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
motion
? 3,The principle of angular momentum
? 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
m
VVC ?
??
??
)22(),,,( ?? tzyx??
?
2-2 velocity field
? 2-2.1 one-,two-,and three-Dimensional flows
),,,( tzyxVV ?? ? kwjviuV ??? ????
Steady flow unsteady flow
0.,,,, ),,( ??? 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:
n?
F??
A??
n? n
?
)7.2(l i m
0 n
t
An A
F
n ?
??
? ?
?
)6.2(lim 0
n
n
An A
F
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
A
xz
x
y
A
xy
x
x
A
xx
A
F
A
F
A
F
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 dA
dF
A
F
y
??
? ?
??
? 0
l i m
dt
d
tr a t end e f o r m a t i o t
?
?
??
?
??
? 0
lim
dy
du
dt
d
y
u
t
yltul
??
??
?
?
?
?
??
???????
2-4.1 Newtonian fluid
Fluid as water,air,and gasoline are Newtonian fluid
dy
du
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
dy
du
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 dydudydudydukdyduk nnyx ?? ??? ?
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
dy
du
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 Static
Absence 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?
d xd yd zgdVgdmgFd B ???? ?? ???
)2()( dyyppyyyppP LL ??????????
2)(
dy
y
ppyy
y
ppP
RL ?
????
?
???
d xd y d zkzpjypixpFd s )???( ???????????
)1.3()( p d x d y d zd x d y d zg r a d pFd s ??????
The gradient of a scalar field gives a vector field
)2.3(
)(
gp
dV
Fd
dVgpFdFdFd sB
?
?
????
?
?
????
??????
For a static fluid,
)3.3(0
00
????
???
gp
amFda ? ???
?
)6.3(
)5.3(00
??
?
????
??
?
?
?
?
?
?
?
?
g
dz
dp
g
z
p
y
p
x
p
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 ???
?? ????? zzpp zzgppg d zdp 00 )( 00 ??
hzz ??0 )7.3(0 ghpp ???
Incompressible liquid:manometers
?
1122
22
11
ghghpp
ghpp
ghpp
CA
CB
BA
??
?
?
???
??
???
3-3.2 Gases (compressible fluid)
The pressure varies with altitude or temperature
)8.3()()1(
)(
/
0
0
/
0
0
0
mRgmRg
T
T
p
T
mz
pp
dz
mzTR
pg
dz
RT
pg
g d zdp
???
?
?????? ?
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 ??
???? s i n000 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 ????? ''' ?????? ???
)12.3('' ?? ?? AA RR x pd AFxan dy pd AFy
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(??? p dAFF RR ?
? ??? A ARR x pd AFxandy pd AFy ''
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 y dAgAppdAF ?? s i n0
The yc is the y coordinate of the centroid of the area A thus
)14.3()(s i n 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(
s i ns i n
??2
2
a
Ay
I
yydAyI
dAygg h d AyAygy
y p d AFy
c
xx
c
A
xx
AA
c
A
R
????
???
??
?
??
?
?????
xxI??
Is the second moment of the area about the centroid axisx?
The same ruler,
)15.3(?? bAyIxx
c
yx
c ???
3-5.3 Hydrostatic force on a curved submerged surface
)16.3(???
)10.3(
)9.3(
zyx RRRR
A
R
FkFjFiF
ApdF
ApdFd
????
??
??
???
??
??
?
??? ????????? xx A xARR p d AiApdiFdiFF ??? ???
In general in the l direction the component of resultant force is,
??? ll A lR p d AF
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.
gVg d VdAghF
ghpp d AFF
ZV
ZVR Z
???
?
???
???
??
?
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 adtdM s y s t e m ?
)1.4( bdVdmM
s y s t e ms y s t e m MMs y s t e m ??
?? ?
4-1.2 Newton’s second law ( Conservation of momentum )
)2.4(
)2.4(
bdVVdmVP
a
dt
Pd
F
s y s t e ms y s t e m MM
s y s t e m
s y s t e m
?? ??
?
?
?
?
?
?
???
?
?
4-1.3 The angular Momentum principle
)3.4(
)3.4(
)3.4(
aTdmgrFrT
bdVVrdmVrH
a
dt
Hd
T
s h a f t
M
s
HM
s y s t e m
s y s t e m
s y s t e m
s y s t e ms y s t e m
??????
?????
?
?
?????
????
?
?
?
?
?
?
?? ?
4-1.4 The first of thermodynamics
)4.4(
2
)4.4(
)4.4(
2
cgz
V
ue
bdVee d mE
a
dt
dE
WQ
dEWQ
s y s t e ms y s t e m
VM
s y s t e m
s y s t e m
???
??
?
?
?
??
??
?? ?
??
??
4-1.5 The second Law of Thermodynamics
)5.4(
)5.4(
1
bdVss d mS
aQ
Tdt
dS
T
Q
dS
s y s t e ms y s t e m
VM
s y s t e m
s y s t e m
?? ??
??
?
?
?
?
?
?
4-2 Relation of system derivatives to the control volume
formulation
dv
s
e
Vr
V
S
E
H
P
M
N
s y s t e m
V
s y s t e m
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
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
dt
dN tcvttIIIIcv
ts y s t e m ?
???? ??
??
? )9.4()lim)lim)lim
0000
000 t
N
t
N
t
NN
dt
dN ttI
t
ttIII
t
tcvttcv
ts y s t e m ?
???? ?? ??
??
??
??
??
??
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
V
t
l
dAldVdN
t
?
?
?
?
???
?? 0
l i m
)c o s( ?????
)11.4(
co s
??
?
??
?
?
??
?
?
???
?
?
?
?
??
?
?
csCV
s
cs
ss
AdVdV
tdt
dN
AdV
t
N
dt
dN
??
??
????
???
4-3 Conservation of mass:(N=m; )
1??
)11.4(?? ??
?
???
?
?
csCV
s
AdVdV
tdt
dN ??????
)12.4(?? ??
?
???
?
?
csCV
s
AdVdV
tdt
dM ????
)13.4(0????? ??
csCV
AdVdVt
??
??
conservation
4-3.1 special cases
incompressible,nondeformable control volume of fixed size
and shape
0???cs AdV ???
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(0???cs AdV ???
A
QVQAdV
A
????
??
At a section,uniform flow,density is constant
nnnA AVAdVn
???? ???? ??
4-4 momentum equation for inertial control volume
)11.4(?? ??
?
???
?
?
csCV
s
AdVdV
tdt
dN ??????
VPN ?? ?? ?
)17.4(?? ??
?
??
???
?
csCV
s
AdVVdVV
tdt
Pd ????
?
??
Newton’s second law for a system moving relative to an inertial
coordinate system
s
BS dt
PdFFF
???
?
???
????
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?
??
??
??
??
?
?
???
??
?
?
???
??
?
?
???
csCV
BSz
csCV
BSy
csCV
BSx
AdVwdVw
t
FFF
AdVvdVv
t
FFF
AdVudVu
t
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,
dAAdVVdpp
AVp
ss
s
???,,,
,,,
?
?
a,Continuity equation
Assumption,Steady,no flow across bounding streamlines,
incompressible flow
Basic equation:
)13.4(0????? ?? csCV AdVdVt ????
0
0
0|))((|||
??
???
?????
ss
sss
sss
A d VdAV
d A d VA d VdAV
dAAdVVAV ??
b,Streamwise component of the momentum equation
Assumption,No friction,pressure forces only
Basic equation,
dA
dp
pdAAdpppAF
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 b
sF
dzdAAgF
dsdzw h e r e
dsdAAgdVgF
dA
dp
d p AF
s
s
s
B
sB
S
)2/(
s i n
)2/)(s i n(
2
???
?
????
???
?
?
???
The momentum flux will be
ss
sssssCS ss
A d VV
dAAdVVdVVAVVdAVu
?
???
?
???????? |)))((|)((|)|(
?
0|))((||| ????? dAAdVVAV sss ??
Where,
continuity
ss A d VVg d A d zg A d zd p d AA d p ??? ????? 2
1
2
1
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 x y zx y zCV x y zSB AdVVdVVtFFF ??????? ??
)26.4(?? ??????
?
?
cs x y zCVs
AdVdVtdtdN ??????
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 x y zx y zCV x y zSB AdVVdVVtFFF ??????? ??
)2.4(
)2.4(
bdVVdmVP
a
dt
Pd
F
s y s t e ms y s t e m VM
s y s t e m
s y s t e m
?? ??
?
?
?
?
?
?
???
?
?
If we denote the inertial reference frame by XYZ,for accelerating
control volume (4.27) is not right
?? ???????
s y s t e ms y s t e m M
X Y Z
M X Y Z
X Y Z dm
dt
VddmV
dt
d
dt
PdF
????
The velocity with respect to the inertial (XYZ) and the control
volume coordinate(xyz) are related by the relative-motion Eq.
)30.4(rfx y zX Y Z VVV ??? ??
Where V is the velocity of the control volume reference frame.
)31.4(rfx y zrfx y zX Y ZX Y Z dtVddtVddtdV ??? ??
??
? ?????
)32.4() sx y zM rf dtPddmF
s y s t e m
?
?? ?? ? ?
)17.4(?? ???????
?
?
csCV
s
AdVVdVVtdt Pd ????
?
??
)34.4(??
??
??
?
?
?
????
cs x y zx y zCV x y z
CV rfSBCV rf
AdVVdVV
t
dVFFdVF
????
?????
??
????
Linear momentum of the system,to derive the control volume
formulation of Newton’s second law
???
???
???
??
?
?
????
??
?
?
????
??
?
?
????
csCVCS
x y zx y zBS
csCVCS
x y zx y zBS
csCVCS
x y zx y zBS
AdVwdVw
t
AdVuFF
AdVvdVv
t
AdVvFF
AdVudVu
t
AdVwFF
zz
yy
xx
????
????
????
???
???
???
)35.4(
4-8 The first law of thermodynamics
)4.4(
2
)4.4(
)4.4(
2
cgz
V
ue
bdVee d mE
a
dt
dE
WQ
s y s t e ms y s t e m
VM
s y s t e m
s y s t e m
???
??
?
?
?
??
?? ?
??
dv
s
e
Vr
V
S
E
H
P
M
N
s y s t e m
V
s y s t e m
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
1
)54.4(?? ??
?
???
?
?
csCV
s
AdVedVe
tdt
dE ????
Since the system and the control volume coincide at t0
u m ec o n t r o l v o ls y s t e m WQWQ ][][ ???? ???
)55.4(?? ??????
csCV
AdVedVetWQ
????
??
4-8.1 rate of work done by a control volume
o t h e rs h e a rn o r m a ls WWWWW ????? ????
1:Shaft work
2,Normal stresses at the control surface
VFdt sdFtwW
tt
????? ????
?? ???? 00 limlim
?
? ??? cs nnn o r m a l AdVW ??? ?
3,shear stresses at the control surface
? ???
?
css he ar
dAVW
dAFd
??
?
??
?
?
Can be expressed as three terms(shaft; solid ports surface)
4 other work
)56.4(o t h e rs h e a rcs nns WWAdVWW ?????? ????? ? ?
4-8.2 control volume equation
??? ????????? cs nncsCVo t h e rs h e a rs AdVAdVedVetWWWQ ???????? ???
pnn ???
?? ??????????? csCVot h e rs h e ars AdVgzVpvudVetWWWQ ?????? ?? )2(
2
4-9 the second law of thermodyhamics
)5.4(1 aQTdtdS
s y s t e m
???
?
?
)5.4( bdVss d mS
s y s t e ms y s t e m MMs y s t e m ??
?? ?
)58.4(?? ??
?
???
?
?
csCV
s
AdVsdVs
tdt
dS ????
dv
s
e
Vr
V
S
E
H
P
M
N
s y s t e m
V
s y s t e m
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
1
)59.4(1 dA
A
Q
T
AdVsdVs
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 ??? ????
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(!2
1
2
dx
x
x
xdxx
????
Neglecting higher order term
22/
x
xdxx
?
?
???
?
???
At the left face
22/
x
xdxx
?
?
???
?
???
22/
x
x
uuu
dxx
?
?
???
? 22/ xxuuu dxx ??????
Statement of conservation of mass is
0?????????????? v ol u m ec o nt r olt hei ns i d e m as sofc h an g eofR a t es u r f a c ec o nt r olt het hr o ug h o utf l u xm as sofr a t eN e t
The first term in Eq,We must consider the mass flux through each
of the six surfaces of the control surface,? ?
cs AdV
???
The net rate of mass flux out through the control surface is given
by
dx dy dzzwy vxu ?
?
?
??
?
?
??
?
??
?
? ???
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
d x d y d zt???
)1.5(0 atzwy vx u ???????????? ????
zkyjxi ?
??
?
??
?
??? ???
)1.5(0 btV ?????? ?? ?
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 ??? ??? ??
?
? ?cs AdV ???
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
dzd r dzVrVrVrV zrr ?????? ? ?????? ?????????
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
d r d zrdt ????
)2.5(01)(1 ???????????? tzVVrr Vrr zr ????? ?
zkrere r ?
??
?
??
?
??? ?1??
??
0?????? tV ?? ?
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?????? yvxu
If a continuous function,,called the stream function,is
defined such that
),,( tyx?
)4.5(xvandyu ??????? ??
The streamline,at given instant,it tangent to the direction of
flow at every point in the flow field,
0
)(?
)??()??(0
??
?
?
?
?
?
??
??
??????
?
??
ddy
y
dx
x
v d xu d y
v d xu d yk
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:
12
2
1
2
1
2
1
???
?
?
?
???
?
?
?
?
?
??
?
??
y
y
y
y
y
y
dQ
dy
y
d
dy
y
udyQ
Along BC,it is same as the side AB
For a 2-D,imcompressible flow in the cylindrical coordinate,
conservation of mass,
x dxd ??? /??
r
V
r
V
t i o ns t r e a m f u n c
V
r
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
t
V
z
V
w
y
V
v
x
V
ua
Dt
VD
t
V
z
V
w
y
V
v
x
V
u
t
V
dtdz
z
V
dtdy
y
V
dtdx
x
V
a
dt
t
V
dz
z
V
dy
y
V
dx
x
V
Vd
dttdzzdyydxxVV
aa n dtzyxVV
p
pppp
pppp
p
tt
p
pptp
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?????
?
??
????
?
?
????
????
?
????
?
??
?
??
///
),,,(
),,,(|
It include,total acceleration of a particle,convective acceleration,
and local acceleration
)10.5()(
)(
t
V
VVa
Dt
VD
z
V
w
y
V
v
x
V
uVV
p
?
?
?????
?
?
?
?
?
?
?
?
???
?
???
?
???
??
5-3.2 Fluid Rotation
A fluid particle moving may rotate about the axes,
zyx kji ???? ??? ???
?
The oa and ob rotate to the position shown during the interval dt
xxvvv oa ?????
x
v
tx
x
v
t
x
t
oa
tt
oa
?
?
?
??
?
?
??
?
??
?
?
?
?
????
?
?
??
?
/
limlim
00
y
u
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
)(
2
1
)(
2
1
)(
2
1
x
w
z
u
z
v
y
w
y
u
x
v
y
x
z
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)14.5(
2
1
)13.5()](?)(?)(?[
2
1
V
y
u
x
v
k
x
w
z
u
j
z
v
y
w
i
??
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
The vorticity is defined as to be twice the rotation??
)15.5(2 V ??? ???? ??In cylindrical coordinates the vorticity is
)16.5()11(?)(?)1(? ?? ??? ???????????????????? rzrzr VrrrVrkrVzVezVVreV ?
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
z
A
z
c
z
dAVdAsdV
yxyx
x
u
x
v
yvxy
x
u
uyx
x
v
vxu
)(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?
)(90 0 ???? ????????
y
u
dt
d
x
v
t
xtxxv
t
x
dt
d
dt
d
dt
d
dt
d
tt
?
?
?
?
?
?
?
?????
?
?
??
?
???
????
?
??
???
)
/)/
(lim
/
lim
00
The rate of angular deformation in the xy plane is
y
u
x
v
dt
d
?
??
?
????? ???
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
Vzwyvxu ????????????
5-4 momentum equation
)2.4(
)2.4(
bdVVdmVP
a
dt
Pd
F
s y s t e ms y s t e m MM
s y s t e m
s y s t e m
?? ??
?
?
?
?
?
?
???
?
?
)22.5)(( zVwyVvxVutVdmDt VDdmdt VddmFd
s y s t e m ?
??
?
??
?
??
?
???
???
?? ???????
5-4.1 forces acting on a fluid particle
d xd yd z
xxx
d xd y
dz
x
d xd y
dz
x
d xd z
dy
x
d xd z
dy
x
d yd z
dx
x
d yd z
dx
x
dF
zxyxxx
zx
zx
zx
zx
yx
yx
yx
yx
xx
xx
xx
xxs
x
)(
)
2
()
2
(
)
2
()
2
(
)
2
()
2
(
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
??
???
?
?
?
?
?
?
?
?
?
?
?
?
)23.5()( ad x d y d zxxxgdFdFdF zxyxxxxSBx
xx ?
??
?
??
?
????? ????
)23.5()( bd x d y d zxxxgdFdFdF zyxyyyySBy
yy ?
??
?
??
?
????? ????
)23.5()( cd x d y d zxxxgdFdFdF zzyzxzzSBz
zz ?
??
?
??
?
????? ????
5-4.2 differential momentum equation
)24.5()(
)24.5()(
)24.5()(
c
z
w
w
y
w
v
x
w
u
t
w
yyy
g
b
z
v
w
y
v
v
x
v
u
t
v
yyy
g
a
z
u
w
y
u
v
x
u
u
t
u
xxx
g
zzyzxz
z
zyyyxy
y
zxyxxx
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
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?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
???
?
?
???
?
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(2
3
2
)25.5(2
3
2
)25.5(2
3
2
)25.5()(
)25.5()(
)25.5()(
f
z
w
Vp
e
y
v
Vp
d
x
u
Vp
c
x
w
z
u
b
z
v
y
w
a
y
u
x
v
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
)]([
)]([)]
3
2
2([
z
u
x
w
z
x
v
y
u
y
V
x
u
xx
p
g
Dt
Du
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
????
?
?
?
?
?
?
?
??
?
????
?
)]([
)]([)]
3
2
2([
y
w
z
v
z
x
v
y
u
x
V
y
v
yy
p
g
Dt
Dv
y
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
????
?
?
?
?
?
?
?
??
?
????
?
)]([
)]([)]
3
2
2([
z
u
x
w
x
y
w
z
v
y
V
z
w
zz
p
g
Dt
Dw
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()(
c
z
w
w
y
w
v
x
w
u
t
w
y
p
g
b
z
v
w
y
v
v
x
v
u
t
v
y
p
g
a
z
u
w
y
u
v
x
u
u
t
u
x
p
g
z
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
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?
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?
?
?
?
?
??
??
??
)3.6(
)2.6())((
)(
Dt
VD
pg
VV
t
V
pg
z
V
w
y
V
v
x
V
u
t
V
pg
?
?
??
?
?
????
?
??
??
??
???
???
?
?
???
?
?
?
?
?
?
?
?
?
?
?
???
)4.6()
1
(
)4.6()
1
(
1
)4.6()(
2
c
z
V
V
V
r
V
r
V
V
t
V
z
p
g
b
r
VV
z
V
V
V
r
V
r
V
V
t
Vp
r
g
a
r
V
z
V
V
V
V
r
V
V
t
V
r
p
g
z
z
zz
r
z
z
r
zr
r
z
rr
r
r
r
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
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?
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?
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?
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??
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??
?
???
?
??
?
?
?
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.
ds d nd xads d nd xgdn dxdssppdn dxdsspp s??? ????????? s i n)2()2(
sags
p ??? ??
?
?? s i n
sas
zg
s
p ?
?
??
?
??
?
s
VV
t
V
Dt
DVa
s ?
??
?
???
)5.6( asVVtVszgsp ???????????? ?
)5.6( bsVVsp ???????
d s d n dxad s d n dxgd s d xdnnppd s d xdnnpp n??? ????????? c os)2()2(
nagn
p ??? ??
?
?? c o s
nan
zg
n
p ?
?
??
?
??
?
R
Va
n
2
??
)6.6(
2
aRVnzgnp ??????? ?
)6.6(
2
bRVnp ????
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
Cgz
Vp
Cgz
Vdp
V d Vgdz
dp
s
V
V
s
z
g
s
p
???
???
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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
0
Vpp ???
The dynamic pressure
2
2
1 V?
?
)(2 0 ppV ??
mdm Qdtdmdm QdtQQ ?? ??? ???
)15.6()(22 122
2
22
1
2
11
dm
QuugzVpgzVp ?
?? ????????
Incompressible
?/121 ?? vv
6-5 unsteady Bernoulli equation-integration of Euler’s equation
along a streamline
)3.6(? DtDVsdDt VDkgp ?????? ?
?
?
)18.6(? dstVdssVVdsDtDVsdDt VDsdkgsdp ??????????????? ?
???
?
)(
)(?
)(
sa l o n gVinc h a n g et h edVds
s
V
sa l o n gzinc h a n g et h edzdsk
sa l o n gp r es s u r einc h a n g et h edpsdp
?
?
?
??
???
?
)19.6(dstVV d Vgdzdp ?????? ?
)20.6(0)(2 2
112
2
1
2
22
1
???????? ?? dstVzzgVVdp?
)21.6(22 2
12
2
22
1
2
11 ?
?
??????? ds
t
VgzVpgzVp
??
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(0
111
)22.6(0
0
2
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
????
??
?
?? rzrz
V
rr
rV
rr
V
z
V
z
VV
r
y
u
x
v
x
w
z
u
z
v
y
w
V
??
6-6.1 Bernoulli equation applied to irrotational flow
CgzVp ??? 1
2
11
2?
)24.6()(
2
1
)(
2
1
?
)()(
2
1
)(
)10.6()(?
2
VVVkg
p
VVVVVV
VVkg
p
??????
?
?
?????????
????
?
?
??
??????
??
?
?
)25.6(
2
2
1
)(
2
1
?
2
2
2
Cgz
Vp
dVg d z
dp
rdVrdkgrd
p
???
???
??????
?
?
?
?
?
???
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(?
1
??
)28.6(
)27.6(
z
V
r
V
r
V
z
e
r
e
r
e
z
w
y
v
x
u
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(
y
v
x
u
x
v
y
u
?
?
??
?
?
??
?
?
??
?
?
?
??
??
Irrotational flow
)31.6(0
)3.5(0
)30.6(0
)22.6(0
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yx
y
v
x
u
eq u a t i o nco n t i n u i t y
yx
y
u
x
v
??
??
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 ??? VDfDV F ?
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
G
or
G
?
?
?????
????
The number m is usually equal to the minimum number of
independent dimensions required to specify the dimensions
of all the parameters
2
4/3
1
6
32
1
5
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
VLLLVAdyduA ???? ??? 2
2)()( LpAp ???
L?
3Lgmg ??
??
? DVDV ??Re
??
? VLVL ??ReRe No.
Pressure coefficient:
2
2
1 V
pCp
?
??
Cavitation phenomena,the pressure express as cavitation
number:
2
2
1 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
22
2
gL
LV
gL
VFr
gL
VFr
?
????
The Weber number is the ratio of inertia forces to surface
tension forces
?
? LVWe 2?
??
vE
V
d
dp
V
c
V
M ???
Compressibility effects
2
22
2
LE
LVM
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
p r o t o t y p eel
VDVD
???
?
???
??
???
?
???
?
?
?
?
?
m o d
p r o t o t y p eel ReRe m o d ?
p r ot ot y p eel DV
F
DV
F
???
?
???
??
???
?
???
?
22
m od
22 ??
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 ? p r o t o t y p eel ReRe m o d ?
The prototype condition Vp = 8.44 ft/s
51099.4Re ???
p
pp
p
DV
?
51099.4Re ???
m
mmm DV
?
sftDV
m
mmm /1 5 7Re ?? ?
Dynamically similar
p r ot ot y p eel DV
F
DV
F
???
?
???
??
???
?
???
?
22
m od
22 ??
l bfDDVVFF
m
p
m
p
m
p
mp 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
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
v
x
v
y
p
g
y
v
v
x
v
u
y
u
x
u
x
p
y
u
v
x
u
u
y
v
x
u
???
??
2
*****
???
????? VppV vvV uuLyyLxx ?
The pressure nondimensional by dividing by
2?V?
)13.7(
)12.7(
)11.7(0
2
*
*2
2
*
*2
2*
*2
*
*
*
*
*
*
2
2
*
*2
2
*
*2
2*
*2
*
*
*
*
*
*
2
*
*
*
*
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
??
y
v
x
v
L
V
y
p
L
V
g
y
v
v
x
v
u
L
V
y
u
x
u
L
V
x
p
L
V
y
u
v
x
u
u
L
V
y
v
L
V
x
u
L
V
?
?
?
?
?
?
?
??
)16.7(
)15.7(
)14.7(0
2
*
*2
2
*
*2
*
*
2*
*
*
*
*
*
2
*
*2
2
*
*2
*
*
*
*
*
*
*
*
*
*
*
*
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
y
v
x
v
LVy
p
V
gL
y
v
v
x
v
u
y
u
x
u
LVx
p
y
u
v
x
u
u
y
v
x
u
?
?
?
?
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??
tVu bc ?s in??
L
Vtt ??*
???
?
???
?
??
??
**
s i n t
V
L
V
uu bc
bc
?
Duplication of the boundary condition requires that
parameter be the same between the two flows,
This parameter is the Strouhal number
?VL/?
?
? VLSt ?
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
D
L 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( aAdVuF csS x ? ?? ???
d x d z
dy
dy
d
dF
d x d z
dy
dy
d
dF
d y d z
dx
x
p
pdF
d y d z
dx
x
p
pdF
yx
yxT
yx
yxB
R
L
)
2
(
)
2
(
)
2
(
)
2
(
?
?
?
?
??
???
?
?
???
?
?
??
)4.8(
2
1
)3.8(0
2
12
1
1
c
c
y
x
p
u
cy
x
p
dy
du
dy
du
cy
x
p
C
x
p
dy
d
dy
d
x
p
yx
yx
yx
yx
???
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
???
??
?
?
?
?
?
?
?
???
?
?
?
??
?
?
?
??
???
?
?
?
According to boundary condition
)5.8()(
2
22
?
?
?
?
?
?
?
?
?
?
??
?
???
?
??
?
?
?
??
a
y
a
y
x
pau
?
Shear stress distribution
)6.8(21)( aayxpayx ?
?
?
??
? ?
?
??
?
???
?
??
?
?
?
???
Volume flow rate
)6.8(
12
1
)(
2
1
3
2
0
ba
x
p
l
Q
dyayy
x
p
AdVQ
a
A
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
??? ??
?
?
??
Flow rate as function of pressure drop
)6.8(
12
3
12
c
L
pa
L
Q
L
p
L
pp
x
p
?
?
?
??
?
?
?
?
?
Average velocity
)6.8(12 112 1 2
3
daxpla laxpAQV ?????? ?????????? ????? ??
Point of maximum velocity
)6.8(
2
3
)(
8
1
2/0
12
)(
2
2
m a x
2
eVa
x
p
uay
dy
du
aa
y
x
pa
dy
du
?
?
?
??????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Transformation of Coordinates
Transform from y=0 at bottom to y=0 at centerline
)7.8(
4
1)(
2
22
?
?
?
?
?
?
?
?
?
?
??
?
???
?
??
?
? ?
?
??
a
y
x
pau
?
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(2 1 212 ccyxpu ???????? ??? ??
From BC,We have
???
????
Uuay
cuy 000 2
)8.8()()(
2
2
1
2
1
2
2
1
12
?
?
?
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
?
?
a
y
a
y
x
pa
a
Uy
u
a
x
p
a
U
ca
c
a
x
p
U
?
?
??
Shear stress distribution
)9.8(21)( aayxpaaUyx ?
?
?
??
? ?
?
??
?
???
?
??
?
?
?
??? ??
Volume flow rate
)9.8(
12
1
2
)(
2
1
[
3
2
0
0
ba
x
pUa
l
Q
dyayy
x
p
a
Uy
l
Q
u l d yAdVQ
a
a
A
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
??
???
?
??
?
?
??
Average Velocity
)9.8(12 12/]12 12[
3
caxpUlala laxpUalAQV ?????? ?????????? ????? ??
Point of Maximum velocity
)/)(/1(
/
2
0
12
)(
2
2
2
xp
aUa
y
dy
du
aa
y
x
pa
a
U
dy
du
??
????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
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
0?xSF
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
dxdrrdr
dr
d
dF
r d xdF
r d rdx
x
p
pdF
r d rpdF
rx
rxO
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(ln
4
2
2
1
)10.8(
)(1
2
1
2
1
1
2
cr
c
x
pr
u
r
c
x
pr
dr
du
dr
du
c
x
pr
rC
x
p
dr
d
r
dr
rd
rdr
d
rx
p
rx
rx
rx
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
?????? ???????????? ??? xpRccxpru ?? 44
2
22
2
)12.8(1
4
22
?
?
?
?
?
?
?
?
?
?
??
?
???
?
??
?
?
?
???
R
r
x
pRu
?
)13.8()(2 axpryx ????
)13.8(
8
2)(
4
1
4
22
0
b
x
pR
Q
r d rRr
x
p
AdVQ
R
A
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
??? ??
?
?
?
?
??
)13.8(
128
4
12
c
L
pD
Q
L
p
L
pp
x
p
?
? ?
?
??
?
?
?
?
?Flow rate as a
function pf
pressure drop
Volume flow
rate
Shear stress distribution
velocity
)13.8(8
2
2 dx
pR
R
QV ?
?
??
?
?
?
????
??
)13.8(2)(
4
00
)(
2
2
2
m a x eVa
x
pR
Uur
dr
du
x
pr
dr
du
?
?
?
???????
?
?
?
?
?
The velocity profile(8.12) can be written in terms of the
maximum velocity as
)14.8()(1 2RrUu ??
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:
0?xBF
?? ??????? csCVBSx AdVudVutFFF 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][ xpRRrrxw ?????? ???
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(vudy udt ur bl am ?????? ?????
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
??? nn
n
U
V
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()(?? ????????? csCVo t h e rs h e a rs AdVpvedVetWWWQ ?????? ??
gzVue ??? 2
2
5-6.1 kinetic Energy Coefficient
Use as the Kinetic energy coefficient
)26.8(
2
)26.8(
222
2
2
222
b
Vm
dAV
V
a
V
mdAV
V
dAV
V
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
nn
n
V
U
????
??
?
???
8-6.2 head loss
Using the definition of,the energy equation can be
written
?
)22()()()(
2
11
2
22
12
12
12
VVmzzgmppmuumQ ??
?? ???????? ????
?
)28.8()()2()2( 122
2
222
1
2
111
dm
QuugzVpgzVp ????????? ?
?
?
?
dm
Q
uu
gz
Vp
?
??
??
)(
)
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
2
222
1
2
111
?
?
?
?
? lhgz
VpgzVp ??????
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
2
222
1
2
111
?
??
?
?
? l
l H
g
hz
g
V
g
pz
g
V
g
p ???????
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 CV ?
2
2?
)31.8()( 1221 lhzzgpp ?????
If the pipe is horizontal
)32.8(21 lhppp ???? ??
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
64
32
32
)4/(1 2 81 2 8
)13.8(
1 2 8
2
4
2
4
4
V
D
L
D
V
D
L
h
D
V
D
L
D
DVL
D
QL
p
c
L
pD
Q
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(21 lhppp ???? ??
),( R e,2 DeDLVh l ??
)35.8(2
2
fDLgVH 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(
Re
64
22Re
64
m i n
22
?
??
?
?
?
?
?
?
arla
l
f
V
D
L
f
V
D
L
h
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 /l o g (0.21 5.05.0 afDef ???
)37.8()]Re 74.57.3 /[ l o g (25.0 29.00 bDef ???
For turbulent flow in smooth pipes,the Blasius correlation
)38.8(Re 3 1 6.0 25.0?f
The wall shear stress is obtained as
)39.8()(0 3 32.0 25.02 VRVw ??? ?
8-7.2 Minor losses(K:loss coefficient from experiments)
)40.8(2)40.8(2
22
bVDLfhaVKh ell 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(
2
1 2
1
12
V
ppC
p
?
??
If the gravity is neglected and 121 ?? ??
mll
hhVpVp ?????
???
)2()2(
2
22
2
11
??
?
??
? ???
pl CA
AVh
m
2
2
1
2
1 )(1(
2
2211 VAVA ?
)42.8()( 11(2 2
2
1 ?
?
?
??
? ???
pl CAR
Vh
m
For frictionless flow
? ? )43.8(
11
2ARCp i ??
0?mlh
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(4 P AD h ?
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
ar
hD
h ?? 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 ??? ionc o n f i g u r a ts y s t e mzeDQLp ???
)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
l o s s esm
pp
Qmuum
pp
W
Qm
p
um
p
uWW
AdVgz
Vp
udVe
t
WQ
in
sin
csCV
s
?
?
????
?
?
????????
?????
?
?
?? ??
?????
?????
??
??
??
??
?
?
?
12
12
12
1
1
2
2
2
)(
))(()(
)
2
(
The losses are determined in terms of the pump efficiency
hpW
pAV
AVppm
pp
W
Wl o s s e s
in
in
in
3 6 8 0 0
)(
11
)1(
12
12
?
?
???
?
?
??
?
??
?
????
?
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
a r g
a r g
s uc t i onedi s c h
pum p
pum p
s uc t i onedi s c h
pum p
gz
Vp
gz
Vp
m
W
h
gz
Vp
gz
Vp
mW
??
?
?
??
?
?
?????
?
?
??
?
?
?????
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?????
?
?
??
?
?
???
??
??
?
?
??
)48.8(22 2
2
2
2
2
1
2
1
1
1
Tlp u m p
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????? ??
csCV
AdVdVt
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?????
?????
2
2
1
2
22
1
2
221
2
2
22
1
2
11
1
2
)(
2
)9.6(
22
V
VV
VVpp
gz
Vp
gz
Vp
??
??
? ? ? ?
? ?
)49.8(
])/(1[
2
1
2
0
2
12
21
2
2
1
2
2
2
21
2
1
2
2
2
1
2211
AA
pp
V
A
AV
pp
A
A
V
V
AVAV
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
??
The theoretical mass flow rate is then given by
)50.8()(2
)/(1 21212
2
22 ppAA
AAVm
lt h e o r e t i c a ???? ???
)52.8()p p(2
1
)/(/
)51.8()p p(2
)/(1
21
4
2
1
4
1
21
2
1
?
?
?
???
?
?
?
?
?
??
?
t
a c t u a l
tt
t
t
a c t u a l
CA
m
AADD
AA
CA
m
?
?
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
C
K
ta c t u a l ??
?
?
?
?
?
For the turbulent flow regime,the discharge coefficient
)55.8(Re
1
n
D
bCC ??
?
The corresponding form for the flow-coefficient
equation is
)56.8(Re
1
1
1
4 nD
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
5.2
81.2
10Re1075.02.0
)57.8(
Re
71.91
1 8 4.00 3 1 2.05 9 5 9.0
1
1
????
????
D
D
C
?
?
??
8-10.2 The flow Nozzle
74
5.0
5.0
10Re1075.025.0
)53.8(
Re
53.6
9 9 7 5.0
1
1
????
??
D
D
C
?
?
a,Pipe installation (K is function of and )
b,Plenum installation 0.95<K<0.99
?
1ReD
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
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
motion
? 3,The principle of angular momentum
? 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
m
VVC ?
??
??
)22(),,,( ?? tzyx??
?
2-2 velocity field
? 2-2.1 one-,two-,and three-Dimensional flows
),,,( tzyxVV ?? ? kwjviuV ??? ????
Steady flow unsteady flow
0.,,,, ),,( ??? 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:
n?
F??
A??
n? n
?
)7.2(l i m
0 n
t
An A
F
n ?
??
? ?
?
)6.2(lim 0
n
n
An A
F
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
A
xz
x
y
A
xy
x
x
A
xx
A
F
A
F
A
F
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 dA
dF
A
F
y
??
? ?
??
? 0
l i m
dt
d
tr a t end e f o r m a t i o t
?
?
??
?
??
? 0
lim
dy
du
dt
d
y
u
t
yltul
??
??
?
?
?
?
??
???????
2-4.1 Newtonian fluid
Fluid as water,air,and gasoline are Newtonian fluid
dy
du
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
dy
du
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 dydudydudydukdyduk nnyx ?? ??? ?
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
dy
du
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 Static
Absence 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?
d xd yd zgdVgdmgFd B ???? ?? ???
)2()( dyyppyyyppP LL ??????????
2)(
dy
y
ppyy
y
ppP
RL ?
????
?
???
d xd y d zkzpjypixpFd s )???( ???????????
)1.3()( p d x d y d zd x d y d zg r a d pFd s ??????
The gradient of a scalar field gives a vector field
)2.3(
)(
gp
dV
Fd
dVgpFdFdFd sB
?
?
????
?
?
????
??????
For a static fluid,
)3.3(0
00
????
???
gp
amFda ? ???
?
)6.3(
)5.3(00
??
?
????
??
?
?
?
?
?
?
?
?
g
dz
dp
g
z
p
y
p
x
p
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 ???
?? ????? zzpp zzgppg d zdp 00 )( 00 ??
hzz ??0 )7.3(0 ghpp ???
Incompressible liquid:manometers
?
1122
22
11
ghghpp
ghpp
ghpp
CA
CB
BA
??
?
?
???
??
???
3-3.2 Gases (compressible fluid)
The pressure varies with altitude or temperature
)8.3()()1(
)(
/
0
0
/
0
0
0
mRgmRg
T
T
p
T
mz
pp
dz
mzTR
pg
dz
RT
pg
g d zdp
???
?
?????? ?
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 ??
???? s i n000 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 ????? ''' ?????? ???
)12.3('' ?? ?? AA RR x pd AFxan dy pd AFy
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(??? p dAFF RR ?
? ??? A ARR x pd AFxandy pd AFy ''
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 y dAgAppdAF ?? s i n0
The yc is the y coordinate of the centroid of the area A thus
)14.3()(s i n 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(
s i ns i n
??2
2
a
Ay
I
yydAyI
dAygg h d AyAygy
y p d AFy
c
xx
c
A
xx
AA
c
A
R
????
???
??
?
??
?
?????
xxI??
Is the second moment of the area about the centroid axisx?
The same ruler,
)15.3(?? bAyIxx
c
yx
c ???
3-5.3 Hydrostatic force on a curved submerged surface
)16.3(???
)10.3(
)9.3(
zyx RRRR
A
R
FkFjFiF
ApdF
ApdFd
????
??
??
???
??
??
?
??? ????????? xx A xARR p d AiApdiFdiFF ??? ???
In general in the l direction the component of resultant force is,
??? ll A lR p d AF
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.
gVg d VdAghF
ghpp d AFF
ZV
ZVR Z
???
?
???
???
??
?
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 adtdM s y s t e m ?
)1.4( bdVdmM
s y s t e ms y s t e m MMs y s t e m ??
?? ?
4-1.2 Newton’s second law ( Conservation of momentum )
)2.4(
)2.4(
bdVVdmVP
a
dt
Pd
F
s y s t e ms y s t e m MM
s y s t e m
s y s t e m
?? ??
?
?
?
?
?
?
???
?
?
4-1.3 The angular Momentum principle
)3.4(
)3.4(
)3.4(
aTdmgrFrT
bdVVrdmVrH
a
dt
Hd
T
s h a f t
M
s
HM
s y s t e m
s y s t e m
s y s t e m
s y s t e ms y s t e m
??????
?????
?
?
?????
????
?
?
?
?
?
?
?? ?
4-1.4 The first of thermodynamics
)4.4(
2
)4.4(
)4.4(
2
cgz
V
ue
bdVee d mE
a
dt
dE
WQ
dEWQ
s y s t e ms y s t e m
VM
s y s t e m
s y s t e m
???
??
?
?
?
??
??
?? ?
??
??
4-1.5 The second Law of Thermodynamics
)5.4(
)5.4(
1
bdVss d mS
aQ
Tdt
dS
T
Q
dS
s y s t e ms y s t e m
VM
s y s t e m
s y s t e m
?? ??
??
?
?
?
?
?
?
4-2 Relation of system derivatives to the control volume
formulation
dv
s
e
Vr
V
S
E
H
P
M
N
s y s t e m
V
s y s t e m
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
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
dt
dN tcvttIIIIcv
ts y s t e m ?
???? ??
??
? )9.4()lim)lim)lim
0000
000 t
N
t
N
t
NN
dt
dN ttI
t
ttIII
t
tcvttcv
ts y s t e m ?
???? ?? ??
??
??
??
??
??
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
V
t
l
dAldVdN
t
?
?
?
?
???
?? 0
l i m
)c o s( ?????
)11.4(
co s
??
?
??
?
?
??
?
?
???
?
?
?
?
??
?
?
csCV
s
cs
ss
AdVdV
tdt
dN
AdV
t
N
dt
dN
??
??
????
???
4-3 Conservation of mass:(N=m; )
1??
)11.4(?? ??
?
???
?
?
csCV
s
AdVdV
tdt
dN ??????
)12.4(?? ??
?
???
?
?
csCV
s
AdVdV
tdt
dM ????
)13.4(0????? ??
csCV
AdVdVt
??
??
conservation
4-3.1 special cases
incompressible,nondeformable control volume of fixed size
and shape
0???cs AdV ???
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(0???cs AdV ???
A
QVQAdV
A
????
??
At a section,uniform flow,density is constant
nnnA AVAdVn
???? ???? ??
4-4 momentum equation for inertial control volume
)11.4(?? ??
?
???
?
?
csCV
s
AdVdV
tdt
dN ??????
VPN ?? ?? ?
)17.4(?? ??
?
??
???
?
csCV
s
AdVVdVV
tdt
Pd ????
?
??
Newton’s second law for a system moving relative to an inertial
coordinate system
s
BS dt
PdFFF
???
?
???
????
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?
??
??
??
??
?
?
???
??
?
?
???
??
?
?
???
csCV
BSz
csCV
BSy
csCV
BSx
AdVwdVw
t
FFF
AdVvdVv
t
FFF
AdVudVu
t
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,
dAAdVVdpp
AVp
ss
s
???,,,
,,,
?
?
a,Continuity equation
Assumption,Steady,no flow across bounding streamlines,
incompressible flow
Basic equation:
)13.4(0????? ?? csCV AdVdVt ????
0
0
0|))((|||
??
???
?????
ss
sss
sss
A d VdAV
d A d VA d VdAV
dAAdVVAV ??
b,Streamwise component of the momentum equation
Assumption,No friction,pressure forces only
Basic equation,
dA
dp
pdAAdpppAF
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 b
sF
dzdAAgF
dsdzw h e r e
dsdAAgdVgF
dA
dp
d p AF
s
s
s
B
sB
S
)2/(
s i n
)2/)(s i n(
2
???
?
????
???
?
?
???
The momentum flux will be
ss
sssssCS ss
A d VV
dAAdVVdVVAVVdAVu
?
???
?
???????? |)))((|)((|)|(
?
0|))((||| ????? dAAdVVAV sss ??
Where,
continuity
ss A d VVg d A d zg A d zd p d AA d p ??? ????? 2
1
2
1
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 x y zx y zCV x y zSB AdVVdVVtFFF ??????? ??
)26.4(?? ??????
?
?
cs x y zCVs
AdVdVtdtdN ??????
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 x y zx y zCV x y zSB AdVVdVVtFFF ??????? ??
)2.4(
)2.4(
bdVVdmVP
a
dt
Pd
F
s y s t e ms y s t e m VM
s y s t e m
s y s t e m
?? ??
?
?
?
?
?
?
???
?
?
If we denote the inertial reference frame by XYZ,for accelerating
control volume (4.27) is not right
?? ???????
s y s t e ms y s t e m M
X Y Z
M X Y Z
X Y Z dm
dt
VddmV
dt
d
dt
????
The velocity with respect to the inertial (XYZ) and the control
volume coordinate(xyz) are related by the relative-motion Eq.
)30.4(rfx y zX Y Z VVV ??? ??
Where V is the velocity of the control volume reference frame.
)31.4(rfx y zrfx y zX Y ZX Y Z dtVddtVddtdV ??? ??
??
? ?????
)32.4() sx y zM rf dtPddmF
s y s t e m
?
?? ?? ? ?
)17.4(?? ???????
?
?
csCV
s
AdVVdVVtdt Pd ????
?
??
)34.4(??
??
??
?
?
?
????
cs x y zx y zCV x y z
CV rfSBCV rf
AdVVdVV
t
dVFFdVF
????
?????
??
????
Linear momentum of the system,to derive the control volume
formulation of Newton’s second law
???
???
???
??
?
?
????
??
?
?
????
??
?
?
????
csCVCS
x y zx y zBS
csCVCS
x y zx y zBS
csCVCS
x y zx y zBS
AdVwdVw
t
AdVuFF
AdVvdVv
t
AdVvFF
AdVudVu
t
AdVwFF
zz
yy
xx
????
????
????
???
???
???
)35.4(
4-8 The first law of thermodynamics
)4.4(
2
)4.4(
)4.4(
2
cgz
V
ue
bdVee d mE
a
dt
dE
WQ
s y s t e ms y s t e m
VM
s y s t e m
s y s t e m
???
??
?
?
?
??
?? ?
??
dv
s
e
Vr
V
S
E
H
P
M
N
s y s t e m
V
s y s t e m
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
1
)54.4(?? ??
?
???
?
?
csCV
s
AdVedVe
tdt
dE ????
Since the system and the control volume coincide at t0
u m ec o n t r o l v o ls y s t e m WQWQ ][][ ???? ???
)55.4(?? ??????
csCV
AdVedVetWQ
????
??
4-8.1 rate of work done by a control volume
o t h e rs h e a rn o r m a ls WWWWW ????? ????
1:Shaft work
2,Normal stresses at the control surface
VFdt sdFtwW
tt
????? ????
?? ???? 00 limlim
?
? ??? cs nnn o r m a l AdVW ??? ?
3,shear stresses at the control surface
? ???
?
css he ar
dAVW
dAFd
??
?
??
?
?
Can be expressed as three terms(shaft; solid ports surface)
4 other work
)56.4(o t h e rs h e a rcs nns WWAdVWW ?????? ????? ? ?
4-8.2 control volume equation
??? ????????? cs nncsCVo t h e rs h e a rs AdVAdVedVetWWWQ ???????? ???
pnn ???
?? ??????????? csCVot h e rs h e ars AdVgzVpvudVetWWWQ ?????? ?? )2(
2
4-9 the second law of thermodyhamics
)5.4(1 aQTdtdS
s y s t e m
???
?
?
)5.4( bdVss d mS
s y s t e ms y s t e m MMs y s t e m ??
?? ?
)58.4(?? ??
?
???
?
?
csCV
s
AdVsdVs
tdt
dS ????
dv
s
e
Vr
V
S
E
H
P
M
N
s y s t e m
V
s y s t e m
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
1
)59.4(1 dA
A
Q
T
AdVsdVs
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 ??? ????
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(!2
1
2
dx
x
x
xdxx
????
Neglecting higher order term
22/
x
xdxx
?
?
???
?
???
At the left face
22/
x
xdxx
?
?
???
?
???
22/
x
x
uuu
dxx
?
?
???
? 22/ xxuuu dxx ??????
Statement of conservation of mass is
0?????????????? v ol u m ec o nt r olt hei ns i d e m as sofc h an g eofR a t es u r f a c ec o nt r olt het hr o ug h o utf l u xm as sofr a t eN e t
The first term in Eq,We must consider the mass flux through each
of the six surfaces of the control surface,? ?
cs AdV
???
The net rate of mass flux out through the control surface is given
by
dx dy dzzwy vxu ?
?
?
??
?
?
??
?
??
?
? ???
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
d x d y d zt???
)1.5(0 atzwy vx u ???????????? ????
zkyjxi ?
??
?
??
?
??? ???
)1.5(0 btV ?????? ?? ?
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 ??? ??? ??
?
? ?cs AdV ???
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
dzd r dzVrVrVrV zrr ?????? ? ?????? ?????????
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
d r d zrdt ????
)2.5(01)(1 ???????????? tzVVrr Vrr zr ????? ?
zkrere r ?
??
?
??
?
??? ?1??
??
0?????? tV ?? ?
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?????? yvxu
If a continuous function,,called the stream function,is
defined such that
),,( tyx?
)4.5(xvandyu ??????? ??
The streamline,at given instant,it tangent to the direction of
flow at every point in the flow field,
0
)(?
)??()??(0
??
?
?
?
?
?
??
??
??????
?
??
ddy
y
dx
x
v d xu d y
v d xu d yk
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:
12
2
1
2
1
2
1
???
?
?
?
???
?
?
?
?
?
??
?
??
y
y
y
y
y
y
dQ
dy
y
d
dy
y
udyQ
Along BC,it is same as the side AB
For a 2-D,imcompressible flow in the cylindrical coordinate,
conservation of mass,
x dxd ??? /??
r
V
r
V
t i o ns t r e a m f u n c
V
r
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
t
V
z
V
w
y
V
v
x
V
ua
Dt
VD
t
V
z
V
w
y
V
v
x
V
u
t
V
dtdz
z
V
dtdy
y
V
dtdx
x
V
a
dt
t
V
dz
z
V
dy
y
V
dx
x
V
Vd
dttdzzdyydxxVV
aa n dtzyxVV
p
pppp
pppp
p
tt
p
pptp
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?????
?
??
????
?
?
????
????
?
????
?
??
?
??
///
),,,(
),,,(|
It include,total acceleration of a particle,convective acceleration,
and local acceleration
)10.5()(
)(
t
V
VVa
Dt
VD
z
V
w
y
V
v
x
V
uVV
p
?
?
?????
?
?
?
?
?
?
?
?
???
?
???
?
???
??
5-3.2 Fluid Rotation
A fluid particle moving may rotate about the axes,
zyx kji ???? ??? ???
?
The oa and ob rotate to the position shown during the interval dt
xxvvv oa ?????
x
v
tx
x
v
t
x
t
oa
tt
oa
?
?
?
??
?
?
??
?
??
?
?
?
?
????
?
?
??
?
/
limlim
00
y
u
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
)(
2
1
)(
2
1
)(
2
1
x
w
z
u
z
v
y
w
y
u
x
v
y
x
z
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)14.5(
2
1
)13.5()](?)(?)(?[
2
1
V
y
u
x
v
k
x
w
z
u
j
z
v
y
w
i
??
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
The vorticity is defined as to be twice the rotation??
)15.5(2 V ??? ???? ??In cylindrical coordinates the vorticity is
)16.5()11(?)(?)1(? ?? ??? ???????????????????? rzrzr VrrrVrkrVzVezVVreV ?
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
z
A
z
c
z
dAVdAsdV
yxyx
x
u
x
v
yvxy
x
u
uyx
x
v
vxu
)(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?
)(90 0 ???? ????????
y
u
dt
d
x
v
t
xtxxv
t
x
dt
d
dt
d
dt
d
dt
d
tt
?
?
?
?
?
?
?
?????
?
?
??
?
???
????
?
??
???
)
/)/
(lim
/
lim
00
The rate of angular deformation in the xy plane is
y
u
x
v
dt
d
?
??
?
????? ???
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
Vzwyvxu ????????????
5-4 momentum equation
)2.4(
)2.4(
bdVVdmVP
a
dt
Pd
F
s y s t e ms y s t e m MM
s y s t e m
s y s t e m
?? ??
?
?
?
?
?
?
???
?
?
)22.5)(( zVwyVvxVutVdmDt VDdmdt VddmFd
s y s t e m ?
??
?
??
?
??
?
???
???
?? ???????
5-4.1 forces acting on a fluid particle
d xd yd z
xxx
d xd y
dz
x
d xd y
dz
x
d xd z
dy
x
d xd z
dy
x
d yd z
dx
x
d yd z
dx
x
dF
zxyxxx
zx
zx
zx
zx
yx
yx
yx
yx
xx
xx
xx
xxs
x
)(
)
2
()
2
(
)
2
()
2
(
)
2
()
2
(
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
??
???
?
?
?
?
?
?
?
?
?
?
?
?
)23.5()( ad x d y d zxxxgdFdFdF zxyxxxxSBx
xx ?
??
?
??
?
????? ????
)23.5()( bd x d y d zxxxgdFdFdF zyxyyyySBy
yy ?
??
?
??
?
????? ????
)23.5()( cd x d y d zxxxgdFdFdF zzyzxzzSBz
zz ?
??
?
??
?
????? ????
5-4.2 differential momentum equation
)24.5()(
)24.5()(
)24.5()(
c
z
w
w
y
w
v
x
w
u
t
w
yyy
g
b
z
v
w
y
v
v
x
v
u
t
v
yyy
g
a
z
u
w
y
u
v
x
u
u
t
u
xxx
g
zzyzxz
z
zyyyxy
y
zxyxxx
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
???
?
?
???
?
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(2
3
2
)25.5(2
3
2
)25.5(2
3
2
)25.5()(
)25.5()(
)25.5()(
f
z
w
Vp
e
y
v
Vp
d
x
u
Vp
c
x
w
z
u
b
z
v
y
w
a
y
u
x
v
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
)]([
)]([)]
3
2
2([
z
u
x
w
z
x
v
y
u
y
V
x
u
xx
p
g
Dt
Du
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
????
?
?
?
?
?
?
?
??
?
????
?
)]([
)]([)]
3
2
2([
y
w
z
v
z
x
v
y
u
x
V
y
v
yy
p
g
Dt
Dv
y
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
????
?
?
?
?
?
?
?
??
?
????
?
)]([
)]([)]
3
2
2([
z
u
x
w
x
y
w
z
v
y
V
z
w
zz
p
g
Dt
Dw
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()(
c
z
w
w
y
w
v
x
w
u
t
w
y
p
g
b
z
v
w
y
v
v
x
v
u
t
v
y
p
g
a
z
u
w
y
u
v
x
u
u
t
u
x
p
g
z
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
??
)3.6(
)2.6())((
)(
Dt
VD
pg
VV
t
V
pg
z
V
w
y
V
v
x
V
u
t
V
pg
?
?
??
?
?
????
?
??
??
??
???
???
?
?
???
?
?
?
?
?
?
?
?
?
?
?
???
)4.6()
1
(
)4.6()
1
(
1
)4.6()(
2
c
z
V
V
V
r
V
r
V
V
t
V
z
p
g
b
r
VV
z
V
V
V
r
V
r
V
V
t
Vp
r
g
a
r
V
z
V
V
V
V
r
V
V
t
V
r
p
g
z
z
zz
r
z
z
r
zr
r
z
rr
r
r
r
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
??
?
???
?
??
?
?
?
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.
ds d nd xads d nd xgdn dxdssppdn dxdsspp s??? ????????? s i n)2()2(
sags
p ??? ??
?
?? s i n
sas
zg
s
p ?
?
??
?
??
?
s
VV
t
V
Dt
DVa
s ?
??
?
???
)5.6( asVVtVszgsp ???????????? ?
)5.6( bsVVsp ???????
d s d n dxad s d n dxgd s d xdnnppd s d xdnnpp n??? ????????? c os)2()2(
nagn
p ??? ??
?
?? c o s
nan
zg
n
p ?
?
??
?
??
?
R
Va
n
2
??
)6.6(
2
aRVnzgnp ??????? ?
)6.6(
2
bRVnp ????
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
Cgz
Vp
Cgz
Vdp
V d Vgdz
dp
s
V
V
s
z
g
s
p
???
???
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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
0
Vpp ???
The dynamic pressure
2
2
1 V?
?
)(2 0 ppV ??
mdm Qdtdmdm QdtQQ ?? ??? ???
)15.6()(22 122
2
22
1
2
11
dm
QuugzVpgzVp ?
?? ????????
Incompressible
?/121 ?? vv
6-5 unsteady Bernoulli equation-integration of Euler’s equation
along a streamline
)3.6(? DtDVsdDt VDkgp ?????? ?
?
?
)18.6(? dstVdssVVdsDtDVsdDt VDsdkgsdp ??????????????? ?
???
?
)(
)(?
)(
sa l o n gVinc h a n g et h edVds
s
V
sa l o n gzinc h a n g et h edzdsk
sa l o n gp r es s u r einc h a n g et h edpsdp
?
?
?
??
???
?
)19.6(dstVV d Vgdzdp ?????? ?
)20.6(0)(2 2
112
2
1
2
22
1
???????? ?? dstVzzgVVdp?
)21.6(22 2
12
2
22
1
2
11 ?
?
??????? ds
t
VgzVpgzVp
??
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(0
111
)22.6(0
0
2
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
????
??
?
?? rzrz
V
rr
rV
rr
V
z
V
z
VV
r
y
u
x
v
x
w
z
u
z
v
y
w
V
??
6-6.1 Bernoulli equation applied to irrotational flow
CgzVp ??? 1
2
11
2?
)24.6()(
2
1
)(
2
1
?
)()(
2
1
)(
)10.6()(?
2
VVVkg
p
VVVVVV
VVkg
p
??????
?
?
?????????
????
?
?
??
??????
??
?
?
)25.6(
2
2
1
)(
2
1
?
2
2
2
Cgz
Vp
dVg d z
dp
rdVrdkgrd
p
???
???
??????
?
?
?
?
?
???
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(?
1
??
)28.6(
)27.6(
z
V
r
V
r
V
z
e
r
e
r
e
z
w
y
v
x
u
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(
y
v
x
u
x
v
y
u
?
?
??
?
?
??
?
?
??
?
?
?
??
??
Irrotational flow
)31.6(0
)3.5(0
)30.6(0
)22.6(0
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yx
y
v
x
u
eq u a t i o nco n t i n u i t y
yx
y
u
x
v
??
??
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 ??? VDfDV F ?
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
G
or
G
?
?
?????
????
The number m is usually equal to the minimum number of
independent dimensions required to specify the dimensions
of all the parameters
2
4/3
1
6
32
1
5
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
VLLLVAdyduA ???? ??? 2
2)()( LpAp ???
L?
3Lgmg ??
??
? DVDV ??Re
??
? VLVL ??ReRe No.
Pressure coefficient:
2
2
1 V
pCp
?
??
Cavitation phenomena,the pressure express as cavitation
number:
2
2
1 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
22
2
gL
LV
gL
VFr
gL
VFr
?
????
The Weber number is the ratio of inertia forces to surface
tension forces
?
? LVWe 2?
??
vE
V
d
dp
V
c
V
M ???
Compressibility effects
2
22
2
LE
LVM
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
p r o t o t y p eel
VDVD
???
?
???
??
???
?
???
?
?
?
?
?
m o d
p r o t o t y p eel ReRe m o d ?
p r ot ot y p eel DV
F
DV
F
???
?
???
??
???
?
???
?
22
m od
22 ??
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 ? p r o t o t y p eel ReRe m o d ?
The prototype condition Vp = 8.44 ft/s
51099.4Re ???
p
pp
p
DV
?
51099.4Re ???
m
mmm DV
?
sftDV
m
mmm /1 5 7Re ?? ?
Dynamically similar
p r ot ot y p eel DV
F
DV
F
???
?
???
??
???
?
???
?
22
m od
22 ??
l bfDDVVFF
m
p
m
p
m
p
mp 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
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
v
x
v
y
p
g
y
v
v
x
v
u
y
u
x
u
x
p
y
u
v
x
u
u
y
v
x
u
???
??
2
*****
???
????? VppV vvV uuLyyLxx ?
The pressure nondimensional by dividing by
2?V?
)13.7(
)12.7(
)11.7(0
2
*
*2
2
*
*2
2*
*2
*
*
*
*
*
*
2
2
*
*2
2
*
*2
2*
*2
*
*
*
*
*
*
2
*
*
*
*
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
??
y
v
x
v
L
V
y
p
L
V
g
y
v
v
x
v
u
L
V
y
u
x
u
L
V
x
p
L
V
y
u
v
x
u
u
L
V
y
v
L
V
x
u
L
V
?
?
?
?
?
?
?
??
)16.7(
)15.7(
)14.7(0
2
*
*2
2
*
*2
*
*
2*
*
*
*
*
*
2
*
*2
2
*
*2
*
*
*
*
*
*
*
*
*
*
*
*
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
y
v
x
v
LVy
p
V
gL
y
v
v
x
v
u
y
u
x
u
LVx
p
y
u
v
x
u
u
y
v
x
u
?
?
?
?
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??
tVu bc ?s in??
L
Vtt ??*
???
?
???
?
??
??
**
s i n t
V
L
V
uu bc
bc
?
Duplication of the boundary condition requires that
parameter be the same between the two flows,
This parameter is the Strouhal number
?VL/?
?
? VLSt ?
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
D
L 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( aAdVuF csS x ? ?? ???
d x d z
dy
dy
d
dF
d x d z
dy
dy
d
dF
d y d z
dx
x
p
d y d z
dx
x
p
yx
yxT
yx
yxB
R
L
)
2
(
)
2
(
)
2
(
)
2
(
?
?
?
?
??
???
?
?
???
?
?
??
)4.8(
2
1
)3.8(0
2
12
1
1
c
c
y
x
p
u
cy
x
p
dy
du
dy
du
cy
x
p
C
x
p
dy
d
dy
d
x
p
yx
yx
yx
yx
???
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
???
??
?
?
?
?
?
?
?
???
?
?
?
??
?
?
?
??
???
?
?
?
According to boundary condition
)5.8()(
2
22
?
?
?
?
?
?
?
?
?
?
??
?
???
?
??
?
?
?
??
a
y
a
y
x
pau
?
Shear stress distribution
)6.8(21)( aayxpayx ?
?
?
??
? ?
?
??
?
???
?
??
?
?
?
???
Volume flow rate
)6.8(
12
1
)(
2
1
3
2
0
ba
x
p
l
Q
dyayy
x
p
AdVQ
a
A
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
??? ??
?
?
??
Flow rate as function of pressure drop
)6.8(
12
3
12
c
L
pa
L
Q
L
p
L
pp
x
p
?
?
?
??
?
?
?
?
?
Average velocity
)6.8(12 112 1 2
3
daxpla laxpAQV ?????? ?????????? ????? ??
Point of maximum velocity
)6.8(
2
3
)(
8
1
2/0
12
)(
2
2
m a x
2
eVa
x
p
uay
dy
du
aa
y
x
pa
dy
du
?
?
?
??????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Transformation of Coordinates
Transform from y=0 at bottom to y=0 at centerline
)7.8(
4
1)(
2
22
?
?
?
?
?
?
?
?
?
?
??
?
???
?
??
?
? ?
?
??
a
y
x
pau
?
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(2 1 212 ccyxpu ???????? ??? ??
From BC,We have
???
????
Uuay
cuy 000 2
)8.8()()(
2
2
1
2
1
2
2
1
12
?
?
?
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
?
?
a
y
a
y
x
pa
a
Uy
u
a
x
p
a
U
ca
c
a
x
p
U
?
?
??
Shear stress distribution
)9.8(21)( aayxpaaUyx ?
?
?
??
? ?
?
??
?
???
?
??
?
?
?
??? ??
Volume flow rate
)9.8(
12
1
2
)(
2
1
[
3
2
0
0
ba
x
pUa
l
Q
dyayy
x
p
a
Uy
l
Q
u l d yAdVQ
a
a
A
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
??
???
?
??
?
?
??
Average Velocity
)9.8(12 12/]12 12[
3
caxpUlala laxpUalAQV ?????? ?????????? ????? ??
Point of Maximum velocity
)/)(/1(
/
2
0
12
)(
2
2
2
xp
aUa
y
dy
du
aa
y
x
pa
a
U
dy
du
??
????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
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
0?xSF
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
dxdrrdr
dr
d
dF
r d xdF
r d rdx
x
p
r d rpdF
rx
rxO
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(ln
4
2
2
1
)10.8(
)(1
2
1
2
1
1
2
cr
c
x
pr
u
r
c
x
pr
dr
du
dr
du
c
x
pr
rC
x
p
dr
d
r
dr
rd
rdr
d
rx
p
rx
rx
rx
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
?????? ???????????? ??? xpRccxpru ?? 44
2
22
2
)12.8(1
4
22
?
?
?
?
?
?
?
?
?
?
??
?
???
?
??
?
?
?
???
R
r
x
pRu
?
)13.8()(2 axpryx ????
)13.8(
8
2)(
4
1
4
22
0
b
x
pR
Q
r d rRr
x
p
AdVQ
R
A
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
??? ??
?
?
?
?
??
)13.8(
128
4
12
c
L
pD
Q
L
p
L
pp
x
p
?
? ?
?
??
?
?
?
?
?Flow rate as a
function pf
pressure drop
Volume flow
rate
Shear stress distribution
velocity
)13.8(8
2
2 dx
pR
R
QV ?
?
??
?
?
?
????
??
)13.8(2)(
4
00
)(
2
2
2
m a x eVa
x
pR
Uur
dr
du
x
pr
dr
du
?
?
?
???????
?
?
?
?
?
The velocity profile(8.12) can be written in terms of the
maximum velocity as
)14.8()(1 2RrUu ??
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:
0?xBF
?? ??????? csCVBSx AdVudVutFFF 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][ xpRRrrxw ?????? ???
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(vudy udt ur bl am ?????? ?????
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
??? nn
n
U
V
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()(?? ????????? csCVo t h e rs h e a rs AdVpvedVetWWWQ ?????? ??
gzVue ??? 2
2
5-6.1 kinetic Energy Coefficient
Use as the Kinetic energy coefficient
)26.8(
2
)26.8(
222
2
2
222
b
Vm
dAV
V
a
V
mdAV
V
dAV
V
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
nn
n
V
U
????
??
?
???
8-6.2 head loss
Using the definition of,the energy equation can be
written
?
)22()()()(
2
11
2
22
12
12
12
VVmzzgmppmuumQ ??
?? ???????? ????
?
)28.8()()2()2( 122
2
222
1
2
111
dm
QuugzVpgzVp ????????? ?
?
?
?
dm
Q
uu
gz
Vp
?
??
??
)(
)
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
2
222
1
2
111
?
?
?
?
? lhgz
VpgzVp ??????
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
2
222
1
2
111
?
??
?
?
? l
l H
g
hz
g
V
g
pz
g
V
g
p ???????
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 CV ?
2
2?
)31.8()( 1221 lhzzgpp ?????
If the pipe is horizontal
)32.8(21 lhppp ???? ??
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
64
32
32
)4/(1 2 81 2 8
)13.8(
1 2 8
2
4
2
4
4
V
D
L
D
V
D
L
h
D
V
D
L
D
DVL
D
QL
p
c
L
pD
Q
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(21 lhppp ???? ??
),( R e,2 DeDLVh l ??
)35.8(2
2
fDLgVH 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(
Re
64
22Re
64
m i n
22
?
??
?
?
?
?
?
?
arla
l
f
V
D
L
f
V
D
L
h
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 /l o g (0.21 5.05.0 afDef ???
)37.8()]Re 74.57.3 /[ l o g (25.0 29.00 bDef ???
For turbulent flow in smooth pipes,the Blasius correlation
)38.8(Re 3 1 6.0 25.0?f
The wall shear stress is obtained as
)39.8()(0 3 32.0 25.02 VRVw ??? ?
8-7.2 Minor losses(K:loss coefficient from experiments)
)40.8(2)40.8(2
22
bVDLfhaVKh ell 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(
2
1 2
1
12
V
ppC
p
?
??
If the gravity is neglected and 121 ?? ??
mll
hhVpVp ?????
???
)2()2(
2
22
2
11
??
?
??
? ???
pl CA
AVh
m
2
2
1
2
1 )(1(
2
2211 VAVA ?
)42.8()( 11(2 2
2
1 ?
?
?
??
? ???
pl CAR
Vh
m
For frictionless flow
? ? )43.8(
11
2ARCp i ??
0?mlh
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(4 P AD h ?
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
ar
hD
h ?? 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 ??? ionc o n f i g u r a ts y s t e mzeDQLp ???
)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
l o s s esm
pp
Qmuum
pp
W
Qm
p
um
p
uWW
AdVgz
Vp
udVe
t
WQ
in
sin
csCV
s
?
?
????
?
?
????????
?????
?
?
?? ??
?????
?????
??
??
??
??
?
?
?
12
12
12
1
1
2
2
2
)(
))(()(
)
2
(
The losses are determined in terms of the pump efficiency
hpW
pAV
AVppm
pp
W
Wl o s s e s
in
in
in
3 6 8 0 0
)(
11
)1(
12
12
?
?
???
?
?
??
?
??
?
????
?
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
a r g
a r g
s uc t i onedi s c h
pum p
pum p
s uc t i onedi s c h
pum p
gz
Vp
gz
Vp
m
W
h
gz
Vp
gz
Vp
mW
??
?
?
??
?
?
?????
?
?
??
?
?
?????
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?????
?
?
??
?
?
???
??
??
?
?
??
)48.8(22 2
2
2
2
2
1
2
1
1
1
Tlp u m p
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????? ??
csCV
AdVdVt
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?????
?????
2
2
1
2
22
1
2
221
2
2
22
1
2
11
1
2
)(
2
)9.6(
22
V
VV
VVpp
gz
Vp
gz
Vp
??
??
? ? ? ?
? ?
)49.8(
])/(1[
2
1
2
0
2
12
21
2
2
1
2
2
2
21
2
1
2
2
2
1
2211
AA
pp
V
A
AV
pp
A
A
V
V
AVAV
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
??
The theoretical mass flow rate is then given by
)50.8()(2
)/(1 21212
2
22 ppAA
AAVm
lt h e o r e t i c a ???? ???
)52.8()p p(2
1
)/(/
)51.8()p p(2
)/(1
21
4
2
1
4
1
21
2
1
?
?
?
???
?
?
?
?
?
??
?
t
a c t u a l
tt
t
t
a c t u a l
CA
m
AADD
AA
CA
m
?
?
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
C
K
ta c t u a l ??
?
?
?
?
?
For the turbulent flow regime,the discharge coefficient
)55.8(Re
1
n
D
bCC ??
?
The corresponding form for the flow-coefficient
equation is
)56.8(Re
1
1
1
4 nD
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
5.2
81.2
10Re1075.02.0
)57.8(
Re
71.91
1 8 4.00 3 1 2.05 9 5 9.0
1
1
????
????
D
D
C
?
?
??
8-10.2 The flow Nozzle
74
5.0
5.0
10Re1075.025.0
)53.8(
Re
53.6
9 9 7 5.0
1
1
????
??
D
D
C
?
?
a,Pipe installation (K is function of and )
b,Plenum installation 0.95<K<0.99
?
1ReD
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
