Unilorin Complete CHE241

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DEPARTMENT OF CHEMICAL ENGINEERING,
UNIVERSITY OF ILORIN,
ILORIN.
LECTURE GUIDE
FUNDAMENTALS OF FLUID MECHANICS
CHE 241 - 3 Credits
2010/2011 SESSION
COURSE: CHE
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS
(3 credits /Compulsory).
Lecturer:
OGUNLEYE,Oladipupo Olaosebikan
B.Tech. Chemical Eng.(Ogbomoso),M.Sc. Industrial Eng. (Ibadan),Ph.D. Industrial Eng.(Ibadan). Reg.
Engr.(COREN), MAiCHE,MSIAM.
Department of Chemical Engineering,
Faculty of Engineering and Technology,
University of Ilorin, Ilorin,
Kwara State, Nigeria..
[email protected]
E-mail:
Office Location: – Room 5, Ground Floor, Department of Chemical Engineering Building.
Consultation Hours: 11.00-1.00pm Mondays and Wednesdays.
Course Content:
Properties of fluids, Fluids Statics, Basic conservation laws, friction effect and losses in laminar and
turbulent follows in ducts and copies. Dimensional analysis and dynamic similitude, principles of
construction and operation of selected hydraulic machinery. Hydropower systems.
45h (T); C
Course Description:
The course is designed to introduce students in the Faculty of
Enginnering and Technology especially to lower level course in
process engineering fluid mechanics, which emphasizes the
systematic
application
of
fundamental
principles
(e.g.,
macroscopic mass, energy, and momentum balances and
economics) to the analysis of a variety of fluid problems of a
practical nature. The scope of coverage includes internal flows of
Newtonian and non- Newtonian incompressible fluids, adiabatic
and isothermal compressible
flows .Applications include
dimensional analysis and scale-up, piping systems with fittings
for Newtonian and non-Newtonian fluids, compressible pipe
flows, flow measurement, and control, pumps, compressors.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
2
Course Justification:
Nigeria being an oil-rich nation transports crude oil and its
allied products through pipe line where so many phenomena
occur due to the nature of the fluid being transported, the
channel of transportation and the forces behind the
transportation. Fluid mechanics gives clear insight into the
understanding of such systems as this. The economics of
crude oil processing will not be complete without an
application of fluid mechanics to imrove pump efficiency in
the course of crude oil transportation thereby minimizing the
cost and maximizing profit. This principle is what many
developed nations had applied to the management of their
pipeline networks whose integrity has been secured through
Artificial Intelligent Systems.
Course Objectives:
The objectives of this course as an integral part for the award of B. Eng. are :
•
For the student to know how the fundamental principles underlying the behavior of fluids can
be applied in an organized and systematic manner to the solution of practical engineering
problems.
•
To provide a ready reference and basic methods for the analysis of a variety of problems
encountered in the movement of fluids through pipes , pumps and all kinds of process
equipment.
Course Requirements:
This is a compulsory course for all students studying Engineering In view of this, students are
expected to participate in all the course activities and have minimum of 75% attendance to be able
to write the final examination.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
3
Methods of grading:
No
1.
2.
3.
Item
Assignment/ Quiz / Monthly Test
Mid Semester Test
Examination
Total
Score %
10
20
70
100
Course Delivery Strategies:
The lecture will be delivered through face-to-face method, lecture guide
(lecture note) will be
provided during lectures. Students will be encouraged and required to read around the topics. The
delivery strategies will also be supported by tutorial sessions.
LECTURES
Week 1 - 2:
General Introduction of Fluid Mechanics
Objective: The students will be able to explain the basic concept of fluid mechanics.
Description: The course outline will be introduced with emphasis on the objectives and delivery
strategies, definitions, the continuum concept , types of fluids, units of measurements,
properties of fluids, Viscosity, Compressibility of fluids, surface tension and capillarity,
vapour pressure.
Week 3:
Dimensional Analysis and Dynamic Similitude
Objective: Students will gain the necessary knowledge of how independent variables are expressed
in terms of their dependent variables without experimentation.
Description: Dimensional Analysis, Techniques of Dimensional Analysis, Interpretation of
Dimensionless Numbers and Dynamic Similitude.
Week 4:
Fluids Statics.
Objective: Student will be able to understand the forces which keep the body of fluid in static
equilibrium.
Description: Concept of stress and pressure, pressure transmission, pressure measurement, Forces on
submerged surfaces,Buoyancy and Static Forces on Solid Boundaries. Student will also
be assessed on the topics covered so far through a short Monthly Quiz
Week 5 - 6:
Basic Conservation Laws.
6:
Laws
Objective: The students will be exposed to the mathematical laws guiding the movement of mass
and energy.
Description: The concept of system, Conservation of Mass, Conservation of Energy, Irreversible
Effects and Conservation of Momentum.
Week 7 - 8 :
Pipe Flows
Objective: Students will be able to understand various flow regimes and their effects on flow
velocities in pipes.
Description: Flow Regimes, Pressure Drop in Laminar Flow, Pressure Drop in Turbulent Flow,
Velocity Profiles in Turbulent Flow, Velocity Distribution, Pressure Losses in Pipe
Fittings and Pipeline Problems. Student will also be assessed on the topics covered so
far through a short Monthly Quiz.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
4
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
Week 9-10:
Compressible Flows
10:
Objective:
Students will understand the effect of pressure and temperature on fluid density.
Description: Gas Properties, Energy equation for compressible flow, Isentropic flow, Compressible
flow in pipes.
Week 11
11: Pumps and Compressors
Objective: Students will understand the working principles of pumps and compressors.
Description: Pumps, Pump Characteristics, Pumping Requirements and Pump Selection, Cavitation
and Net Positive Suction Head (NPSH) , Compressors.
Week 12:
12: Mid Semester Test
Objective: To assess students’ mastery of the course.
Description: Students will be tested on all the topics taken so far. This is to serve as a preparation for
the final examination at the end of the semster
Week 13
13 -14:
14: Flow Measurement and Control.
Objective: Students will have the understanding of the construction and operation of the hydraulic
machineries.
Description: The Pitot Tube, The Venturi and Nozzle, The Orifice Meter, Loss Coefficient, Orifice
Problems and Control Valves.
Week 15: Revision/ Tutorial Exercises
Objective: Student will have opportunity to ask questions on all the topics covered in the course.
Description: A general overview of the course will be made. Students are expected to seek
explanation on any difficult concept or topic treated during the course.
LIST OF BOOKS FOR FURTHER READING:
1 Ron Darby (2001) .Chemical Engineering Fluid Mechanics. Second Edition. Marcel Dekker,
Inc. New York.
2 McCabe,W. L., Smith,J.C. and Harriott, P.(1993) .Unit Operations of Chemical Engineering.
Fifth Edition. McGraw Hill Co. Singapore.
3 Ogboja ,O. (2005). Fluid Mechanics .UNESCO, Nairobi Kenya.
4 Coulson, J.M. and Richardson, J.F., Backhurst,J.R. and Harker, J.H. (2001 ) Chemical
Engineering Volume 1, Sixth Edition, Butterworth Heinmann Inc. Oxford.
5 Douglas,J.F. and Gasiorek, J. M. (1997) Fluid Mechanics. Third Edition,Longman Press,
Singapore.
6 Rajput, R.K. (1998) . A Textbook of Fluid Mechanics in SI Unit. First Edition. S. Chand &
Company Ltd. New Delhi.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
5
MODULE 1
1.0
1.1
General Introduction of Fluid Mechanics
Definitions
Fluid mechanics is the study of the behaviour of fluids under the influence of forces and it can
be studied under two broad topics: fluid statics and fluid dynamics. Fluid statics is the study of the
forces that keep fluids in static equilibrium while fluid dynamics deals with the motion of fluids and
the forces that keep them in motion.
A fluid is a substance that undergoes continuous deformation when subjected to a shear stress.
Liquids and gases are called fluids. An attempt to change the shape of a mass of fluid results in layers
of fluid sliding over one another until a new shape is attained. During the change in shape, shear
stresses exist, the magnitude of which depends upon the viscosity of the fluid and the rate of sliding,
but when a final shape has been reached, all shear stresses will disappear. A fluid in equilibrium is free
from shear stresses. It is easier to define these materials in terms of how they respond (i.e. deform or
flow) when subjected to an applied force in a specific situation such as the simple shear situation
illustrated in Figure 1. A fluid is bounded by two large parallel plates, of area A, separated by a small
distance H. The bottom plate is held fixed. Application of a force F to the upper plate causes it to move
at a Velocity V. The fluid continues to deform as long as the force is applied, unlike a solid, which
would undergo only a finite deformation. The material is assumed to adhere to the plates, and its
properties can be classified by the way the top plate responds when the force is applied.
Figure 1.1 Deformation of a fluid subjected to a shear stress.
1.2
The Continuum Concept
Although gases and liquids consist of molecules, it is possible in most cases to treat them as
continuous media for the purposes of fluid flow calculations. On a length scale comparable to the mean
free path between collisions, large rapid fluctuations of properties such as the velocity and density
occur. However, fluid flow is concerned with the macroscopic scale: the typical length scale of the
equipment is many orders of magnitude greater than the mean free path. Even when an instrument is
placed in the fluid to measure some property such as the pressure, the measurement is not made at a
point-rather, the instrument is sensitive to the properties of a small volume of fluid around its
measuring element. Although this measurement volume may be minute compared with the volume of
fluid in the equipment, it will generally contain millions of
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
6
molecules and consequently the instrument measures an average value of the property. In almost all
fluid flow problems it is possible to select a measurement volume that is very small compared with the
flow field yet contains so many molecules that the properties of individual molecules are averaged
out.
It follows from the above facts that fluids can be treated as continuous media with continuous
distributions of properties such as the pressure, density, temperature and velocity. Not only does this
imply that it is unnecessary to consider the molecular nature of the fluid but also that meaning can be
attached to spatial derivatives, such as the pressure gradient dP/dx, allowing the standard tools of
mathematical analysis to be used in solving fluid flow problems.
Two examples where the continuum hypothesis may be invalid are low pressure gas flow in which the
mean free path may be comparable to a linear dimension of the equipment, and high speed gas flow
when large changes of properties occur across a (very thin) shock wave.
1.3
Types of Fluids
So many researchers have classified fluids based of their physical and reohological properties.
Fluids may be classified as ideal and real or non-ideal fluids. The ideal fluids does not exhibit viscous
properties and cannot sustain frictional and shear stresses when in motion. Its motion is being purely
sustained by pressure forces and so, it cannot dissipate mechanical energy into heat. The real fluid
possesses viscous properties, sustains frictional and shear stresses and dissipates mechanical energy
into heat. In practice, the ideal fluid does not exist, but the flow of many real fluids can be analyzed by
assuming that they are ideal especially if their viscosities are low.
1.3.1
Classification of Fluids
Fluids may also be classified in two different ways:
1
According to their behaviour under the action of externally applied pressure.
Incompressible fluids are those whose volume of the element is independent of its pressure
and temperature, but if its volume changes, it is said to be Compressible.
2
According to the effect produced by the action of a shear stress.
The most important physical properties affecting the stress distribution within the fluid is its
viscosity. In many problem involving the flow of gas or liquid, the viscous stress are important and
give rise to velocity gradients within the fluid, and dissipation of energy occurs as a result of the
frictional forces set up. In gases and most pure liquids where the ratio of the shear stress to the rate of
shear is constant and equal to the viscosity of the fluid, such fluids are said to be Newtonian in their
behaviour.
However, in some liquids, particularly those containing a second phase in suspension, the ratio
is not constant and the apparent viscosity of the fluid is a function of the rate of shear. The fluid is said
to be Non-Newtonian and to exhibit rheological properties. Fluids that exhibit a nonlinear relationship
between stress and strain rate are termed non-Newtonian fluids. Many common fluids that we see
everyday are non-Newtonian. Paint, peanut butter, and toothpaste are good examples. High viscosity
does not always imply non-Newton behavior. Honey is viscous and Newtonian while 5W30 motor oil
is not very viscous, but it is non-Newtonian. There are several types of non-Newtonian fluids. Figure
1.2 shows several of the more common types.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
7
Figure 1.2 : Types of Non- Newtonian Fluid
Non-Newtonian fluids include those for which a finite stress τy is required before continuous
deformation occurs; these are called yield-stress materials.
1 The Bingham plastic fluid is the simplest yield-stress material; its rheogram has a constant
slope µ∞, called the infinite shear viscosity. Highly concentrated suspensions of fine solid
particles frequently exhibit Bingham plastic behavior.
1.1
2
Shear-thinning fluids are those for which the slope of the rheogram decreases with increasing
shear rate. These fluids have also been called pseudoplastic, but this terminology is outdated
and discouraged. Many polymer melts and solutions, as well as some solids suspensions, are
shear-thinning. Shear-thinning fluids without yield stresses typically obey a power law model
over a range of shear rates.
1.2
The apparent viscosity is
1.3
The factor K is the consistency index or power law coefficient, and n is the power law
exponent. The exponent n is dimensionless, while K is in units of kg/(m .s2 − n). For shearthinning fluids, n < 1. The power law model typically provides a good fit to data over a range
of one to two orders of magnitude in shear rate; behavior at very low and very high shear rates
is often Newtonian. Shear-thinning power law fluids with yield stresses are sometimes called
Herschel-Bulkley fluids. Numerous other rheological model equations for shear-thinning fluids
are in common use.
3 Dilatant, or shear-thickening, fluids show increasing viscosity with increasing shear rate. Over
a limited range of shear rate, the power law model with n > 1 may describe them. Dilatancy is
rare, observed only in certain concentration ranges in some particle suspensions.
4 Time Dependent Fluids :Time dependent fluids are fluids that increase or decrease in viscosity
over time at a constant shear rate. The construction industry uses a cement slurry that thins with
time at a constant pump rate because it is easier to pump and fills forms easily. There are two
types of time dependent fluids: Rheopectic and Thixotropic.
Rheopectic Fluids
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
8
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
Rheopectic fluids increase in viscosity with time at constant shear rate. There are few examples
of rheopectic behavior. In the British literature, rheopectic behavior is called anti-thixotropic
behavior. Rheopectic behavior has been observed in bentonite sols, vanadium pentoxide sols,
and gypsum suspensions in water as well as in some polyester solutions.
Thixotropic Fluids
These fluids lose viscosity over time at a constant shear rate. Dispersing agents for cements
tend to make them thixotropic. Viscosity is recovered after the cessation of shear. In some
systems, the time it takes to recover is so short that a sheared sample cannot be poured out of a
container before it gets too thick to flow. Examples of thixotropic fluid include mayonnaise,
clay suspensions used as drilling muds, and some paints and inks, that show decreasing shear
stress with time at constant shear rate.
Viscoelastic Fluids
These fluids exhibit elastic recovery from deformation when stress is removed. Polymeric
liquids comprise the largest group of fluids in this class. A property of viscoelastic fluids is the
relaxation time, which is a measure of the time required for elastic effects to decay.
Viscoelastic effects may be important with sudden changes in rates of deformation, as in flow
startup and stop, rapidly oscillating flows, or as a fluid passes through sudden expansions or
contractions, where accelerations occur. In many fully developed flows where such effects are
absent, viscoelastic fluids behave as if they were purely viscous. In viscoelastic flows, normal
stresses perpendicular to the direction of shear are different from those in the parallel direction.
These give rise to such behaviors as the Weissenberg effect, in which fluid climbs up a shaft
rotating in the fluid, and die swell, where a stream of fluid issuing from a tube may expand to
two or more times the tube diameter. A parameter indicating whether viscoelastic effects are
important is the Deborah number, which is the ratio of the characteristic relaxation time of the
fluid to the characteristic time scale of the flow. For small Deborah numbers, the relaxation is
fast compared to the characteristic time of the flow, and the fluid behavior is purely viscous.
For very large Deborah numbers, the behavior closely resembles that of an elastic solid.
1.4
Units of Measurements
Fluid motion and behaviour can be described in units of mass, length and time. These are the basic
units from which a number of derived units can be obtained as listed in Table 1.1. Several different sets
of units are used in both scientific and engineering systems of dimensions. These can be classified as
either metric - Système International d’Unites (SI and cgs) or English – foot-pound-seconds (fps).
Although the internationally accepted standard is the SI scientific system, English engineering units are
still very common and will probably remain so for some time. Therefore, it is necessary for students
master these two systems and become adept at converting them from one to the other. Typical
conversion factors are given in the Table 1.2 below.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
9
Table 1.1 Derived Units
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
10
Table 1.2: Units Conversion Factor
Source: Chemical Engineering Fluid Mechanics by Ron Darby. Marcel Dekker, Inc. New York,2nd
Edition, 2001
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
11
1.5
1.5.1
Properties of Fluids
Density
Density is an extremely important property of matter. The density of a material can be
considered continuous except at the molecular level. Density can also be thought of as the constant that
relates mass to volume. It is defined as mass per unit volume.
1.4
1.5.2
Specific Weight
The specific weight is a quantity that is used frequently in fluid mechanics. In the
American Engineering Series (AES) of units, it is numerically equal to the density. The units are lbf
rather than lbm. It is defined as weight per unit volume.
w
=
1.5
1.5.3 Specific Gravity
Specific gravity is used instead of density to tabulate data for different materials. Using the specific
gravity, the density in any set of units may be found by picking the reference density in the desired
units. Note: When the reference density is expressed in g/cm3 , the density and
specific gravity have the same numerical value.
1.6
1.6
Viscosity
It is a measure of the fluids resistance to flow because of the cohesive forces of particles. All
real fluids resist any force tending to cause one layer to move over another. Consequently, adjacent
layers experience shearing stresses when the fluid is flowing. The shearing stresses are similar to the
frictional forces experienced by two solids sliding on each other. This resistance to the movement of
one layer of the fluid over an adjacent one is called the viscosity of the fluid.
It is the ratio of shear stress to shear rate is the viscosity. The SI units of viscosity are kg/(m.s) or Pa.s
(pascal second). The cgs unit for viscosity is the poise; 1 Pa.s equals 10 poise or 1000 centipoise (cP)
or 0.672 lbm/(ft.s).
Consider Figure 1.2 where fluid flows over a solid boundary. Due to the tendency of fluid
particles to cling to the surface of the solid, the particle closest to the boundary are brought to rest such
that the fluid velocity increases gradually from zero on the wall to the free stream velocity U far away
from the wall. The velocity distribution is as shown.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
12
Figure 1.2 : The Velocity Distribution
Source: Fluid Mechanics by Ogboja ,Olu. UNESCO, Nairobi Kenya,2005.
Consider adjacent two thin layer of fluid moving together at a distance of dy apart and a velocity
difference of dV. As result of difference in their velocities, friction or shear stress will develop between
the two layers. The shear stress between the two layers can be obtained as follows. Assuming the
velocity profile given in Figure 1.3 below,the flow is effected by subjecting the top plate to a force F
which consequently moves with velocity U. The fluid layer adjacent to the plate will adhere to it and
move with velocity U while the lower layer will move with velocity lower than U. The opposing action
between the two layers where the upper accelerates and the lower retards results in shearing of the
fluid. The velocity gradually reduces towards the lower plate where it finally assume zero.
Figure 1.3 : Shearing of Fluid.
Source: Fluid Mechanics by Ogboja , Olu. UNESCO, Nairobi Kenya, 2005.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
13
According to Newton, the force F is directly proportional to the product of the area of the moving
plate, A ,and the velocity gradient :
U
FαA
1.7
d
U
F = µA
1.8
d
Where µ is a constant of proportionality known as coefficient of dynamic viscosity or simply
viscosity .The equation can be rearranged as
F
U
τ= =µ
1.9
A
d
τ is the shear stress at the plate/liquid interface. This equation is valid if d is small and the velocity
profile is assumed to be linear .If d is large and the velocity profile is therefore parabolic, the
differential form of the equation should be used:
dV
1.10
τ=µ x
dy
The above equation is valid when the origin of the coordinate, which is traversed to the direction of
flow, is chosen such that the velocity increases as y increases. Otherwise, it is reversed as:
dV
1.11
τ = −µ x
dy
Viscosity of a fluid can be determined through:
(i)
Cup-and-Bob (Couette) Viscometer and
(ii)
(ii) Tube Flow (Poiseuille) Viscometer.
Kinematic viscosity is sometimes used and it is the ratio of dynamic viscosity to the density:
µ
υ=
1.12
ρ
The unit of kinematic viscosity is m2/s.
1.7
Compressibility of Fluids
The nature of particles of fluids does not allow their handling like that of solid that can be
compressed or stretched .The volume of a given mass of fluid can be changed by increasing or
decreasing the pressure on it. If the fluid is gas ,its volume can also be altered by changing its
temperature. The new state of the perfect gas can be predicted using the perfect gas law. For many
practical purposes, the liquid can be regarded as incompressible. When subjected to large pressure,
the compressibility of the liquid cannot be ignored.
If a liquid of volume V is subjected to a sudden increase in pressure, dp with a corresponding
decrease in volum of dV .The compressibility of the liquid is given as :
dp
κ=−
1.13
dV
V
Where κ is the bulk modulus of elasticity .The negative sign takes care of the fact that dV is negative
since κ cannot be negative.
In terms of density, if m = ρV is differentiated then , we have
dm = ρdV + Vdρ
1.14
Since m is constant, dm =0 and
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
14
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
−
dV dρ
=
V
ρ
1.15
dρ
dV
for
in the initial equation yields
ρ
V
dp
κ=
1.16
dρ
ρ
κ varies with temperature and at any particular temperature, the higher the compression pressure ,the
higher the resistance to compression and the higher the value of κ .The variation of κ with
temperature are available in literature for different liquids.
Substitute
1.8
1.8.1
Surface Tension and Capillarity
Surface Tension
Surface tension, ( σ ), is a property of a liquid surface. It describes the strength of the surface
interactions. The units on surface tension are Force/length . Surface tension is the driving force for
water beading on a waxy surface and free droplets of liquid assuming a spherical shape.
Lowering of surface tension can be accomplished by adding species that tend collect at the surface.
This breaks up the interactions between the molecules of the liquid and reduces the strength of the
surface. Surface tension varies with temperature. Water has a surface tension at 200 is 0.074N/m and at
1000C it is 0.059N/m. Surface tension can be reduced by adding surface-active agents (surfactants) to
the liquid.
A relationship can be derived between surface tension, σ ,and the excess pressure above
atmospheric pressure, p, by considering the force balance on the fluid under consideration. For
instance, a cylinder whose length is l, the surface tension force and the excess pressure force are 2 σ l
and 2prl respectively. The force balance equation is :
2rl (p j − p a ) = 2σl
1.17
And so,
σ
p = (p j − p a ) =
1.18
r
Capillary rise in liquids with adhesive properties and capillary depression in liquids with cohesive
properties are caused by surface tension. Cohesion is the intermolecular attraction between molecules
of the same liquid while adhesion is the attraction between molecules of a liquid and the molecules of
solid bounded surface in contact with the liquid. If the adhesive force between molecules of a
particular liquid and a particular solid is greater than the cohesive force between the liquid molecules,
the liquid tends to stick to the solid and the area of contact between them tend to increase. This
explains why water wets glass but mercury does not and water will not also wet wax or greasy surface.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
15
1.8.2
Capillarity
Figure 1.3: Capillary rise and capillary depression
When a glass tube with small diameter is partially immersed in water and held vertically, the
surface tension will cause the liquid level in the tube to rise above the level of the liquid in the
container. The liquid surface in the tube will form a fairly parabolic meniscus. If the liquid wets glass,
the meniscus is concave. The surface tension force is in the direction which is inclined to the tube at an
angle, θ ,the angle between the tangent to the meniscus and the tube wall. θ is called contact angle.
The capillary rise can be calculated as follows:
2σ cos θ
h=
1.19
ρgr
A parameter for determining when surface tension is important in the flow is call capillary constant,
a. Surface tension effect cannot be ignored when capillary rise is of the order of a.
2σ cos θ
a = rh =
1.20
ρg
1.9
Vapour Pressure
Evaporation occurs when a molecule on the surface of liquid gathers enough energy to break
away from the liquid into the space above it. As molecules escape into the space and random
movement continues, pressure is exerted on the liquid surface. This pressure is called vapour
pressure. When the rate of evaporation is equal to the rate of reabsorption, equilibrium is said to be
reached and vapour pressure at that point in time is called saturation vapour pressure. Evaporation of
is a molecular activity which is promoted by temperature .Therefore, vapour pressure and saturation
vapour pressure depends on the liquid temperature and increase with it.
When the pressure above a liquid at any given temperature is equal to the vapour pressure of
the liquid, the liquid boils and the temperature is called the boiling point. Boiling can occur at any
temperature as long as the pressure above the liquid can be made equal the vapour pressure.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
16
1.10
1
2
3
4
5
6
Problems.
Convert a force of 17 dynes to Newton, poundals and lbf.
A gas whose molecular weight is 28 is kept under a pressure of 200kN/m2 and a
temperature of 600C .Calculate the density of the gas.
The viscosity of an organic liquid is 15cP. Calculate the corresponding values on the fps
and SI system.
Two plates with area 10m2 are 2mm apart. The lower plate is fixed while the upper plate is
moved at a constant velocity of 1m/s by a force of 50N. Determine the viscosity of the
liquid between the plates.
A bubble of a liquid whose surface tension is 0.0289N/m has a radius of 1mm.Calculate the
pressure within the bubble if the atmospheric pressure is 101kN/m2.
Two glass tubes diameter 2 and 4 mm respectively, are attached to the side of a water tank
to measure the level inside the tank,( σ = 0.074N / m). Use this information to express the
capillary rise in the tube in the form h= mr + c where m and c are constants and r is the
tube radius and hence determine the ideal tube diameter.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
17
MODULE 2
2.0
Dimensional Analysis and Dynamic Similitude
2.1
Dimensional Analysis
In building a complex and giant devices in engineering, the usual practice is to start by
designing and building a scale model of the device. The scale model is geometrically exact replica of
the prototype but it may be smaller of bigger in size depending on the convenience. The model is
studied fro performance and efficiency, making alterations until the performance parameters are
satisfactory. The dimensions and performance of the prototype are then deduced from those of the
model by using dimensional anlaysis and similarity criteria which require that the model and prototype
be similar geometrically, kinematically and dynamically. Dimensional analysis is used to determine
which groupings of the flow properties affect the performance of the model and prototype. This
approach saves energy and cost, and tremendously increases the chance of producing a near perfect
design of the prototype at first attempt.
The procedure for obtaining the dimensionless parameters is called Dimensional Analysis .
Dimensional analysis helps to reduce the number of variables by grouping them into dimensionless
parameters. It is then easier to determine the functional relationship between the parameters by
experiment. Where there are many parameters, the relative influence of each parameter can be
determined from experiment; less influential parameters can be dropped while the more influential
parameters can be related empirically using appropriate coefficients and exponents. Many approaches
have been proposed for converting a functional relationship in dimensional variables to a functional
relationship in nondimensional or dimensionless variables. Among them are , the step-by-step method,
Buckingham II Theorem.
2.1.1
Units and Dimensions
The dimensions of a quantity identify the physical character of that quantity, e.g., force (F),
mass (M), length (L), time (t), temperature (T), electric charge (e), etc. On the other hand, Units
identify the reference scale by which the magnitude of the respective physical quantity is measured.
Many different reference scales (units) can be defined for a given dimension; for example, the
dimension of length can be measured in units of miles, centimeters, inches, meters, yards, angstroms,
furlongs, light years, kilometers, etc.
Dimensions can be classified as either fundamental or derived. Fundamental dimensions
cannot be expressed in terms of other dimensions and include length (L), time (t), temperature (T),
mass (M), and/or force (F) (depending upon the system of dimensions used). Derived dimensions can
be expressed in terms of fundamental dimensions, for example, area ([A] = L2), volume ([V] =L3),
energy ([E]=FL=ML2/t2), power ([HP]=FL/t=ML2/t3),viscosity ([µ]=Ft/L2=M/Lt) e.t.c.
2.1.2
Conservation of Dimensions
It states that the terms in an equation which is used to express a relationship between physical
quantities must be dimensionally homogenous. For any equation to be valid, every term in the
equation must have the same physical character, i.e., the same net dimensions (and consequently the
same units in any consistent system of units).. This principle was enunciated by Fourier in 1922.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
18
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
The principle simply implies that if :
a = b+c
2.1
a, b and c must have the same dimensions of if:
d=ef
2.2
then the dimensions of d must equal to the product of those of e and f. Note that what gives the
homogeneity of dimension here is the combined dimensions of e and f and not individual
dimensions. Take for instance, the equation which expresses the distance moved by a body falling
under gravity:
1
s = gt 2
2.3
2
Since g is a constant, one can rewrite the equation as :
s = kt 2
2.4
Where k is a constant that must be dimensional constant having the same dimension as g since s is not
equal in dimension to t2.
2.2
Techniques of Dimensional Analysis
2.2.1 Step-by- Step Method
In this procedure, dimensions of variables of functional relationship are rendered dimensionless in
mass, length and time by step by step. At each step, one of the variables is taken and combined with it
to eliminate it and with others to render them dimensionless in whatever dimensions one desire. It may
be necessary to use multiples of the variables in rendering some of the others dimensionless.
For example, the pressure drop ∆p experienced by a fluid in turbulent motion over a length l of a
smooth pipe has been found to be a function of the fluid velocity V, the pipe diameter D, the fluid
density ρ, and the fluid viscosity µ. Use the step-by-step method to reduce the functional relationship to
that of dimensionless terms
The relationship can be expressed as follows
∆p
= f (V , D, ρ, µ )
l
The dimensions of the variables are:
M
 ∆p 
=
 l  L2 T 2
[V ] = L
T
[D] = L
2.5
[ρ] = M3
L
[µ ] = M
LT
We can render the variables dimensionless in mass (M) by dividing them through by ρ to obtain

∆p
ρ µ
= f1  V , D, , 
2.6
lρ
ρ ρ

And since ρ/ρ = 1,
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
19

∆p
µ
= f1  V , D, 
ρ
lρ

The dimensions of the terms in this equation are:
 ∆p  L
 lρ  = 2
  T
[V ] = L
T
[D] = L
2.7
2.8
 µ  L2
ρ  = T
 
Next ,we render the equation dimensionless in length L by dividing ∆p/lρ ,V, and D by D and µ/ρ by
D2 to obtain
V µ 
∆p
= f 2  , 2 
2.9
lρD
 D ρD 
With the dimensions
 ∆p 
1
 lρD  = 2

 T
V 1
2.10
D = T
 
 µ  1
 2=
 ρD  T
And finally, we render this equation dimensionless in time (T) by dividing the left term by (µ/ρD2)2
and the right hand terms by just µ/ρD2 to obtain
 ρVD 
∆pρD 3

2.11
= f 3 
2
lµ
 µ 
Inspection will show that both terms of this equation are dimensionless. Note that in the last step, we
used µ/ρD2 to render the equation dimensionless in T; we could have used V/D and obtain
 µ 
∆pD 2


2.12
=
f
4
lρDV 2
 ρVD 
Note that the final results are not the same, only as a result of the choice of the variable we have used
in nondimensionalising in T. Both approaches are correct and experiment can be employed to prove
this point. In obtaining the dimensionless parameter in the above example, we adopted the order of
M,L and T. Do note that there is no rigidity as to which order to adopt. We ended up with two
parameter having carried out exercise on five variables employing three basic dimensions.
2.2.2 Buckingham Π Theorem
The theorem proves the conclusion we reached at the end of the last section that is, if a functional
relationship contains m variables with a total of n basic dimensions, on the application of dimensional
analysis, the relationship will contain m-n groups of dimensionless groups . In the step –by-step
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
20
procedure, the dimensionless groups are determined together whereas in this procedure ,the groups are
determined one by one and each is designated as a Π .The procedure is as follows as illustrated in the
following example.
∆p
Example: Solve the problem
= f (V, D, ρ, µ, g ) using Buckingham’s procedure.
l
Solution:
The problem can be stated as
 ∆p

f 
, V, D, ρ, µ, g  = 0
2.13

1 l
There are 6 variables and 3 basic dimensions M, L and T. Therefore, the solution should yield 3 Π
terms. In selecting the repeating variables, it has been found to be helpful to the interpretation of the
functional relationship to select one to reflect the rate of flow of the fluid, another to reflect mass of
the fluid and the last one to reflect the characteristic dimension of the conduit. Consequently, for this
problem ,V, ρ and D are selected. The Π terms are therefore :
∆p
Π 1 = V a1ρ a 2 D a3
l
b1 b 2 b 3
Π2 = V ρ D µ
2.14
Π 3 = V c1ρ c 2 D c3 g
Substituting dimensions for the variables, we have,
a1
a2
L  M
M L T =   3
T L 
0
0
b1
L  M 
M L T =   3
T L 
0
0
(L )a3 
M 

 L T2 
0
b2
0
L
M 0 L0 T 0 =  
T
c1
M
 3
L 
c2
2
(L )b3 
M 

L T 
2.15
(L )c3 
L 

 T2 
Equating the exponents for M, L and T in the first equation above gives:
M:
a2 +1 =0
L:
a1 -3a2 +a3 -2 =0
T:
-a1 -2=0
Solving these three equations gives : a1= -2 ,a2 = -1 and a3=1
Therefore,
∆p D
Π1 =
l ρV 2
Equating the exponents for M, L and T in the second equation above gives:
M:
b2+1 =0
L:
b1-3b2+b3-1 =0
T:
- b1-1=0
Solving these three equations gives : b1 =-1, b2 =-1 and b3 = -1
Therefore
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
2.16
21
µ
ρVD
Equating the exponents for M, L and T in the third equation above gives:
M:
c2 = 0
L:
c1-3c2+c3+1 = 0
T:
- c1 - 2 = 0
Solving these three equations gives :c1 = 2; c2 = 0 and c3 =1
And consequently,
gD
Π3 = 2
V
Finally, the resulting dimensionless equation is
 ∆p D
µ gD 
f 2 
,
, 2  = 0
2
 l ρV ρVD V 
Π2 =
2.3
2.17
2.18
2.19
Interpretation of Dimensionless Numbers
Table 2.1 lists some dimensionless groups that are commonly encountered in fluid mechanics
problems. The name of the group, and its symbol, definition, significance, and most common area of
application are given in the table. Wherever feasible, it is desirable to express basic relations (either
theoretical or empirical) in dimensionless form, with the variables being dimensionless groups,
because this represents the most general way of presenting results and is independent of scale or
specific system properties.
Table 2.1 Dimensionless Groups in Fluid Mechanics
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
22
Source: Chemical Engineering Fluid Mechanics by Ron Darby. Marcel Dekker, Inc. New York,2nd
Edition, 2001
2.4 Similitude
Similitude is the science of predicting the behaviour of a large object from the behaviour of the
smaller object which is geometrically similar. The former is called prototype and the later model.
The model ids built and subjected to various forces that the prototype would encounter. The
information thereby obtained is used to improve the design. Model studies can be carried out on
components to further improve the performance of the prototype. In order to achieve total
similarity between the prototype and the model, there must be kinematic and dynamic similarity in
addition to geometrical similarity.
Kinematic similarity implies that the ratio of velocity and the ratios of quantities derived from
velocity at corresponding points of the model and prototype must be constant while the dynamic
similarity implies that the ratio of dynamic forces at corresponding points of the prototype and
model must be constant. The dynamic similarity implies that the dimensionless numbers in Table
2.1 must be the same for corresponding points of the model and the prototype.
2.5 Application of Similarity Theory
2.5.1
Submerged Bodies
The behaviour of submerged bodies are studied in water and wind tunnels;
water being preferred where very high velocities are required. Tunnel studies are used
to determine drag characteristics and compressibility characteristics where very high air
velocities are involved. The Reynolds model is used to achieve dynamic similarity
between prototype and model flows.
Example: The drag characteristics of a new brand of motor car is to be studied in a
wind tunnel .The speed of the car in still air is to be limited to 150km/hr. A scale model
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
23
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
of 1:10 is to be used. What is the maximum speed that should be employed in the
tunnel?
Solution :
The flow characteristics as depicted by Reynolds number should be used in the study.
Therefore Re should be the same in both model and prototype :
Rem = Rep
ρ m Vm L m ρ p Vp L p
=
µm
µp
Vm = Vp
Lp ρpµ m
Lm ρmµ p
= 150 x 10 x 1 x1
= 1,500km/hr.
2.5.2 Pipe Flow
A typical flow of incompressible fluids in closed conduit is the flow in pipes. For such flows,
gravity, surface tension and compressibility forces have no effect and the flow is controlled solely by
pressure and viscosity. Therefore, the Reynolds model is applicable for the analysis of the flow. The
flow velocity varies across the pipe, being zero at the wall and maximum at the centre. Depending on
the mean velocity through the pipe, the mode of flow can be laminar or turbulent. For Re ≤ 2000, the
flow is laminar and viscous forces predominate. When Re > 2000, inertia forces predominate.
Complete similarity is achieved between the model and the prototype for laminar flow under all
conditions. However, for turbulent flow, there has to be geometric similarity between the roughness of
the pipes and their laminar sub layer before dynamic similarity can be attained between the model and
the prototype flows.
Example : A venture meter is to be used to measure the flow through a pipe of 1m diameter.
The throat of the prototype is 0.1m in diameter and the fluid velocity is 5m/s.Determine the
discharge through the model for a scale of 1:10.
Solution
 ρVD 
 ρVD 
 = 

The equation 
 µ m  µ p
Reduces to (VD )m = (VD )p If the same liquid is used in both model and prototype.
Substituting the given values yields
Vm x 0.01=0.1 x 5
Vm = 50m/s
The discharge through the model is
πd 2
Qm =
Vm
4
π
= x0.012 x50
4
= 0.004m3/s
2.6 Problems
1 An ethylene storage tank in your plant explodes. The distance that the blast wave travels from the
blast site (R) depends upon the energy released in the blast (E), the density of the air ( ρ ), and
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
24
2
3
4
5
6
time (t). Use dimensional analysis to determine the dimensionless group(s) that can be used to
describe the relationship between the variables in the problem
Determine the number of the dimensionless parameters for the following functional relationship:
∆h
= f 6 (V , D, ρ, µ, g ) using step –by step method
l
In general, the pressure drop ∆p of a fluid in motion in a conduit depends on the length of the pipe l
,its diameter D, the fluid velocity V, density ρ, viscosity µ, surface tension σ, bulk modulus k ,
gravitational acceleration g and pipe roughness e. Determine the Π term for the functional
relationship
Water is used to simulate the flow characteristics of oil in an industrial pipeline. The diameter of the
test pipeline is 1/10th of that of the industrial pipeline. The viscosities of water and oil are 3.0 x10-4
and 2.9 x 10-2 N.s/m2 while their densities are 1000 and 1720 kg/m3 respectively. If the water
velocity in the test pipeline is 10m/s, what is the corresponding oil velocity in the industrial
pipeline?
The head on a spillway is 1.5m and the discharge is 100m3/min. What is the head on the scale
model of 1:80. Determine the model discharge.
Water discharge through a pipe of length 50m and diameter 0.5m at 3m3/s. For this flow, f = 0.025
. Determine the head loss through the model pipe for a scale ratio of 1:10.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
25
MODULE 3
3.0
Fluids Statics.
3.1 Concept of Stress and Pressure
Fluid statics is the study of the forces that keep a body of fluid in static equilibrium. Static
equilibrium implies that the resultant of all forces does not result in translation or rotation of the body.
Consequently, the resultant moment must be zero. Linear and angular acceleration are therefore zero
and shear stresses are also zero. Since there is no rotation, the body of fluid is in equilibrium under the
influence of normal force. In the absence of mechanical, electrical ,thermal, magnetic and gravitational
forces, the normal forces are hydrostatic forces which are directly proportional to the weight of the
fluid above the body.
The most important characteristics of static fluid is the pressure it exerts on the body floating on
or immersed in it. The pressure on a given plane in a fluid is defined as the fluid force acting
perpendicularly to the plane, that is
F
p=
3.1
A
Where F is the force which is acting perpendicular ly on area A. This equation is based on the
assumption that force F will be the same all over the points of the area A. If the force is not uniformly
distributed, than expression is for the average pressure on A. The pressure at any point on A can be
determined by noting the force ∆F on a small area ∆A surrounding the point. When ∆A tends to a
point, the ration of ∆F and ∆A yields pressure at a point as expressed below:
∆F
3.2
∆A
Force is a vector quantity, which consists of normal surface force, and body force that is due to the
weight of the element, the hydrostatic pressures in the fluid acts in the three Cartesian coordinates
while the normal pressure may act in any direction normal to the point of action. Since the hydrostatic
pressures in various directions are mutually perpendicular to each other but not to the normal pressure,
it is then concluded that the three directions represent arbitrary directions and therefore the pressure at
a point acts with equal magnitude in all directions. Conclusively, pressure is therefore taken as a scalar
quantity.
p = lim ∆A → 0
3.1.1 Pressure Head
The pressure at a point in a fluid is usually expressed in terms of the height h of the fluid above
the point. The height h is usually referred to as the head of the liquid above the point. The relationship
between pressure at a point and the height of fluid above that point can be deduced as follows:
Consider the column of fluid of arbitrary cross sectional area ∆A and height h standing
perpendicularly on area ∆A in the fluid as shown in figure 3.1.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
26
h
∆A
Figure 3.1 : Pressure Head
If the fluid acting on the upper bounding area of the column is ignored, the force on ∆A is solely that
due to the weight of the elemental column which is ρ(h∆A )g .Therefore ,the pressure due to this
weight is :
ρ(h∆A )g
p=
= ρhg
3.3
∆A
Since g is constant for a given location on the earth’s surface and ρ is constant for a given fluid, p can
be computed easily once h is known and vice versa.
3.1.2 Pressure Variation with Depth
Consider Figure 3.2 below, the effective force on the element are hydrostatic pressure forces in the
z –direction and the weight of the element. Hydrostatic pressure acts inwards. Since the fluid is at rest,
shear forces are zero and the forces on opposite side of the pipe in the direction x and y directions are
equal and opposite and therefore cancel out. Since the element is in static equilibrium , the sum of the
forces on it must be equal to zero.
Figure 3.2 : Pressure Variation with Depth
δp 

p∆x∆y −  p +
∆z ∆x∆y − ρg∆x∆y∆z = 0
δz 

Which reduces to
δp
= − ρg
δz
3.4
3.5
Since p is a function of z only, we can write
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
27
dp = −ρgdz
3.6
This is the hydrostatic equation and it is the general expression for pressure variation in static fluids.
For fluids that are incompressible, ρ is constant and therefore the general equation can be integrated to
give :
p z = − ρgz + p 0
3.7
If the fluid is a liquid with a free surface whose surface is open to the atmosphere, p0 then represents
air pressure on the free surface and pz the pressure at depth –z from the free surface. If h is substituted
for –z and p for pz - p0 ,we have
p = ρgh
3.8
Note that h is measured downwards while z is measured upwards. The above equation is the
hydrostatic law of pressure variation in static fluids and it is used to determine the pressure
difference between two points in a static fluids.
For pressure variation in compressible fluid, the density varies with position. If we assume
the fluid to be a perfect gas, we can make use of the equation of state to relate pressure and density:
p
= RT
3.9
ρ
Combining this equation with the general hydrostatic law equation gives:
dp
g
=−
dz
3.10
p
RT
If we assume an isothermal case (T= constant) and integrate between limits:
z = z 0 , p = p0
z=z,p=p
where p0 is a place whose pressure is known such as earth’s surface, we obtain
p dp
g Z
=−
3.11
∫
∫ dz
RT Z0
p0 p
which yields
 p 
g
g
(z − z 0 )
(z − z 0 )}
ln  = −
or
p = p 0 exp{−
3.12
RT
RT
 p0 
3.2 Pressure Transmission
Compressible fluids decrease in volume when subjected to increased pressure while incompressible
fluids do not change noticeably when subjected to increased pressure. When an incompressible fluid
experiences increased pressure at any point, the increase in pressure is transmitted equally in all
directions to the other points such that the difference in pressure between any tow points in the fluids is
that due to hydrostatic head. Consider figure 3.3 in which a tank containing liquid is open to the
atmosphere. The surface pressure is the atmospheric pressure pa .which is pressure at point A.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
28
Figure 3.3 Pressure Transmission
The pressure at depth h below the surface is higher than the pressure at the surface by an amount due
to the head of liquid of height h. The pressure at point B and C at the same depth h is given as :
3.13
p B = p C = p A + ρhg = p a + ρhg
This is also applicable to figure 3.3 b for point B and P:
p B = p P = p a + ρHg
3.14
This is applicable in determining the pressure transmitted by an hydraulic jack as shown on figure 3.4.
Suppose pressure pA due to a weight of mass m is applied to the surface of the liquid at A to produce
pressure pD at D to raise the bigger mass M.
Figure 3.4 . The Hydraulic Jack
From the earlier derivations,
p B = p A + ρhg
p C = p D + ρHg
Since B and C are on the same level in the liquid, pB = pC and therefore
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
3.15
29
p A + ρhg = p D + ρHg
3.16
If the head terms are small or if h ≈H, then
pA = pD
3.17
This equation can be used to calculate the effort required to raise a given load.
3.3 Pressure Measurement
Pressure is a measure of the force of the bombardment of the fluid particles in a space on the
wall containing the fluid particles. When the space is completely empty of particle, it is called a
vacuum and its pressure is zero pressure. Pressure measured with reference to zero pressure is
called absolute pressures. Atmospheric pressure is the pressure of the atmosphere measured
with reference to zero pressure. In most practical situations, such pressures measured with respect
to the atmospheric pressures are called gauge pressures. Many devices are available for
measuring pressures
3.3.1 Barometers : These devices are used to measure atmospheric pressure. The simplest of
them is the mercury barometer. Aneroid barometers are used to measure absolute
pressure. The Bourdon gauges are used to measure gauge pressure.
Figure 3.5 : The Bourdon Gauge
3.3.2
Manometers : are devices which employ change in elevation to determine pressure or
pressure differences. Simple manometers measure gauge pressures while differential
manometers measure pressure difference.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
30
Figure 3.6 : Simple Open-end manometer.
Figure 3.7 : Differential manometer.
3.4 Forces on Submerged Surface
Every point on a submerged surface experiences an amount of pressure, which is directly
proportional to the head of fluid above it. The total fluid force on a submerged surface can be obtained
by summing the forces on the elemental areas that make up the surface. The resultant force can be
assumed to act through a point on the surface called the centre of pressure.
3.4.1 Horizontal Plane Surface
Consider the figure 3.8 below where z- axis is assumed to be the fluid’s free surface. If the
surface is at depth y below the surface of the fluid and the surface on z –axis is horizontal where
all the elements of the surface experiences same pressure as discussed before, then the total or
resultant force is therefore normal to the surface and is given as :
F = ∫ pdA = pA
A
Where A is the area of the surface. The line of action passes through the centre of pressure which is
located at the centre of the area or centroid. The distance of the line of action form y-axis can be
obtained by taking moments about the y-axis.
Fx = pAx = ∫ pxdA
3.18
A
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And since p is constant
x=
1
∫ xdA
AA
3.19
Figure 3.8 Fluid force on horizontal Surface
3.4.2
Vertical Surface
Consider fluid force on a vertical surface AB which is perpendicular to x-axis as shown in
figure 3.9. Points on the surface are at different depth and hence experiences different pressures.
The coordinate axes have to be chosen such that the h axis is parallel to AB and both axes intersect
at the liquid surface at O. The centre of area (centroid) is represented by C and the centre of
pressure is P. The pressure at any point h is
p = ρhg
3.20
And the force on the elemental area δA is
δF = ρhgδA
3.12
The total forces on the surface is therefore
F = ρ g ∫ h δA
3.22
A
The term in integral is defined as the first moment of area about the x-axis which is usually written
as hCA ,with hC being the depth of the centroid of the area and consequently,
F = ρgh C A
3.23
But the expression for pressure at the centroid is
p C = ρgh C
3.24
Therefore, F = p C A
3.25
The resultant hydrostatic force on the vertical surface is the product of the pressure at the centroid
and the area of the surface
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Figure 3.9 : Vertical Submerged Surface
3.4.3 Inclined Surface
The pressure distribution on an inclined surface AB is as shown on figure 3.10.The force on an
elemental area δA is
δF = pδA = ρghδA = ρgy sin αδA
3.26
Where h is the depth of the element and y its distance from O. The total force on the surface after
integration is given as :
F = ρgAy C sin α
3.27
But from the notation in the figure, the depth of the centroid is:
h C = y C sin α
3.28
Therefore , F = ρgh C A = p C A
3.29
Where pC is the pressure at the cenroid of the surface. This is the same as the result obtained for
vertical surface. The y coordinate of the centre of pressure can be determined by equating the
moments of the resultant force about x-axis to those of the distributed forces:
y P F = ∫ ypdA
3.30
A
On substituting ρgAy C sin α for F and ρgy sin α for p in the above equation
1
Ix
2
yp=
3.31
∫ y dA =
yCA A
yCA
Where Ix is the second moment of area of the plane about the x-axis. By the application of the
parallel axes theorem, the equation becomes:
I
yp= C + yC
3.32
yCA
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Figure 3.10: Force on an Inclined Surface
3.4.4 Curved Surface
The analysis of the hydrostatic forces on a curved surface can be simplified by considering the
horizontal and Vertical components of the forces separately. Consider the curved surface shown
on figure 3.11a.The fluid force on the surface has horizontal and vertical components FH and FV
respectively. It is convenient to consider FH as acting on the surface CD which is the projection of
the curved surface on the vertical plane. Similarly, FV can be considered as acting on the plane ED
which is the projection of the surface on the horizontal plane. The magnitudes and line of action of
these components can be determined as has been done for plane surfaces and the magnitude and
line of action of their resultant forces can then be determined from vector consideration.
Figure 3.11: Force on Curved Surface
Since we are considering a static situation, the surface must develop a reaction to the fluid
forces whose components are RH and RV respectively. For equilibrium to occur, FH and RH must
be equal ,opposite and have same line of action. This must also be true for FV and RV .
FH = ρg x area of CD x depth of centroid of CD
3.33
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FH acts through centre of pressure of the vertical projection of the curved surface while FV is equal to
the weight of the fluid and acts through its centre of gravity. Figure 3.11b can be treated in similar
manner.
3.5
Buoyancy
When a body is immersed in a fluid, it experiences two fluid forces in each of the three
coordinates directions. The two forces in each of the to horizontal directions are equal and opposite
.There are therefore no net effects in these directions. The downward and upward forces in the vertical
direction are not equal and therefore there is net upward force in this direction. The force is called
Buoyant Force.
Figure 3.12: Upthrust on a submerged body
Considering figure 3.12 above, the two forces are opposite and the net upward fluid force is given by :
F = (p 2 − p1 ) A
3.34
The net upward force F is the weight of the fluid corresponding to the volume of fluid displaced. This
is Archimedes Principle. If the body is partially immersed the upthrust is equal to the volume of the
fluid displaced.
F = ρVg
3.35
Where V is the volume of the fluid displaced, ρV its mass and ρVg its weight.
The upthrust acts upwards through the centroid of the displaced volume of fluid. The cenroid can be
determined by taking moments about an axis normal to the vertical projection of the volume. Consider
figure 3.13, the upthrust on the elemental volume whose cross sectional area is δA is given by:
δF = (p 2 − p1 )δA = ρghδA = ρgδV
3.36
The upthrust on the whole volume is obtained by integrating the equation above:
F = ∫ dF = ∫ ρgdV =ρg ∫ dV = ρgV
3.37
V
V
V
We can now equate the sum of the moments due to all the elemental volumes which make up the
object about the axis through point O to that of the total upthrust F:
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Fx = ρg ∫ xdV
3.38
V
Where x is the distance of the centroid from the selected axis. If we substitute for F and rearrange:
1
x = ∫ xdV
3.39
VV
The centroid of the displaced fluid is called centre of buoyancy
Figure 3.13: Determination of Centre of Buoyancy
3.5.1 Hydrometry
The principle of buoyancy is applied to the subject of hydrometry for measuring the specific
gravity of liquids. The device is call hydrometer and is as shown in figure 3.14. The stem of
hydrometer cam be calibrated to measure specific gravity directly. The calibration equation is
derived as follows . When immersed in pure water, the upthrust on the hydrometer is given as
F = W = ρ w Vw g
3.40
When immersed in another liquid then,
F = W = ρ l Vl g
3.41
Let the point on the hydrometer stem which coincide with the water surface be marked h=0.
Suppose the second liquid is denser so that the hydrometer will sink less. Let h denote the
distance of the new liquid level from the h=0 mark. If a is the cross sectional area of the stem, the
displaced volume can be related as
Vl = Vw − ah.
3.42
Combining the equations givens:
ρ l (Vw − ah) = ρ w Vw
3.43
If we make h the subject of the equation and writes s=ρl/ρw , the specific gravity of the liquid is :
V 1 V
h=− w + w
3.44
a s
a
This equation is a straight line equation of the form y= mx+c. Therefore -Vw/a represents the
gradient of the plot of h versus 1/s and Vw/a its intercept. This equation can be used to calibrate an
hydrometer by measuring Vw and a and then determining h and s for various liquids. The specific
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gravity values rater than h can then be marked on the stem of the hydrometer. This way the
hydrometer can be used to measure specific gravities directly.
Figure 3.14: The Hydrometer
Figure 3.15 : Calibration graph for the Hydrometer
3.6 Static Forces on Solid Boundaries
A floating object whether partially of fully immersed ia said to be stable if it tends to return to its
original position after a small displacement. Such a body may be linearly or rotationally displaced.
The upthrust on a fully immersed floating body is the weight of the fluid displaced by the volume
of the body. Since the volume is constant, the upthrust will remain the same no matter the depth at
which the body is. Displacing the body linearly does not change the equilibrium between the two
forces since neither of the two is changed in magnitude by the displacement, but it only assumes new
position. However, when a body is rotationally displaced, its ability to restore itself back to the original
position after a small rotational displacement depends on the relative positions of the centre of
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buoyancy and gravity. Figure 3.16 shows various positions of a loaded sphere to illustrated the effect
of centre of buoyancy and gravity on its stability.
(a) Stable
(b) Neutral
Figure 3.16: A fully –Immersed Floating body.
(c) Unstable
3.6 Problems
1. If an egg can be crushed by pressure of 200kPa, at what depth will it burst in sean water of
specific gravity 1.1?
2. Predict the atmospheric pressure at a height of 1000m above sea level on a day when the
temperature is 300C.Assume that the air density at sea level is 1.15kg/m3 and take the gas
constant as 287J/kg.K. State other assumptions you make.
3. If the effort and load arm areas of a hydraulic jack are in the ratio 1:25, determine the effort
required to raise a load of one tonne mass.
4. A gate of a reservoir is hinged at a point and it is prevented from opening by the force in the
spring. The gate is 3m high and 1m wide. The level of water is 2m above th top of the gate.
Determine the force in the spring if the gate is just about to open.
5. The wall of a reservoir on which the slide gate is located is inclined at 300 to the horizontal.
The gate is hinged at its upper end and a massive weight is pushed against its lower end to keep
it closed. Determine the reaction of the weight and its direction if the gate has a circular cross
section of diameter 3m.
6. A hydrometer whose mass is 25g is made up of a bulb of diameter 2cm and height 5cm and a
stem of diameter 1cm and height 10cm. Determine the range of specific gravities that the
hydrometer can be employed to measure.
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MODULE 4
4.0
BASIC CONSERVATION LAWS
4.1 THE SYSTEM
The basic principles that apply to the analysis and solution of flow problems include the
conservation of mass, energy, and momentum in addition to appropriate transport relations for these
conserved quantities. For flow problems, these conservation laws are applied to a system, which is
defined as any clearly specified region or volume of fluid with either macroscopic or microscopic
dimensions (also referred to as a ‘‘control volume’’) as illustrated in Figure 4.1.
Figure 4.1: A system with inputs and outputs.
The general conservation law is :
4.1
where X is the conserved quantity, i.e., mass, energy, or momentum. In the case of momentum,
because a ‘‘rate of momentum’’ is equivalent to a force (by Newton’s second law), the ‘‘rate in’’ term
must also include any (net) forces acting on the system. It is emphasized that the system is not the
‘‘containing vessel’’ (e.g., a pipe, tank, or pump) but is the fluid contained within the designated
boundary. The following sections will show how this generic expression is applied for each of these
conserved quantities.
4.2 CONSERVATION OF MASS
The continuity equation is a mathematical statement of the law of conservation of mass, which
states that the mass of a system is constant. If the system is a open one, for example, a control volume
into and from which fluid flows, then for conservation of mass, the rate of mass inflow from the
control volume minus the rate of mass outflow from the control volume must equal the rate of increase
of mass within the control volume.
4.2.1 Macroscopic Balance
For a given system (e.g., Fig. 4.1), each entering stream (subscript i) will carry mass into the system
& i ), and each exiting stream (subscript o) carries mass out of the system (at rate m
& o ). Hence,
(at rate m
the conservation of mass, or continuity equation for the system is:
4.2
where ms is the mass of the system.
For each stream,
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4.3
that is, the total mass flow rate through a given area for any stream is the integrated value of the local
mass flow rate over that area. Note that mass flow rate is a scalar, whereas velocity and area are
vectors. Thus it is the scalar (or dot) product of the velocity and area vectors that is required. (The
‘‘direction’’ or orientation of the area is that of the unit vector that is normal to the area.). The
corresponding definition of the average velocity through the conduit is :
4.4
r
Where
is the volumetric flow rate and the area A is the projected component of A that is
r
r
r
normal to V (i.e., the component of A whose normal is in the same direction as V ). For a system at
steady state, Equation 4.1 reduces to
or
4.5
4.2.2 Microscopic Balance
The conservation of mass can be applied to an arbitrarily small fluid element to derive the
‘‘microscopic continuity’’ equation, which must be satisfied at all points within any continuous fluid.
This can be done by considering an arbitrary (cubical) differential element of dimensions dx, dy, dz,
with mass flow components into or out of each surface, e.g.,
4.6
Dividing by the volume of the element (dx, dy, dz) and taking the limit as the size of the element
shrinks to zero gives
4.7
This is the microscopic (local) continuity equation and must be satisfied at all points within any
flowing fluid. For a steady state flow, the right hand side becomes zero. If the fluid is incompressible
(i.e. constant ρ), Equation 4.7 reduces to
4.8
Example 4.1 : Water is flowing at a velocity of 7 ft/s in both 1 in. and 2 in. ID pipes, which are joined
together and feed into a 3 in. ID pipe, as shown in Figure below. Determine the water velocity in the 3
in. pipe.
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Solution:
Because the system is at steady state, Equation 4.5 applies:
For constant density, this can be solved for V3:
Example 4.2 : Water flows through a divergent conduit at the rate of 1000l/min. Determine the
velocities at the ports if their diameters are 100mmand 150mm respectively.
Solution
1 3
m /s
60
= 16.67 x10 −3 m 3 / s
Q = 1000x10 − 3 x
π
x0.12 m 2 = 0.0079m 2
4
π
A 2 = x0.152 m 2 = 0.0177m 2
4
From Equation 4.4 V1=Q/A1 and V2=Q/A2 ,therefore
A1 =
16.67 x10 −3
= 2.11m / s
0.0079
16.67 x10 − 3
V2 =
= 0.94m / s
0.0177
Example 4.3 The x- component of the velocity vector of a flow field is Vx= xy2-2y+x2. If the velocity
vector satisfies the continuity equation for steady incompressible flow, determine the expression for Vy
V1 =
SOLUTION
For 2 –dimensional incompressible steady flow, the continuity equation is
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∂Vx ∂Vy
=0
+
∂x
∂y
Therefore , Vy = ∫
∂Vx
∂y + C
∂x
Where C is constant of integration which may be a function of time and y. From the equation for Vx
provided,
∂Vx
= y 2 + 2x
∂x
1

Consequently, Vy = − y 3 + 2xy  + C
3

If it is assumed that y =0, Vy = 0, then C = 0 and
1

Vy = −  y 3 + 2xy 
3

From which we obtain,
∂Vy
∂y
(
= − y 2 + 2x
)
Hence ,
∂Vx ∂Vy
=0
+
∂x
∂y
4.3 CONSERVATION OF ENERGY
Energy can take a wide variety of forms, such as internal (thermal), mechanical, work, kinetic,
potential, surface, electrostatic, electromagnetic, and nuclear energy. We will consider only internal
(thermal), kinetic, potential (due to gravity), mechanical (work), and heat forms of energy.
For the system illustrated in Figure 4.1,a unit mass of fluid in each inlet and outlet stream may
contain a certain amount of internal energy (u) by virtue of its temperature, kinetic energy (V2/2) by
virtue of its velocity, potential energy (gz) due to its position in a (gravitational) potential field, and
‘‘pressure’’ energy (P/ρ). The pressure energy is sometimes called the flow work, because it is
associated with the amount of work or energy required to inject a unit mass of fluid into the system or
eject it out of the system at the appropriate pressure. In addition, energy can cross the boundaries of the
system other than with the flow streams, in the form of heat (Q) resulting from a temperature
difference and shaft work (W). Shaft work is so named because it is normally associated with work
transmitted to or from the system by a shaft, such as that of a pump, compressor, mixer, or turbine.
The sign conventions for heat (Q) and work (W) are arbitrary and consequently vary from one
authority to another. Heat is usually taken to be positive when it is added to the system, so it would
seem to be consistent to use this same convention for work (which is the convention in most scientific
references). However, engineers, being pragmatic, use a sign convention that is directly associated
with value. That is, if work can be extracted from the system (e.g., to drive a turbine) then it is positive,
because a positive asset can be sold to produce revenue. However, if work must be put into the system
(such as from a pump), then it is negative, because it must be purchased (a negative asset). This
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convention is also more consistent with the driving force interpretation of the terms in the energy
balance, as will be shown later.
The conservation of energy equation for any system is as follows:
4.9
Here,
is the enthalpy per unit mass of fluid. Note that the inlet and exit streams include
enthalpy (i.e., internal energy, u, and flow work, ), whereas the system energy includes only the
internal energy but no
flow work (for obvious reasons). If there are only one inlet stream and one
exit stream
and the system is at steady state, the energy balance becomes:
4.10
Where
are the heat added to the system and
work done by the system, respectively, per unit mass of fluid. This expression also applies to a system
comprising the fluid between any two points along a streamline within a flow field. Specifically, if
these two points are only an infinitesimal distance apart, the result is the differential form of the energy
balance:
4.11
The d( ) notation represents a total or exact differential and applies to
Where
those quantities that are determined only by the state (T, P) of the system and are thus point properties.
The δ( ) notation represents quantities that are inexact differentials and depend upon the path taken
from one point to another.
Note that the energy balance contains several different forms of energy, which may be
generally classified as either mechanical energy, associated with motion or position, or thermal
energy, associated with temperature. Mechanical energy is useful, in that it can be converted
directly into useful work, and includes potential energy, kinetic energy, flow work and shaft work. The
thermal energy terms, i.e., internal energy and heat, are not directly available to do useful work unless
they are transformed into mechanical energy, in which case it is the mechanical energy that does the
work.
Let us take a closer look at the significance of enthalpy and internal energy, because these
cannot be measured directly but are determined indirectly by measuring other properties such as
temperature and pressure.
4.3.1 Internal Energy
An infinitesimal change in internal energy is an exact differential and is a unique function of
temperature and pressure (for a given composition). Since the density of a given material is also
uniquely determined by temperature and pressure (e.g., by an equation of state for the material), the
internal energy may be expressed as a function of any two of the three terms T, P and ρ
Hence,
4.12
By making use of classical thermodynamic identities, this is found to be equivalent to
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4.13
Where
4.14
is the specific heat at constant volume (e.g., constant density).
For an ideal gas,
4.15
Thus Equation 4.13 reduces to
4.16
which shows that the internal energy for an ideal gas is a function of temperature only.
For a non-ideal gas, Equation 4.15 is not valid, so
4.17
Consequently, the last term in Equation4.13 does not cancel as it did for the ideal gas, which means
that
4.18
The form of the implied function, fn(T, P), may be analytical if the material is described by a
non-ideal equation of state, or it could be empirical such as for steam, for which the properties are
expressed as data tabulated in steam tables.
For solids and liquids, ρ≈constant (or dυ=0), so
4.19
This shows that the internal energy depends upon temperature only ( just as for the ideal gas, but for an
entirely different reason).
4.3.2 Enthalpy
The enthalpy can be expressed as a function of temperature and pressure:
4.20
which, from thermodynamic identities, is equivalent to
4.21
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Here,
4.22
is the specific heat of the material at constant pressure. Let us again consider some special cases.
For an ideal gas
4.23
Thus Equation 4.21 for the enthalpy becomes
4.24
which shows that the enthalpy for an ideal gas is a function of temperature only (as is the internal
energy).
For a non ideal gas
4.25
which, like ∆u, may be either an analytical or an empirical function. All gases can be described as ideal
gases under appropriate conditions (i.e., far enough from the critical point) and become more nonideal
as the critical point is approached. That is, under conditions that are sufficiently far from the critical
point that the enthalpy at constant temperature is essentially independent of pressure, the gas should be
adequately described by the ideal gas law.
For solid and liquids υ =1/ ρ ≈ constant, so that (∂υ/∂T)p= 0 and
Therefore,
4.26
Or
4.27
This shows that for solids and liquids the enthalpy depends upon both temperature and pressure. This
is in contrast to the internal energy, which depends upon temperature only.
4.3.3 Bernoulli Equation
The Bernoulli equation states that the total energy of a flowing inviscid and frictionless fluid is
constant
p V2
+
+ z = cons tan t
4.28
ρ g 2g
The energies of two points in the fluid can be related as
p1 V1 2
p
V 2
+
+ z1 = 2 + 2 + z 2
ρ g 2g
ρg
2g
Equation 4.29 is a form of energy equation 4.9 where:
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
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4.29
45
•
The fluid must not be receiving or delivering shaft work between the two points that are being
considered.
• There must be no heat transfer and,
• The flow must be inviscid and frictionless.
Equation 4.29 holds for ideal fluids. But real fluid experiences fluid friction which is represented by
hL. Hence, the equivalent Bernoulli equation for real fluid is
p1 V1 2
p
V 2
+
+ z1 = 2 + 2 + z 2 + e f
4.30
ρ g 2g
ρg
2g
In other words, the Bernoulli equation for real fluids is the same as energy equation for case of no heat
transfer and no work addition or rejection between the two points under consideration.
4.4 CONSERVATION OF MOMENTUM
A macroscopic momentum balance for a flow system must include all equivalent forms of
momentum. In addition to the rate of momentum converted into and out of the system by the entering
and leaving streams, the sum of all the forces that act on the system (the system being defined as a
specified volume of fluid ) must be included. This follows from Newton’s second law, which provides
an equivalence between force and the rate of momentum. The resulting macroscopic conservation of
momentum thus becomes
4.31
Note that because momentum is a vector, this equation represents three component equations, one for
each direction in three-dimensional space.
. If the system is also
If there is only one entering and one leaving stream, then
at steady state, the momentum balance becomes
4.32
4.4.1 One-Dimensional Flow in a Tube
Consider the steady state momentum balance to a fluid in plug flow in a tube, as illustrated in Figure
4.2. (The ‘‘stream tube’’ may be bounded by either solid or imaginary boundaries; the only condition
is that no fluid crosses the boundaries other that through the inlet and outlet planes.) The shape of the
cross section does not have to be circular; it can be any shape.
Figure 4.2 : Momentum balance on a slice in a stream tube.
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The fluid element in the slice of thickness dx is our system, and the momentum balance equation on
this system is
4.33
The forces acting on the fluid result from pressure (dFP), gravity (dFg), wall drag (dFw), and external
shaft work (
):
4.34
is the stress exerted by the fluid on the wall (the reaction to the stress exerted on the fluid by
Here,
the wall), and Wp is the perimeter of the wall in the cross section that is wetted by the fluid (the
‘‘wetted perimeter’’). After substituting the expressions for the forces from Equation . 4.34 into the
momentum balance equation, Equation 4.32, and dividing the result by
where
the
result is
4.35
where
is the work done per unit mass of fluid. Integrating this expression from the
inlet (i) to the outlet (o) and assuming steady state gives
4.36
Comparing this with the Bernoulli equation shows that they are identical, provided
4.37
For steady flow in uniform conduit
4.38
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4.39
is called the hydraulic diameter. Note that this result applies to a conduit of any cross-sectional shape.
For a circular tube, for example, Dh is identical to the tube diameter D.
4.4.2 Conservation of Angular Momentum
In addition to linear momentum, angular momentum (or the moment of momentum) may be
conserved. For a fixed mass (m) moving in the x direction with a velocity Vx, the linear x-momentum
(Mx) is mVx. Likewise, a mass m rotating counterclockwise about a center of rotation at an angular
velocity
has an angular momentum
equal to
where R is the distance from the center of rotation to m. Note that the angular momentum has
dimensions of ‘‘length times momentum,’’ and is thus also referred to as the ‘‘moment of
momentum.’’ If the mass is not a point but a rigid distributed mass (M) rotating at a uniform angular
velocity, the total angular momentum is given by
4.40
where I is the moment of inertia of the body with respect to the center of rotation.
For a fixed mass, the conservation of linear momentum is equivalent to Newton’s second law:
4.40
The corresponding expression for the conservation of angular momentum is
4.41
is the moment (torque) acting on the system and
is the angular
where
acceleration.
For a flow system, streams with curved streamlines may carry angular momentum into and/or out of
the system by convection. To account for this, the general macroscopic angular momentum balance
applies:
4.42
For a steady-state system with only one inlet and one outlet stream, this becomes
4.43
This is known as the Euler turbine equation, because it applies directly to turbines and all rotating fluid
machinery. We will find it useful later in the analysis of the performance of centrifugal pumps.
4.4.3 Moving Boundary Systems and Relative Motion
One can sometimes encounter a system that is in contact with a moving boundary, such that the
fluid that composes the system is carried along with the boundary while streams carrying momentum
and/or energy may flow into and/or out of the system. Examples of this include the flow impinging on
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
48
a turbine blade (with the system being the fluid in contact with the moving blade) and the flow of
exhaust gases from a moving rocket motor. In such cases, we often have direct information concerning
the velocity of the fluid relative to the moving boundary (i.e., relative to the system), Vr, and so we
must also consider the velocity of the system, Vs, to determine the absolute velocity of the fluid that is
required for the conservation equations. For example, consider a system that is moving in the x
direction with a velocity Vs a fluid stream entering the system with a velocity in the x direction relative
to the system of Vri, and a stream leaving the system with a velocity Vro relative to the system. The
absolute stream velocity in the x direction Vx is related to the relative velocity Vrx and the system
velocity Vsx by
4.44
The linear momentum balance equation becomes
4.45
4.5 Problems
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
49
MODULE 5
5.0 FLOW IN PIPES
5.1 FLOW REGIMES
In 1883, Osborn Reynolds conducted a classical experiment, illustrated in Figure 5.1, in which
he measured the pressure drop as a function of flow rate for water in a tube. He found that at low flow
rates the pressure drop was directly proportional to the flow rate, but as the flow rate was increased a
point was reached where the relation was no longer linear and the scatter in the data increased
considerably. At still higher flow rates, the data became more reproducible, but the relationship
between pressure drop and flow rate became almost quadratic instead of linear.
To investigate this phenomenon further, Reynolds introduced a trace of dye into the flow to
observe what was happening. At the low flow rates where the linear relationship was observed, the dye
was seen to remain a coherent, rather smooth thread throughout most of the tube. However, where the
data scatter occurred, the dye trace was seen to be rather unstable, and it broke up after a short
distance. At still higher flow rates, where the quadratic relationship was observed, the dye dispersed
almost immediately into a uniform cloud throughout the tube. The stable flow observed initially was
termed laminar flow, because it was observed that the fluid elements moved in smooth layers relative
to each other with no mixing. The unstable flow pattern, characterized by a high degree of mixing
between the fluid elements, was termed turbulent flow. Although the transition from laminar to
turbulent flow occurs rather abruptly, there is nevertheless a transition region where the flow is
unstable but not thoroughly mixed.
Careful study of various fluids in tubes of different sizes has indicated that laminar flow in a
tube persists up to a point where the value of the Reynolds number (Re =DVρ
ρ/µ
µ) is about 2000, and
turbulent flow occurs when Re is greater than about 4000, with a transition region in between. A good
knowledge of the flow regimes in a given pipe is important for reasons of accurate flow analysis and
pipe design. Laminar flows can be analysed mathematically whereas turbulent flows require both
mathematical and experimental analysis because of their complexity.
Figure 5.1 : Reynolds’ Experiment
5.2 PRESSURE DROP IN LAMINAR FLOW
The pressure varies from point to point in the flowing fluid in a pipe. This variation is caused by the
existence of shear stress in the fluid and consequently, the fluid pressure decreases in the direction of
the flow. The expression for the mean velocity in laminar flow in a circular pipe is given as :
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
50
D2  d
(p + ρgz )
−
5.1

32µ  dx

Where D is the pipe diameter,-dp is the pressure drop over a short length dx of the pipe and z is the
vertical distance between the inlet and exit of length dx. If equation 5.1 is integrated between the two
sections, an expression for the pressure drop can be obtained as follows:
32µVm
p1 − p 2
= (z 2 − z 1 ) +
L
5.2
ρg
ρgD 2
Where L is the length of pipe (x2-x1).
If equation 5.2 is rewritten in a way similar to energy equation then
p1
p
5.3
+ z1 = 2 + z 2 + h f
ρg
ρg
32µVm
hf =
L
5.4
ρgD 2
hf represent the frictional term.
Vm =
Example 5.1
Water flows in a pipe of diameter 25mm.What is the maximum mean velocity if the flow should
always remain laminar? Calculate the head loss per metre length of the pipe. Take the viscosity of
water as 1.2cP(1.2x10-3 N.s/m2)
Solution
For laminar flow, Re should not exceed 2000 and therefore the maximum mean velocity is given by
ρVmax D
= 2000
µ
2000x1.2x10 − 3
= 0.096m / s
1000x 25x10 − 3
The head loss in laminar flow can be obtained by substituting values into equation 5.4
32µVm
32x1.2x10 −3 x0.096
hf =
L
=
= 6.01x10 − 4 m
2
3
−3 2
ρgD
10 x9.81x(25x10 )
Vmax =
5.3 PRESSURE DROP IN TURBULENT FLOW
The pressure drop over a length L of pipe of diameter D and surface roughness ∈ can be expressed
in functional terms as
∆p = f1(D,ρ,V,µ,L,∈)
5.5
Using dimensional Analysis and other mathematical manipulations:
The frictional factor f is given by
∆p D 2g
f=
. .
5.6
ρg L V 2
The frictional head loss hf is given by
L V2
hf = f. .
5.7
D 2g
A chart of f versus Re with ∈ as a parameter has been developed by Moody as presented in Figure
5.2.For a pipe of given roughness ∈ and diameter D with flow regime Re, the frictional factor can be
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
51
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
read from the chart and the pressure head loss calculated from equation 5.7.For non circular pipes, D
should be replaced by the equivalent diameter of hydraulic diameter De, which is defined as
4x flow area
5.8
De =
wetted perimeter
Figure 5.2 : Moody Diagram
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
52
Example 5.2
The relative roughness (∈/D) of a steel pipe is 6x10-5 and the pipe diameter is 0.82m.The pipe is
carrying crude oil of specific gravity 0.86 and kinematic viscosity 6x10-5 m2/s.Determine the head loss
per kilometer length of the pipe for discharge of 1m3/s.
Solution
π
.0.82 2 = 0.528m 2
4
Q
1
Velocity : V = =
= 1.89m / s
A 0.528
VD 1.89x0.82
µ
Re =
=
= 25,830 recall
υ=
−5
υ
ρ
6x10
Using the values of Re and ∈/D given, f obtained from Moody’s chart is 0.025.Substituting into
equation 5.7 gives
L V2
1000 1.89 2
hf = f. .
= 0.025.
x
= 5.6m
D 2g
0.82 2x9.81
Pipe Area : A =
5.4 PRESSURE LOSSES IN PIPE FITTINGS
Head loss in straight pipes can be calculated from equation 5.7. However, head losses in fittings
like reducers, bends, valves and so on requires some similar means of measuring them. Some available
methods of calculating losses in pipe fittings are presented as follows:
5.4.1 Sudden Expansion
Fluids experiences sudden expansion when the diameter of the pipe in which it is flowing increases
abruptly. For an incompressible flow, the head loss (hl) due to sudden expansion (enlargement) in the
flow cross section is given as
2
V2
A 
V2
h l = 1  1 − 1  = K l 1
2g  A 2 
2g
5.9
2

A 
5.10
Where V1 is the velocity in the smaller pipe and k l =  1 − 1 
A2 

5.4.2 Sudden Contraction
A sudden reduction in the cross section of the conduit causes a reduction in the cross section of the
flow area.The reduction continues until the smallest flow area called the vena contracta is
achieved.The head loss due to contraction is given as
V2 2
hc =
2g
Where hc
2
2
 A2


V2 2  1
V2 2


− 1 =
− 1 = K c
2g  C c
2g
 Ac


denotes head loss due to sudden contraction, Cc =Ac /A
5.11
2
 1

K c = 
− 1
5.12
C
c


Cc is described as coefficient of contraction,its values has been found experimentally to lie between
0.6 and 1.0. The values of Kc are also available in the literature.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
53
5.4.3 Pipe Entrance Loss
The flow situation at the pipe entrance is similar to that of sudden contraction. The head loss is:
V2
he = K e
5.13
2g
Where Ke = 0.50. A rounded entrance prevents the formation of vena contracta and reduces the value
of Ke to about 0.05.If the discharge pipe projects into the reservoir ,the entrance is called a re-entrant
entrance .Such an arrangement increases the energy loss considerably. For projection of one pipe
diameter or more, Ke is between 0.8 and 1.0.
5.4.4 Losses in Pipe Fittings
In pipe networks, fluids losses pressure as a result of expansion or contraction through fittings such as
couplings, bends, tees and reducers. Analytical expression for hf are only possible for a few of the
fittings and therefore ,empirical approaches are commonly used. Two empirical approaches are
available:
• The equivalent length of the pipe fittings is determined from a chart and then hf is calculated
from
Le V 2
hf = f . .
5.14
D 2g
Where Le denotes equivalent length of the pipe fittings
• Here, hf is calculated from an equation of the form
V2
hf = K
5.15
2g
Where K is a loss coefficient .For both methods the values of Le/D and K are available in the
literature based on the works of researchers.
Solving for Le from equations 5.14 and 5.15 gives
KD
Le t =
5.16
f
If there are n fittings in a line, the total equivalent length can be written as
D n
Le = ∑ K i
5.17
f i =1
The total length of the line is given by
D n
L t = L + ∑ K i = L + Le t
5.18
f i =1
The equation for total frictional loss can be written as
2
 L + Le t  V 2  L n
V
= f + ∑Ki 
5.19

∑ h f = f .
 D  2g  D i =1  2g
The equation for head loss for sudden enlargement uses the upstream velocity while contraction
uses downstream. In order to avoid confusion when using equation 5.19, it is better to adopt
V 2
L V2 n
+ ∑ K i fi
5.20
∑ hf = f. .
D 2g i =1
2g
Where Vfi is the velocity through the device or fitting number I and Ki is the corresponding loss
coefficient.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
54
5.5 PIPELINE PROBLEMS
5.5.1 Pipes in Parallel
When main pipeline at upstream point divides into two or more parallel pipes that again join
downstream and continues as a main line, such pipes are said to be in parallel as shown in figure
5.3 below
Q1, f1, D1, L1
Q
Q
Q2, f2, D2, L2
.Figure 5.3: A System of Pipes in Parallel
The frictional losses are the same for all the branches while the discharge is the sum of the
discharges in each of the branches. Therefore,
h f =h f 1 = h f 2
5.21
Q = Q1 + Q 2
5.22
Consequently, equation 5.21 can be written as
L1 V1 2
L 2 V2 2
f1 . .
= f2.
.
D1 2g
D 2 2g
This equation yields the velocity ratio as :
5.23
1/ 2
V2  f1 L1 D 2 

= .
.
V1  f 2 L 2 D1 
Equation 5.22 can also be expressed in terms of flow velocities as
V 
π 
π
Q = D12 V1 + D 2 2 V2
or Q = V1  D12 + D 2 2 2 
4 
V1 
4
Generally, if there are n pipes in parallel, equations 5.21 to 5.25 can be modified as
L V2
hf = fi . i . i
D i 2g
Where I denote the ith pipe in the parallel which contains n pipe
(
)
Q = Q1 + Q 2 + ....... + Q n
Q=
V
V 
π  2
V1  D1 + D 2 2 2 + ..... + D n 2 n 
4 
V1
V1 
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
5.24
5.25
5.26
5.27
5.28
55
1/ 2
V2  f1 L1 D 2 

.
= .
V1  f 2 L 2 D1 
V3  f1 L1 D 3 

= . .
V1  f 3 L 3 D1 
.
1/ 2
5.29
1/ 2
Vn  f1 L1 D n 

.
= .
V1  f n L n D1 
5.5.2 Pipes in Series
Consider a case where three pipes as shown in figure 5.4 are connected in series to two reservoirs.
The length and diameters are L1, L2, L3, D1, D2 and D3 .Let D2 < D1 and D2 < D3
1
H
2
HA
3
HB
Figure 5.4: Pipes in Series
As the rate of flow through each pipes is same, therefore
Q = A1V1 = A 2 V2 = A 3 V3
5.30
Q = V1D12 = V2 D 2 2 = D 2 3 V3
Also the difference in the liquid surface levels equals sum of various head losses in the pipe
V3 2
H = H A − H B = he + hf1 + hc + hf 2 + hl + hf 3 +
5.31
2g
If we assume that the minor losses are negligible compared to frictional losses, then
H A − H B = h f1 + h f 2 + h f 3
5.32
L 3 V3 2
L1 V1 2
L 2 V12
H = H A − H B = f1 . .
+ f2.
.
+ f3.
.
D1 2g
D 2 2g
D 3 2g
Using equation 5.30 in equation 5.33 yields
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
5.33
56
4
4
 L
 f . 1 . + f . L 2 . D1  + f . L 3 . D1  
5.34
2
3
 1 D1
D 2  D 2 
D 3  D 3  


5.5.3 Equivalent Pipe
An equivalent pipe is defined as the pipe of uniform diameter having loss of head and
discharge equal to the loss of head and discharge of a compound pipe consisting of several pipes
of different lengths and diameter. The uniform diameter of the equivalent pipe is known as the
equivalent diameter of the series or compound pipe. Consider the series of pipes in figure 5.4, the
equivalent single pipe must have the same total losses with the system.
h fe = h fs = H
5.35
The loss in the equivalent pipe can also be written as
L e Ve 2
h fe = f e .
.
5.36
D e 2g
The equivalent length of the pipes is obtained as
2
4
4
 V1  D e  L1
L 3  D1  
L 2  D1 


 + f 3 .
5.37
L e = 
f1 . . + f 2 .
.
.


V
f
D
D
D
D
D
1
2 
2 
3  3  
 e e 
From equation 5.37,the length of a pipe equivalent to a pipe of length L1 and diameter D1 is
V2
H= 1
2g
2
 V  D e  L1 
 f1 . . 
L e =  1 
 Ve  f e  D1 
D 
V
Since 1 =  e 
Ve  D1 
5.38
2
5
f D 
L e = L1 1  e 
5.39
f e  D1 
5.5.4 Pipe Network
In calculating the flow rates through the pipe network, the following conditions must be satisfied by
the network:
(i)
Continuity must be satisfied at each junction of the network, that is the total flow into a
junction must be equal to the total flow out of the junction.
(ii)
The algebraic sum of pressure head drops around each circuit must be zero.
(iii) The Darcy-Weisbach equation (equation 5.7) or any equivalent exponential formula for
head loss calculation must be satisfied in each pipe
5.6 Problems
1 Classify the following flows as laminar or turbulent
a) Water at 200C flowing in a 0.30m diameter pipe at 1.1m/s.
b) Oil of specific gravity 1.1, viscosity 0.05kg/m.s at 200C flowing in a 0.15m diameter pipe at
0.5m/s
2. A liquid of viscosity 3.2N.s/m2 and specific gravity 1.1 flows in a pipe of diameter 125mm.The
pressure drop over a meter is 150N/m2. Calculate the frictional factor for the pipe if the flow is in the
laminar regime.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
57
3 A pipe whose average roughness is 0.02mm is employed to raise water through a height of 10m.The
pipe length is 100m and its diameter is 300mm.Calculate the power required for a discharge rate of
30,000l/min.
4 Three pipes of diameter 300mm, 200mm and 400mm and lengths 450mm, 255mm and 315mm
respectively are connected in series. The difference in water surface levels in two tanks is 18m.
Determine the rate of flow of water if coefficients of friction are 0.0075, 0.0078 and 0.0072
respectively considering cases of minor losses and neglecting minor losses.
5 When a sudden contraction is introduced in a horizontal pipe line from 500mm diameter to 250mm
diameter, the pressure changes from 105kN/m2 to 69kN/m2.If the coefficient of contraction is
assumed to be 0.65, calculate the water flow rate. Following this, if there is a sudden enlargement
from 250mm to 500mm and if the pressure at the 250mm section is 69kN/m2, what is the pressure at
the 500mm enlarged portion.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
58
MODULE 6
COMPRESSIBLE FLOW
When a compressible fluid, i.e a gas, flows from a region of high pressure to one of low
pressure it expands and its density decreases. It is necessary to take this variation of density into
account in compressible flow calculations. In a pipe of constant cross-sectional area, the falling density
requires that the fluid accelerate to maintain the same mass flow rate. Consequently, the fluid’s kinetic
energy increases.
It is found convenient to base compressible flow calculations on an energy balance per unit
mass of fluid and to work in terms of the fluid’s specific volume V rather than the density ρ. The
specific volume is the volume per unit mass of fluid and is simply the reciprocal of the density:
6.0
6.1
6.1 Energy Relationships
The total energy E per unit mass of fluid is given by the following equations:
6.2
Or equation 6.2 can also be written as
6.3
Where
are the internal, potential, pressure and kinetic energies per unit
mass respectively.
Consider unit mass of fluid flowing in steady state from a point 1 to a point 2. Between these
two points, let a net amount of heat energy q be added to the fluid and let a net amount of work W be
done on the fluid. An energy balance for unit mass of fluid can be written either as
6.4
For steady flow in a pipe or tube the kinetic energy term can be written as u /(2α) where u is
the volumetric average velocity in the pipe or tube and α is a dimensionless correction factor which
accounts for the velocity distribution across the pipe or tube. Fluids that are treated as compressible are
almost always in turbulent flow and α is approximately 1 for turbulent flow. Thus for a compressible
fluid flowing in a pipe or tube, equation 6.4 can be written as
2
6.5
where in SI units each term is in J/kg.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
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Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
59
Since the enthalpy per unit mass of a fluid H is defined by the equation
6.6
Equation 6.5 can be written in the alternative form
6.7
The work term W in equations 6.5 and 6.7 is positive if work is done on the fluid by a pump or
compressor. W is negative if the fluid does work in a turbine. W is often referred to as shaft work since
it is transmitted into or out of a system by means of a shaft. The differential form of equation 6.5 is
6.8
For a reversible change, the first law of thermodynamics can be expressed by the equation
6.9
where dU is the increase in internal energy per unit mass of fluid and PdV is the work of expansion on
the fluid layers ahead for a net addition of heat dq to the system.
In flow, energy is required to overcome friction. The effect of friction is to generate heat in a
system by converting mechanical to thermal energy. Thus where friction is involved, equation 6.9 can
be written as
6.10
where dF is the energy per unit mass required to overcome friction. Substituting equation 6.10 into
equation 6.8 gives
6.11
Equation 6.11 can be integrated between states 1 and 2 to give
6.12
where in SI units each term is in J/kg.
Equations 6.5, 6.7 and 6.12 all relate to the energy changes involved for a fluid in steady turbulent
flow. The most appropriate equation is selected for each particular application: equation 6.12 is a
convenient form from which a basic flow rate-pressure drop equation will be derived.
Due to the change in the average velocity u, it is more convenient in calculations for compressible flow
in pipes of constant cross-sectional area to work in terms of the mass flux G. This is the mass flow rate
per unit flow area and is sometimes called the mass velocity. If the mass flow rate is constant, as will
usually be the case, then G is constant when the area is constant. In SI units G is in kg/(m2s). The
relationship between G and u is given by
6.13
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
60
Writing equation 6.11 in terms of G, noting that d(V2/2) = VdV, gives
6.14
Given that the pressure drop ∆Pf due to friction in a pipe of length L and inside diameter di is given by
6.15
where f is the Fanning friction factor.
For an element of length dx of the pipe, equation 6.15 can be written as
6.16
The corresponding energy required to overcome friction is df = VdPf . Thus equation 6.16 gives dF as
6.17
where advantage has been taken of equations 6.1 and 6.13. Substituting for dF in equation 6.14 gives
6.18
Dividing equation 6.18 throughout by V2 and integrating between states 1 and 2 over a length L of pipe
gives
6.19
In integrating the frictional term it has been assumed that the value of the friction factor is constant:
this is a good approximation because the Reynolds number will usually be very high, a condition for
which f is independent of Re.
In almost all cases, the change in potential energy will be negligible for gas flow. Also, it is
convenient to treat a compressor separately from flow in the pipe, ie equation 6.19 will be applied to a
section of pipe in which no shaft work is done. Consequently, 6.19 can be written in a reduced form:
6.20
Equation 6.20 is that required for most calculations involving compressible flow in a pipe. The three
terms represent, respectively, changes in pressure energy, kinetic energy and the conversion of
mechanical energy to thermal energy by frictional dissipation. The terms in the square brackets are
necessarily positive so the pressure energy term must be negative: this reflects the fact that the pressure
falls in the direction of flow.
In most cases the kinetic energy term will be negligible compared with the frictional term. This
is useful when calculating the pressure drop for a given flow rate because in this case one of the
pressures and therefore the corresponding specific volume will be unknown. An approximate
calculation can be made neglecting the kinetic energy term then, when the pressures are known, the
value of that term can be calculated to check whether it was in fact negligible.
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In equation 6.20 , and all other equations in this module, P denotes the absolute pressure.In order to
make use of equation 6.20 or equation 6.19 it is necessary to know the relationship between the
pressure P and the specific volume V so that terms such as
relationship between P and V is known as the equation of state.
can be evaluated. The
6.2 Equations of State
An ideal or perfect gas obeys the equation
6.21
where R is the universal gas constant, T the absolute temperature and RMM the relative molecular
mass converson factor for the gas. In SI units R = 8314.3 J/(kmol K) and T is in K. The conversion
factor RMM has the numerical value of the relative molecular mass and the units kg/kmol in the SI
system. Equation 6.21, which is a combination of Boyle's and Charles's laws, will be more familiar in
the form
6.22
where is the molar volume of the gas. The relative molecular mass has to be introduced in equation
6.21 because V is the specific volume, ie the volume per unit mass. It is convenient to define a specific
gas constant R' by
6.23
so that equation 6.21can be written as
6.24
It is essential to remember that in equation 6.24 both V and R' are values per unit mass of gas and they
must not be confused with the molar equivalents. The value of R' is different for gases of different
relative molecular masses.
Many gases obey equation 6.24 up to a few atmospheres pressure. At high pressures it is
necessary to modify equation 6.24 by introducing the compressibility factor Z:
6.25
The compressibility factor is a function of the reduced pressure Pr, and the reduced temperature Tr, of
the gas. Pr, is the ratio of the actual pressure P to the critical pressure Pc of the gas:
6.26
and Tr, is the ratio of the actual temperature T to the critical temperature Tc, of the gas:
6.27
Plots of Z against P, at constant T, are available [Perry (1984), Smith and Van Ness (1987)l.
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When ideal gases are compressed or expanded they obey the following general equation:
6.28
Thus, for two states 1 and 2, equation 6.28 gives
6.29
6.30
Combining equation 6.30 with 6.24 gives the relationship between pressure and temperature:
6.31
Equation 6.31 shows that, in general, expansion or compression of a gas is accompanied by a change
of temperature.
A change of state according to equation 6.28 is called a polytropic change. Two special cases
are the isothermal change and the adiabatic change.As the name implies, an isothermal change takes
place at constant temperature. This requires that the process be relatively slow and heat transfer
between the gas and the surroundings be rapid. An isothermal change corresponds to k = 1 and
equation 6.28 for an ideal gas becomes:
6.32
The other extreme case is the adiabatic change, which occurs with no heat transfer between the gas and
the surroundings. For a reversible adiabatic change,
where
, the ratio of the
specific heat capacities at constant pressure (Cp,) and at constant volume (Cv,). For areversible
adiabatic change of an ideal gas, equation 6.28 becomes
6.33
Then, 6.30 becomes
6.34
And 6.31 becomes
6.35
In a reversible adiabatic change the entropy remains constant and therefore this type of change is called
an isentropic change. Although not rigorously valid for irreversible changes, equations 6.33 to 6.35
are good approximations for these conditions.
Approximate values of at ordinary temperatures and pressures are 1.67 for monatomic gases
such as helium and argon, 1.40 for diatomic gases such as hydrogen, carbon monoxide and nitrogen,
and 1.30 for triatomic gases such as carbon dioxide. Gases and vapours of complex molecules can have
significantly lower values of , for example 1.05 for n-heptane and 1.03 for n-decane.
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6.3 Isothermal flow of an ideal gas in a horizontal pipe
For steady flow of a gas between points 1 and 2, distance L apart, in a horizontal pipe of constant
cross-sectional area in which no shaft work is done, the energy relationships are given by equation
6.20. Using Isothermal equation of state in equation 6.20 gives the following working equation for
isothermal flow of an ideal gas:
6.36
As noted previously, the first term is negative, as must be the case because P1 > P 2 . Equation 6.36 is
the basic form of the energy equation to be used for isothermal conditions, however it is instructive to
write the equation in a slightly different form that allows easy comparison with incompressible flow.
The pressure energy term can be written in the following form:
6.37
where Pm, = (P2 + P1)/2 is the arithmetic mean pressure in the pipe. Then
6.38
So that
6.39
where Vm, is the specific volume at the mean pressure P,. Thus, equation 6.36 can be written as
6.40
As noted previously, the kinetic energy term is usually negligible compared with the frictional term
and this is certainly true when the pressure drop ∆P' = P1 - P2 is small compared with P1. In this case,
equation 6.40 can be approximated by
6.41
Or
6.42
where ρm and um, are the density and average velocity at the mean pressure Pm,. Equation 6.42 will be
recognized as being of the same form as incompressible flow, except that it is written in terms of
average properties.
Thus, when the pressure drop is small compared with the meanpressure in the pipe, the gas
flow may be treated as incompressible flow. For large values of the pressure drop it is necessary to use
equation 6.36.
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In order to maintain isothermal flow it is necessary for heat to be transferred across the pipe wall. From
equation 6.7, for flow in a section with no shaft work and negligible change in elevation, the energy
equation takes the form
6.43
For an ideal gas under isothermal conditions, the enthalpy remains constant and hence it follows from
equation 6.43 that the required heat leak into the pipe is equal to the increase in kinetic energy. This is
usually a small quantity and therefore flow in long, uninsulated pipes will be virtually isothermal.
Example 6.1
Hydrogen is to be pumped from one vessel through a pipe of length 400 m to a second vessel, which is
at a pressure of 20 bar absolute. The required flow rate is 0.2 kg/s and the allowable pressure at the
pipe inlet is 25 bar. The flow conditions are isothermal and the gas temperature is 25°C. If the friction
factor may be assumed to have a value of 0.005, what diameter of pipe is required?
Solution : Recall that
For Isothermal conditions
Equation of state
Substituting these values into equation 6.20 gives
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It may be anticipated that 0.223 << 4/di, i.e di <<17.9 m. Thus the calculation may be simplified by
neglecting the kinetic energy term, so that
6.4 Adiabatic Flow of an Ideal Gas in a Horizontal Pipe
For adiabatic flow in a horizontal pipe with no shaft work, equation 6.7 reduces to
6.44
The differential form of equation 6.44 with u expressed in terms of G and V is
6.45
The enthalpy change can be found from the following two fundamental thermodynamic relationships
which, in the case of ideal gases, are valid for irreversible processes as well as reversible ones:
6.46
From equation 6.6
6.47
Thus
6.48
And
6.50
Then
6.51
Thus equation 6.45 can be written as
6.52
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Integrating equation 6.52 gives the desired relationship between P and V:
6.53
Thus,
6.54
and the integral in equation 6.20 is readily found by integrating equation 6.54:
6.55
Substituting this result in equation 6.20 gives
6.56
Where from equation 6.53
6.57
Calculations for adiabatic flow require the use of equations 6.56 and 6.57. For example, if the upstream
conditions P1 and V1are known, and G and d, are specified, C can be calculated from equation 6.57
then V2 from equation 6.57. Substituting this value of V2 in equation 6.57 gives P2. If the logarithmic
term is not negligible, an iterative calculation will be needed to determine V2 from equation 6.56.
6.5 Gas Compression and Compressors
Compressors are devices for supplying energy or pressure head to a gas. For the most part,
compressors like pumps can be classified into centrifugal and positive displacement types. Centrifugal
compressors impart a high velocity to the gas and the resultant kinetic energy provides the work for
compression. Positive displacement compressors include rotary and reciprocating compressors
although the latter are the most important for high pressure applications.
From equation 6.12, the shaft work of compression W required to compress unit mass of gas
from pressure P1 to pressure P2 in a reversible frictionless process, in which changes in potential and
kinetic energy are negligible, is
6.58
Although isothermal compression is desirable, in practice the heat of compression is never removed
fast enough to make this possible. In actual compressors only a small fraction of the heat of
compression is removed and the process is almost adiabatic.
When ideal gases are compressed under reversible adiabatic conditions they obey equation
6.33, which can be written as
6.59
so that the specific volume is given by
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6.60
Substituting V into equation 6.58 and integrating gives
6.61
Equation 6.60 gives the theoretical adiabatic work of compression from pressure PI to pressure P2.
Compression is often done in several stages with the gas being cooled
between stages. For two-stage compression from P1 to P2to P3, with the gas cooled to the initial
temperature T1 at constant pressure, equation 6.61 becomes
6.62
In the case of compression from pressure P1 to pressure P2 through n stages each having the same
pressure ratio (P2/Pl)1/n the compression work is given by
6.63
Equations 6.61 to 6.63 give the work required to compress unit mass of the gas. It should be noted that
the work required depends on the pressure ratio so that compression from 1 bar to 10 bar requires as
much power as compressing the same mass of gas, with the same initial temperature, from 10 bar to
100 bar.
In practice it is possible to approach more nearly isothermal compression by carrying out the
compression in a number of stages with cooling of the gas between stages. When ideal gases are
compressed under reversible adiabatic conditions the temperature rise from T1 to T2 is given by
equation 6.35:
6.64
So far only reversible adiabatic compression of an ideal gas has been considered. For the irreversible
adiabatic compression of an actual gas, the shaft work W required to compress the gas from state 1 to
state 2 can be obtained from equation 6.7, which in this case becomes
6.65
where H is the enthalpy per unit mass of gas.The actual work of compression is greater than the
theoretical work because of clearance gases, back leakage and friction.
Example 6.2
Calculate the theoretical work required to compress 1 kg of a diatomic ideal gas initially at a
temperature of 200 K adiabatically from a pressure of 10000Pa to a pressure of 100000Pa in (i) a
single stage, (ii) a compressor with two equal stages and (iii) a compressor with three equal stages. The
relative molecular mass of the gas is 28.0 and the ratio of specific heat capacities is 1.40.
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Solution
(i)
For a single stage compression,
6.61
From given values
Therefore
Equation of state
Therefore
Also
Substituting these values into 6.61 gives
(ii) For adiabatic compression of an ideal gas in n equal stages
For n = 2
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Since
And as before
It follows that
(iii) Repeating the above calculation for n = 3 gives
6.6 Problems
1
2
3
4
An ideal gas in which the pressure P is related to the volume V by the equation PV = 75 m2/s2
flows in steady isothermal flow along a horizontal pipe of inside diameter di, = 0.02 m. The
pressure drops from 20000 Pa to 10000 Pa in a 5 m length. Calculate the mass flux assuming that
the Fanning function factorf= 9.0 X10-3.
Ethylene flows through a pipeline 10 km long to a receiving station A. At a point 3 km from A, a
spur leads off the main pipeline and runs 5 km to a receiving station B. The internal diameter of
the main pipeline is 0.20 m and that of the spur is 0.15 m. The flow rates into A and B are
regulated by valves at these locations. If the pressure immediately upstream of valve A is 3.88 bar
(absolute) and that at B is 3.69 bar when the flow rate into B is 0.63 kg/s, calculate the pressure at
the beginning of the main pipeline, assuming that flow in the pipeline is isothermal at a
temperature of 20°C.
Data: specific volume of ethylene at 200C, 1 bar = 0.870 m3/kg,
Fanning friction factor = 0.0045.
Calculate the air velocity in m / s required to cause a temperature drop of 1 K on a conventional
thermometer given that for the air at atmospheric pressure and 373 K, the thermal capacity per
unit mass at constant pressure Cp, = 1006 J/(kg K).
An ideal gas flows in steady state adiabatic flow along a horizontal pipe of inside diameter di, =
0.02 m. The pressure and density at a point are P = 20000 Pa and ρ = 200 kg/m3 respectively. The
density drops from 200kg/m3 to 100kg/m3 in a 5m length. Calculate the mass flux assuming that
the Fanning friction factor f = 9.0 x 10-3 and the ratio of heat capacities at constant pressure and
constant volume = 1.40.
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MODULE 7
7.0 PUMPS
Pumps are devices for supplying energy or head to a flowing liquid in order to overcome head
losses due to friction and also, if necessary, to raise the liquid to a higher level. The head imparted to a
flowing liquid by a pump is known as the total head ∆h. If a pump is placed between points 1 and 2 in
a pipeline, the heads for steady flow are related by equation:
7.1
In equation 7.1, z, P/(ρg), and u /(2gα) are the static, pressure and velocity heads respectively and hf is
the head loss due to friction. The dimensionless velocity distribution factor α is 0.5 for laminar flow
and approximately 1 for turbulent flow.
For a liquid of density ρ flowing with a constant mean velocity u through a pipeline of circular
cross section and constant diameter between points 1 and 2 separated by a pump, equation 7.1 can be
written as
2
7.2
There exist a wide variety of pumps that are designed for various specific applications.
However, most of them can be broadly classified into two categories: positive displacement and
centrifugal. The most significant characteristics of each of these will be described later.
7.1 System Heads
The important heads to consider in a pumping system are the suction, discharge, total and
available net positive suction heads. The following definitions are given in reference to the typical
pumping system shown in Figure 7.1 where the arbitrarily chosen base line is the centre-line of the
pump.
Figure 7.1: Typical Pumping System
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Sunction Head:
7.3
Discharge Head:
7.4
In equation 7.3, hfs is the head loss due to friction, zs is the static head and Ps is the gas pressure above
the liquid in the tank on the suction side of the pump. If the liquid level on the suction side is below the
centre-line of the pump, zs, is negative. In equation 7.4, hfd is the head loss due to friction, zd is the
static head and Pd is the gas pressure above the liquid in the tank on the discharge side of the pump.
hs and hd are the values of
at the suction flange and at the discharge
flange respectively. Equations 7.3. and 7.4 are obtained by applying Bernoulli’s equation between the
supply tank and the suction flange, and between the discharge flange and the receiving tank,
respectively. On the suction side, the frictional loss hfs reduces the total head at the suction flange but
on the discharge side, hfd increases the head at the discharge flange.
The total head ∆h which the pump is required to impart to the flowing liquid is the difference
between the discharge and suction heads:
7.5
Equation 7.5 can be written in terms of 7.3 and 7.4 as:
7.6
The head losses due to friction are given be equations :
7.7
And
7.8
where
are the total equivalent lengths on the suction and discharge sides of the
pump respectively.
The suction head hs decreases and the discharge head hd increases with increasing liquid flow
rate because of the increasing value of the friction head loss terms hfs and hfd . Thus the total head ∆h
which the pump is required to impart to the flowing liquid increases with the liquid pumping rate.
It is clear from equation 7.3 that the suction head hs can fall to a very low value, for example
when the suction frictional head loss is high and the static head z, is low. If the absolute pressure in the
liquid at the suction flange falls to, or below, the absolute vapour pressure Pv of the liquid, bubbles of
vapour will be formed at the pump inlet. Worse still, even if the pressure at the suction flange is
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slightly higher than the vapour pressure, Cavitation - the formation and subsequent collapse of vapour
bubbles will occur within the body of the pump because the pressure in the pump falls further as the
liquid is accelerated. In order that cavitation may be avoided, pump manufacturers specify a minimum
value by which the total head at the suction flange must exceed the head corresponding to the liquid's
vapour pressure. The difference between the suction head and the vapour pressure head is known as
the Net Positive Suction Head, NPSH:
7.9
Substituting for hs from equation 7.3, the available NPSH is given by
7.10
The available NPSH given by equations 7.9 and 7.10 must exceed the value required by the pump and
specified by the manufacturer. The required NPSH increases with increasing flow rate.
7.2 Positive Displacement Pumps
The term positive displacement pump is quite descriptive, because such pumps are designed to
displace a more or less fixed volume of fluid during each cycle of operation. They include piston,
diaphragm, screw, gear, progressing cavity, and other pumps. The volumetric flow rate is determined
by the displacement per cycle of the moving member (either rotating or reciprocating) times the cycle
rate (e.g., rpm). The flow capacity is thus fixed by the design, size, and operating speed of the pump.
The pressure (or head) that the pump develops depends upon the flow resistance of the system in
which the pump is installed and is limited only by the size of the driving motor and the strength of the
parts. Consequently, the discharge line from the pump should never be closed off without allowing for
recycle around the pump or damage to the pump could result.
In general positive displacement pumps have limited flow capacity but are capable of relatively
high pressures. Thus these pumps operate at essentially constant flow rate, with variable head. They
are appropriate for high pressure requirements, very viscous fluids, and applications that require a
precisely controlled or metered flow rate.
7.3 Centrifugal Pumps
The term ‘‘centrifugal pumps’’ is also very descriptive, because these pumps operate by the
transfer of energy (or angular momentum) from a rotating impeller to the fluid, which is normally
inside a casing. A sectional view of a typical centrifugal pump is shown in Figure 7.2. The fluid enters
at the axis or ‘‘eye’’ of the impeller (which may be open or closed and usually contains radial curved
vanes) and is discharged from the impeller periphery. The kinetic energy and momentum of the fluid
are increased by the angular momentum imparted by the high-speed impeller. This kinetic energy is
then converted to pressure energy (‘‘head’’) in a diverging area (the ‘‘volute’’) between the impeller
discharge and the casing before the fluid exits the pump. The head that these pumps can develop
depends upon the pump design and the size, shape, and speed of the impeller and the flow capacity is
determined by the flow resistance of the system in which the pump is installed. Thus, these pumps
operate at approximately constant head and variable flow rate, within limits, of course, determined by
the size and design of the pump and the size of the driving motor.Centrifugal pumps can be operated
in a ‘‘closed off’’ condition (i.e., closed discharge line), because the liquid will recirculate within the
pump without causing damage. However, such conditions should be avoided, because energy
dissipation within the pump could result in excessive heating of the fluid and/or the pump or unstable
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operation, with adverse consequences. Centrifugal pumps are most appropriate for ‘‘ordinary’’
(i.e.,low to moderate viscosity) liquids under a wide variety of flow conditions and are thus the most
common type of pump.
Figure 7.2 Sectional view of a typical centrifugal pump.
The performance of a centrifugal pump for a particular rotational speed of the impeller and liquid
viscosity is represented by plots of total head against capacity, power against capacity, and required
NPSH against capacity. These are known as characteristic curves of the pump.
The relationship between ∆P and ∆h is given by :
7.11
Combining equations 7.6,7.7 and 7.8 gives the total head as
7.12
The mean velocity u of the liquid is related to the volumetric flow rate or capacity Q by
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7.13
Using 7.13 in 7.12 gives
7.14
For laminar flow, the Fanning friction factor f is given by equation:
7.15
Substituting for fin equation 7.14, the total head for laminar flow can be written as
7.16
Or
7.17
The system ∆h against Q curve can be plotted using equation 7.14 to calculate the values of the system
total head ∆h at each volumetric flow rate of liquid or capacity Q. Equation 7.17 shows that for laminar
flow the total head ∆h increases linearly with capacity Q. Thus for laminar flow, the system ∆h against
Q curve is a straight line.
The available NPSH in a system can be calculated from equation 7.10 having substituted for hfs:
7.18
Equation 7.18 shows that the available NPSH in a system decreases as the liquid throughput increases
because of the greater frictional head losses.
Example 4.1
Calculate the values for a system total head against capacity curve for the initial conditions of the
system shown in Figure 7.1 given the following data:
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Solution
From f against Re graph
Repeating the calculations for other values of u gives the following results:
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Table 7.1
7.4 Centrifugal pump relations
Holland and Chapman ( 1966)] developed an approach through dimensional analysis to
express the power PE required in an ideal centrifugal pump as a function of the liquid density ρ, the
impeller diameter D and the rotational speed of the impeller N.
7.19
7.20
7.21
7.22
Also, Consider a centrifugal pump with an impeller diameter D1 operating at a rotational speed N1 and
developing a total head ∆hl. Consider an homologous pump with an impeller diameter D2 operating at
a rotational speed N2 and developing a total head ∆h2.
Equations 7.19 to 7.22 can be manipulated to
7.23
Similarly’
7.24
and by analogy with equation 7.23 the net positive suction heads for the two homologous pumps can
be related by the equation
7.25
Equations 7.23 to 7.25 are the affinity laws for homologous centrifugal pumps.
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For a particular pump where the impeller of diameter D1 is replaced by an impeller with a slightly
different diameter D2, the following equations hold [Holland and Chapman (1966):
7.26
If the characteristic performance curves are available for a centrifugal pump operating at a given
rotation speed, equations in 7.26 enable the characteristic performance curves to be plotted for other
operating speeds and for other slightly different impeller diameters.
Example 7.2
A volute centrifugal pump with an impeller diameter of 0.02 m has the following performance data
when pumping water at the best efficiency point:
Evaluate the performance characteristics of a homologous pump with twice the impeller diameter
operating at half the impeller speed.
Solution
Let subscripts 1 and 2 refer to the first and second pumps respectively.
Given: The ratio of impeller spreeds N1/ N2 = 2 and the ratio of impeller diameters D1/D2 = 1/2. The ratio of
capacities is given by
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7.5 Pumping Efficiencies
The liquid power PE can be defined as the rate of useful work done on the liquid. It is given by the
equation
7.27
If the volumetric flow rate Q is in m3/s and the pressure developed by the pump ∆P is in Pa or N/m2,
the liquid power PE is in Nm/s or W. The pressure developed by the pump ∆P is related to the total
head developed by the pump ∆h by equation
7.28
7.29
Or
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7.30
since M = ρQ is the mass flow rate. If M is in kg/s, ∆h is in m and the gravitational accelerating g =
9.81m/s2, PE is in W.
The brake power PB can be defined as the actual power delivered to the pump by the prime mover. It is
the sum of liquid power and power lost due to friction and is given by the equation
7.31
is the mechanical efficiency expressed in per cent.
The mechanical efficiency decreases as the liquid viscosity and hence the frictional losses
increase. The mechanical efficiency is also decreased by power losses in gears, bearings, seals etc. In
rotary pumps contact between the rotor and the fixed casing increases power losses and decreases the
mechanical efficiency. These losses are not proportional to pump size. Relatively large pumps tend to
have the best efficiencies whilst small pumps usually have low efficiencies. Furthermore, high speed
pumps tend to be more efficient than low speed pumps. In general, high efficiency pumps have high
NPSH requirements. Sometimes a compromise may have to be made between efficiency and NPSH.
Another efficiency which is important for positive displacement pumps is the volumetric
efficiency. This is the delivered capacity per cycle as a percentage of the true displacement per cycle. If
no slip occurs, the volumetric efficiency of the pump is 100 per cent. For zero pressure difference
across the pump, there is no slip and the delivered capacity is the true displacement. The volumetric
efficiency of a pump is reduced by the presence of entrained air or gas in the pumped liquid. It is
important to know the volumetric efficiency of a positive displacement pump when it is to be used for
metering.
7.6 Factors In Pump Selection
The selection of a pump depends on many factors which include the required rate and
properties of the pumped liquid and the desired location of the pump.
In general, high viscosity liquids are pumped with positive displacement pumps. Centrifugal
pumps are not only very inefficient when pumping high viscosity liquids but their performance is very
sensitive to changes in liquid viscosity. A high viscosity also leads to high frictional head losses and
hence a reduced available NPSH. Since the latter must always be greater than the NPSH required by
the pump, a low available NPSH imposes a severe limitation on the choice of a pump. Liquids with a
high vapour pressure also reduce the available NPSH. If these liquids are pumped at a high
temperature, this may cause the gears to seize in a close clearance gear pump.
If the pumped liquid is shear thinning, its apparent viscosity will decrease with an increase in
shear rate and hence pumping rate. It is therefore an advantage to use high speed pumps to pump shear
thinning liquids and in fact centrifugal pumps are frequently used. In contrast, the apparent viscosity of
a shear thickening liquid will increase with an increase in shear rate and hence pumping rate. It is
therefore an advantage to use large cavity positive displacement pumps with a low cycle speed to
pump shear thickening liquids.
Some liquids can be permanently damaged by subjecting them to high shear in a high speed
pump. For example, certain liquid detergents can be broken down into two phases if subject to too
much shear. Even though these detergents may exhibit shear thinning characteristics they should be
pumped with relatively low speed pumps.
Wear is a more serious problem with positive displacemeni pumps than with centrifugal pumps.
Liquids with poor lubricating qualities increase the wear on a pump. Wear is also caused by corrosion
and by the pumping of liquids containing suspended solids which are abrasive.
where
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
80
In general,-centrifugal pumps are less expensive, last longer and are more robust than positive
displacement pumps. However, they are unsuitable for pumping high viscosity liquids and when
changes in viscosity occur.
7.7 Problems
1
Calculate the available net positive section head NPSH in a pumping system if the liquid
density ρ = 1200 kg/m3. The liquid dynamic viscosity µ = 0.4 Pa.s, the mean velocity u = 1
m/s, the static head on the suction side zs = 3m, the inside pipe diameter di, = 0.0526m, the
gravitational acceleration g = 9.81 m/s2 , and the equivalent length on the suction side
2
The liquid is at its normal boiling point. Neglect entrance and exit losses.
A volute centrifugal pump has the following performance data at the best efficiency point:
Evaluate the performance of an homologous pump which operates at an impeller speed of 29.2
rev/s but which develops the same total head ∆h and requires the same NPSH.
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
81
MODULE 8
FLOW MEASUREMENT
8.1 The Pitot Tube
As previously discussed, the volumetric flow rate of a fluid through a conduit can be determined by
integrating the local (‘‘point’’) velocity over the cross section of the conduit:
8.1
If the conduit cross section is circular, this becomes
8.2
The pitot tube is a device for measuring v(r), the local velocity at a given position in the conduit, as
illustrated in Figure 8.1 The measured velocity is then used to determine the flow rate in equation 8.2.
It consists of a differential pressure measuring device (e.g., a manometer, transducer, or DP cell) that
measures the pressure difference between two tubes. One tube is attached to a hollow probe that can be
positioned at any radial location in the conduit, and the other is attached to the wall of the conduit in
the same axial plane as the end of the probe. The local velocity of the streamline that impinges on the
end of the probe is v(r). The fluid element that impacts the open end of the probe must come to rest at
that point, because there is no flow through the probe or the DP cell; this is known as the stagnation
point. The Bernoulli equation can be applied to the fluid streamline that impacts the probe tip:
8.3
where point 1 is in the free stream just upstream of the probe and point 2 is just inside the open end of
the probe (the stagnation point). Since the friction loss is negligible in the free stream from 1 to 2, and
v2 =0 because the fluid in the probe is stagnant, Equation 8.3 can be solved for v1 to give
8.4
The measured pressure difference ∆P is the difference between the ‘‘stagnation’’ pressure in the
velocity probe at the point where it connects to the DP cell and the ‘‘static’’ pressure at the
corresponding point in the tube connected to the wall.
Figure 8.1 : The Pitot Tube
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
82
8.2 The Venturi And Nozzle
There are other devices, however, that can be used to determine the flow rate from a single
measurement. These are sometimes referred to as obstruction meters, because the basic principle
involves introducing an ‘‘obstruction’’ (e.g., a constriction) into the flow channel and then measuring
the pressure drop across the obstruction that is related to the flow rate. Two such devices are the
venturi meter and the nozzle, illustrated in Figure. 8.2 and 8.3 respectively. In both cases the fluid
flows through a reduced area, which results in an increase in the velocity at that point. The
corresponding change in pressure between point 1 upstream of the constriction and point 2 at the
position of the minimum area (maximum velocity) is measured and is then related to the flow rate
through the energy balance. The velocities are related by the continuity equation,and the Bernoulli
equation relates the velocity change to the pressure change:
8.5
Figure 8.2: Venturi Meter
Figure 8.3 : Nozzle
For constant density
8.6
And the Bernoulli Equation is
8.7
where plug flow has been assumed. Using Equation 8.6 to eliminate V1 and neglecting the friction loss,
Equation 8.7 can be solved for V2:
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
83
8.8
Where
(where d2 is the minimum diameter at the throat of the
venturi or nozzle). To account for the inaccuracies introduced by assuming plug flow and neglecting
friction, Equation 8.8 is written
8.9
where Cd is the ‘‘discharge’’ or venturi (or nozzle) coefficient and is determined by calibration as a
function of the Reynolds number in the conduit.
8.3 The Orifice Meter
The simplest and most common device for measuring flow rate in a pipe is the orifice meter,
illustrated in Figure 8.4 This is an ‘‘obstruction’’ meter that consists of a plate with a hole in it that is
inserted into the pipe, and the pressure drop across the plate is measured. The major difference
between this device and the venturi and nozzle meters is the fact that the fluid stream leaving the
orifice hole contracts to an area considerably smaller than that of the orifice hole itself. This is called
the vena contracta, and it occurs because the fluid has considerable inward radial momentum as it
converges into the orifice hole, which causes it to continue to flow ‘‘inward’’ for a distance
downstream of the orifice before it starts to expand to fill the pipe. If the pipe diameter is D, the orifice
diameter is d, and the diameter of the vena contracta is d2, the contraction ratio for the vena contracta is
defined as
For highly turbulent flow, Cc≈ 0:6.
The complete Bernoulli equation, as applied between point 1 upstream of the orifice where the
diameter is D and point 2 in the vena contracta where the diameter is d2, is
8.10
Figure 8.4 :Orifice Meter
As for the other obstruction meters, when the continuity equation is used to eliminate the upstream
velocity from Equation 8.10, the resulting expression for the mass flow rate through the orifice is
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
84
8.11
where
is the orifice coefficient:
Co is obviously a function of β and the loss coefficient Kf (which depends on Re).
For incompressible flow,equation 8.11 becomes
8.12
8.13
While equation 8.11 becomes 8.14 for compressible flow
8.14
It is more convenient to express this result in terms of the ratio of Equation 8.14 to the corresponding
incompressible equation, Equation 8.13, which is called the expansion factor Y:
8.15
where the density ρ1 is evaluated at the upstream pressure (P1). For convenience, the values of Y are
shown as a function of ∆P/P1 and β for the square-edged orifice, nozzles, and venturi meters for values
of
of 1.3 and 1.4 .
CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION
Department of Chemical Engineering, University of Ilorin.
Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010
85
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