Fluid Flow Notes - The University of...

Fluid Flow

Thomas Rodgers

2013

Contents

List of Figures iii

Nomenclature v

1 Introduction to Fluid Flow 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 What is a Fluid? . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.3 Fluid Flow in Chemical Engineering Applications . . . . . . . . 5

1.3 Properties of Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Hydrostatics 92.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.1 Variation of Pressure with Position in a Fluid at Rest . . . . . . . 142.3 Archimedes’ Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4 Pascal’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5 Pressure Measurement Devices . . . . . . . . . . . . . . . . . . . . . . . 18

2.5.1 The Barometer . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.5.2 Manometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5.3 The Bourdon Gauge . . . . . . . . . . . . . . . . . . . . . . . . 20

2.6 Pressure Heads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 Flow Dynamics — Ideal Fluids 253.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Flow Regimes and Reynolds Number . . . . . . . . . . . . . . . . . . . 273.3 Mass Balance in a Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.3.1 Streamlines and Streamtubes . . . . . . . . . . . . . . . . . . . . 323.3.2 The Continuity Equation . . . . . . . . . . . . . . . . . . . . . . 32

3.4 Energy Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.4.1 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 343.4.2 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4.3 Pressure Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4.4 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.4.5 Thermal Losses . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

i

CONTENTS

3.4.6 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . 37

4 Applications of Bernoulli’s Equation 394.1 Flow from a Tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.1.1 Tanks with Constant Height . . . . . . . . . . . . . . . . . . . . 414.1.2 Tank with Variable Height . . . . . . . . . . . . . . . . . . . . . 42

4.2 Flow Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.2.1 Velocity Profile Measurement – The Pitot Tube . . . . . . . . . . 444.2.2 The Orifice Meter . . . . . . . . . . . . . . . . . . . . . . . . . . 454.2.3 The Venturi Meter . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5 Momentum Balance 535.1 Derivation of the Momentum Equation . . . . . . . . . . . . . . . . . . . 555.2 Applications of the Momentum Equation . . . . . . . . . . . . . . . . . . 57

5.2.1 Flow through a contracting section of pipework or through a nozzle 575.2.2 Flow around a contracting pipe bend . . . . . . . . . . . . . . . . 60

5.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6 Friction Losses - Ideal Fluids 656.1 Force Balance on Fluid Element . . . . . . . . . . . . . . . . . . . . . . 676.2 Other Pipe Fittings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

ii

List of Figures

1.1 Example of a simple chemical process . . . . . . . . . . . . . . . . . . . 5

2.1 Pressure and shear stress . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Relationships between gauge and absolute pressures . . . . . . . . . . . . 142.3 Rectangular co-ordinate system, and differential volume element . . . . . 142.4 Archimedes’ Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.5 Diagram of a barometer . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.6 Diagram of a U-tube manometer . . . . . . . . . . . . . . . . . . . . . . 192.7 Diagram of a compound manometer . . . . . . . . . . . . . . . . . . . . 202.8 Diagram of a Bourdon gauge . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1 Fluid flow in a pipe between tanks at different pressures . . . . . . . . . . 273.2 Velocity of fluid flow versus pressure difference . . . . . . . . . . . . . . 283.3 Flow profiles in pipes under laminar, turbulent and ideal flow . . . . . . . 293.4 Flow patterns observed on injection of a stream of dye into a flowing fluid 303.5 Streamlines in steady straight flow and approaching a constriction . . . . 323.6 Flow through a streamtube . . . . . . . . . . . . . . . . . . . . . . . . . 333.7 Diagram of fluid flow along a pipe . . . . . . . . . . . . . . . . . . . . . 353.8 Piston acting against a pressure P1 over a distance d l at a velocity u . . . 36

4.1 Liquid discharging through a hole in the base of a tank . . . . . . . . . . 414.2 Pitot tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.3 Orifice meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.4 Venturi meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.1 Fluid flowing into a tank . . . . . . . . . . . . . . . . . . . . . . . . . . 565.2 Contracting pipe, through which fluid is flowing . . . . . . . . . . . . . . 575.3 Flow through a reducing pipe bend . . . . . . . . . . . . . . . . . . . . . 60

6.1 Force balance over a control volume . . . . . . . . . . . . . . . . . . . . 676.2 Surface roughness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.3 Moody Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

iii

Nomenclature

RomanA Area m2

a Acceleration m s−2

CD Discharge coefficient −D Diameter mF Force Nf Friction factor −g Acceleration due to gravity 9.81 m s−2

h Height mK Number of velocity heads ml Distance mm Mass kgM Mass flow rate kg s−1

n Moles molP Pressure Pap Vapour pressure PaQ Volumetric flow rate m3 s−1

Q Heat loss WR Gas constant 8.314 J mol−1 K−1

Re Reynolds number Duρ/µT Temperature Kt Time su Velocity m s−1

V Volume m3

v Velocity m s−2

W Work Wx x-direction my y-direction mz z-direction mGreekε Surface roughness cmµ Viscosity Pa sρ Density kg m−3

σ Surface tension N m−1

τw Wall shear stress Pa

v

Chapter 1Introduction to Fluid Flow

Contents1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 What is a Fluid? . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.3 Fluid Flow in Chemical Engineering Applications . . . . . . . 5

1.3 Properties of Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1

1.1. INTRODUCTION 3

1.1 Introduction

Chemical engineers are frequently interested in systems involving the flow of fluids. Thisinterest occurs at two levels:

1. the practical level, in which many of the methodologies for the design of chemicalengineering processes and operations require fluid flow calculations; and

2. the conceptual level, in which fluid flow illustrates one of the distinctive and defin-ing skills of a chemical engineer, that of being able to take a fundamental under-standing of the physical universe and apply it to solve practical problems in a waythat is both effective and elegant.

Practical examples of fluid flow calculations include pipe network design (including siz-ing of pumps and pipelines for transferring fluids within processes) and fluid flow effectson rates of heat and mass transfer, on reaction rates and on separation systems.

Conceptually, fluid flow is an example of a transport process, in which the rate of transferof matter or energy depends on physical factors affecting the transfer, such as the physicalproperties of the material under investigation and the geometry of the system.

Some of the key fundamental ideas needed for fluid flow are the Conservation Laws, ofwhich the three most important ones are:

Conservation of Mass

Conservation of Energy

Conservation of Linear Momentum

(There are others, notably Conservation of Angular Momentum and Conservation of Elec-tric Charge.) Thanks to Einstein, Mass and Energy are now recognised as equivalent andinterchangeable, such that it is more correct to speak of Conservation of (Mass and En-ergy), but in the absence of nuclear reactions, mass and energy are separately conserved.So in most cases, and certainly in the case of Fluid Flow, the chemical engineer canassume the laws of Conservation of Mass and Conservation of Energy to apply individu-ally.

These three conservation laws will form the basis for developing our fundamental under-standing of Fluid Flow, as later on we will employ these fundamental ideas in order toderive the three major mathematical descriptions of Fluid Flow: the Continuity Equation,Bernoulli’s Equation, and the Momentum Equation. We will derive these firstly for thesimplified case of Ideal Fluids, then extend the equations to allow us to do useful cal-culations involving Real Fluids. We will illustrate the use of fluid flow in the contextof chemical engineering using examples relating to flow in pipes and vessels. Before wemove to flowing fluids, we need to have some understanding of the static behaviour of flu-ids, which is the study of Hydrostatics; and before we begin with that, we need to formallyidentify what fluids are and be introduced to some of their important properties.

4 CHAPTER 1. INTRODUCTION TO FLUID FLOW

1.2 Fluids

1.2.1 What is a Fluid?

Generally, matter exists in many one of four fundamental states (though there are manymore): solid, liquid, gas, or plasma (although materials with intermediate properties exist,in particular, materials that exhibit both solid-like and fluid-like properties e.g. liquidcrystals). A fluid is a liquid or a gas, which differ from solids in that

FluidsFluids deform continuously under the action of an applied shear stress.

Table 1.1: The four common states of matter.

Solid Liquid Gas PlasmaFluid

A shear stress is a force that acts on an area tangential to (as opposed to perpendicularto) the direction of the force. If a shear force is applied to a solid, it may deform a littlefrom its initial position, but it will not continue to do so, and when the force is removed,it will recover its initial shape. By contrast, a fluid will continue to deform as long as theshearing action is applied, and when the shear force is removed, it will not bounce back toits original position. (A viscoelastic material is in between – it will flow to some extent onapplication of a shear force, and will bounce back to some extent, but not completely, onremoval of the applied force. Blu-tack and bread doughs are both examples of viscoelasticmaterials.)

Fluids may be subdivided into liquids and gases (and plasma), which differ from eachother in several distinctive ways. The molecules in liquids stick closely together such thatliquids have a definite volume and will not fill a container having a larger volume. Thevolume occupied by a liquid varies only slightly with temperature and pressure, such thatliquids are essentially incompressible. Liquids form interfaces with gases, including theirown vapours, and sometimes with other liquids (e.g. oil-water mixtures). By contrast, themolecules of gases will always move apart so as to completely fill an enclosing container;only then is a gas in equilibrium.

With respect to their fluid behaviour, gases differ from liquids in two significant respects.Firstly, while liquids are essentially incompressible for most practical purposes, gasesby contrast are easily compressible. Secondly, liquids have much greater densities thangases, such that the weight of a liquid plays a major role in its behaviour, while the effectsof weight can usually be ignored when dealing with gases.

1.2. FLUIDS 5

1.2.2 Fluid Mechanics

Fluid mechanics is the application of the fundamental principles of mechanics and ther-modynamics – such as conservation of mass, conservation of energy and Newton’s lawsof motion – to the study of liquids and gases, in order to explain observed phenomenaand to be able to predict behaviour. Fluid mechanics can be sub-divided into Fluid Statics(or Hydrostatics) – the study of fluids at rest – and Fluid Dynamics (or Hydrodynamics),the study of fluids in motion. This course is entitled Fluid Flow, emphasising the issuesof fluid behaviour under dynamic conditions, because chemical engineering is concernedprincipally with processes, and processes imply change and are inherently dynamic. Weneed to understand fluid statics (or hydrostatics) as well.

1.2.3 Fluid Flow in Chemical Engineering Applications

Figure 1.1 illustrates a simple chemical process incorporating a Feed stream into a Re-actor, in which an exothermic reaction results in formation of a desired Product and anunwanted Byproduct, and causes an increase in the temperature of the contents of theReactor. The stream from the Reactor is cooled and then subject to an initial Separationoperation in order to recover the unreacted Feed, which is recycled back to the Reactorin order to increase the conversion of Feed to Product. A second Separation system thenseparates the desired Product from the undesired Byproduct.

Feed Reactor

Unreacted FeedProductByproduct

Separator Separator

Unreacted Feed Product

Byproduct

Figure 1.1: Example of a simple chemical process.

This example allows us to identify the relevance of fluid flow to numerous aspects ofchemical process design:

Pipes – streams are conveyed between processes via numerous pipes, which contributea large fraction of the capital costs of a chemical plant, and so need to be sizedappropriately.

Pumps – make the fluids flow along the pipes! They need to be sized to deliver therequired flowrates, while their power consumption contributes to the operating costsof the plant.

Reactor – the nature of the fluid flow patterns determines the way that molecules meeteach other and react, and hence the yield, throughput and selectivity of the process.

6 CHAPTER 1. INTRODUCTION TO FLUID FLOW

Heat exchangers – only one shown here (a cooling operation), but there are many in atypical process, and the rate of heat transfer, and therefore the size and cost of theheat exchanger, depends strongly on the fluid flow patterns.

Separation – probably via distillation, in which flow patterns of the liquid and vapourphases affect mass transfer between them.

Fluid flow also has an enormously wide range of application outside of chemical engi-neering, including in the other engineering disciplines (in particular, civil and mechanicalengineering), in meteorology and oceanography (weather patterns and ocean currents),and even in medicine (e.g. blood flow). Mastery of fluid flow can open up a wide rangeof employment and research opportunities both within the chemical industries and be-yond.

1.3 Properties of Fluids

Density (ρ) – mass per unit volumeρ =

m

V(1.3.1)

In SI units, density has the units of kg m−3, because it is compressible, the densityof a gas depends on its pressure. Relative density or specific gravity is the ratio ofthe density of a material to the density of water at 4◦C. Because it is a ratio, it doesnot have units.

Pressure (P ) – Fluids have pressure, and fluids flow because of pressure differences. Anunderstanding of what pressure is and how it relates to fluids is key to understandingthe subject of fluid flow. A key equation is:

∆P = ρg∆h (1.3.2)

where ∆P is a pressure difference, ρ we have already met, g is the acceleration dueto gravity (9.81 m s−2 in SI units), and ∆h is the height over which the pressuredifference is being measured. By considering the units, we can see that the units ofpressure are:

kgm3× m

s2×m =

kgm s2

Pressure is often presented in other units as well. We’ll consider this importantsubject more fully in the next lecture.

Vapour Pressure (p) – Liquids exhibit a vapour pressure, which contributes to the totalpressure above the liquid. Vapour pressures are not considered within this course.

Viscosity (µ) – This is the fluid property responsible for resistance to applied forces.Honey has high viscosity, water has much lower viscosity, and air has an evenlower viscosity. We’ll define viscosity formally later on this in the course. For nowwe’ll note that its SI units are kg m−1 s−1 or Pa s.

Surface Tension (σ) – Liquids in contact with gases (e.g. water in contact with air) forman interface. The liquid molecules at the interface are attracted to each other more

1.3. PROPERTIES OF FLUIDS 7

than to the gas molecules, to they tend to pull sideways. This has the effect ofminimising the size of the interface. For this reason, bubbles are spherical, becausesurface tension aims to minimise the interface between the liquid that makes up thebubble wall and the air inside and outside of the bubble, and a sphere is the shapethat has the smallest surface area for a given volume. However, surface tensioneffects are outside the scope of this course.

Chapter 2Hydrostatics


2.2 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.1 Variation of Pressure with Position in a Fluid at Rest . . . . . . 14

2.3 Archimedes’ Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Pascal’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.5 Pressure Measurement Devices . . . . . . . . . . . . . . . . . . . . . 18

2.5.1 The Barometer . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.5.2 Manometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.5.3 The Bourdon Gauge . . . . . . . . . . . . . . . . . . . . . . . 20

2.6 Pressure Heads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

9

2.1. INTRODUCTION 11

2.1 Introduction

Hydrostatics is the study of fluids at rest, for which there are no shear (tangential) forces.However, fluids at rest are still subject to pressure, as pressure arises as the result ofinnumerable molecular collisions. Pressure is therefore where our study of hydrostaticsbegins. Then, as we will discover, hydrostatics is also the basis for measurement ofpressure.

2.2 Pressure

We noted earlier that an understanding of what pressure is and how it relates to fluidsis key to understanding the subject of fluid flow. The aim of this part of the course isto develop your understanding of pressure, including what it is, how it behaves, what itsrelevance is, and how to measure it.

Pressure is Force per unit Area:

Pressure

P =F

A(2.2.1)

Pressure therefore has SI units of Newtons per metre squared, N m−2, which we sim-plify to Pascals, Pa. 1 Pa = 1 N m−2. This is a very small value, approximately equiva-lent to the force exerted by a 2 g weight spread over the surface of your hand. The unitkPa (103 Pa) is therefore more often used. Standard atmospheric pressure has a value of101.325 kPa.

Interestingly, shear stress is also a force per unit area, and also has units of Pa. So whatis the difference between shear stress and pressure? The difference is the direction of theforce. In the case of pressure, the direction of the force is perpendicular to (or normal to)the area, while in the case of shear stress, the direction of the force is tangential to thearea.

Area, A

Force, F

(a) Pressure = F/A (units: Pa)

Area, A

Force, F

(b) Shear stress = F/A (units: Pa)

Figure 2.1: Pressure and shear stress.

Here’s an interesting thought: Pressure is a scalar quantity (has magnitude but no direc-tion), while Force is a vector (has both magnitude and direction). So how can Pressure,which has no direction, give rise to a Force which has a direction?

The answer is that when the Pressure acts on an Area, this has an orientation (i.e. a direc-tion), which determines the direction of the resulting Force.

12 CHAPTER 2. HYDROSTATICS

We sometimes say, slightly imprecisely, that the pressure at a particular point in a fluid atrest is the same in all directions. But because pressure is a scalar quantity, it is meaninglessto talk of the pressure having a particular direction. It is more precise to say that themagnitude of the force due to the pressure is independent of the orientation of the planeover which the force acts. This is in effect the same as saying that the pressure at apoint is the same in all directions (provided we understand that we actually mean that themagnitude of the force arising from the pressure at that point is the same in all directions).This is strictly true only for fluids at rest, but close enough to the truth even for fluidsin motion. Another way of putting this idea is that the magnitude of the pressure isindependent of the orientation of the surface used to define it.

Let’s have another look at that definition of pressure, and focus on the units:

P =F

A; units

Nm2

Let’s multiply top and bottom by a length dimension (m):

Nm2× m

m=

N mm3

Newtons × metres is like Force × distance, which equals Work, which is a form of En-ergy. So the top part has units of energy (which are Joules, J). The bottom part has unitsof Volume (m3), so the whole indicates that the units of pressure, as well as indicatingforce per unit area, have an equivalent meaning of energy per unit volume (J m−3). Thisis also apparent when we look at the units of the ideal gas law:

PV = nRT (2.2.2)

The units of the universal gas constant, R, are J mol−1 K−1. On the right hand side of theequation, this is multiplied by the number of moles (mol) and by the temperature (K) togive J, which is the unit of energy. The left hand side must also therefore multiply to giveunits of energy. V is volume, in m3, so the pressure term, P , must represent in some senseenergy per unit volume, with units of J m−3:

Jm3×m3 = J = mol× J

mol K× K

So pressure appears to be a form of energy. Now, we have to have a slight word ofcaution, that some textbooks insist pressure is not a type of energy that a fluid possesses,and that it is incorrect to speak of “pressure energy”. However, while recognising thatpressure does not precisely mean “Energy per unit Volume”, at the same time it is quitehelpful sometimes to think of pressure in this way, instead of our usual way of thinkingof pressure as Force per unit Area, and to recognise that the units of pressure, N m−2, areequivalent to J m−3.

Two complications arise when dealing with the practical measurement and interpretationof pressure. The first is that, although the SI units of pressure are Pa (or kPa), many otherunits for measuring and reporting pressure are commonly in use. These include poundsper square inch (psi), Torr, mm Hg, atmosphere (atm), bar, m H2O. Get used to seeingpressure expressed in these various alternative units, and get used to converting them toPa using your conversion tables or appropriate equations.

2.2. PRESSURE 13

Of these, the bar needs a special mention. It has a value of 105 Pa. This makes it anunusual measurement in the context of the SI system, which usually uses multiples andsubmultiples of the fundamental units which vary by powers of 3 (e.g. kilo, mega andgiga, 103, 106 and 109, respectively, or milli, micro and nano, 10−3, 10−6, and 10−9, re-spectively). However, the bar is convenient in that it is very close to atmospheric pressure(1.01325 × 105 Pa), such that 1 bar is approximately equal to 1 atmosphere. Hence thebar is commonly used in science and engineering (as is the millibar, which is 10−3 bar or102 Pa).

The second complication regarding pressure is that it is commonly measured relative toone of two different datum points– and when we are dealing with pressure measurements,we need to be aware which datum point is being employed or assumed. The two datumpoints are

• relative to atmospheric pressure; and

• relative to an absolute pressure of zero (absolute vacuum).

The first of these is convenient, as we are generally measuring the pressure of thingsthat are situated on the ground and surrounded by the atmosphere, and we want to knowwhether they are under pressure or perhaps under vacuum, relative to the surroundingatmosphere. Things that are at the same pressure as the atmosphere we would say arenot under pressure, and that their pressure is zero (meaning, zero relative to the surround-ing atmospheric pressure). Things that are above atmospheric pressure are under positivepressure, and things that are below are under a partial vacuum or a negative pressure.Such a pressure measurement is called a “gauge pressure”. However, in other circum-stances, we want to know what the absolute value of the pressure is, not just the pressurerelative to atmospheric pressure. Otherwise, for example, how would we know that thevalue of standard atmospheric pressure is 101.325 kPa? This value only has meaning rel-ative to an absolute pressure of zero. Atmospheric pressure in fact changes, as we know,with geographical position and over time. This is why we need to define “standard” at-mospheric pressure. But measuring a quantity relative to something that changes withtime or is different in different places is not a very solid basis for science or engineering.Absolute vacuum, a pressure of zero, does not change, so measuring pressures relative tothis constant datum point is much less ambiguous and a much sounder basis. Pressuresmeasured and reported relative to absolute vacuum are called “absolute pressures”, in con-trast to “gauge pressures”. This may be made explicit in the reporting (“50 kPa gauge” or“20 kPa abs”, “1 barg” (bar gauge) or “2 bara” (bar absolute), “15 psig” or “30 psia”), ormay be omitted, with the engineer expected to know from the context whether a gaugeor absolute pressure is being used. Clearly, the numerical values of absolute pressuresare larger than those of gauge pressures, by an amount equal to the absolute atmosphericpressure (in whichever units are being used):

Pressure Types

Absolute pressure = Gauge pressure + Atmospheric pressurePabs = Pgauge + Patm (2.2.3)

So, both practices are used, and you need to be aware of both and of the relationshipbetween them, and very aware of which is being used in the particular context of the cal-culation you are performing or the problem you are addressing. Note that many properties


of gases are functions of the absolute pressure, and so in problems concerning gases, orin using the ideal gas equation PV = nRT , absolute pressures should be used.

Figure 2.2 illustrates the relationship between gauge and absolute pressures, and theequivalence or approximate equivalence of several common units for reporting pressure.

2 atm = 202650 Pa= 1520 mm Hg≈ 2 bar≈ 30 psi (all absolute)= 1 atm = 101235 Pa

Above atmospheric = 760 mm Hg≈ 1 bar≈ 15 psi (all absolute)pressure

1 atm = 101325 PaAtmospheric pressure = 760 mm Hg ≈ 1 bar ≈ 15 psi (all absolute)

0 pressure (in any units) gauge

Below atmosphericpressure - vacuum or

suction0 pressure (in any units) absolute−1 atm = −101325 Pa

Absolute vacuum = −760 mm Hg ≈ −1 bar ≈ −15 psi (all gauge)

Figure 2.2: Relationships between gauge and absolute pressures, and between differentunits used to report pressures.

2.2.1 Variation of Pressure with Position in a Fluid at Rest

Consider an element of fluid at rest within a rectangular co-ordinate system defined bythree axes, x, y, and z, as shown in Figure 2.3.

x

y

z

δ x

δ z

δ y

Fz,z+δ z

Fz,z

Gravity force, Fg

Figure 2.3: Rectangular co-ordinate system, and differential volume element.

The element is at rest (not moving), so there must be no overall (net) force acting on it– the various forces must be in balance. These forces include that arising from pressure

2.2. PRESSURE 15

acting downward on the top face of element, that arising from pressure acting upward onthe bottom face of the element, and the (downward) force due to gravity acting on themass of the element. (As there is no sideways movement, the sideways forces must alsoadd up to a net force of zero; as there is no component in addition to the force arising fromthe pressure on each side face, these cancel out and are not included in this analysis). Justto clarify the nomenclature, Fz,z means the force in the z-direction at the point z; Fz,z+δ zmeans the force in the z-direction at the point z + δ z. The force in the z-direction at thepoint z is given by multiplying the pressure at that point, Pz, by the area over which thepressure operates, which is δ x× δ y:

Fz,z = Pzδ xδ y –upward on the bottom face (2.2.4)

Similarly, the force in the z-direction at the point z + δ z is given by

Fz,z+δ z = Pz+δ zδ x.δ y –downward on the top face (2.2.5)

The “body force” (the downward force due to gravity) is given by (mass) × (accelerationdue to gravity), where (mass) = (density × volume):

Fg = mg = ρgδ x.δ y.δ z (2.2.6)

A force balance then gives:

Fz,z+δ z + Fg = Fz,z

Pz+δ zδ x.δ y + ρgδ x.δ y.δ z = Pzδ x.δ y (2.2.7)

Rearranging and dividing through by volume (δ x.δ y.δ z):

Pz+δ zδ x.δ y − Pzδ x.δ y = − ρgδ x.δ y.δ zPz+δ z − Pz

δ z= − ρg (2.2.8)

Setting the limit on the left hand side as δ z → 0 gives the definition of the derivative ofP with respect to z:

dP

d z= −ρg (2.2.9)

In other words, the change in pressure, P , with height, z, is negative – pressure reduceswith height (or increases with depth – as any diver knows). Then, to find the pressureat a particular position relative to the pressure at the surface, we integrate from the sur-face: ∫ P

P0

dP = −∫ z

0

ρgd z (2.2.10)

If the density of the fluid is constant with z-position, and if g does not change with height(which it does not, over most changes in height of practical interest), then:

P − P0 = −ρgz (2.2.11)

Note that z is negative as we go downwards – let’s replace this with h, depth, which ispositive as we go downwards, i.e. h = −z:

P − P0 = ρgh

P = P0 + ρgh (2.2.12)


Alternatively,∆P = ρg∆h (2.2.13)

which we met previously, where ∆P is the difference in pressure between two points ina liquid separated by a horizontal distance ∆h.

Note that this means that within a continuous expanse of the same fluid at rest, the pressureat any height is constant, or in other words, the pressure is the same across any horizontalplane.

Note that if P0 = 0, for example, if we were considering pressure relative to the ambientatmosphere, and set atmospheric pressure equal to zero, then our equation would indicatea gauge pressure and would become:

Pgauge = ρgh (2.2.14)

2.3 Archimedes’ Principle

Archimedes’ principle states that the buoyant force on a submerged object is equal to theweight of the fluid it displaces. Our understanding of the relationship between pressureand depth can be used to demonstrate this principle, as shown in Figure 2.4.

V

P1

P2

h

A

Figure 2.4: Archimedes’ Principle.

The pressure on the top of the object is P1, and the pressure on the bottom is P2 =P1 + ρgh. This means that if the top and bottom of the object each have area A, the forceon the top is P1A downward, while the force on the bottom is P2A upwards. The net forceon the object is then

F = (P2 − P1)A = ρghA = ρgV = mfg

where m is the mass of the water which is displaced by the object. This is Archimedes’Principle for the object shown; the same goes for any shape of object. Thus the buoyancyforce is given by the weight of the fluid displaced,

2.4. PASCAL’S PRINCIPLE 17

Archimedes’ PrincipleFB = mfg (2.3.1)

QuestionIf a 100 kg chest, of volume 0.03 m3, is submerged under water. What is the forceneeded to pull the chest up to the water surface? What is the force needed to lift thechest from the water?Water ρ = 1000 kg m3, Air ρ = 1.2 kg m3

2.4 Pascal’s Principle

There is one big difference between liquids and gases. The density of a gas is easy tochange. However, fluids are usually incompressible. Incompressiblility means that thedensity of a fluid is independent of the pressure. This is not perfectly true: fluids do con-tract and expand a little, but not much at all: this expansion and contraction can easily beneglected. We’ve already used fluid incompressibility. For example, the formula for howthe pressure depends on the depth of the fluid assumed that the density remained constant,even though the pressure increases. Pascal’s Principle also depends on it. Pascal’s princi-ple says that if you push at one end of the fluid, the pressure increases everywhere. If thefluid were compressible, what would just happen is that part of the fluid would becomemore dense. This is what happens to a solid (or the solid just moves). A gas, on the otherhand, will compress uniformly. Stated precisely, Pascal’s principle is:

Pascal’s PrincipleAny change in the pressure applied to a completely enclosed fluid is transmittedundiminished to all parts of the fluid and hence to the walls.

This is how a hydraulic pump works, to say lift your car up at the garage. You have a twopistons, one narrower than than the other, each pushing on the same water. The wider onesupports the car. The pressure inside the fluid always obeys the formula P2 = P1 + ρgh.Say the two pistons are at the same height. Then P1 = P2. Now say you push one pistondown, thus increasing the pressure P1. Thus P2 increases as well. Then P1 = P2 impliesthat F1/A1 = F2/A2, where A1 and A2 are the areas of the two pistons. Thus

F2 = F1A2

A1

If A2 > A1, then the force at the second piston is larger than that at the other. This is whyyou put the car on the larger piston: the force you apply is magnified. If the two pistonsaren’t at the same height, you need to take into account the fact that the pressure at thetwo places is different. This you can easily do by using the formula P2 = P1 +ρgh.

A hydraulic pump seems almost like a miracle – it can magnify a force as much as youwant. Of course, you don’t get something for nothing. Remember that work is forcetimes distance. The work you do must be the same as the work done on the car. Thus ifA2 = 10A1, then F2 = 10F1. However, W1 = W2, so the distance you must push is 10times as far as the car moves: d2 = d1/10.


2.5 Pressure Measurement Devices

In practice, pressure is always measured by measuring a pressure difference. If the differ-ence is between that of the fluid in question and that of a vacuum, then we are measuringthe absolute pressure of the fluid. If, as is more usual, the difference is between the pres-sure of the fluid and that of the surrounding atmosphere, then this is a gauge pressure (socalled because that’s what pressure gauges normally measure). In this section we willconsider several common devices for measuring pressure: the barometer, the manometer,refinements of the manometer for measuring small pressure differences more accurately,and the Bourdon gauge.

2.5.1 The Barometer

Figure 2.5 illustrates the basis of operation of a barometer. The sealed tube contains asuitable liquid such as mercury (which has a high density of around 13560 kg m−3 and soallows short tubes to be used, and has a low vapour pressure, such that the space at thetop of the tube is very close to a perfect vacuum). With the presence of a vacuum in thetop of the sealed tube, the height of the column depends on the external pressure; to putit another way, the external pressure can only push the liquid column up so high and nohigher.

Container of Mercury

P1

P0

∆h

Figure 2.5: Diagram of a barometer, showing how the external atmospheric pressureforces liquid up the tube.

The difference in pressure between the base of the column of liquid and the top surfaceis (atmospheric − absolute vacuum), and the difference in height is given by the equationwe derived above,

P0 = ��0

P1 + ρg∆h

For standard atmospheric pressure, the height of the column can be calculated in SI unitsas:

∆h =∆P

ρg=

101325

13560× 9.81= 0.762 m

If water were used, with a density of 1000 kg m−3, clearly the equivalent height would be10.33 m, if a perfect vacuum could be maintained in the space at the top of the tube. But

2.5. PRESSURE MEASUREMENT DEVICES 19

unlike mercury, water has a significant vapour pressure, so that at 15 ◦C, for example, theheight would be about 0.18 m less than this.

Clearly, the actual height of the column of mercury in a barometer varies with the ambientpressure, and also to some extent with other weather conditions (in particular temperature,which causes thermal expansion of both the mercury and the casing).

2.5.2 Manometers

A manometer is a device in which a column of a suitable liquid is arranged such that theheight difference between its two ends indicates the difference in pressure at the two ends.Figure 2.6 illustrates this.

P1 P2

h

Figure 2.6: A U-tube manometer, in which the higher pressure P1 pushes the column ofliquid towards the lower pressure P2.

At equilibrium, the difference in height indicates the difference between the two pressures,according to ∆P = ρg∆h. Knowing the density of the liquid, the pressure differencecan be calculated from the height difference. If one of the ends of the manometer is opento the atmosphere, then the pressure difference indicates the gauge pressure at the otherend of the column of liquid. Alternatively, neither end may be at atmospheric pressure,as in the use of a manometer to relate pressure differences along a pipe to fluid flowrates,which we’ll meet later.

In order to improve the sensitivity of manometers, several refinements have been intro-duced. The simplest is the inclined manometer, in which the manometer is simply in-clined at an angle θ to the horizontal. This does not alter the height difference at all, but itmeans this height difference is now extended over a longer length (given by l = h/ sin θ),such that changes in length, and thus changes in pressure, can be measured more pre-cisely.

Alternatively, the sensitivity of a manometer can be improved by using an intermediateliquid with its top surface in a vessel of wide cross sectional area compared with thatof the U-tube, to amplify the change in heights produced by a small pressure difference.Figure 2.7 illustrates one arrangement that would achieve this.

Assume that the two pressures P1 and P2 act at the surface indicated by h3 (i.e. P1 and P2

don’t vary much with height, as would be the case for a “light” fluid of very low density ρ1such as air), and that differences in h3 are very small because of the large cross sectionalarea of this part of the manometer. The pressure at position h1 on both sides of the U-tube


P1 P2

h1

h2

heavy fluid ρ3

h3

intermediate fluid ρ2

light fluid ρ1

Figure 2.7: A compound manometer, ρ3 > ρ2 > ρ1.The sensitivity is ×5 or more ifρ2 ∼ ρ3.

must be the same. On the left hand side this is given by P1 plus the contribution of thecolumn of liquid of density ρ2. This is equal to the pressure at position h1 on the righthand side of the column, which is given by P2 plus the contribution of the column ofliquid of density ρ2 and the contribution of the column of liquid of density ρ3:

LHS: P1 + ρ2g (h3 − h1) = P2 + ρ2g (h3 − h2) + ρ3g (h2 − h1) :RHS

Now we can rearrange this to relate the pressure difference between P1 and P2 to thevarious heights:

P1 − P2 = ρ2g [(h3 − h2)− (h3 − h1)] + ρ3g (h2 − h1)= (ρ3 − ρ2) g (h2 − h1) (2.5.1)

Clearly ρ3 must be greater than ρ2, otherwise it would float rather than stay in the bot-tom of the manometer. However, if ρ2 and ρ3 are close in value such that (ρ3 − ρ2) issmall, then a small difference in pressure will correspond to a large difference in height,(h2 − h1). This becomes clearer if we rearrange the equation to express it in terms of theheight difference resulting from a given pressure difference:

Manometers

∆h =∆P

(ρ3 − ρ2) g(2.5.2)

The difference in height for a given ∆P is larger than would be obtained if ρ2 = 0,which is essentially the case in the ordinary manometer in which the “intermediate” fluidis air.

2.5.3 The Bourdon Gauge

The most common type of pressure gauge is the Bourdon gauge, which is compact, robustand easy to use, and which relies on the deformation of an elastic solid as its basis for

2.6. PRESSURE HEADS 21

indicating pressure. It comprises a curved tube of elliptical cross section closed at oneend and free to move at this end, while the other end is fixed in position but open to entryof the fluid whose pressure is being measured. When the pressure exceeds the externalpressure, the tube cross section tends to become more circular, causing the tube to uncurlslightly. The movement of the free end of the tube is transmitted to a scale via a suitablepointer and mechanical linkage, as illustrated in Figure 2.8.

Figure 2.8: Diagram of a Bourdon gauge.

2.6 Pressure Heads

It will be clear from the above examples of pressure measurement that pressure is fre-quently measured in terms of the height of a column of liquid. In particular, the relation-ship between gauge pressure and height is directly proportional, such that pressure can infact be visualised in terms of the height of a column of liquid of density ρ. This is termedthe pressure head, a term used frequently in fluid flow, corresponding to

h =P

ρg= the pressure head corresponding to P (2.6.1)

2.7. PROBLEMS 23

2.7 Problems

1. A tank contains a 2.1 m deep layer of seawater (density = 1030 kg m−3), with a0.6 m layer of oil (density = 880 kg m−3) floating on top. The pressure in theheadspace above the liquids is 1 bar gauge. Calculate the pressure (in kPa gauge)at the bottom of the tank.

2. A barometer consists of a length of glass tube with an inner diameter of 3 mm,sealed at one end, and positioned vertically with its open end in a trough of mercury.Above the mercury in the tube is a vacuum, and the length of tube taken up by thisis 150 mm. Some air is introduced into the tube above the mercury. The air wouldoccupy 1 ml at atmospheric pressure. Calculate the new height of mercury in thetube and the pressure of the air in the tube.

[0.499 m; 66.6 kN m−2]

3. Answer all parts of this question.

(a) A scientist carries a thermometer, barometer, pan and gas camping stove upa mountain. At a certain point he collects water from a stream, boils it andmeasures the temperature of the steam condensing on the thermometer to be98 ◦C. What is the height of the column of mercury (density = 13560 kg m−3)in the barometer?

(b) Estimate the altitude of the scientist relative to sea level, assuming the densityof air to be 1.186 kg m−3.

(c) The scientist notices that it takes about 3 minutes to bring about 2 litres ofwater just to the boiling temperature. Assuming the thermal capacity of thewater dominates (i.e. ignoring the thermal capacity of the pan), estimate therate of heat transfer delivered by the camping stove to the pan of water.

4. A beach ball (diameter 50 cm) has an internal pressure of 0.2 atm gauge.

(a) To what depth has the ball to be taken in the sea before it starts to change insize? Discuss how the size of the ball changes with pressure and depth.

(b) What is (theoretically, assuming the ball is extremely flexible and contains anideal gas) the diameter of the ball at a depth of 10 km?

[h > 2.1 m; 0.054 m]

Chapter 3Flow Dynamics — Ideal Fluids


3.2 Flow Regimes and Reynolds Number . . . . . . . . . . . . . . . . . 27

3.3 Mass Balance in a Pipe . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.3.1 Streamlines and Streamtubes . . . . . . . . . . . . . . . . . . . 32

3.3.2 The Continuity Equation . . . . . . . . . . . . . . . . . . . . . 32

3.4 Energy Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.4.1 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.4.2 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.4.3 Pressure Energy . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.4.4 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.4.5 Thermal Losses . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.4.6 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . 37

25

3.1. INTRODUCTION 27

3.1 Introduction

Flow of fluids is always associated with pressure differences. As pressure is a form ofenergy, and energy can be converted into other forms, the relationship between pressureand flow is in some ways not always obvious or intuitive. One form of energy is thermalenergy (heat), such that a flowing fluid converts some of its energy into heat. This isdue to the internal friction of the flowing fluid, i.e. its viscosity. In this chapter we willfirstly introduce the different types of flow or “flow regimes”. As we’ll see, the nature offluid flow depends, among other things, on the fluid’s viscosity. However, we’ll start byconsidering “ideal” fluids that have no viscosity and that convert none of their energy intoheat as a result of flow, in order to introduce and understand some fundamental concepts.We will do this by applying the law of conservation of energy in order to relate energychanges and different forms of energy (potential, kinetic and pressure) to fluid flow. (We’llalso make use of the law of conservation of mass.) Later we will extend this fundamentalknowledge to deal with real fluids in order to take account of the effects of viscosity.

3.2 Flow Regimes and Reynolds Number

Consider two tanks connected by a pipe, as shown in Figure 3.1. Tank A is at higherpressure (in this case atmospheric pressure), while Tank B is kept at a lower pressurethrough the action of a pump. Because there is a pressure difference between the tanks,fluid flows along the pipe from Tank A to Tank B at a velocity u. What is the relationshipbetween the pressure drop, (PA − PB) causing the fluid to flow along the pipe, and thevelocity with which it flows?

u

A B

PBPA = Patm

Figure 3.1: Fluid flow in a pipe between tanks at different pressures.

The effect of pressure drop on flow velocity depends on the type of flow patterns inthe pipe and the flow regime. Figure 3.2 illustrates possible relationships between uand (PA − PB) and defines different regimes corresponding to different patterns of flow.

From this we can make a number of observations:

• Gases can achieve much higher flow velocities than liquids; ugases >> uliquids.

• Region I corresponds to laminar incompressible flow, in which flow velocity isproportional to pressure drop; u ≈ ∆P . Both gases and liquids can undergo RegionI flow.

28 CHAPTER 3. FLOW DYNAMICS — IDEAL FLUIDS

Velocity, u

Pressure difference, PA − PB

Liquid

I II

PB = vapour pressure

Gas

I

II

III

IVus

PA − PBPA

∼ 0.3

Re = 2100

Figure 3.2: Velocity of fluid flow versus pressure difference.

• Region II corresponds to turbulent incompressible flow, in which u ≈ ∆P 0.5. Bothgases and liquids can undergo Region II flow.

• Region III corresponds to compressible flow. This applies to gases only, and indi-cates that the pressure change is sufficiently large that the density of the gas changessignificantly as it flows and can no longer be considered constant. In this region,u ≈ ∆P n where n < 0.5.

• Region IV applies to gases only and describes choking flow. This occurs whenthe gas is flowing at the speed of sound, us, and cannot flow any faster, no matterwhat size of pressure drop is applied. Thus under choking flow, u is independentof ∆P . Choking flow typically occurs when the magnitude of the pressure drop,(PA − PB), is about 30% of the initial pressure of the gas.

Let’s consider the flow patterns of the fluid within the pipe linking Tanks A and B. Itsoverall average velocity is constant, but the velocity at different points in the pipe is likelyto vary – for example, the velocity along the centre of the pipe is likely to be higherthan that near the walls of the pipe, where the stationary walls slow the fluid down. Sovelocity depends on radial position within the pipe, and across the pipe there is a velocityprofile.

If the fluid had no viscosity, then there would be no friction to slow the fluid down, andall the elements of fluid would flow at the same velocity. Under this ideal, frictionlessflow, u (r) = constant = u = umax. A plot of velocity versus radial position would showconstant velocities across the pipe, as illustrated below in Figure 3.3(c). This never occursin reality, but it is approximated, as we’ll see in a moment, by turbulent flow.

If the fluid has viscosity, then the shear experienced by the fluid at the wall is transmittedthrough the fluid, such that the fluid near the wall travels more slowly than the fluid in thecentre. In fact, the fluid right at the pipe wall is considered to have zero velocity, whilethe velocity increases towards the centre of the pipe. In this case, there are two extremepossibilities for the nature of the fluid flow: laminar flow, and turbulent flow.

Under laminar flow, the flow is smooth, steady and ordered such that fluid elements flow

3.2. FLOW REGIMES AND REYNOLDS NUMBER 29

u

umax

r

(a) Laminar Flow

u

umax

r

(b) Turbulent Flow

u

umax

r

(c) Ideal (Frictionless) Flow

Figure 3.3: Flow profiles in pipes under laminar, turbulent and ideal flow.

along individual layers (laminae) or streamlines that do not cross each other. Flow isdominated by viscosity, leading to a parabolic flow profile as illustrated in Figure 3.3(a).For a circular pipe, it can be shown that the maximum velocity, which occurs at the centreof the pipe, is twice the average velocity: u = 0.5u. Laminar flow occurs at low flowrates in which the fluid has little inertia.

Turbulent flow, by contrast, is wild and complex. Under turbulent flow, the paths of indi-vidual elements of fluid are no longer straight and steady, but are erratic and intercrossing.Thus, only the average motion of the fluid is in the direction of flow (parallel to the axis ofthe pipe). Thus the velocity profile illustrated in Figure 3.3(b) shows the average velocityat different radial positions. Turbulent flow is found at high flow rates in which the fluid’sinertia dominates over its viscosity in determining the flow patterns. Turbulent flow ex-hibits a high shear stress at the wall and steep velocity gradient near the wall. Under fullydeveloped turbulent flow, the velocity profile away from the walls is quite flat, approach-ing that of ideal flow, with the average velocity in a circular pipe approximately 82% ofthe maximum velocity: u ≈ 0.82u.


The fluid near the wall flows more slowly because the wall is stationary and the viscosityof the fluid transmits momentum between the stationary wall and the flowing fluid. Underturbulent flow, the layer near the wall, where the velocity profile is affected by the pres-ence of the stationary surface, is called the laminar sublayer or viscous sublayer.

Returning to Figure 3.2, evidently there is a transition from laminar (Region I) flow toturbulent (Region II) flow – although it is not a sudden or clear transition. The transitionfrom laminar to turbulent flow was first studied in the early 1880s by Osborne Reynolds,Professor of Engineering at the University of Manchester.

Reynolds constructed an apparatus in which the flowrate of water through a glass tubecould be carefully controlled. He then introduced a fine stream of dye into the flow andobserved the patterns of motion of the stream of dye. Figure 3.4 shows some images ofhis observations. At low flowrates (Figure 3.4(a)), the dye filament remained steady andunbroken as it flowed along with the water, indicating laminar flow. At higher flowrates((b) and (c) in Figure 3.4), the flow became turbulent such that the dye filament began tooscillate and eventually broke up into eddies which dispersed the dye right across the tube.These experiments clearly demonstrated the transition from laminar to turbulent flow, andallowed the conditions leading to either laminar or turbulent flow to be identified.

(a) Laminar Flow

(b) Transitional Flow

(c) Turbulent Flow

Figure 3.4: Flow patterns observed on injection of a stream of dye into a flowingfluid.Water (a) velocity 11 cm s−1, Re = 1.5 × 103, (b) velocity 17 cm s−1, Re =2.34 × 103, and (c) velocity 54 cm s−1, Re = 7.5 × 103 pipe ID 14 mm, dye injectionmethod.

Experiments with a variety of fluids of varying physical properties and of a range ofpipe diameters and flowrates demonstrated that the nature of the flow depended on thefluid’s density, ρ, and viscosity, µ, its average velocity through the pipe (u, but we tendjust to write u, as it’s more convenient), and the pipe diameter, D. Larger pipes, fastervelocities and greater densities tend to give more inertia to the flow and make it turbulent,while small pipes, slow flow and high viscosities tend to give laminar flows. Further, itwas discovered that these parameters could be arranged to form a dimensionless group,which was later called the Reynolds Number, Re, in honour of Osborne Reynolds andhis seminal work. The Reynolds Number determines the nature of the fluid flow and is

3.2. FLOW REGIMES AND REYNOLDS NUMBER 31

defined as

Re =Duρ

µunits:

m× ms× kg

m3

kgm s

= no units (3.2.1)

At values of Re less than about 2100, flow is laminar. For Re > 10, 000, flow is almostalways turbulent. For 2100 < Re < 10000, the flow is transitional between laminar andturbulent.

The Reynolds Number is an example of a dimensionless number. Dimensionless numbersare important in chemical engineering, and you’ll meet many others in due course. Theyhave the great benefit that they can bring together a wide range of experimental condi-tions and allow them to be described in simple terms, and that they allow similarities ofbehaviour at different scales to be identified. In this case, while many different conditionsof fluid flow might exist, in terms of the density and viscosity of the fluid and the veloc-ity and geometry of the flow, all these conditions can be made comparable by describingthem in terms of the Reynolds Number. The Reynolds Number is fundamental to fluidflow; the equation for the Reynolds Number is one that you must memorise.

Dimensionless numbers frequently achieve their dimensionless status by representing theratio of two things.

Reynolds NumberThe Reynolds Number, the ratio is between inertial forces, which favour turbulentflow, and viscous forces, which favour laminar flow.

Re =Duρ

µ

QuestionCalculate the Reynolds Numbers for water at 20 ◦C flowing at 2 m s−1, and for airflowing at 20 m s−1, along a 1 inch pipe. Before you do the calculation, visualisethe situation and have a guess – do you think the flow would be turbulent or laminarin each case?Water — Density = 1000 kg m3, Viscosity - 0.001 Pa sAir — Density = 1.207 kg m3, Viscosity - 1.81× 10−5 Pa s

In most practical engineering situations, flows are turbulent. Turbulent flow is importantin chemical engineering, as it gives rapid heat and mass transfer.

Some useful definitions:

• Velocity profile: u = u (r), i.e. velocity depends on, and is a function of, radialposition.

• Boundary condition: The value of velocity or its derivatives at the edge of a regionof interest.

• “No-slip” boundary condition: The value of velocity at a solid surface has beenshown experimentally to be zero. Mathematically, u = 0 at r = R or u (R) = 0.


3.3 Mass Balance in a Pipe – Derivation of the ContinuityEquation

The fact that (for practical, fluid flow purposes) mass is conserved means that what flowsinto one end of a pipe must flow out the other end and/or accumulate. Similarly, in othergeometries (fluid flows in other contexts, just not in pipes), the flow of fluid must obeythe law of conservation of mass. Expressing this mathematically is therefore an importantpart of fluid flow calculations. Our mathematical descriptions of this physical reality willbe more accurate and powerful if we have a good understanding of the idea that fluidsflow along streamlines.

3.3.1 Streamlines and Streamtubes

When fluids flow, the velocity over a plane at right angles to the flow is not normallyuniform, as we saw in Figure 3.3. The variation of velocity can be shown by drawingstreamlines, such that the velocity vector is always tangential to the streamline. Thus, bydefinition, there cannot be a net flow of fluid across a streamline (because the velocityvectors do not cross streamlines but run tangential to them). The flowrate between twostreamlines must therefore always be constant (as fluid cannot enter or leave the spacebetween the streamlines – there is no escape). If the fluid speeds up, we show this bymaking the streamlines closer together (such that the area between them stays constant).To put it another way, if a given quantity of fluid were flowing through a smaller pipe(equivalent to streamlines being closer together), it would need to flow faster. Figure 3.5illustrates parallel streamlines in steady flow in a straight tube, and streamlines comingcloser together as the fluid enters a constriction and speeds up.

Figure 3.5: Streamlines in steady straight flow and approaching a constriction.

In laminar flow, sometimes called streamline flow, the fluid flows along streamlines anddoes not cross them. In turbulent flow, circulations and eddies mean that fluid elementsdo cross streamlines, but there is no net flow across a streamline, and the average velocityis along the streamline.

A group of streamlines can be visualised as forming a streamtube, and the whole area offlow can be visualised as bundles of streamtubes. Describing fluid flow along streamlinesand within streamtubes is the basis for the mathematical description of the flow.

3.3.2 The Continuity Equation

Consider flow of fluid through a streamtube as shown in Figure 3.6, in which the crosssectional area of the tube is sufficiently small that the fluid can be considered to have a

3.3. MASS BALANCE IN A PIPE 33

uniform velocity across the tube. Fluid flows in at point 1 with a velocity u1 through across sectional area A1, and out at point 2 with velocity u2 and cross sectional area A2.By conservation of mass, under steady flow the mass flow rate into the tube at point 1equals the mass flow rate out of the tube at point 2.

1

A1

u1

2

A2

u2Streamline

Streamtube

Streamline = path tracedby particle of fluid

Streamtube = imaginary tubewith streamlines as walls

Figure 3.6: Flow through a streamtube.

Mass flowrate in at point 1 = Mass flowrate out at point 2

Mass = density× volume, and the volumetric flowrate at each point is equal to the crosssectional area times the velocity, so:

Continuity Equationρ1u1A1 = ρ2u2A2 = M (3.3.1)

where M is the mass flowrate. This is the Continuity Equation, which indicates that themass flowrate must be constant along a streamtube, and that changes in the area of flowmust be balanced by changes in the fluid velocity and/or density.

If density is constant (i.e. incompressible flow, as for a liquid or for a gas flowing under asmall pressure drop), then

u1A1 = u2A2 = Q (3.3.2)

where Q is the volumetric flowrate. (Ideally, we would write Q, to underline that it is aflowrate and to be consistent with M . However, we use Q to indicate heat transfer rate inour Heat Transfer course. Don’t get these two different uses ofQ confused! To help avoidconfusion, but at risk of being inconsistent in our notation, we’ll use Q in this course toindicate volumetric flowrate of fluid, and Q in the Heat Transfer course to indicate heattransfer rate.)

Averaging across an entire cross section comprising bundles of streamtubes, we need touse the average velocity at each point:

u1A1 = u2A2 (3.3.3)


Frequently we just write u instead of u; when we see u, we generally assume it means theaverage velocity, unless the context indicates otherwise.

Understanding the requirement for continuity of flow can lead us to an alternative versionof the Reynolds Number. Frequently we are given the mass flowrate of a fluid flowingthrough a pipe, from which we want to calculate the Reynolds Number in order to de-termine the nature of the flow. To do so, we must convert the mass flowrate, M , to avolumetric flowrate, Q, via the fluid density, then calculate the fluid’s average velocityfrom Q = uA. Thus,

u =Q

A= MρA (3.3.4)

and

Re =Duρ

µ=DMρ

ρAµ=

DM

πD2

4µ

=4M

πDµ(3.3.5)

This version allows the Reynolds number to be calculated directly from the mass flow rate,without having to look up the density or to calculate the fluid’s average velocity.

QuestionWater at 50 ◦C flows at along a 1.5 inch pipe at a flowrate of 900 kg per hour.Calculate the Reynolds number using both versions of the equation. The density ofwater at 50 ◦C is 988 kg m−3, and the viscosity is 548× 10−6 Pa s.

Note that this raises an interesting question. For a given flowrate of fluid, would theReynolds Number increase or decrease if the pipe diameter was increased? Why?

3.4 Energy balance – Derivation of Bernoulli’s Equation

Conservation of energy applies to flowing fluids, but as the fluid flows, energy may beconverted from one form to another – pressure energy, potential energy, kinetic energy,thermal energy and mechanical work can be interconverted. The relationships betweenthese different forms of energy can be found by performing an energy balance.

Consider fluid flowing within a pipe as illustrated in Figure 3.7, in which the cross sec-tional area of the pipe changes, so the fluid velocity changes, and the height of the fluidin the pipe also changes, i.e. the fluid is having to be pushed “uphill”, for example. Themass flowrate of the fluid is M , and the fluid is undergoing steady flow. We will considerthe energy of the system bounded by the cross sectional planes at points 1 and 2.

An energy balance tells us that the rate at which energy enters the system at point 1 mustbe equal to the rate at which energy leaves the system at point 2, plus the rate of workdone by the fluid,W , which leaves the system as mechanical work, plus the rate of thermallosses, Q.

3.4.1 Potential Energy

At point 1, the potential energy of the fluid is equal to,

Epot,1 = Mgh1

3.4. ENERGY BALANCE 35

h2

h1

W

1

2

Figure 3.7: Diagram of fluid flow along a pipe, as a basis for performing an energy balancebetween points 1 and 2.

The rate at which potential energy is added into the system at point 1 is therefore equalto,

Epot,1 = Mgh1 (3.4.1)

Similarly, the rate at which potential energy leaves the system at point 2 is Mgh2.

3.4.2 Kinetic Energy

At point 1, the kinetic energy of the fluid is,

Ekin,1 = 0.5Mu21

The rate of addition of kinetic energy to the system at point 1 is therefore,

Ekin,1 = 0.5Mu21 (3.4.2)

Similarly, the rate at which kinetic energy leaves the system at point 2 is 0.5Mu22. Therelationship between u1 and u2 depends on the cross sectional areas at points 1 and 2 andis, of course, given by the Continuity Equation.

3.4.3 Pressure Energy

The rate at which energy is added to the system due to the pressure at point 1, P1, is foundby considering a piston acting on the fluid at that point, as illustrated in Figure 3.8.

The force exerted by the piston, F , is equal to the pressure times the cross sectionalarea,

F = P1A1


d l

u P1

Figure 3.8: Piston acting against a pressure P1 over a distance d l at a velocity u.

Work = force× distance moved in the direction of the force, therefore,

W = Fd l = P1A1d l

= pressure× volume displaced by the piston

Rate of energy input = work per unit time, therefore,

Epress =dW

d t= P1A1

d l

d t= P1A1u1 =

P1M

ρ1(3.4.3)

as u = d l/d t, the fluid’s velocity. So rate of energy input to the system is P1M1/ρ1. Sim-ilarly, rate of energy done by the system due to the pressure at point 2 is P2M2/ρ2.

3.4.4 Work

The fluid may be doing mechanical work at a certain rate, W , e.g.

• by flowing through a turbine to produce power — in which case W would have apositive sign, as work done by the system; or

• there may be a pump imparting energy to the fluid — in which case W would havea negative sign, as work done on the system.

3.4.5 Thermal Losses

The fluid may also gain/lose energy as it flows along the pipe, Q (not to be confused withthe volumetric flowrate of the fluid, Q, as noted above!), e.g.

• The fluid would convert some of its energy to thermal energy as it flows along (dueto the internal friction of the fluid, i.e. its viscosity), and more so if there are bendsand fittings in the pipe. For the fluid to stay at the same temperature (and hencesame internal energy) this heat is lost to the surroundings.

• There could also energy inputs to the fluid via heat transfer or chemical reaction.

3.4. ENERGY BALANCE 37

3.4.6 Bernoulli’s Equation

Having defined the various contributing terms, we can now perform the energy bal-ance:[

Pressure energyinput at point 1

]+

[Kinetic energyinput at point 1

]+

[Potential energyinput at point 1

]=[

Pressure energyinput at point 2

]+

[Kinetic energyinput at point 2

]+

[Potential energyinput at point 2

]+[

Rate of work doneby the fluid

]+

[Thermal lossesdue to viscosity

]Using Equations 3.4.1, 3.4.2, and 3.4.3, produces

P1M

ρ1+Mu21

2+ Mgh1 =

P2M

ρ2+Mu22

2+ Mgh2 + W + Q (3.4.4)

Now, the mass flowrate, M , is constant, so can be divided through. It is also conventionalto divide through by the acceleration due to gravity, g:

P1

ρ1g+u212g

+ h1 =P2

ρ2g+u222g

+ h2 +W

Mg+

Q

Mg(3.4.5)

If no work is done on or by the system, then the second to last term disappears. If no heatis transferred to/from the system and there is no chemical reaction, then the last term isonly due to the losses due to viscosity which are generally described by the term ∆hf .Therefore the equation becomes:

Bernoulli’s EquationP1

ρ1g+u212g

+ h1 =P2

ρ2g+u222g

+ h2 + ∆hf (3.4.6)

If the fluid is inviscid, i.e. has no viscosity, such that there is no conversion of energy tothermal energy, then the final term disappears, and we get

P

ρg+u2

2g+ h = Z = constant (3.4.7)

This is Bernoulli’s Equation, named in honour of the Swiss mathematician Daniel Bernoulli(1700-1782) who did the essential original mathematical and experimental work underly-ing the equation. Bernoulli’s Equation indicates that as a fluid flows, its pressure energy,kinetic energy, and potential energy can be interconverted, from one form to another andback again, but the total energy of the fluid stays constant. Bernoulli’s Equation applies toincompressible, frictionless (inviscid) fluids under steady flow along a streamline.

It can be extended to describe the flow of real fluids (i.e. to take account of viscosity),by including the term ∆hf to account for thermal losses. We will extend Bernoulli’sEquation to real fluids later and learn how to calculate the ∆hf term then.


Looking at the units of the various terms in Bernoulli’s Equation:

P

ρg:

kg m−1 s−2

kg m−3 m s−2= m

u2

2g:

(m s−1)2

m s−2= m

h : m

Clearly, the terms must have consistent units, and as it happens, in this form of the equa-tion, the units are in metres. This has led to the use of the term head to describe thedifferent terms. The first term is the pressure head (in fact, we introduced this term ear-lier, in the context of hydrostatics). The second term is the velocity head, while the thirdterm is the potential head. Thus the total head, Z, is constant. We can also now give aname to the term ∆hf , which is the head loss due to friction.

In order to get a deeper understanding about the relationship between the three terms inBernoulli’s Equation, we could multiply each term by g, to give:

P

ρ+u2

2+ gh = constant (3.4.8)

Working backwards, the last term is reminiscent of mgh, which is potential energy. Thusthe combination gh on its own indicates potential energy per unit mass of fluid. Similarly,the second term is reminiscent of the equation for kinetic energy, 0.5mu2, and indicatesthe kinetic energy per unit mass of fluid. By extension, then, the first term must indicatethe pressure energy per unit mass of fluid. Remembering that pressure has units of energyper unit volume, and that density is mass per unit volume, then clearly pressure dividedby density does indeed have units of energy per unit mass. So in this form, Bernoulli’sEquation indicates that the energy per unit mass of fluid stays constant as the fluid flowsalong.

We can also multiply through by the fluid’s density (reminding ourselves as we do thatBernoulli’s Equation only applies for fluids of constant density, i.e. incompressible flow):

P +ρu2

2+ ρgh = constant (3.4.9)

Now all the terms have units of energy per unit volume, instead of energy per unitmass.

You need to be familiar with these different forms of Bernoulli’s Equation and to beable to use it in these different forms. Bernoulli’s Equation has wide applicability andis extremely useful in understanding the behaviour of flowing fluids and in designingpipelines and equipment. In many cases several of the terms cancel or are negligible,such that the equation becomes much simpler.

Chapter 4Applications of Bernoulli’s Equation

Contents4.1 Flow from a Tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.1.1 Tanks with Constant Height . . . . . . . . . . . . . . . . . . . 41

4.1.2 Tank with Variable Height . . . . . . . . . . . . . . . . . . . . 42

4.2 Flow Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2.1 Velocity Profile Measurement – The Pitot Tube . . . . . . . . . 44

4.2.2 The Orifice Meter . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2.3 The Venturi Meter . . . . . . . . . . . . . . . . . . . . . . . . 48

4.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

39

4.1. FLOW FROM A TANK 41

4.1 Flow from a Tank

How long does your bath take to empty? How rapidly will fluid flow from a tank in whichthe fluid has a certain height and flows from the tank through a hole of a certain diameter?Bernoulli’s Equation can be employed to give insights into these situations, in terms ofrelationships between the relevant parameters, and to estimate the answers.

4.1.1 Tanks with Constant Height

Consider a tank of cross sectional area A1 containing a liquid to a height h1, which flowsfrom the bottom of the tank through a small hole of area A2, as shown in Figure 4.1.The height of the liquid is kept constant by adding liquid at an appropriate rate. Find anexpression for the rate of discharge of liquid.

h1

•1, A1

•2, A2

Figure 4.1: Liquid discharging through a hole in the base of a tank.

Apply Bernoulli’s Equation between points 1 and 2.

P1

ρ1g+u212g

+ h1 =P2

ρ2g+u222g

+ h2 + ∆hf (4.1.1)

Assumptions:

• Ideal (frictionless) flow; ∆hf = 0

• u1 = 0 (because of the large area of A1 compared with A2, the velocity at point 1is negligible compared with the velocity at point 2)

• h2 = 0 (datum point is chosen at h2).

• P1 = P2 = Patm. Both the top surface and the liquid discharging through the holeare at atmospheric pressure.

Therefore,

��7

Patmρ1g

P1

ρ1g+��0

u212g

+ h1 =��7

Patmρ1g

P2

ρ2g+u222g

+��0

h2 +��*0

∆hf

h1 =u222g

u2 =√

2gh1 (4.1.2)

42 CHAPTER 4. APPLICATIONS OF BERNOULLI’S EQUATION

The volumetric flowrate of fluid through the hole is then given by:

Q = u2A2 = A2

√2gh1 (4.1.3)

and the mass flowrate by:M = ρu2A2 = ρA2

√2gh1 (4.1.4)

In practice, friction effects arising from the fluid’s viscosity will not be negligible, suchthat ∆hf 6= 0 and the velocity of the fluid discharging from the hole will be smaller. Thisis allowed for in practice by including an empirically (i.e. experimentally) determineddischarge coefficient, CD, where CD < 1:

u = = CD√

2gh1

Q = CDA2

√2gh1

M = CDρA2

√2gh1

4.1.2 Tank with Variable Height

In the situation in which the tank is emptying and the liquid discharging through the hole isnot replaced, such that h1 is no longer constant, the varying flowrate can be described, andthe time to empty the tank determined. Note that this is now an unsteady state situation,compared with the steady state system described above. We would need to assume thatthe velocity of the exiting liquid has the same relation to the liquid’s height in the unsteadysituation as it did in the steady situation, and check our assumption by experiment. Evenso, such an analysis indicates the broad relationships between the factors determining thetime to empty the tank, and gives a basis for an initial estimate of that time.

Mass balance:

Mass flow in−Mass flow out = Rate of change of mass in tank

0− ρu2A2 =d (ρA1h1)

d t

Based on our previous analysis for the steady state situation, assuming that the dischargevelocity has the same relation to h even though h is now changing from h1 initially to h2finally, we can say:

u2 = CD√

2gh

Substituting this expression for u2 gives:

A1dh

d t= − A2CD

√2gh

dh√h

= − A2CD√

2g

A1

d t

4.2. FLOW MEASUREMENT 43

For the tank to empty, h varies from h1 initially (at time t = 0) to h2 finally (at time t = t).So, we can integrate both sides of the equation:∫ h2

h1

dh√h

= − A2CD√

2g

A1

∫ t

0

d t[2√h]h2h1

= − A2CD√

2g

A1

t

t =2A1

A2CD√

2g

(√h1 −

√h2

)

If h2 = 0, then

t =2A1

A2CD√

2g

√h1 (4.1.5)

In other words, the time to empty the tank varies with the square root of the initial heightof the liquid in the tank. Doubling the amount of liquid in the tank initially will lead to a41% increase in the time to empty the tank (as

√2 ≈ 1.414).

QuestionHow does the design of the wing of a plane enable it to overcome gravity and fly?Apart from the risk of being hit by the train, why are you advised to stand awayfrom the platform edge when a high speed train passes through a station?An on-demand hot water boiler detects the flow of the water in order to initiate heat-ing. How might a device be designed that uses knowledge of Bernoulli’s equationto detect the flow?Sometimes to assist filtration in a lab, an experimenter will attach one end of a tubeto a flowing stream of water in a pipe, and the other end to the bottom of the filterfunnel. Why?

4.2 Flow Measurement

Bernoulli’s Equation can be applied to address a critical issue of interest to chemical andpetroleum engineers – that of measuring the flowrate of fluid in a pipe. The principle isthat the flow of fluid is either accelerated or retarded (i.e. change in velocity), and theresulting change in pressure measured (as kinetic and pressure energy interconvert), withBernoulli’s Equation then used to interpret the change and relate it to the fluid flowrate.Devices that employ this principle include:

The Pitot tube, in which a small element of fluid is brought to rest, such that its kineticenergy is converted into pressure energy – by measuring the increase in the pressure,the original kinetic energy and hence the velocity can be determined

The Orifice meter, in which fluid passes through a sudden constriction, increasing itsvelocity and hence its kinetic energy, which is then related to the reduction in pres-sure

The Venturi meter, in which the fluid passes through a gradual constriction to increaseits velocity and reduce its pressure.


4.2.1 Velocity Profile Measurement – The Pitot Tube

The principle of the Pitot tube was first used by Henri Pitot in 1732 for measuring veloci-ties of water in the River Seine. Figure 4.2 shows a diagram of a modern Pitot tube, whichconsists of two concentric tubes arranged to join the two arms of a manometer. Fluid isflowing at point 1 with a velocity u1. At point 2, the presence of the nose of the Pitot tubecauses a stagnation point, at which the velocity is zero, because fluid is unable to flowinto the inner tube. Thus the pressure at point 2 is higher than at point 1, as the kineticenergy has been converted into pressure. At point 3, the fluid is flowing with (close to) itsoriginal velocity, so its pressure is lower than the pressure at point 2.

Figure 4.2: Pitot tube, used for measuring the local flow velocity in a fluid.

The position of the hole at point 3 is critical (in fact, several holes are drilled around thecircumference of the outer tube at that point). The presence of the head of the Pitot tubedisrupts the flow slightly and affects the measured pressure, as does the presence of thestem. On a standard Pitot tube, the distance from the head to the stem is 14 tube diameters,and the holes are located 6 diameters from the head and 8 diameters from the stem – atthis point, the effects on the measured pressure are equal and opposite and cancel eachother out. A well-designed and correctly operated Pitot tube is accurate to within about1%.

In order to relate the measured height difference in the manometer to the velocity, weapply Bernoulli’s Equation between points 2 and 3:

P2

ρ2g+u222g

+ h2 =P3

ρ3g+u232g

+ h3 + ∆hf (4.2.1)

Assumptions:

• Ideal (frictionless) flow; ∆hf = 0.


• u2 = 0 (fluid is brought to a stagnation point).

• h2 = h3 (the Pitot tube is horizontal).

• u3 = u (the general fluid velocity).

• ρ2 = ρ3 = ρ (incompressible fluid).

Therefore,

P2

ρg+��0

u222g

+��0

h2 =P3

ρg+u2

2g+��

0h3 +��

�*0∆hf

P2

ρg=P3

ρg+u2

2g

P2 − P3 =ρu2

2(4.2.2)

The pressure difference is indicated by the hydrostatic height difference in the manometer(where ρF is the density of the manometer fluid). This is found by considering the pres-sure at the lower datum level (indicated by the lower dashed line) in the two arms of themanometer. In the left hand arm, which corresponds to point 2, the pressure is:

P = P2 + ρgh2

This is equal to the pressure at the same height in the right hand arm, given by:

P = P3 + ρg (h2 −∆h) + ρFg∆h

So

P2 + ρgh2 = P3 + ρg (h2 −∆h) + ρFg∆h

P2 − P3 = − ρg∆h+ ρFg∆h = (ρF − ρ) g∆h =ρu2

2(4.2.3)

Therefore, the velocity is given by:

Pitot Tube

u =

√2 (ρF − ρ) g∆h

ρ(4.2.4)

Note that this is the local velocity of the fluid at the particular location in the pipe (not theaverage velocity across the whole pipe). By moving the Pitot tube across a pipe (or river,or other fluid flow system of interest), one can measure the velocity profile.

4.2.2 The Orifice Meter

In contrast with the Pitot tube, which slows the flow and measures the resulting increase inpressure, orifice and Venturi meters cause the fluid to flow through a restriction in orderto increase the flow velocity, and measure the resulting decrease in pressure. Also in


contrast with the Pitot tube, which measures the local velocity at a particular point withinthe flow, these two meters give a measure of the average flow velocity across the wholecross sectional area of the pipe.

Figure 4.3 illustrates the orifice meter. An orifice plate is inserted across the cross sectionof the flow. The hole in the orifice plate is smaller than the pipe diameter, and (accord-ing to continuity) the flow velocity must increase in order to get the fluid through theconstriction.

Figure 4.3: Orifice meter, in which an orifice plate inserted into the flow increases thefluid velocity, causing a decrease in pressure, which is measured with a manometer.

Upstream of the orifice, the flow cannot suddenly turn the corner when it meets the ori-fice plate. Instead, the streamlines converge, to form a cone of streamlines as they ap-proach the orifice. Again, these streamlines cannot suddenly change direction as theypass through the orifice, so they continue to converge for some distance past the orifice,to form what is called a vena contracta, a jet of fast moving fluid surrounded by slowlymoving eddies near the wall. As this jet mixes with the slower moving fluid downstream,it slows down and the streamlines diverge to fill the entire pipe cross section once again.The flow downstream of the orifice is highly turbulent, such that most of the kinetic en-ergy of the jet is dissipated as heat and is not recoverable as pressure energy. Thus thepressure drop over an orifice meter is high (which may not be acceptable in some circum-stances).

The orifice plate has a sharp edge on its upstream side, rather than a right angle, so thatthe flow through the plate itself is not affected by the walls of the plate. Similarly, to avoidflow effects from upstream bends and fittings, the orifice plate should be located at least30 pipe diameters (and preferably at least 50D) downstream of any such fittings.

The jet velocity is at its fastest when the vena contracta is at its narrowest, which occursabout half a pipe diameter downstream of the orifice. The pressure at this point willtherefore be lowest, and this is where one arm of the manometer is located, with the other


arm placed one pipe diameter upstream of the plate. (The velocity profile across the jetis also constant at this point.) Labelling the upstream point as point 1, and the minimumof the vena contracta as point 2, and noting that h1 = h2, applying Bernoulli’s Equation(once again ignoring viscosity effects, although we’ll take account of these in a moment)gives:

P1

ρ1g+u212g

+ h1 =P2

ρ2g+u222g

+ h2 + ∆hf (4.2.5)

Assumptions:

• Ideal (frictionless) flow; ∆hf = 0.

• h1 = h2 (the orifice is horizontal).

• ρ2 = ρ3 = ρ (incompressible fluid).

Therefore,

P1

ρg+u212g

+��0

h1 =P2

ρg+u222g

+��0

h3 +��*0

∆hf

P1

ρg+u212g

=P2

ρg+u222g

P1 +ρu212

= P2 +ρu222

P1 − P2 =ρ

2

(u22 − u21

)(4.2.6)

As before for the Pitot tube, the pressure difference is related to the height difference inthe manometer:

P1 − P2 = (ρF − ρ) g∆h =ρ

2

(u22 − u21

)By continuity:

u1 = u2A2

A1

Thus

(ρF − ρ) g∆h =ρ

2u22

(1−

(A2

A1

)2)

Therefore,

u2 =

√√√√√√2 (ρF − ρ) g∆h

ρ

(1−

(A2

A1

)2)

The flow rate is therefore,

Q = u2A2 = A2

√√√√√√2 (ρF − ρ) g∆h

ρ

(1−

(A2

A1

)2)

Now, the problem is we don’t know the cross-sectional area of the vena contracta, A2. Wedo know, of course, the cross-sectional area of the orifice itself, which we’ll call Ao. We


can define a coefficient of contraction, Cc, which is the ratio of the cross-sectional areasof the vena contracta and the orifice: A2 = CcAo.

In addition, frictional effects due to viscosity cause the velocity in the vena contracta tobe less that the velocity predicted by Bernoulli’s equation. A coefficient of velocity, Cv,can be defined as the ratio of the actual mean velocity to the ideal velocity in the venacontracta.

In combination, these two effects – the contraction of the area of the jet and the slowingeffects of friction – mean that the discharge (the volumetric flowrate through the orifice)is less than one might expect based on Bernoulli’s Equation for a given area of the orifice.Combining the two effects, we can define a coefficient of discharge, CD, as the ratio ofthe actual discharge to the ideal discharge, such that CD includes the effects of Cc and Cv.It also includes the necessary conversions from A2 to Ao and any calibration factors, togive:

Orifice Meter

Qactual = CDAo

√√√√√√2 (ρF − ρ) g∆h

ρ

(1−

(A2

A1

)2) = CDAo

√√√√√√2 (P1 − P2)

ρ

(1−

(A2

A1

)2) (4.2.7)

The value of the discharge coefficient depends on the specific design of the orifice plate(e.g. the ratio of orifice diameter to pipe diameter, and the precise positioning of thepressure tappings) and on the value of the Reynolds Number in the orifice (where thediameter D used in the calculation of the Reynolds Number is the orifice diameter). ForRe > 104, CD typically has a value of about 0.61. For Re < 104, the relationship iscomplex, rising with increasing Re to a maximum, and then decreasing towards a steadyvalue of 0.61.

Orifice plates have the advantages that they are cheap and take up little space. They havethe disadvantage that they incur a large permanent pressure drop.

QuestionA fluid of relative density 0.86 and viscosity 0.003 Pa s flows through a pipe of12 cm diameter. The flow rate is measured using an orifice plate with a 6 cm diam-eter orifice, with pressure tappings connected to a differential U-tube manometerusing mercury (density = 13600 kg m−3) as the manometer fluid. The coefficientof discharge of the orifice meter is 0.62. The difference in mercury levels in themanometer is 100 mm. Calculate the pressure drop across the orifice plate and themass flowrate of the fluid. Then calculate the Reynolds Number based on the orificediameter.[12500 Pa, 8.39 kg s−1, 59347]

4.2.3 The Venturi Meter

In contrast to an orifice meter, which uses an orifice plate to give a sudden contractionthrough which the fluid must flow, a Venturi meter uses a gradual contraction followed bya gradual expansion. Figure 4.4 illustrates a Venturi meter, which consists of a converging


section of pipe, leading to a short parallel-sided throat, followed by a diverging sectionknown as a diffuser.

Figure 4.4: Venturi meter, in which a converging section of pipe increases the fluid ve-locity, causing a decrease in pressure, which is measured with a manometer. The gentlydiverging section then recovers most of the increased velocity head as pressure.

As for the orifice meter, the constriction causes the fluid velocity to increase, and thepressure difference is measured with a manometer. However, the gradually convergingsection means that there is no vena contracta, so the coefficient of contraction, Cc, is unity.Similarly, the gradually diverging section avoids high turbulence and the formation ofeddies, so recovers most of the kinetic energy as pressure. Thus Venturi meters incur onlya small pressure drop, which is attractive when there is only a small pressure drop drivingthe flow in the first place. They are also more accurate than an orifice meter. However,they take up more space than an orifice plate and are more expensive to construct. Thecoefficient of discharge, CD, of a Venturi meter is typically around 0.96− 0.98 for fluidsof low viscosity. As with the orifice meter, they must be installed in a straight piece ofpipe with no bends or fittings at least 30 pipe diameters upstream and 5 pipe diametersdownstream of the meter.

The angle of divergence in the diffuser determines the extent of pressure recovery, lowangles giving more recovery, but requiring a longer meter. In practice an angle of 5 − 7◦

gives a good compromise. Larger angles are sometimes used, to reduce the length andcost of the meter, but at the cost of a larger pressure drop.

The ratio of throat diameter to pipe diameter is typically 0.5, giving a ratio of areas of 0.25,and a ratio of velocities of 4. A smaller ratio gives a greater fluid velocity in the throatand hence a greater and more accurately measured pressure difference. However, thesubsequent dissipation of kinetic energy as heat in the diffuser is also greater (less pressurerecovery), while very low pressures in the throat can cause dissolved gases to come outof solution or vaporisation to occur, both of which would affect the measurement.

Applying Bernoulli’s Equation across a Venturi meter gives the same result as for anorifice plate, but with a different value of the discharge coefficient, and with A2, thethroat diameter, replacing the orifice diameter:


Venturi Meter

Qactual = CDA2

√√√√√√2 (ρF − ρ) g∆h

ρ

(1−

(A2

A1

)2) = CDA2

√√√√√√2 (P1 − P2)

ρ

(1−

(A2

A1

)2) (4.2.8)

Table 4.1 summarises the differences between orifice and Venturi meters.

A flow nozzle is another design of flow meter, which uses a curved entrance to an orificein order to accelerate the fluid, but does not have a diverging zone to recover the velocityhead. It has a high coefficient of discharge (∼ 0.99), but a high pressure loss. It is cheaperthan a Venturi meter.

Table 4.1: Comparison of Orifice and Venturi meter.

Comparison Orifice Plate VenturimeterCost Cheaper More ExpensiveAdaptability (change area ratio) Versatile Fixed area ratioPrecision Less accurate More accuratePermanent pressure loss High LowCD (typical) 0.61− 0.65 0.96− 0.98

QuestionWater flows through a 280 mm diameter pipe at a flowrate of 195 m3 h−1. A Venturimeter is installed in the pipe, with a throat diameter of 140 mm, and connectedto a differential U-tube manometer using mercury as the manometer fluid. Thecoefficient of discharge of the Venturi meter is 0.97. Calculate (to the nearest mm)the difference in height of the mercury in the manometer.[50 mm]

4.3. PROBLEMS 51

4.3 Problems

1. Show that the alternative version of the Reynolds Number,

Re =4M

πDµ

is dimensionless.

2. The jugular vein in a giraffe’s neck is approximately 25 mm in diameter, and theflowrate approximately 0.6 kg s−1. Estimate the Reynolds number, assuming bloodviscosity of 4 mPa s. Comment on your answer.

3. Steam at a pressure of 2 bar gauge flows along a 1 inch internal diameter pipe at aflowrate of 18 kg h−1. Calculate the velocity of the steam in the pipe. The steamexits to atmospheric pressure through an 8 mm diameter nozzle. Use the continuityequation to calculate the exit velocity of the steam. (Use steam tables to determinethe density of steam at 2 bar gauge and at atmospheric pressure.

4. Depicted in the Figure below are a number of systems on which we need to do flowcalculations. Write down for each system the assumptions you would make to solvethe problems. Discuss in your group your answers.

5. A vertical pipe carrying water changes from a cross sectional area of 0.3 m2 at apoint ’A’ to 0.15 m2 at a point ’B’. Point ’B’ is 6 m above point ’A’. If the liquidvelocity at point ’A’ is uniform and has a value of 1.8 m s−1, and the pressure at thispoint is 1.17 × 105 N m−2 gauge, then determine the pressure at ’B’ ignoring anyfrictional effects.


6. Oil, with a density of 800 kg m−3, is pumped through a horizontal pipe that narrowsfrom a diameter of 20 cm to a diameter of 10 cm, before it discharges the oil intoa tank at atmospheric pressure. If the flow rate through the pipe is 0.1 m3 s−1,calculate the fluid velocity and pressure in the wide and narrow parts of the pipe.

7. A siphon is used to empty a cylindrical water tank with a diameter of 2 m and afluid height of 3 m. The arrangement, depicted in the Figure below, incorporates asiphon with a bore of 50 mm and the discharge occurs 1 m below the base of thetank. Calculate the time taken to empty the tank and the initial velocity of the waterexiting the siphon.

8. A venturimeter has inlet and throat diameter of 150 mm and 75 mm respectively,and is used to measure a flow of water through a horizontal pipe. If the actual flowrate is 0.05 m3 s−1 and Cd = 0.97, calculate:

(a) the pressure difference between inlet and throat,

(b) the difference in the liquid levels of a water/mercury manometer used to mea-sure the flow,

(c) the pressure difference and difference in liquid levels if the pipe is not hori-zontal but vertical. The distance between the points at which the pressure ismeasured is 150 mm.

Chapter 5Momentum Balance

Contents5.1 Derivation of the Momentum Equation . . . . . . . . . . . . . . . . 55

5.2 Applications of the Momentum Equation . . . . . . . . . . . . . . . 57

5.2.1 Flow through a contracting section of pipework or through anozzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.2.2 Flow around a contracting pipe bend . . . . . . . . . . . . . . . 60

5.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

53

5.1. DERIVATION OF THE MOMENTUM EQUATION 55

5.1 Derivation of the Momentum Equation

We started this course by noting that the study of fluids must necessarily take as its start-ing point Physics, and that some of the key fundamental ideas of Physics include theConservation Laws, of which the three most important ones in relation to Fluid Flow areConservation of Mass, Conservation of Energy and Conservation of Linear Momentum.We have used Conservation of Mass to derive the Continuity Equation and Conservationof Energy to derive Bernoulli’s Equation, and have noted that these equations have numer-ous practical applications, of which we’ve scratched the surface by analysing flow fromtanks and devices for measuring flow. For this chapter, we are going to use the final Con-servation Law, that of Conservation of Linear Momentum, to look at the forces exertedon and by fluids as they flow.

To start with, it seems timely to remind ourselves of Newton’s Three Laws of Motion:These laws depends on conservation of momentum, so we need to remind ourselves thatmomentum is the product of mass × velocity. Then, the laws are:

Newton’s Laws of MotionFirst Law If no net force acts on a particle, then it is possible to select a set of non-

accelerating reference frames, called inertial reference frames, observed fromwhich the particle moves without any change in velocity. This law is oftensimplified to “An object will stay at rest or continue at a constant velocityunless acted upon by an external unbalanced force”.

Second Law Observed from an inertial reference frame, the net force on a particleis equal to the rate of change with time of its linear momentum:

F =d (mv)

d t

This law is often stated as F = ma (the force on an object is equal to its massmultiplied by its acceleration). This is because, if mass is constant, then theequation becomes

F = mdv

d t= ma

Third Law Whenever a particle A exerts a force on another particle B, B simulta-neously exerts a force on A with the same magnitude in the opposite direc-tion. This arises because momentum is conserved – the change in momentumof one body is experienced as a force, an equal but opposite force then beingexperienced by the other body, which results in a change in its momentumequal but opposite to the change experienced by the first body, such that mo-mentum is conserved. This law is often simplified into the sentence “Everyaction has an equal and opposite reaction”.

Clearly if momentum is decreasing with time, then d (mv) /d t is negative and the forcecausing the change is negative. This is the force acting on the body causing its momentumto change. The equal and opposite force is exerted by the body.

In applying conservation of momentum to fluid flow, we will be struck by two difficulties,compared with applying conservation of mass and energy. Firstly, momentum as a con-served quantity is simply less familiar to us than mass or energy – we will have slightlygreater difficulty visualising what is going on. Secondly, and more importantly, mass and

56 CHAPTER 5. MOMENTUM BALANCE

energy are both scalar quantities that have magnitude only, while momentum is a vectorquantity that has both magnitude and direction. Momentum is conserved in a particulardirection, so we must be sensitive to the directions in which various forces are acting andcomponents moving, and we must (in more advanced applications) be able to perform themathematics in several directions simultaneously.

Returning to Newton’s Second Law, applying the product rule gives a form more appro-priate for the analysis of fluid flow:

F =d (mv)

d t= m

dv

d t+ v

dm

d t= ma + mv (5.1.1)

where m is the rate of change of mass (i.e. the rate at which mass enters or leaves acontrol volume). Mostly we are concerned with steady flow, i.e. no accelerations, so a iszero and F = mv. As noted above, the momentum force exerted by the fluid is of thesame magnitude, but in the opposite direction.

Consider a tank into which fluid is flowing at a flowrate m and velocity ux−in in the x-direction, as shown in Figure 5.1.

m, ux−in

x

zy

Figure 5.1: Fluid flowing into a tank at a flowrate m and velocity uin.

The tank is stationary, so the fluid has to lose its momentum. The rate at which it losesits momentum clearly depends on its mass and velocity, and on the time taken for a givenmass to enter the tank, i.e. the mass flowrate. This is equal to the force that it exerts onthe tank, which is clearly in the positive x-direction:

Fmx = mux−in

The tank exerts the opposite force on the fluid, in order to bring it to rest, and in so doinggains the equal amount of momentum (which, because the tank is bonded tightly to theearth, it transmits to the earth, which moves slightly in response! However, whatever isimparting the momentum to the fluid in the first place is also ultimately connected to theearth and pushing it in the other direction, so the momentum transfer to the earth cancelsout.)

Now, let’s imagine our tank has a hole on the other side, from which fluid is exiting, againwith a flowrate , but this time with a velocity ux−out. This forms a jet of water exiting fromthe side of the tank in the x-direction, which clearly pushes the tank in the other direction,i.e. creates a force on the tank in the negative x-direction. The total force exerted on thetank in the x-direction is then equal to the positive force of the water entering the tankand coming to rest, minus the force exerted on the tank in the negative x-direction due tothe jet of water spurting out:

Fmx = mux−in − mux−out = m (ux−in − ux−out)

5.2. APPLICATIONS OF THE MOMENTUM EQUATION 57

This is the momentum equation, which gives momentum force of the fluid flow on thetank. Clearly, whether the net force is positive or negative depends on the relative magni-tudes of ux−in and ux−out. (Note, this is the force exerted by the fluid on the system. Theforce exerted on the fluid, in order to change its momentum, is the opposite.)

The mass flowrate is given by m = ρAu; therefore the momentum equation can also bewritten as

Momentum EquationFmx = ρinAinuinux−in − ρoutAoutuoutux−out (5.1.2)

5.2 Applications of the Momentum Equation

5.2.1 Flow through a contracting section of pipework or through anozzle

Now, let’s replace our “tank” with a section of pipe of varying cross section, as shown inFigure 5.2. A control volume is defined by the dotted line, over which we can perform aforce balance. Fluid flows into the control volume at point 1 with velocity u1, and flowsout at point 2 with velocity u2. What will be the magnitude and direction of the overallforce on the section of pipe?

x

Frx

A1

P1

u1

A2

P2

u2

Figure 5.2: Contracting pipe, through which fluid is flowing with mass flowrate m.

The control volume is not accelerating (in fact it is stationary), so a force balance saysthat the sum of the forces on the system must be equal to zero.∑

F = 0

Considering just forces in the x-direction, these include the net momentum force exertedby the fluid flowing into and out of the control volume, as well as forces resulting fromgravity, pressure, shear (drag) and the force holding the pipework in place, which we callthe reaction force:

Fmx + Fgx + Fpx + Fsx + Frx = 0 (5.2.1)


As the pipe is horizontal, there is no gravity force in the x-direction. Also, for frictionlessflow, we assume there is no shear (drag) force. So

Fmx + Fpx + Frx = 0

Therefore, if we know the forces due to the pressure and momentum, we can calculate thereaction force, Frx, i.e. the force needed to hold the pipework in place.

As noted already, the momentum force due to the fluid flowing into and out of the pipe atdifferent velocities is given by

Fmx = mu1 − mu2 = ρA1u21 − ρA2u

22

(Note, as m is constant, u2 is clearly greater than u1 the net momentum force is in thenegative x-direction. The water pushes back on the system.)

From continuity:

u2 = u1A1

A2

so

Fmx = ρA1u21 − ρA2u

21

(A1

A2

)2

= ρA1u21 − ρA1u

21

A2

A1

(A1

A2

)2

= ρA1u21

(1− A1

A2

)

(Note that this is still negative, as A1 > A2.)

Consider now the overall force on the control volume due to pressure (using gauge pres-sures to simplify things). P1 operates in the x-direction, P2 in the negative x-direction,such that:

Fpx = P1A1 − P2A2 (5.2.2)

Let’s assume, for the sake of example, that the jet is exiting at point 2 to atmosphericpressure, such that p2 is zero. (This would describe a nozzle on the end of a hose, forexample.) Then the pressure force is

Fpx = P1A1

5.2. APPLICATIONS OF THE MOMENTUM EQUATION 59

From Bernoulli’s Equation

P1 +ρu212

= ��0

P2 +ρu222

u22 = u21

(A1

A2

)2

P1 +ρu212

=ρ

2u21

(A1

A2

)2

P1 =ρ

2u21

((A1

A2

)2

− 1

)

P1A1 =ρ

2A1u

21

((A1

A2

)2

− 1

)

Fpx =ρ

2A1u

21

((A1

A2

)2

− 1

)

=ρ

2A1u

21

(A1

A2

− 1

)(A1

A2

+ 1

)

(Note that this is positive, as A1 > A2. The pressure force pushes on the nozzle in thepositive x-direction, trying to push if off the end of the hose.)

So, the reaction force in the x-direction is given by,

Frx = − Fmx − Fpx

= − ρA1u21

(1− A1

A2

)− ρ

2A1u

21

(A1

A2

− 1

)(A1

A2

+ 1

)= ρA1u

21

(A1

A2

− 1

)− ρ

2A1u

21

(A1

A2

− 1

)(A1

A2

+ 1

)(5.2.3)

The reaction force is negative overall. This is because the pressure force is larger thanthe momentum force. By expressing both forces in terms of A1 and u1, it is possible tocompare the two forces directly and hence to see this. From the above equation, in whichthe first term is the momentum force and the second is the pressure force, clearly the latteris larger than the former because (A1/A2 + 1) > 2, if A1 > A2.

By collecting the like terms together, we can simplify the equation further:

Frx = ρA1u21

(A1

A2

− 1

)(2−

(A1

A2

+ 1

))= ρA1u

21

(A1

A2

− 1

)(1− A1

A2

)= − ρA1u

21

(A1

A2

− 1

)2

= − ρA1

(u22 − u21

)Again, it is evident from this final equation that the reaction force is negative with respectto the x-direction. Both the term outside the bracket and that inside are positive (becausethey both contain squared terms, if for no other reason), so the overall product is negativebecause of the negative sign at the front.


(Note that the analysis applies equally if A2 > A1. And the reaction force would still benegative! This seems strange! It occurs because the pressure and momentum forces bothreverse sign but also swap relative magnitudes. Equivalently, if the direction of flow isreversed, the reaction force is still in the negative x-direction, for the same reason.)

This derivation has brought together the Continuity Equation (based on conservation ofmass), Bernoulli’s Equation (based on conservation of energy) and the Momentum Equa-tion (based on conservation of momentum) in order to perform a sophisticated analysis ofthe fluid flow.

5.2.2 Flow around a contracting pipe bend

Figure 5.3 shows a 45◦ pipe bend with a reducing diameter from 600 mm at the inletto 300 mm at the outlet. The gauge pressure at the inlet is 140 kPa and the volumetricflowrate of water is 0.425 m3 s−1. As the fluid flows around the bend, it will exert a forcewhich will need to be counterbalanced in order to hold the pipework in place. Ignoringfriction forces, and assuming the pipe bends in the horizontal plane, develop an expressionfor the magnitude and direction of the force exerted by the fluid on the pipe.

P1, A1, u1

P2, A2, u2

45◦

x

y

Figure 5.3: Flow through a reducing pipe bend.

As before, excluding gravity and shear forces, and performing the force balance firstly inthe x-direction:

Fpx + Fmx + Frx = 0

Fmx = ρA1u1ux1 − ρA2u2ux2

At point 1, the velocity of flow is entirely in the x-direction and is equal to

u1 =Q

A1

=0.425(π × 0.62

4

) = 1.503 m s−1

At point 2,

u2 =Q

A2

=0.425(π × 0.32

4

) = 6.01 m s−1

5.3. SUMMARY 61

The velocity in the x-direction at point 2 is equal to u2×cos θ = 6.01 cos 45 = 4.25 m s−1.Therefore the momentum force on the pipe bend in the x-direction is,

Fmx = ρA1u21 − ρA2u

22 cos θ

= ρQ (u1 − u2 cos θ)

= 1000× 0.425 (1.503− 4.25)

= − 1167 N

Turning now to the pressure force in the x-direction, From Bernoulli’s Equation

P2 = P1 +ρ

2

(u21 − u22

)= 140 000 +

1000

2

(1.5032 − 6.012

)= 123 069 Pa

Fpx = P1A1 − P2A2 cos θ

= 140 000× 0.2827− 123 069× 0.0707 cos 45

= 39 584− 6153

= 33 431 N

The total force exerted by the water on the bend in the x-direction is therefore 33 431 −1167 = 32 264 N. The reaction force, required to hold the pipe bend in place so that itdoesn’t move, is 32 264 N in the negative x-direction.

Turning to the y-direction, the momentum force is

Fmy = ρA1u1uy1 − ρA2u2uy2

= 0− ρQu2 sin θ

= 0− 1000× 0.425× 6.01 sin 45

= − 1806 N

The pressure force is

Fpy = P1Ay1− P2A2 sin θ

= 0− 123 069× 0.0707 sin 45

= − 6153 N

So the total force exerted by the water in the y-direction is−1806−6153 = −7958 N.

The total net force is found by adding the two force vectors, as shown below. The magni-tude is given by

√32 2642 + 79582 = 33 231 N. The direction is found from tan−1 (−7958

/32 264) = −13.86◦, i.e. at an angle 13.86◦ anticlockwise from the x-direction. Draw thison Figure 5.3. This is the direction of the force exerted by the water as it travels aroundthe pipe bend. The reaction force required to hold the pipe bend stationary is equal inmagnitude to this and opposite in direction.

5.3 Summary

Applying conservation of mass, energy and momentum has allowed us to derive equationsto describe fluid flow in a wide variety of situations, of which we have given examples of


a very few – there are many more! In order to introduce and illustrate the principles, wehave largely omitted the effects of viscosity. Viscosity causes friction within the flowingfluid and between the fluid and the walls of pipes and vessels. This causes some energy tobe dissipated as heat, which must be accounted for in the extended version of Bernoulli’sEquation. It also causes stresses to be transmitted from the walls of pipes and vesselsthrough the fluid, giving rise to velocity profiles. Velocity profiles, i.e. differences invelocity with spatial position, represent differences in momentum. Momentum transferwithin fluids therefore arises as a result of fluid viscosity.

5.4. PROBLEMS 63

5.4 Problems

1. A water sprinkler with four arms is used to water the garden. If the total flow ofwater is 100 ml s−1, and the ends of the sprinkler have holes in them with a diameterof 1 mm, calculate the momentum force on the end of each sprinkler arm. What isthe velocity of each jet?

2. I am using a garden hose to wash my car. The hose has an inner diameter of 1 cm,and the volumetric flow rate is 500 ml s−1.

(a) Calculate the momentum force exerted on the end of the hose pipe by theliquid.

(b) I have put a nozzle on the hose pipe which reduces the diameter at the end ofthe hose pipe to half its former size. Assuming the volumetric flow rate staysthe same, calculate:

i. The force exerted on the nozzle due to the reduction of pipe diameter,

ii. The new momentum force on the hose pipe end.

3. Water flows steadily through the 90◦ reducing elbow shown in the Figure below. Atplane 2, the water discharges to atmospheric pressure with a velocity of 16 m s−1,the cross sectional plane being 0.0025 m2. The area of plane 1 is 0.01 m2 and thebend is in the horizontal plane. Show that, if there are no losses in the elbow,the pressure at plane 1 will be 120 kN m−2 gauge. Determine the magnitude anddirection of the force required to hold the elbow in place. Neglect static pressurechanges with height.

Chapter 6Friction Losses - Ideal Fluids

Contents6.1 Force Balance on Fluid Element . . . . . . . . . . . . . . . . . . . . 67

6.2 Other Pipe Fittings . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

6.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

65

6.1. FORCE BALANCE ON FLUID ELEMENT 67

6.1 Force Balance on Fluid Element

Fg

τw

1 2

P1 P2

u1 u2x

y

Non-idealfluid

Figure 6.1: Force balance over a control volume.

Remembering the force balance on the control volume (Equation 5.2.1), we have the netmomentum force exerted by the fluid flowing into and out of the control volume, as wellas forces resulting from gravity, pressure, shear (drag) and the reaction force:

Fmx + Fgx + Fpx + Fsx + Frx = 0 (6.1.1)

The momentum force, Fmx, is given by (Equation 5.1.2),

Fmx = ρA1u1ux−1 − ρA2u2ux−2 (6.1.2)

However from the continuity equation we know that ρA1u1 = ρA2u2, and as A1 = A2,then u1 = u2, which means that,

Fmx = 0

The pressure force, Fpx, is given by (Equation 5.2.2),

Fpx = P1A1 − P2A2 (6.1.3)

The force due to gravity, Fgx, is 0 as the pipe is horizontal, and we know that the reaction,Frx, is 0 as it is a horizontal section of pipe with no area change. This means that,

��*0

Fmx +��>

0Fgx +�

��>[(P1 − P2)A]

Fpx + Fsx +��*0

Frx = 0

Fsx = − (P1 − P2)A (6.1.4)

The shear force is defined as the wall shear stress τw times the area of the wall, so for ourfluid element,

Fsx = −τwπdL

Therefore,

Fsx = − (P1 − P2)A

−τwπdL = − (P1 − P2)A

= − (P1 − P2)π

4d2

P1 − P2 = τw4L

d

68 CHAPTER 6. FRICTION LOSSES - IDEAL FLUIDS

From the extended form of Bernoulli’s equation (Equation 3.4.6), with no energy in-put,

P1

ρ1g+��0

u212g

+��0

h1 =P2

ρ2g+��0

u222g

+��0

h2 + ∆hf

Which means that,P1 − P2

ρg= ∆hf

Substituting in for P1 − P2 from above gives,

∆hf =τwρg

4L

d

As the friction loss is related to the kinetic energy of the fluid, i.e. the fluid velocity, it istypical to include the kinetic energy term (or velocity head),

∆hf =τwρg

4L

d

u2

2g

2g

u2

=τw

0.5ρu24L

d

u2

2g

The τw/0.5ρu2 term is often referred to as the friction factor, f , so,

Friction Loss

∆hf = f4L

d

u2

2g(6.1.5)

known as the Darcy equation, with the fanning friction factor, f .

The fanning friction factor has a complex relationship which depends on the Reynold’snumber and the dimensionless surface roughness, Figure 6.2. The dimensionless surfaceroughness is given by the surface roughness, ε, divided by the pipe diameter,

Dimensionless surface roughness =ε

d(6.1.6)

Figure 6.2: Diagram of the surface roughness, ε.

6.1. FORCE BALANCE ON FLUID ELEMENT 69

This complex relationship has been determined from experiments and is shown by theMoody chart, Figure 6.3.

Some key facts about the fanning friction factor, f .

• In the laminar zone – f decreases as Re increases

• transition zone – uncertainty – not possible to predict

• Beyond 4000, for a given Re, as the relative roughness term ε/d decreases (lessrough), friction factor decreases

• For given relative roughness, friction factor decreases with increasing Reynoldsnumber till the zone of complete turbulence

• Within the zone of complete turbulence – Reynolds number has no affect

• As relative roughness decreases – the boundary of the zone of complete turbulenceincreases

For convenience people have correlated the fanning friction factor to make calculationeasier. For laminar flow, this can actually be solved analytically from first principles togive,

f =16

Re(6.1.7)

For turbulent flow, the Haaland equation,

1√f

= −3.6 log

ε

d3.7

1.11

+6.9

Re

(6.1.8)

Table 6.1 shows some typical values of the surface roughness.

Table 6.1: Some typical roughness values.

Material ε / cmDrawn tubing (brass, lead, glass) 0.00015Commercial steel (wrought iron) 0.0045Asphalted Cast iron 0.012Galvanized iron 0.015Cast iron 0.025Wood stave 0.018-0.09Concrete 0.03-0.30Riveted Steel 0.09-0.90

Friction FactorIn this course the friction factor used is the fanning friction factor, be careful not toconfuse this with the Darcy-Weisbach friction factor which is four times larger

fDarcy-Weisbach = 4ffanning

The Darcy-Weisbach friction factor is often given as λ but this is not always true.

70 CHAPTER 6. FRICTION LOSSES - IDEAL FLUIDS

Figure6.3:M

oodychartforthe

fanningfriction

factorf.

6.2. OTHER PIPE FITTINGS 71

6.2 Other Pipe Fittings

Actual pipework is not straight and has lots of appendages (e.g. valves), and it changes di-ameter, therefore for these Equation 6.1.5 is simplified in terms of the number of velocityheads, K,

∆hf = Ku2

2g

which can be added together for all the components,

Friction Losses

∆hf =∑i

Kiu2i2g

Table 6.2 shows some typical values of the number of velocity heads for common pipefittings.

Table 6.2: K values.

Pipe fitting K45◦ elbow 0.3590◦ elbow 0.7090◦ square elbow 0.90Tee (i) entry through branch 1.50

(ii) entry through side 0.40Couplings 0.04

6.3. PROBLEMS 73

6.3 Problems

1. Calculate the head loss per meter due to friction for water flowing at 0.2 m3 s−1

through a 200 mm cast iron pipe.

ρwater = 1000 kg m−3, µwater = 1× 10−3 Pa s

2. Oil (ρ = 950 kg m−3, µ = 0.25 Pa s) flows through a 20 cm diameter horizontalpipe made of drawn tubing with 2 90◦ elbows at a velocity of 4 m s−1. Calculate thepressure drop over the pipe if it is 100 m long.

3. Water enters a 0.3 m diameter vertical pipe at atmospheric pressure where it de-scends 8 m takes a 90◦ turn and travels along 50 m of horizontal pipe. Calculate thewater exit velocity and flow rate if the pipe has a dimensionless surface roughnessof 0.0002 and 20 couplings.

ρwater = 1000 kg m−3, µwater = 1× 10−3 Pa s

4. Water at 10 ◦C (ρ = 999.7 kg m−3, µ = 1.308 mPa s) flows from a large reservoirto a small one through a 5 cm diameter cast iron piping system shown in the Figurebelow. Determine the elevation z1 for a flow rate of 6 l s−1.

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