kempf notes
TRANSCRIPT
1M FluidsSupplementary Information
Andreas M Kempf
2010-2011
ii
Disclaimer: This document summarises some of the information required for
the 1M Fluids course. It can help you study and prepare, but it is not offi-
cially related to the course. The official documentation for 1M Fluids is Fox,
McDonald, Pritchard, Introduction to Fluid Mechanics.
Some Notes for 1M Fluids
With contributions by: Bharat Lad, Konstantinos Zarogoulidis
Proofreading and corrections by:
Mark Dean, David Farrell, Marc Hinken, Josef Huwiler, Hassan Joudi, Bharat
Lad, Rolf Lechner, Sam Weeks, David Muller-Wiesner, Francesco Ferroni,
Ailing Wang, Chern Sim, Victor Luzzato, Konstantinos Zarogoulidis, Ruili
Chen, Jianwei Zhang, Mazda Rustomji, ... and many others.
Dr. Andreas M. Kempf
Department of Mechanical Engineering
Imperial College London
Room 642
Typeset with LATEX.The LATEX system is freely available as tetex for Unix-platformslike Linux and Mac OS-X, and as miktex for windows.
Contents
1 Introduction 1
2 Basic Treatment of Fluid Flow 3
2.1 Continuum Assumption . . . . . . . . . . . . . . . . . . . . 3
2.2 Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.3 Definition of a Fluid . . . . . . . . . . . . . . . . . . . . . . 7
2.4 Systems vs. Control Volumes . . . . . . . . . . . . . . . . . 8
2.5 Velocity Field . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.6 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.7 Classification of Fluid Motion . . . . . . . . . . . . . . . . . 12
2.7.1 Laminar vs. Turbulent . . . . . . . . . . . . . . . . . 12
2.7.2 Incompressible vs. Compressible . . . . . . . . . . . . 13
2.7.3 Newtonian vs. non-Newtonian Fluids . . . . . . . . . 13
2.7.4 Single-phase vs. multi-phase . . . . . . . . . . . . . 13
2.7.5 Sub-sonic vs. super-sonic and trans-sonic . . . . . . . 14
2.7.6 Reactive vs. non-reactive flows . . . . . . . . . . . . 14
3 Fluid Statics 15
3.1 Basic Equation of static fluids . . . . . . . . . . . . . . . . . 16
3.2 Pressure Variation in fluids of constant density . . . . . . . . 17
3.3 Hydrostatic Force on Submerged Surfaces . . . . . . . . . . 18
3.4 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4 Flow Rates 21
5 Basic Laws for Systems and Control-Volumes 25
5.1 Specific conservation principles for systems . . . . . . . . . . 26
5.2 General conservation principle for systems . . . . . . . . . . 27
iii
iv CONTENTS
5.3 Relation between Systems and Control Volumes . . . . . . . 28
5.4 Conservation of Mass for a Control Volume . . . . . . . . . 30
5.4.1 Conservation of mass for constant density . . . . . . 31
5.5 Conservation of Energy for a Control Volume . . . . . . . . . 32
5.6 Conservation of Momentum for a Control Volume . . . . . . 33
6 Specific Laws and Special Cases 35
6.1 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . 35
6.1.1 Bernoulli’s Equation Derived from the Energy Equation 35
6.2 Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . 37
6.3 Laminar Pipe-Flow . . . . . . . . . . . . . . . . . . . . . . . 38
7 Final notes 41
Chapter 1
Introduction
Recommended Books
The book by Fox, McDonald and Pritchard is the official documentation for
the 1M Fluids course.
• Fox, McDonald, Pritchard, Introduction to Fluid Mechanics, Wiley &
Sons 2004, ISBN 0-471-20231-2. This sixth edition may no longer be
available, but any later edition will be suitable.
• Hardalupas, Y., Crane, R., Formulas and Tables, Imperial College 2007
Lectures and Tutorials 2010/2011
The tutorial questions given in the following table are recommended for self
study beyond the normal compulsory tutorials. The numbers refer to the sixth
and seventh edition of ’Introduction to Fluid Mechanics’ by Fox, McDonald,
Pritchard in both the US and SI editions. Starred questions may have changed
from the sixth edition.
1
2 CHAPTER 1. INTRODUCTION
# Lecture, Topic FMP 6 FMP 7 (US) FMP 7 (SI)
1 What is Fluid Mechanics? 1.5, 1.6, 1.7 1.5, 1.6, 1.7 1.5, 1.6, 1.7
2 Velocity Field. Stresses. 2.1, 2.2, 2.3 2.1, 2.2, 2.3 2.1, 2.2, 2.3
3 Viscosity and Fluid Motion. 2.29, 2.41, 2.42 2.35, 2.51, 2.54 2.34, 2.50, 2.53
4 Hydro Statics. 3.1, 3.5, 3.20 3.1, 3.5, 3.20 3.1, 3.5, 3.20
5 Hydro Static Forces. 3.43, 3.46, 3.48 3.48, 3.51, 3.52 3.48, 3.51, 3.52
6 Examples Class. 3.23, 3.31, 3.55 3.26, 3.34, 3.62 3.48, 3.51, 3.52
7 Fluxes and flow rates. 4.10, 4.20 4.10, 4.20 4.10, 4.20
8 Systems, Control Volumes. 4.13, 4.21 4.13, 4.30 4.13, 4.30
9 Conservation of Mass. 4.19, 4.25, 4.27 4.28, 4.34, 4.36 4.28, 4.34, 4.36
10 Conservation of Energy. 4.183, 4.184 4.198, 4.199 4.198, 4.199
11 Introduction to Fluids Labs. 4.185, 4.189 *, 4.205 *, 4.205
12 Energy, Bernoulli’s Equation. 6.44, 6.45, 6.46 6.51*, 6.52*, 6.53 6.51*, 6.52*, 6.53
13 Conservation of Momentum. 4.11, 4.13 4.11, 4.13 4.11, 4.13
14 Examples Class. 4.112, 4.115 4.127, 4.130* 4.127, 4.130*
15 Boundary Layers. – – –
16 Laminar Pipe Flows. 8.44, 8.52 8.45, 8.49 8.45, 8.49
17 Losses in Pipe Flows. 8.84, 8.86 8.84, 8.87* 8.84, 8.87*
18 Pipe Flow Problems. 8.141, 8.142 8.154*, 8.153* 8.154*, 8.153*
19 Flow Measurements. –
20 Revision. many...
21 Revision for exams. ....more
Chapter 2
Basic Treatment of Fluid
Flow
2.1 Continuum Assumption
→ Fox et al. [1]: 2-1
Fluids and solids consist of a very large number of small particles, the molecules.
Each molecule is a distinct, individual particle, but taken together, they be-
have like the continuous fluids and solids we experience in daily life: they form
a continuum.
In most engineering applications of solid or fluid mechanics, we make use
of this continuum assumption, as we are not interested in the behaviour of
individual molecules. (If we could determine the behaviour of each individual
molecule, we would end up with much irrelevant information.)
The continuum assumption implies that the property of a fluid remains con-
stant regardless of how much fluid we examine. It also implies that a fluid’s
behavior does not change within the local region of any given point. This is
not the case if we were to consider individual particles, where properties at a
point would differ depending on whether it was being occupied by a particle
or not.
3
4 CHAPTER 2. BASIC TREATMENT OF FLUID FLOW
Figure 2.1: Air seen as a group of many individual molecules or as a contin-
uum. If we consider a sufficiently large number of molecules, only statistical
properties matter (the behaviour of individual molecules can be ignored) and
the material can be treated as a continuum.
2.2 Stresses→ Fox et al. [1]: 2-3
In many cases, mechanical forces are distributed over an area. For example,
consider a steel cable carrying a weight. We know that the cable will break if
the force is too large, but a cable with a larger cross-section would support a
greater force.
Figure 2.2: A cable of cross sectional area A undergoing stress σ due to the
force F .
In the end, it is not the total force F that determines whether a structure will
break, but it is the force per area A of cross section. This force per area of
cross section is called a stress, and we usually use the symbol σ to refer to
it.
σ = F/A (2.1)
A similar situation occurs if we consider the force a wheel exerts on a road.
Here, the stresses are important, which we would calculate as the weight W
2.2. STRESSES 5
(per wheel) divided by the contact area A:
σ = W/A (2.2)
Since the weight force acts normal (perpendicular) to the surface, we call
this a normal stress. If the weight of the vehicle exerts too great a stress,
the surface beneath it can be destroyed. This can be very relevant for large
aircraft, which use many wheels to increase the contact area and thus reduce
stress. But what happens if the vehicle accelerates, for example when braking
or when cornering? The force F required for acceleration acts in a direction
parallel (tangential) to the road, and the load is also taken by the contact
area A. The resulting stress F/A is very different from the normal stress σ, as
F/A acts parallel to the surface. We call such parallel stresses shear stresses
and refer to them as τ .
Figure 2.3: A wheel supporting the weight W with an accelerating force F
acting on contact area A. These forces lead to normal stresses σ = W/A
and shear stresses τ = F/A.
In the example of the stresses in a cable, we did not consider stresses on a
real surface, but just in a cross-section through the cable. We could certainly
consider any cross-section through the cable, even one that is aligned along
the length of the cable: in this plane, we would expect the stresses to be
zero.
To elaborate on normal and shear stresses, consider a solid cube on a flat
surface. If we cut the cube in a top and a bottom half, and consider the
6 CHAPTER 2. BASIC TREATMENT OF FLUID FLOW
forces on the cut surface, a force balance will yield a normal stress that equals
the weight of the cube’s top half divided by the cut area. If we cut the cube
into a left and a right half, we will not observe any forces between the two
halves. However, if we cut the cube diagonally, the force-balance will yield
both normal stresses σ and shear stresses τ .
Figure 2.4: A solid cube on a flat surface, cut in different directions. The
only force on the cube is gravity, leading to normal stresses σ and shear
stresses τ that depend on the cutting plane.
Therefore, the stresses we see across a cut plane depend on the direction of the
cross sectional area. So for any stress, we have to mention the direction which
the surface faces. As a result it is important to understand the conventional
notation used for normal and shear stresses.
Figure 2.5: Stress notations for normal (σ) and shear (τ) stresses.
Consider three independent directions x, y and z, to which the surfaces are
perpendicular. For each direction, we can encounter a normal stress σ corre-
sponding to a force that points in the same direction, and two independent
shear stresses τ pointing parallel to the surface. For example, on the surface
facing the x direction, we can encounter the normal stress σxx and the shear
stresses τxy and τxz, where as on the y surface: σyy, τyx and τyz and on the
2.3. DEFINITION OF A FLUID 7
z surface: σzz, τzx and τzy. Applying the notational convention to the cut
cube we discussed earlier, the stresses on the cut face can be described like
so:
Figure 2.6: Notation of stresses. The first index denotes the direction normal
to the face, whilst the second index denotes the direction of the stress: "Face
first, stress second"
All together, we have a total of nine stresses (3 normal and 6 shear), and
we follow the convention that the first index denotes the direction in which a
surface faces (normal to the surface), whereas the second index refers to the
direction of the force.
"Face first, stress second."
The stresses for each direction have the property of a vector (3 components
in x, y and z). This means that stresses at a point can be described by "a
vector of vectors", a matrix that is usually called the "stress tensor":
σxx τxy τxzτyx σyy τyzτzx τzy σzz
(2.3)
2.3 Definition of a Fluid→ Fox et al. [1]: 1-2
We already have an idea of what distinguishes fluids (liquids or gases) from
solids. We can define a fluid by relating its behaviour in response to an applied
8 CHAPTER 2. BASIC TREATMENT OF FLUID FLOW
stress. According to Fox, McDonald and Pritchard [1], we can define a fluid
as:
"A substance that deforms continuously under the application of a shear stress
no matter how small the stress may be."1
2.4 Systems vs. Control Volumes
→ Fox et al. [1]: 1-5
In mechanics, it is usually necessary to compute the forces acting on an
object in order to understand or predict its response. There are two distinct
approaches for doing this. We could compute the acceleration of a point-mass
due to a force, or we could compute the forces acting on a frame surrounding
the point mass.
In continuum mechanics, we must define exactly what a force acts on. In
fluid mechanics, we will often consider fluids that move through or around an
object like a pump, an aeroplane, a blood-vessel or a jet-engine. We can then
define a volume in physical space through which the fluid will flow, and can
do a force-balance for this volume. Such a volume is called a control volume
(CV). The size, location and shape of the CV is up to us, but we would usually
specify it such that the computation will be as simple as possible.
A concept similar to control volumes is often applied in thermo-dynamics: a
system. Here again, we can decide what shall be part of the system, but there
is an important difference between control volumes and systems: a system
consists of a fixed mass, and no mass can cross the system boundaries. A
system can deform, absorb heat, or be accelerated, but it will always consist of
the original material (or of the original molecules). In contrast, mass can cross
the boundaries of a control volume, but the control volume cannot change its
size, shape or location.
2.5. VELOCITY FIELD 9
Figure 2.7: Systems (left) and Control Volumes (right). Systems are defined
as an amount of mass, with no exchange of mass across the system boundaries.
A control volume is a fixed volume in space that neither moves nor changes
its shape, but mass can pass through it.
2.5 Velocity FieldFox et al. [1]: 2-2
In mathematics, functions describe the relations between different quantities.
A simple example would be the function f that relates x to y, for exam-
ple y = f(x). f(x) could describe a range of different relations including
f(x) = 2x, x2 or x3+2x2+3x etc. One common example used in electrical
engineering relates a current I to voltage U over a resistor R by the function
fr: I = fr(U,R) = U/R
A function that depends on a position in physical space describes a field. If the
function depends on one coordinate only (e.g. x), the field is one-dimensional.
If it depends on two coordinates (e.g. x, y), the field is two-dimensional.
Accordingly the field is three dimensional if the function depends on all three
coordinates (e.g. x, y, z).
One example of a field could be a map of the wind-velocity across the UK, as it
is frequently seen in weather-forecasts: This velocity-field is two dimensional
(Realistically the velocity field of wind will be 3D however this s simplified
to 2D for the weather forecast). As the wind has magnitude and direction,
it as a vector: in contrast to a scalar-field (such as temperature which has
magnitude only and no direction), the wind-field is a vector-field.
In fluid mechanics, we are concerned with the forces on fluids and with the
motion of fluids. This motion of fluids is described by three independent
velocity components u, v, w at each point x, y, z, in the velocity-vector-field
1The distinction between fluids and solids is not always clear, but for now, and in
many technical applications, we can ignore borderline cases such as toothpaste, sand, solid
emulsion paint, shaving gel, tar, or dough.
10 CHAPTER 2. BASIC TREATMENT OF FLUID FLOW
~V .2 All three velocity components u, v, w can change at different locations
x, y, z, so each component u, v, w of the velocity vector-field is a function of
x, y, z:
u = u(x, y, z)
v = v(x, y, z) (2.4)
w = w(x, y, z).
In vector notation, we could also write:
~V =
u
v
w
=
u(x, y, z)
v(x, y, z)
w(x, y, z)
. (2.5)
Writing vector fields in this notation is laborious as we always need three
different lines. However, we can use a notation that is based on the unity
vectors3 i, j, k pointing in x, y, z direction respectively.4
~V = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k. (2.6)
2.6 Viscosity→ Fox et al. [1]: 2-4
So far we have looked at describing fluids as continua, what kind of stresses
exist within fluids and how we can describe the velocity in terms of a field
within a volume of fluid. In section 2.3 we defined a fluid as "a substance
that deforms continuously under the application of a shear stress no matter
how small the stress may be"[1]. However, we have not yet discussed how
quickly it will deform when a stress is applied.
Sir Isaac Newton showed that for many fluids, the rate of deformation is pro-
portional to the stress applied. For these fluids the constant of proportionality
2To simplify our life, we usually assume that the velocity component u describes motion
in the x direction, v describes motion in the y direction, and w describes motion in the z
direction.3A unity vector is a vector of magnitude one.
4The unity vectors i, j, k are defined as: i =
1
0
0
; j =
0
1
0
; k =
0
0
1
2.6. VISCOSITY 11
U top
Ubottom
x
y
Figure 2.8: In between two moving plates, the fluid velocity profile will be
linear between the velocities of the upper and bottom plates due to the no-slip
condition.
is viscosity. So the fact that different fluids (e.g. water, oil or honey) flow
down an inclined surface at very different speed is due to the viscosity of the
fluid.
Let us consider the example of a fluid between two horizontal parallel plates
that slide over each other in the x direction. We introduce a second coordinate
y perpendicular to the surface of the plates. We also define that at a wall,
fluids stick to the surface (this is called the no-slip condition). Therefore
the fluid-velocity on the surface of the top-plate equals the velocity of the
top-plate, whereas the velocity on the bottom-plate equals the velocity of the
bottom plate. In between the plates, we just encounter a linear velocity profile
that ranges from the velocity of the bottom plate to the velocity of the top
plate.
We can define the deformation rate as the change of surface-parallel velocity
u, per unit change of normal distance y i.e. dudy . Sliding the plates over
each other requires a force. This will produce the shear stress τyx within the
fluid. Hence the fluid encounters a shear stress τyx that is proportional to the
12 CHAPTER 2. BASIC TREATMENT OF FLUID FLOW
deformation rate dudy :
τyx ∝ du
dy(2.7)
∴ τyx = cdu
dy(2.8)
The proportionality constant determines the magnitude of stress required to
produce a given deformation rate. As explained previously the proportionality
constant is viscosity (µ) and is a property of the fluid. Viscosity is a property
of the fluid, and Newton discovered its importance. So Newton’s famous law
of viscosity is:
τyx = µdu
dy(2.9)
The viscosity of different fluids varies significantly, for example, air at 0◦C and
atmospheric pressure has a viscosity of only µ = 1.72 · 10−5 Nsm−2, whereas
water at 0◦C has a viscosity of µ = 1.76 · 10−3 Nsm−2. This means water
requires a greater shear stress applied than air for a given deformation rate.
We know this to be true from what we experience in our daily life.
2.7 Classification of Fluid Motion→ Fox et al. [1]: 2-6
In the next chapter, we will try to solve problems of fluid-mechanics math-
ematically. However the equations that accurately describe fluid-motion are
generally too complicated to solve directly, even for a computer. But in some
special cases, terms in the equations can be neglected to an extent where even
solutions using pen and paper become viable. To identify situations where we
can neglect certain terms, it is useful to classify the different types of fluid
motion.
2.7.1 Laminar vs. Turbulent
In the late 19th century, Osborne Reynolds discovered an important change in
the way a fluid behaves depending on how fast it flows, its viscosity, its density,
and on the size of the flow-domain (e.g. the diameter of a pipe). For low
velocities, all the fluid-particles will move parallel along a highly predictable,
2.7. CLASSIFICATION OF FLUID MOTION 13
smooth path – very often even along a straight line. However, if the velocity or
density are sufficiently high (or if the viscosity is sufficiently low), an instability
occurs which deflects the fluid-particles from their simple straight paths –
inducing a chaotic helical motion or turbulence. Turbulence can be seen in
the shape of clouds ("cauliflower-like"), in the chaotic motion of cigarette
smoke, in large flames and even in a cup when stirring tea and milk. From
observing a turbulent flow it quickly becomes clear that it’s prediction is much
harder than that of a non-turbulent (i.e. laminar) flow.
2.7.2 Incompressible vs. Compressible
A fluid is incompressible if its density does not change with pressure. In real-
ity, all fluids are compressible, but the degree of compressibility varies. When
a change in pressure does not induce a significant change in density, the fluid
can be treated as incompressible (this will introduce only minor errors). The
difference between an incompressible and compressible fluid can be highlighted
by comparing a gas such as air and a liquid such as water. Whereas air can be
compressed with ease, water requires very large pressures for only minor com-
pression. However, even gases can be treated as incompressible if the change
in pressure is so small that the change in volume remains negligible.5
2.7.3 Newtonian vs. non-Newtonian Fluids
A fluid is called "Newtonian" if Newton’s law of viscosity (2.9) applies. This
is the case for gases, water, many oils and most solvents. In other fluids,
viscosity is no longer constant but changes with the shear rate, for example in
tooth-paste, gels and even in blood. Such fluids are called "non-Newtonian"
fluids, as their rate of strain is disproportional to the applied stress.
2.7.4 Single-phase vs. multi-phase
In some situations, multi-phase flows occur, where a liquid phase is combined
with a gaseous or solid phase, for example in the case of a water jet in air or
5We will later see that pressure increases as the velocity of a fluid increases. This
means that a fluid can be treated as incompressible if the velocity remains low – typically
significantly lower than the speed of sound in this fluid.
14 CHAPTER 2. BASIC TREATMENT OF FLUID FLOW
of sand particles suspended in sea-water. Such multi-phase flows are much
harder to understand and can lead to very interesting phenomena.
2.7.5 Sub-sonic vs. super-sonic and trans-sonic
In super-sonic flows, the fluid moves faster than the speed of sound. The fluid
is compressed and expanded significantly and tends to behave very differently
to sub-sonic flows. If the fluid-velocity is close to the speed of sound, one
refers to trans-sonic speeds. Civil aircraft typically cruise at 0.8 times the
speed of sound, close to the transonic region. However, the flow can be
trans-sonic on parts of the airframe.
2.7.6 Reactive vs. non-reactive flows
If a fluid undergoes chemical reaction, one refers to a reacting flow. A typical
example would be a flame, where gaseous fuel reacts chemically with an
oxidiser to form combustion-products.
Chapter 3
Fluid Statics
→ Fox et al. [1]: 3
In section 2.6 we introduced Newton’s law of viscosity, according to which
shear stresses in a fluid are related to the rate of deformation. The shear
stresses are zero if the fluid is at rest – if the fluid is static. This means that
in static fluids, not only we can ignore motion but also the shear stresses.
Furthermore, in static fluids, all the normal stresses σxx, σyy, σzz will be the
same, equaling the pressure p, which means that the mechanics of a static
fluid will be much simpler1. To conclude, in static fluids all six shear stresses
are zero and the three normal stresses equal the pressure: the entire stress
field can be described by the scalar "pressure" only.
Diving, flying or climbing we all have experienced that the air or water pres-
sure drops with increasing altitude and grows with depth. The cause for the
pressure on a submerged surface is the weight of the fluid above it. Let us
consider a cube (with sides of 1m length) supported in water so that the
top-surface is flush with the water surface. If we submerge that cube 1m
under the water level, the pressure force on the top-surface will increase by
9,810N, which is precisely the weight of the thousand liters of water above
the cube. One meter under the water surface, the top-surface of the cube
will experience a pressure that is the sum of the atmospheric pressure and the
1The equality of all three normal stresses can be seen from the following thought-
experiment: Let us assume a static cube of fluid could experience the stresses σxx 6= 0,
σyy = σzz = 0. If we were to consider a cross-section that is not perpendicular to the x
direction, a force-balance would yield non-zero shear-stresses. But according to Newton’s
law, the shear stresses are zero if no deformation occurs. This contradiction shows that
the normal stresses in a static fluid can not be different, they all must be the same.
15
16 CHAPTER 3. FLUID STATICS
Figure 3.1: Submerged cube, experiencing 9,810N pressure force from
shaded region
pressure due to the weight of the water above.
The force from the atmospheric pressure is equal to the weight of the air above
the surface. In this case using an atmospheric pressure value of 1.013 · 105Nm−2 for a 1m2 surface, corresponds to 1.013 · 105 N (corresponding to
approximately ten tons of air above each square meter – at sea level).
3.1 Basic Equation of static fluids→ Fox et al. [1]: 3-1
Let us consider the effect of gravity g using a fluid element of height dh
density ρ and cross-sectional area ∆A. The change in pressure down the
height of the element due to the weight of the fluid will then be dp. This
means that the pressure at the bottom of the fluid-element is dp higher than
at the top. We can compute the change in pressure dp for the co-ordinate z
in the vertical direction from the weight W of the fluid:
∆Adp = ∆W = −∆Aρg dz
As we are only interested in the pressure, we can discard the area ∆A and
obtain the basic equation of static fluids:
dp = −ρgdz (3.1)
3.2. PRESSURE VARIATION IN FLUIDS OF CONSTANT DENSITY 17
Figure 3.2: The change in pressure results from the weight of the fluid.
3.2 Pressure Variation in fluids of constant den-
sity
Figure 3.3: Integration of the change in pressure
→ Fox et al. [1]: 3-3
Equation 3.1 can be solved if we know how ρ and g depend on z. For a fluid
of constant density ρ and for constant gravity g, we can just integrate eq.
3.1:∫
dp =
∫
−ρg dz (3.2)
If we integrate from a reference level z0 to any level z1, and from the reference
pressure p0 at z0 to the pressure p1 at z1, we obtain:
p∫
po
dp =
z∫
z0
−ρg dz
18 CHAPTER 3. FLUID STATICS
Evaluating the integral leads to the basic equation of hydrostatics for constant
density and gravity:
p1 = p0 − ρg(z1 − z0) (3.3)
3.3 Hydrostatic Force on Submerged Surfaces→ Fox et al. [1]: 3-5
In the introduction to fluid statics, we have seen that static fluids only excert
normal stresses which are of the same magnitude in each direction (σxx =
σyy = σzz = p). This means that the force on any static surface element
d ~A = A~n equals the pressure p on the surface times the surface area A, and
that the force acts in the negative direction of the surface normal unity vector
~n:~F = p d ~A = ~np dA (3.4)
The force on a surface A is then the integral of the pressure over the entire
surface A:~F =
∫
A
~np dA (3.5)
For example, we could compute the total force on a dam. Let us assume the
dam’s length l is 100 m, its height h is 10 m, and that the reservoir is filled
with water. On the water-surface and on the dry side of the dam we encounter
atmospheric pressure. The horizontal force Fh on the dam is then:
Fh =
∫
Ap0 − ρg(z − z0) dA (3.6)
The atmospheric pressure acting on the water-surface is cancelled by the
atmospheric pressure acting on the dam, so p0 can be removed from 3.3.
Then, we introduce a co-ordinate system for which z = z0 = 0 is located at
the surface of the water. The integral hence simplifies to∫
A−ρgz dA. As
the dams’ length remains constant with depth, dA = l dz. This leads to a
simplified integral for the horizontal force:
Fh =
∫ 10m
0ρgzl dz (3.7)
(Remember, this equation is only valid for a constant dam-length l, the inte-
gral can easily be evaluated by the reader.)
3.4. BUOYANCY 19
Figure 3.4: A dam of height h = 10 m and length l = 100 m. The pressure
of the water in the reservoir increases linearly with the depth z. Left: Cross-
section through dam. Right: Frontal view of dam
3.4 Buoyancy→ Fox et al. [1]: 3-6
Buoyancy is the net vertical force a fluid exerts on an immersed object. To
compute the buoyancy force on any object floating in a fluid of density ρf ,
we can compute the pressure force acting on the top of the object (at z1,
pressing the object down) and on the bottom of the object (at z2, pressing the
object up). We consider the floating object to be made up of small elements
of a constant cross-section dA and a height h that corresponds to the local
height of the object, as shown in fig. 3.5.
The buoyancy force in positive vertical direction on such an element is then
dFB = ρfg(z2 − z1) dA = ρfgh(x, y) dA (3.8)
where h(x,y) is the height of the element at the coordinates (x,y). We can
integrate this equation over the entire cross-sectional area A of the floating
object to compute the buoyancy force FB on the whole object. Here again,
we assume that density and gravity are constant:
FB =
∫
A
ρfgh(x, y) dA = ρg
∫
A
h(x, y)dA (3.9)
Integrating the local height h(x, y) over the cross-sectional area A yields the
volume V of the floating object, and we obtain the equation for the buyancy
20 CHAPTER 3. FLUID STATICS
Figure 3.5: A three-dimensional object floating under the surface of a fluid.
The resulting force will equal the weight of the displaced fluid.
force in a fluid of density ρf :
FB = ρfgV (3.10)
This equation simply expresses that the buoyancy force on a floating object
is as large as the weight of the displaced fluid. If the buoyancy force equals
an objects weight, it will stay suspended at the same level, otherwise, it will
rise or sink.
Chapter 4
Flow Rates
In the previous chapter, we looked at the properties of a static fluid. In this
chapter we consider fluids in motion and define some of the basic properties
necessary to analyse fluid systems. We begin by defining a flow rate.
Flow rate: The rate of a (fluid) flow over a given area (control surface).
Flow Rates for Constant Velocity
We will often be interested in the amount of fluid flow over an area called the
control surface. For example, we may ask for the volume of air that leaves
the exit-plane of a balloon-noozle every second (m3/s). Or we may want to
determine the rate at which water flows into a bucket (kg/s). To calculate the
heat-loss over a chimney, one might be interested in the flow rate of thermal
energy over the chimney’s exit plane (kJ/s). And as a final example, we could
calculate the thrust of a jet-engine as the flow rate of momentum over the
nozzle exit plane (kg·m/s2 = N).
Consider calculating the volume flow rate of air over the exit-plane of an air-
system. If the air-system channel has a cross-section of 1 m×2 m, and the
mean flow-speed is 2 m/s, then we know that a volume of V =1 m ·2 m ·2m/s ·1s = 4 m3 will leave the channel every second, as illustrated in fig. 4.1.
The volume flow rate of air is then Fv =4 m3/s. In general, the volume flow
21
22 CHAPTER 4. FLOW RATES
rate Fv over an area A at a constant flow-velocity u normal to the area is
simply given as:
Fv = Au (4.1)
Figure 4.1: Flow rate through the outlet of a channel. The volume indicated
by the dotted line will leave the channel within one second.
For the previous example shown in fig. 4.1, we could also calculate the flow
rate of mass (Fm ) from the channel. The mass M of the volume V of air that
leaves the channel per second can be calculated with the density ρair = 1.293
kg/m3 as M = ρairV = 5.172 kg. Hence, the mass flow rate is Fm = 5.172
kg/s. Generally, the mass flow rate Fm over a surface A can be calculated
for a constant density ρ and constant velocity u normal to the surface of area
A according to:
Fm = ρAu = ρFv (4.2)
Finally, let us consider FP , the flow rate of momentum1 P from the channel.
The volume indicated in fig. 4.1 comprises fluid of the mass M = ρairV
moving with the velocity u, so its momentum is P = Mu = 10.344 kg·m/s.
This momentum will cross the outflow plane every second, so that the mo-
mentum flow rate from the channel is FP =10.344 kg·m/s2 =10.344 N. (If
1Momentum and momentum flow rate are vectors, just like forces. For now, we only
consider the component normal to the control surface, but the next section will consider
the complete vectors.
23
the fluid leaving the channel stems from a large pressure vessel, then the ’sys-
tem’ would gain a momentum of -10.344 kg·m/s every second. This means
that the nozzle would create a thrust T = FP =10.344 N.)
In general, the momentum flow rate FP normal to a surface A for constant
density ρ and velocity u is given as:
FP = ρAu2 = ρuFv (4.3)
Flow Rates for Non-Constant Velocity
In general, the velocity will not be constant over the entire control surface, and
in some cases (for example in flames), the density will not even be constant
over the control surface. We will now generalise the equations for volume flow
rate, mass flow rate and momentum flow rate. (This will involve integration
over areas, a mathematical technique you are expected to apply in the end-
of-year exams.)
To calculate the volume flow rate Fv accross a control surface CS, we can
just integrate the velocity component u normal to the surface over the surface
area A, using surface elements dA:
Fv =
∫
CS
u dA (4.4)
The mass flow rate Fm and the momentum flow rate FP are given accord-
ingly:
Fm =
∫
CS
ρu dA (4.5)
FP =
∫
CS
ρu2 dA (4.6)
In many cases, the flow-velocity will not be normal to the control surface, and
we have to calculate the normal component u first. With the surface normal
24 CHAPTER 4. FLOW RATES
vector ~n and the velocity vector ~V , the surface normal velocity component u
is the dot product of ~V and ~n:
u = ~V · ~n (4.7)
With this equation, we get the following expressions for the volume and mass
flow rates:
Fv =
∫
CS
~V · ~ndA (4.8)
Fm =
∫
CS
ρ~V · ~n dA (4.9)
For momentum flow rate, the situation is more complicated, as momentum
is a vector itself, and the flow rate is also a vector. However, the flow rate
results from the velocity component normal to the surface u = ~V · ~n and
the specific momentum ρ~V of the fluid. Skipping a detailed derivation, the
momentum flow rate is given as:
~Fm =
∫
CS
ρ~V (~V · −→n ) dA (4.10)
Chapter 5
Basic Laws for Systems and
Control-Volumes
In chapters 3 and 4 we looked at defining some fundamental properties of
both static and moving fluids. In this section we use these definitions to
define key properties and laws of fluid systems. → Fox et al. [1]: 4
We introduced systems in section 2.4 as an entity that consists of a fixed
amount of mass. A system is an object, it can deform, absorb heat, be
accelerated, but it will always consist of the original material (or of the original
molecules). This implies that the conservation principles for the following
quantities apply:
• Conservation of mass
• Conservation of momentum (Newton’s second law)
• Conservation of angular momentum
• Conservation of energy (1st Law of Thermodynamics)
• Growth of entropy (2nd Law of Thermodynamics)
The following section will provide the mathematical formulation for these prin-
ciples and show a generalised formulation. For fluid mechanics, the conserva-
tion of mass and momentum are most relevant, and students are expected to
know these laws.
25
26CHAPTER 5. BASIC LAWS FOR SYSTEMS AND CONTROL-VOLUMES
5.1 Specific conservation principles for systems→ Fox et al. [1]: 4-1
Conservation of mass means that a systems’ mass Msystem cannot change –
which is satisfied from the definition of a system. We can express this law by
requiring that the time-derivative of the systems’ mass is zero:
dMsystem
dt= 0 (5.1)
The mass of a system can be computed by integrating over the mass-elements
dm that make up the system mass M(system). This integration is equivalent
to integrating the density over the volume elements dV that make up the
system volume V (system):
Msystem =
∫
M(system)
dm =
∫
V (system)
ρ dV (5.2)
The momentum ~P of a system is also conserved, as the momentum of in-
dividual particles making up the system is also conserved. To change the
momentum of a system, an external force ~F is required. To compute the
momentum of the system, we can integrate over the mass-elements dm or
the volume elements dV again:
~F =d~Psystem
dtwith: (5.3)
~Psystem =
∫
M(system)
~V dm =
∫
V (system)
~V ρ dV (5.4)
Similarly to linear momentum ~P , angular momentum ~H is also conserved 1,
and angular momentum can only change if a torque ~T is applied. The torque
can stem from an individual force ~F , from gravity ~g, or from a shaft.
~T =d ~Hsystem
dtwith: (5.5)
~Hsystem =
∫
M(system)
~r × ~V dm =
∫
V (system)
~r × ~V ρ dV (5.6)
and: ~T = ~r × ~F +
∫
M(system)
~r × ~g dm+ ~Tshaft (5.7)
1Angular momentum is a vector of which the direction denotes the axis of rotation,
whereas the magnitude denotes the speed of rotation
5.2. GENERAL CONSERVATION PRINCIPLE FOR SYSTEMS 27
The first law of thermodynamics provides a conservation principle for energy,
which we can formulate for a system. The energy of a system will only change
if heat Q is added or if the system performs mechanical work W :
Q− W =dEsystem
dtwith: (5.8)
~Esystem =
∫
M(system)
e dm =
∫
V (system)
eρ dV (5.9)
And finally, we consider the second law of thermodynamics, according to
which a change in entropy2 S will be at least as large as the ratio of heat flow
Q over the temperature T of an object:
Q
T≤ dSsystem
dtwith: (5.10)
Ssystem =
∫
M(system)
s dm =
∫
V (system)
sρ dV (5.11)
5.2 General conservation principle for systems→ Fox et al. [1]: 4-1
We have discussed the conservation principles for the most relevant quantities
in thermofluids. In general, a conserved property of a system stays constant
unless external forcing triggers a change. This forcing can be due to forces,
moments, heat flow rate, work, etc.
In the previous section, we have also seen absolute values that describe an
entire system globally as opposed to other quantities that describe a prop-
erty locally, usually a field. We call the global quantities extensive, as they
describe a system as seen from outside, examples being a systems mass M ,
it’s momentum ~P , angular momentum ~H, energy E or entropy S. In turn
intensive quantities desribe the local state inside the system, as given by the
velocity field ~V , the angular velocity field ~r × ~V , energy e or entropy s. The
relation of an intensive quantity η and an extensive quantity N is given by
2It does not really matter whether you understand what entropy really is. But in the
end, it is a quantity that is useful for calculations of thermodynamic processes, and so we
mention it here.
28CHAPTER 5. BASIC LAWS FOR SYSTEMS AND CONTROL-VOLUMES
the following equation:
Nsystem =
∫
M(system)
η dm =
∫
V (system)
ηρ dV (5.12)
The following table shows extensive properties and the related intensive prop-
erties, and confirms again that intensive properties relate an extensive quan-
tity to the unit mass of 1 kg. An intensive porperty describes the state of a
point in a field, whereas an extensive property describes the state of an entire
system.
Table 5.1: Extensive properties and related intensive properties
mass momentum angular energy entropy
momentum
Extensive kg kg m/s kg m2/s J J/K
Property M ~P ~H E S
Intensive 1 ~V ~r × ~V e s
Property - m/s m2/s J/kg J/K/kg
5.3 Relation between Systems and Control Vol-
umes→ Fox et al. [1]: 4-2
A control volume (CV) is a fixed volume in physical space through which mass
can flow. The CV shape, volume and position cannot change, but its content
will due to flow rates over its surface. In the previous section, we determined
the change in a systems’ extensive properties N with time t and have related
it to the intensive properties η according to:
dNsystem
dt=
d
dt
∫
V (system)
ηρ dV (5.13)
To relate the change in a system to the change in a control volume, we need
to consider the distinction between systems and CVs: a system is moving, a
CV is stationary. This means that a system can move through a CV, leading
to flow rates over the system surface.
5.3. RELATION BETWEEN SYSTEMS AND CONTROL VOLUMES 29
We consider a CV that (at one instant) is collocated with the moving system
and provide the resulting equation:3
dNsystem
dt=
∂
∂t
∫
CV
ηρ dV
︸ ︷︷ ︸
accumulation
+
∫
CS
ηρ ~V ·~n dA
︸ ︷︷ ︸
flux
(5.14)
This equation relates the change in a system to the change in a control
volume. It is arranged such that the terms for the system are on the left
hand side, the terms for the CV are on the right. The accumulation term
describes the change of the CV content, whereas the flow rate term describes
the transport over the control volume surface CS.
Equation (5.14) is a very general equation for the change of an extensive
quantity N for a control volume CV. The equation is of great importance and
is often called "Reynolds Transport Theorem". Unfortunately, eq. (5.14) is
relatively hard to understand, but we will apply it to different conservation
principles throughout the following sections to make things clear. So...
Figure 5.1: ...some general advice on fluid-mechanics... [2]
3We do not derive eq. (5.14). Instead, we refer the reader to the textbook by Fox,
McDonald and Pritchard [1] section 4-2.
30CHAPTER 5. BASIC LAWS FOR SYSTEMS AND CONTROL-VOLUMES
5.4 Conservation of Mass for a Control Volume
In the previous section, 5.3, we said that the Reynolds Transport Theorem eq.
(5.14) can be applied to any extensive quantity. We now apply it to the mass
M , and according to table 5.1, the corresponding intensive quantity is 1. We
know that the mass of the system cannot change, so that dMsystem/(dt) = 0
(see eq. 5.1), and we obtain the following equation:
∂
∂t
∫
CV
ρ dV +
∫
CS
ρ ~V ·~n dA = 0 (5.15)
Equation 5.15 describes the conservation of mass formulated for a control
volume.
This equation is one of the most important equations of fluid mechanics, and
we have easily obtained it from Reynolds Transport Theorem. But is there
another way to obtain this equation? A way that is easier to understand?
The next few paragraphs try to outline a simpler derivation.
From daily life, we know that mass cannot be created or destroyed: neutrons,
protons, and electrons cannot just be destroyed or created 4. This is expressed
in eq. (5.1), according to which the mass of a system cannot change in time.
However, mass can be moved, and it can flow over surfaces into control
volumes and out off control volumes. According to eq. (5.2), the mass in a
control volume CV can be computed as
MCV =
∫
V
ρ dV (5.16)
so that the change of this mass in time results as:
∂MCV
∂t=
∂
∂t
∫
V
ρ dV (5.17)
Mass leaving a control volume over its surface will reduce the mass content
of the CV: the reduction of the mass inside the CV must equal the mass flow
rate M out of the CV, and we obtain:
−∂MCV
∂t= − ∂
∂t
∫
V
ρ dV = M (5.18)
4This is no longer true for high-energy physics considering velocities close to the speed
of light – but such cases are rarely relevant for engineering fluid-mechanics.
5.4. CONSERVATION OF MASS FOR A CONTROL VOLUME 31
The mass flow rate over the control volume surface CS is given as:
M =
∫
CS
ρ ~V ·~n dA (5.19)
From equations (5.18, 5.19) we obtain the conservation of mass in control
volume formulation (5.15) again:
∂
∂t
∫
CV
ρ dV
︸ ︷︷ ︸
accumulation
+
∫
CS
ρ ~V ·~n dA
︸ ︷︷ ︸
flux
= 0
5.4.1 Conservation of mass for constant density
In section 2.7.2 we introduced a distinction between compressible and incom-
pressible fluids. Compressible fluids can change their density, but for incom-
pressible fluids, the density is constant5. We will now exploit the fact that
the density is constant to simplify the equation (5.15) for the conservation of
mass.
The equation of mass conservation has an accumulation term (with a volume
integral) and a flow rate term (with a surface integral). If the density field is
homogenous6 (i.e. the density is the same everywhere), the mass of the fluid
in the control volume cannot change in time, so that the accumulation term
will be zero.
This means that for incompressible flows, the equation of mass-conservation
can be written as:
ρ = const. ⇒∫
CS
ρ ~V ·~n dA = 0 (5.20)
Equation (5.20) expresses that if mass enters a CV at one point, the same
amount of mass must leave the CV at another point.
5The density of incompressible fluids can actually vary if temperature varies, if chemical
reactions occur (e.g. combustion), or if we consider mixtures of different fluids.6A scalar field is homogeneous if its values are constant in space.
32CHAPTER 5. BASIC LAWS FOR SYSTEMS AND CONTROL-VOLUMES
5.5 Conservation of Energy for a Control Vol-
ume
In the previous sections, we applied the Reynolds Transport Theorem eq.
(5.14) to obtain the conservation of mass eq. (5.15). We will now apply
the Reynolds Transport Theorem to obtain the equation for the conserva-
tion of energy. The conservation of energy is identical to the first law of
Thermodynamics, only now formulated for a control-volume rather than for
a system.
For a system, the first law of thermodynamics has the form
Q− W =dEsystem
dt(5.21)
where E denotes the energy, Q the rate at which heat is added to the system,
and W the rate at which the system performs work on its environment. With
the specific energy e, we can write eq. (5.21) in terms of the content of the
control volume CV and the flow rate over the control surface CS:
Q− W =∂
∂t
∫
CVeρ dV+
∫
CSeρ ~V·d~A (5.22)
If we specify the work-term W in greater detail, and expand the specific energy
e according to e = u+ V 2/2 + gz, we obtain:
Q− Ws − Wshear − Wother = (5.23)
∂
∂t
∫
CV(u+
V 2
2+ gz)ρ dV +
∫
CS
(
u + pυ +V2
2+ gz
)
ρ ~V·d~A (5.24)
In this equation, Ws denotes the shaft power that acts on the control volume’s
surrounding: in a gas turbine, Ws would be positive, in a compressor negative.
The power through shear on the surface of the control volume, and any other
type of work/power, are described as Wshear and Wother. On the right hand
side of the equation, u, p and v denote the specific internal energy, pressure
and specific volume v = 1/ρ, where u+pv = h is the enthalpy. The equation
also features a kinetic energy term with the velocity V and a potential energy
term gz as known from hydrostatics.
5.6. CONSERVATION OF MOMENTUM FOR A CONTROL VOLUME 33
5.6 Conservation of Momentum for a Control Vol-
ume
In the previous section, we applied the Reynolds Transport Theorem eq. (5.14)
to the extensive property mass and energy, yielding the equation of conser-
vation for mass and energy. But we can also apply the Reynolds Transport
Theorem to the extensive quantity momentum ~P and the related intensive
quantity velocity ~V to obtain an equation describing the conservation of mo-
mentum.
~FS + ~FB =∂
∂t
∫
CV
ρ~V dV +
∫
CS
ρ~V ~V ·~n dA (5.25)
This equation describes how the momentum content of a control volume
changes with flow rates and due to surface forces ~FS and volume or body
forces ~FB . In most situations, volume or body forces comprise gravity, but
sometimes also magnetic or electro-static forces. The forces ~Fs acting on
the control surface can be due to shear stresses resulting from deformation
of the fluid, and can be introduced through solid surfaces within the control
volume.
Chapter 6
Specific Laws and Special
Cases
6.1 Bernoulli’s Equation
Bernoulli’s equation is probably the best-known equation of fluid mechanics. It
is very useful in many cases, and at the same time so simple that most people
can apply it. However, this leads to the danger that the equation is used in
many situations where it must not be used, and the following derivations try
to point out where it can be used, and when it must not be used.
6.1.1 Bernoulli’s Equation Derived from the Energy Equa-
tion
Bernoulli’s equation can be considered as a simplified energy equation for
a stream-tube1. We will derive Bernoulli’s equation from the energy equa-
tion:
Q︸︷︷︸
1: Heat flux
− Ws︸︷︷︸
2: Shaft power
− Wshear︸ ︷︷ ︸
3: Shear power
− Wother︸ ︷︷ ︸
4: Other power
= (6.1)
∂
∂t
∫
CVeρ dV
︸ ︷︷ ︸
5: Change in energy content
+
∫
CS
u︸︷︷︸
6: spec. inner energy
+pv +V 2
2+ gz
ρ ~V ·d ~A
1A stream-tube is an infinitely thin tube of which the axis is defined by a stream line.
35
36 CHAPTER 6. SPECIFIC LAWS AND SPECIAL CASES
If no heat is added to the stream-tube CV, term 1 (heat flow rate) will be
zero. If no drive-shaft performs work in the control-volume, or is driven by the
control volume, term 2 (shaft work) will also be zero. Term 3 (shear power)
will be zero, if the shear forces on the surface of the control volume is zero,
for example if there is no shear or if the viscosity is zero: term 3 (shear power)
can be discarded if there is no friction or if there are no losses. Certainly, if
no other work is performed, term 4 (other power) can also be discarded. For
steady flows, energy cannot accumulate within the control volume, so that
term 5 (change in energy content) will also be zero. Finally, term six can be
neglected if the specific inner energy of the fluid cannot change, which is the
case if terms 1 to 5 are all zero and if the fluid is also incompressible.
We now have a much simplified energy equation of the form:
∫
CS
(
pv +V 2
2+ gz
)
ρ ~V ·d ~A = 0 (6.2)
We will now evaluate this integral for a stream-tube. No flow rate occurs over
the shell of the stream tube, as the shell consists of stream lines. This means
that we only have to evaluate this integral at the control surfaces CS1 and
CS2 at the beginning and the end of the stream tube:
∫
CS1
(
pv +V 2
2+ gz
)
ρ ~V ·d ~A =
∫
CS2
(
pv +V 2
2+ gz
)
ρ ~V ·d ~A (6.3)
The diameter of a stream tube is infinitesimally small, so that p, v, V, g and z
can be treated as constants over each of the surfaces CS1 and CS2. According
to the equation of conservation of mass, the mass flow rate m =∫
CS ρ~V ·d ~Aover the inflow surface CS1 is equal to the mass flow rate over CS2. The spe-
cific volume v equals 1/ρ, and we obtain Bernoulli’s famous equation:
p+ ρV 2
2+ ρgz = const. (6.4)
However, when using Bernoulli’s equation, it is most important to remember
the limits of its validity. Bernoulli’s equation can only be applied if:
1. No heat is added to the flow.2. No (shaft work) is added to the flow.3. No shear forces act in the fluid (i.e. no friction or no shear).4. Energy is not added to the CV in any other form.5. The flow is steady.
6.2. BOUNDARY LAYERS 37
6. The flow is incompressible.
7. The reference points are located on the same streamline.
6.2 Boundary Layers
A fluid flowing around an obstacle is typically affected in two ways:
(1) The fluid is deflected around the object, as the fluid cannot flow through
the surface of the object. This deflection typically alters the pressure field,
and the change in velocity along a (little curved) streamline can typically be
calculated from Bernoulli’s equation.
(2) The fluid is also decelerated by friction. At the surface of the object, the
fluid has zero tangential velocity (the fluid sticks to the object, this is often
referred to as the no-slip condition). This friction only decelerates fluid that
is close to the surface, the effect of friction is spatially limited.
Let us consider the example of a (infinitesimally thin) plate in a laminar flow
as shown in fig. 6.1. The plate is so thin that the fluid is not deflected,
however, friction will still slow it down. At the leading edge of the plate, fluid
particles arrive with the constant free stream velocity U0. However because
of the no-slip condition the velocity of the fluid particles at the plate will be
zero. This leads to very high velocity gradients ∂u/∂z at the plate surface,
which in turn leads to large shear stresses τzx. These shear stresses decelerate
the fluid particles next to the plate. As fluid particles next to the plate slow
down, velocity gradients ∂u/∂z develop between these fluid particles and their
neighbouring fluid particles further from the plate surface still travelling at U0.
The resulting stresses in turn slows down fluid further from the plate, that
has not yet decelerated. This process continues, propagating away from the
plate as the fluid travels along the plate. Eventually, a region develops close
to the plate where the fluid has been decelerated by friction, this region is
called the boundary layer. Outside the boundary layer, the effect of friction
can be neglected and Bernoulli’s equation can usually be applied.
38 CHAPTER 6. SPECIFIC LAWS AND SPECIAL CASES
x
u(x,r) u(x,r)
u(x,r) u(x,r)
z
Uo
Figure 6.1: Development of the boundary layer along a thin plate.
6.3 Laminar Pipe-Flow
In section 6.2, we have seen how the thickness of a boundary layer grows
along a flat plate. Such a growing boundary layer also exists at the inner
surface of a pipe, as illustrated in fig. 6.2. Near the inlet, the flow inside
the pipe will hardly be affected by friction, whereas further downstream, the
boundary layer has grown so that even the fluid on the centre-line is affected.
Figure 6.2 illustrates this development of axial velocities inside a pipe. Far
downstream of the inflow, the velocity profile no longer changes, and we call
such a flow fully developed.
Figure 6.2: Flow at the inlet section of a pipe. The boundary layers grow
until they merge, eventually creating a fully developed pipe-flow.
For a laminar flow, it is relatively easy to calculate the velocity profile u(r)
from a momentum balance. We consider a cylindrical control volume CV of
length dx and radius r that is located inside the pipe of inner radius R, as
illustrated in fig. 6.3. The flow is driven by the pressure, and a pressure drop
occurs along the pipe due to the friction.
We start with the conservation of momentum
~F = ~FS + ~FB =∂
∂t
∫
CV
~Uρ dV +
∫
CS
~Uρ~U · d ~A (6.5)
6.3. LAMINAR PIPE-FLOW 39
Figure 6.3: A cylindrical control volume CV inside a section of a pipe.
but, as we are just interested in the steady pipe-flow with no body forces (e.g.
gravitation), we neglect ~FB . The transient term is expressed as:
~FS =
∫
CS
~Uρ~U · d ~A (6.6)
The control surface CS consists of an inflow surface Sin, an outflow surface
Sout and a lateral surface Sl. For a fully developed pipe-flow, there is no flow
in the radial direction, so that the flow rate over the lateral surface is zero and
we only have to consider the velocity component u in axial direction:
FS =
∫
CSuρ~U · d ~A = −
∫
Sin
ρu2 dA+
∫
Sout
ρu2 dA (6.7)
As the u velocity profile is the same over Sin and Sout, the flow rates over
both surfaces cancel, and the surface force FS in equation (6.7) must be
zero.
The surface force FS in equation (6.7) results from (a) the shear stresses τrxon the lateral surface of area 2πr dx and from (b) the pressure forces on the
inflow and outflow surfaces of area πr2:
FS = pinπr2 − poutπr
2 + τrx2πr dx = 0 (6.8)
The pressures pin and pout for a control volume at the position x can be
calculated by the following linearisation
pin = p− ∂p
∂x
dx
2; pout = p+
∂p
∂x
dx
2(6.9)
40 CHAPTER 6. SPECIFIC LAWS AND SPECIAL CASES
that can be inserted into equation 6.8:
FS = −∂p
∂xdxπr2 + τrx2πr dx = 0 (6.10)
⇔ τrx =r
2
∂p
∂x(6.11)
Inserting the shear stress τrx = µ (du)/(dr) into equation (6.12) leads to a
differential equation that can be solved for u(r):
du
dr=
r
2
∂p
∂x(6.12)
⇔ du =r
2µ
∂p
∂xdr (6.13)
⇔ u = C +r2
4µ
∂p
∂x(6.14)
The value of the constant C can be determined from the no-slip condition at
the surface of the pipe, u(R) = 0:
u =1
4µ
∂p
∂x(r2 −R2) (6.15)
Note that in a pipe-flow in positive x-direction (u > 0), the pressure drops in
x direction, i.e. (∂p)/(∂x) < 0.
The total volume flow rate Q in a laminar pipe can be calculated by integration
over the radius r:
Q = −πR4
8µ
(∂p
∂x
)
(6.16)
The volume flow rate depending on the total pressure drop ∆p in a laminar
pipe of length L and diameter D can then be calculated as:
Q =π∆pD4
128µL(6.17)
Chapter 7
Final notes
The reader is reminded that this document is just a draft of notes for the 1M
Fluid Mechanics Course at the Department of Mechanical Engineering, Impe-
rial College 2010-2011. The official course documentation consists of the book
"Introduction to Fluid Mechanics" [1] by Fox, McDonald and Pritchard. The
present notes are meant to help you study and understand fluid-mechanics,
but they cannot provide the depth of information available from the textbook.
Like many drafts, the present document will likely contain errors, please share
them with me – and be reminded that the present document is not the official
course documentation. Even more welcome than corrections are suggestions
and recommendations for the improvement of the document. I very much
look forward to your input!
I wish you all the best for your studies in fluid-mechanics and hope that you
will be successful in the exams!
Andreas Kempf
London, 01/10/2009
Forum for corrections:
http://moodlepilot.imperial.ac.uk/mod/forum/view.php?id=480
41
Bibliography
[1] R. W. Fox, A. T. McDonald, P. J. Pritchard, Introduction to Fluid Me-
chanics. Wiley & Sons 2004, ISBN 0-471-20231-2
[2] N. Gaiman, Don’t Panic: Douglas Adams and the "Hitchhiker’s Guide to
the Galaxy". Titan Books, 1993, ISBN 1-85286-411-7
43
“Hall of Fame”
This appendix provides a quick overview over the most important, most basic
equations. By the end of the course, you should have understood these
equations and be able to use them. I would even recommend that you try
to remember these equations so you need not waste time for looking them
up.
Newton’s law of viscosity:
τyx ∼ µdu
dy(1)
The basic equation of fluid statics:
dp = −ρgdz (2)
The force on a submerged surface:
~F =
∫
A
pd ~A (3)
The conservation of mass for control volumes CV bounded by a control surface
CS:
∂
∂t
∫
CV
ρ dV +
∫
CS
ρ ~V ·d ~A = 0 (4)
The conservation of momentum for control volumes CV bounded by a control
surface CS. A change in momentum is only possible due to surface forces ~FS
and body forces (volume forces) ~FB :
~FS + ~FB =∂
∂t
∫
CV
ρ~V dV +
∫
CS
ρ~V ~V ·d ~A = 0 (5)
45
46 BIBLIOGRAPHY
The conservation of energy for control volumes CV bounded by a control
surface CS. A change in energy is only possible due to the addition of heat
Q, the work being performed on the environment through a shaft Wshaft,
through shear on the control volume surface Wshear and through other forms
of work Wother:
Q− Ws − Wshear − Wother = (6)
∂
∂t
∫
CVeρ dV+
∫
CS
(
u + pv +V2
2+ gz
)
ρ ~V·d~A (7)
Bernoulli’s famous equation relates pressure to velocity to height. The equa-
tion is particularly useful, easy to remember, and easy to use.
p+ ρV 2
2+ ρgz = const. (8)
However, when using Bernoulli’s equation, it is most important to remember
the limits of its validity:
1. No heat is added to the flow.2. No (shaft work) is added to the flow.3. No shear forces act in the fluid (i.e. no friction or no shear).4. Energy is not added to the CV in any other form.5. The flow is steady.6. The flow is incompressible.7. The reference points are located on the same streamline.
Of great importance is also the equation for the velocity profile of a laminar
pipe flow that you should be able to derive when given enough time:
u =1
4µ
∂p
∂x(r2 −R2) (9)
Terminology
Boundary Layer The thin layer of fluid close to an obstacle that is affected
by friction. Within the boundary layer, Bernoulli’s equation does not
hold.
Control Volume A volume in physical space through which a fluid will flow
but cannot change its size, shape or location.
Fluid "A fluid is a substance that deforms continuously under the application
of a shear stress no matter how small the stress may be." [1]
Homogenous A scalar or vector field is homogeneous if the scalar or vector
values do not change with space, i.e. they are the same everywhere.
(The values can still change with time, but synchronously everywhere)
Pathline The path along which a particle would travel in a flow. For steady
flow fields, a pathline is identical to a streakline and to a streamline.
Steady Flow A flow whose variables (i.e. velocity, density, pressure, etc.)
do not change in time.
Streakline The line connecting all the particles that started at the same point
in space at different times. For steady flows, a streakline is identical to
a streamline or pathline.
Streamline At any given time t, the streamline is the line along which a
particle would travel if the flow-field was steady.
System Similar to the Control Volume, but no mass can cross its boundaries.
A system can deform, absorb heat, or be accelarated but it will always
consist of the original material.
Unsteady Flow A flow that has some of its variables (i.e. velocity, density,
47
48 BIBLIOGRAPHY
pressure, etc.) change in time.
Tutorial Questions
#1 Fluids, Continua and Scope of Fluid Mechan-
ics
1.1 What are fluids: Which of the following substances are fluids, and why?
- water, oil, air, methane, tar, surf-wax, glass, honey, grease
1.2 Estimate the mass of air in a car-tire, the cabin of an Airbus A380, an
empty oil-tanker.
Car tyre dimensions vary, taking a rough estimate:
tyre width: 0.25m
outer radius: 0.25m
rim (inner) radius: 0.2m
V = π(0.252 − 0.22)× 0.25 = 0.02m3
ρair = 1.2kgm−3
M = ρair × V = 0.02kg
Fuselage dimensions of Airbus A380 = 3.5m radius × 50m length (assume a
circular cross section)
The cabin usually occupies roughly 2/3 the fuselage, with the rest dedicated
to luggage, electrics, landing gear and fuel.
Vcabin = π(3.5m)2 × 50m× 23 = 1280m3
Mair = V × ρair = 1300m3 × 1.2kgm−3 = 1539kg
49
50 BIBLIOGRAPHY
Note that during take off, flight and landing, the cabin pressurise changes so
the mass will also change.
A Very Large Crude Carrier classification oil tanker can hold up to 320,000
tonnes of crude oil
ρ = 820kgm−3 (depending on composition)
Voil = Vair
Moil
ρoil= Mair
ρair
Mair = 1.2kgm−3 × 320×106kg820kgm−3 = 4.68 × 105kg
1.3 Calculate the mass of air in an empty gas tank (height 50m, diameter
50m) and inside a tutorial room (4m x 8m x 6m).
Answer:
Mass in gas tank: 117,800kg
Mass in tutorial room: 230kg
Working:
ρair = 1.2kgm−3
M = V × ρ
1.4 Calculate the minimum volume of the hydrogen fuel tank in a bus
that can carry 50 kg of hydrogen. How would you store hydrogen fuel for a
bus and why?
Answer: V = 714m3
Working:
ρH = 0.07kgm−3
V = Mρ
Ask your tutor to sign off your work. _______
BIBLIOGRAPHY 51
#2 Velocity Field and Stresses
2.1 Homogeneous fields: Which of the following vector fields are homoge-
neous?
a) ~V = 2i, b) ~V = 2ix, c) ~V = 12 j+ k, d) ~V = 2jx− 3kx, e)
~V = (4xi)x+ 3k − (2x)2 i− 1/3j
Answer: a, c, e
Reason:
Homogeneous fields are independant of the location (x, y, z) .
2.2 Dimensionality: Which of the following vector fields are one-dimensional
(1D), two-dimensional (2D), or three dimensional (3D) - and which ones are
steady or unsteady?~V = 2ix, ~V = 2jx− 3kx+ t, ~V = 2i+ 3xj − 2xk,~V = 1
2xi−2yj−4eztk, ~V = (2+x)(~i−2~j), ~V = (2y+x)(~3i)
Answer:
~V = 2ix, - 1D steady
V = 2jx− 3kx+ t, - 1D unsteady
~V = 2i+ 3xj − 2xk, - 1D steady
~V = 12xi− 2yj − 4eztk, - 3D unsteady
~V = (2 + x)(~i− 2~j), - 1D steady
~V = (2y + x)(~3i) - 2D steady
The dimension depends on how many of the 3 co-ordinates x, y, z the field
relies on. If the field depends on time t, it is unsteady.
2.3 The velocity field between two plates (vertical spacing d = 0.1 mm) is
given as ~V = 1dUszi . The top plate slides in the x direction with velocity Us.
What are the dimensions of this velocity field? How does the fluid behave
on the surface of the top and bottom plates? Does this behavior appear
realistic?
52 BIBLIOGRAPHY
Answer:
The velocity vectors only depend on the z coordinate, the field is one dimen-
sional
at z = 0 (surface of the bottom plate), ~V = 0
at z = d (surface of the top plate), ~V = Us
This is realistic because, we expect a fluid’s velocity to match the velocity of
the respective surface (known as the no slip condition).
2.4 Calculate the (vertical) normal stress in a rope (diameter D = 10mm)
that supports a weight of 1.0 t (metric ton).
Answer: 125MPa
Working:
F = Mg = 1000kg × 9.81ms−2 = 9810N
A = (π ×D2)/4 = 7.85 × 10−5m2
σ = F/A = 125MPa
2.5 Calculate the shear stress between tarmac and the front wheel of a
motor-bike for a contact area A = 50 cm2 and a brake-force F = 1 kN.
Answer: 200kNm−2
Working:
A = 50cm2 = 5× 10−3m2
τ = F/A = 200kNm−2
2.6 Calculate the shear stress between an object and an inclined surface on
which the object rests. (Contact area A=1 m2, weight w=1 MN, inclination
of 45º)
Answer = 0.7MN
Working:
BIBLIOGRAPHY 53
F = 1× sin(45◦) = 0.7MN
τ = F/A = 0.7MNm−2
45°
1MN
Ask your tutor to sign off your work. _______
#3 Fluid Motion, Viscosity
3.1 The velocity field between two plates (vertical spacing d) is given
as ~V = 1dUszi . The top plate slides in x direction with the velocity Us.
Calculate the shear stress at the top-plate as a function of d, Us and the
viscosity µ. Calculate the shear stress at the top-plate for the case d=0.01
mm, Us=1 m/s, µoil =0.1 Ns/m2. Calculate the shear stress at the bottom
plate for the same case.
z
x
Us
d
Answer: 10.0kPa
Working:
~V = 1dUszi → u = 1
dUsz
∂u∂z = 1
dUs
τ = µ∂u∂z = 10.0kN/m2
54 BIBLIOGRAPHY
As ∂u∂z is independent of z for this case, the shear stress is the same at the
top and bottom plate.
3.2 The velocity field near the wall of a pipe can (almost) be approximated
as ~V = U0z1
7 i , where z is the distance from the wall. Calculate the shear
stress at the wall. Is this realistic?
Calculate the shear stress at z = 0.1 m forUs=1 m/s, µ =0.1 Ns/m2.
Answer: τ = 0.1Nm−2
Working:
~V = U0z1
7 → u = U0z1
7
∂u∂z = 1
7U0z−
6
7
τ = µ∂u∂z = 1
7µU0z−
6
7
At the wall, z → 0; τ → ∞ which is certainly not realistic, the assumed
velocity profile can therefore not be correct right next to the wall.
At z = 0.1m, U0 = 1m/s, µ = 0.1Ns/m2:
τ = 0.1Nm−2
3.3 The drive-shaft (d=1 m) of a container ship rotates (60 RPM ) inside
a journal bearing (length l=1 m). The gap between the bearing and the shaft
has a width of w = 1 mm and is filled with oil (µ=0.2 Ns/m2). Calculate
the torque that the bearing excerts on the shaft. (The torque is the moment
about the shaft centre-line.)
BIBLIOGRAPHY 55
Answer: 987Nm
Working:
Shaft spins at 60RPM=1RPS
Circumferential velocity of shaft= 2πr/∆t = πms−1
dudr = πms−1
0.001m
shear stress τ = µdudr = 0.2kgsm−2 × πms−1
0.001m = 200πNm−2
A = 2πr × w = πd×w = πm2
F = τ ×A = 200π2N
T = r × F = 100π2 = 987Nm
Ask your tutor to sign off your work. _______
56 BIBLIOGRAPHY
#4 Hydro Statics
4.1 The Airbus A380 has a fuselage diameter of D=7.14 m. At cruise al-
titude, the cabin pressure is approximately 85 KPa, while the outside pressure
is 15 KPa. Assume that the pressure cabin is a cylindrical pressure vessel of a
length of 60 m, with a constant wall-thickness of 4 mm. Calculate the mass
of air inside the cabin during cruise. Calculate the biggest normal stress in the
wall (i.e. σφφ, where φ is the coordinate in circumferential direction).
Answer: 2426.4kg, 62.5MNm−2
Working:
Cabin temperature = 293◦K, Universal gas constant for air: 287.05Jkg−1K−1
ρc =PRT = 85000Pa
287.05Jkg−1K−1×293K
= 1.01kgm−3
Vc =πD2l4 = 2402.36m3
m = ρc × Vc = 2426.4kg
Normal stress:
Calculate Force acting normal to cut plane shown above:
F = ∆P ×A = (85kPa− 15kPa) × 7.14m× 60m = 30MN
Area subjected to stress:
A = 2× l × t = 2× 60m× 4× 10−3m = 0.48m2
BIBLIOGRAPHY 57
σ=FA = 62.5MNm−2
4.2 A floatation device (cube, 0.2m x 0.2 m x 0.2 m) of styro-foam (density
100 kg/m3) is held partially submerged in a swimming pool. Calculate the
hydro-static force on the top and on the bottom surfaces of the cube, if the
cube’s bottom is [0.1 m, 0.2 m, 1.0 m] beneath the water surface. Calculate
the static lift experienced by the cube. – A large quantity of oil is added to the
water, forming an oil-layer (thickness: 1 m, density: 800 kg/m3). Calculate
the static lift force that the floatation device can supply if it’s top is 1.0 m
below the fluid surface.
Answer:
Submerged cube:
d = 0.1m; Ftop = 0N, Fbottom = 39.24N
d = 0.2m; Ftop = 0N, Fbottom = 78.48N
d = 1.0m; Ftop = 313.92N, Fbottom = 392.40N
Flift = 70.63N
Working:
Excluding the atmospheric pressure in the followin work, the hydrostatic pres-
sure experienced at a point is given by:
P = ρgh
So, for the bottom surfaces, at the three depths given, the hydrostatic force
of water is:
F0.1 = P0.1×A = ρghA = 1000kgm−3×9.81ms−2×0.1m×0.04m2
F0.1 = 39.24N
F0.2 = P0.2×A = ρghA = 1000kgm−3×9.81ms−2×0.2m×0.04m2
F0.2 = 78.48N
F1.0 = P1.0×A = ρghA = 1000kgm−3×9.81ms−2×1.0m×0.04m2
F1.0 = 392.40N
58 BIBLIOGRAPHY
the hydrostatic force on the top surface is 0N unless the top surface is sub-
merged, this occurs only at d = 1.0m:
F1.0 = ρghA = 1000kgm−3 × 9.81ms−2 × (1.0 − 0.2)m × 0.04m2
F1.0 = 313.92N
The static lift force will be the resultant force acting on the cube:
W = ρc × V × g = 100kgm−3 × 0.23m3 × 9.81ms−2 = 7.85N
Ftop = ρoilgh = 800kgm−3 ×9.81ms−2×1.0m×0.22m2 = 313.92N
Fbottom = Ftop + ρwghwA
Fbottom = 313.92N+1000kgm−3×9.81ms−2×0.2m×0.04m2 = 392.4N
∴ Flift = 392.4N − 313.92N − 7.85N = 70.63N
4.3 The tube shown is filled with mercury at 20◦C. Calculate the force
applied to the piston. (question based on FMP 3.5)
Answer:
F = 47.9N (book case)
F = 377kN (given case)
Working:
Without the piston weight, the mercury level would be the same on both
sides of the tube. The piston weight alone displaces the mercury level by 1
inch. Application of a further force increases the the diplacement by 7 inches.
BIBLIOGRAPHY 59
We can use the hydrostatic pressure equation to calculate the displacement
caused by the force:
P = ρg∆h = 13550kgm−3×9.81ms−2×(7in.×0.0254) = 23634.15Nm−2
Multiply by the contact area of the piston:
F = P ×A = 23634.15kgm−3 × 0.25 × π(2in.× 0.0254)2 = 47.9N
Following the same working for the given case:
P = ρg∆h = 13550kgm−3 × 9.81ms−2 × 0.9m = 120kNm−2
F = P ×A = 120kNm−2 × 0.25 × π × (2m)2 = 377kN
Ask your tutor to sign off your work. _______
#5 Hydro Static Forces on Submerged Surfaces
5.1 A door 1m wide and 1.5m high is located in a plane vertical wall of
a water tank. The door is hinged along its upper edge, which is 1m below
the water surface. Atmospheric pressure acts on the outer surface of the door
and at the water surface. (a) Determine the magnitude and line of action of
the total resultant force from all fluids acting on the door. (b) If the water
surface gage pressure is raised to 0.3 atm, what is the resultant force and
where is its line of action? (c) Plot the ratios F/F0 and y′/yc for different
values of the surface pressure ratio Ps/Patm. (F0 is the resultant force when
Ps = Patm). (Question from FMP 3.44)
Answer:
F=25.7kN
line of action =1.86m
with surface presure added:
F=71.3kN
60 BIBLIOGRAPHY
line of action=1.79m
5.2 A triangular access port must be provided in the side of a form contain-
ing curing concrete. Using the coordinates and dimensions shown, determine
the resultant force that acts on the port and it’s point of application. Do the
same for b = [0.1 m, 0.2 m, 0.4 m, 0.5 m]. (Question based on FMP 3.45).
Answer:
Force = 376N
line of action = 0.3m
For the various values of b:
F0.1 = 125.6N
F0.2 = 251.1N
F0.3 = 376.7N
F0.4 = 502.3N
F0.5 = 627.8N
Working:
F =∫
A P dA =∫ h0 ρgzw(z) dz
BIBLIOGRAPHY 61
The width of the triangle is dependent on the position (z) from the top of
the access ports, using simple trigonometry the width at z is given by:
w = 2z × tan(20.556◦) = 0.75z
F =∫ h0 2400 × 9.8 × 0.75 × z2 dz = 5880z3
∣∣∣
0.4
0= 376.3N
The moment arm of this force can then be found by:
M = F × d
The resultant moment of the hydrostatic pressure from the concrete can be
calculated using:
M =∫
A Pz dA =∫ h0 ρgz2w(z) dz
M =∫ h0 2400 × 9.8 × 0.75 × z3 dz = 4410z4
∣∣∣
0.4
0
0.4 = 112.9Nm
Hence the moment arm is given by:
d = M/F = 113.32Nm/376.90N = 0.3m
For the different width of access port tanθ changes accordingly:
F =∫
A P dA =∫ h0 ρgzw(z) dz
and as tanθ = b2α ∴ w = z b
α
F =∫ h0 ρg b
αz2 dz = 1
3ρgbαz
3∣∣∣
α
0= 1255.7b
The various moments can be calculated in a similar way:
M =∫ h0 ρgz2w(z) dz =
∫ h0 ρg b
hz3 dz = 0.25ρgh3b = 376.3b
Hence the moment arm for each is:
d = MF = 376.3b
1254.4b = 0.3m
and remains constant.
so the force for various values of b:
F0.1 = 125.6N
F0.2 = 251.1N
F0.3 = 376.7N
62 BIBLIOGRAPHY
F0.4 = 502.3N
F0.5 = 627.8N
Ask your tutor to sign off your work. _______
#6 Revisions
Use the time to understand tutorial questions you struggled with before. Try
to solve the recommended questions in FMP.
Ask your tutor to sign off your work. _______
#7 Flow Rates
7.1 In the days before the credit crunch, a bored, drunk hedge-fund
manager throws £20 notes from a roof-top bar, on average one note every
3.21 s. Calculate the average cash-flow rate. Calculate the average mass-flow
rate if a £20 note has a weight of 0.9 g.
Answer:
Cash flow rate : 6.23£/s
mass flow rate : 0.28g/s
Working:
cash flow rate : unit cash × rate of dispense: 20× 13.21 = 6.23£s−1
mass flow rate : unit mass × rate of dispense: 0.9× 13.21 = 0.28gs−1
7.2 The circular opening of a wind-sock has a diameter of 1m. Calculate
the volume flow rate, mass flow rate and momentum flow rate of air through
the windsock if the wind-speed is 10 m/s. (Assume that the air is not slowed
down inside the wind-sock)
Answer:
Volume flow rate: 7.85m3s−1
BIBLIOGRAPHY 63
mass flow rate: 9.62kgs−1
momentum flow rate: 96.2N
Working:
A = πD2/4 = 0.785m2
Volume flow rate: u×A = 7.85m3s−1
mass flow rate: ρuA = 9.62kgs−1
momentum flow rate: ρu2A = 96.2N
7.3 The fixture of the wind-sock from question 7.2 gets jammed, so that
the wind blows into the wind-sock at an angle of 45º. Calculate the air
volume flow rate, mass flow rate and momentum flow rate normal to the
surface.
Answer:
volume flow rate: 5.6m3s−1
mass flow rate: 6.8kgs−1
momentum flow rate: 48.1N
Working:
A = πD2/4 = 0.785m2
consider only the velocity normal to the wind sock:
u = 10cos(45◦)ms−1
Volume flow rate: u×A = 5.6m3s−1
mass flow rate: ρuA = 6.8kgs−1
momentum flow rate: ρu2A = 48.1N
7.4 On a rainy day, every cubic meter of air contains 2 g of rain water.
Calculate the water volume flow rate, mass flow rate and momentum flow
rate through the wind-sock assuming that the droplets are suspended in the
air.
64 BIBLIOGRAPHY
Answer:
volume flow rate: 1.57 × 10−5m3s−1
mass flow rate: 1.57 × 10−2kgs−1
momentum flow rate: 0.157N
Working:
volume of water per cubic metre of air: mρ = 2× 10−6m3
Volume flow rate: u×A× Vratio = 1.57 × 10−5m3s−1
mass flow rate: ρwuAVratio = 1.57 × 10−2kgs−1
momentum flow rate: ρwu2AVratio = 0.157N
Ask your tutor to sign off your work. _______
#8 Systems and Control Volumes
8.1 Wind is blowing through a football goal (height 2.44 m, width 7.32 m)
in normal direction. At the ground, the air is at rest, but with growing distance
z from the ground, the wind-speed U increases according to U =√zc.
Calculate the volume and mass-flow rate of air through the goal. Calculate
the mass-flow rate if c = 5√m/s.
Answer:
Volume flow rate: 93m3s−1
mass flow rate: 114kgs−1
Working:
Volume flow rate:∫
A u dA =∫ 2.440 uw dz
=∫ 2.440 cwz1/2 dz = 2
3cwz3/2
∣∣∣
2.44
0= 93m3s−1
BIBLIOGRAPHY 65
mass flow rate:∫
A ρu dA = 114kgs−1
8.2 The wind-direction changes with altitude. Calculate the mass-flow
rate through the goal of question 8.1 if the wind-field is given as ~V =
az cos(z)i + bz sin(z)j where i is the direction normal to the goal and j
is parallel to the crossbar (a = 2 m/s, b = 0.2 m/s).
Hint:∫x cos(x)dx = cos(x)+x sin(x),
∫x sin(x)dx = sin(x)−x cos(x).
Answer: −3.32kgs−1
Working:
we are concerned only with the velocity normal to the goal posts u = az cos (z):
mass flow rate:∫
A ρu dA =∫ 2.440 ρuw dz =
∫ 2.440 ρawzcos(z) dz
mass flow rate: ρaw [cos(z) + zsin(z)]2.440 = −3.32kgs−1
8.3 A wind turbine with rotor diameter D (50 m) converts kinetic energy
of the air into mechanical energy (torque on the shaft) and eventually into
electricity. Calculate the power output of a hypothetical wind-turbine that
converts half of the kinetic energy of the air flowing through the rotor disk.
The air flows at U = 20 m/s.
Answer: 0.24MW
Working:
A = 0.πD2/4 = 1963.5m2
Power: kinetic enery transferred per second = 12ρU
2 × U ×A× η
where η is the efficiency.
P = 4.8MW
8.4 The velocity distribution for laminar flow in a long circular tube of
radius R is given by the one-dimenional expression,
~V = ui = umax
[
1−(rR
)2]
i
66 BIBLIOGRAPHY
For this profile obtain expressions for the volume flow rate and the momentum
flow rate through a section normal to the pipe axis. (Question from FMP 4.14)
Answer:
Volume flow rate: 12πumaxR
2
Momentum flow rate: 13R
2πρu2max
Working:
Volume flow :∫
A u dA =∫ R0 u2πr dr
=∫ R
2πumax
[
r − r3
R2
]
dr = πumax
[
r2 − r4
2R2
]∣∣∣
R
0
= πumax
[
R2 − R4
2R2
]
= πumax
[R2 − 1
2R2]= 1
2πumaxR2
mass flow rate=∫
A ρu2 dA =∫ R0 ρu22πr dr
=∫ R
2πρu2maxr[
1− r2
R2
]2dr =
∫ R2πρu2max
[
r − 2 r3
R2 + r5
R4
]
dr
= 2πρu2max
[12r
2 − r4
2R2 + r6
6R4
]∣∣∣
R
0
= ρu2max2π[12R
2 − 12R
2 + 16R
2]= 1
3R2πρu2max
Ask your tutor to sign off your work. _______
#9 Conservation of Mass
9.1 Consider steady, incompressible flow through the device shown. De-
termine the magnitude and direction of the volume flow through the de-
vice shown. Determine the magnitude and direction of the volume flow rate
through port 3. (Question from FMP 4.17)
BIBLIOGRAPHY 67
Answer:−→V3A3 = |V1A1| − |V2A2|, V3 = 25m/s flowing into the vol-
ume
Working:
sum all flow rates:
|V1A1|+ |V2A2|+ |V3A3| = 0
V1 flows into the control volume, therefore it has a negative sign. So:
−→V3A3 = |V1A1| − |V2A2| = −5m3/s
V3 = −25m/s
9.2 Fluid with 1050Kg/m3 density is flowing steadily through the rectan-
gular box shown.
Given A1 = 0.05m2, A2 = 0.01m2,A3 = 0.06m2, ~V1 = 4im/s, and ~V2 =
−8jm/s, determine velocity ~V3. (Question based on FMP 4.18)
68 BIBLIOGRAPHY
Answer: 4.04i − 2.33jms−1
Working:
Sum all flows into control volume:
|V1A1|+ |V2A2|+−→V3A3 = 0
V1 and V2 flow into the flow rate, so have negative signs:
−→V3A3 = |V1A1|+ |V2A2|
∴ V3 = 4.67ms−1
−→V3 = V3Sin(60)i + V3Cos(60)j
−→V3 = 4.04i − 2.33j ms−1
9.3 The velocity profile for laminar flow in an annulus is given by
u(r) = − ∆p4µL
[
R2o − r2 +
R2o−R2
i
ln(Ri/Ro)lnRo
r
]
where ∆p/L = −10kPa/m is the pressure gradient, µ is the viscosity (SAE
10 oil at 20◦C), and Ro = 5mm and Ri = 1mm are the outer and inner radii.
Find the volume flow rate, the average velocity, and the maximum velocity.
Plot the velocity distribution. (Question from FMP 4.24)
Answer:
Q = 10.45 × 10−6m3s−1
BIBLIOGRAPHY 69
U = 0.139ms−1
Umax = 0.213ms−1
Working:
Q =∫ Ro
Ri2πru(r) dr
Q = −2π∆p4µL
∫ Ro
RiR2
or − r3 +(
R2o−R2
i
ln(Ri/Ro)
)
rln(Ro
r
)dr
Q = −2π∆p4µL
[12R
2or
2 − 14r
4]∣∣∣
Ro
Ri
− 2π∆p4µL
(R2
o−R2
i
ln(Ri/Ro)
) ∫ Ro
Rirln
(Ro
r
)dr
Q = −2π∆p4µL
(14R
4o − 1
2R2oR
2i +
14R
4i
)− 2π∆p
4µL
(R2
o−R2
i
ln(Ri/Ro)
) ∫ Ro
Rirln
(Ro
r
)dr
Evaluate∫ Ro
Rirln
(Ro
r
)dr by parts:
u = ln(Ro
r
), dv = r dr
dudr = du
da × dadr , where a = Ro
r
dudr = 1
a ×−Ro
r2 = rRo
×−Ro
r2 = −1r
du = −1r dr
v = 12r
2
∴
∫ Ro
Rirln
(Ro
r
)dr = uv −
∫v du
∫ Ro
Rirln
(Ro
r
)dr = 1
2r2ln
(Ro
r
)∣∣Ro
Ri+
∫ Ro
Ri
12r dr
∫ Ro
Rirln
(Ro
r
)dr =
(12R
2oln(1) +
14R
2o
)−
(12R
2i ln
(Ro
Ri
)
+ 14R
2i
)
Q = −2π∆p4µL
(14R
4o − 1
2R2oR
2i +
14R
4i
)
−2π∆p4µL
(R2
o−R2
i
ln(Ri/Ro)
) [14R
2o −
(12R
2i ln
(Ro
Ri
)
+ 14R
2i
)]
µ = 0.1Nms−1
∴ Q = 10.45 × 10−6m3s−1
U = QA = Q
π(R2o−R2
i )= −0.139ms−1
70 BIBLIOGRAPHY
Find position of max velocity:
u(r) = − ∆p4µL
[
R2o − r2 +
R2o−R2
i
ln(Ri/Ro)lnRo
r
]
∂u(r)∂r = − ∆p
4µL
[
−2r − R2o−R2
i
ln(Ri/Ro)1r
]
= 0
−2r2 =R2
o−R2
i
ln(Ri/Ro)
r = 2.73 × 10−3m
Umax = U(0.00273) = 0.00273R2o−0.002732+
(R2
0−R2
i
ln(Ro/Ri)
)
ln(
Ro
0.00273
)
Umax = 0.213ms−1
Ask your tutor to sign off your work. _______
#10,11 Conservation of Energy
X.1 Air at standard conditions enters a compressor at 75m/s and leaves at
an absolute pressure and temperature of 200kPa and 345K, respectively, and
speed V = 125m/s. The flow rate is 1kg/s. The cooling water circulating
around the compressor casing removes 18kJ/kg of air. Determine the power
required by the compressor.(Question from FMP 4.183)
Answer: Win = 80.23kW
Working:
Starting from the first law of thermodynamics, noting that enthalpy h =
u+ pv:
Q−WS−WShear−Wother =∂∂t
∫
CV eρ dV+∫
CS
[
h+ V 2
2 + gz]
ρ−→V d
−→A
Q− WS =(
h+ V 2
2
)
m∣∣∣2−
(
h+ V 2
2
)
m∣∣∣1
where, h = CPT, CP = 1004Jkg−1K−1
enter the properties given:
V1 = 75ms−1, V2 = 125ms−1
m1 = m2 = m = 1kgs−1
BIBLIOGRAPHY 71
T1 = 288◦K, T2 = 345◦K
Q−WS =(
h+ V 2
2
)
m∣∣∣2−(
h+ V 2
2
)
m∣∣∣1= mCp(T2−T1)+
12(V
22 −V 2
1 ) =
62kW
∴ Ws = −18kJkg−1 · 1kg/s − 62kW = −80kW
negative sign represents rate of work input to the system, hence Win =
80kW
X.2 Compressed air is stored in a pressure bottle with a volume of 10ft3,
at 3000psia and 140◦F . At a certain instant a valve is opened and mass
flows from the bottle at m = 0.105lbm/s. Find the rate of change of tem-
perature in the bottle at this instant - Provide an equation only – no values
needed.(Question based on FMP 4.184)
Answer: ∂∂tT = − Pm
V CV ρ2
Working:
Starting from the first law of thermodynamics:
Q− W = ∂∂t
∫
CV
[
u+ V 2
2 + gz]
ρ dV +∫
CS
[
h+ V 2
2 + gz]
ρ−→V d
−→A
0 = ∂∂t
∫
CV uρ dV +∫
CS hρ−→V d
−→A
0 = ∂∂tmu+ hm = m ∂
∂tu+ u ∂∂tm+ hm
where, u = CV T andh = u + pv. To simplify the equation, consider the
continuity:
∂∂t
∫
CV ρ dV +∫
CS ρ−→V �
−→n dA = 0
∴∂∂tm+ m = 0
Using this, we can write:
0 = m ∂∂tu+ hm− um
m ∂∂tu = um− (u+ pv)m
where u = CV T
72 BIBLIOGRAPHY
∂∂tT = − pvm
CV m .
The specific volume v = 1ρ and mass m = ρV :
∂∂tT = − pm
ρ2CV V.
X.3 A centrifugal water pump with a 4in. diameter inlet and a 4in. diam-
eter discharge pipe has a flow rate of 300gpm. The inlet pressure in 8in.Hg
vacuum and the exit pressure is 35psig. The inlet and outlet sections are lo-
cated at the same elevation. The measured power input is 9.1hp. Determine
the pump efficiency - Provide an equation only – no values needed.(Question
based on FMP 4.185)
Answer: η = Q(P2−P1)Pin
Working:
Starting from the first law of thermodynamics:
Q− W = ∂∂t
∫
CV
[
u+ V 2
2 + gz]
ρ dV +∫
CS
[
h+ V 2
2 + gz]
ρ−→V d
−→A
−W = pvm|2 − pvm|1 = (p2 − p1)mρ
η = WPin
= Q(p2−p1)Pin
X.4 A pump draws water from a reservoir through a 150mm diameter
suction pipe and delivers it to a 75mm diameter discharge pipe. The end of
the suction pipe is 2m below the free surface of the reservoir. The pressure
gage on the discharge pipe (2m above the reservoir surface) reads 170kPa.
The average speed in the discharge pipe is 3m/s. If the pump efficiency is
75%, determine the power required to drive it. (Question from FMP 4.189)
Answer: Wactual = 3.43kW
Working:
Starting from the first law of thermodynamics:
Q− W = ∂∂t
∫
CV
[
u+ V 2
2 + gz]
ρ dV +∫
CS
[
h+ V 2
2 + gz]
ρ−→V d
−→A
BIBLIOGRAPHY 73
−W =(
pv + V 2
2 + gz)
m∣∣∣2− (pv + gz) m|1
where pv|1 = ρg|z|1ρ = g|z|
∴ −W =(
pv + V 2
2 + gz)
m∣∣∣2
m = ρV A|2 = 13.25kgs−1
Wideal = 2572W
Considering the efficiency of the pump:
Wactual =Win
η = 3.4kW
Ask your tutor to sign off your work. _______
Happy Holidays!
#12 Energy and Bernoulli’s Equation
12.1 Water flows in a cirular duct. At one section the diameter in 0.3m,
the static pressure is 260kPa (gage), the velocity is 3m/s, and the elevation
is 10m above ground level. At a section downstream at ground level, the
duct diameter is 0.15m. Find the gage pressure at the downstream section if
frictional effects may be neglected. (Question from FMP 6.43)
Answer: P = 291kPa (gage)
Working:
Start from Bernoulli’s equation:
p+ ρV 2
2 + ρgz∣∣∣1= p+ ρV 2
2 + ρgz∣∣∣2
p2 = p1 + ρV 2
1−V 2
2
2 + ρg (z1 − z2)
V2 can be found from mass conservation:
m2 = m1
V2 =V1A1
A2=
V1d21d22
= 12ms−1
74 BIBLIOGRAPHY
∴ p2 = 291kPa
12.2 A fire nozzle is coupled to the end of a hose with inside diameter
D = 75mm. The nozzle is contoured smoothly and has outlet diameter
d = 25mm. The design inlet pressure for the nozzle is P1 = 689kPa (gage).
Evaluate the maximum flow rate the nozzle could deliver. (Question from
FMP 6.49)
Answer: Q = 66m3hr−1
Working:
Start from Bernoulli’s equation:
p+ ρV 2
2 + ρgz∣∣∣1= p+ ρV 2
2 + ρgz∣∣∣2
Assume no change in height and P2 = 0Pa (gage).
m1 = m2, ∴ V1 =V2A2
A1
p1 +12ρV
22
(A2
A1
)2= ρ
V 2
2
2
12ρV
22
[
1−(A2
A1
)2]
= p1
V2 =√
p1
1
2ρ
[
1−(
A2
A1
)2] = 39.37ms−1
Q = A2V2 =πd2
4 V2 = 0.0183m3s−1 = 1.1m3min−1 = 66m3hr−1
12.3 An Indianapolis racing car travels at 98.3 m/s along a straightaway.
The team engineer wishes to locate an air inlet on the body of the car to obtain
cooling air for the driver’s suit. The plan is to place the inlet at a location
where the air speed is 25.5 m/s along the surface of the car. Calculate the
static pressure at the proposed inlet location. Express the pressure rise above
ambient as a fraction of the freestream dynamic pressure.(Question from FMP
6.51)
BIBLIOGRAPHY 75
1 2
Answer:
∆p = 5.54kPa
∆ppd2
= 0.93
Working:
Start from Bernoulli’s equation:
p+ ρV 2
2 + ρgz∣∣∣1= p+ ρV 2
2 + ρgz∣∣∣2
where 1 is on the car surface and 2 in the free stream
p+ ρV 2
2
∣∣∣1= p+ ρV 2
2
∣∣∣2
p1 =ρ2
(V 22 − V 2
1
)= 5.54kPa (gage)
using mass conservation:
m1 = m2
A1V1 = A2V2
A2
A1= V1
V2= 0.26
V1 =A2V2
A1= 0.26V2
∴ p1 − pa = ρ2
(V 22 − V 2
1
)= ρ
2
(
V 22 − (0.26V2)
2)
= 0.93(12ρV
22
)
∆ppd2
= 0.93
Ask your tutor to sign off your work. _______
76 BIBLIOGRAPHY
#13 Conservation of Momentum
13.1 Wings create lift by deflecting air ’downwards’. An Airbus A380
requires 5 MN of lift in flight. Assume (falsely) that the wing deflects all
the air travelling through a cross-section of 70 m width (wingspan) and 10 m
height (value without any basis). Calculate the vertical velocity component
of the deflected air and the deflection angle for steady flight at sea level,
at speeds of 100 m/s and 200 m/s. How does lift (for constant deflection
angles) depend on air-speed? (Bear in mind that this is an over-simplified
estimate only, your answers may well deviate by 10%.)
Answer:
For 100m/s:
v2 ≈ 60ms−1
θ ≈ 33◦
For 200m/s:
v2 ≈ 30ms−1
θ ≈ 8◦
Fy ≈ ρA1V21 sin θ ≈ ρA1V
21 tan θ
Bear in mind that this is an over-simplified estimate only, your answers may
well deviate by 10%.
Working:
Start with the conservation of momentum:
BIBLIOGRAPHY 77
−→F = ∂
∂t
∫
CV ρ−→V dV +
∫
CS ρV−→V dA
For the vertical component of force:
Fy = v2 ρV A|2 − v1 ρV A|1v1 = 0ms−1
V2 = V1 (assuming there is no loss to friction over the wing surface)
v2 = V1sinθ
From mass conservation:
V2A2 = A1V1
∴ Fy = (ρA1V1)V1 tan θ
Using different assumptions, you may end up with:
∴ Fy = (ρA1V1)V1 tan θ
13.2 Consider incompressible steady flow of standard air in a boundary
layer on the length of porous surface shown. Asume the boundary layer at
the downstream end of the surface has an approximately parabolic veloc-
ity profile u/U∞ = 2(y/δ) − (y/δ)2. Uniform suction is applied along the
porous surface, as shown. Calculate the volume flow rate across surface cd,
through the porous suction surface, and across surface bc. (Question from
FMP 4.41)
Answer:
Qcd = 4.5× 10−3m3s−1
78 BIBLIOGRAPHY
Qbc = 1.65 × 10−3m3s−1
Working:
Qcd = UcdAcd = w∫ δ0 u dy
Ucd = U∞
[
2yδ −
(yδ
)2]
Qcd =wU∞
δ
∫ δ0 2y − y2
δ dy = wU∞
δ
[y2 − 1
3δy3]δ
0
Qcd =wU∞
δ
[δ2 − 1
3δ2]= 2
3U∞wδ = 4.5× 10−3m3s−1
Qad = UadAad = 6× 10−4m3s−1
Qbc = Qab −Qad−Qcd = 3× 1.5× 1.5× 10−3m3s−1− 0.6× 10−3m3s−1−4.5× 10−3m3s−1
Qbc = 1.65 × 10−3m3s−1
13.3 A farmer purchases 675kg of bulk grain from the local co-op. The
grain is loaded into his pickup truck from a hopper with an outlet diameter of
0.3m. The loading operator determines th payload by observing the indicated
gross mass of the truck as a function of time. The grain flow from the hopper
(m = 40kg/s) is terminated when the indicated scale reading reaches the
desired gross mass. If the grain density is 600kg/m3, determine the true
payload. (Question from FMP 4.55)
Answer: mactual = 671.15kg
Working:
The reading on the scale w is given by:
mreadingg = mactualg + ρu2A = mactualg + mu
∴ mactual =mreadingg−mu
g
u = mρA = 0.9431ms−1
∴ mactual = 671.15kg
BIBLIOGRAPHY 79
13.4 The Eurofighter Typhoon aircraft is powered by two EJ200 engines,
providing 90 kN of thrust each (with reheat). The reheat outlet temperature
is 1500ºC, the outlet diameter is 0.8 m, and you can assume that the thrust
equals the momentum flow rate from the outlet. Calculate the jet-velocity
at the outlet. Calculate the volume flow rate and the mass flow rate at
the outlet. Calculate the total fuel mass flow rate into both engines for an
assumed air/fuel [mass] ratio of 20:1. For how long can reheat be engaged
based on a maximum fuel capacity of 4,000kg?
Answer:
outlet velocity: 948.7ms−1
volume flow rate: 474.34m3s−1
mass flow rate: 94.9kgs−1
total fuel mass flow rate: 9.5kgs−1
Time: 7min. 2sec.
Working:
area of engine outlet: 0.25πD2 = 0.5m2
outlet temperature: 1773.15◦K
density of air at outlet: PRT = 1.01325×105N/m2
287Jkg−1K−1×1773.15◦K
= 0.2kgm−3
Thrust = momentum flow rate
∴ ρu2A = 90 × 103 ⇒ u =√
90×103
ρA = 948.7ms−1
volume flow rate: uA = 474.34m3s−1
mass flow rate: ρuA = 94.9kgs−1
mass flow rate for both engines: 189.7kgs−1
fuel mass flow rate: 189.7kgs−1/20 = 9.5kgs−1
Tank capacity of the Eurofighter: 4, 000kg
Time: 4000kg9.5kgs−1 = 420s = 7min 2s.
Ask your tutor to sign off your work. _______
80 BIBLIOGRAPHY
#14 Examples Class
14.1 A water jet pump has jet area 0.01m2 and jet speed 30m/s. The jet
is within a secondary stream of water having speed Vs = 3m/s. The total
area of the duct (the sum of the jet and secondary stream areas) is 0.075m2.
The water is thoroughly mixed and leaves the jet pump in a uniform stream.
The pressure of the jet and secondary stream are the same at the pump inlet.
Determine the speed at the pump exit and the pressure rise p2−p1. (Question
from FMP 4.72)
Answer:
V2 = 6.6ms−1
∆p = 84.2kPa
Working:
Mass conservation:
ms + mj = m2
V2 =VjAj+Vs(As−Aj)
As= 6.6ms−1
Apply the conservation of momentum:
−→F = ∂
∂t
∫
CV ρ−→V dV +
∫
CS ρV−→V dA
F = ρ(V 22 A2 − V 2
1 A1
)= ρ
(
V 22 A2 − V 2
j Aj − V 2s (As −Aj)
)
p1A1 − p2A2 = −6.318kN
A1 = A2 = As
BIBLIOGRAPHY 81
∴ p2 − p1 = 84.2kPa
14.2 A reducer in a piping system is shown. The internal volume of the
reducer is 0.2m3 and it’s mass is 25kg. Evaluate the total force that must
be provided by the surrounding pipes to spport the reducer. The fluid is
gasoline.(Question from FMP 4.71)
Answer:−→F = −4.69kNi + 1.66kNj
Working:
Vertical force:
wr + wgas = g (mr + ρgasVr) = 1.66kN
where ρgas = 720kgm−3
Horizontal force, apply the conservation of momentum, considering difference
in static pressure:
−→F = p2A2 − p1A1 +
∂∂t
∫
CV ρ−→V dV +
∫
CS ρV−→V dA
F = p2A2 − p1A1 + ρ(V 22 A2 − V 2
1 A1
)= −7.13kN + 2.44kN
Note that p2 is an absolute pressure reading, so the ambient pressure must
be accounted for.
∴ F = −4.69kNi + 1.66kNj
14.3 The circular dish, whose cross section is shown, has an outside diam-
82 BIBLIOGRAPHY
eter of 0.20m. A water jet with speed 35 m/s strikes the dish concentrically.
The dish moves to the left at 15 m/s. The jet diameter is 20 mm. The dish
has a hole at its center that allows a stream of water 10 mm in diameter
to pass through without resistance. The remainder of the jet is deflected
and flows along the dish. Calculate the force required to maintain the dish
motion.(Question from FMP 4.111)
Answer: F = −167N
Working:
min = ρUrπD2
4 = ρ(V − U)πD2
4 = 6.284kgs−1
mout = ρUrπd2
4 = 1.57kgs−1
mdeflect = min − mout = 4.713kgs−1
Total force:
Ftotal = Fin + Fout + Fdeflect
Ftotal = −minUr + moutUr − mdeflectUdeflectcos (40◦)
Assume that no loss occurs from friction between the deflected stream and
dish. Hence Udeflect = Ur = (V − U) = 20ms−1
∴ Ftotal = −167N
Ask your tutor to sign off your work. _______
BIBLIOGRAPHY 83
#15 Boundary Layers
15.1 Air at standard conditions flows along a flat plate. The undisturbed
free stream speed is U0 = 10m/s. At L = 145mm downstream from the
leading edge of the plate, the boundary-layer thickness is δ = 2.3mm. The
velocity profile at this location is
uU0
= 32yδ − 1
2
[yδ
]3
Calculate the horizontal componet of force per unit witdh required to hold
the plate stationary. (Question from FMP 4.85)
Answer: Fw = 0.04Nm−1
Working:
Apply the conservation of momentum:
−→F = ∂
∂t
∫
CV ρ−→u dV +∫
CS ρu−→u dA
F = ρ(u23A3 + u22A2 − u21A1
)
Calculate mass flow rate at 3:
m3 = m1 − m2
m1 = ρU0δw = 0.0283w kgs−1
m2 = ρw∫ δ0 u dy = ρw
∫ δ0 U0
[32yδ − 1
2
(yδ
)3]
dy
m2 = ρwU01δ
[34y
2 − 18y4
δ2
]δ
0= 0.625ρwU0δ
m2 = 0.0177w kgs−1
84 BIBLIOGRAPHY
∴ m3 = 0.0106w kgs−1
Calculate momentum flow rate at 2:
m2u2 = ρw∫ δ0 u2 dy = ρδwU2
0
∫ δ0
[32yδ − 1
2
(yδ
)3]2
dy
m2u2 = ρwU20
∫ δ0
[94
(yδ
)2 − 32
(yδ
)4+ 1
4
(yδ
)6]
dy
m2u2 = ρwU20
[912
(y3
δ2
)
− 310
(y5
δ4
)
+ 128
(y7
δ6
)]δ
0= 0.486ρδwU2
0
m2u2 = 0.137wN
∴ F = m3u3 + m2u2 − m1u1 = −0.04wN
So the plate will require an opposite force per unit width of the same magni-
tude:
Fplate
w = 0.04Nm−1
15.2 A viscous oil flows steaily between stationary parallel plates. The
flow is laminar and fully developed. The total gap width between the plates
is h = 5mm. The oil viscosity is 0.5N · s/m2 and the pressure gradient is
−1000N/m2/m. Find the magnitude and direction of the shear stress on
the upper plate and the volume flow rate through the channel, per meter of
width. (Question from FMP 8.8)
Answer:
τyx = −2.5Nm−2
Qw = 2.08 × 10−5m2s−1
Working:
∂τyx∂y = ∂p
∂x
integrate and evaluate for bottom plate:
τyx =∫ h/20
∂p∂x dy = ∂p
∂xh2 = −2.5Nm−2
τyx = µ∂u∂y ∴ u =
∫ τyxµ dy
BIBLIOGRAPHY 85
from above; τyx = ∂p∂xy + c1
so u =∫
1µ∂p∂xy +
1µc1 dy
u = 12µ
∂p∂xy
2 + 1µc1y + c2
find c1 and c2:
at y = 0, u = 0 so c2 = 0
∴ u = 12µ
∂p∂xy
2 + 1µc1y
at y = h, u = 0 so
c1 = −h2∂p∂x
∴ u = 12µ
∂p∂xy
2 − h2µ
∂p∂xy
Q = w∫ h0 u dy = w
∫ h0
12µ
∂p∂xy
2 − h2µ
∂p∂xy dy
Q = w 1µ∂p∂x
[16y
3 − h4y
2]h
0= 2.08 × 10−5wm3s−1
Qw = 2.08 × 10−5 m2s−1
Ask your tutor to sign off your work. _______
#16 Laminar Pipe Flows
16.1 A viscosity measurement setup for an undergraduate fluid mechanics
laboratory is to be made from flexible plastic tubing; the fluid is to be water.
Assume the tubing diameter is D = 2.5 mm ± 0.2 mm and the length is l =
20 m. Evaluate the maximum volume low rate at which laminar flow would
be expected and the corresponding pressure drop. Estimate the experimental
uncertainty in viscosity measured using this apparatus. How could the setup
be improved? (Question based on FMP 8.47)
Answer:
Qmax = 5.57× 10−6m3s−1
86 BIBLIOGRAPHY
∆p = 97.36kPa
Working:
Q = uA
Re = ρuDµ = ρu4πD2
µ4πD = 4ρQπDµ
Q = ReµπD4ρ
Qmax = 2300×1.14×10−3π×2.7×10−3
4×999 = 5.57× 10−6m3s−1
∆p = 128µLπD4 Q = 97.36kPa
uncertainty analysis:
µ = ∆pπD4
128LQ
uµ = ±[(
∆pµ
∂µ∂∆pu∆p
)2+
(Dµ
∂µ∂DuD
)2+
(Lµ
∂µ∂LuL
)2+
(Qµ
∂µ∂QuQ
)2]1/2
∆pµ
∂µ∂∆p = ∆p
µπD4
128LQ = 1
Dµ
∂µ∂D = D
µ4πD3
128LQ = 4
Lµ
∂µ∂L = L
µ (−1) πD4
128L2 Q= −1
Qµ
∂µ∂Q = Q
µ (−1) πD4
128LQ2 = −1
Hence,
uµ = ±[
u2∆p + (4uD)2 + (−uL)
2 + (−uQ)2]1/2
since
ud = δDD = ±0.2
2.5 = ±8%
uµ ∼ 4ud = 32%
The setup could be therefore be improved by reducing the uncertainty in
the tube diameter by using a larger diameter or a more uniform-diameter
tube.
16.2 Consider fully devloped laminar flow in a circular pipe. Use a cylindri-
cal control volume as shown. Indicate the forces acting on the control volume.
BIBLIOGRAPHY 87
Using the momentum equation, develop an expression for the velocity distri-
bution. (Question from FMP 8.51)
Answer: u = 14µ
∂p∂x
(r2 −R2
)
Working:
Form expressions for the pressure and shear stresses acting on the control
volume:
Pressure force on left face:
FPL=
(
p− ∂p∂x · ∆x
2
)
πr2
Pressure force on right face:
FPR= −
(
p+ ∂p∂x · ∆x
2
)
πr2
Shear stress at pipe center:
Fτ0 = 0
Shear stress at ∆r:
Fτr = τrxπr2∆x
Sum all forces
FPL+ FPR
+ Fτr + Fτ0 = − ∂p∂x∆xπr2 + ∂τrx
∂r∆x 2πr = 0
∴ τrx = 12r
∂p∂x
substitute τrx = µ∂u∂r :
∂u∂r = 1
2µ∂p∂xr
integrate over r:
u(r) = 14µ
∂p∂xr
2 + C1
Solve for C1 with boundary condition:
u(R) = 0
C1 = −R2
4µ∂p∂x
88 BIBLIOGRAPHY
∴ u = 14µ
∂p∂x
(r2 −R2
)
16.3 The velocity profile for fully developed flow between stationary parallel
plates is given by u = a(h2/4− y2), where a is a constant, h is the total gap
width between plates, and y is the distance measured from the center of the
gap. Determine the ratio V /umax. (Question from FMP 8.6)
Answer: Vumax
= 23
Working:
maximum velocity will be at mid-distance between the plates:
umax = u(0) = ah2
4
mean velocity:
V = QA = 1
A
∫u dA = 1
wh
∫ h/2−h/2 uw dy = 1
h
∫ h/2−h/2 a
(h2
4 − y2)
dy = ah
(h2
4 y − y3)∣∣∣
h/2
−h/2
V = 16ah
2
∴V
umax= 2
3
#17 Losses in Pipe Flows
17.1 Water is pumped at the rate of 0.6m3/s from a reservoir 6m above
a pump to a free discharge 27m above the pump. The pressure on the
intake side of the pump is 34kPa and the pressure on the discharge side is
0.3MPa. all pipes are commercial steel of 15cm diamter. Determine a) the
head supplied by the pump and (b) the total head loss between the pump and
the point of free dischrge. (Question from FMP 8.77)
BIBLIOGRAPHY 89
Answer:
∆hpump = 266.3Jkg−1
hlt = 35.43Jkg−1
Working:
∆hpump =Win
m =[(
p3ρ +
V 2
3
2 + gz3
)
−(p2ρ +
V 2
2
2 + gz2
)]
V3 = V2 (mass conservation)
z2 = z3 (negligible)
∆hpump =1ρ (p3 − p2) = 266.27Jkg−1
Head loss:
pρ + V 2
2 + ρgz∣∣∣3= p
ρ + V 2
2 + gz∣∣∣4+ hlt
p4 = 0Pa (gage)
V3 = V4 (mass conservation)
z3 = 0m
hlt =p3ρ − gz4 = 35.4Jkg−1
17.2 A small-diameter capillary tube made from drawn aluminium is used
in place of an expansion valve in a home refrigerator. The inside diameter
is 0.5mm. Calculate the corresponding relative roughness. Comment on
90 BIBLIOGRAPHY
whether this tube may be considered “smooth” with regard to fluid flow.
(Question from FMP 8.79)
Answer: eD = 3× 10−3
Working:
e (for drawn tubing) = 1.5 ×−6 m
∴eD = 1.5×10−6
0.5×10−3 = 3× 10−3
Looking at the Moody diagram, it is clear that this tube cannot be considered
smooth for turbulent flow through the tube. For laminar flow (Re<2300) the
relative roughbess has no effect.
17.3 A smooth, 75mm diameter pipe carries water (65◦c) horizontally.
When the mass flow rate is 0.075kg/s, the pressure drop is measured to be
7.4Pa per 100m of pipe. Based on these measurements, what is the friction
factor? What is the Reynolds number? Does this Reynolds number generally
indicate laminar or turbulent flow? Is the flow actually laminar or turbulent?
(Question from FMP 8.80)
Answer:
f = 0.039
Re ≈ 3000 - indicates turbulent flow
Working:
ρ = 980kgm−3
A = 4.42 × 10−3m2
pρ + V 2
2 + ρgz∣∣∣1= p
ρ + V 2
2 + gz∣∣∣2+ hlt
∆pρ = hlt = f L
DV 2
2
∴ f = ∆pρ
DL
2V 2 = ∆p
ρDL 2
(ρAm
)2
f = 0.038
BIBLIOGRAPHY 91
V = mρA = 0.0173ms−1
Re = ρV Dµ = 980×0.0173×0.075
4.31×10−4 ≈ 3000
indicating turbulent flow.
#18 Solution of Pipe Flow Problems
18.1 Two reservoirs are connected by three clean cast-iron pipes in series,
L2 = 600m, D2 = 0.3m, L3 = 900m, D3 = 0.4m, L4 = 1500m, and
D4 = 0.45m. When the discharge is 0.11m3/s of water at 15◦C, determine
the difference in elevation between the reservoirs. (Question from FMP 8.112)
L =600m
L =900m
L =1500m
D =0.3m
D =0.4m
D =0.45m
2
2
3
3
4
4
1
2
z2
Δz
Answer: ∆z = 8.21m
Working:
µ = 1.1× 10−3Pa · spρ + V 2
2 + ρgz∣∣∣1= p
ρ + V 2
2 + gz∣∣∣2+ hlt
g∆z = f2L2
D2
V 2
2
2 + f3L3
D3
V 2
3
2 + f4L4
D4
V 2
4
2
V3A3 = V2A2 ⇒ V3 = V2A2
A3= 9
16V2
V4A4 = V2A2 ⇒ V4 = V2A2
A4= 4
9V2
92 BIBLIOGRAPHY
V2 =QA2
= 1.56ms−1 : Re2 = 4.25 × 105 : eD = 8.67× 10−4
V3 = 0.875ms−1 : Re3 = 3.18 × 105 : eD = 6.5× 10−4
V4 = 0.69ms−1 : Re4 = 2.83 × 105 : eD = 5.78 × 10−4
f2 = 0.02 : f3 = 0.019 : f4 = 0.0188
∆z = 1g
[f22
L2
D2+ f3
L3
D3
0.322 + f4
L4
D4
0.22
]
V 22
∆z = 9.81−1 [20 + 6.84 + 6.27] 1.562 = 8.21m
18.2 Water, at volume flow rate Q = 20L/s, is delivered by a fire hose
and nozzle assembly. The hose (L = 80m, D = 75mm, and e/D = 0.004)
is mae up of four 20m sections joined by couplings. The entrance is square-
edged; the minor loss coefficient for each coupling is Kc = 0.5, based on
mean velocity through the hose. The nozzle loss coefficient is Kn = 0.02,
based on velocity in the exit jet, of d = 25mm diameter. Estimate the supply
pressure required at this flow rate.(Question from FMP 8.113)
Answer: p1 = 1166kPa
Working:
pρ + V 2
2 + ρgz∣∣∣1= p
ρ + V 2
2 + gz∣∣∣2+ hlt
∆z = 0m : V1 = 0ms−1 : P2 = 0Pa (gage)
p1 = ρV 2
3
2 + ρhlt = ρV 2
3
2 + ρ [hle + 4hlc + hln + hl]
p1 = ρV 2
3
2 + ρ[12keV
22 + 2kcV
22 + 0.01V 2
3 + f LD
V 2
2
2
]
V3A3 = V2A2 : V3 = V2A2
A3= 9V2
p1 = ρ81V 2
2
2 + ρ[12keV
22 + 2kcV
22 + 0.01V 2
2 81 + f LD
V 2
2
2
]
BIBLIOGRAPHY 93
p1 = ρ[812 + 1
2ke + 2kc + 0.01 × 81 + f LD
12
]V 22
ke = 0.5
V2 =QA2
= 4.53ms−1
Re = 3.39 × 105
Calculate f from:
1f0.5 = −2log
[e/D3.7 + 2.51
Ref0.5
]
f = 0.0287
p1 = ρ[812 + 0.25 + 1 + 0.81 + 15.3
]4.532
p1 = 1166kPa
18.3 Gasoline flows in a long, underground pipline at a constant temperature
of 15◦C. Two pumping stations at the same elevation are located 13km
apart. the pressure drop between the sections is 1.4MPa. The pipeline is
made from 0.6m diameter pipe. Although the pipe is made from commercial
steel, age and corrosion have raised the pipe roughness to approximately that
for galvnized iron. Compute the volume flow rate. (Question from FMP
8.118)
Answer: Q = 1m3s−1
Working:
pρ + V 2
2 + ρgz∣∣∣1= p
ρ + V 2
2 + gz∣∣∣2+ hlt
∆p = hlt = ρf LD
V 2
2
2
V2 =√
2D∆pρfL : 1
f0.5 = −2log[e/D3.7 + 2.51
Ref0.5
]
Begin with an approximate value of f and V2, say 0.004 and 1ms−1
- Calculate Re based on approximate V2
- Calculate f for approximate Re
- Calculate V2 for next iteration
Repeat process until values of f and V2 converge.
94 BIBLIOGRAPHY
f = 0.014 : V2 = 3.498ms−1
∴ Q = 1m3s−1
#19 Flow Measurements
18.1 You are asked to size a pump for installation in the water supply of the
Sears Tower in Chicago. The system requires 100gpm of water pumped to a
reservoir at the top of the tower 340m above the street. City water pressure
at the street-level pump inlet is 400kPa (gage). Piping is to be commercial
steel. Determine the minimum diameter required to keep the average velocity
below 3.5m/s in the pipe. Calculate the pressure rise required across the
pump. Estimate the minimum power needed to drive the pump. (Question
from FMP 8.144)
Answer:
D = 0.048m
∆p = 3.84MPa
w = 24.2kW
Working:
pρ + V 2
2 + ρgz∣∣∣1= p
ρ + V 2
2 + gz∣∣∣2+ hlt
p2 = 0Pa (gage)
BIBLIOGRAPHY 95
V1 = V2 = 3.5ms−1
A = QV ⇒ D =
√4QV π = 0.048m
eD = 4.6×10−5
0.048 = 9.6× 10−4
Re = 1.68 × 105
1f0.5 = −2log
[e/D3.7 + 2.51
Ref0.5
]
⇒ f = 0.021
p1 = ρ [g∆z + hl] = ρ[
g∆z + f LD
V 2
2
]
= 4.243MPa
∆p = p1 − p0 = 4.243MPa − 400kPa = 3.84MPa
w = m∆p = 6.3× 10−3m3s−1 × 3.84MPa = 24.2kW
18.2 A venturi meter with a 75mm diameter throat is placed in a 150mm
diameter line carrying water at 25◦C. The pressure drop between the up-
stream tap and the venturi throat is 300mm of mercury. compute the rate
of flow. (Question from FMP 8.161)
Answer: Q = 0.04m3s−1
Working:
mactual =CAt√1−β4
√2ρ∆p
β = 75150 = 0.5 : C = 0.99 : At =
π(0.075)2
4 = 4.42 × 10−3m2
∆p = ρHgg∆h = 39.88kPa
∴ mactual = 40.36kgs−1
Q = ˙mactual
ρ = 0.04m3s−1
#20 Revision
#21 TBD