Fluids mechanics (a letter to a friend) part 1 . .

Musadoto for felician deus
AE 218
IWRE 210
SUMMARIES +QUESTION WITH
ANSWERS
SOKOINE UNIVERSITY
OF AGRICULTURE
musadoto © 2018

FUNDAMENTALS AND
APPLICATIONS
Summaries, lecture notes and problems with solutions
Acknowledgement to professors:
ÇENGEL: Department of Mechanical Engineering, University of
Nevada, Reno.
JOHN M. CIMBALA Department of Mechanical and Nuclear
Engineering, The Pennsylvania State University.
LOWA UNIVERSITY
DR. MBUNGU lecture notes 2018 + assignments.
MR. MATERU lecture notes 2017 + Counter Attacks.
SEMESTER 4
© 2018 Musadoto

This book is special to
Agricultural engineering student 2
This course is ‚different‛ – very physics based. Bro!
• Fluids is HARD - lots of new concepts/equations
Based on 4 balances:
1. Force,
2. momentum,
3. mass,
4. energy
You are not in this alone! Bro…
Musadoto with his family
Classmates
Heavenly Father & the Spirit.
4.0 GPA VS DELL

HARD WORK + THE SPIRIT BRO. DEUS
Face the future with optimism. I believe we are standing on the
threshold of a new era of growth, prosperity, and abundance.
Barring a calamity or unexpected international crisis, I think
the next few years will bring a resurgence in the economy as new
discoveries are made in communication, medicine, energy,
transportation, physics, computer technology, and other fields of
endeavor. Many of these discoveries, as in the past, will be the
result of the Spirit whispering insights into and enlightening
the minds of truth-seeking individuals. Many of these discoveries
will be made for the purpose of helping to bring to pass the
purposes and work of God and the quickening of the building of
His kingdom on earth today. With these discoveries and advances
will come new employment opportunities and prosperity for those
who work hard and especially to those who strive to keep the
commandments of God. This has been the case in other significant
periods of national and international economic growth.
-Elder M. Russell Ballard

Table of contents
1.The background of Fluid Mechanics
2.Fields of Fluid mechanics
3.Introduction and Basic concepts
4.Properties of Fluids
5.Pressure and fluid statics
6.Hydrodynamics

1. BACKGROUND
Fluid mechanics is an exciting and fascinating subject with unlimited practical
applications ranging from microscopic biological systems to automobiles, airplanes,
and spacecraft propulsion. Yet fluid mechanics has historically been one of the most
challenging subjects for undergraduate students. Unlike earlier freshman- and
sophomore-level subjects such as physics, chemistry, and engineering mechanics,
where students often learn equations and then ‚plug and chug‛ on their calculators,
proper analysis of a problem in fluid mechanics requires much more. Oftentimes,
students must first assess the problem, make and justify assumptions and/or
approximations, apply the relevant physical laws in their proper forms, and solve
the resulting equations before ever plugging any numbers into their calculators.
Many problems in fluid mechanics require more than just knowledge of the subject,
but also physical intuition and experience.
HISTORY OF FLUID MECHANICS
Fluid mechanics has a history of erratically occurring early achievements, then an
intermediate era of steady fundamental discoveries in the eighteenth and nineteenth
centuries. Ancient civilizations had enough knowledge to solve certain flow
problems. Sailing ships with oars and irrigation systems were both known in
prehistoric times. The Greeks produced quantitative information. Archimedes and Hero
of Alexandria both postulated the parallelogram law for vector addition in the third
century B.C.. Archimedes (285-212 B.C.) formulated the laws of buoyancy and applied
them to floating and submerged bodies, actually deriving a form of the differential
calculus as part of the analysis. Up to the Renaissance, there was a steady
improvement in the design of such flow systems as ships, canals, and water conduits,
but no recorded evidence of fundamental improvements in flow analysis. Then Leonardo
da Vinci (1452-1519) derived the equation of conservation of mass in one-dimensional
steady flow. Leonardo was an excellent experimentalist, and his notes contain
accurate descriptions of waves, jets, hydraulic jumps, eddy formation, and both low-
drag (streamlined) and high-drag (parachute) designs. A Frenchman, Edme Mariotte
(1620-1684), built the first wind tunnel and tested models in it. In 1687, Isaac
Newton (1642-1727) postulated his laws of motion and the law of viscosity of the
linear fluids now called newtonian. The theory first yielded to the assumption of a
‚perfect‛ or frictionless fluid, and eighteenth-century mathematicians (Daniel
Bernoulli, Leonhard Euler, Jean d’Alembert, Joseph-Louis Lagrange, and Pierre-Simon
Laplace) produced many beautiful solutions of frictionless-flow problems. Euler
developed both the differential equations of motion and their integrated form, now
called the Bernoulli equation. D’Alembert used them to show his famous paradox: that
a body immersed in a frictionless fluid has zero drag. These beautiful results
amounted to overkill, since perfect-fluid assumptions have very limited applications
in practice and most engineering flows are dominated by the effects of viscosity.
Engineers began to reject what they regarded as a totally unrealistic theory and
developed the science of hydraulics, relying almost entirely on experiment. Such
experimentalists as Chézy, Pitot, Borda, Weber, Francis, Hagen, Poiseuille, Darcy,
Manning, Bazin, and Weisbach produced data on a variety of flows such as open
channels, ship resistance, pipe flows, waves, and turbines. At the end of the
nineteenth century, unification between experimental hydraulics and theoretical
hydrodynamics finally began. William Froude (1810-1879) and his son Robert (1846-
1924) developed laws of model testing, Lord Rayleigh (1842-1919) proposed the
technique of dimensional analysis, and Osborne Reynolds (1842-1912) published the
classic pipe experiment in 1883 which showed the importance of the dimensionless
Reynolds number named after him. Meanwhile, viscous-flow theory was available but

unexploited since Navier (1785-1836) and Stokes (1819-1903) had successfully added
the newtonian viscous terms to the governing equations of motion. Unfortunately, the
resulting Navier-Stokes equations were too difficult to analyze for arbitrary flows.
In 1904, a German engineer, Ludwig Prandtl (1875-1953), published perhaps the most
important paper ever written on fluid mechanics. Prandtl pointed out that fluid
flows with small viscosity (water and air flows) can be divided into a thin viscous
layer, or boundary layer, near solid surfaces and interfaces, patched onto a nearly
inviscid outer layer, where the Euler and Bernoulli equations apply. Boundary-layer
theory has proven to be the single most important tool in modern flow analysis. The
twentieth-century foundations for the present state of the art in fluid mechanics
were laid in a series of broad-based experiments by Prandtl and his two chief
friendly competitors, Theodore von Kármán (1881-1963) and Sir Geoffrey I. Taylor
(1886-1975).
Source: http://majdalani.eng.auburn.edu/courses/02_fluids/handout_f01_history.pdf
2. FIELDS OF FLUID MECHANICS

3. INTRODUCTION
Mechanics is the oldest physical science that deals with both stationary and moving
bodies under the influence of forces. The branch of mechanics that deals with bodies
at rest is called statics, while the branch that deals with bodies in motion is
called dynamics.
The subcategory fluid mechanics is defined as the science that deals with the
behavior of fluids at rest (fluid statics) or in motion (fluid dynamics), and the
interaction of fluids with solids or other fluids at the boundaries. Fluid mechanics
is also referred to as fluid dynamics by considering fluids at rest as a special
case of motion with zero velocity
Fluid mechanics itself is also divided into several categories. The study of the
motion of fluids that are practically incompressible (such as liquids, especially
water, and gases at low speeds) is usually referred to as hydrodynamics. A
subcategory of hydrodynamics is hydraulics, which deals with liquid flows in pipes
and open channels. Gas dynamics deals with the flow of fluids that undergo
significant density changes, such as the flow of gases through nozzles at high
speeds. The category aerodynamics deals with the flow of gases (especially air) over
bodies such as aircraft, rockets, and automobiles at high or low speeds. Some other
specialized categories such as meteorology, oceanography, and hydrology deal with
naturally occurring flows.
WHAT IS A FLUID?
From physics point of views that a substance exists in three primary phases: solid,
liquid, and gas. (At very high temperatures, it also exists as plasma.) A substance
in the liquid or gas phase is referred to as a FLUID. Distinction between a solid
and a fluid is made on the basis of the substance’s ability to resist an applied
shear (or tangential) stress that tends to change its shape. A solid can resist an
applied shear stress by deforming, whereas a FLUID deforms continuously under the
influence of shear stress, no matter how small. In solids stress is proportional to
strain, but in fluids stress is proportional to strain rate. When a constant shear
force is applied, a solid eventually stops deforming, at some fixed strain angle,
whereas a fluid never stops deforming and approaches a certain rate of strain.
SIGNIFICANCE OF FLUID MECHANICS
Fluids omnipresent
Weather & climate
Vehicles: automobiles, trains, ships, and planes, etc.
Environment
Physiology and medicine
Sports & recreation
There Many other examples!

Field of Fluid Mechanics can be divided into 3 branches:
1. Fluid Statics: mechanics of fluids at rest
2. Kinematics: deals with velocities and streamlines with considering forces or
energy
3. Fluid Dynamics: deals with the relations between velocities and accelerations
and forces exerted by or upon fluids in motion
STREAMLINES
A streamline is a line that is tangential to the instantaneous velocity
direction (velocity is a vector that has a direction and a magnitude)
Mechanics of fluids is extremely important in many areas of engineering and science.
Examples are:
1. Biomechanics
 Blood flow through arteries
 Flow of cerebral fluid
2. Meteorology and Ocean Engineering
 Movements of air currents and water currents
3. Chemical Engineering
 Design of chemical processing equipment
4. Mechanical Engineering
 Design of pumps, turbines, air-conditioning equipment, pollution-
control equipment, etc.
5. Civil Engineering
 Transport of river sediments
 Pollution of air and water
 Design of piping systems
 Flood control systems
DIMENSION AND UNITS
Two primary sets of units are used:
1. SI (System International) units
2. English units
Instantaneous streamlines in flow around a cylinder

UNIT TABLE
Quantity SI Unit English Unit
Length (L) Meter (m) Foot (ft)
Mass (m) Kilogram (kg) Slug (slug) = lb*sec2
/ft
Time (T) Second (s) Second (sec)
Temperature ( ) Celcius (o
C) Farenheit (o
F)
Force Newton (N)=kg*m/s2
Pound (lb)
1 Newton – Force required to accelerate a 1 kg of mass to 1 m/s2
1 slug – is the mass that accelerates at 1 ft/s2
when acted upon by a force
of 1 lb
To remember units of a Newton use F=ma (Newton’s 2nd
Law)
[F] = [m][a]= kg*m/s2
= N
To remember units of a slug also use F=ma => m = F / a
[m] = [F] / [a] = lb / (ft / sec2
) = lb*sec2
/ ft
1 lb is the force of gravity acting on (or weight of ) a platinum standard
whose mass is 0.45359243 kg
Weight and Newton’s Law of Gravitation
 Weight
 Gravitational attraction force between two bodies
 Newton’s Law of Gravitation
F = G m1m2/ r2
Where
 G - universal constant of gravitation
 m1, m2 - mass of body 1 and body 2, respectively
 r - distance between centers of the two masses
 F - force of attraction
WEIGHT
 m2 - mass of an object on earth’s surface
 m1 - mass of earth
 r - distance between center of two masses

 r1 - radius of earth
 r2 - radius of mass on earth’s surface
 r2 << r1, therefore r = r1+r2 ~ r1
 Thus, F = m2 * (G * m1 / r2
)
 Weight (W) of object (with mass m2) on surface of earth (with mass m1) is
defined as
W = m2g ; g =(Gm1/r2
) gravitational acceleration
g = 9.31 m/s2
in SI units
g = 32.2 ft/sec2
in English units
Now #TBT
What is Fluid
 Difn: Fluid is a ‚substance which conforms continuously under the action of
shearing forces‛.
 Difn: Fluid mechanics is the ‚science and technology of fluids either at rest
(fluid statics) or in motion (fluid dynamics) and their effects on boundaries
such as solid surfaces or interfaces with other fluids‛.
To understand this,lets remind ourselves of what a shear force is: Application and
effect of shear force on a book(AE 211 Strength of materials for beginners 1st
ed,2018 by musadoto)
https://www.slideshare.net/musadoto/strength-of-materials-for-beginners
Definition Applied to Static Fluids
 ‚If a fluid is at rest there can be no shearing forces acting and therefore
all forces in the fluid must be perpendicular to the planes in which they act‛.
 Note here that we specify that the fluid must be at rest. This is because, it
is found experimentally that fluids in motion can have slight resistance to
shear force. This is the source of viscosity.
Definition Applied to Fluids in Motion
• For example, consider the fluid shown below flowing along a fixed surface. At
the surface there will be little movement of the fluid (it will ‘stick’ to the
surface), whilst further away from the surface the fluid flows faster (has
greater velocity):

 No slip condition: no relative motion between fluid and boundary, i.e., fluid
in contact with lower plate is stationary, whereas fluid in contact with upper
plate moves at speed U.
 If one layer of is moving faster than another layer of fluid, there must be
shear forces acting between them. For example, if we have fluid in contact with
a conveyor belt that is moving we will get the behavior shown above.
FLUID IN MOTION
NEWTON’S LAW OF VISCOSITY
 When fluid is in motion, any difference in velocity between adjacent layers has
the same effect as the conveyor belt does.
 Therefore, to represent real fluids in motion we must consider the action of
shear forces. Consider the small element of fluid shown, which is subject to
shear force and has a dimension s into the page. The force F acts over an area
A = BC×s. Hence we have a shear stress applied:
Shear stress =
=
Any stress causes a deformation,or strain and a shear stress causes a shear strain.
This shear strain is measured by angle .Remember that a continuously deforms

when under the action of shear.This is different to solid :a solid have a single
value of for each value of
Shear stress is directly proportional to the rate of shear strain. There is a need
to understand the rate of shear strain .from the above diagram
RATE OF SHEAR STRAIN =
Suppose that the particles of fluid at E moves in distance x and time t for small
angles ,then
THE RATE OF SHEAR STRAIN IS
Where is the velocity of Fluid
This is change in velocity ( ) with height ( ), But when we consider the
infinitesimally small change in height ( we can write the rate of shear strain in
form of du/dy (noting that shear stress is proportional to the rate of shear strain)
Then
This is newton’s law of viscosity
is the property of fluid called its dynamic viscosity ,it is dynamic because the
fluid is in motion then the viscosity resists the shear stress.
NON NEWTONIAN FLUIDS
Fluids for which the shear stress is not linearly related to the shear strain rate
are called non-Newtonian fluids. Examples include slurries and colloidal
suspensions, polymer solutions, blood, paste, and cake batter. Some non-Newtonian
fluids exhibit a ‚memory‛—the shear stress depends not only on the local strain
rate, but also on its history. A fluid that returns (either fully or partially) to
its original shape after the applied stress is released is called viscoelastic.
Some non-Newtonian fluids are called shear thinning fluids or pseudo plastic fluids,
because the more the fluid is sheared, the less viscous it becomes. A good example
is paint. Paint is very viscous when poured from the can or when picked up by a
paintbrush, since the shear rate is small. However, as we apply the paint to the
wall, the thin layer of paint between the paintbrush and the wall is subjected to a
large shear rate, and it becomes much less viscous. Plastic fluids are those in
which the shear thinning effect is extreme. In some fluids a finite stress called
the yield stress is required before the fluid begins to flow at all; such fluids are
called Bingham plastic fluids. Certain pastes such as acne cream and toothpaste are
examples of Bingham plastic fluids. If you hold the tube upside down, the paste does
not flow, even though there is a nonzero stress due to gravity. However, if you
squeeze the tube (greatly increasing the stress), the paste flows like a very
viscous fluid. Other fluids show the opposite effect and are called shear thickening
𝝉 𝝁
𝒅𝒖
𝒅𝒚

fluids or dilatant fluids; the more the fluid is sheared, the more viscous it
becomes.
Non-Newtonian fluids follow the generalized law of viscosity expressed below but
when plotted they show much different behavior than Newtonian fluids
Where A ,B and n are constants to be found experimentally.
In this graph the Newtonian fluid is represent by a straight line, the slope of
which is μ . Some of the other fluids are:
 Plastic: Shear stress must reach a certain minimum before flow commences.
 Pseudo-plastic: No minimum shear stress necessary and the viscosity decreases
with rate of shear, e.g. substances like clay, milk and cement.
 Dilatant substances; Viscosity increases with rate of shear, e.g. quicksand.
 Viscoelastic materials: Similar to Newtonian but if there is a sudden large
change in shear they behave like plastic.
 Solids: Real solids do have a slight change of shear strain with time, whereas
ideal solids (those we idealize for our theories) do not.
Bingham plastics: Bingham plastics are those materials that have a linear shear
stress vs shear rate characteristics. These have to overcome a threshold value of
shear stress before they begin to flow. As long as the shear stress being applied
lies below the threshold value for that material, it retains its original solid
state.
4. PROPERTIES OF FLUID
Property is any characteristic of a system. Some familiar properties are:

Pressure P
Temperature T
Volume V
Mass m.
The list can be extended to include less familiar ones such as:
Viscosity,
Thermal conductivity,
Modulus of elasticity,
Thermal expansion coefficient,
Electric resistivity,
Velocity
Elevation.
PROPERTIES are considered to be either INTENSIVE or EXTENSIVE.
Intensive properties are those that are independent of the mass of a system, such as
Temperature
Pressure
Density
Extensive properties are those whose values depend on the size—or extent—of the
system.
Total mass
total volume
total momentum
The state of a system is described by its properties. In order to understand
mechanics of any fluid, it is important to understand its properties first:-
1. Viscosity
2. Density
3. Specific weight
4. Surface tension
5. Vapor pressure
6. Compressibility
7. Specific gravity
VISCOSITY
Viscosity, , is the property of a fluid, due to cohesion and interaction between
molecules, which offers resistance to shear deformation. Different fluids deform
at different rates under the same shear stress. The ease with which a fluid pours
is an indication of its viscosity. Fluid with a high viscosity such as syrup
deforms more slowly than fluid with a low viscosity such as water. The viscosity
is also known as dynamic viscosity.
Units: N.s/m2
or kg/m/s
Typical values:
Water = 1.14x10-3
kg/m/s; Air = 1.78x10-5
kg/m/s
KINEMATIC VISCOSITY, 
Definition: is the ratio of the viscosity to the density;

Will be found to be important in cases in which significant viscous and
gravitational forces exist.
Units: m2
/s
Typical values:
Water = 1.14x10-6
m2
/s; Air = 1.46x10-5
m2
/s;
In general,
viscosity of liquids decrease with increase in temperature, whereas
viscosity of gases increases with decrease in temperature.
DENSITY
Density of a fluid, , Defined as mass per unit volume, it is slightly affected by
changes in temperature and pressure.  = mass/volume = m/
Units: kg/m3
Typical values:
Water = 1000 kg/m3
; Air = 1.293 kg/m3
QUIZ
Densities of gases are variable, but for most liquid densities are constant (why?)
Expected answer bro! deus.
‚ Density is NOT constant, but for liquids and solids it doesn't change very much
with temperature or pressure. Most gases obey (at least approximately) the ideal gas
law: PV = nRT , which can be rearranged to show that the density of a gas is
proportional to pressure and inversely proportional to temperature: P/RT = n/V.
Liquids and solids expand only slightly with increasing temperature, and they
compress only slightly with increasing pressure, so their densities are
approximately constant under most ordinary conditions‛.
Any other scientific answer in your brain bro ,deus ? now click HERE to verify your
answer.
SPECIFIC WEIGHT
Specific weight of a fluid, 

• Definition: weight of the fluid per unit volume
• Arising from the existence of a gravitational force
• The relationship  and g can be found using the following:
Since  = m/
therefore  = g
Units: N/m3
Typical values:
Water = 9814 N/m3
; Air = 12.07 N/m3
QUESTIONS
1. Fluid A and B has specific weight of 7000N/m3
and 9000N/m3
respectively, which
fluid is heavier? Explain
2. Specific weight of fluid is 8.2KN/m3
calculate its density (Answer =836kg/m3
)
(usiangalizie mzee baba deus , chemsha kichwa) click HERE to translate in English.
SPECIFIC GRAVITY
The specific gravity (or relative density) can be defined in two ways:
Definition 1: A ratio of the density of a substance to the density of water at
standard temperature (4C) and atmospheric pressure, or
Definition 2: A ratio of the specific weight of a substance to the specific weight
of water at standard temperature (4C) and atmospheric pressure.
Unit: dimensionless.
Try this example bro!
A reservoir of oil has a mass of 825 kg.The reservoir has a volume of 0.917 m3
.
Compute the density, specific weight, and specific gravity of the oil.
Solution:
BULK MODULUS
All fluids are compressible under the application of an external force and when
the force is removed they expand back to their original volume.The
compressibility of a fluid is expressed by its bulk modulus of elasticity, K,
which describes the variation of volume with change of pressure, i.e.
Cw
s
Cw
s
SG


4@4@ 




Thus, if the pressure intensity of a volume of fluid, , is increased by Δp and
the volume is changed by Δ, then
,
Typical values: Water = 2.05x109
N/m2
; Oil = 1.62x109
N/m2
VAPOR PRESSURE
A liquid in a closed container is subjected to a partial vapor pressure in the
space above the liquid due to the escaping molecules from the surface; It reaches
a stage of equilibrium when this pressure reaches saturated vapor pressure.
Since this depends upon molecular activity, which is a function of temperature,
the vapor pressure of a fluid also depends on its temperature and increases with
it. If the pressure above a liquid reaches the vapor pressure of the liquid,
boiling occurs; for example if the pressure is reduced sufficiently boiling may
occur at room temperature.
Engineering significance of vapor pressure
In a closed hydraulic system, Ex. in pipelines or pumps, water vaporizes
rapidly in regions where the pressure drops below the vapor pressure.
There will be local boiling and a cloud of vapor bubbles will form.
This phenomenon is known as cavitations, and can cause serious problems, since
the flow of fluid can sweep this cloud of bubbles on into an area of higher
pressure where the bubbles will collapse suddenly.
If this should occur in contact with a solid surface, very serious damage can
result due to the very large force with which the liquid hits the surface.
Cavitations can affect the performance of hydraulic machinery such as pumps,
turbines and propellers, and the impact of collapsing bubbles can cause local
erosion of metal surface.
Cavitations in a closed hydraulic system can be avoided by maintaining the
pressure above the vapor pressure everywhere in the system.
Quiz , mzee baba Deus.
In the space provided explain how does a pressure cooker works?

Hints: “The steam and water will both increase in temperature and pressure‛
Click HERE to verify your answer Bro! Deus
Wow!!!! bro , Tell your friend Masika that now you know what is VAPOR PRESSURE take
time for SURFACE TENSION next page.
SURFACE TENSION
Liquids possess the properties of cohesion and adhesion due to molecular
attraction. Due to the property of cohesion, liquids can resist small tensile
forces at the interface between the liquid and air, known as surface tension, .
Surface tension is defined as force per unit length, and its unit is N/m. The
reason for the existence of this force arises from intermolecular attraction. In
the body of the liquid (Figure a), a molecule is surrounded by other molecules
and intermolecular forces are symmetrical and in equilibrium.
At the surface of the liquid (Figure b), a molecule has this force acting only
through 180. This imbalance forces means that the molecules at the surface tend
to be drawn together, and they act rather like a very thin membrane under
tension. This causes a slight deformation at the surface of the liquid (the
meniscus effect).
A steel needle floating on water, the spherical shape of dewdrops, and the rise
or fall of liquid in capillary tubes is the results of the surface tension.
Surface tension is usually very small compared with other forces in fluid flows
(e.g. surface tension for water at 20C is 0.0728 N/m).
Surface tension,, increases the pressure within a droplet of liquid.The internal
pressure, P, balancing the surface tensional force of a spherical droplet of
radius r, is given by
ANSWER

The above figure showing the force acting on one-half of a liquid drop.
CAPILLARITY
The surface tension leads to the phenomenon known as capillarity, where a column
of liquid in a tube is supported in the absence of an externally applied
pressure.
Rise or fall of a liquid in a capillary tube is caused by surface tension and
depends on the relative magnitude of cohesion of the liquid and the adhesion of
the liquid to the walls of the containing vessels.Liquid rise in tubes if they
wet a surface (adhesion > cohesion), such as water, and fall in tubes that do not
wet (cohesion > adhesion), such as mercury.
Capillarity is important when using tubes smaller than 10 mm (3/8 in.).For tube
larger than 12 mm (1/2 in.) capillarity effects are negligible.
where h = height of capillary rise (or depression)
 = surface tension
 = wetting (contact) angle
 = specific weight of liquid
r = radius of tube
Worry NOT bro! deus, see example below…………………………..

Water has a surface tension of 0.4 N/m. In a 3-mm diameter vertical tube, if the
liquid rises 6 mm above the liquid outside the tube, calculate the wetting angle.
Solution
Capillary rise due to surface tension is given by;
 = 83.7
Try this simple question below bro! Deus…..the ANSWER is 14.8mm
Find the capillary rise in the tube shown in figure (next page), the air- water-
glass interface angle is 0 and tube radius is 1mm. Given that the surface tension
of water is 0.0728N/m .
COMPRESSIBILITY OF LIQUIDS
The compressibility (change in volume due to change in pressure) of a liquid is
inversely proportional to its volume modulus of elasticity, also known as the bulk
modulus. This modulus is defined as
( )
Where v = specific volume and p = pressure. As v/dv is a dimensionless ratio, the
units of Ev and p are identical. The bulk modulus is analogous to the modulus of
elasticity for solids; however, for fluids it is defined on a volume basis rather
than in terms of the familiar one-dimensional stress–strain relation for solid
bodies.
QN you must know bro!

If Ev is large, the compressibility is higher or low?
GENERALLY , its known that the Large values for the bulk modulus indicate that
the fluid is relatively incompressible therefore Need higher pressure to change
small volume hence In most cases liquid are considered as incompressible.
HEAVY DUTY QUIZ bro,Deus! ( solve by yourself)- answer 200MPa
A liquid compressed in a cylinder has a volume of 1000cm3
at 1MN/m2
and volume of
995cm3
at 2MN/m2
. What is bulk modulus of elasticity?
OVER ALL SIMPLE QUESTIONS YOU MUST SOLVE BRO, DEUS!
1. At a depth of 8 km in the ocean the pressure is 81.8 MPa. Assume that the
specific weight of seawater at the surface is 10.05 kN/m3
and that the average
volume modulus is 2.34 109 N/m2
for that pressure range.
(a) What will be the change in specific volume between that at the
surface and at that depth?
(b) What will be the specific volume at that depth?
(c) What will be the specific weight at that depth?
Solution (see next page bro!)
2. A rigid cylinder, inside diameter 15 mm, contains a column of water 500 mm long.
What will the column length be if a force of 2 kN is applied to its end by a
frictionless plunger? Assume no leakage. (home work broo!)
3. A 1-in-wide space between two horizontal plane surfaces is filled with SAE 30
Western lubricating oil at 80°F. What force is required to drag a very thin
plate of 4-ft2
area through the oil at a velocity of 20 ft/min if the plate is
0.33 in from one surface?
solution
(a) 1/𝝆 𝟏 g/𝜸 𝟏
𝟗.𝟖𝟏
𝟏𝟎𝟎𝟓𝟎
𝟎. 𝟎𝟎𝟎𝟗𝟕𝟔 𝒎 𝟑
/𝒌𝒈
𝒗
𝟎.𝟎𝟎𝟎𝟗𝟕𝟔 𝟖𝟏.𝟖 𝟏𝟎 𝟔
𝟐.𝟑𝟒 𝟏𝟎 𝟗
= -34.1 x 10E-6m3
/kg ans.
(b) 𝒗 𝟐 𝒗 𝟏 𝒗 0.000 942m3
/kg ans.
(c) 𝜸 𝟐
𝒈
𝒗 𝟐
𝟗.𝟖𝟏
𝟎.𝟎𝟎𝟎𝟗𝟒𝟐
𝟏𝟎𝟒𝟏𝟎𝑵
𝒎 𝟑 ans

4. Water at 10°C stands in a clean glass tube of 2-mm diameter at a height of 35
mm. What is the true static height? [ ans 35.00 - 15.14 = 19.86 mm]
5. Tap water at 68°F stands in a glass tube of 0.32-in diameter at a height of
4.50 in.What is the true static height?
6. Distilled water at 20°C stands in a glass tube of 6.0-mm diameter at a height
of 18.0 mm. What is the true static height?
7. Compute the capillary depression of mercury at 68°F(ϴ = 140°) to be expected in
a 0.05-in-diameter tube.
8. Compute the capillary rise in mm of pure water at 10°C expected in an 0.8-mm
diameter tube.
9. Compute the capillary rise of water to be expected in a 0.28-in-diameter tube.
Assume pure water at 68°F.
10. Consider water initially at 20°C and 1 atm. Determine the final density of
water (a) if it is heated to 50°C at a constant pressure of 1 atm, and (b) if it
is compressed to 100-atm pressure at a constant temperature of 20°C. Take the
isothermal compressibility of water to be a = 4.80 x105
atm-1
.
Ans. (988.0 kg/m3 , 1002.7 kg/m3)) this question gives you general knowledge
Do you remember AE 213 /BPE 211 ?
LIQUIDS VS. GASSES
Although liquids and gasses behave in much the same way and share many similar
characteristics, they also possess distinct characteristics of their own.
Specifically
1. A liquid is difficult to compress and often regarded as being incompressible. A
gas is easily to compress and usually treated as such - it changes volume with
pressure.
2. A given mass of liquid occupies a given volume and will occupy the container it
is in and form a free surface (if the container is of a larger volume). A gas
has no fixed volume, it changes volume to expand to fill the containing vessel.
It will completely fill the vessel so no free surface is formed.
SYSTEM AND CONTROL VOLUME

A system refers to a fixed, identifiable quantity of mass which is separated from
its surrounding by its boundaries. The boundary surface may vary with time however
no mass crosses the system boundary. In fluid mechanics an infinitesimal lump of
fluid is considered as a system and is referred as a fluid element or a particle.
Since a fluid particle has larger dimension than the limiting volume (refer to
section fluid as a continuum). The continuum concept for the flow analysis is valid.
control volume is a fixed, identifiable region in space through which fluid flows.
The boundary of the control volume is called control surface. The fluid mass in a
control volume may vary with time. The shape and size of the control volume may be
arbitrary.
HYDROSTATIC FLUIDS
Pressure is defined as a normal force exerted by a fluid per unit area. We speak of
pressure only when we deal with a gas or a liquid. The counterpart of pressure in
solids is normal stress. Since pressure is defined as force per unit area, it has
the unit of newtons per square meter (N/m2), which is called a pascal (Pa).
That is, 1Pa = 1N/M2
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the
force of gravity. Is the branch of hydraulics that deals with pressure and forces of
fluid at rest. The pressure at a point on a plane surface always acts normal to the
surface because there is no shear stress in fluid at rest Always pressure is due to
weight of fluid.
PRESSURE AT A POINT
Pressure is the compressive force per unit area, and it gives the impression of
being a vector. However, pressure at any point in a fluid is the same in all
directions. That is, it has magnitude but not a specific direction, and thus it is a
scalar quantity
PASCAL'S LAW
The Pascal's law states that the pressure at a point in a fluid at rest is the same
in all directions. Simply pressure in a fluid at rest is the same at all points. In
other words a liquid exerts pressure equally in all directions.The law has several
application like Heavy load lifter and Car brakes
HYDRAULIC LIFT (HYDRAULIC PRESS)

Lifting a car easily by one arm, as shown in Figure below. Noting that P1 = P2 since
both pistons are at the same level (the effect of small height differences is
negligible, especially at high pressures), the ratio of output force to input force
is determined to be
The area ratio A2 /A1 is called the ideal mechanical advantage of the hydraulic
lift. Using a hydraulic car jack with a piston area ratio of A2 /A1 = 10,
For example,
A person can lift a 1000-kg car by applying a force of just 100 kg (= 908 N).
Solve this example below using the above principle bro! deus (Answer 1.5x103
N )
A car is lifted by compressed fluid due to force F1 on a small piston having a
small radius of 5 cm. This pressure is transmitted to a second piston of
radius of 15cm. Assume the mass of a car to be lifted is 1.35tons. What is F1?
PRESSURE VARIATION WITH DEPTH
Usually pressure increase with the increase of depth (Linearly for incompressible
fluid). If the liquid is in equilibrium (at rest), then the pressure at all point in
the same level (depth) must be equal , example bro deus! as you go deeper in
swimming pool, the more pressure on you because…
A small force of
1500N is usedt to
lift 1.35T

Absolute pressure (P2) is greater than the atmospheric pressure (P1) by an amount
ρgh
Huhu! Why worrying bro! deus to try this example below? (Answer 1.99x105
Pa ~ 2atm)
QN! What is the pressure on a swimmer 10m below the surface of a lake?
Bro can you evaluate? This is almost twice the pressure on surface!!!
Stop here bro! and note down the following in real life.
At a depth of 1km the pressure is (100atm)….. 100 times the
pressure on the surface. Hence it is dangerous for swimmers
because……
( Boyle’s law)-do you this law? If no click HERE if yes continue
next page bro! Deus

If you hold your breath on ascent, your lung volume would increase
by a factor of 100, which may cause embolism and/or death.
Submarines are designed to withstand this kind of pressure, it can
operate at 1km depth.
HYDROSTATIC PRESSURE DIFFERENCE BETWEEN TWO POINTS
Pressure in a fluid at rest is independent of the shape or cross section of the
container. It changes with the vertical distance, but remains constant in other
directions. Therefore, the pressure is the same at all points on a horizontal plane
in a given fluid. The Dutch mathematician Simon Stevin (1548–1620) published in 1586
the principle illustrated in figure below. Note that the pressures at points A, B,
C, D, E, F, and G are the same since they are at the same depth, and they are
interconnected by the same static fluid. However, the pressures at points H and I
are not the same since these two points cannot be interconnected by the same fluid
(i.e., we cannot draw a curve from point I to point H while remaining in the same
fluid at all times), although they are at the same depth. (Can you tell at which
point the pressure is higher?) Also, the pressure force exerted by the fluid is
always normal to the surface at the specified points.
A consequence of the pressure in a fluid remaining constant in the horizontal
direction is that the pressure applied to a confined fluid increases the pressure
throughout by the same amount
QUIZ
Does the shape of container matters in hydrostatic pressure? IF NO ,Pressure at
point 1, 2 & are the same regardless the shape of containers (why??)
Hahaha!! see figures below.
Pressure in layered fluid.
 If fluids of different densities are layered, then hydrostatics equation must
be used twice, once in each of the liquid

PRESSURE MEASUREMENT
BAROMETERS
The first mercury barometer was constructed in 1643-1644 by Torricelli. He
showed that the height of mercury in a column was 1/14 that of a water
barometer, due to the fact that mercury is 14 times more dense that water. He
also noticed that level of mercury varied from day to day due to weather
changes, and that at the top of the column there is a vacuum.
MANOMETRY
Manometry is a standard technique for measuring pressure using liquid columns in
vertical or include tubes. The devices used in this manner are known as
manometers.
The operation of three types of manometers are here for you bro!:
𝑷 𝟏 𝑷 𝒂𝒕𝒎 𝝆 𝒘𝒂𝒕𝒆𝒓 𝐠 𝒁 𝟏
𝑷 𝟐 𝑷 𝒂𝒕𝒎 𝝆 𝒘𝒂𝒕𝒆𝒓 𝐠 𝒁 𝟏 𝑷 𝒎𝒆𝒓𝒄𝒖𝒓𝒚 𝐠 𝒁 𝟐
Note, often pvapor
is very
small, 0.0000231 psia at
68° F, and patm
is 14.7 psi,
thus:

1) The Piezometer Tube
2) The U-Tube Manometer
3) The Inclined Tube Manometer
The fundamental equation for manometers since they involve columns of fluid at
rest is the following:
h is positive moving downward, and negative moving upward, that is pressure in
columns of fluid decrease with gains in height, and increase with gain in depth.
PIEZOMETER TUBE
Piezometer is simple and accurate. Piezometer is only suitable if the pressure in
the container is greater than atmospheric pressure. Fluid in the container in which
the pressure is measured must be a liquid rather than gas.
Disadvantages:
1)The pressure in the container has to be greater than atmospheric pressure.
2) Pressure must be relatively small to maintain a small column of fluid.
3) The measurement of pressure must be of a liquid.
U-TUBE MANOMETER
Note: pA = p1
because they are
at the same level

Note: in the same fluid we can ‚jump‛ across from 2 to 3 as they are at the sam
level, and thus must have the same pressure.
The fluid in the U-tube is known as the gage fluid. The gage fluid type depends on
the application, i.e. pressures attained, and whether the fluid measured is a gas or
liquid.
Final notes to know bro Deus!:
1)Common age fluids are Hg and Water, some oils, and must be immiscible.
2)Temperature must be considered in very accurate measurements, as the gage fluid
properties can change.
3) Capillarity can play a role, but in many cases each meniscus will cancel.
Then the equation for the pressure difference in the container is the following:
Helo! Bro because you know the U-tube manometer with its formula
please try the following unsolved examples next page>>>hahahahaa!
1. A manometer is used to measure the pressure in tank as
shown. If the local atmospheric pressure is 96 kPa,
determine the absolute pressure within the tank.
2. A closed tank contains compressed air and oil (SGoil=0.90)
as Is shown in the Figure. h1=36 in., h2=6 in, h3=9 in.
Determine the pressure reading (in psi) of the gage using
the U-tube Hg manometer
If the fluid in the container is a gas, then
the fluid 1 terms can be ignored:

DIFFERENTIAL U-TUBE MANOMETER.
This suitable to measure the pressure difference between two
points.
>>>>
INCLINED-TUBE MANOMETER
This type of manometer is used to measure small pressure changes.

Thus, for the length of the tube we can measure a greater
pressure differential.
QUIZ 1
QUIZ 2
The fuel gauge in the gasoline tank in a car reads proportional to the bottom gauge
pressure as in figure below.If the tank is 30cm deep and accidentally contains 2cm
of water plus gasoline,how many centimeters of air remain at the top when the gauge
erroneously reads ‚FULL‛?
2
2
sin
l
h
 sin22 lh 

Bro Deus! use the following HINTs below to solve quiz 2
QUIZ 3
(offer with final answer h = 0.487m =48.7cm)
The gauge pressure of the air in the tank shown in the figure is measured to be
65kPa,Detemine the differential height h of the mercury column
REVISION EXERCISES BRO TO MAKE YOU PERFECT
1. In figure below pressure gage A reads 1.5 kPa (gage). The fluids are at 20o
C.
Determine the elevations z, in meters, of the liquid levels in the open
piezometer tubes B and C (p2.11.frank m white 7ed).

2. A closed tank contains 1.5 m of SAE 30 oil, 1 m of water, 20 cm of mercury, and
an air space on top, all at 20o
C. The absolute pressure at the bottom of the
tank is 60 kPa. What is the pressure in the air space? (p2.10)
3. For the three-liquid system shown, compute h1 and h2. Neglect the air density
(p2.14).
4. The U-tube in Figure below has a 1-cm ID and contains mercury as shown. If 20
cm3
of water is poured into the right hand leg, what will the free-surface
height in each leg be after the sloshing has died down? (p2.19)
5. At 20o
C gage A reads 350 kPa absolute. What is the height h of the water in cm?
What should gage B read in kPa absolute?(p2.21)

6. In Figure below all fluids are at 20o
C. Determine the pressure difference (Pa)
between points A and B. (p2.31)
7. For the inverted manometer of Figure below, all fluids are at 20o
C. If pB - pA =
97 kPa, what must the height H be in cm? (p2.32)
8. In Figure below the pressure at point A is 25 lbf/in2
. All fluids are at 20o
C.
What is the air pressure in the closed chamber B, in Pa? (p2.33)

9. Water flows upward in a pipe slanted at 30o
C, as in Figure below. The mercury
manometer reads h = 12 cm. Both fluids are at 20o
C. What is the pressure
difference p1 - p2 in the pipe? (p2.35)
10. In Figure below both the tank and the tube are open to the atmosphere. If
L = 2.13 m, what is the angle of tilt ϴ of the tube? (p2.36)
11. If the pressure in container A in Figure below is 150 kPa, compute the
pressure in container B.(p2.38)
12. In Figure below , determine the gage pressure at point A in Pa. Is it higher
or lower than atmospheric?

SOKOINE UNIVERSITY OF AGRICULTURE
COLLEGE OF AGRICULTURE
DEPARTMENT OF ENGINEERING SCIENCES AND TECHNOLOGY
IWRE 210- FLUID MECHANICS
INSTRUCTOR: DR MBUNGU
DATE OF SUBMISSION: 17TH
MAY 2018
Honor code pledge.
“As an Engineering Student, I will conduct myself with honor and integrity at all times.
I will not lie, cheat, or steal, nor will I accept the actions of those who do.”
I have neither given nor received unauthorized assistance on this assignment.
Your signature
-----------------------------------------------------
DOTO, MUSA GESE IWR/D/2016/0011
QUESTIONS WITH SOLUTIONS
1. Oil of specific gravity 0.750 flows through the nozzle shown in Fig. 2-10 and
deflects the mercury in the U-tube gage. Determine the value of h if the
pressure at A is 20.0 psi.

SOLUTION
GIVEN
Specific gravity of oil ( ) = 0.750
Pressure at A (PA) = 20.0 Psi
Asked height (h)
Consider the figure 2-10 above , Pressure at B = pressure at C
.
. . .
=
. .
. . . .
.
.
.
Therefore the value of h if pressure at A is 20.0 Psi is .
2. For a gage pressure at A of 10.89 kPa, find the specific gravity of the gage
liquid B in Fig. 2-11.
SOLUTION
GIVEN
Gauge pressure at A (Pgauge) = -10.89 KPa
Asked the specific gravity of gauge liquid in B
From figure 2-11 above, Pressure at C = Pressure at D
PA + = PD
-10.89 +(1.60 x9.79)(3.200 – 2.743) = - 3.73 KPa = PD
The weight of air can be neglected without any introduction of significant errors
Then PD = PG - 3.73 KPa and PE = PF = 0
Pressure at G = Pressure at E = P of (3.429-3.028) m of
gauge liquid.
PG = PE – (sp gr x 9.79)(3.429 – 3.028)
-3.73 = 0 – (0.382)(9.79 sp gr) ; sp gr =1.000002681
Mercuryoil

The specific gauge of liquid B is ≈ 1.00
3. For a gage reading at A of —2.50 psi, determine (a) the elevations of the
liquids in the open piezometer columns E, F, and G and (b) the deflection of
the mercury in the U-tube gage in Fig, 2-12.
SOLUTION
DATA GIVEN
Pressure of gauge A (P) = - 2.50 Psi
Specific gravity of Air = 0.700
Specific gravity of water = 1.600
Elevations :
At H = 49.00ft At K = 38.00ft Of water = 26.00ft
At h1 = 14.00ft
(a) Elevations of liquids in the open piezometer column E,F and G.
Since the unit weight of the air is very small compared with that of
the liquids, the pressure at elevation 49.00 may be considered to be
-2.50 psi without introducing significant error in the calculations
For column E
Pressure at K = pressure at L
PK = PL : PH + = 0
Then -2.50 X 144 + (0.700 X 62.4)h = 0
h = 8.24ft
E = H – h = (49.00 – 8.24)ft
Elevation at E = 40.76 ft
For column F

P 38.00 = P 49.00 + P 11.00 ,sg of 0.700
.
. . . .
= 0.837 Psi
Pressure at M =
.
.
= 1.93 ft of water
Elevation at F = 38.00 + 1.93 = 39.93ft (will rise to N)
For column G
Pressure at 26.00 = pressure at 38.00 + pressure of 12 ft of water
= .
.
=1.93 ft of water
Pressure at R =
.
. .
=8.71ft
Elevation at G = 26.00 + 8.71 = 34.71 ft (will be at Q).
(b) Deflection of mercury
Pressure head at D = pressure head at C
(13.57)h = PH38.00 + PH 24.00 (water)
(13.57)h = 1.93 + 24.00
h = 1.9108 ft
4. Find the pressure difference between A and B for the setup shown in Fig. 2-
17.
SOLUTION
DATA GIVEN
Specific gravity of oil = 0.8
Asked Pressure difference between A and B
Then
PB = PA –(9.79)Y – (0.8 x 9.79 x 0.70) + [(9.79(Y-0.80) ]
PB = PA -13.3144 KPa
PA – PB = 13.3144 KPa
The pressure difference between A and B is 13.3144KPa

5. A glass U-tube open to the atmosphere at both ends is shown in Fig. 2-19. If
the U-tube contains oil and water as shown, determine the specific gravity of
the oil.
SOLUTION
DATA GIVEN
Depth of Oil (hO) = 0.35m
Depth of water (hW) = 0.30m
Asked the specific gravity of Oil (sp gr )
There is no change in pressure between Oil and Water
Then PO – PW =0 :
(sp gr x 9.79 x hO) – 9.79 hW =0
(sp gr x 9.79 x 0.35) – 9.79 x 0.30 =0
Sp gr = 0.841716103
The specific gravity of Oil is ≈ 0.842
6. A differential manometer is attached to two tanks as shown in Fig. 2-18.
Calculate the pressure difference between chamber A and chamber B.
SOLUTION
DATA GIVEN
Depth of chamber A (SAE 30 Oil) = 1.1 m

Depth of chamber B (Carbon Tetrachloride) = 0.8 m
Required the Pressure difference between chamber A and B
Recalling PA – PB = P carbon - P mercury - PSAE
= 𝑚 𝑚
= [(1.59 x 9.79 x 0.8 ) –(13.6 x 9.79 x 0.3)- (0.89 x 9.79 x 1.1)]kpa
PA – PB = -37.07473 KPa
The Pressure differences is -37.07473 KPa
CONTROL VOLUME
Isaac Newton proposed the following three laws of motion:
1. A body in motion continues to stay in motion unless acted upon by a net external
force.
2. The net force on the body is equal to the mass times the acceleration.
3. When a body exerts a force on another body, the other body exerts an equal and
opposite force.
Clearly, these laws of mechanics and other related laws of conservation such as
conservation of mass and angular momentum (also one can include energy and electric
charge conservation) are all strongly bounded to the material/body under
investigation. Therefore, the foremost thing to note while doing a control volume
analysis is that the laws of mechanics have little to do with the choice of the
fictitious dashed border that one draws to identify the control volume. It is the
material contained in it that holds the conserved quantity. If the control volume
confines the same material at all times, then the laws of mechanics would not get
altered. However, if there is flux of material in and/or out of the chosen control
volume, then we need to modify the laws of mechanics to correctly identify the rate
of change of the conserved quantity associated with the material inside the control
volume at a given time.
Control volume is the same as free body diagrams that is used in engineering
mechanics
Is used to study fluid in motion.
LAWS OF MECHANICS
1. Law of conservation of mass
Mass can neither be created nor destroyed Within control volume, the net mass
must remain constant such that Mass in = Mass out

Example
2. Law of conservation of energy
Energy can neither be created nor destroyed.Energy flowing into control
volume is equal to energy flowing out of it. The law is the basis for
the derivation of Bernoulli equations.
Net work done
Net Kinetic energy
Net potential energy

Work done= kinetic energy + potent Energy
This is Bernoulli equation.
HYDRAULIC GRADE LINE (HGL) AND ENERGY LINE (EL)
A useful visual interpretation of Bernoulli’s equation is to sketch two grade lines
of a flow.
The energy grade line (EGL) shows the height of the total Bernoulli constant h0 = z
+ p/ = V2
/(2g). In frictionless flow with no work or heat transfer, the EGL has
constant height.
The hydraulic grade line (HGL) shows the height corresponding to elevation and
pressure head z = p/ that is, the EGL minus the velocity head V2
/(2g). The HGL is
the height to which liquid would rise in a piezometer tube attached to the flow. In
an open-channel flow the HGL is identical to the free surface of the water.
Figure below illustrates the EGL and HGL for frictionless flow at sections 1 and 2
of a duct. The piezometer tubes measure the static pressure head z = p/ and thus
outline the HGL. The pitot stagnation-velocity tubes measure the total head z = p/
= V2
/(2g), which corresponds to the EGL. In this particular case the EGL is
constant, and the HGL rises due to a drop in velocity.
In more general flow conditions, the EGL will drop slowly due to friction losses and
will drop sharply due to a substantial loss (a valve or obstruction) or due to work
extraction (to a turbine). The EGL can rise only if there is work addition (as from
a pump or propeller). The HGL generally follows the behavior of the EGL with respect
to losses or work transfer, and it rises and/or falls if the velocity decreases
and/or increases. As mentioned before, no conversion factors are needed in
computations with the Bernoulli equation if consistent SI or BG units are used, as
the following examples will show.
In all Bernoulli-type problems in this text, we consistently take point 1 upstream
and point 2 downstream.

NOTE BRO! Velocity head changes due to change in fluid velocity caused by change in
diameter of flow As velocity head change, HGL also shifts
 At the pipe outlet, the pressure head is zero so that the pipe elevation and
hydraulic grade line coincide….
MOMENTUM ANALYSIS OF FLOW SYSTEMS
For a rigid body of mass m, Newton’s second law is expressed as

Therefore, Newton’s second law can also be stated as the rate of change of the
momentum of a body is equal to the net force acting on the body.
The product of the mass and the velocity of a body is called the linear momentum.
Newton’s second law  the linear momentum equation in fluid mechanics
The momentum of a system is conserved when it remains constant  the conservation
of momentum principle.
Momentum is a vector. Its direction is the direction of velocity.
Momentum =
Consider a stream tube and assume steady non-uniform flow
LINEAR MOMENTUM EQUATION
Newton’s second law for a system of mass m subjected to a force F is expressed as

During steady flow, the amount of momentum within the control volume remains
constant. The net force acting on the control volume during steady flow is equal to
the difference between the rates of outgoing and incoming momentum flows.
DONT SKIP THE FOLLOWING BRO DEUS!
In time δt a volume of the fluid moves from the inlet
at a distance v1δt, so
volume entering the stream tube = area x distance
= A1 x v1δt
The mass entering,
mass entering stream tube = volume x density
= ρ1A1v1 δt
And momentum
momentum entering stream tube = mass velocity
= ρ1A1v1 δt v1
Similarly, at the exit, we get the expression:
momentum leaving stream tube = ρ2A2v2 δt v2
By Newton 2nd law
Force = rate of change of momentum
F = (ρ2A2v2 δt v2 - ρ1A1v1 δt v1)/ δt
We know from continuity that
Q= A1v1 = A2v2
And if we have fluid of constant density, ρ1 = ρ2 = ρ, then
F = Qρ (v2-v1)
An alternative derivation
From conservation of mass
mass into face 1 = mass out of face 2
we can write
rate of change of mass = m= dm/dt
= ρ1A1v1 = ρ2A2v2
The rate at which momentum enters face 1 is ρ1A1v1 v1 = mv1
The rate at which momentum leaves face 2 is ρ2A2v2 v2 = mv2
Thus the rate at which momentum changes across the stream tube is
ρ2A2v2 v2 - ρ1A1v1 v1 = mv2 - mv1
Force = rate of change of momentum
F = m(v2-v1)

So, we know these two expression. Either one is known as momentum equation:
F = m(v2-v1)
F = Qρ (v2-v1)
The momentum equation:This force acts on the fluid in the direction of the flow of
the fluid
HYDRODYNAMICS
FLOW CONCEPT
Fluid flow in circular and noncircular pipes is commonly encountered inpractice. The
hot and cold water that we use in our homes is pumped through pipes. Water in a city
is distributed by extensive piping networks. Oil and natural gas are transported
hundreds of miles by large pipelines. Blood is carried throughout our bodies by
arteries and veins. The cooling water in an engine is transported by hoses to the
pipes in the radiator where it is cooled as it flows. Thermal energy in a hydronic
space heating system is transferred to the circulating water in the boiler, and then
it is transported to the desired locations through pipes.
Fluid flow is classified as external and internal, depending on whether the fluid is
forced to flow over a surface or in a conduit. Internal and external flows exhibit
very different characteristics. In this chapter we consider internal flow where the
conduit is completely filled with the fluid, and flow is driven primarily by a
pressure difference. This should not be confused with open-channel flow where the
conduit is partially filled by the fluid and thus the flow is partially bounded by
solid surfaces, as in an irrigation ditch, and flow is driven by gravity alone.
When the fluids in motion i.e. in pipe, the individuals molecules moves in velocity
of different magnitude and directions Depending on different factors, the molecules
may moves in straight line (streamline) or random manner. Because of this we have
different TYPES of fluid flows
LAMINAR AND TURBULENT FLOWS
If you have been around smokers, you probably noticed that the cigarette smoke
rises in a smooth plume for the first few centimeters and then starts fluctuating
randomly in all directions as it continues its rise. Other plumes behave
similarly.Likewise, a careful inspection of flow in a pipe reveals that the fluid
flow is streamlined at low velocities but turns chaotic as the velocity is
increased above a critical value, as shown in Fig. 8–4. The flow regime in the
first case is said to be laminar, characterized by smooth streamlines and highly
ordered motion, and turbulent in the second case, where it is characterized by
velocity fluctuations and highly disordered motion. The transition from laminar to
turbulent flow does not occur suddenly; rather, it occurs over some region in which
the flow fluctuates between laminar and turbulent flows before it becomes fully
turbulent. Most flows encountered in practice are turbulent. Laminar flow is
encountered when highly viscous fluids such as oils flow in small pipes or narrow
passages.We can verify the existence of these laminar, transitional, and turbulent
flow regimes by injecting some dye streaks into the flow in a glass pipe, as the
British engineer Osborne Reynolds (1842–1912) did over a century ago. We observe
that the dye streak forms a straight and smooth line at low velocities when the
flow is laminar (we may see some blurring because of molecular diffusion), has
bursts of fluctuations in the transitional regime, and zigzags rapidly and randomly
when the flow becomes fully turbulent. These zigzags and the dispersion of the dye
are indicative of the fluctuations in the main flow and the rapid mixing of fluid
particles from adjacent layers. The intense mixing of the fluid in turbulent flow
as a result of rapid fluctuations enhances momentum transfer between fluid
particles, which increases the friction force on the surface and thus the required

pumping power. The friction factor reaches a maximum when the flow becomes fully
turbulent
REYNOLDS NUMBER
The transition from laminar to turbulent flow depends on the geometry, surface
roughness, flow velocity, surface temperature, and type of fluid, among other
things. After exhaustive experiments in the 1880s, Osborne Reynolds discovered that
the flow regime depends mainly on the ratio of inertial forces to viscous forces in
the fluid. This ratio is called the Reynolds number and is expressed for internal
flow in a circular pipe as
Re
WHERE
Vavg = average flow velocity (m/s),
D = characteristic length of the geometry (diameter in this case, in m)
V = = kinematic viscosity of the fluid (m2
/s).
Note that bro Deus!
The Reynolds number is a dimensionless quantity. Also, kinematic viscosity has the
unit m2
/s, and can be viewed as viscous diffusivity or diffusivity for momentum.
After several experiments, he came up with expression and He concluded that
 Re < 2000 - the flow is Laminar

 Re > 4000 – the flow is Turbulent flow
 2000 < Re < 4000 – Transitional flow
For flow through noncircular pipes, the Reynolds number is based on the hydraulic
diameter Dh defined as
Dh
UNIFORM AND NON UNIFORM FLOW
1.Uniform flow
Occurs when flow characteristics do not change from point to point Flow
characteristics include velocity, discharges, cross section etc
WHERE C is flow characteristics & S is path
2.Non uniform flow
Occurs when flow characteristics change from point to point Most flow
are non-uniform because
C is flow characteristics & S is path
STEADY AND UN-STEADY FLOWS
3.Steady flow
Steady flow-flow characteristics at a given point do not change with
time…
4.Unsteady flow

Unsteady flow –flow characteristics at given point change with time
 Let’s say we’re not dealing with a system open to the atmosphere
(e.g., a pipe vs. a pond).There’s no storage potential, so Q1 = Q2, a
mass balance equation. For essentially incompressible fluids such as
water, the equation becomes
V1A1 = V2A2,; where V = velocity (m/s) and A = area (m2)
This Can be used to estimate flow velocity along a pipe, especially
where constrictions are concerned.
Self-check bro Deus!
Solve the following self-check question using the above concept
(answer V2 = 5m/s)
QN If one end of a pipe has a diameter of 0.1 m and a flow rate of 0.05
m/s, what will be the flow velocity at a constriction in the other end
having a diameter of 0.01 m? (hints: V1A1 = V2A2)
PIPE NETWORKS
‘Pipe flow’ generally refers to fluid in pipes and appurtenances
flowing full and under pressure .
Examples are Water distribution in homes, industry, cities; irrigation
SYSTEM COMPONENTS
1.Pipes
2.Valves
3.Bends
4.Pumps and turbines
5.Storage (often unpressurized, in reservoirs, tanks, etc.)

ENERGY RELATIONSHIPS IN PIPE SYSTEMS
Energy equation between any two points:
Analysis involves writing expressions for hL in each pipe and for each
link between pipes (valves, expansions, contractions), relating

velocities based on continuity equation, and solving subject to system
constraints (Q, p, or V at specific points).
MAJOR LOSES IN PIPE SYSTEMS
Source:
1.Frictions due to pipe material
2.Length of the system
3.Diameter/Cross-sectional Area
ENERGY LOSSES IN PIPING SYSTEMS
A quantity of interest in the analysis of pipe flow is the pressure drop
∆P since it is directly related to the power requirements of the fan or
pump to maintain flow. We note that dP/dx = constant, and integrating
from x = x1 where the pressure is P1 to x = x1 + L where the pressure is P2
gives
………………………….eqn1
Substituting Eqn1 into the Vavg expression (see below) the pressure drop
can be expressed as
Laminar flow
The symbol ∆ (is typically used to indicate the difference between the
final and initial values, like (∆y = y2 - y1). But in fluid flow, ∆P is
used to designate pressure drop, and thus it is P1 - P2. A pressure drop
due to viscous effects represents an irreversible pressure loss, and it
is called PRESSURE LOSS
∆PL to emphasize that it is a loss (just like the head loss hL, which is
proportional to it).
NOTE that, from the ∆P equation above the pressure drop is proportional
to the viscosity of the fluid, and ∆P would be zero if there were no
friction. Therefore, the drop of pressure from P1 to P2 in this case is
due entirely to viscous effects, and ∆P Eqn above represents the pressure
loss ∆PL when a fluid of viscosity flows through a pipe of constant
diameter D and length L at average velocity Vavg.
In practice, it is found convenient to express the pressure loss for all
types of fully developed internal flows (laminar or turbulent flows,
circular or noncircular pipes, smooth or rough surfaces, horizontal or
inclined pipes)
PRESSURE LOSS:
Where
is the dynamic pressure and f is the Darcy friction factor

IMPORTANT bro!
It is also called the Darcy–Weisbach friction factor, named after the
Frenchman Henry Darcy (1803–1858) and the German Julius Weisbach (1806–
1871),the two engineers who provided the greatest contribution in its
development. It should not be confused with the friction coefficient Cf
[also called the Fanning friction factor, named after the American
engineer John Fanning (1837–1911)], which is defined as
Cf = = f /4.
Setting P2 and ∆P eqns equal to each other and solving for f gives the
friction factor for fully developed laminar flow in a circular pipe,
Circular pipe, laminar
This equation shows that in laminar flow, the friction factor is a
function of the Reynolds number only and is independent of the roughness
of the pipe surface.
In the analysis of piping systems, pressure losses are commonly expressed
in terms of the equivalent fluid column height, called the HEAD LOSS hL.
Noting from fluid statics that ∆P = gh and thus a pressure difference of
∆P corresponds to a fluid height of h =∆P/ g, the pipe head loss is
obtained by dividing ∆PL by g to give
OR
Where
hL = Head loss due to friction, m [ft]
f = Moody friction factor
L = Pipe length, m [ft]
V = Velocity, m/s [ft/sec]
g = Gravitational acceleration, 9.81 m/sec2
[32.2 ft/sec2
]
D = Inside diameter, m [ft]
THE MOODY CHARTThe friction factor in fully developed turbulent pipe flow depends on the Reynolds
number and the relative roughness  /D. which is the ratio of the mean height of
roughness of the pipe to the pipe diameter. The functional form of this dependence
cannot be obtained from a theoretical analysis, and all available results are
obtained from painstaking experiments using artificially roughened surfaces (usually
by gluing sand grains of a known size on the inner surfaces of the pipes). Most such
experiments were conducted by Prandtl’s student J. Nikuradse in 1933, followed by

the works of others. The friction factor was calculated from the measurements of the
flow rate and the pressure drop.
The experimental results obtained are presented in tabular, graphical, and
functional forms obtained by curve-fitting experimental data. In 1939, Cyril F.
Colebrook (1910–1997) combined the available data for transition and turbulent flow
in smooth as well as rough pipes into the following implicit relation known as the
Colebrook equation:
We note that the logarithm in Equation above is a base 10 rather than a
natural logarithm. In 1942, the American engineer Hunter Rouse (1906–
1996) verified Colebrook’s equation and produced a graphical plot of f as
a function of Re and the product √ . He also presented the laminar flow
relation and a table of commercial pipe roughness.
Two years later, Lewis F. Moody (1880–1953) redrew Rouse’s diagram into
the form commonly used today. The now famous Moody chart is shown below.
It presents the Darcy friction factor for pipe flow as a function of the
Reynolds number and e/D over a wide range. It is probably one of the most
widely accepted and used charts in engineering. Although it is developed
for circular pipes, it can also be used for noncircular pipes by
replacing the diameter by the hydraulic diameter.

OBSERVATIONS FROM THE MOODY CHART
• For laminar flow, the friction factor decreases with increasing
Reynolds number, and it is independent of surface roughness.
• The friction factor is a minimum for a smooth pipe and increases with
roughness. The Colebrook equation in this case ( = 0) reduces to the
Prandtl equation.
• The transition region from the laminar to turbulent regime is
indicated by the shaded area in the Moody chart. At small relative
roughnesses, the friction factor increases in the transition region
and approaches the value for smooth pipes.
• At very large Reynolds numbers (to the right of the dashed line on
the Moody chart) the friction factor curves corresponding to
specified relative roughness curves are nearly horizontal, and thus
the friction factors are independent of the Reynolds number. The flow
in that region is called fully rough turbulent flow or just fully
rough flow because the thickness of the viscous sublayer decreases
with increasing Reynolds number, and it becomes so thin that it is
negligibly small compared to the surface roughness height. The
Colebrook equation in the fully rough zone reduces to the von Kármán
equation.
Note Bro!
At very large Reynolds numbers, the friction factor curves on the
Moody chart are nearly horizontal, and thus the friction factors are
independent of the Reynolds number.

IMPORTANT NOTE BRO!
In calculations, we should make sure that we use the actual internal
diameter of the pipe, which may be different than the nominal diameter.
TYPES OF FLUID FLOW PROBLEMS
1.Determining the pressure drop (or head loss) when the pipe length
and diameter are given for a specified flow rate (or velocity)
2.Determining the flow rate when the pipe length and diameter are
given for a specified pressure drop (or head loss)
3.Determining the pipe diameter when the pipe length and flow rate
are given for a specified pressure drop (or head loss)

To avoid tedious iterations in head loss, flow rate, and diameter
calculations, these explicit relations that are accurate to
within 2 percent of the Moody chart may be used.
QUIZ 1
Compare the velocity and pressure heads for typical
conditions in a street main: V = 1.5 m/s; D = 0.5 m;
p = 500 kPa
(hints)
Answer
If f = 0.02, hL for each 0.5 m of pipe is 2% of the
velocity head, or 0.0023 m, corresponding to 0.0045%
of the pressure head.
QUIZ 2
A 20-in-diameter galvanized pipe (e = 0.0005 ft) 2 miles long carries 4
cfs at 60o
F. Find hL using (a) the Moody diagram and (b) the Colebrook
eqn.(use the mood chart in above notes)
 
 
22
2
1.5 m/s
0.115 m
2 2 9.8 m/s
V
g
    2
3
500 kPa 1000 N/m kPa
51.0 m
9800 N/m
p

 
  

a)
B) Colebrook equation
TYPICAL PIPE FLOW PROBLEM >>(repeated, unique anyway)
• Type II: Pipe properties (e, D, l) and hL known, find V.
1 2.71
2log
3.7 Re
D
f f
 
    
 

• Guess V, determine f and hL as in Type I, iterate until hL equals
known value, or
• Solve Colebrook and DW eqns simultaneously to eliminate V, yielding:
SOLVING TYPE II PIPE PROBLEMS
Iterative Approach
Rearranged D-W eqn:
2 2.51
2 log
3.7 2
L
L
gDh D l
V
l D gDh
  
    
 

Example
For the pipe analyzed in the preceding QUIZ 2, what is the
largest flow rate allowable if the total frictional head loss
must remain <8 ft?
SOLUTION
Substituting known values,
Type III.
ANOTHER SIMPLE QUIZ bro! DEUS
QN What diameter galvanized pipe would be required in the preceding
QUIZ 2 if a flow rate of 10 cfs was needed, while keeping the total
frictional headloss at <8 ft?
(HINTS SOLUTION)
2 2.51
2 log
3.7 2
L
L
gDh D l
V
l D gDh
  
    
 

DEPENDENCE OF HL ON D AND V
1. In laminar region:
2. In turbulent region, when f becomes constant
Under typical water distribution conditions, hL in a given
pipe can be expressed as kQn
with n slightly <2.
QUIZ 2 CONTINUES……..
For the systems analyzed in the QUIZ 2, what value of n
causes the data to fit the equation hL = kQn
?
ALTERNATIVE EQUATIONS FOR FLOW
Headloss Relationships in Turbulent Pipe Flow
 Hazen-Williams equation – widely used for hL as function
of flow parameters for turbulent flow at typical
velocities in water pipes:

NOTE: Coefficients shown are for SI units; for BG units, replace
0.849 by 1.318 and 10.7 by 4.73.
COMPARISON OF EQUATIONS FOR TRANSITIONAL AND TURBULENT CURVES ON
THE MOODY DIAGRAM
*
Coefficients shown are for SI units (V in m/s, and D and Rh in
m); for BG units (ft/s and ft), replace 0.849 by 1.318; 0.354 by
0.550; 0.278 by 0.432; 10.7 by 4.73; 1/n by 1.49/n; 0.397 by
0.592; 0.312 by 0.465; and 10.3 by 4.66.
ENERGY LOSSES IN BENDS, VALVES, AND OTHER TRANSITIONS
(‘MINOR LOSSES’)
Minor headlosses generally significant when pipe sections are
short (e.g., household, not pipeline) Caused by turbulence
associated with flow transition; therefore, mitigated by
modifications that ‘smooth’ flow patterns Generally much greater
for expansions than for contractions.
Often expressed as multiple of velocity head:

Where, K is the ratio of energy lost via friction in the device
of interest to the kinetic energy of the water (upstream or
downstream, depending on geometric details)
ENERGY LOSSES IN CONTRACTIONS
Note bro!: all pictures by
Fluid Mechanics With Engineering Applications10th
Edition

By E. Finnemore and Joseph Franzini
Copyright: 2002
ENERGY LOSSES IN EXPANSIONS

ENERGY LOSSES IN PIPE
FITTINGS AND BENDS
Bro! Deus….Use the idea above to solve following the
following EXAMPLE
A 5-in-diameter pipe with an estimated f of 0.033 is 110
feet long and connects two reservoirs whose surface
elevations differ by 12 feet. The pipe entrance is
flushed, and the discharge is submerged.
a. Compute the flow rate.
b. How much would the flow rate change if the last 10
ft of the pipe were replaced with a smooth conical
diffuser with a cone angle of 10o
?

Any idea bro? about the question
SOLUTION
From graph, for a smooth, 10o
cone, kcone = 0.175

BRO!if your following, Mr Materu’s slides ‘I’ end here.
See questions next pages……………..
REVISION QUESTIONS
1. A mountain lake has a maximum depth of 40m, the barometric pressure
at the surface is 598mm Hg. Determine the absolute pressure (in
Pascal) at the deepest part of the lake. Given that the density of
water and mercury are 1000kg/m3
and 13558kg/m3
respectively
(Answer=471.9KPa)
2. A manometer is attached to a tank containing three different fluids
as shown in figure below, what will be the difference in elevation
of mercury column in the manometer (y)? (Answer =0.626m)
3. Determine the new differential reading along the inclined leg of
the mercury manometer of Figure below, if the pressure in pipe A
is decreased 10 kPa and the pressure in pipe B remains unchanged.

The fluid in A has a specific gravity of 0.9 and the fluid in B is
water. (Answer=0.212m)
4. Viscosity of liquid decrease with increase in temperature while for
a gas the viscosity increase with the increase in temperature.
Explain why???
HINTS bro! to use
1.Molecular structure
2.Cohesive force
3.Momentum exchange (gas)
5. What is the viscous force of the fluid on a 30.48m length pipe of
0.305m diameter if shear stress is 0.0262N/m2
?
See hints below
6. Determine the Torque and power required to run a 300 mm diameter
shaft at 400 rpm in journals with uniform oil thickness of 1 mm.
Two bearings of 300 mm width are used to support the shaft. The
dynamic viscosity of oil is 0.03N-s/m2
(Answer Torque = 15.995Nm and Power =670W)
BRO! (KIM DE FEL) ATTACK THE FOLLOWING ADDITIONAL QUESTIONS
Question 1
a)Find the capillary rise in the tube shown in figure F1 below, the
air- water-glass interface ( 0o
  ) and tube radius is 1mm at 20ºC
temperature. Given that the surface tension of water at 20ºC is
0.0728N/m (Answer 14.8mm)
b)Assume the liquid is mercury with air-mercury-glass interface is 130º
and density of mercury is 13 570kg/m3
while surface tension of
mercury is 0.514N/m, calculate capillary rise, explain why the
results is negative and draw the figure to represent the result
(Answer -5mm)

Figure F1
Question 2
a)A liquid compressed in a cylinder has a volume of 1000cm3 at 1MN/m2
and volume of 995cm3 at 2MN/m2. What is bulk modulus of elasticity
(answer 200MPa)
b)The glass tube in figure F2 is used to measure Pressure (P1) in the
water tank, the tube diameter is 1mm and water surface tension is
0.0712N/m. the tube is reading 17cm of height, what is the true
height of water after correcting the effect of surface tension
(density of water is 1000kg/m3
) [answer 2.9cm]
Figure F2
Question 3
Water flows into larger tank as shown in figure F3 at the rate of
0.11m3
/s, the water leaves the tank through 20 holes in the bottom of the
tank, each hole has a diameter of 10mm. Determine equilibrium height (h)
for the steady state operation (answer 2.50m)

Figure F3
Question 4
The volume of fluid is found to be 0.00015m3
, if the specific gravity of
this fluid is 2.6. Calculate the weight of fluid (Answer 3.82N)
Question 5
If the specific weight of a substance is 8.2KN/m3
, calculate its density
[Answer: 836kg/m3
]
Question 6
A vertical cylindrical tank with a diameter of 12m and depth of 4m is
filled to the top with water at 20°C. If water is heated to 50°C, how
much water will spill over? Given that density of water at 20°C and 50°C
is 999kg/m3
and 989kg/m3
respectively. [Answer 4.6m3]
Question 7
If bulk modulus of elasticity for water is 2.2GPa, what pressure is
required to reduce a volume of water by 0.6% [Answer: 13.2MPa]
Question 8
The mercury manometer of Figure indicates a differential reading of 0.30
m when the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure
in pipe B. (ANS;4171.28Pa)

Figure
Question 9
For the stationary fluid shown in Figure , the pressure at point B is 20
kPa greater than at point A. Determine the specific weight of the
manometer fluid (Answer = 7100N/m3
)
Figure
Question 10
Water flows through the pipe contraction as shown in Figure . For the
given 0.2m difference in manometer level, determine the flow rate as a
function of the diameter of the small pipe, D. [Answer Q=1.56D2
where Q
(m3
/s) and D (m)]
Figure
Question 11
The fluid in figure M8 is water, determine the manometer reading (h)
[Answer=0.37m)]

Figure
Question 11
Determine the elevation difference h between the water levels in the two
open tanks shown in figure M9 [Answer = 0.040m]
Figure
Question 12
Determine the flowrate through the pipe shown in figure below
(Answer = 0.0111m3
/s)
Figure
Question 13

A swimming pool is 18 m long and 7m wide, determine the magnitude and
location of the resultant force of the water on the vertical end of the
pool where depth is 2.5m (Answer, FR= 214KN, YR= 1.67)
Question 14
The vertical cross section of closed storage tank as shown in figure
contains ethyl alcohol, the air pressure is 40KPa. Determine the
magnitude of the resultant fluid force acting on one end of the tank.
Given that specific weight of ethyl alcohol is 7.74 KN/m3
(Answer=FR=847KN)
Figure1
Question 15
The rectangular gate CD of figure Q2 is 1.8m wide and 2m long, assuming
the material of the gate to be homogenous and neglecting friction at the
hinge C, determine the weight of the gate necessary to keep the gate
shunt until the water level rises to 2m above the hinge (Answer = 180KN)
Figure

Question 16
A 4m long curved gate is located in the side of the reservoir containing
water as shown in figure Q3 below. Determine the magnitude of the
horizontal, vertical and resultant forces of water on the gate. Will
resultant force pass through point O? Explain! (Answer, FH=882KN,
FV=983.67KN, FR=1321.8KN)
Figure
Question 17
The rigid gate OAB of figure G1 below is hinged at O and rests against a
rigid support at B. What minimum horizontal force P is required to hold
the gate closed if its widht is 3m? Neglet the weight of the gate and
friction in the hinge, the back of the gate exposed to the atmosphere.
(Answer 436KN)
Figure

Question 18
A dam of 20m long retain 7m of water as shown in figure below, find the
total resultant force acting on the dam and location of the centre of
pressure. Given that the angle between water and dam at the surface is
60° (Answer 5550.6KN, Centre of pressure is 4.667m below the surface)
Question 19
An inclined circular gate with water on one side shown in figure G2
below, Determine the total resultant force acting on the gate and
location of the Centre of pressure (Zcp). Assume specific weight of water
at 20°C is 9.79KN/m3
(Answer Resultant force =14.86KN and Zcp =2.26m)
Figure
Question 20
The 4-m-diameter circular gate in figure G3 below is located in the
inclined wall of a large reservoir containing water. The gate is mounted
on a shaft along its horizontal diameter. For a water depth of 10 m above
the shaft determine
(a) The magnitude of the resultant force exerted on the gate by water
(Answer= 1.23MN)
(b) Location in y-axis (YR) of the resultant force (Answer 11.6m)

Figure
Question 21
Consider figure G4 below, if atmospheric pressure is 101.03KPa and
absolute Pressure at the bottom of the tank is 231.3KPa. What is the
density of olive Provided that the density of mercury, water and SAE-30
oil are 13570kg/m3
, 1000kg/m3
and 800kg/m3
respectively?
Figure
Question 22
Determine the pressure heads at A and B in meter of water in figure G5
below and explain your answer (Answer HA= -2.38m H2O and HB= -0.51m H2O)

Figure
Question 23
A large open tank contains a layer of oil floating on water as shown in
figure M2, the flow is steady and fluid is incompressible, Determine
a)Height, h, to which the water will rise
b)Water velocity in the pipe
c)Pressure in the horizontal pipe.
Figure
Question 24
Determine the flow rate through the pipe in M3 below (Answer = 0.0111m3
/s)
Figure
Question 25
The specific gravity of the manometer fluid shown in figure , determine
the volume flowrate, Q, if the flow is incompressible and the flowing
fluid is water (density of water is 1000kg/m3
)

Figure
Question 26
Determine the elevation difference h between the water levels in the two
open tanks shown in figure F6 [Answer = 0.040m]
Figure
BRO! ARE YOU TIRED IS YES STOP HERE FOR TODAY? IF YES SEE MY GIFT TO YOU
SAMPLE QUESTIONS ON FLUID FLOW IN CLOSED CONDUITS
Question 1
Lubricating Oil at a velocity of 1 m/s (average) flows through a pipe of
100 mm; determine whether the flow is laminar or turbulent. Also
determine the friction factor (f) and the pressure drop over 10 m length
in Pa or N/m2
. Given that Density = 930 kg/m3
and Dynamic viscosity μ= 0.1
Ns/m2
Solution
Given
Velocity (v) =1m/s
Pipe diameter (D) =100mm = 0.1m
Pipe length = 10m
Density = 930kg/m3
Dynamic viscosity (µ) = 0.1Ns/m2
Type of flow: 930*1*0.1
R 930
0.1
e
vD

  
Since Re<2000, then the flow is laminar

Friction factor: since the flow is laminar, then 64 64
0.06882
Re 930
f   
Pressure drop:
2 2
0.06882*10*1
0.351
2 2*0.1*9.81
930*9.81*0.351
3200
f
fLV
h m
Dg
P gh
P Pa

  
  
 
Question 2
Use mood diagram to find friction factor for the following data: Diameter
of the pipe =0.305m, Kinematic viscosity of the fluid = 1.3 x 10-6
,
Velocity =0.043m/s and Internal pipe roughness =0.00061m
Solution
Relative roughness (e/D) 0.00061
0.002
0.305
 
Reynolds number 6
0.305*0.043
10000
4.3 10
DV
v 
  

From moody diagram, f = 0.034 (check!! IT BRO! DEUS F)
QUESTION 3
Calculate the energy head loss due to friction in a pipe of length 1000m,
diameter of 0.25m and roughness of 0.0005m given that the fluid of
kinematic viscosity of 1.306x10-6
m2
/s flow in the pipe at the rate of
0.051m3
/s.
Solution
Given
Pipe length =1000m
Diameter = 0.25m
Roughness = 0.0005m
Kinematic viscosity= 1.306x10-6
m2
/s
Flow rate = 0.051m3
/s
Friction head loss
2
( )
2
f
fLV
h a
Dg
            
Velocity
2
2
4
( )
4*0.051
( ) 1.039 /
*(0.25)
Q Q
V
A D
V m s


 
 
Friction factor
6
0.25*1.039
Re 200000
1.036 10
0.0005
0.002
0.25
DV
v
e
D

   

  
From moody diagram, f=0.0245 (check!! BRO!)

Head loss (from equation a)
2 2
0.0245*1000*(1.039)
2 2*0.25*9.81
5.39
f
f
fLV
h
Dg
h m
 

Hence head loss due to friction is 5.39m
Question 4
Oil (specific weight = 8900N/m3
& viscosity = 0.10Ns/m2
flows through
horizontal 23mm diameter tube as shown in figure Q1 below, a differential
U-tube manometer is used to measure the pressure drop along the tube,
determine the range of value for h for laminar flow (Answer h<0.509m)
Figure
Question 5
Oil of SG = 0.87 and kinematic viscosity (v) = 2.2 x10-4
m2
/s flows through
a vertical pipe as shown in figure at flow rate of 4x10-4
m3
/s. Determine
the manometer reading h, also determine the magnitude and direction of
flow rate which will cause h to be zero. (Answer h=18.5m and Q=0)
Figure
Question 6
Oil with density of 900 kg/m3
and kinematic viscosity of 0.0002 m2
/s
flows upward through an inclined pipe as shown in figure Q3. The pressure
and elevation are known at sections 1 and 2, 10 m apart. Assuming the
flow is steady laminar, calculate

a)Friction head loss between (hf)point 1 and 2
b)Discharge (Q)
c)Velocity (V)
d)Reynolds number (Re)
[Answers: hf=4.9m, Q=0.0076m3/s, V=2.7m/s and Re=810]
Figure
Question 7
From figure Q4 below, find the diameter of the pipe which connects two
reservoirs given that the length of the pipe is 304.8m, the flow is
0.013m3
/s, roughness (e) is 0.001m and kinematic viscosity is 1.31x10-
6m2
/s (Answer=0.15m)
Figure
Solution
This is the kind of engineering problem that you are suppose to find the
diameter of pipe to accommodate a given fluid flow with other available
information. Find everything in term of diameter

 
2
22
2
Relative roughness in term of D
0.001
Velocity in term of D
4
4*0.013 0.017
Reynolds number in term of D
0.017 *
Re
1
e
D D
Q Q
V
A D
V
DD
DVD D
v





 
 

 
 
 
6
2
2
2
4
1
.31 10
12977
Re
Head loss in term of D
2
15.2
2
0.017304.8*
15.2
2*9.81*
0.088
15.2
19.62
0.295
L
D
fLV
h
Dg
fLV
Dg
f
D
D
f
D
D
D f






 
  

 5
f can be solved by trial and error, try different values of f until the
trial value of f converges with f values of moody chart as shown in table
below
f-trial D e/D Re f-chart Remarks
0.025 0.0141 0.007 92035 0.034 Try again
0.034 0.15 0.0067 86513 0.034 Converges
Since the f-trial converges with f from moody chart, then the diameter of
the pipe is 0.15m
Question 8
Oil with density of 900 kg/m3
and kinematic viscosity of 0.00001m2
/s,
flows at 0.2 m3
/s through 500m of 200-mm diameter cast-iron pipe.
Determine (a) the head loss and (b) the pressure drop provided that the
pipe slopes down at 10° in the flow direction and caste iron pipe has
roughness of 0.26mm (Answers: hf=117 and change in pressure =265KPa)
Question 9
Oil with density of 950kg/m3
/s, flow
through 30cm diameter pipe 100m long with a head loss of 8m. The relative
roughness (e/D) of the pipe is 0.00002. Find the average velocity and
flow rate (Answer V=4.84m/s and Q=0.342m3
/s)

Solution
This type of problems requires iteration process or computer software
(solvers) because velocity (or flow rate) appears in both the ordinate
and the abscissa on the Moody chart, iteration for turbulent flow is
nevertheless quite fast, because f varies so slowly with Re.
First solve V in term of f
2
2
2
2
2
2
8*2*0.3*9.81
100
0.471
0.471
f
f
fLV
h
Dg
h Dg
f
LV
f
V
fV
V
f





To get started, you only need to guess f, compute V from equation above,
then get Re (Re=VD/v), compute a better f from the Moody chart and
repeat. The process converges fairly rapidly. A good first guess is to
assume the flow is ‚fully rough‛ e.g. the value of f for (e/D) 0.0002 is
f =0.014. Calculation can be done as shown in table below
f-trial Velocity (V) Re (VD/v) f-chart Remarks
0.014 5.80 87000 0.0195 Try again
0.0195 4.91 73700 0.0201 Try again
0.0201 4.84 72600 0.0201 Converges
Since the f-trial converges with f from moody chart, then the velocity of
flow is 4.84m/s
Flow rate
2
2
3
Flow rate (Q)
Q=AV
Q=
4
*(0.3) *4.84
0.342 /
4
D V
Q m s


 
Hence the flow is 0.342m3
/s
Question 10
Oil with density of 950kg/m3
/s, flow
through a pipe of unknown diameter, the length of pipe is 100m long with
a head loss of 8m. The relative roughness (e/D) of the pipe is 0.00002
and the flow is 0.342m3/s. calculate the pipe diameter (Answer: 30cm)

Question 11
Consider figure Q5, the total pressure drop PA-PB = 150,000 Pa, and the
elevation drop ZA- ZB=5 m. The pipe data are shown in table 1 below. The
fluid is water, with density of 1000kg/m3
and kinematic viscosity of
1.02x10-6m2
/s. Calculate the flow rate Q in m3
/h through the system.
Figure
Table : Pipes data
REST FOR A WHILE BRO KIM DE FEL BEFORE TAKING
YOU TO PUMPS

PUMP
PUMP is machine used to lift fluid from one point to another.It converts
mechanical energy into hydraulic energy. Pump may need to lift liquids
from some heights below the pump and push them to some height above the
pump.
PUMP CLASSIFICATION
 Pumps are classified according to its working principles.
 Pumps are classified into three main groups:
Positive displacement pumps (Static pumps)
Kinetic pumps
Electric pumps (electromagnetic pumps)
Basic groups are sub-divided more into several types.

POSITIVE DISPLACEMENT PUMPS (STATIC TYPE)
Fluid is pumped into & out of a chamber by changing the volume of the
chamber. Pressures & works done are a result of static forces rather
than dynamic effects. Common examples are tire pump, human heart,
gear pump etc
KINETIC PUMPS
It involve a collection of blades, buckets, flow channels around
an axis of rotation to form a rotor. It is dominated by dynamic
force. Rotor’s rotation produces a dynamic effect that adds energy
to the fluid. A good example of kinetic pump is the one that used
in deep wells (i.e. centrifugal pumps). Depending on direction of
the fluid motion relative to the rotor’s, kinetic pump classified
into axial-flow, mixed-flow & radial-flow
1. Radial flow pump
It involves a substantial radial-flow component at the rotor
inlet or exit. Dominated by the action of centrifugal forces

2. AXIAL FLOW PUMPS
The pumped fluid maintains a significant axial-flow direction
from the inlet to outlet. The flow entering the pump inlet
axially and discharge nearly axially. These pumps has low head
(H) but with larger discharges (Q)
CENTRIFUGAL PUMPS
Is the radial flow pump. Has two main components which are an
impeller & stationary casing (housing).As the impeller rotates,
fluid is sucked in through the hub (eye)

The casing shape (increase in area in direction of flow) is
designed to reduce the velocity so as increase in pressure.
PUMP CHARACTERISTICS
There is theoretical head (hi) –provided by manufactures. There is
actual head (ha) – field performance
This is because of Losses (hL)
hL=(friction losses, minor losses, other losses)
ACTUAL HEAD (SHOULD BE USED FOR DESIGN)
 Actual head should be obtained for design. Engineer should find
it from available information e.g. flow rate. Sometimes pumps
characteristics are obtained by experiments

Energy at the inlet is less than the energy at exit, the difference
is energy added by pump/energy gained by fluid (ha)
This (ha) is the net head rise, Can be converted to Power gained by
fluid by (in Watts) or (in horse power)
In watts
In horsepower:
CHARACTERISTIC PERFORMANCE CURVE

Is the graph which gives the information about the
characteristics of pump
The characteristics includes:-
a.Discharge
b.Efficiency
c.Pressure requirements (head)
d.Power consumption etc.
All are important information to engineer. All characteristics are
drawn Vs flow rate.
DISCHARGE (Q)
Is the volume of liquid pumped per unit time. Head curve raises
as the flow rate (Q) decrease. To lift water up to higher
building the flow rate should be small
ACTUAL HEAD
Is the net work done on a unit weight of water –done by pump Head
at zero discharge is called the shutoff head No flow.Efficiency is
zero.
OPERATING POINT OF PUMP
Is the point which gives the head and flow rate that satisfies both
system equation and pump equation. Can be obtained by plotting both
curves on the same graph. The intersection point is operating point
STOP HERE AND PRACTISE THE FOLLOWING EXAMPLE BRO!
QN Water is pumped from a deep well to students hostel’s tanks, the
centrifugal pump used for pumping has characteristics which is given by
equation H = 22.9+10.7Q-111Q3
,but one Agricultural engineering students
tried to find the system demand and she obtained the equation
H = 15 +85Q2
.Where Q is the flow in m3
/s and H is head in m. Determine
the operating point of the pump (Flow rate and head)
[Answer: Q=0.23 and H=19.49]
POWER
Is the rate at which work is done on a liquid by pump. Is when a
unit volume of liquid is raised through a given height.

Is power required by water to be lifted.As the discharge increase
then the power requirement also increases
Efficiency
1.Pumps receives power from motors through shaft (BP)
2.Motor receives power from electricity (LP)
3.Pump develops power into fluid (fluid required power)
Neither motor (M) nor Pump (P) operates in 100% efficiency.
Efficiency of the motor (ηm)
Efficiency of the Pump (ηp)
SIMPLE QUIZ BRO! DEUS
QN The pump is used to increase the pressure of water flow rate of
0.2m3
/s from 200KPa to 600KPa. If the overall efficiency of the
pump is 85%, how much electrical power is required to pump the

water? The suction tank is 10cm below the centre line of the pump
and delivery tank is 10cm above the centerline of the pump. Assume
the inlet and exit diameters are equal and velocities at suction
and delivery can be neglected. Also minor and friction losses can
be neglected. [Answer P=187KW]
The overall pump efficiency is affected by the
1.Hydraulic losses in the pump due to friction loss and
minor losses..
2.Mechanical losses in the bearings and seals
3.Volumetric losses due to leakage of the fluid.
Efficiency is important characteristics of pump performance.
NET POSITIVE SUCTION HEAD (NPSH)
Is expression of the suction capability of the pump. It is used to
calculate the inlet pressure needed at the pump to avoid
cavitation. Inlet pressure must be equal or higher than designed
(Requirements)
There are two types of NPSH
1.NPSHR-(Required NPSH)
 Is required suction head of the chosen pump
 It is given by the manufacturer
2.NPSHA-(Available NPSH)
This is true (available) suction head at the pump location Can be
estimated mathematically or experimentally. NPSHA must be greater
than NPSHR –otherwise it may cause cavitation.
Consider the figure below.
Total head at the suction side
Liquid vapour pressure
To avoid cavitation

NOTE THE FOLLOWING,BRO! from the above ,last expression.
1.If z1 is increased, the NPSHA is decreased
2.There is some critical value of Z1 which the pump cannot
operate without cavitation
3.What is the effect to NPSHA if source is above the pump?
Bro! Use the above idea to solve the following example
QN The pump is installed to pump water from Mazimbu well as shown
in figure F1 below, determine the critical elevation (Z1) where the
pump can be situated above water surface of suction without
experiencing cavitation. Given that the diameter of the pump is
240m, pumping rate is 250m3
/hr and NPSH value for discharge is
7.4m. Use atmospheric pressure of 101Kpa and Vapour pressure of
1666Pa. [Answer = 2.72m]

These are the Laws which gives the relations between the following…
i. Volumetric flow rate (Q)
ii. Head (H)
iii. Power requirements
iv. Diameter (D)
v. Shaft rotation speed (N)
There are two affinity laws.
Affinity law 1
In this case, the point of interest is to investigate how change in
operating speed (N) affects pumps characteristics like
1.Discharge (Q)
2.Head (H)
3.Required Power (P)
Discharge versus rotational speed
Head versus rotational speed
Power versus rotational speed
Affinity law 2
In this case, the point of interest is to investigate how change
in diameter (D) affects pumps characteristics like…
1.Discharge (Q)
2.Head (H)
3.Required Power (P)

Test your IQ bro!
Is it possible to increase impeller diameter?
Discharge versus Impeller diameter
Head versus Impeller diameter
Power versus Impeller diameter
PUMP INSTALLATION
Sometime one pump is not enough to meet discharge demand /Head
required. Hence two or more pumps may be connected in series or
parallel. The choice of connection depend on weather you want to
increase discharge or head.
PUMPS IN SERIES
 The discharge form first pump is piped into the inlet side of
the second pump.
 Each pump adding more energy to the fluid.
 Only head is increased.
 Discharge remains the same.
PUMPS IN SERIES
 Applicable in deep wells pumping or higher building water
pumping….
 The combined head (total head) is equal to the sum of
individual heads…

PUMPS IN PARALLEL
Two or more pumps draw water from the sources and individual flows
are discharged into a single pipeline. Pumps in parallel operates
approx. in the same head. The total discharge is equal to the sum
of individual discharge
QUIZLET
QN The pump of Figure below is to increase the pressure of 0.2 m3
/s of water from 200 kPa to 600 kPa. If the pump is 85%
efficient, how much electrical power will the pump require? The
exit area is 20 cm above the inlet area. Assume inlet and exit
areas are equal.
Do you like 3D view answers? Hahahaha! See below

Try the following important question
1.A water pump has one inlet and two outlets as shown in Figure below,
all at the same elevation. What pump power is required if the pump is
85% efficient? Neglect pipe losses.
2.A vehicle with a mass of 5000 kg is traveling at 900 km/h. It is
decelerated by lowering a 20-cm wide scoop into water a depth of 6 cm
(Figure below). If the water is deflected through 180°, calculate the
distance the vehicle must travel for the speed to be reduced to 100
km/h.

Fluids mechanics (a letter to a friend) part 1 . .

Fluids mechanics (a letter to a friend) part 1 . .

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