1. LECTURE NOTES ON FLUID MECHANICS
Version 1.1
Ming-Jyh Chern, D.Phil. Oxon
Department of Mechanical Engineering
National Taiwan University of Science and Technology
43 Sec. 4 Keelung Road
Taipei 10607 Taiwan
2.
3. PREFACE
Fluid mechanics is one of important subjects in engineering science. Although it has been developing for
more than one hundred years, the area which fluid mechanics covers is getting wider, e.g. biomechanics
and nanofluids. I started to write up this manuscript when I was assigned to give lectures on fluid
mechanics for senior undergraduate students. The main purpose of this lecture is to bring physics of
fluid motion to students during a semester. Mathematics was not addressed in the lecture. However,
students were also required to learn use mathematics to describe phenomena of fluid dynamics when
they were familiar with physics in this subject. As I finished this book, I do hope that readers can get
something from this book. Meanwhile, I wold like to express my graditude to those who helped me finish
this book.
Ming-Jyh Chern
Associate Professor
Department of Mechanical Engineering
National Taiwan University of Science and Technology
mjchern@mail.ntust.edu.tw
May 29, 2007
I
9. Chapter 1
INTRODUCTION
1.1 Why study FLUID MECHANICS?
Fluid mechanics is highly relevant to our daily life. We live in the world
full of fluids!
Fluid mechanics covers many areas such as meteorology, oceanography,
aerodynamics, biomechanics, hydraulics, mechanical engineering, civil en-
gineering, naval architecture engineering, and etc.
It does not only explain scientific phenomena but also leads industrial
applications.
1.2 What is a fluid?
The main difference between fluid and solid is their behaviour when shear
forces acting on them. A certain amount of displacement is found when
a shear force is applied to a solid element. The displacement disappears
as the shear force is released from the solid element. A fluid deforms
continuously under the application of a shear force. Liquids and gases are
both regarded as fluids.
1
10. 1.3 Approaches to study Fluid Mechanics
• Analytical Methods
• Experiments
• Computations
1.3.1 Analytical Methods
Using advanced mathematics, we can solve governing equations of fluid
motions and obtain specific solutions for various flow problems. For ex-
ample: pipe flows.
1.3.2 Expenmental Fluid Mechanics
This approach utilities facilities to measure considered flow fields or uses
various visualization methods to visualize flow pattern. For example: LDA
(Laser Doppler Anemometer), hot wire, wind-tunnel test.
1.3.3 Computation Fluid Dynamics (CFD)
For most of flow problems, we cannnot obtain an analytical solution.
Hence, we can adopt numerical methods to solve governing equations.
The results are so-called numerical solutions. On the other hands, costs
of experiments become very expensive. Numerical solutions proides an al-
ternative approach to observe flow fields without built-up a real flow field.
For example: finite volume method, finite element method.
1.4 History of Fluid Mechanics
• Archmides (207-212 B.C.): buoyance theory.
·2· INTRODUCTION
11. • Leodnado da Vinci (1452-1519): He described wave motions, hydraulic
jump, jet and vortex motion.
• Torricelli (1608-1647): He is well known for measuring atmospheric
pressure.
• Newton (1643-1727): He explained his famous second law in ” Philosophiae
Naturalis Principia Mathematica”. This is one of main laws governing
fluid motions. He also provided the idea of linear viscosity describing
the relationship between fluid deformation and shearing forces.
• Bernoulli (1700-1782): Bernoulli equation.
• Euler (1707-1783): Euler equation.
• Reynolds (1842-1919): Pipe flows, Reynolds stress, turbulence theory.
• Prandtl (1875-1953), Boundary layer theory.
Y
Volume V
of mass m
Y0 Volume δ V
of mass, δ m
C
X0
X
Z0
Z
Figure 1.1: Concept of a continuum.
1.4 History of Fluid Mechanics ·3·
12. δm
δV
ρ =δlimδ V ’ δ m
V δV
δV
δV ’
Figure 1.2: Variation of a physical property with respect to the size of a continuum. Density is used as
an example.
1.5 Fluid as a continuum
The concept of a continuum is the basis of classic fluid mechanics. The
continuum assumption is valid in treating the behaviour of fluids under
normal conditions. However, it breaks down whenever the mean free path
of the magnitude as the smallest characteristic dimension of the problem.
In a problem such as rare fied gas flow (e.g. as encountered in flights into
the upper reaches of the atmosphere), we must abandon the concept of a
continuum in favor of the microscopic and statistical points of view.
As a consequence of the continuum, each fluid property is assumed to
have a definite value at every point in space. Thus fluid properties such as
density, temperature, velocity, and so on, are considered to be continuous
functions of position and time.
There exists a nondimensional number which is utilizd to judge whether
·4· INTRODUCTION
13. DISCRETE COLLISIONLESS
PARTICLE OR BOLTZMANN EQUATION BOLTZMANN
MOLECULAR EQUATION
MODEL
CONSERVATION EQUATIONS
CONTINUUM EUL.ER NAVER-STOKES DO NOT FROM A
MODEL EQS. EQUATIONS CLOSED SET
0 0.01 0.1 1 10 100 00
INVISCID FREE-MOLECULE
LIMIT LOCAL KNUDSEN NUMBER LIMIT
Figure 1.3: Knusden number and continuum.
fluids are continuous or not. Its definition is
ℓ
Kn = , (1.1)
L
where ℓ is the free mean path of a fluid molecule and L is the smallest
characteristic length of a flow field. Kn is the so-called Knusen number.
1.5 Fluid as a continuum ·5·
14. 1.6 Macroscopic physical properties of fluids
1.6.1 density, ρ
kg · m−3
Air 1.204
Water 998.2
Sea Water 1025
Mercury 13550
1.6.2 specific gravity, SG
density of substance
SG = (1.2)
density of water
Air 0.001206
Oil 0.79
Ice 0.917
1.6.3 specific volume, ν
1
ν= (1.3)
ρ
1.6.4 specific weight, γ
γ = ρg (1.4)
1.6.5 Compressibility of fluids
When fluids are pressurized, the total volume V is changed. The amount
of volume change is the compressibility of fluids. In fluid mechanics, we
use bulk modulus which is denoted as
dP dP
Ev = −V =ρ , (1.5)
dV dρ
·6· INTRODUCTION
15. A high bulk modulus means that fluids are not easy to be compressed.
Hence, fluids with a high bulk modulus are incompressible. Units and
dimensions of bulk modulus are as same as pressure.
For most of liquids, they have very large bulk moduluses (109 in S.I.).
It means liquids are incompressible. For most of gases, they are regarded
as compressible fluids due to their small bulk moduluses.
1.7 Ideal gas law
The ideal gas law describes the relationship among pressure, density, and
temperature for an ideal gas. It can be shown that P = ρRT where R is
the gas constant. For air
R = 287.03 m2s−2 K−1 = 1716.4 ft2 s−2R−2 (1.6)
1.8 Pascal’s law
The Pascal’s law indicates that pressure transmission does not decrease
within a closed container filled with fluids. As shown in Fig. 1.4, pressure
at point A and point B are equal in terms of Pascal law. Therefore, if we
apply a force to the area A, it will produce a force on B and the force is
larger than the force on A.
1.9 Speed of sound
When disturbances are intorduced into fluid, they are propagated at a
finite velocity. The velocity depends on the compressibility of considered
fluids. It is called the acoustic velocity or the speed of sound, C. It is
defind as
1.7 Ideal gas law ·7·
16. A B
Figure 1.4: Concept of Pascal’s law.
dP Ev
C= =
dρ ρ
For ideal gases,
d(ρRT ) √
C= = RT
dρ
Example: Determine acoustic velocities of air and water where the tem-
perature is 20o C.
Ev 2.19 × 109 N · m−2
Cwater = = −3
= 1480 m · s−1 (1.7)
ρ 998.2 kg · m
Consider air as an ideal gas
√
Cair = RT = 290 m · s−1 (1.8)
It implies that sound in incompressible fluids propagates faster than in
compressible fluids.
·8· INTRODUCTION
17. ¢¢
u t
y
¢ y
¡
¢ x
x
Figure 1.5: Deformation of a fluid experiencing shear stress.
1.9.1 Viscosity, µ ν
Newtonian fluids
Consider fluids are full of two parallel walls. A shear stress, τ , is applied
to the upper wall. Fluids are deformed continuously because fluids can-
not support shear stresses. The deformation rate, however, is constant.
Furthermore, if the deformation rate or the so-called rate of strain is pro-
portional to the shear stress, then the fluid will be classified as a Newtonian
fluid, i.e.
dγ
τ∝ , (1.9)
dt
where γ is shear angle or
dγ
τ =µ . (1.10)
dt
In addition,
dγ du
= . (1.11)
dt dy
Hence,
du
τ =µ . (1.12)
dy
Again, the relationship between shear stress acting on a Newtonian fluid
and rate of strain (or velocity gradient) is linear. If it is not linear, then
1.9 Speed of sound ·9·
18. the fluid will be called a non-Newtonian fluid. µ is the so-called dynamic
viscosity. Its units are dyne · cm2 or Poise (cP). In addition, lb· s 2 or Ryne
s
in
in B.G. 1 microRyne = 0.145 µ (cP)
Another definition of viscosity is the kinematic viscosity which is ν = µ
ρ
2
Its units are cm or Stoke(cS) in S.I. In addition, in or Newt in B.G. 1
2
s s
Newt = 0.00155 (cS).
Example: Determine the shear stress exerted on the bottom.
Solution:
U = 10 cm/s
oil ( = 0.036 N·s/m2)
y d =5.0 mm
u(y)
x
According to Newton’s viscosity law, we have
du
τb = µ . (1.13)
dy y=0
The velocity profile is available by a non-slip boundary condition, i.e.
U
u = y
d
0.1 m · s−1
= y
0.005 m
= 20y . (1.14)
In addition, the velocity gradient on the bottom can be obtained by
· 10 · INTRODUCTION
19. du U
= = 20 . (1.15)
dy y=0 d
Therefore, the shear stress is
τb = 0.036 × 20 = 0.72 N · m−2. (1.16)
Saybolt viscometer
When we try to measure the viscosity for a fluid, we do not measure the
shear stress, and the volocity gradient but another variable, time.
Saybolt viscometer is designed to measure the viscosity of a fluid in
constant temperature. The principle of a fluids drain from a container in
constant temperature and we measure the total time till it takes for 60 ml
of fluids. Then we use empirical formulae to evaluate kinematic viscosity,
ν. The time, measured in second, is the viscosity of the oil in offficial units
called Saybolt Universal Seconds (SOS).
195
ν(cS) = 0.226t − , t ≤ 100 SOS (1.17)
t
135
ν(cS) = 0.22t − , t ≥ 100 SOS (1.18)
t
(temperature= 1500 F )
1.10 Hooke’s law and Newton’s viscosity law
Hooke’s law for a solid element
δ
σ = Eǫ = E , (1.19)
L
where σ is stress, ǫ is strain and E is the so-called Young’s modulus.
1.10 Hooke’s law and Newton’s viscosity law · 11 ·
20. Sample
temperature
is constant
60ml
Figure 1.6: Saybolt viscosmeter
Newton’s viscosity law
du
τ = µǫ = µ
˙ (1.20)
dy
solid σ δ E
fluid τ u µ
In solid mechains, we utilize displacement to describe solid motions or
respons. Velocity, however, is employed in fluid motions instead of dis-
placement. It is because fluid deformation under shear stress is continu-
ous, so it is hard to find a displacement to indicate the magnitude of a
fluid motion.
1.11 Categories of Fluid Dynamics
Hydrodynamics Hydraulics
· 12 · INTRODUCTION
23. Chapter 2
FLUID STATICS
In fluid statics, fluids at rest are considered. No relative motion between
adjacent fluid particles. Since there is no relative motion between fluids,
viscous stress shoud not exist. Otherwise, fluids would not be at rest.
Weight of fluids is the only force in fluid statics. To keep static equilibrium,
resultant forces must be zero. Therefore pressure should be included to
keep equilibrium.
2.1 Review of Taylor Expansion
For a continuous function, f (x), it can be expanded in a power series in
the neighborhood of x = α . This is the so-called Taylor Expansion given
by
f ′(α) f ′′ (α) 2 f n (α)
f (x) = f (α)+ (x−α)+ (x−α) +. . .+ (x−α)n +. . . (2.1)
1! 2! n!
2.2 Pressure
Pressure is continuous throughout a flow field in terms of continuum con-
cept. Pressure is isotropic. In other words, pressure is independent of
15
24. direction. Positive pressure means compression. On the other hand, neg-
ative pressure means tension. It is opposite to a normal stress. Pressure
can be regarded as a scalar.
z
z P1dA
g
dz ds
y P2dydz
ρgdxdydz/2
P3dxdy
x x
dA = ds · dy =dy · dz/sin
Figure 2.1: Fluid element in a static fluid domain.
F=0 (2.2)
Fx = P2 dydz − P1 dA sin θ = 0 (2.3)
dz
P2 dydz = P1 dy sin θ (2.4)
sin θ
P2 = P1 (2.5)
1 dx
Fz = P3 dydx = ρgdxdydz + P1 dy cos θ (2.6)
2 cos θ
1
P3 = P1 + ρgdz (2.7)
2
dz → 0, P3 = P1 (2.8)
∴ P1 = P2 = P3 (2.9)
units of pressure
S.I.
1 N · m−2 = 1 Pascal(Pa) = 0.01 mbar(mb) (2.10)
· 16 · FLUID STATICS
25. B.G.
1 lb · in−2 = 1 psi = 144 psf(lbf · ft−2) (2.11)
2.3 The Hydrostatic Equation
Consider a fluid particle at rest shown in Figure 2.2. The centroid of the
z
z
¡ O
x
y
y
x
Figure 2.2: Concept of a fluid element.
fluid element is at the original point O. The fluid element has a small
volume δV = ∆x∆y∆z . Furthermore, the fluid is at static equilibrium,
so resultant forces acting on the fluid element should be zero, i.e.
F=0 . (2.12)
No shear stresses should exist owing to static equilibrium. Therefore, we
can just consider resultant forces in the z-direction, i.e.
Fz = 0 . (2.13)
Resultant forces in the z-direction include the weight of the fluid and sur-
face forces caused by pressure. The weight of the fluid particle can be
given by
W = ρgδV = ρg∆x∆y∆z . (2.14)
2.3 The Hydrostatic Equation · 17 ·
26. Subsequently, surface forces acting on the fluid element can be given by
Fs = (P2 − P1 )∆x∆y , (2.15)
where P1 and P2 are pressures on the top and the bottom respectively. P1
and P2 can be expanded using Taylor Expansion, i.e.
2
P ′ (0) ∆z P ′′ (0) ∆z
P1 = P (0) + + + + + ... (2.16)
1! 2 2! 2
and
2
P ′ (0) ∆z P ′′ (0) ∆z
P2 = P (0) + − + − + ... (2.17)
1! 2 2! 2
Substituting formulae above into the surface force, the surface force be-
comes
3
∆z
′ ∆z
Fs = −2 P (0) + P ′′′ (0) + . . . ∆x∆y . (2.18)
2 2
Consider static equilibrium again, then we find
3
∆z′ ∆z
Fz = Fs +W = −2 P (0) + P ′′′ (0) + . . . ∆x∆y−ρg∆x∆y∆z = 0
2 2
(2.19)
3
∆z ∆z
2 P ′ (0) + P ′′′ (0) + . . . = −ρg∆z (2.20)
2 2
In terms of continuum concept, ∆z should be very small (not zero), so we
can negelect high order terms in the formula, i.e.
P ′ (0)∆z = −ρg∆z (2.21)
or
dP
= −ρg . (2.22)
dz z=0
We can use a notation directional gradient to show the equation again, i.e.
∇P = ρg . (2.23)
This is called the hydrostatic equation.
· 18 · FLUID STATICS
27. 2.4 Pressure variation in incompressible fluids
Density is constant throughout an incompressible fluid domain. Hence, we
can evaluate the pressure difference between two points(z = z1 and z2 ),
i.e.
2
dP
∆P |2
1 = dz
1 dz
2
= −ρgdz
1
2
= −ρg dz
1
= −ρg (z2 − z1 ) . (2.24)
∆P
ρg is called a pressure head and equal to −∆z .
2.5 Pressure variation in compressible fluids
Density is not constant throughout a compressible fluid domain. In other
words, density may be affected by temperature and pressure. If we consider
a perfect gas, then the equation of state for a perfect gas can be used:
P = ρRT (2.25)
Substituting the perfect gas law to the hydrostatic equation, we obtain
dP Pg dP g
= −ρg = − ⇒ =− dz (2.26)
dz RT P RT
In addition, the pressure difference between two points (z = z1 and z2 )
can be evaluated by integrating the hydrostatic equation:
2 2
dP g
= − dz (2.27)
1 P 1 RT
g
⇒=lnP|2=- RT (z2 − z1 )
1
2.4 Pressure variation in incompressible fluids · 19 ·
28. P2 g
⇒=ln P1 =- RT (z2 − z1 )
g
P2 =P1 exp[- RT (z2 − z1 )]
g
△P |2 =P2 -P1 =-P1 1 − exp − RT (z2 − z1 )
1
Example: Determine the pressure at the gasoline-water interface, and at
the bottom of the tank (see Fig. 2.3). Gasoline and water can be both
open
17ft
gasoline S.G.=0.68
P1
water 3ft
P2
Figure 2.3: Problem of hydrostatic force on bottom of a tank.
regarded as incompressible fluids. Hence,
P1 = γgasoline · h + P0 (2.28)
If we assume P0 =0, then
P1 = 0.68 · 62.4 lb/ft3 · 17 = 721 psf (2.29)
In addition, the pressure at the bottom is determined by
P2 = γwater · 3 + P1
= 62.4 · 3 + 721
= 908 psf . (2.30)
2.6 Standard Atmosphere
Sea level conditions of the U.S. Standard Atmosphere.
· 20 · FLUID STATICS
29. 50
z(km) 40
20
10
surface
-60 -40 -20 0 20 40 40 80 120
Temperature Pressure
T = T0- (z-zo)
= 6.5Kkm-1
Figure 2.4: Variation of atmospheric pressure.
Table 2.1: sea level condition
S.I. B.G.
Temperature 15o C 59oC
Pressure 101.33 kPa 2116.2 psf
Density 1.225 kg/m3 0.002377 slug/ft3
Homework: Derive the formula for the pressure variation within the con-
vection layer. Remember pressure and temperature are both functions of
elevation.
Ans:
g/αR
α(z − z0 )
P = P0 1− (2.31)
T0
α = 6.5 Kkm−1 (2.32)
R = 287 Jkg−1K−1 (2.33)
g = 9.8 ms−2 (2.34)
2.6 Standard Atmosphere · 21 ·
30. 2.6.1 Absolute pressure
Pressure measured relative to an absolute vacuum.(Pb)
2.6.2 Gauge pressure
Pressure measured relative to atmospheric pressure.(Pg )
Pa
d
h . Pressure caused by fluid
weight.
z Pressure caused by
atmospheric.
Figure 2.5: Variation of static pressure.
Pb = Pg + Pa , (2.35)
(Pa : atmospheric pressure)
Consider fluids shown in Fig. 2.5. Its depth is h. If we evaluate pressure
at z = h − d, pressure at z = h − d should include two components,
atmospheric pressure and static pressure, i.e.
Pz = Pa + ρgd = Pa + ρg(h − z) . (2.36)
The resultant force acting on a small area dA at z can be given by
dF = Pz dA = Pa + ρg(h − z)dA . (2.37)
If we evaluate the resultant force on the bottom, then we obtain
F = (Pa + ρgh)dA . (2.38)
· 22 · FLUID STATICS
31. 2.7 Facilities for pressure measurement
2.7.1 Manometers
P1 P2
B
h
A
Z2
Z1
Figure 2.6: Schematic of a manometer.
Manometers are utilized to measure pressure difference between two
points,
∆P = P1 − P2 = ρgδh . (2.39)
2.7.2 Barometers
Barometers are devices designed to measure absolute pressure,
¦¥¥¤£¢
¡ h
Figure 2.7: Schematic of a barometer.
Pb = ρg∆h . (2.40)
2.7 Facilities for pressure measurement · 23 ·
32. 2.8 Inclined-tube Manometer
The main purpose of an inclined-tube manometer is to improve its resolu-
tion. Therefore, if a small pressure change is expected in an experiment,
then an inclined-tube manometer should be considered.
γ 3
h2 B
γ 1 γ 2
A h1 l2
Figure 2.8: Inclined manometer.
P1 = P2 + γ2(l2 sin θ) (2.41)
PA + γ1h1 = PB + γ3h2 + γ2 (l2 sin θ) (2.42)
PA − PB = γ3 h3 + γ2(l2 sin θ) − γ1 h1 (2.43)
If we ignore γ1 and γ3, then
PA − PB = γ2 l2 sin θ (2.44)
and
PA − PB
l2 =
. (2.45)
γ2 sin θ
If PA -PB and γ2 are constant, l2 is quite large as θ is small.
2.9 Hydrostatic force on vertical walls of constant width
dF = Pb wdz (2.46)
· 24 · FLUID STATICS
33. Pa
dF
dz
h
z
Figure 2.9: Hydrostatic force exerted on a vertical gate.
Pb = Pa + ρg(h − z) (2.47)
dF = [Pa + ρg(h − z)]wdz (2.48)
For the whole vertical wall, the resultant force is
F = dF
h
= [Pa + ρg(h − z)]wdz
0
h h
= Pa wdz + ρg(h − z)wdz (2.49)
0 0
Pawh ρgh2
w
2
If we just consider pressure caused by the weight of fluids, then the force
will be
2.9 Hydrostatic force on vertical walls of constant width · 25 ·
34. ρgh2
Fs = w . (2.50)
2
The force exerts a moment at point z = 0 and the moment is given by
dM0 = zdFs = z · ρg(h − z)wdz (2.51)
and then
M0 = dM0
h
= ρg(h − z)wzdz
0
h
hz 2 z 3
= ρgw −
2 3 0
3 3
h h
= ρgw −
2 3
ρgh3 w
= . (2.52)
6
We can evaluate the moment arm z , i.e.
¯
ρgh3 w
M0 6 h
z=
¯ = ρgh2 w
= . (2.53)
F 3
2
2.10 Hydrostatic force on an inclined surface
Consider an inclined surface shown in Fig. 2.10, then
dF = ρghdA, h = y sin θ
= ρgy sin θdA, dA = wdy (2.54)
· 26 · FLUID STATICS
35. O
θ
Y
h
dF
w X
dA
Y
Figure 2.10: Hydrostatic force exerted on an inclined gate.
and
F = dF
= ρgy sin θdA
= ρg sin θ ydA . (2.55)
ydA is the first moment of the area with respect to the x-axis, so we can
say
ydA = yc A, (2.56)
where yc is the centroid of the area. Furthermore, the resultant force
becomes
F = ρg sin θyc A
= ρghc A (2.57)
We consider the moment caused by the resultant force with respect to
2.10 Hydrostatic force on an inclined surface · 27 ·
36. the original point O. First of all,we know
dM = ydF (2.58)
and then
M = dM
= ydF
= ρgy 2 sin θdA . (2.59)
y 2 dA is called the second moment of the area with respect to the x-axis,
Ix. We know
M = F · yR (2.60)
and
M ρg sin θ y 2 dA Ix
yR = = = , (2.61)
F ρg sin θyc A yc A
where yR is the acting point of the resultant force or so-called the centre
of pressure.
Example: Consider a dam of width 100 m and depth 6 m. Determine the
resultant hydrostatic force and the moment with respect to A.
· 28 · FLUID STATICS
37. h
A
Figure 2.11: Problem of hydrostatic force exerted on a dam.
Sol:
F = γhc A
h
= γ A
2
= 1000 × 9.8 × 0.5 × 6 × (6 × 100)
= 17660 kN
M = F · hf
1
= F· h
3
= 35320 kN-m (2.62)
2.11 Hydrostatic force on a curved surface
Consider a curved surface shown in Fig. 2.12. The resultant force acting
2.11 Hydrostatic force on a curved surface · 29 ·
38. F Fx
h
dF Fz α
Z dA
X
Figure 2.12: Hydrostatic force exerted on a curved surface.
on a small element of the curved surface is given by
dF = P n · dA
= ρg(h − z)n · dA (2.63)
The resultant force in the x-direction, Fx , can be denoted as
dFx = ρg(h − z) sin αdA, (2.64)
where α is the angle between the z-axis and the normal direction of the
small area. In addition,
Fx = dFx
= ρg(h − z) sin αdA
= ρg (h − z) sin αdA
= ρg (h − z)dAv , (2.65)
· 30 · FLUID STATICS
39. where dAv is the project area of dA on the z-axis. In terms of the formula,
the resultant force in the x-axisis equal to the force acting on a vertical
plane. On the other hand, the resulatant force in the z-axis is given by
dFz = −ρg(h − z) cos αdA (2.66)
In addition,
Fz = dFz
= −ρg(h − z) cos αdA
= −ρg (h − z)dAh , (2.67)
where dAh is the project area of dA on the x-axis. In terms of this formula,
Fz is equal to the weight of liquids above the curved surface. The resultant
force F can be given by
|F| = Fx + Fz2 .
2 (2.68)
2.12 Buoyance
It is well-knoen that Archimede provided the buoyance principle to eval-
uate the buoyant force acting on a submerged solid body. In fact, we can
derive the buoyance principle from the hydrostatic equation. Let us con-
sider a submerged body shown in Fig. 2.13. The resultant force caused by
pressure on the small wetted area is given by
dF = P2 dA − P1 dA = (−ρgz2 + ρgz1 )dA (2.69)
and
F = dF = ρg (z1 − z2 )dA = −ρgV . (2.70)
2.12 Buoyance · 31 ·
40. P1
Z
Z1 dA
Z2
P2
Figure 2.13: Schematic of buoyance exerted on an immersed body.
Therefore, we know the resultant force caused by static pressure or called
the buoyant force is equal to the weight of liquids of volume equal to the
submerged body. In addition, the point where the buoyant force exerts is
called the centre of buoyance.
· 32 · FLUID STATICS
41. Chapter 3
INTRODUCTION TO FLUID
MOTION I
The chapter demonstrates basic concepts of fluid kinematics and funda-
mental laws which fluids conserve.
3.1 Lagrangian and Eulerian Systems
When we describle physical quantities, such as density, pressure, and so
on, of adynamic problem, we usually chose either Lagrangine or Eulerian
system. In terms of Lagrangine system, we move with the considered
system or particles, so physical quantities, say φ, is only a function of
time, i.e.
φ = φ(t) = φ(x(t), y(t), z(t), t) . (3.1)
Its coordinates are also functions of time. Lagrangian system is often
employed in solid dynamic. On the other hand, we fix a point in space and
observe the variation at this point in terms of Eulerian system. Therefore
physical quantities, φ, are not only functions of time but also functions of
33
42. space, i.e.
φ = φ(x, y, z, t) , (3.2)
where x, y, z, and t are independent. Eulerian system is commonly used in
fluid dynamics. It may be because lots of fluid particles are involved in a
considered flow. It contains different fluid particles at the observed point
as time goes in Eulerian system. Hence it is hard to describe a system or
its physical quantities in terms of a specified fluid particle. Therefore, we
utilize Eulerian system to describe a system.
3.2 Control Volume
In addition, we utilize a control volume concept to describe a fluid flow
problem. Coupled with Eulerian system, a control volume is a fixed region
with artifical boundaries in a fluid field. A control volume contains lots of
and various fluid particles as time goes. Fluid flows in and out through its
control surface and then physical quantities in a control volume change.
3.3 Steady and Unsteady flow
If physical quantities of a flow field are independent of time, then the flow
will be called steady. Otherwise, it is unsteady.
3.3.1 Streamlines
A steamline is defined as a line that is everywhere tangential to the in-
stantaneous velocity direction, i.e.
dy v dy v dx u
= , = , and = . (3.3)
dx u dz w dz w
Streamlines cannot cross.
· 34 · INTRODUCTION TO FLUID MOTION I
43. 3.3.2 Pathlines
A pathline is defined as the path along which a specified fluid particle
flows. It is a Lagrangine concept. Hence, coordinates of a pathline are
functions of time.
3.3.3 Streaklines
A streakline is the line traced out by particles that pass through a partic-
ular point.
3.3.4 Streamtubes
A streamtube is formed by steamlines. Since streamlines cannot cross,
they are parallel in a streamtube.
3.3.5 Definition of 1-D flows
1
2
Figure 3.1: 1-D flow
1-D flows are idealizd flows (see Fig. 3.1). It means physical properties
of flows are only functions of a spatial variable. The spatial variable can
be coordinates of an axis, such as x, or along a streamline. For example,
3.3 Steady and Unsteady flow · 35 ·
44. density ρ, for 1-D flows can be given:
ρ = ρ(x) . (3.4)
In addition, 1-D flows can be steady or unsteady, so it may be
ρ = ρ(x, t) . (3.5)
3.4 Variation of physical properties in a control volume
Consider a control volume in a flow field (see Fig. 3.2). The rate of
variation of a physical property in a control volume shall be equal to the
sum of the flux through its control surface and the surface of the physical
property.
uφ source of φ
Figure 3.2: Control volume
d ∂φ
φdV = φu · dA + dV (3.6)
dt control surface ∂t
φ: physical property in a unit volume. For example, mass in a unit volume
is density. ( m = ρ)
V
· 36 · INTRODUCTION TO FLUID MOTION I
45. 3.5 Mass conservation of 1-D flows
When fluids move, the mass conservation law should be satisfied through-
out a flow field. In terms of a control volume, the change rate of mass in
a control volume should be zero, i.e.
˙
m=0 . (3.7)
Consider a 1-D flow like the figure and fluids move along a streamline. If
we consider the control volume between point 1 and point 2 and the mass
conservation law should be satisfied in the control volume. If we donot
consider any mass source or sink in the control volume, then the rest will
be mass flux on the surface 1 and 2, i.e.
mc = m1 + m2 = 0 .
˙ ˙ ˙ (3.8)
m1 = −m2
˙ ˙ (3.9)
In addition,
m = ρu · A
˙ (3.10)
and then
ρ1 u1 A1 = ρ2 u2A2 , (3.11)
where u1 and u2 are average velocities at points 1 and 2, respectively. If
density of fluids are the same at surface 1 and 2, i.e.
Q = u 1 A1 = u 2 A2 , (3.12)
where Q is the volumetric flow rate. In terms of the mass conservation
law, we find that average velocity on a small area is higher than one on a
large area.
3.5 Mass conservation of 1-D flows · 37 ·
46. 3.6 Momemtum conservation for 1-D flows
According to Newton’s second law, an object should retain the same ve-
locity or be at rest if the resultant force exerted on it is zero. That means
the change rate of momentum in the object should be zero. We look into
the control volume concept again. If a control volume is not accelerated,
then the resultant force should be zero in the control volume. i.e.
F=0 , (3.13)
or
d
(mu) = 0 . (3.14)
dt
If we donot consider any force source in a control volume for a 1-D flow
like Fig. 3.2, then only momentum fluxes on surface 1, 2 are considered,
i.e.
d
F= (m1u1 + m2 u2) = 0 (3.15)
dt
or
d
(ρ1A1 u1 · u1 + ρ2 A2u2 · u2) = 0 (3.16)
dt
If the 1-D flow is steady, then we can remove the total derivative, i.e.
ρ1 A1(u1 · u1 ) + ρ2 A2 (u2 · u2) = 0 (3.17)
or
ρ1 A1 u2 = ρ2 A2u2 .
1 2 (3.18)
If we consider other forces acting on the control volume, then
d
F = 0 = F0 + (mu) = 0 (3.19)
dt
· 38 · INTRODUCTION TO FLUID MOTION I
47. d
F0 + (ρ1A1u1 · u1 + ρ2 A2u2 · u2) = 0 . (3.20)
dt
This is consistent with Newton third law. F can be divided into two parts:
1. body forces such as gravity forces, magnetic forces; 2. surface forces
such as pressure.
3.6 Momemtum conservation for 1-D flows · 39 ·
49. Chapter 4
INTRODUCATION TO FLUID
MOTION II
4.1 The Bernoulli equation
Consider a steady inviscid flow. If we apply Newton’s second law along a
stream line, we will obtain the Bernoulli equation
1 1
P1 + ρu2 + ρgz1 = P2 + ρu2 + ρgz2 = const . (4.1)
2 1 2 2
The detailed deviation of the Bernoulli equation will be given later. The
Bernoulli equation above is based on four assumptions:
1. along a same streamline
2. steady flow
3. same density
4. inviscid
41
50. 4.2 Derive the Bernoulli equation
Consider a steady flow shown in Fig. 4.1. For a fluid particle in the
streamline A, the momentum should be conserved. Assume the volume of
the fluid is ∆x∆n∆s. The total force along the streamline should be
Z
g
A
∂ P ds dndx n
( P+ )
∂s 2 s
∆s
β
∆n
∆n
β β
( P- ∂P ds ) dndx
∂s 2
ρg∆x∆n∆s
Y
Figure 4.1: Force balance for a fluid element in the tangential direction of a streamline.
∂P ds ∂P ds
ΣFs = P− − P+ dndx − ρg∆x∆n∆s sin β
∂s 2 ∂s 2
∂P
= − dsdndx − ρg∆x∆n∆s sin β . (4.2)
∂s
· 42 · INTRODUCATION TO FLUID MOTION II
51. The momentum change along the streamline should be
∂ 1 ∂u
(Σmu) = − (ρ∆x∆n∆s) (u) + (ρ∆x∆n∆s) u + ds
∂t ∆t ∂s
1 ∂u
= ρ(∆x∆n∆s) ds
∆t ∂s
∂u
= ρ (∆x∆n∆s) u , (4.3)
∂s
where u is the tangential velocity component. Let us consider Newton’s
second law, i.e.
∂
ΣFs = (Σmu) (4.4)
∂t
Substitution of Eq. (4.2) into (4.3) gives
∂P ∂u ∂z
− − ρg sin β = ρu , sin β = (4.5)
∂s ∂s ∂s
and then
∂P ∂z ∂u
− − ρg = ρu . (4.6)
∂s ∂s ∂s
This is the so-called Euler equation along a streamline in a steady flow. If
the Euler equation is multiplied by ds, it will become
−dP − ρgdz = ρudu (4.7)
Futhermore, we integrate the whole equation and obtain the Bernoulli
equation, i.e.
P 1
+ u2 + gz = constant . (4.8)
ρ 2
The Euler equation refers to force balance along a streamline, so the prod-
uct of the Euler equation and ds can be regarded as work done by a fluid
along the streamline. The integral of the resultant equation is constant
along a streamline. It turns out that the Bernoulli equation refers to en-
4.2 Derive the Bernoulli equation · 43 ·
52. P
ergy conservation along a streamline. ρ + gz can be regarded as potential
u2
energy and 2, of course, is the kinetic energy.
Moreover, we consider force balance across a streamline. The resultant
force should be
∂ P dn dsdx
( P+ )
∂n 2
∆n
β
( P-∂P dn ) dsdx
∂n 2
W
Figure 4.2: Force balance of a fluid element in the normal direction of a streamline.
∂P dn ∂P dn
ΣFn = P− dsdx − P + dsdx − ρg∆x∆s∆n cos ρ .
∂n 2 ∂n 2
(4.9)
Its momentum change across a streamline should be
∂ u2
Σmun = −ρ ∆x∆s∆n , (4.10)
∂t R
where un is the velocity component normal to a streamline and R is the
· 44 · INTRODUCATION TO FLUID MOTION II
53. curvature radius. Let us consider Newton’s second law again.
∂
ΣFn = Σmun (4.11)
∂t
Substitution of Eq. (4.9) into (4.10) gives
∂P u2
− dndsdx − ρg∆x∆s∆n cos β = −ρ ∆x∆s∆n (4.12)
∂n R
∂z
cos β = (4.13)
∂n
and then
∂P ∂z u2
+ ρg =ρ . (4.14)
∂n ∂n R
This is the Euler equation across a streamline. If the Euler equation is
multiplied by dn and integrated along the normal direction, it will become
u2
− dP − ρgdz = ρ dn . (4.15)
R
It is the Bernoulli equation along the normal direction of a stream.
Example: Determine the pressure variation along the streamline from
z
A a x
O B
3
u=u0(1+a3 )
x
Figure 4.3: 2-D flow past a circle.
point A to point B.
4.2 Derive the Bernoulli equation · 45 ·
54. Solution:
From the Bernoulli equation along a streamline,
−dP − ρgdz = ρudu (4.16)
Since point A and B are at the horizontal streamline, dz = 0 Hence
−dP = ρudu . (4.17)
In additions,
O O
dP = ρudu .
A A
We know that
du = u0a3 (−3)x−4dx
a3
= −3u0 4 dx . (4.18)
x
As a result,
O
a3 a3
PO − PA = ρ u0 1 + −3u0 4 dx
A x3 x
O
2 a3 a6
= ρ −3u0 + dx
A x4 x7
O
a3 a6
= ρu2
0 +
x3 2x6 A
O
u2
0 1
= ρa3 1+ . (4.19)
x3 2x3 A
The x-coordinate of point B is -a, so
1 u2
0 1
PB − PA = −ρu2
0 1− 3 − ρa 3 3
1+ . (4.20)
2a xA 2x3A
4.3 Stagnation Pressure and Dynamic Pressure
Consider fluids flow toward a horizontal plate far upstream. Fluids moves
at u∞ and pressure is P∞ upstream. Because fluids cannot pass through a
· 46 · INTRODUCATION TO FLUID MOTION II
55. P ∞
u ∞
P0
stagnation point
stagnation streamline
Figure 4.4: Stagnation point
plate, fluids must flow along the plate. Subsequently we can find a point
where fluids are at rest. This is the so-called stagnation point. Further-
more, we can find a stagnation steamline which leads to the stagnation
point. Owing to no variation of altitude in the whole flow, pressure and
velocity are considered in the Bernoulli equation. If we apply the Bernoulli
equation along the stagnation line, we will find
P∞ u2 P0
+ ∞= , (4.21)
ρ 2 ρ
where P0 is called the stagnation pressure or total pressure, P∞ is called
2
ρu∞
the static pressure, and 2 is called dynamic pressure which is distincted
from the pressure due to hydrostatic pressure, P∞ .
4.3 Stagnation Pressure and Dynamic Pressure · 47 ·
56. Pressure coefficient is defined as
P − P∞ u
Cp = 1 2
= 1 − ( )2 . (4.22)
2 ρu∞
u∞
Its means the ratio of pressure difference to inertia force. At a stagnation
point, Cp = 1, that means all of kinetic energy is transfered to pressure
energy. Cp is zero far upstream. It means no kinetic energy is transfered
to pressure energy.
4.4 Mass conservation in channel flows
Consider fluid flow in a channel with various cross section areas show in
Fig. 4.5. Fluids connot accumulate at any cross sections. In other words,
1 2
Figure 4.5: Mass conservation in 1-D flow.
mass must be conserved at any cross section. Hence mass flowrates, the
amount of mass passing a cross section per unit time, must be equal at
every cross section, i.e.
m = m1 = m2 ,
˙ ˙ ˙ (4.23)
· 48 · INTRODUCATION TO FLUID MOTION II
57. where m is the mass flow rate in the channel. In addition,
˙
m = ρQ ,
˙ (4.24)
where ρ is fluid density and Q is volumeric flowrate. Then,
ρ1 Q1 = ρ2 Q2 (4.25)
or
ρ1 u1 A1 = ρ2 u2A2 , (4.26)
where u1 and u2 are average velocity at cross sections 1 and 2, A1 and
A2 are cross sectional areas. For incompressible fluids, ρ1 = ρ2 and conse-
quently u1 A1 = u2A2.
4.5 Relationship between cross area, velocity ana pressure
Consider a steady flow in a channel with varied cross sectional areas. In
terms of the continuity equation, velocity decreases as its cross sectional
area diverages for incompressible fluids. In addition, pressure increases
as velocity decreases in terms of the Bernoulli equation. For a converged
channel, cross sectional area decreases so velocity increases. Subsequently,
pressure decreases owing to increasing velocity.
4.6 Applications of Bernoulli equation
4.6.1 Pitot tube
1 1
P∞ + ρa u2 + ρa gz∞ = PO + ρa u2 + ρa gzO
∞ O (4.27)
2 2
z∞ = zO , uO = 0 (4.28)
1
∴ P∞ + ρa u2 = PO
∞ (4.29)
2
4.5 Relationship between cross area, velocity ana pressure · 49 ·
58. A A
V V
P P
Figure 4.6: Variations of velocity and pressure in converged and diverged channels.
1
(PO − P∞ ) = ρa u2
∞ (4.30)
2
PO − P∞ = ρℓ g∆h (4.31)
1
ρℓ g∆h = ρa u2
∞ (4.32)
2
ρℓ
u2 = 2 g∆h
∞ (4.33)
ρa
4.6.2 Siphon(ÞÜ)
A siphon is a device transfering fluids from a lower level to a higher level.
Consider a siphon shown in Fig. 4.8. The free surface in the tank is
assumed to be still owing to the flow rate to the siphon is very slow.
Hence the velocity is zero at the free surface. Furthermore, the Bernoulli
equation is applied to analyze the flow in a siphon. Consider conditions
at points 1 and 3 and
Pa Pa u2
+ 0 + gz1 = + 3 + gz3 , (4.34)
ρ ρ 2
· 50 · INTRODUCATION TO FLUID MOTION II
59. O
P∞
u∞
z0
∞
z∞
∆h
ρl
Figure 4.7: Schematic of Pitot tube.
where z1 = 0, z3 = −h3 . Velocity at point 3 is obtained from the equation
i.e.
u3 = 2gh3 . (4.35)
Another interesting location is at point 2. In terms of Bernoulli equation,
we find
Pa P2 u2
+ 0 + gz1 = + 2 + gz2 , (4.36)
ρ ρ 2
where z1 = 0, z2 = h2 . Then we find pressure at point 2 is
P2 Pa u2
= − 2 − gh2 , (4.37)
ρ ρ 2
u2
where 2
2
and gh2 must be positive. It turns out that P2 should be less
than the atmospheric pressure. If point 2 is high enough to let pressure at
point 2 less than vapor presure, then gas in fluids will form bubbles. These
bubbles will move with fluids. If pressure around bubbles increases and is
higher than vapor pressure, then bubbles will burst. The phenomenon is
4.6 Applications of Bernoulli equation · 51 ·
60. 2
z h2
1
h3
3
Figure 4.8: Schematic of siphon tube.
called cavitation. Cavitation is often found in flow fields around a inside
propeller or fluid machinery.
4.6.3 Torricelli’s Theorem
1 Pa
H
Pa
2
Figure 4.9: Torricelli’s theorem.
Consider a liquid tank of high H. There is a hole, shown in Fig. 4.9, near
the ground. Liquids drain from the hole. It is assumed that the tank is
quite large, so the location of the free surface is almost still. Hence, u1 = 0.
Moreover, pressure at the hole is assumed to be equal to the atmospheric
· 52 · INTRODUCATION TO FLUID MOTION II
61. pressure. Now we can apply Bernoulli equation to point 1 and 2, i.e.
P1 u21 P2 u2
+ + gz1 = + 2 + gz2 , (4.38)
ρ 2 ρ 2
where P1 = P2 = Pa , u1 = 0, and (z1 − z2 ) = H. It then becomes
Pa Pa u2
+ gH = + 2 . (4.39)
ρ ρ 2
It turns out that
u2 = 2gH . (4.40)
This is the Torricelli’s Theorem.
4.6.4 vena contracta effect
dj
dh
Figure 4.10: Vena contracta effect
contraction coefficient
Aj (dj )2
Cc = = (4.41)
Ah (dh)2
4.6 Applications of Bernoulli equation · 53 ·
62. 1
h
l
2
Figure 4.11: Free jet
4.6.5 Free jets
Consider fluids in a tank. A nozzle is arranged at the bottom of the
tank. Fluids flow through the nozzle due to the gravitational force and
consequently a jet is observed. Suppose no energy loss in the nozzle.
Bernoulli equation can be utilized to determine the jet condition at the
exit of the nozzle. The free surface of the tank is assumed to be still if
the tank is large enough. Therefore, u1 = 0. According to the Bernoulli
equation, the total energy along a streamline from the free surface to the
exit should be the same, i.e.
P1 u21 P2 u2
+ + gz1 = + 2 + gz2 = constant . (4.42)
ρ 2 ρ 2
We know u1 = 0, P1 = P2 = Pa and (z1 − z2 ) = h + l. The equation
becomes
u2
2
= g(h + l) (4.43)
2
· 54 · INTRODUCATION TO FLUID MOTION II
63. or
u2 = 2g(h + l) . (4.44)
The result is as same as Torricelli’s Theorem.
However, if the nozzle is not designed well, then there will be energy
loss at the nozzle. As a result, Bernoulli equation has to be modified.
4.6.6 Venturi tube
A
B
Figure 4.12: Venturi tube.
4.6 Applications of Bernoulli equation · 55 ·
64. u A AA = u B AB
AA
uB = uA
AB
AA AB
uA uB
2
PA uA PB u2
+ = + B
ρ 2 ρ 2
PA − PB uB − u2
2
A
=
ρ 2
AA
u2 ( AB ) − u2
A A
=
2
2 2
ua AA
= −1 (4.45)
2 AB
A Venturi tube is a device made up of a contraction followed by a diverging
section. Fluids moving toward the contraction are speeded up according
to the continuity equation. In addition, pressure decreases as velocity
increases in terms of the Bernoulli equation. A famous application of a
Ventui tube is a carburetor. A carburetor is shown in Fig. 4.13. Fuel is
sucked into the throat due to the low pressure at the throat. Subsequently,
fuel is mixed with air at the throat. Venturi tube is a facility to measure
the flow rate in a pipe. Fluids flow a contraction part and then a expansion
part in a Venturi tube.
· 56 · INTRODUCATION TO FLUID MOTION II
65. Q(Air)
Butterfly
Valve
Throat of
Venturi
FUEL
Air-Fuel
Mixture
Q
Figure 4.13: Schematic of caburetor.
4.6.7 Flowrate pass through a sluice gate
Form the Bernoulli equation,
P1 u21 P2 u2
+ + z1 = + 2 + z2
r 2g r 2g
P1 = P2 = Pa
u2
1 u2
+ z1 = 2 + z2 (4.46)
2g 2g
Form mass conservation
u1z1 = u2 z2
z2
u1 = u2 . (4.47)
z1
4.6 Applications of Bernoulli equation · 57 ·
66. Substituting into Bernoulli equation
2
u2
2 z2 u2
+ z1 = 2 + z2
2g z1 2g
2
u2
2 z2
−1 = z2 − z1
2g z1
z2 − z1
u2 = 2g z2 . (4.48)
( z1 )2 − 1
The flowrate pass through the sluice gate must be
Q = u2 · z2
2g(z2 − z1 )
= z2 z2 . (4.49)
( z1 )2 − 1
· 58 · INTRODUCATION TO FLUID MOTION II
67. Chapter 5
EQUATIONS OF MOTION IN
INTEGRAL FORM
We consider one-dimensional flows in Chapter 3 and 4. Conservation laws
of mass, momentum and energy are obtained for one-dimensional flows.
Most of fluid flows, however, cannot be simplified as one-dimensional flows.
Therefore, we have to look into conservation laws again and derive gov-
erning equations for general fluid flows.
These equations for fluid flows can be either in integral form or in differ-
ential form. Equations in integral form are derived in terms of the control
volume concept. Equations in integral form do not give any information
throughout a flow field, but they can provide resultant forces acting on a
control volume. On the other hand, equations in differential form provide
details regarding variations in a flow field, so we can get values of physical
variables throughout a flow field.
In this chapter, we consider governing equation of fluid flows in integral
form first.
59
68. 5.1 Flux
We mentioned the control volume concept in Chapter 3. A control volume
is bounded artificially in a flow field. Physical properties in a control
volume may vary in space or in time, because fluids with various physical
properties flow in and out a control volume and it causes variations of
physical properties in a control volume. The amount of a physical property
cross an unit surface per second is called flux.
A flux can be revealed as b(u · A), where b is a physical property per
unit volume, u is the velocity over the area and A is the area vector. We
may use nA instead of A and n is the unit vector in the normal direction
of the area. Physical properties considered in this chapter can be mass,
momentum or energy, so we have different fluxes:
mass flux : ρ(u · n)A (5.1)
momentum flux : ρu(u · n)A (5.2)
energy flux : e(u · n)A (5.3)
e : energy contained in a unit volume, (5.4)
i.e., specific energy (5.5)
It should be noticed that n is positive in the outward direction of the
area but negative in the inward direction.
· 60 · EQUATIONS OF MOTION IN INTEGRAL FORM
69. 5.2 Reynolds’ Transport Theorem
2
1
III
II
I
Figure 5.1: Flow through a control volume.
We consider a control volume I+II in a flow field. Fluids contained in
the control volume at t = t will flow, so the control volume containing
same fluids at t = t+δt will be II+III. The rate of change of a physical
property in the control volume can be shown in DtD
c.v. ραdV where α
is the amount of the physical property per unit mass. In terms of Fig. 5.1,
we know the rate of change in the control volume can be divided into two
parts. The first is the local chang at the region II, which can be shown
∂
in ∂t II ραdV . The second is the net flux including the flux from the
region I to the region II and the flux from the region II to the region III,
so we have c.s.1 ρα(u · n)dA and c.s.2 ρα(u · n)dA. We can combine
5.2 Reynolds’ Transport Theorem · 61 ·
70. fluxes across two surfaces and get c.s. ρα(u · n)dA. As δt → 0, we will
have
D ∂
ραdV = ρα(u · n)dA + ραdV (5.6)
Dt c.v. c.s. ∂t c.v.
At t = t0
Bsys = BI (t) + BII (t) . (5.7)
At t = t0 + ∆t
Bsys = BII (t + δt) + BIII (t + δt) (5.8)
∆Bsys DBsys
lim = (5.9)
∆t→0 ∆t Dt
or
∆Bsys BII (t + δt) + BIII (t + δt) − BII (t) − BI (t)
= (5.10)
∆t ∆t
BII (t + δt) − BII (t) ∂BII
lim = (5.11)
∆t→0 ∆t ∂t
−BI (t)
∆t is the flux flow through in C.S.1 and is denote as
ρα(u · dA) (5.12)
C.S.1
BIII (t+δt)
In addition, ∆t
is the flux flow out C.S.2 and is denoted as
ρα(u · dA) (5.13)
C.S.2
ρα(u · dA) + ρα(u · dA) = ρα(u · dA) (5.14)
C.S.1 C.S.2 C.S.
Besides,
lim (C.V.I +C.V.II ) = lim (C.V.III +C.V.II ) = C.V.II = C.V. = ραdV
∆t→0 ∆t→0 C.V.
(5.15)
· 62 · EQUATIONS OF MOTION IN INTEGRAL FORM
71. As a result
DBsys D
= ραdV (5.16)
Dt Dt c.v.
and
∂Bc.v.II ∂
= ραdV (5.17)
∂t ∂t c.v.
5.3 Continuity Equation
If we consider mass variation in a control volume, then we will have α = 1.
In terms of Reynold’s transport theorem, the conservation of mass can be
revealed as
D ∂
ρdV = ρ(u · n)dA + ρdV = 0 (5.18)
Dt c.v. c.s. ∂t c.v.
This is the continuity equation in integral form.
5.4 Momentum Equation
Subsequntly, we consider momentum in a control volume, then α will be
u. The momentum equation in integral form then is denoted as
D ∂
ρudV = ρu(u · n)dA + ρudV . (5.19)
Dt c.v. c.s. ∂t c.v.
Moreover, the rate of momentum is equal to the resultant force acting on
the control volume, i.e.
D
ρudV = F = Fbody + Fsurface + Fext . (5.20)
Dt c.v.
If we consider gravity in body force, then we will have
Fbody = ρgdV . (5.21)
c.v.
5.3 Continuity Equation · 63 ·
