Kempf Notes

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

[email protected]

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


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


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.


~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


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.


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


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


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.


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.


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.


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.


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)


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

[email protected]

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


#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


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


#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


#6 Revisions

Use the time to understand tutorial questions you struggled with before. Try

to solve the recommended questions in FMP.


#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



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


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


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


#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


#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


#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:


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:


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


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:


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:


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


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


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 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.


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


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


#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

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