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Fluid Mechanics
Lectuer:Sun Gang
Introduction• 1-2 Definition of a Fluid• The solid object will no change
inside the a closed container• The liquid will change its shape
to conform to that of the container and will take on the same boundaries as the container up to the maximum depth of the liquid
• Fluid mechanics:the behavior of fluids at rest and in motion
• A fluid is a substance that deforms continuously under the application of a shear(tangential) stress no matter how small the shear stress may be
• A solid deforms when a shear stress is applied does not continue to increase with the time
• Dye maker to outline a fluid element
AF /
Introduction• The deformation of solid• Experience a Deformation
• Finite(solid)
• Continuously increasing• Shear stress is proportional• To the rate of change of • The deformation
At the atomic and molecular level:
Solid:the molecular are packed so closely together that their nuclei and electrons form a rigid geometric structure,”glued”together by powerful intermolecular forces.
Liquid:the space between molecular is large,the intermolecular forces allow enough movement of the molecules to give the liquid its “fluidity”
Gas:the spacing between molecular is much larger, the influence of the intermolecular forces is much weaker,and the motion of the molecules occurs rather freely throughout the gas
Introduction• 1-4 Basic Equation• The ideal gas equation of state
• The Basic laws governing the flow motion include:
• 1、 The conservation of mass• 2、 Newton’s second law of moti
on• 3、 The principle of angular momentu
m• 4、 The first law of theromdynamics• 5、 The second law of theromdynamics
)1.1(RTp
Introduction• 1-5 Methods of analysis • The system that you are
attempting to analyze Basic mechanics : free-body
diagram thermodynamics: closed
system(terms: system and control volume)
1-5.1 System and Control Volume
• A system is defined as a fixed, identifiable quantity of mass; the system boundaries separate the system from the surroundings(fixed or movable), no mass crosses the system boundaries.
• A control volume is an arbitrary volume in space through which fluid flows. The geometric boundary of the comtrol volume is called the control surface.(include real or imaginary)
1-5.2 Differential versus Integral Approach
• The basic laws can be formulated in terms of infinitesimal or finite systems and control volumes.
• The first case the resulting equation are differential equation. • The integral formulations of basic laws are easier to treat analytically,
for deriving the control volume equation , we need the basic laws of mechanics and thermodynamics ,formulated in terms of finite systems
1-5.3 Methods of Description
• Use of the basic equations applied to a fixed , identifiable quantity of mass, keep track of identifiable elements of mass(in particle mechanics: the Lagrangian method of description)
• Example: th eapplication of Newton’s second law to a particle of fixed mass
• Consider a fluid to be composed of a very large number of particle whose motion must be described
• With control volume analyses, the Eulerian on the properties of a flow at a given point in space as a function of time
1-6 Dimension and Unit
• The physical quantities of engineering problems include units: length,time,mass, and temperature as dimension
• The primary quantities(length,time,mass, and temperature as dimension )and secondary quantities (dimensions are expressible in terms of the dimension of the primary quantities)
1-6 Dimension and Unit
• 1-6.1 System of Dimension • a) mass(M), length(L), time(t), temperature(T)(MLtT)• b) Force(F), length(L), time(t), temperature(T)(FLtT)• c) Force(F),mass(M) ,length(L), time(t), temperature(T) (FMLtT)
1-6.2 system of unit
• a. SI, MLtT(primary) unit of mass(kg) length (meter) time (second) s temperature (kelvin) Absolute metric system of unit
Secondary dimension Force(N) 1N=1kg. m/s2
1dyne=1.g.cm/s2
• FLtT, British Gravitational system:
• force(1bf); length(ft);• time(second);temperat
ure(degree Rankine)
• 1 slug = 1lbf.s2/ft
• FMLtT (English Engineering system)
• force(1bf) mass(lbm) length(foot) time (second) temperature (degree Rankine)
• gc=32.2 ft.lbm/(lbf. S2)
4. State the three basic system of dimension
5. The typical units of physical quantities in the SI, British Gravitational, and English Engineering system of units
SI: 1N=1kg. m/s2
FLtT: 1bf Mass: 1slug = 1lbf.s2/ft
FMLtT: gc=32.2 ft.lbm/(lbf. S2)
Chapter 2 Fundamental Concept
Mechanics and thermodynamics
2-1 Fluid as a continuum
• The average or macroscopic effects of many molecules……….continuum(classic fluid mechanics)(p, ,T,V is continuous function of position and time)
• the mean free path of the molecules is same order of magnitude as the smallest significant characteristic dimension of the problem(rarefied gas flow)
For determine the density at a point , In fig.2.1 point C(x,y,z)’s density is defined as mass per unit volume, the mean density within volume V would be given by = m/V, at point C
)12(lim'
V
mVV
C
)22(),,,( tzyx
2-2 velocity field
• 2-2.1 one-,two-,and three-Dimensional flows
),,,( tzyxVV
kwjviuV ˆˆˆ
Steady flow unsteady flow0.....),,(
t
Vp 0.....),,(
t
Vp
2-2.2 Timelines, Pathlines, streaklines, and Streamlines
• Visual representation of a flow field: TL; PL, STKL, STML
• Timeline:a number of adjacent fluid particles in the flow field are marked at given instant, they form a line in the field
• Pathline: the path or trajectory traced out by moving fluid particle, the line traced out by the particle
Streakline:a number of identifiable fluid particles in the flow passed through one fixed location in space , the line joining these fluid particles is defined as a streakline
Streamline:are lines drawn in the flow field so that at a given instant they are tangent to the direction of flow at every point in the flow field. No flow across a streamline
2-3 Stress Field
• Surface and body forces:• body force:gravitational body force• Stress force (nine quantities)to specify the state of
stress in a fluid• Imagine any surface within a flowing fluid, and
consider the contact force applied to the fluid on one side by that on the other,surrounding point C, the surface
dVg
A
the unit normal vector outwardly , the force, , acting on nay be resolved in two components, one normal to and the other tangent to the area. The normal stress and a shear stress are then defined as:
n̂F
A
n n
)7.2(lim0
n
t
An AF
n
)6.2(lim0
n
n
An AF
n
the rectangular coordinates the stress acting on the planes whose outwardly drawn normals are in the x,y, or z directions, The first subscript indicate the plane on which the stress acts. The second one indicates the direction in which the stress acts
)8.2(limlim
lim
00
0
x
z
Axzx
y
Axy
x
x
Axx
AF
AFAF
xx
x
The stress at a point is specified by the nine components
zzzyzx
yzyyyx
xzxyxx
denote the normal stress and shear stress
2-4 viscosity• We have defined a fluid as a substance that
continues to deform under the action of a shear stress. Consider the behavior of a fluid element between the two infinite plates
y
x
y
x
Ayx dAdF
AF
y
0lim
dtd
tratendeformatio
t
0lim
dydu
dtd
yu
t
yltul
2-4.1 Newtonian fluid
Fluid as water,air, and gasoline are Newtonian fluid
dydu
yx
The different Newtonian fluid will deform at different rates under the action of the same applied shear stress; the water, glycerin exhibits a much larger resistance to deformation than water
Newton’s law of viscosity is given for one-dimensional flow by
dydu
yx
The dimension, [F/L2],du/dy are [1/t], [Ft/L2]; in the SI system, the unit of viscosity are kg/(ms) or Pas( 1Pas = 1Ns/m2)(page 26)
The kinematic viscosity( )[L2/t]is represented. Viscosity data for a number of common Newtonian fluid are given in Appendix A. Note that for gases, viscosity increases with temperature, whereas for liquids, viscosity decreases with increasing temperature.
/
2-4.2 Non-Newtonian fluids
• Fluids in which shear stress is not directly proportional to deformation rate are non-Newtonian flow: toothpaste and Lucite paint
)12,11.2()()( 1
dydu
dydu
dyduk
dyduk nn
yx
2-5 surface Tension• Surface tension is the apparent interfacial tensile
stress(force per unit length of interface) that acts whenever a liquid has a density interface, such as when the liquid contacts a gas, vapor,second liquid, or solid
• Contact angle between the liquid and solid is defined When the contact angle is less than 900, the liquid tends to wet the solid surface as shown in fig2.10a, and the tensile stress due to surface tension tends to pull the liquid free surface up near the solid, forming a curved meniscus.
The contact angle>90, the liquid can not wet the solid; surface tension tend sto pull the liquid free surface down along the solid.
The magnitude and direction of surface tension against a solid surface depend on the liquid and solid
2-6 Description and classification of fluid motions
• Continuum fluid mechanics• Inviscid-(compressible and incompressible)• Viscous-laminar(internal and external)• -turbulent(internal and external)
•2-6.1 Viscosity and Inviscid flow
dydu
yx
2-6.2 Laminar and Turbulent flows
• The basis of flow structure
• Smooth motion in laminae or layers
• Random, three-D motions of fluid particles in addition to the mean motion
2-6.3 Compressible and Incompressible Flows
• The variations in density are negligible are termed incompressible(liquid)
• Density variations within a flow are not negligible, the flow is called compressible(gas)
• M=V/c; M<0.3, M>0.3
2-6.4 Internal and External Flows• Flows completely bounded by solid surfaces called
internal or duct flows. Flows over bodies immersed in an unbounded fluid are termed external flows.
• The incompressible flow through a pipe, the nature of the flow(laminar and turbulent) is determined by the Reynolds number:the ratio between inertial force and viscous force
• Incompressible flow through pipe• Laminar flow: Re<2300• The flow over a semi-infinite flat plate,
laminar:Re<500000
/Re DV
Chapter 3 Fluid StaticAbsence of shear stresses, fluid either at rest or in “rigid-
body” motion are able to sustain only normal stresses, fluid element do not deform.
3-1 The Basic equation of fluid statics• Newton’s second law to a fluid element of mass dm=• Body forces(gravity) and surfaces forces are applied to
fluid element(no shear stress, include pressure force) • The body force is• Pressure is scalar field, p=p(x,y,z),using Taylor series
expansion, the pressure of left face of the element is:
• The right face:
dV
dxdydzgdVgdmgFd B
)2
()( dyyppyy
yppP LL
2)( dy
yppyy
yppP RL
dxdydzkzpj
ypi
xpFd s )ˆˆˆ(
)1.3()( pdxdydzdxdydzgradpFd s
The gradient of a scalar field gives a vector field
)2.3(
)(
gpdV
Fd
dVgpFdFdFd sB
For a static fluid,
)3.3(000
gpamFda
)6.3(
)5.3(00
gdzdp
gzp
yp
xp
Restriction (1) Static fluid
(2) Gravity is the only body force
(3) The z axis is vertical and upward
3-2 The standard atmosphere
3-3 Pressure variation in a static fluid
3-3.1 incompressible liquid:Manometers
gdzdP
z
z
p
pzzgppgdzdp
00
)( 00
hzz 0)7.3(0 ghpp
Incompressible liquid:manometers
•
1122
22
11
ghghppghppghpp
CA
CB
BA
3-3.2 Gases (compressible fluid)
The pressure varies with altitude or temperature
)8.3()()1(
)(
/
00
/
00
0
mRgmRg
TTp
Tmzpp
dzmzTR
pgdzRTpggdzdp
mzTT 0
3-5 Hydrostatic force on submerged surfaces
Determine the resultant force acting on a submerged surface we must specify:
(1) The magnitude of the force
(2) The direction of the force
(3) The line of action of the force
3-5.1 hydrostatic force on a plane submerged surface
The hydrostatic force on any element of the surface must act normal to the surface )9.3(ApdFd
The resultant force acting on the surface is found by summing the contributions of the infinitesimal forces over the entire area.
)10.3(AR ApdF
sin000 yhghpgdhppgdhdp h
The point of application of the resultant force(the center of pressure) must be such that the moment of the resultant force about any axis is equal to the moment of the distributed force about the same axis, the position vector is designated as 'r
)11.3(' ApdrFdrFr R
Substitute into Eq.(3.11) gives
kdAAdjyixrjyixr ˆˆˆˆˆ '''
)12.3('' AA RR xpdAFxandypdAFy
1. The resultant force is the sum of the infinitesimal forces (3.10)
2. The moment of the resultant force about any axis is equal to the moment of the distributed force about the same axis(3.12)
In evaluating the hydrostatic force acting on a plane submerged surface, the resultant force and moment is:
1. The magnitude of is given by
2. The direction of is normal to the surface
3. For a surface in the xy plane, the line of passes through the point x,y ( the center of pressure), where
RF
RF
RF
)13.3( pdAFF RR
A ARR xpdAFxandypdAFy ''
3-5.2 Computing equation for pressure force and point of application on a plane submerged surface
The pressure distribution on the lower surface is uniform ambient pressure p0 , on the upper surface is given by
ghpp 0
The magnitude of the resultant force on the upper surface is
AAR ydAgAppdAF sin0
The yc is the y coordinate of the centroid of the area A thus
)14.3()(sin 00 ApAghpAygApF CccR
To find the expressions for coordinates of the center of pressure , the moment of the resultant force about any axis must be equal to the moment of the distributed force about the same axis
)15.3(
sinsin
ˆˆ2
2
aAyIyydAyI
dAygghdAyAygy
ypdAFy
c
xxcAxx
AAc
AR
xxI ˆˆ Is the second moment of the area about the centroid axisx̂
The same ruler:
)15.3(ˆˆ bAyI
xxc
yxc
3-5.3 Hydrostatic force on a curved submerged surface
)16.3(ˆˆˆ
)10.3(
)9.3(
zyx RRRR
AR
FkFjFiF
ApdF
ApdFd
x
x A xARR pdAiApdiFdiFF ˆˆˆ
In general in the l direction the component of resultant force is:
l
l A lR pdAF
With the free surface at atmospheric pressure, the vertical component of the resultant hydrostatic force on a curved submerged surface is equal to the total weight of the liquid directly above the surface.
gVgdVdAghF
ghppdAFF
ZV
ZVRZ
Basic equation in integral form for a control volume
4-1 Basic laws for a system
4-1.1 Conservation of mass
)1.4(0 adt
dMsystem
)1.4( bdVdmMsystemsystem MMsystem
4-1.2 Newton’s second law ( Conservation of momentum )
)2.4(
)2.4(
bdVVdmVP
adtPdF
systemsystem MMsystem
system
4-1.3 The angular Momentum principle
)3.4(
)3.4(
)3.4(
aTdmgrFrT
bdVVrdmVrH
adtHdT
shaftMs
HMsystem
system
system
systemsystem
4-1.4 The first of thermodynamics
)4.4(2
)4.4(
)4.4(
2
cgzVue
bdVeedmE
adtdEWQ
dEWQ
systemsystem VMsystem
system
4-1.5 The second Law of Thermodynamics
)5.4(
)5.4(1
bdVssdmS
aQTdt
dSTQdS
systemsystem VMsystem
system
4-2 Relation of system derivatives to the control volume formulation
dv
seVr
V
SEHPM
NsystemVsystem
1
The control volume is fixed in space relative to coordinate system xyz,during t+dt-t time the system has been chosen so that the mass within region I enters the control volume during interval dt, and the mass in region III leaves the control volume during the same interval.
)8.4(
)(lim 00
0 t
NNNN
dtdN tcvttIIIIcv
tsystem
)9.4(
)lim
)lim
)lim 0000
000 tN
tN
t
NN
dtdN ttI
t
ttIII
t
tcvttcv
tsystem
The surface of control volume include: surface of the flow enter the control volume and the surface of the flow leaves the control volume and he surface no flow pass it
Vtl
dAldVdN
t
0lim
)cos(
)11.4(
cos
csCVs
csss
AdVdVtdt
dN
AdVtN
dtdN
4-3 Conservation of mass:(N=m; )1
)11.4(
csCVs
AdVdVtdt
dN
)12.4(
csCVs
AdVdVtdt
dM
)13.4(0
csCVAdVdV
t
conservation
4-3.1 special cases
incompressible, nondeformable control volume of fixed size and shape
0csAdV
AdV
The integral of over a section of the control surface is commonly called the volume flow rate or volume rate of flow. For incompressible flow , the volume flow rate into a fixed control volume must be equal to the volume flow rate out of the control volume
)16.4(0csAdV
AQVQAdV
A
At a section, uniform flow, density is constant
nnnAAVAdV
n
4-4 momentum equation for inertial control volume
)11.4(
csCVs
AdVdVtdt
dN
VPN
)17.4(
csCV
s
AdVVdVVtdt
Pd
Newton’s second law for a system moving relative to an inertial coordinate system
s
BS dtPdFFF
The sum of all forces (surface and body) acting on a nonaccelerating control volume is equal to the sum of the rate of change of momentum inside the control volume and the net rate of flux of momentum out through the control surface
CV ASB ApdFdvBF
We denote the body force per unit mass as B
csCVBSz
csCVBSy
csCVBSx
AdVwdVwt
FFF
AdVvdVvt
FFF
AdVudVut
FFF
zz
yy
xx
4-4.1 Differential Control volume analysis
Application of the basic equations to a differential control volume leads to differential equations describing the relationships among properties in the flow field (property variations) For the case Steady, incompressible, frictionless flow along a streamline, integration of one such differential equation leads to a useful relationship among speed, pressure, and elevation in a flow field,
The control volume is bounded by streamlines, flow across the bounding surfaces occurs only at the end section,
dAAdVVdppAVp
ss
s
,,,,,,
a. Continuity equation
Assumption: Steady,no flow across bounding streamlines, incompressible flow
Basic equation: )13.4(0
csCVAdVdV
t
00
0|))((|||
ss
sss
sss
AdVdAVdAdVAdVdAV
dAAdVVAV
b. Streamwise component of the momentum equation
Assumption: No friction, pressure forces only
Basic equation:
dAdppdAAdpppAF
AdVuAdVVFF
s
ss
S
cs scsBS
)2
())((
Where is the pressure force acting in the s direction on the bounding stream surface of the control volume
bsF
dzdAAgFdsdzwhere
dsdAAgdVgF
dAdpdpAF
s
s
s
B
sB
S
)2/(sin
)2/)(sin(2
The momentum flux will be
ss
sssssCS ss
AdVV
dAAdVVdVVAVVdAVu
|)))((|)((|)|(
0|))((||| dAAdVVAV sss Where : continuity
ss AdVVgdAdzgAdzdpdAAdp 21
21
dpdA and dAdz are negligible compared with the remaining term
)244(0)2
( gdzVddp s
For incompressible flow
)25.4(2
2
CgzVp
For an infinitesimal stream tube control volume, steady, incompressible flow without friction. We can get formation(4.25)
, the Bernoulli equation
4-4.2 Control volume moving with constant velocity
The previous equation based on the stationary control volume. A control volume (fixed relative to reference frame xyz) moving with constant velocity reference frame XYZ, is also inertial , since it has no acceleration with respect to XYZ
)27.4(
cs xyzxyzCV xyzSB AdVVdVV
tFFF
)26.4(
cs xyzCVs
AdVdVtdt
dN
4-5 Momentum equation for control volume with rectilinear acceleration
For an inertial control volume (having no acceleration relative to a stationary frame of reference xyz)
)27.4(
cs xyzxyzCV xyzSB AdVVdVV
tFFF
)2.4(
)2.4(
bdVVdmVP
adtPdF
systemsystem VMsystem
system
If we denote the inertial reference frame by XYZ,for accelerating control volume (4.27) is not right
systemsystem MXYZ
M XYZXYZ dm
dtVddmV
dtd
dtPdF
The velocity with respect to the inertial (XYZ) and the control volume coordinate(xyz) are related by the relative-motion Eq.
)30.4(rfxyzXYZ VVV
Where V is the velocity of the control volume reference frame.
)31.4(rfxyzrfxyz
XYZXYZ
dtVd
dtVd
dtdV
)32.4()sxyz
M rf dtPd
dmFsystem
)17.4(
csCV
s
AdVVdVVtdt
Pd
)34.4(
cs xyzxyzCV xyz
CV rfSBCV rf
AdVVdVVt
dVFFdVF
Linear momentum of the system, to derive the control volume formulation of Newton’s second law
csCVCS xyzxyzBS
csCVCS xyzxyzBS
csCVCS xyzxyzBS
AdVwdVwt
AdVuFF
AdVvdVvt
AdVvFF
AdVudVut
AdVwFF
zz
yy
xx
)35.4(
4-8 The first law of thermodynamics
)4.4(2
)4.4(
)4.4(
2
cgzVue
bdVeedmE
adtdEWQ
systemsystem VMsystem
system
dv
seVr
V
SEHPM
NsystemVsystem
1
)54.4(
csCVs
AdVedVetdt
dE
Since the system and the control volume coincide at t0
umecontrolvolsystem WQWQ ][][
)55.4(
csCV
AdVedVet
WQ
4-8.1 rate of work done by a control volume
othershearnormals WWWWW 1:Shaft work
2. Normal stresses at the control surface
VFdt
sdFt
wWtt
00limlim
cs nnnormal AdVW
3. shear stresses at the control surface
csshear dAVW
dAFd
Can be expressed as three terms(shaft; solid ports surface)
4 other work
)56.4(othershearcs nns WWAdVWW
4-8.2 control volume equation
cs nncsCVothershears AdVAdVedVe
tWWWQ
pnn
csCVothershears AdVgzVpvudVe
tWWWQ
)2
(2
4-9 the second law of thermodyhamics
)5.4(1 aQTdt
dS
system
)5.4( bdVssdmSsystemsystem MMsystem
)58.4(
csCVs
AdVsdVstdt
dS
dv
seVr
V
SEHPM
NsystemVsystem
1
)59.4(1 dAAQ
TAdVsdVs
t cscsCV
Introduction to differential analysis of fluid motion
The chapter 4 describe the basic equation in integral form for control volume. This chapter present the
differential equations in terms of infinitesimal systems and control volume
5-1 Conservation of mass
5-1.1 rectangular coordinate system
the control volume chosen is an infinitesimal cube with sides of length dx,dy,dz as shown in Fig5.1. The parameter of flow, density,velocity,pressure etc. is defined at center O point
wkvjuiV ˆˆˆ
To evaluate the properties at each of the six faces of the control surface, we use a Taylor series expansion about point O, at right
If we define:
22
2
2/)
2(
!21
2dx
xx
xdxx
Neglecting higher order term
22/
xxdxx
At the left face22/
xxdxx
22/
xxuuu
dxx
22/
xxuuu
dxx
Statement of conservation of mass is
0
volumecontroltheinside
massofchangeofRatesurfacecontrolthethrough
outfluxmassofrateNet
The first term in Eq. We must consider the mass flux through each of the six surfaces of the control surface:
csAdV
The net rate of mass flux out through the control surface is given by
dxdydzzw
yv
xu
The mass inside the control volume at any instant is the product of the mass per unit volume and the volume, dxdydz. The rate of
Change of mass inside the control volume is given by
dxdydzt
)1.5(0 atz
wyv
xu
zk
yj
xi
ˆˆˆ
)1.5(0 bt
V
5-1.2 Cylindrical Coordinate System
A suitable differential control volume for cylindrical coordinate is in Fig.5.2, also define desity, velocity, at the control volume center O
zrr VkVeVeV ˆˆˆ
csAdV
The mass flux through each of the six faces of control surface, from the Taylor series expansion about point O.
The net rate of mass flux out through the control surface is given by
dzdrdzVrV
rVrV zr
r
The mass inside the control volume at any instant is the product of the mass per unit volume and the volume drdzrd
The rate of change of mass inside the control volume is given by
drdzrdt
)2.5(01)(1
tz
VVrr
Vrr
zr
zk
re
rer
ˆ1ˆˆ
0
t
V
5-2 Stream function for 2-D incompressible flow
Relation between the streamlines and the statement of conservation of mass, for 2-D incompressible flow in the xy plane
0
yv
xu
If a continuous function , , called the stream function, is defined such that
),,( tyx
)4.5(x
vandy
u
The streamline ,at given instant, it tangent to the direction of flow at every point in the flow field,
0
)(ˆ)ˆˆ()ˆˆ(0
ddyy
dxx
vdxudy
vdxudyk
dyjdxivjuirdV
Where the time is defined at t0, the volume flow rate, Q, between streamlines and can be evaluated by consider the flow across AB or across BC. For a unit depth, the flow rate across AB
Is:
122
1
2
1
2
1
y
y
y
y
y
y
dQ
dyy
d
dyy
udyQ
Along BC, it is same as the side AB
For a 2-D, imcompressible flow in the cylindrical coordinate, conservation of mass,
xdxd /
rV
rV
tionstreamfunc
Vr
rV
r
r
1
0)(
5-3 Motion of a fluid element(kinematics)
5-3.1Fluid translation:Acceleration of a fluid particle in a velocity field
tV
zVw
yVv
xVua
DtVD
tV
zVw
yVv
xVu
tVdtdz
zVdtdy
yVdtdx
xVa
dttVdz
zVdy
yVdx
xVVd
dttdzzdyydxxVV
aandtzyxVV
p
pppp
pppp
pttp
pptp
///
),,,(
),,,(|
It include: total acceleration of a particle, convective acceleration, and local acceleration
)10.5()(
)(
tVVVa
DtVD
zVw
yVv
xVuVV
p
5-3.2 Fluid Rotation
A fluid particle moving may rotate about the axes,
zyx kji ˆˆˆ
The oa and ob rotate to the position shown during the interval dt
xxvvv oa
xv
txxv
tx
t
oa
ttoa
/limlim00
yu
ob
Similar, the angular velocity of line ob
The rotation of the fluid element about z axis is the average angular velocity of the two mutually perpendicular line elements
)(21
)(21
)(21
xw
zu
zv
yw
yu
xv
y
x
z
)14.5(21
)13.5()](ˆ)(ˆ)(ˆ[21
V
yu
xvk
xw
zuj
zv
ywi
The vorticity is defined as to be twice the rotation
)15.5(2 V
In cylindrical coordinates the vorticity is
)16.5()11(ˆ)(ˆ)1(ˆ
rzrzr
Vrr
rVr
kr
Vz
Vez
VVr
eV
The circulation is defined as the line integral of the tangential velocity component about a closed curve fixed in the flow
)17.5( C
sdV
For closed curve oacb
A zA zc
z
dAVdAsdV
yxyxxu
xv
yvxyxuuyx
xvvxu
)(2
2)(
)()(
5-3.3 fluid Deformation
a. Angular deformation:The angular changes between two mutually perpendicular line segments in the fluid. Fig.5.9 in the xy plane the rate of decrease of angle between lines oa and ob
)(900
yu
dtd
xv
txtxxv
tx
dtd
dtd
dtd
dtd
tt
)/)/(lim/lim00
The rate of angular deformation in the xy plane is yu
xv
dtd
b. Linear deformation
The element change length in the x direction only if du/dx=0, dv/dy, dw/dz, changes in the length of the sides may produce changes in volume of the element.
Volume dilation rate =
For incompressible flow, the rate of volume dilation is zero
Vzw
yv
xu
5-4 momentum equation
)2.4(
)2.4(
bdVVdmVP
adtPdF
systemsystem MMsystem
system
)22.5)((zVw
yVv
xVu
tVdm
DtVDdm
dtVddmFd
system
5-4.1 forces acting on a fluid particle
dxdydzxxx
dxdydzx
dxdydzx
dxdzdyx
dxdzdyx
dydzdxx
dydzdxx
dF
zxyxxx
zxzx
zxzx
yxyx
yxyx
xxxx
xxxxsx
)(
)2
()2
(
)2
()2
(
)2
()2
(
)23.5()( adxdydzxxx
gdFdFdF zxyxxxxSBx xx
)23.5()( bdxdydzxxx
gdFdFdF zyxyyyySBy yy
)23.5()( cdxdydzxxx
gdFdFdF zzyzxzzSBz zz
5-4.2 differential momentum equation
)24.5()(
)24.5()(
)24.5()(
czww
ywv
xwu
tw
yyyg
bzvw
yvv
xvu
tv
yyyg
azuw
yuv
xuu
tu
xxxg
zzyzxzz
zyyyxyy
zxyxxxx
5-4.3 Newtonian fluid :N-S equation
For a Newtonian fluid the viscous stress is proportional to the rate of shearing strain(angular deformation rate). The stresses may be expressed in terms of velocity gradients and fluid properties in rectangular coordinates as follow:
)25.5(232
)25.5(232
)25.5(232
)25.5()(
)25.5()(
)25.5()(
fzwVp
eyvVp
dxuVp
cxw
zu
bzv
yw
ayu
xv
zz
yy
xx
xzzx
zyyz
yxxy
If the expression for the stresses are introduced into the differential equations of motion(eqs 5.24) we obtain
)]([
)]([)]322([
zu
xw
z
xv
yu
yV
xu
xxpg
DtDu
x
)]([
)]([)]322([
yw
zv
z
xv
yu
xV
yv
yypg
DtDv
y
)]([
)]([)]322([
zu
xw
x
yw
zv
yV
zw
zzpg
DtDw
z
Chapter 6 Incompressible inviscid flow
Many flow cases is reasonable to neglect the effect of viscosity , no shear stresses are present in inviscid flow, normal stress are considered as the negative of the thermodynamic pressure -p
6-1 momentum equation for frictionless flow: Euler’s equations
)1.6()(
)1.6()(
)1.6()(
czww
ywv
xwu
tw
ypg
bzvw
yvv
xvu
tv
ypg
azuw
yuv
xuu
tu
xpg
z
y
x
)3.6(
)2.6())((
)(
DtVDpg
VVtVpg
zVw
yVv
xVu
tVpg
)4.6()1(
)4.6()1(1
)4.6()(2
cz
VVVr
Vr
VVt
Vzpg
brVV
zVVV
rV
rVV
tVp
rg
ar
Vz
VVVVr
VVt
Vrpg
zz
zzr
zz
rzr
rz
rrr
rr
6-2 Euler’s equation in streamline coordinates
The motion of a fluid particle in a steady flow, “streamline coordinates” also may be used to describe unsteady flow streamline in unsteady flow give a graphical representation of the instantaneous velocity field.
dsdndxadsdndxgdndxdssppdndxds
spp s
sin)2
()2
(
sagsp
sin
saszg
sp
sVV
tV
DtDVas
)5.6( asVV
tV
szg
sp
)5.6( bsVV
sp
dsdndxadsdndxgdsdxdnnppdsdxdn
npp n
cos)2
()2
(
nagnp
cos
nanzg
np
RVan
2
)6.6(2
aR
Vnzg
np
)6.6(2
bR
Vn
p
6-3 Bernoulli equation-integration of Euler’s equation along a streamline for steady flow
For incompressible inviscid flow
6-3.1 Derivation using streamline coordinates
Along streamline:
)9.6(2
)8.6(2
1
2
2
CgzVp
CgzVdp
VdVgdzdpsVV
szg
sp
c
6-3.3 Static, Stagnation, and Dynamic Pressure
The static pressure is that pressure which would be measured by an instrument moving with flow.
The stagnation pressure is obtained when a flowing fluid is decelerated to zero speed by a frictionless process. Neglecting elevation difference,
CVp
2
2
2
2
0Vpp
The dynamic pressure 2
21 V
)(2 0 ppV
mdmQ
dtdm
dmQ
dtQQ
)15.6()(22 122
222
1
211
dmQuugzVpgzVp
Incompressible /121 vv
6-5 unsteady Bernoulli equation-integration of Euler’s equation along a streamline
)3.6(ˆDtDVsd
DtVDkgp
)18.6(ˆ dstVds
sVVds
DtDVsd
DtVDsdkgsdp
)(
)(ˆ)(
salongVinchangethedVdssV
salongzinchangethedzdsk
salongpressureinchangethedpsdp
)19.6(dstVVdVgdzdp
)20.6(0)(2
2
112
21
222
1
dstVzzgVVdp
)21.6(22
2
12
222
1
211
ds
tVgzVpgzVp
Restrictions (1) incompressible
(2) Frictionless flow
(3) Flow along a streamline
6-6 irrotational flow
The fluid element moving in the flow field without any rotation
)23.6(0111
)22.6(0
021
rzrz Vrr
rVrr
Vz
Vz
VVr
yu
xv
xw
zu
zv
yw
V
6-6.1 Bernoulli equation applied to irrotational flow
CgzVp 1
211
2
)24.6()(21)(
21ˆ
)()(21)(
)10.6()(ˆ
2VVVkgp
VVVVVV
VVkgp
)25.6(2
21
)(21ˆ
2
2
2
CgzVp
dVgdzdp
rdVrdkgrdp
Since the dr was an arbitrary displacement Eq.6.25 is valid between any two points in a steady, incompressible, inviscid flow that is also irrotational.
6-6.2 Velocity potential
We formulated the stream function which relates the streamlines and mass flow rate in 2-D , incompressible flow.
we can formulate a relation called the potential function for a velocity field that is irrotational.
curl(grad ) = = 0 (6.26)
)29.6(1
)21.3(ˆ1ˆˆ
)28.6(
)27.6(
zV
rV
rV
ze
re
re
zw
yv
xu
V
zr
zr
The velocity potential exists only for irrotational flow. The stream function satisfies the continuity equation for incompressible flow; the stream function is not subject to the restiction of irrotational flow.6-6.3 stream function and velosity potential for 2-D , irrotational , incompressible flow: Laplace’s equation
)28.6(
)4.5(
yv
xu
xv
yu
Irrotational flow
)31.6(0
)3.5(0
)30.6(0
)22.6(0
2
2
2
2
2
2
2
2
yx
yv
xu
equationcontinuityyx
yu
xv
Along streamline c
Along a line of constant , d = 0
We see that the slope of a constant streamline at any point is the negative reciprocal of the slope of constant velocity potential line at that point; lines of constant stream and constant velocity potential are orthogonal
6-6.4 Elementary Plane Flow
A variety of potential flows can be constructed by superposing elementary flow patterns, five elementary 2-D flows-----a uniform flow, a source, a sink, a vortex, and a doublet----are summaried in Tablet 6.1
6-6.5 superposition of elementary plane flows
Both stream function and velocity potential satisfy Laplace’s equationfor flow that is both incompressible and irrotational . Sincer Laplace’s equation is a linear, homogeneous partial differential equation , solution may be superposed to develop more complex and interesting patterns of flow
Chapter 7 Dimensional Analysis And Similitude
The real physical flow situation is approximated with a mathematical model that is simple enough to yield a solution, then experimental measurements are made to check the analytical results. Experimental measurement is very time-consuming and expensive. When experimental testing of a full-size prototype is either impossible or prohibitively expensive. The model flow and the prototype flow must be related by known scaling laws.
7-1 Nature of dimensional analysis
The physical parameters can write the symbolic equation: ),,,( VDfF
We need do many experiments for determining the parameter (diameter, velocity,density,fluid viscosity), through the use of dimensional analysis, we can get very useful formulation ( example 7.1)
)(12
VDf
DVF
The Buckingham Pi Theorem is a statement of the relation between a function expressed in terms of dimensional parameters and a related function expressed in terms of nondimensional parameter
7-2 Buckingham Pi Theorem
The dependent parameter is a function of n-1 independent parameters, we may express the relationship among the variables in functional form as
),...,,( 211 nqqqfq
0),...,,,( 221 nqqqqg),,,( VDfF 0),,,,( VDFg
The n parameters may be grouped into n-m independent dimensionless ratios, or II parameters, expressible functional form by
),....,,(
0),....,,(
3211
21
mn
mn
GorG
The number m is usually equal to the minimum number of independent dimensions required to specify the dimensions of all the parameters
2
4/31
632
15
3
2
or
7-3 Determining the II groups
The six steps listed below outline a recommended procedure for determining the II parameters
Step1. List all the dimensional parameters involved
Step 2. Selected a set of fundamental dimensions MLt ..
Step 3. List the dimensions of all parameters in terms of
primary dimensions
Step 4. Select a set of r dimensional parameters that
include all the primary dimensions
Step 5. Set up dimensional equations, combining the
parameters selected in step 4 with each of the
Other parameters in turn , to form dimensionless
group
Step 6. Check to see that each group obtained is dimensionless
7-4 Significant Dimensionless groups in fluid mechanics
In flow field, we use physical force such as interia, viscous, pressure, gravity, surface tension, and compressibility
Viscous force
Pressure force
Gravity force
Surface tension force
VLLLVA
dyduA 2
2)()( LpAp
L
3Lgmg
DVDV
Re
VLVL
ReRe No.
Pressure coefficient:2
21 V
pCp
Cavitation phenomena, the pressure express as cavitation number:
2
21 V
pp v
Froude number was significant for flows with free surface effects which may be interpreted as the ratio of inertia force to gravity forces.
3
222
gLLV
gLVFr
gLVFr
The Weber number is the ratio of inertia forces to surface tension forces
LVWe
2
vE
V
ddpV
cVM Compressibility effects
2
222
LELVM
v
As a ratio of inertia forces to forces due to compressibility
7-5 Flow similarity and model studies
Geometric similarity..model and prototype have same shape and both flow are kinematically similar
Kinematically similar: velocities at corresponding points are in the same direction and are related in magnitude by a constant scale factor, the streamline patterns related by a constant scale factor
Kinematic similarity requires that the regimes of flow be the same for model and prototype.
Then the dependent parameter is duplicated between model and prototype
prototypeel
VDVD
mod
prototypeel ReRemod
prototypeel DVF
DVF
22
mod22
And the result determined from the model study can be used to predict the drag on the full-scale prototype. As long as the Reynolds numbers are matched. The actual force on the object due to the fluid have the value of its dimensionless group.
Effects are absent from the model test.)(122
VDfDV
F prototypeel ReRemod
The prototype condition Vp = 8.44 ft/s
51099.4Re p
ppp
DV
51099.4Re m
mmm
DV
sftD
Vm
mmm /157Re
Dynamically similar
prototypeel DVF
DVF
22
mod22
lbfDD
VV
FFm
p
m
p
m
pmp 9.532
2
2
2
7-6 Nondimensionalizing the basic differential equation
Use the Buckingham Pi theorem , a more rigorous and broader approach to determine the conditions under which two flows are similar is to use the governing differential equations and boundary conditions. Two physical phenomena are governed by differential equations and boundary conditions that have the same dimensionless forms. Dynamic similarity is guaranteed by duplicating the dimensionless coefficients of the equations and boundary conditions between prototype and model.
Nondimensionalizing the basic differential equation, steady incompressible 2-D flow in the xy plane
)9.7(
)8.7(
)7.7(0
2
2
2
2
2
2
2
2
yv
xv
ypg
yvv
xvu
yu
xu
xp
yuv
xuu
yv
xu
2*****
Vpp
Vvv
Vuu
Lyy
Lxx
The pressure nondimensional by dividing by 2V
)13.7(
)12.7(
)11.7(0
2*
*2
2*
*2
2*
*2
*
**
*
**
2
2*
*2
2*
*2
2*
*2
*
**
*
**
2
*
*
*
*
yv
xv
LV
yp
LVg
yvv
xvu
LV
yu
xu
LV
xp
LV
yuv
xuu
LV
yv
LV
xu
LV
)16.7(
)15.7(
)14.7(0
2*
*2
2*
*2
*
*
2*
**
*
**
2*
*2
2*
*2
*
*
*
**
*
**
*
*
*
*
yv
xv
LVyp
VgL
yvv
xvu
yu
xu
LVxp
yuv
xuu
yv
xu
The differential equations for two flow system will be identical if the quantities
Are the same for both flows. Thus , model studies to determine the drag force on a surface ship require duplication of both the Froude number and the Reynolds number to ensure dynamically similar flows.
Emphasize that in addition to identical nondimensional equation, the nondimensional boundary conditions also must be identical if the two flow are to be kinematically similar . The periodic flow define the velocity on the boundary:
Nondimensionalize time:
2// VgLandLV
t V ubc sin
LVtt *
* *sint
VL
Vu u
bcbc
Duplication of the boundary condition requires that parameter be the same between the two flows. This parameter is the Strouhal number
VL /
V
LSt
Chapter 8 Internal incompressible viscous flow
Flow completely bounded by solid surfaces are called internal flows: pipes, nozzles, diffusers, sudden contractions and expansions, valves, and fittings.Laminar and turbulent flow, some laminar flow may be solved analytically, the case of turbulent flow we must rely heavily on semi-empirical theories and on experimental data. The flow regime is primarily a function of the Reynolds number.
8-1 introduction
The pipe flow regime(laminar or turbulent) is determined by the Reynolds number, the qualitative For laminar flow, the entrance length,L, is the function of Reynolds number,
DV
DL 06.0
Part A Fully Developed Laminar Flow
8-2 Fully developed laminar flow between infinite parallel plates
8-2.1 Both plates stationary
Boundary at y=0 u=0; y=a u=0. u=u(y)(v=w=0)
For analysis we select a differential control volume of size dV=dxdydz and apply the x component of the momentum equation
Assumption (1) steady flow (2) fully developed flow
(3) FBx = 0
Basic equation
For fully developed flow, the net momentum flux through the control surface is zero. FSx = 0
The next step is to sum the forces acting on the control volume in the x direction. We recognize that the normal forces(pressure forces) act on the left and right aces and tangential forces(shear forces) act on the top and bottom faces faces.
)19.4( aAdVuFcsSx
dxdzdydy
ddF
dxdzdydy
ddF
dydzdxxppdF
dydzdxxppdF
yxyxT
yxyxB
R
L
)2
(
)2
(
)2
(
)2
(
)4.8(21
)3.8(0
212
1
1
ccyxpu
cyxp
dydu
dydu
cyxpC
xp
dyd
dyd
xp
yx
yxyx
yx
According to boundary condition
)5.8()(2
22
ay
ay
xpau
Shear stress distribution
)6.8(21)( a
ay
xpayx
Volume flow rate
)6.8(12
1
)(21
3
2
0
baxp
lQ
dyayyxpAdVQ
a
A
Flow rate as function of pressure drop
)6.8(12
3
12
cLpa
LQ
Lp
Lpp
xp
Average velocity
)6.8(12
112
1 23
daxp
lala
xp
AQV
Point of maximum velocity
)6.8(23)(
812/0
12)(2
2max
2
eVaxpuay
dydu
aay
xpa
dydu
Transformation of Coordinates
Transform from y=0 at bottom to y=0 at centerline
)7.8(41)(
2
22
ay
xpau
8-2.2 Upper plate moving with constant speed, U
The boundary condition
u=0 at y=0; u=U at y=a
(8-4) is equally valid for the moving plate case , velocity distribution is given by
)4.8(21
212 ccy
xpu
From BC. We have
Uuaycuy 000 2
)8.8()()(2
21
21
22
112
ay
ay
xpa
aUyu
axp
aUcaca
xpU
Shear stress distribution
)9.8(21)( a
ay
xpa
aU
yx
Volume flow rate
)9.8(12
12
)(21[
3
2
0
0
baxpUa
lQ
dyayyxp
aUy
lQ
uldyAdVQ
a
a
A
Average Velocity
)9.8(12
12
/]12
12
[3
caxpUla
lala
xpUal
AQV
Point of Maximum velocity
)/)(/1(/
20
12)(2 2
2
xpaUay
dydu
aay
xpa
aU
dydu
8-3 Fully developed laminar flow in a pipe
For a fully developed steady flow, the x component of momentum equation applied to the differential control volume, reduce to
0xSF
On the control volume in the x direction. Normal forces (pressure surface) acting on the left and right ends of the control volume, and that tangential forces(shear forces) act on the inner and outer cylindrical surfaces as well
dxdrrdrdr
ddF
rdxdF
rdrdxxppdF
rdrpdF
rxrxO
rxl
R
L
)(2)(
2
2)(
2
Pressure force on the left
Pressure force on the right
Shear force on the inner cylindrical
On the outer cylindrical
)11.8(ln4
2
21
)10.8()(1
21
2
1
1
2
crcxpru
rc
xpr
drdu
drdu
cxprrC
xp
drd
r
drrd
rdrd
rxp
rx
rxrx
rxrxrx
Boundary condition u=0 r=R, and the physical considerations that the velocity must be finite at r=0, the only way that this can be true is for c1 to be zero
xpRcc
xpru
44
2
22
2
)12.8(14
22
Rr
xpRu
)13.8()(2
axpr
yx
)13.8(8
2)(41
4
22
0
bxpRQ
rdrRrxpAdVQ
R
A
)13.8(128
4
12
cL
pDQ
Lp
Lpp
xp
Flow rate as a
function pf pressure drop
Volume flow rate
Shear stress distribution
velocity
)13.8(8
2
2 dxpR
RQV
)13.8(2)(4
00
)(2
22
max eVaxpRUur
drdu
xpr
drdu
The velocity profile(8.12) can be written in terms of the maximum velocity as
)14.8()(1 2
Rr
Uu
The maximum velocity is on the point
The average velocity
Part B Flow in Pipes and Ducts
This section is to evaluate the pressure changes from the flow velocity and from friction.
To develop relations for major losses due to friction in constant-area ducts,
8-4 Shear stress distribution in fully developed pipe flow
In fully developed steady flow in a horizontal pipe, be it laminar or turbulent, the pressure drop is balanced only by shear forces at the pipe wall.
Assumption: Horizontal pipe,
Steady flow, incompressible flow, Fully developed flow
The x component of the momentum equation:
0xBF
csCVBSx AdVudVu
tFFF
xx
The shear stress on the fluid varies linearly across the pipe, from zero at the centerline to a maximum at the pipe wall, at the surface of the pipe
)16.8(2
][xpR
Rrrxw
To relate the shear stress field to the mean velocity field, we could determine analytically the pressure drop over a length of pipe for fully developed flow for laminar flow.
In turbulent flow, no simple relation exists between the shear stress field and the mean velocity field. For fully developed turbulent pipe flow, the total shear stress is:
)17.8(vudyud
turblam
The profile fits the data close to the centerline, it fails to give zero slope there. It give adequate results in many calculation.
For Re>2x104 : n=-1.7+1.8logReU (8.23)
A
AdVQandAQV
/
The ratio of the average velocity to the centerline velocity
)24.8()12)(1(
2 2
nnn
UV
8-6 Energy considerations in pipe flow
By applying the momentum equation for a control volume with the formulation of conservation of mass, we have derived all the results. About conservation of energy-the first law of the thermodynamics, we can get insight into the nature of the pressure losses in internal viscous flows can be obtained from energy equation
)57.4()(
csCVothershears AdVpvedVe
tWWWQ
gzVue 2
2
5-6.1 kinetic Energy Coefficient
Use as the Kinetic energy coefficient
)26.8(2
)26.8(222
2
2
222
bVm
dAVV
aV
mdAVV
dAVV
A
AA
For laminar flow in a pipe, = 2.0
In turbulent pipe flow, the velocity profile is quite flat(fig.8.11), substitute the power-law velocity profile into(8.26b)
)27.8()23)(3(
2 23
nnn
VU
8-6.2 head loss
Using the definition of , the energy equation can be written
)22
()()()(2
112
2212
1212
VVmzzgmppmuumQ
)28.8()()2
()2
( 122
2222
1
2111
dmQuugzVpgzVp
dmQuu
gzVp
)(
)2
(
12
2
The mechanical energy per unit mass at a cross section
The difference in mechanical energy per unit mass between section(1) and (2), it represents the conversion of mechanical energy at section (1) to unwanted thermal energy(u2-u1)and the loss of energy via heat transfer. We identify this group of terms as the total energy loss per unit mass and designate it by the symbol
lh
)29.8()2
()2
( 2
2222
1
2111
lhgzVpgzVp
For incompressible, frictionless flow, there is no conversion of mechanical energy to internal energy
)2
(2
gzVp
For viscous flow in a pipe, one effect of friction may be to increase the internal energy of the flow Eq.(8.28)
As the empirical science of hydraulics developed, it was common practice to express the energy balance in terms of energy per unit weight of flowing liquid rather than energy per unit mass
)30.8()2
()2
( 2
2222
1
2111
l
l Hgh
zgV
gpz
gV
gp
Equation(8.29) and (8.30) can be used to calculate the pressure difference between any two points in a piping system, provided the head loss. (or )
lh
lH
8-7 Calculation of head loss
lh total head loss is regarded as the sum of major loss,
, due to frictional effects in fully developed flow in constant-area tubes, and minor losses due to entrances , fittings, area changes, and so on
lh
mlh
8-7.1 Major losses: friction factor
For fully developed flow = 0 and
Eq.(8.29) becomes
mlh C
V
2
2
)31.8()( 1221
lhzzgpp
If the pipe is horizontal )32.8(21lhppp
Since head loss represents the energy converted by frictional effects from mechanical to thermal energy , head loss for fully developed flow in a constant-area duct depends only on the details of the flow through the duct. Head loss is independent of pipe orientation.
a. Laminar flow
)33.8(2Re
6432
32)4/(128128
)13.8(128
2
4
2
4
4
VDL
DV
DLh
DV
DL
DDVL
DQLp
cL
pDQ
l
b. Turbulent flow
The pressure drop can not be evaluate, we get it from the experimental results and use dimensional analysis. The pressure drop in fully developed turbulent flow due to friction is depended on pipe diameter, D, pipe length, L, pipe roughness, e, average flow velocity, V, fluid density, and fluid viscosity
)32.8(21lhppp
),(Re,2 De
DL
Vhl
)35.8(2
2
fDL
gVH l
The friction factor is determined experimentally(Fig 8.13)
To determine head loss for fully developed flow with known conditions, the Reynolds number is evaluated first, Roughness, e, is obtained from Table8.1 the friction factor f is read from the appropriate curve in Fig8.13,at the known values of Re and e/D.
For laminar flow, the friction factor from (8.33) and (8.34)
)36.8(Re64
22Re64
min
22
arla
l
f
VDLfV
DLh
e/D>0.001 Re>Re(transition), the friction factor is greater than the smooth pipe value.
In general, the Re number is increased, the friction factor decreases as long as the flow remain laminar. At transition f increases sharply. In the turbulent flow regime, the friction factor decreases gradually and finally levels out at a constant value for large Reynolds number .
The Colebrook formula for friction factor
)37.8()Re
51.27.3
/log(0.215.05.0 a
fDe
f
)37.8()]Re
74.57.3
/[log(25.0 29.00 bDef
For turbulent flow in smooth pipes, the Blasius correlation
)38.8(Re
316.025.0f
The wall shear stress is obtained as
)39.8()(0332.0 25.02
VRVw
8-7.2 Minor losses(K:loss coefficient from experiments)
)40.8(2
)40.8(2
22
bVDLfhaVKh e
ll mm
Where Le is an equivalent length of straight pipe.
a. Inlets and Exits
b.Enlargements and Contraction
Fig 8.15 gives the results for sudden expansion and constraction
Losses in diffusers depend on a number of geometric and flow variables. Diffuser data are in terms of a pressure recovery coefficient defined as the static pressure rise to inlet dynamic pressure
)41.8(
21 2
1
12
V
ppC p
If the gravity is neglected and 121
mll hhVpVp
)
2()
2(
222
211
pl C
AAVh
m
2
2
12
1 )(1(2
2211 VAVA
)42.8()(
11(2 2
21
pl C
ARVh
m
For frictionless flow
)43.8(11 2AR
Cpi
0ml
h
Applying the Bernolli equation with the mass conservation for frictionless flow through the diffuser, the head loss for flow through an actual diffuser maybe written
)44.8(2
)(2
1VCCh ppl im
c.pipe Bends
d. Valves and Fittings
Table 8.4(p367)
8-7.3 Noncircular Ducts
If the square or rectangular cross section may be treated if the ratio of height to width is less than about 3 or 4. The correlation for turbulent pipe flow are extended for use with noncircular geometries by introducing the hydraulic diameter, defined as
)45.8(4PADh
A is cross-sectional area, and P is wetted perimeter.
For a rectangular duct of width b and height h, A=bh and P=2(b+h), and the aspect ratio ar=h/b, then
arhDh
1
2
8-8 solution of pipe flow problem
From the total head loss, pipe flow problem can be solved using the energy equation Eq.8.29 . Consider single-path pipe flow problem.
8-8.1 single-path system
The pressure drop through a pipe system is a function of flow rate,elevation change,and total head loss.(1. Major losses due to friction in constant-area section(Eq.8.34) and minor losses due to fittingd,area changes, and so forth (Eq.8.40). The pressure drop
For the fixed pipe flow(incompressible,roughness,elevation change)
),,,,,,,(3 ionconfiguratsystemzeDQLp
)46.8(),,(4 DQLp
(a) L,Q,and D known, unknown
friction factor from fig8.13, the total head loss is computed from Eqs.8.34 and 8.40, Eq.29 used to calculate the (Example 8.5)
(b) ,Q, and D known, L unknown
Eq8.29 for head loss. Friction factor calculate from Re and e/D, Eq.8.34 for unknown length(Example8.6)
© , L,and D known, Q unknown
(8.29) and lead loss; the result is an expression for average V(or Q) in terms of the friction factor f (see P369)
(d) , L,and Q known, D unknown
the problem is to determine the smallest pipe size that can deliver the desired flow rate. Since the pipe diameter is
p
p
p
p
p
unknown, neither the Reynolds number nor relative roughness can be computed directly, and an iterative solution is required. Example8.8
lossesmppQmuumppW
QmpumpuWW
AdVgzVpudVet
WQ
in
sin
csCVs
1212
12
11
22
2
)(
))(()(
)2
(
The losses are determined in terms of the pump efficiency
hpW
pAVAVppmppW
Wlosses
in
in
in
36800
)(11
)1(
1212
8-8.2 Pumps in fluid systems
The driving force causing the fluid motion was explicitly stated as either a pressure difference or as an elevation difference. The energy per unit mass added by the pump is calculated
)47.8(22
22
arg
arg
suctionedisch
pumppump
suctionedischpump
gzVpgzVpm
Wh
gzVpgzVpmW
)48.8(22 2
22
22
1
21
11
Tlpump hhgzVpgzVp
Part C Flow measurement
8-9 direct methods
For determine flow rate (volume or mass of liquid collected)
8-10 Restriction flow meters for internal flows
Restriction flow meter are based on acceleration of a fluid stream through some form of nozzle. Flow separation at the sharp edge of the nozzle throat causes a recirculation zone to form , as shown by the dashed line lines downstream from the nozzle. The mainstream flow continues to acceleration from the nozzle throat to form a vena contracta at section (2) and then decelerates again to fill the duct. At the vena contracta, the flow area is minimum, the flow streamlines are essentially straight, and the pressure is uniform across the channel section
Assumption: steady, incompressible, flow along a streamline, no friction, uniform velocity at section,pressure is uniform across sections and z1=z2
)13.4(0
csCVAdVdV
t
2
2
12
221
2221
2
222
1
211
12
)(2
)9.6(22
VVVVVpp
gzVpgzVp
)49.8(])/(1[
2
12
0
212
212
2
1
22
221
2
1
2
2
2
1
2211
AAppV
AAVpp
AA
VV
AVAV
The theoretical mass flow rate is then given by
)50.8()(2)/(1
21212
222 pp
AA
AAVm ltheoretica
)52.8()p p(21
)/(/
)51.8()p p(2)/(1
214
21
41
2121
tactual
tt
t
tactual
CAm
AADD
AA
CAm
The above formulation is adjusted for Reynolds number and diameter ratio by defining an empirical discharge coefficient such that
The velocity-of-approach factor and The discharge coefficient are combined into a single flow coefficient
)54.8()(2
)53.8(1
21
4
ppKAm
CK
tactual
For the turbulent flow regime, the discharge coefficient
)55.8(Re
1
nD
bCC
The corresponding form for the flow-coefficient equation is
)56.8(Re1
1
14 n
D
bKK
8-10.1 The Orifice Plate
The thin plate , the pressure gaps for orifices may be placed in several locations, the location of the pressure taps influences the empirically determined flow coefficient, you need select handbook values of C or K consistent with the location of pressure gaps
74
75.0
5.281.2
10Re1075.02.0
)57.8(Re71.91184.00312.05959.0
1
1
D
D
C
8-10.2 The flow Nozzle
74
5.0
5.0
10Re1075.025.0
)53.8(Re53.69975.0
1
1
D
D
C
a. Pipe installation (K is function of and )
b. Plenum installation 0.95<K<0.99
1
ReD
8-10.3 The Venturi
Venturi meters are heavy,bulky,and expensive. The conical diffuser section downstream from the throat gives excellent pressure recovery; overall head loss is low. The discharge coefficients for venturi meters range from 0.980 to 0.995 at Re >2x10e+5. C=0.99