222

2 ftmin/N00689.0)ft1

m3048.0(sec60

min1lbf1

N45.4m

seclbf1msec/lbf1 ⋅=×××⋅

=⋅

…………………………………………………………………………………..

Calculate (px - py) in FSS units. 3w ft/lbf4.62=γ

【Answer】

yWWWx p)12/30()8.0)(12/10()12/60(p =γ+γ+γ−

psf4.114ft/lbf4.114)ft/lbf4.62(]ft)12/38(ft5[pp 23yx ==−=−

63 A

如圖所示，一個 U 形管中有三種靜止液體，已知乙流體與丙之比重為 0.7

與 0.8，試求甲流體之比重(S)。

【Answer】

Method 1:

0)08.0()7.0()06.0()8.0()06.0(S0 WWW =γ−γ−γ⋅+

Method 2:

4.10)6)(8.0()8)(7.0()6(S =+=

73.16

4.10S ==

3公分

6公分

8公分

流體甲

流體乙

流體丙

64 A

一個長度（垂直紙面方向）為 1 米圓柱之兩側有兩靜止液體：分別為油

（比重 0.8）與水。試求圓柱體受流體作用之總力（牛頓）。

【Answer】

3w m/N9810=γ ；

3OIL m/N7848)9810(8.0 ==γ

m/N156962

)2)(7848(F2

1H == （→）

m/N49052

)1)(9810(F2

2H == （←）

m/N10791490515696FH =−=∑ （→）

m/N12321)7848(4

)2()14.3(21F

2

1V == （↑）

m/N7701)9810(4

)2()14.3(41F

2

2V == （↑）

m/N20022770112321FV =+=∑ （↑）

m/N22745)20022()10791(F 22 =+=∑

與水平線夾角 o1 68.61)

1079120022(tan ==θ − ( )

1 米 水

2 米 油

65 A Water flows between two flat plates ( sm /101 26−×=ν ) as shown. The maximum speed (at the centerline) is s/m1 . (Hint: assume Newtonian Fluid.)

(1) Find the mean velocity.

cbyayv 2 ++= ；

0c)0y(v === ； 1ba)1y(v =+== ； 0ba2)1y(dy/dv =+==

Then y2yv 2 +−= ； ∫ =+−=+−=1

0

210

232 s/m34)yy

31(2dy)y2y(2q

s/m667.02qV == #

(2) Calculate the shear stress (in 2/ mN ) at m5.0y = .

2y2ydvd

+−= ； at y = 0.5 s/112)5.0(2ydvd

=+−= 。

236 m/N10)1)(10)(1000( −− ==τ #

…………………………………………………………………………………

Use M (Mass), L (Length) and t (time) to represent absolute viscosity.

【Answer】

]tLM[]L/t/LtLM[]

dy/dustress[][ 11

21−−

−−

===µ

y

2 m U(y)

66 A

在一個與垂直紙面方向無關的二維(two-dimensional)水槽問題中，截面

之入流量為每秒 3 平方米（垂直紙面 1 米寬計），截面與之速度分佈

形狀分別為拋物線形與三角形（最大速度均位於出口中心且均為每秒 1米）。在入/出流保持不變的情況下，試求出水槽液面上昇/下降之速度（米

/秒；上昇為正值， 下降為負值）

【Answer】

cxbxaV 22 ++= ； bxa2

xdVd 2 +=

0)0(V2 = 得 0c =

1)1(V2 = 得 1ba =+

0)1(xd

Vd 2 = 得 0ba2 =+ ； 1a −= ， 2b =

s/m333.1)131(2dx)x2x(2q 2

1

0

22 =+−=+−= ∫

s/m12

)2)(1(q 23 ==

s/m667.01333.13q 2=−−=∑

s/m111.06667.0VS == （上昇）

2 m

q1=3 m2/sec

1 m/sec

1 m/sec

2 m

6 m

67 A

在一個位於 yx − 水平面之二維(two-dimensional)流場中，定義 u 與 v

分別為沿x與y座標方向之速度分量，且

4yx2x3u 2 +−= ， 2yx6yv 2 −−= 。

(1) 試証此為 incompressible 與 rotational 流場。

(2) 試求在 )1,1( 位置 x 方向加速度之大小。

【Answer】

(1)

0)x6y2()y2x6(yv

xu

≡−+−=∂∂

+∂∂

(incompressible flow)

0)x2()y6(yu

xv

≠−−−=∂∂

−∂∂

(rotational flow)

(2)

yuv

xuua X ∂

∂+

∂∂

=

34)2()7()4()5()2()261()26()423( =−−+=−−−+−+−=

68 B

Examples (Chapter 4 -- 5)

Water, with a density of 1000 Kg/m3, flows steadily through a contracting round duct. The diameters of the upstream and downstream cross-sections are 2 m and 1 m. The water deflection (h) in the mercury (S.G.=13.6) manometer is 1 cm. Calculate the volumetric flow rate of the duct in m3/sec. (Note：Use 1-D analysis and neglect friction)

From manometer readings:

)01.0()6.13(p)01.0()V5.0p( W1W222 γ+=γ+ρ+

)01.0()6.13(p)V5.0p( WW1222 γ−γ=−ρ+

or m126.0)01.0)(6.12(p)g2

Vp(W

122

W

2 ==γ

−+γ

…..………

1-D Bernoulli’s equation: g2

Vpg2

Vp 21

W

122

W

2 +γ

=+γ

(z1=z2) ………….. ;

From , m126.0g2

V21 =

s/m572.1)126.0)(81.9)(2(V1 == ;

s/m235.1)1)(4

()572.1(Q 32 =π

= #

h

1 m 2 m Water

Hg

69 B

Form manometer readings:

)5.0()zp()zp( BBAA γ=γ+−γ+

Or m5.0)zp()zp( BB

AA =+

γ−+

γ

Then m5.0)zz(ppBA

BA =−+γ−

…

From continuity:

2B

2A )100(V)150(V =

AB V25.2V =

1-D Bernoulli’s equation:

g2Vzp

g2Vzp 2

BB

B2

AA

A ++γ

=++γ

)125.2(g2

Vg2VV)zz(pp 22

2A

2A

2B

BABA −=

−=−+

γ−

…

Thus, from and 5.0V)81.9(2

0625.4 2A = m

One gets s/m553.1VA =

s/m02745.0)553.1()15.0(4

Q 32 =π

=

70 B

Water flows steadily in a round pipe. Calculate the velocity in the 3” cross-section. The specific weight of mercury is 13.57 .

0p3 = (The exit is open to the atmosphere.)

Manometer: 0)07.2()07.2(])10(V2

p[ HgWW2

33 =γ−γ+γ+ρ

+

)07.2(57.13)07.12(g2

V23 =+ ; s/ft12.32)02.16)(2.32)(2(V3 ==

…………………………………………………………………………………. Water flows steadily through a gate as shown. Neglect friction and use a

1-D analysis to calculate the mean velocity at the upstream section (in m/s).

From mass conservation )1(V)2(V 21 = , so 12 V2V =

Assume at sections 1 & 2, streamlines are straight & parallel ( yzγp

=+ )

Bernoulli’s: g2

Vyg2

Vy22

2

21

1 +=+ , or g2)V2(1

g2V2

21

21 +=+

g2V31

21= ; s/m56.2

3)81.9(2V1 == #

……………………………………………………………………………………………….

Air Water

Gate

2 m 1 m

Datum

z

z

71 B Oil with a specific gravity of 0.90 flows downwards through a Venturi meter as shown. If the manometer deflection (h) is 28 in, calculate the volumetric flow rate.

From continuity: AB V4V = …..

Manometer:

hzph)zp( HgBOilBOilAOilA γ+γ+=γ+γ+

Or

hγ)hz(γp)zγp( HgBOilBAOilA +−+=+

ft770.32)1228()

9.09.054.13(

h)()zz(pp

Oil

OilHgBA

Oil

BA =−

=γγ−γ

=−+γ− ……

Bernoulli’s relationship:

g2Vzp

g2Vzp 2

BB

Oil

B2

AA

Oil

A ++γ

=++γ

Or g32

V15])41(1[

g2V)zz(pp 2

B222B

BAOil

BA =−=−+γ− ………

From and

g32V15770.32

2B= , s/ft446.47

15)2.32)(32)(770.32(VB ==

22B ft02182.0)

121)(1416.3(A ==

)s/ft(cfs035.1)2182.0(446.47Q 3==

z

72 C

Examples (Chapter 6 -- 8)

Fx

Fy

o2

o21x

o22

o2211 60cos)V()

2Q(ρ60cosV)

2Q(ρVQρF60cosAp60cosApAp −++−=−+−

p1 = p2 = 0 (atmospheric) then 1x VQρF −=− = -1767 N

In the y-direction: (p1 = p2 = 0)

)60sinV()2Q(60sinV)

2Q()0(Q

F60sinAp60sinAp

o2

o2

yo

22o

22

−ρ+ρ+ρ−=

+−

Then, Fy = 0 Therefore, the force on the plate is -1767 N (→) Noted that the force acting on the fluid is +1767 N (←)

73 C

Z1=0

Z

×

×

F

(p2A2)

(p1A1)

(→+)

(↑+)

74 C

(exposed to atmosphere)

75 C

76 C

77 C

78 C

A shallow circular dish has a sharped-edged orifice at its center. A horizontal water jet, with a speed of V, strikes the dish concentrically. Evaluate the horizontal force (in Newtons) to hold the dish in place.

V = 2 m/s , D = 6 cm , d = 2 cm

【Solution】

22D m00283.0)03.0(A =π= ， 22

d m000314164.0)01.0(A =π=

s/m00565.0)00283.0)(2(Q 31 == ， s/m0006283.0)000314164.0)(2(Q 3

2 == ，

s/m00503.0QQQ 3213 =−=

∑ ρ−−ρ+ρ=−= VQ)45sinV(QVQFF 1o

32X

N16.17)]2)(00565.0()7071.0)(2)(00503.0()2)(0006283.0([)1000( −=−−=

Thus F = 17.16 N (←)

F

Q3

Q3

Q1 Q2

79 C

y22

y2

n 3n 1 22

(y ) 0.19y3.021

+ +=

80 C The flow around a two-dimensional bluff body is investigated. It is known that vortices are generated behind the body at a shedding frequency (ω), which depends on D, H, velocity (V), fluid density (ρ ), and absolute viscosity (µ ). Related quantities are listed as follows:

Variables Prototype Model

D 20 cm 2 cm H 0.5 m 0.05 m

V 10 Km/hr Vm

ρ 1.23 Kg/m3 (air) 998 Kg/m3 (water) µ 4 x 10-7 lbf⋅sec/ft2 2.4 x 10-5 lbf⋅sec/ft2

ω ω p 1 Hz

(1) By selecting "D" as one of the repeated variables (you need to properly choose the other two),

present the relationship in a dimensionless form. (15%)

(2) For similarity, determine Vm (in Km/hr) for the model. (8%)

(3) Compute ω p (in Hz). (7%)

【Solution】

0),,V,H,D,(f1 =µρω

n = 6 ， k = 3 (M, L, t) ， 故有 (6 – 3) = 3 個無因次參數。

取 ρ、V、D 為重覆變數， 得 0)DV,DH,

VD(f2 =

µρω

或 )DV,DH(f

VD

3 µρ

=ω

……………..(1)

本題中，已具幾何相似（101

LL

p

mL ==λ ）。為達到完全相似，雷諾數應相等（動力相似）。

故 pm )DV()DV(µ

ρ=

µρ

， …………………(1)

hr/Km38.7)10()04.04.2()

110()

100023.1(V)()

DD

()(V pp

m

m

p

m

pm ==

µµ

ρρ

= ……….(2)

當完全相似時， mp )VD()

VD( ω

=ω

，

故 Hz1355.0)1()38.7

10()101()

VV

()DD( m

m

p

p

mp ==ω=ω ……………….(3)

81 C A Venturi meter is a device for the measurement of the flow rate in a pipe. It is known that the

pressure difference (∆p) between points “A” and “B” depends on the pipe diameter (D), the pipe speed (V), the fluid density (ρ), the dynamic viscosity (µ) and the gravitational acceleration (g).

Prototype Model D (m) 3 1 V (ft/s) 10 ? ρ (slug/m3) 1.94 3.0 µ (lbf·s/ft2) 2 × 10-5 4 × 10-5 g (m/s2) 9.81 9.81 ∆p (psi) ? 2

(1) Find the dimensionless form of the relationship between ∆p and the related variables. (10%) (2) Assume that the gravitational force is comparatively not important, what is “V” in model

(Vm) equal to ? (8%)

(3) Again, as the gravitational force is not important, calculate (∆p)P. (7%)

【Solution】 n = 6 (6 個變數)； k = 3 (M, L, t)；故有 (n-k) = 3 個 π numbers.

0)g,,,V,D,p(f =µρ∆ , Choose repeated variables: ρ , V, D

)D,V,,p(f11 ρ∆=π , 得 21 Vp

ρ∆

=π

)D,V,,(f22 ρµ=π , 得 µ

ρ=π

VD2

)D,V,,g(f33 ρ=π , 得 Dg

V3 =π ; ==> )

DgV,VD(F

Vp

2 µρ

=ρ∆ ……(1)

To get complete similarity, Reynolds numbers are the same. Thus pm )VD()VD(µ

ρ=

µρ ,

s/ft8.38)10)(24)(

13)(

394.1(V))(

DD

)((V pp

m

m

p

m

pm ==

µµ

ρ

ρ= ……(2)

Then m2p2 )Vp()

Vp(

ρ∆

=ρ∆ , psi0859.0)2()

8.3810)(

394.1()p()

VV

)(()p( 2m

2

m

p

m

pp ==∆

ρ

ρ=∆

82 C The power (P) needed for the operation of an axial pump depends on the fluid density (ρ ), the

rotating speed (N), the blade diameter (D), the head loss (H) and volumetric flow rate (Q). Related quantities are listed below:

Variables Prototype Model

ρ (Kg/m3) 1000 800

N (turn/min) 300 900

D (cm) 15 5

H (ft) 30 Hm

Q (ft3/s) 27 Qm

P (HP) PP 2

(1) By selecting "N" and “D” as two of the repeated variables (you need to properly choose the third), present the relationship in a dimensionless form. (13%)

(2) For similarity, determine Qm for the model. (6%) (3) Predict PP. (6%)

【Solution】n = 6 (6 個變數)； k = 3 (M, L, t)；故有 (n-k) = 3 個 π numbers.

Repeated variables: ρ、N、D

)H,D,N,(f11 ρ=π ， 得 1HD

π =

)Q,D,N,(f22 ρ=π ， 得 32 DNQ

=π

)P,D,N,(f33 ρ=π ， 得 533 DNP

ρ=π ==> )

DNQ,

DH(f

DNP

353 =ρ

….……(1)

P3m3 ]DN

Q[]DN

Q[ = ， s/ft3)27()155(

300900Q)

DD(

NNQ 33

P3

P

m

P

mm === ………….(2)

當達到完全相似時， m53P53 ]DN

P[]DN

P[ρ

=ρ

成立。故，

HP5.22)2()5

15()900300(

8001000P)

DD()

NN(P 53

m5

m

P3

m

P

m

PP ==

ρρ

= ………….………(3)