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Michael V. Lurie

Modeling of Oil Product andGas Pipeline Transportation

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Michael V. Lurie

Modeling of Oil Product andGas Pipeline Transportation

The Author

Prof. Dr. Michael V. LurieRussian State Universityof Oil and GasMoscow, Russian Federation

Translation

Emmanuil G. SinaiskiLeipzig, Germany

Cover Picture

Trans-Alaska Pipeline

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2008 WILEY-VCH Verlag GmbH & Co.KGaA, Weinheim

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ISBN: 978-3-527-40833-7

V

In memory of the Teacher – academician Leonid I. Sedov

Modeling of Oil Product and Gas Pipeline Transportation. Michael V. LurieCopyright 2008 WILEY-VCH Verlag GmbH & Co. KGaA, WeinheimISBN: 978-3-527-40833-7

VII

Foreword

This book is dedicated first and foremost to holders of a master’s degree andpostgraduate students of oil and gas institutes who have decided to specializein the field of theoretical problems in the transportation of oil, oil products andgas. It contains methods of mathematical modeling of the processes takingplace in pipelines when transporting these media.

By the term mathematical model is understood a system of mathematicalequations in which framework a class of some processes could be studied. Thesolution of these equations provides values of parameters without carrying outmodel and, especially, full scale experiments.

Physical laws determining the dynamics of fluids and gases in pipes arepresented. It is then shown how these laws are transformed into mathematicalequations that are at the heart of one or another mathematical model. Inthe framework of each model, are formulated problems with the aim ofinvestigating concrete situations. In doing so there are given methods of itssolution.

The book is self-sufficient for studying the subject but the text is outlined insuch a way that it impels the reader to address oneself to closer acquaintanceof considered problem containing in special technical literature.

Professor Michael V. LurieMoscow


IX

Contents

Dedication Page V

Foreword VII

Preface XIII

List of Symbols XV

1 Fundamentals of Mathematical Modeling of One-DimensionalFlows of Fluid and Gas in Pipelines 1

1.1 Mathematical Models and Mathematical Modeling 11.1.1 Governing Factors 31.1.2 Schematization of One-Dimensional Flows of Fluids and Gases

in Pipelines 41.2 Integral Characteristics of Fluid Volume 51.3 The Law of Conservation of Transported Medium Mass.

The Continuity Equation 71.4 The Law of Change in Momentum. The Equation of Fluid Motion 91.5 The Equation of Mechanical Energy Balance 111.5.1 Bernoulli Equation 151.5.2 Input of External Energy 161.6 Equation of Change in Internal Motion Kinetic Energy 171.6.1 Hydraulic Losses (of Mechanical Energy) 181.6.2 Formulas for Calculation of the Factor λ(Re, ε) 201.7 Total Energy Balance Equation 221.8 Complete System of Equations for Mathematical Modeling

of One-Dimensional Flows in Pipelines 29

2 Models of Transported Media 312.1 Model of a Fluid 312.2 Models of Ideal and Viscous Fluids 322.3 Model of an Incompressible Fluid 342.4 Model of Elastic (Slightly Compressible) Fluid 34


X Contents

2.5 Model of a Fluid with Heat Expansion 342.6 Models of Non-Newtonian Fluids 362.7 Models of a Gaseous Continuum 382.7.1 Model of a Perfect Gas 392.7.2 Model of a Real Gas 392.8 Model of an Elastic Deformable Pipeline 42

3 Structure of Laminar and Turbulent Flows in a Circular Pipe 453.1 Laminar Flow of a Viscous Fluid in a Circular Pipe 453.2 Laminar Flow of a Non-Newtonian Power Fluid in a Circular Pipe 473.3 Laminar Flow of a Viscous-Plastic Fluid in a Circular Pipe 493.4 Transition of Laminar Flow of a Viscous Fluid to Turbulent Flow 513.5 Turbulent Fluid Flow in a Circular Pipe 523.6 A Method to Control Hydraulic Resistance by Injection of

Anti-Turbulent Additive into the Flow 623.7 Gravity Fluid Flow in a Pipe 65

4 Modeling and Calculation of Stationary Operating Regimes of Oiland Gas Pipelines 73

4.1 A System of Basic Equations for Stationary Flow of anIncompressible Fluid in a Pipeline 73

4.2 Boundary Conditions. Modeling of the Operation of Pumps andOil-Pumping Stations 75

4.2.1 Pumps 754.2.2 Oil-Pumping Station 784.3 Combined Operation of Linear Pipeline Section and Pumping

Station 814.4 Calculations on the Operation of a Pipeline with Intermediate

Oil-Pumping Stations 844.5 Calculations on Pipeline Stationary Operating Regimes in

Fluid Pumping with Heating 874.6 Modeling of Stationary Operating Regimes of Gas-Pipeline Sections 924.6.1 Distribution of Pressure in Stationary Gas Flow in a Gas-Pipeline 944.6.2 Pressure Distribution in a Gas-Pipeline with Great Difference

in Elevations 964.6.3 Calculation of Stationary Operating Regimes

of a Gas-Pipeline (General Case) 974.6.4 Investigation of Thermal Regimes of a Gas-Pipeline Section 984.7 Modeling of Blower Operation 100

5 Closed Mathematical Models of One-Dimensional Non-StationaryFlows of Fluid and Gas in a Pipeline 109

5.1 A Model of Non-Stationary Isothermal Flow of a Slightly CompressibleFluid in a Pipeline 109

5.2 A Model of Non-Stationary Gas Flow in a Pipeline 112

Contents XI

5.3 Non-Stationary Flow of a Slightly Compressible Fluidin a Pipeline 113

5.3.1 Wave Equation 1135.3.2 Propagation of Waves in an Infinite Pipeline 1155.3.3 Propagation of Waves in a Semi-Infinite Pipeline 1175.3.4 Propagation of Waves in a Bounded Pipeline Section 1195.3.5 Method of Characteristics 1215.3.6 Initial, Boundary and Conjugation Conditions 1245.3.7 Hydraulic Shock in Pipes 1275.3.8 Accounting for Virtual Mass 1345.3.9 Hydraulic Shock in an Industrial Pipeline Caused by Instantaneous

Closing of the Gate Valve 1355.4 Non-Isothermal Gas Flow in Gas-Pipelines 1385.5 Gas Outflow from a Pipeline in the Case of a Complete Break

of the Pipeline 1465.6 Mathematical Model of Non-Stationary Gravity Fluid Flow 1495.7 Non-Stationary Fluid Flow with Flow Discontinuities

in a Pipeline 152

6 Dimensional Theory 1576.1 Dimensional and Dimensionless Quantities 1576.2 Primary (Basic) and Secondary (Derived) Measurement Units 1586.3 Dimensionality of Quantities. Dimensional Formula 1596.4 Proof of Dimensional Formula 1616.5 Central Theorem of Dimensional Theory 1636.6 Dimensionally-Dependent and Dimensionally-Independent

Quantities 1646.7 Buckingham �-Theorem 168

7 Physical Modeling of Phenomena 1737.1 Similarity of Phenomena and the Principle of Modeling 1737.2 Similarity Criteria 1747.3 Modeling of Viscous Fluid Flow in a Pipe 1757.4 Modeling Gravity Fluid Flow 1767.5 Modeling the Fluid Outflow from a Tank 1787.6 Similarity Criteria for the Operation of Centrifugal Pumps 179

8 Dimensionality and Similarity in Mathematical Modelingof Processes 183

8.1 Origination of Similarity Criteria in the Equations of aMathematical Model 183

8.2 One-Dimensional Non-Stationary Flow of a Slightly CompressibleFluid in a Pipeline 184

8.3 Gravity Fluid Flow in a Pipeline 1868.4 Pipeline Transportation of Oil Products. Batching 187

XII Contents

8.4.1 Principle of Oil Product Batching by Direct Contact 1888.4.2 Modeling of Mixture Formation in Oil Product Batching 1898.4.3 Equation of Longitudinal Mixing 1928.4.4 Self-Similar Solutions 194

References 199

Appendices 201

Author Index 205

Subject Index 207

XIII

Preface

This book presents the fundamentals of the mathematical simulation ofprocesses of pipeline transportation of oil, oil products and gas. It is shownhow the basic laws of mechanics and thermodynamics governing the flow offluids and gases in pipelines are transformed into mathematical equationswhich are the essence of a certain mathematical model and, in the frameworkof a given physical problem, appropriate mathematical problems are formulatedto analyze concrete situations.

The book is suitable for graduate and postgraduate students of universitieshaving departments concerned with oil and gas and to engineers and researchworkers specializing in pipeline transportation.

Beginners will find in this book a consecutive description of the theory andmathematical simulation methods of stationary and non-stationary processesoccurring in pipelines. Engineers engaged in the design of and calculations onpipelines will find a detailed theoretical and practical text-book on the subjectof their work. Graduate and postgraduate students and research workers willbecome acquainted with situations in the theory and methods in order togeneralize and develop them in the future.

The author of the book, Professor Dr. M. Lurie, is a great authority in Russiain the field of the hydromechanics of oil and gas pipeline transportation.

Prof. Emmanuil SinaiskiLeipzig

‘‘. . . No human investigation could be referred to as true when it is not supportedby mathematical proof’’

Leonardo da Vinci


XV

List of Symbols

Symbol Definitiona radius of the flow corea dimensionless constanta parameter of the (Q − �H) characteristicA proportionality factorA+ value of parameter A to the left of the discontinuity frontA− value of parameter A to the right of the discontinuity front[A] = A+ − A− jump of parameter A at the discontinuity frontdAin elementary work of internal forcedAex elementary work of external forceb parameter of the (Q − �H) characteristicc velocity of wave propagation in a pipelinec sound velocity in gasC2 integration constantCf friction factorCp heat capacity at constant pressureCSh Chezy factorCv heat capacity at constant volumecP centipoise, 0.01 PcSt centistokes, 0.01 St = 10−6 m2 s−1

d pipeline internal diameter�d diameter incrementd0 nominal internal diameter of pipeline; cylinder internal

diameterD pipeline external diameterD velocity of hydraulic shock wave propagation in a pipelineD velocity of discontinuity front propagation in the positive

direction of the x-axisDim diameter of impellerDp diameter of pump impellerD∗ Joule–Thompson factorein internal energy density; specific internal energyekin kinetic energy density; specific kinetic energy


XVI List of Symbols

E elastic modulus in extension and compression,Young’s modulus

Ein internal energyEkin kinetic energyEi(z) Euler functionf1 dimensionless factorfτ(Q)− friction forceF restoring forcedFn elementary forceFr Froude numberg acceleration due to gravityg0, g1 dimensionless constantsh piezometric headh(S) depth of pipeline cross-section filling with fluidhc head losses in station communicationshcr critical depthhπ. head before PLPhn normal depth of gravity flow in the pipeH head�H differential head�H = F(Q) head-discharge (Q − H) characteristic of a pumpH1 hydraulic headH1 hydraulic headHe Hedstroem numberi hydraulic gradienti0 hydraulic gradientI momentumI Ilyushin numberI1, I2 Riemann invariantsJ gas enthalpyk factor of string elasticityk factor of power, Ostwald fluidk parameter of non-Newtonian fluidκ factor; heat-transfer factor; empirical factor 1/Kκ Karman constantk dimensionless constantk kinematic consistencyK heat transfer factorK elastic modulus of fluid, PaK factor of longitudinal mixing of oil productlc length of the mixture regionL length of a pipeline or a pipeline sectionM mass flow rateM0 initial mass flow rateOPS oil-pumping station

List of Symbols XVII

n factor of power, Ostwald fluidn exponent in Ostwald rheological lawn exponentn number of revolutions of centrifugal blower shaftn unit normal vectorn0 nominal number of revolutions of blower shaftnin specific power of internal friction forcesN power consumption, kWNmech power of external mechanical devicesNus useful power of mechanical force acting on gasN/ρe specific powerp pressure�p difference between internal and external pressures, pressure

dropp0 nominal pressure, initial pressure, normal pressure, pressure

at the beginning of the pipeline sectionpen pressure of gas at the entrance of compressor station

and blowerpcr critical pressurepex external pressure; pressure at initial cross-section

of the pipeline sectionpin internal pressurepL pressure at the end of the pipeline sectionpl, pe, pπ pressure at the pressure line of pumps (PLP)pr reduced pressurepst standard pressure, pst = 101 325 Papu pressure before oil-pumping stationpu head before stationpv saturated vapor tension (pressure)[p] pressure jump[pinc] incident pressure wave amplitude[prefl] reflected pressure wave amplitude[ptrans] transmitted pressure wave amplitudeP poise, 0.1 kg m s−1

Ps wetted perimeterPa pascal (SI unit), kg m−1 s−2

Pe Peclet numberqh specific heat fluxqex heat inflow (qex > 0) to gas; heat outflow (qex < 0) from gasqM specific mass flow rateqn external heat fluxQ volume flow rate; fluid flow rateQe flow rate of gas at the entrance to the compressor stationQk commercial flow rate of gasQM mass flow rate

XVIII List of Symbols

Qv volume flow rate of gas at pipeline cross-sectionr radial coordinater0 pipeline radiusR gas constant (R = R0/µg)R0 universal gas constantRh hydraulic radiusRim radius of the impellerRr reduced gas constantRe Reynolds numberRecr critical Reynolds numberRe∗ generalized Reynolds numberS area of a cross-section; area of pipeline cross-section part filled

with fluidS0 area of pipeline cross-section; nominal (basic) areaSt stokes, 10−4 m2 s−1

t timeT absolute temperatureT0 nominal temperature; initial temperature; temperature of fluid

at normal conditionTav average temperature over pipeline section lengthTcr critical temperatureTB temperature of gas at the entrance to the compressor stationTex temperature of external mediumTL temperature at the end of pipeline sectionTm mean temperatureTr reduced temperatureTst standard absolute temperatureu(y) velocity distribution over cross-sectionumax maximum value of velocityuw fluid velocity at pipe wallu∗ dynamic velocityv velocity averaged over cross-sectionv mean flow rate velocityvcr critical velocityV volume[v] fluid velocity jumpw accelerationx coordinate along the pipeline axisx1 coordinate of gravity flow section beginningx2 coordinate of gravity flow section endy coordinate transverse to the pipeline axis; direction of a normal

to the elementary surface dσ

z(x) elevation level of a pipeline cross-section x(z1 − z2) geometrical height differences of sections 1 and 2Z over-compressibility factor

List of Symbols XIX

Zav average over-compressibility factorZ = Zr reduced gas over-compressibility

α angle of inclination of the pipeline axis to the horizontalακ, ακ factorsαv volume expansion factorαT thermal expansion factorβ compressibility factorγ adiabatic indexγ ratio between the hydraulic gradient of pipeline section

completely filled with fluid and the absolute value of thegravity flow section with slope αp to the horizontal

γ shear rate, s−1

δ pipeline wall thickness� absolute equivalent roughness; roughness of wall surfaceε relative roughness; compression ratio; thickness ratioς(t) local resistance factorη dimensionless radiusη(%) efficiencyθ function of temperature; concentration; parameter of over-

compressibility factor; parameter of state equation of real gas;concentration of anti-turbulent additive

λ hydraulic resistance factorλeff effective factor of hydraulic resistanceµ dynamic viscosity factor kg m−1 s−1

µg molar mass of gas; molecular weightµt turbulent dynamic viscosityµ apparent viscosity of power Ostwald fluidδ kinematic viscosity factor m2 s−1

ν0 kinematic viscosity factor at temperature T0

ν1 kinematic viscosity factor at temperature T1

νP Poisson ratioνt turbulent kinematic viscosityξ factor of volumetric expansion, K−1; self-similar coordinate;

dimensionless coordinate� dimensionless parameter; similarity criterion�(x) initial pressure distributionρ densityρ−, v−, p−, S− values of parameters before hydraulic shock waveρ+, v+, p+, S+ values of parameters after hydraulic shock waveρ0 nominal density; fluid density at p0; density of fluid under

normal conditionsρst gas density under standard conditionsσ area of suction branch pipe cross-section; hoop stress; degree

of pipe filling; circumferential stress

XX List of Symbols

dσ elementary surface area; surface elementτ tangential (shear) stress

tangential friction stressτ0 critical (limit) shear stressτw tangential (shear) stress at the pipeline internal surfaceυ specific volumeυcr critical specific volumeϕ angle of inclination of a straight line to the abscissa; central

angle�(x) initial fluid velocity distributionω frequency of rotor rotation; angular velocity of impeller

rotation

1

1Fundamentals of Mathematical Modeling ofOne-Dimensional Flows of Fluid and Gas in Pipelines

1.1Mathematical Models and Mathematical Modeling

Examination of phenomena is carried out with the help of models. Each modelrepresents a definite schematization of the phenomenon taking into accountnot all the characteristic factors but some of them governing the phenomenaand characterizing it from some area of interest to the researcher.

For example, to examine the motion of a body the material point model isoften used. In such a model the dimensions of the body are assumed to beequal to zero and the whole mass to be concentrated at a point. In other wordswe ignore a lot of factors associated with body size and shape, the materialfrom which the body is made and so on. The question is: to what extent wouldsuch a schematization be efficient in examining the phenomenon? As we allknow such a body does not exist in nature. Nevertheless, when examining themotion of planets around the sun or satellites around the earth, and in manyother cases, the material point model gives brilliant results in the calculationof the trajectories of a body under consideration.

In the examination of oscillations of a small load on an elastic spring wemeet with greater schematization of the phenomenon. First the load is takenas a point mass m, that is we use the material point model, ignoring bodysize and shape and the physical and chemical properties of the body material.Secondly, the elastic string is also schematized by replacing it by the so-calledrestoring force F = −k · x, where x(t) is the deviation of the material pointmodeling the load under consideration from the equilibrium position and k isthe factor characterizing the elasticity of the string. Here we do not take intoaccount the physical-chemical properties of the string, its construction andmaterial properties and so on. Further schematization could be done by takinginto account the drag arising from the air flow around the moving load andthe rubbing of the load during its motion along the guide.

The use of the differential equation

md2x

dt2= −k · x, (1.1)


2 1 Mathematical Modeling of One-Dimensional Flows of Fluid and Gas in Pipelines

expressing the second Newtonian law is also a schematization of thephenomenon, since the motion is described in the framework of Euclidiangeometry which is the model of our space without taking into account therelativistic effects of the relativity theory.

The fact that the load motion can begin from an arbitrary position with anarbitrary initial velocity may be taken into account in the schematization byspecifying initial conditions at

t = 0 : x = x0; v =(

dx

dt

)0

= v0. (1.2)

Equation (1.1) represents the closed mathematical model of the consideredphenomenon and when the initial conditions are included (1.2) this is theconcrete mathematical model in the framework of this model. In the given casewe have the so-called initial value (Cauchy) problem allowing an exact solution.This solution permits us to predict the load motion at instants of time t > 0and by so doing to discover regularities of its motion that were not previouslyevident. The latest circumstance contains the whole meaning and purpose ofmathematical models.

It is also possible of course to produce another more general schematizationof the same phenomenon which takes into account a great number ofcharacteristic factors inherent to this phenomenon, that is, it is possible, inprinciple, to have another more general model of the considered phenomenon.

This raises the question, how can one tell about the correctness orincorrectness of the phenomenon schematization when, from the logicalpoint of view, both schematizations (models) are consistent? The answer is:only from results obtained in the framework of these models. For example,the above-outlined model of load oscillation around an equilibrium positionallows one to calculate the motion of the load as

x(t) = x0 · cos

(√k

m· t

)+

√m

k·v0 · sin

(√k

m· t

)having undamped periodic oscillations. How can one evaluate the obtainedresult? On the one hand there exists a time interval in the course of whichthe derived result accords well with the experimental data. Hence the modelis undoubtedly correct and efficient. On the other hand the same experimentshows that oscillations of the load are gradually damping in time and cometo a stop. This means that the model (1.1) and the problem (1.2) do not takeinto account some factors which could be of interest for us, and the acceptedschematization is inadequate.

Including in the number of forces acting on the load additional forces,namely the forces of dry −f0 · sign(x) and viscous −f1 · x friction (where thesymbol sign(x) denotes the function x− sign equal to 1, at x > 0; equal to −1,at x < 0 and equal to 0, at x = 0), that is using the equation

md2x

dt2= −k · x − f0 · sign(x) − f1 · x (1.3)

1.1 Mathematical Models and Mathematical Modeling 3

instead of Eq. (1.1), one makes the schematization (model) more complete.Therefore it adequately describes the phenomenon.

But even the new model describes only approximately the model underconsideration. In the case when the size and shape of the load strongly affectits motion, the motion itself is not one-dimensional, the forces acting on thebody have a more complex nature and so on. Thus it is necessary to use morecomplex schematizations or in another words to exploit more complex models.Correct schematization frequently represents a challenging task, requiringfrom the researcher great experience, intuition and deep insight into thephenomenon to be studied (Sedov, 1965).

Of special note is the continuum model, which occupies a highly importantplace in the following chapters. It is known that all media, includingliquids and gases, comprise a great collection of different atoms andmolecules in permanent heat motion and with complex interactions.By molecular interactions we mean such properties of real media ascompressibility, viscosity, heat conductivity, elasticity and others. Thecomplexity of these processes is very high and the governing forcesare not always known. Therefore such seemingly natural investigationof medium motion through a study of discrete molecules is absolutelyunacceptable.

One of the general schematization methods for fluid, gas and otherdeformable media motion is based on the continuum model. Because eachmacroscopic volume of the medium under consideration contains a greatnumber of molecules the medium could be approximately considered as ifit fills the space continuously. Oil, oil products, gas, water or metals may beconsidered as a medium continuously filling one or another region of thespace. That is why a system of material points continuously filling a part of spaceis called a continuum.

Replacement of a real medium consisting of separate molecules by acontinuum represents of course a schematization. But such a schemati-zation has proved to be very convenient in the use of the mathemat-ical apparatus of continuous functions and, as was shown in practice,it is quite sufficient for studying the overwhelming majority of observedphenomena.

1.1.1Governing Factors

In the examination of different phenomena the researcher is always restrictedby a finite number of parameters called governing factors (parameters) withinthe limits of which the investigation is being studied. This brings up thequestion: How to reveal the system of governing parameters?

It could be done for example by formulating the problem mathematicallyor, in other words, by building a mathematical model of the considered


phenomenon as was demonstrated in the above-mentioned example. In thisproblem the governing parameters are:

x, t, m, k, f0, f1, x0, v0.

But, in order to determine the system of governing parameters, there is noneed for mathematical schematization of the process. It is enough to be guided,as has already been noted, by experience, intuition and understanding of themechanism of the phenomenon.

Let us investigate the decrease in a parachutist’s speed v in the air whenhis motion can be taken as steady. Being governed only by intuition itis an easy matter to assume the speed to be dependent on the mass ofthe parachutist m, acceleration due to g, the diameter of the parachutecanopy D, the length L of its shroud and the air density ρ. The viscosity ofthe air flowing around the parachute during its descent can be taken intoaccount or ignored since the force of viscous friction is small compared toparachute drag. Both cases represent only different schematizations of thephenomenon.

So the function sought could be assumed to have the following general formv = f (m, g, D, L, ρ). Then the governing parameters are:

m, g, D, L, ρ.

The use of dimensional theory permits us to rewrite the formulated dependencein invariant form, that is, independent of the system of measurement units(see Chapters 6 and 7)

v√gD

= f

(m

ρD3,

L

D

), ⇒ v = √

gD · f

(m

ρD3,

L

D

).

Thus, among five governing parameters there are only two independentdimensionless combinations, m/ρD3 and L/D, defining the sought-fordependence.

1.1.2Schematization of One-Dimensional Flows of Fluids and Gases in Pipelines

In problems of oil and gas transportation most often schematization of theflow process under the following conditions is used:• oil, oil product and gas are considered as a continuum continuously filling

the whole cross-section of the pipeline or its part;• the flow is taken as one-dimensional, that is all governing parameters

depend only on one space coordinate x measured along the pipeline axisand, in the general case, on time t;

• the governing parameters of the flow represent values of the correspondingphysical parameters averaged over the pipeline cross-section;

1.2 Integral Characteristics of Fluid Volume 5

• the profile of the pipeline is given by the dependence of the height of thepipeline axis above sea level on the linear coordinate z(x);

• the area S of the pipeline cross-section depends, in the general case, on xand t. If the pipeline is assumed to be undeformable, then S = S(x). If thepipeline has a constant diameter, then S(x) = S0 = const.;

• the most important parameters are:ρ(x, t) – density of medium to be transported, kg m−3;v(x, t) – velocity of the medium, m s−1;p(x, t) – pressure at the pipeline axis, Pa = N m−2;T(x, t) – temperature of the medium to be transported, degrees;τ(x, t) – shear stress (friction force per unit area of the pipeline internalsurface), Pa = N m−2;Q(x, t) = vS – volume flow rate of the medium, m3 s−1;M(x, t) = ρvS – mass flow rate of the medium, kg s−1 and other.

Mathematical models of fluid and gas flows in the pipeline are basedon the fundamental laws of physics (mechanics and thermodynamics) of acontinuum, modeling a real fluid and a real gas.

1.2Integral Characteristics of Fluid Volume

In what follows one needs the notion of movable fluid volume of the continuumin the pipeline. Let, at some instant of time, an arbitrary volume of themedium be transported between cross-sections x1 and x2 of the pipeline(Figure 1.1).

If the continuum located between these two cross-sections is identifiedwith a system of material points and track is kept of its displacement intime, the boundaries x1 and x2 become dependent on time and, togetherwith the pipeline surface, contain one and the same material points ofthe continuum. This volume of the transported medium is called themovable fluid volume or individual volume. Its special feature is that italways consists of the same particles of the continuum under consideration.If, for example, the transported medium is incompressible and the pipelineis non-deformable, then S = S0 = const. and the difference between thedemarcation boundaries (x2 − x1) defining the length of the fluid volumeremains constant.

Figure 1.1 Movable fluid volume of thecontinuum.


Exploiting the notion of fluid or individual volume of the transportedmedium in the pipeline one can introduce the following integral quantities:

M =∫ x2(t)

x1(t)ρ(x, t) · S(x, t) dx − mass of fluid volume (kg);

I =∫ x2(t)

x1(t)ρ(x, t) · v(x, t) · S(x, t) dx − momentum of fluid volume

(kg m s−1);

Ekin =∫ x2(t)

x1(t)αk

ρv2

2S(x, t) dx − kinetic energy of the fluid volume (J),

where αk is the factor;

Ein =∫ x2(t)

x1(t)ρ(x, t) · ein(x, t) · S(x, t) dx − internal energy of the fluid

volume, where ein is the density of the internal energy (J kg−1), that is theinternal energy per unit mass.

These quantities model the mass, momentum and energy of a material pointsystem.

Since the main laws of physics are often formulated as connections betweenphysical quantities and the rate of their change in time, we ought to adducethe rule of integral quantity differentiation with respect to time. The symbolof differentiation d()/ dt denotes the total derivative with respect to time,associated with individual particles of a continuum whereas the symbol ∂()/∂tdenotes the local derivative with respect to time, that is the derivative of aflow parameter with respect to time at a given space point, e.g. x = const. Thelocal derivative with respect to time gives the rate of flow parameter change ata given cross-section of the flow while, at two consecutive instances of time,different particles of the continuum are located in this cross-section.

The total derivative with respect to time is equal to

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx.

From mathematical analysis it is known how an integral containing aparameter, in the considered case it is t, is differentiated with respect tothis parameter, when the integrand and limits of integration depend on thisparameter. We have

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx =

∫ x2(t)

x1(t)

∂

∂t[A(x, t) · S(x, t)] dx

+ A(x, t) · S(x, t)|x2(t) · dx2

dt− A(x, t) · S(x, t)|x1(t) · dx1

dt.

1.3 The Law of Conservation of Transported Medium Mass. The Continuity Equation 7

First, at frozen upper and lower integration limits, the integrand isdifferentiated (the derivative being local) and then the integrand calculated atthe upper and lower integration limits is multiplied by the rates of change ofthese limits dx2/ dt and dx1/ dt, the first term having been taken with a plussign and the second with a minus sign (see Appendix B).

For the case of the fluid volume of the medium the quantities dx2/ dt anddx1/ dt are the corresponding velocities v2(t) and v1(t) of the medium in theleft and right cross-sections bounding the considered volume. Hence

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx =

∫ x2(t)

x1(t)

∂

∂t[A(x, t) · S(x, t)] dx

+ A(x, t) · v(x, t) · S(x, t)|x2(t) − A(x, t) · v(x, t) · S(x, t)|x1(t).

If, in addition, we take into account the well-known Newton–Leibniz formula,according to which

A(x, t) · v(x, t) · S(x, t)|x2(t) − A(x, t) · v(x, t) · S(x, t)|x1(t)

=∫ x2(t)

x1(t)

∂

∂x[A(x, t) · v(x, t) · S(x, t)] dx,

we obtain

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx =

∫ x2(t)

x1(t)

(∂AS

∂t+ ∂ASv

∂x

)dx. (1.4)

1.3The Law of Conservation of Transported Medium Mass. The Continuity Equation

The density ρ(x, t), the velocity of the transported medium v(x, t) and thearea of the pipeline cross-section S(x, t) cannot be chosen arbitrarily sincetheir values define the enhancement or reduction of the medium mass inone or another place of the pipeline. Therefore the first equation would beobtained when the transported medium is governed by the mass conservationlaw

d

dt

∫ x2(t)

x1(t)ρ(x, t) · S(x, t) dx = 0, (1.5)

This equation should be obeyed for any fluid particle of the transportedmedium, that is for any values x1(t) and x2(t).

Applying to Eq. (1.4) the rule (1.5) of differentiation of integral quantity withregard to fluid volume, we obtain∫ x2(t)

x1(t)

(∂ρS

∂t+ ∂ρvS

∂x

)dx = 0.


Since the last relation holds for arbitrary integration limits we get the followingdifferential equation

∂ρS

∂t+ ∂ρvS

∂x= 0, (1.6)

which is called continuity equation of the transported medium in the pipeline.If the flow is stationary, that is the local derivative with respect to time is

zero (∂()/∂t = 0), the last equation is simplified to

dρvS

dx= 0 ⇒ M = ρvS = const. (1.7)

This means that in stationary flow the mass flow rate M is constant along thepipeline.

If we ignore the pipeline deformation and take S(x) ∼= S0 = const.,from Eq. (1.7) it follows that ρv = const. From this follow two importantconsequences:

1. In the case of a homogeneous incompressible fluid (sometimes oil andoil product can be considered as such fluids) ρ ∼= ρ0 = const. and theflow velocity v(x) = const. Hence the flow velocity of a homogeneousincompressible fluid in a pipeline of constant cross-section does not changealong the length of the pipeline.

Example. The volume flow rate of the oil transported by a pipeline withdiameter D = 820 mm and wall thickness δ = 8 mm is 2500 m3 h−1. It isrequired to find the velocity v of the flow.

Solution. The internal diameter d of the oil pipeline is equal to

d = D − 2δ = 0.82 − 2 · 0.008 = 0.804 m;

v = 4Q/πd2 = const.

v = 4 · 2500/(3600 · 3.14 · 0.8042) ∼= 1.37 m s−1.

2. In the case of a compressible medium, e.g. a gas, the density ρ(x)

changes along the length of pipeline section under consideration. Sincethe density is as a rule connected with pressure, this change representsa monotonic function decreasing from the beginning of the section toits end. Then from the condition ρv = const. it follows that the velocityv(x) of the flow also increases monotonically from the beginning of thesection to its end. Hence the velocity of the gas flow in a pipeline withconstant diameter increases from the beginning of the section betweencompressor stations to its end.

Example. The mass flow rate of gas transported along the pipeline(D = 1020 mm, δ = 10 mm) is 180 kg s−1. Find the velocity of the gas flow

1.4 The Law of Change in Momentum. The Equation of Fluid Motion 9

v1 at the beginning and v2 at the end of the gas-pipeline section, if thedensity of the gas at the beginning of the section is 45 kg m−3 and at theend is 25 kg m−3.

Solution. v1 = M/(ρ1 S) = 4 · 180/(45 · 3.14 · 12) ∼= 5.1 m s−1;v2 = M/(ρ2 S) = 4 · 180/(25 · 3.14 · 12) ∼= 9.2 m s−1, that is the gas flowvelocity is enhanced by a factor 1.8 towards the end as compared with thevelocity at the beginning.

1.4The Law of Change in Momentum. The Equation of Fluid Motion

The continuity equation (1.6) contains several unknown functions, hence theuse of only this equation is insufficient to find each of them. To get additionalequations we can use, among others, the equation of the change in momentumof the system of material points comprising the transported medium. This lawexpresses properly the second Newton law applied to an arbitrary fluid volumeof transported medium

dI

dt= d

dt

∫ x2(t)

x1(t)v · ρS dx = (p1 S1 − p2 S2) +

∫ x2(t)

x1(t)p∂S

∂xdx

−∫ x2(t)

x1(t)πd · τw dx −

∫ x2(t)

x1(t)ρg sin α(x) · S dx. (1.8)

On the left is the total derivative of the fluid volume momentum of thetransported medium with respect to time and on the right the sum of allexternal forces acting on the considered volume.

The first term on the right-hand side of the equation gives the differencein pressure forces acting at the ends of the single continuum volume.The second term represents the axial projection of the reaction forcefrom the lateral surface of the pipe (this force differs from zero whenS �= const.). The third term defines the friction force at the lateral surfaceof the pipe (τw is the shear stress at the pipe walls, that is the frictionforce per unit area of the pipeline internal surface, Pa). The fourth termgives the sliding component of the gravity force (α(x) is the slope of thepipeline axis to the horizontal, α > 0 for ascending sections of the pipeline;α < 0 for descending sections of the pipeline; g is the acceleration due togravity).

Representing the pressure difference in the form of an integral over thelength of the considered volume

p1 S1 − p2 S2 = −∫ x2(t)

x1(t)

∂pS

∂xdx


and noting that

−∫ x2(t)

x1(t)

∂pS

∂xdx +

∫ x2(t)

x1(t)p∂S

∂xdx = −

∫ x2(t)

x1(t)S

∂p

∂xdx,

we obtain the following equation

d

dt

∫ x2(t)

x1(t)ρvS dx =

∫ x2(t)

x1(t)

(−S

∂p

∂x− S · 4

dτw − Sρg sin α(x)

)dx.

Now applying to the left-hand side of this equation the differentiation rule offluid volume∫ x2(t)

x1(t)

(∂ρvS

∂t+ ∂ρv2 S

∂x

)dx

=∫ x2(t)

x1(t)

(−S

∂p

∂x− S · 4

dτw − Sρg sin α(x)

)dx.

As far as the limits of integration in the last relation are arbitrary one candiscard the integral sign and get the differential equation

∂ρvS

∂t+ ∂ρv2 S

∂x= S ·

(− ∂p

∂x− 4

dτw − ρg sin α(x)

). (1.9)

If we represent the left-hand side of this equation in the form

v

(∂ρS

∂t+ ∂ρvS

∂x

)+ ρS

(∂v

∂t+ v

∂v

∂x

)and take into account that in accordance with the continuity equation (1.6) theexpression in the first brackets is equal to zero, the resulting equation may bewritten in a more simple form

ρ

(∂v

∂t+ v

∂v

∂x

)= − ∂p

∂x− 4

dτw − ρg sin α(x). (1.10)

The expression in brackets on the left-hand side of Eq. (1.10) represents thetotal derivative with respect to time, that is the particle acceleration

w = dv

dt= ∂v

∂t+ v

∂v

∂x. (1.11)

Now the meaning of Eq. (1.10) becomes clearer: the product of unit volumemass of transported medium and its acceleration is equal to the sum of allforces acting on the medium, namely pressure, friction and gravity forces. SoEq. (1.10) expresses the Newton’s Second Law and can therefore also be calledthe flow motion equation.

Remark. about the connection between total and partial derivatives with respectto time. The acceleration w = dv/ dt is a total derivative with respect to time(symbol d()/ dt), since we are dealing with the velocity differentiation of oneand the same fixed particle of the transported medium moving from one

1.5 The Equation of Mechanical Energy Balance 11

cross-section of the pipeline to another one, whereas the partial derivative withrespect to time (symbol ∂()/∂t) has the meaning of velocity differentiation ata given place in space, that is at a constant value of x. Thus such a derivativegives the change in velocity of different particles of the transported mediumentering a given cross-section of the pipeline.

Let a particle of the medium at the instant of time t be in the cross-sectionx of the pipeline and so have velocity v(x, t). In the next instant of time t + �tthis particle will transfer to the cross-section x + �x and will have velocityv(x + �x, t + �t). The acceleration w of this particle is defined as the limit

w = dv

dt= lim

�t⇒0

v(x + �x, t + �t) − v(x, t)

�t= ∂v

∂t

∣∣∣∣x

+ ∂v

∂x

∣∣∣∣t

· dx

dt.

Since dx/ dt = v(x, t) is the velocity of the considered particle, from the lastequality it follows that

dv

dt= ∂v

∂t+ v · ∂v

∂x. (1.12)

A similar relation between the total derivative ( d/ dt), or as it is also calledthe individual or Lagrangian derivative, and the partial derivative (∂/∂t), or asit is also called the local or Eulerian derivative, has the form (1.12) no matterwhether the case in point is velocity or any other parameter A(x, t)

dA(x, t)

dt= ∂A(x, t)

∂t+ v · ∂A(x, t)

∂x.

1.5The Equation of Mechanical Energy Balance

Consider now what leads to the use of the mechanical energy change law asapplied to the system of material points representing a fluid particle of thetransported medium. This law is written as:

dEkin

dt= dAex

dt+ dAin

dt(1.13)

that is the change in kinetic energy of a system of material points dEkin is equalto the sum of the work of the external dAex and internal dAin forces acting onthe points of this system.

We can calculate separately the terms of this equation but first we shoulddefine more exactly what meant by the kinetic energy Ekin. If the transportedmedium moves in the pipeline as a piston with equal velocity v(x, t) over thecross-section then the kinetic energy would be expressed as the integral

Ekin =∫ x2(t)

x1(t)

ρv2

2S dx.


But, in practice, such a schematization is too rough because, as experimentsshow, the velocity of the separate layers of the transported medium (fluid orgas) varies over the pipe cross-section. At the center of the pipe it reaches thegreatest value, whereas as the internal surface of the pipe is approached thevelocity decreases and at the wall itself it is equal to zero. Furthermore, if ata small velocity of the fluid the flow regime is laminar, with an increase invelocity the laminar flow changes into a turbulent one (pulsating and mixingflow) and the velocities of the separate particles differ significantly from theaverage velocity v of the flow. That is why models of the flow are, as a rule,constructed with regard to the difference in flow velocity from the averagevelocity over the cross-section.

The true velocity u of a particle of the transported medium is given as the sumu = v + �u of the average velocity over the cross-section v(x, t) and the additiveone (deviation) �u representing the difference between the true velocity andthe average one. The average value of this additive �u is equal to zero, butthe root-mean-square (rms) value of the additive (�u)2 is non-vanishing.The deviation characterizes the kinetic energy of the relative motion of thecontinuum particle in the pipeline cross-section. Then the kinetic energy of thetransported medium unit mass ekin may be presented as the sum of two terms

ekin = v2

2+ (�u)2

2

namely the kinetic energy of the center of mass of the considered point systemand the kinetic energy of the motion of these points relative to the center ofmass. If the average velocity v �= 0, then

ρv2

2+ ρ(�u)2

2= ρv2

2·(

1 + (�u)2

v2

)= αk · ρv2

2

where αk = 1 + (�u)2/v2 > 1. For laminar flow αk = 4/3, while for turbulentflow the value of αk lies in the range 1.02–1.05.

Remark. It should be noted that in one-dimensional theory, as a rule, the casesv = 0 and (�u)2 �= 0 are not considered.

With regard to the introduced factor the kinetic energy of any movablevolume of transported medium may be represented as

Ekin =∫ x2(t)

x1(t)αk · ρv2

2· S dx.

Let us turn now to the calculation of the terms in the mechanical energyequation (1.13). Let us calculate first the change in kinetic energy

dEkin

dt= d

dt

(∫ x2(t)

x1(t)αk · ρv2

2S dx

).


Employing the rule of integral quantity integration with reference to the fluidvolume, that is an integral with variable integration limits, we get

dEkin

dt=

∫ x2(t)

x1(t)

[∂

∂t

(αk · ρv2

2S

)+ ∂

∂x

(αk · ρv2

2S · v

)]dx.

The work of the external forces (in this case they are the forces of pressure andgravity), including also the work of external mechanical devices, e.g. pumps ifsuch are used, is equal to

dAex

dt= (p1Sv1 − p2Sv2) −

∫ x2(t)

x1(t)ρg sin α · v · S dx + Nmech

= −∫ x2(t)

x1(t)

∂

∂x(pSv) dx −

∫ x2(t)

x1(t)ρg sin α · v · S dx + Nmech.

The first term on the right-hand side of the last expression gives the workperformed in unit time or, more precisely, the power of the pressure forceapplied to the initial and end cross-sections of the detached volume. Thesecond term gives the power of the gravity force and the third term Nmech thepower of the external mechanical devices acting on the transported mediumvolume under consideration.

The work of the internal forces (pressure and internal friction) executed inunit time is given by

dAin

dt=

∫ x2(t)

x1(t)p∂(Sv)

∂xdx +

∫ x2(t)

x1(t)nin · ρS dx.

The first term on the right-hand side gives the work of the pressure force inunit time, that is the power, for compression of the particles of the medium,the factor ∂(Sv)/∂x · dx giving the rate of elementary volume change. Thesecond term represents the power of the internal friction forces, that is theforces of mutual friction between the internal layers of the medium, nin

denoting specific power, that is per unit mass of the transported medium.In what follows it will be shown that this quantity characterizes the amountof mechanical energy converting into heat per unit time caused by mutualinternal friction of the transported particles of the medium.

Gathering together all the terms of the mechanical energy equation we get∫ x2(t)

x1(t)

[∂

∂t

(αk · ρv2

2S

)+ ∂

∂x

(αk · v2

2ρvS

)]dx

= −∫ x2(t)

x1(t)ρSv

[(1

ρ

∂p

∂x

)+ g sin α

]dx +

∫ x2(t)

x1(t)nin · ρS dx + Nmech.

If the transported medium is barotropic, that is the pressure in it depends onlyon the density p = p(ρ), one can introduce a function P(ρ) of the pressure


such that dP = dp/ρ, P(ρ) = ∫dp/ρ and 1

ρ

∂p∂x = ∂P(ρ)

∂x . If, moreover, we takeinto account the equality sin α(x) = ∂z/∂x, where the function z(x) is referredto as the pipeline profile, the last equation could be rewritten in the simpleform ∫ x2(t)

x1(t)

[ρS

∂

∂t

(αkv2

2

)+ ρvS

∂

∂x

(αkv2

2+ P(ρ) + gz

)]dx

=∫ x2(t)

x1(t)nin · ρS dx + Nmech. (1.14)

If we assume that in the region [x1(t), x2(t)] external sources of mechanicalenergy are absent. Then Nmech = 0 and we can go from the integral equality(1.14) to a differential equation using, as before, the condition of arbitrarinessof integration limits x1(t) and x2(t) in Eq. (1.14). Then the sign of the integralcan be omitted and the corresponding differential equation is

ρS∂

∂t

(αkv2

2

)+ ρvS

∂

∂x

(αkv2

2+ P(ρ) + gz

)= ρS · nin (1.15)

or

∂

∂t

(αkv2

2

)+ v · ∂

∂x

(αkv2

2+

∫dp

ρ+ gz

)= nin. (1.16)

This is the sought differential equation expressing the law of mechanical energychange. It should be emphasized that this equation is not a consequence of themotion equation (1.10). It represents an independent equation for modelingone-dimensional flows of a transported medium in the pipeline.

If we divide both parts of Eq. (1.16) by g we get

∂

∂t

(αkv2

2g

)+ v · ∂

∂x

(αkv2

2g+

∫dp

ρg+ z

)= nin

g.

The expression

H = αkv2

2g+

∫dp

ρg+ z (1.17)

in the derivative on the left-hand side of the last equation has the dimension oflength and is called the total head. The total head at the pipeline cross-sectionx consists of the kinetic head (dynamic pressure) αkv2/2g, the piezometric head∫

dp/ρg and the geometric head z. The concept of head is very important in thecalculation of processes occurring in pipelines.


1.5.1Bernoulli Equation

In the case of stationary flow of a barotropic fluid or gas in the pipeline thederivative ∂()/∂t = 0, hence the following ordinary differential equations apply

vd

dx

(αkv2

2g+

∫dp

ρg+ z

)= nin

g

or

d

dx

(αkv2

2g+

∫dp

ρg+ z

)= nin

gv= i, (1.18)

where i denotes the dimensional quantity nin/gv called the hydraulic gradient

i = dH

dx= nin

gv.

Thus the hydraulic gradient, defined as the pressure loss per unit length ofthe pipeline, is proportional to the dissipation of mechanical energy into heatthrough internal friction between the transported medium layers (i < 0).

In integral form, that is as applied to transported medium located betweentwo fixed cross-sections x1 and x2, Eq. (1.18) takes the following form(

αkv2

2g+

∫dp

ρg+ z

)1

−(

αkv2

2g+

∫dp

ρg+ z

)2

= −∫ x2

x1

i dx. (1.19)

This equation is called the Bernoulli equation. It is one of the fundamentalequations used to describe the stationary flow of a barotropic medium ina pipeline.

For an incompressible homogeneous fluid, which under some conditions canbe water, oil and oil product, ρ = const.,

∫dp/ρg = p/ρg + const. Therefore

the Bernoulli equation becomes(αkv2

2g+ p

ρg+ z

)1

−(

αkv2

2g+ p

ρg+ z

)2

= −∫ x2

x1

i dx.

If in addition we take i = −i0 = const. (i0 > 0), then(ακv2

2g+ p

ρg+ z

)1

−(

ακv2

2g+ p

ρg+ z

)2

= i0 · l1−2 (1.20)

where l1−2 is the length of the pipeline between cross-sections 1 and 2.This last equation has a simple geometric interpretation (see Figure 1.2). This

figure illustrates a pipeline profile (heavy broken line); the line H(x) denotingthe dependence of the total head H on the coordinate x directed along theaxis of the pipeline (straight line) with constant slope β to the horizontal(i = dH/ dx = tgβ = const.) and three components of the total head at an


Figure 1.2 Geometric interpretation of the Bernoulli equation.

arbitrary cross-section of the pipeline: geometric head z(x), piezometric headp(x)/ρg and kinetic head αkv2(x)/2g.

The line H(x) representing the dependence of the total head H on thecoordinate x along the pipeline axis is called the line of hydraulic gradient.

It should be noted that if we neglect the dynamic pressure (in oil andoil product pipelines the value of the dynamic pressure does not exceed thepipeline diameter, e.g. at v ≈ 2 m s−1, αk ≈ 1.05 then v2/2g ∼= 0.25 m), andthe length of the section between the pipeline profile and the line of hydraulicgradient multiplied by ρg gives the value of the pressure in the pipeline cross-section x. For example, when the length of the section AA (see Figure 1.2) is500 m and diesel fuel with density ρ = 840 kg m−3 is transported along thepipeline, then

p

840 · 9.81= 500 ⇒ p = 500 · 840 · 9.81 = 4 120 200 (Pa)

or 4.12 MPa (≈42 atm).

1.5.2Input of External Energy

In fluid flow in the pipeline the mechanical energy is dissipated into heat andthe pressure decreases gradually. Devices providing pressure restoration orgeneration are called compressors.

Compressors installed separately or combined in a group form the pumpingplant destined to set the fluid moving from the cross-section with lesserpressure to the cross-section with greater pressure. To do this it is required toexpend, or deliver from outside to the fluid, energy whose power is denoted byNmech.

Let index 1 in the Bernoulli equation refer to parameters at the cross-sectionx1 of the pump entrance (suction line) and index 2 at the cross-section x2 of the

1.6 Equation of Change in Internal Motion Kinetic Energy 17

pump exit (discharge line). Since ρvS = const., the Bernoulli equation (1.14)may be written as:∫ x2

x1

d

dx

[ρvS ·

(αkv2

2+ p

ρ+ gz

)]dx =

∫ x2

x1

nin · ρS dx + Nmech.

Ignoring the difference between the kinetic and geometric heads we get

ρvS · p2 − p1

ρ−

∫ x2

x1

nin · ρS dx = Nmech.

Denoting by �H = (p2 − p1)/ρg the differential head produced by the pump orpumping plant and taking into account that ρvS = ρQ = const. and nin = gv · i,we obtain

Nmech = ρgQ · �H −∫ x2

x1

ρgQ · i dx = ρgQ · �H ·(

1 −∫ x2

x1

i

�Hdx

).

The expression in parentheses characterizes the loss of mechanical energywithin the pump. Usually this factor is taken into account by insertion of thepump efficiency η

η =(

1 −∫ x2

x1

i/�H dx

)−1

< 1

so that

Nmech = ρgQ · �H

η(Q). (1.21)

The relation (1.21) is the main formula used to calculate the power of thepump generating head �H in fluid pumping with flow rate Q .

1.6Equation of Change in Internal Motion Kinetic Energy

At the beginning of the previous section it was noted that the totalkinetic energy of the transported medium consisted of two terms – thekinetic energy of the center of mass of the particle and the kinetic energyof the internal motion of the center of mass, so that the total energyof a particle is equal to αkρv2/2, where αk > 1. Now we can derive anequation for the second component of the kinetic energy, namely the kineticenergy of the internal or relative motion in the flow of the transportedmedium.

Multiplication of motion equation (1.10) by the product vS yields

ρSd

dt

(v2

2

)= − ∂p

∂x· vS − 4

dτw · vS − ρgvS · sin α(x).


Subtracting this equation term-by-term from the Bernoulli equation (1.15),one obtains

ρSd

dt

[(αk − 1)

v2

2

]= 4

dτw · vS + ρS · nin.

Introduction of nin = −gv · i0 gives

ρSd

dt

[(ακ − 1)

v2

2

]=

(4

dτw · v

)S − ρgvS · i0. (1.22)

This is the sought equation of change in kinetic energy of internal motion ofone-dimensional flow of the transported medium. Its sense is obvious: thepower of the external friction forces (4τw · vS/d) in one-dimensional flow minus thepower ρgS(v · i0) of internal friction forces between the particles causing transitionof mechanical energy into heat is equal to the rate of change of internal motionkinetic energy in the flow of the transported medium.

For stationary flow ( d/ dt = 0 + v · ∂/∂x) of the transported mediumEq. (1.22) gives

d

dx

[(αk − 1)

v2

2

]= 4

d

τw

ρ− g · i0. (1.23)

If v ∼= const., which for the flow of an incompressible medium in a pipelinewith constant diameter is the exact condition, the left-hand part of the equationvanishes. This means that the tangential friction tension τw at the pipelinewall and the hydraulic gradient i0 are connected by

τw = ρgd

4· i0. (1.24)

It must be emphasized that in the general case, including non-stationary flow,such a connection between τw and i0 is absent (see Section 4.1).

1.6.1Hydraulic Losses (of Mechanical Energy)

The quantity nin entering into Eq. (1.16) denotes the specific power of theinternal friction force, that is per unit mass of transported medium. Thisquantity is very important since it characterizes the loss of mechanical energyconverted into heat owing to internal friction between layers of the medium.In order to derive this quantity theoretically one should know how the layersof transported medium move at each cross-section of the pipeline but this isnot always possible. In the next chapter it will be shown that in several cases,in particular for laminar, flow such motion can be calculated and the quantitynin can be found. In other cases, such as for turbulent flows of the transported


medium, it is not possible to calculate the motion of the layers and othermethods of determining nin are needed.

The quantity of specific mechanical energy dissipation nin has the followingdimension (from now onwards dimension will be denoted by the symbol [ ])

[nin] = W

kg= J

s kg= N m

s kg= kg m s−2 m

s kg= m2

s3=

[v3

d

].

So the dimension of nin is the same as the dimension of the quantity v3/d,hence, without disturbance of generality, one can seek nin in the form

nin = −λ

2· v3

d(1.25)

where λ is a dimensional factor (λ > 0), the minus sign shows that nin < 0,that is the mechanical energy decreases thanks to the forces of internal friction.The factor 1/2 is introduced for the sake of convenience.

The presented formula does not disturb the generality of the considerationbecause the unknown dependence of nin on the governing parameters of theflow is accounted for by the factor λ. This dependence is valid for any mediumbe it fluid, gas or other medium with complex specific properties, e.g. waxycrude oil, suspension or even pulp, that is a mixture of water with large rigidparticles.

For stationary fluid or gas flow one can suppose the factor λ to be dependenton four main parameters: the flow velocity v (m s−1), the kinematic viscosityof the flow ν (m2 s−1), the internal diameter of the pipeline d (m) and themean height of the roughness of its internal surface � (mm or m), so thatλ = f (v, ν, d, �). The density of the fluid ρ and the acceleration due to gravityg are not included here because intuition suggests that the friction betweenfluid or gas layers will be dependent on neither their density nor the force ofgravity.

Note that the quantity λ is dimensionless, that is its numerical valueis independent of the system of measurement units, while the parametersv, ν, d, � are dimensional quantities and their numerical values depend onsuch a choice. The apparent contradiction is resolved by the well-knownBuckingham I-theorem, in accordance with which any dimensionless quantitycan depend only on dimensionless combinations of parameters governing thisquantity (Lurie, 2001). In our case there are two such parameters

v · d

ν= Re and

�

d= ε,

the first is called the Reynolds number and the second the relative roughness ofthe pipeline internal surface. Thus

λ = λ(Re, ε).


The formula (1.25) acquires the form

nin = −λ(Re, ε) · 1

d· v3

2. (1.26)

The factor λ in this formula is called the hydraulic resistance factor, oneof the most important parameters of hydraulics and pipeline transportation.Characteristic values of λ lie in the range 0.01–0.03. More detailed informationabout this factor and its dependence on the governing parameters will bepresented below.

Turning to the hydraulic gradient i0, one can write

i0 = −nin

gv= λ · 1

d· v2

2g. (1.27)

Characteristic values of the hydraulic slope are 0.00005–0.005.If we substitute Eq. (1.27) into the Bernoulli equation (1.20), we obtain(

αkv2

2g+ p

ρg+ z

)1

−(

αkv2

2g+ p

ρg+ z

)2

= λ(Re, ε) · l1−2

d

v2

2g. (1.28)

The expression hτ = λ · l1–2/d · v2/2g on the right-hand side of this equationis called the loss of head in Darcy-Veisbach form.

Using Eq. (1.27) in the case of stationary flow of the transported mediumpermits us to get an expression for the tangential friction stress τw at thepipeline wall. Substitution of Eq. (1.27) into Eq. (1.24), yields

τw = ρgd

4· i0 = ρgd

4·(

λ1

d

v2

2g

)= λ

4· ρv2

2= Cf · ρv2

2, (1.29)

Cf (Re, ε) = λ(Re, ε)

4

where the dimensional factor Cf is called the friction factor of the fluid on theinternal surface of the pipeline or the Funning factor (Leibenson et al., 1934).

1.6.2Formulas for Calculation of the Factor λ(Re, ε)

Details of methods to find and calculate the factor of hydraulic resistance λ inEqs. (1.26)–(1.29) and one of the primary factors in hydraulics and pipelinetransportation will be given in Chapter 3. Here are shown several formulasexploiting the practice.

If the flow of fluid or gas in the pipeline is laminar, that is jetwise or layerwise(the Reynolds number Re should be less than 2300), then to determine λ theStokes formula (see Section 3.1) is used

λ = 64

Re. (1.30)


As the Reynolds number increases (Re > 2300) the flow in the pipelinegradually loses hydrodynamic stability and becomes turbulent, that is vortexflow with mixing layers. The best known formula to calculate the factor λ inthis case is the Altshuler formula:

λ = 0.11 ·(

ε + 68

Re

)1/4

(1.31)

valid over a wide range of Reynolds number from 104 up to 106 and higher.If 104 < Re < 27/ε1.143 and Re < 105, the Altshuler formula becomes the

Blasius formula:

λ = 0.31644√

Re(1.32)

having the same peculiarity as the Stokes formula for laminar flow, whichdoes not consider the relative roughness of the pipeline internal surface ε.This means that for the considered range of Reynolds numbers the pipelinebehaves as a pipeline with a smooth surface. Therefore the fluid flow in thisrange is flow in a hydraulic smooth pipe. In this case the friction tension τw atthe pipe wall is expressed by formula

τw = −λ

4· ρv2

2= −0.0791

4√

vd/ν· ρv2

2≈ v1.75

signifying that friction resistance is proportional to fluid mean velocity to thepower of 1.75.

If Re > 500/ε, the second term in parentheses in the Altshuler formula canbe neglected compared to the first one. Whence it follows that at great fluidvelocities the fluid friction is caused chiefly by the smoothness of the pipelineinternal surface, that is by the parameter ε. In such a case one can use thesimpler Shiphrinson formula λ = 0.11 · ε0.25. Then

τw = −λ

4· ρv2

2= −0.11 · ε1/4

4· ρv2

2≈ v2.

From this it transpires that the friction resistance is proportional to the squareof the fluid mean velocity and hence this type of flow is called square flow.

Finally, in the region of flow transition from laminar to turbulent, thatis in the range of Reynolds number from 2320 up to 104 one can use theapproximation formula

λ = 64

Re· (1 − γ•) + 0.3164

4√

Re· γ∗, (1.33)

where γ∗ = 1 − e−0.002·(Re−2320) is the intermittency factor (Ginsburg, 1957). Itis obvious that the form of the last formula assures continuous transfer fromthe Stokes formula for laminar flow to the Blasius formula for turbulent flowin the zone of hydraulic smooth pipes.


To calculate the hydraulic resistance factor λ of the gas flow in a gas main,where the Reynolds number Re is very large and this factor depends only onthe condition of the pipeline internal surface, Eq. (1.34) is often used.

λ = 0.067 ·(

2�

d

)0.2

(1.34)

in which the absolute roughness � is equal to 0.03–0.05 mm.

Exercise 1. The oil (ρ = 870 kg m−3, ν = 15 s St) flows along the pipeline(D = 156 mm; δ = 5 mm; � = 0.1 mm) with mean velocity v = 0.2 m s−1.Determine through the Reynolds criterion the flow regime; calculate factors λ

and Cf .

Answer. Laminar; 0.033; 0.0083.

Exercise 2. Benzene (ρ = 750 kg m−3, ν = 0.7 s St) flows along the pipeline(D = 377 mm; δ = 7 mm; � = 0.15 mm) with mean velocity v = 1.4 m s−1.Determine through the Reynolds criterion the flow regime; calculate factors λ

and Cf .

Answer. Turbulent; 0.017; 0.0041.

Exercise 3. Diesel fuel (ρ = 840 kg m−3, ν = 6 s St) flows along the pipeline(D = 530 mm; δ = 8 mm; � = 0.25 mm) with mean velocity v = 0.8 m s−1.Determine the flow regime; calculate factors λ and Cf .

Answer. Turbulent; 0.022; 0.0054.

1.7Total Energy Balance Equation

Besides the law (1.13) of mechanical energy change of material points,applied to an arbitrary continuum volume in the pipeline there is one morefundamental physical law valid for any continuum – the law of total energyconservation or, as it is also called, the first law of thermodynamics. This lawasserts that the energy does not appear from anywhere and does not disappearto anywhere. It changes in total quantity from one form into another. Asapplied to our case this law may be written as follows

d(Ekin + Ein)

dt= dQex

dt+ dAex

dt(1.35)

that is the change in total energy (Ekin + Ein) of an arbitrary volume of thetransported medium happens only due to the exchange of energy withsurrounding bodies owing to external inflow of heat dQex and the workof external forces dAex.

1.7 Total Energy Balance Equation 23

In Eq. (1.35) Ein is the internal energy of the considered mass of transportedmedium, unrelated to the kinetic energy, that is the energy of heat motion,interaction between molecules and atoms and so on. In thermodynamicsreasons are given as to why the internal energy is a function of state, thatis at thermodynamic equilibrium of a body in some state the energy has awell-defined value regardless of the means (procedure) by which this statewas achieved. At the same time the quantities dQex/ dt and dAex/ dt arenot generally derivatives with respect to a certain function of state but onlyrepresent the ratio of elementary inflows of heat energy (differential dQex) andexternal mechanical energy (differential dAex) to the time dt in which theseinflows happened. It should be kept in mind that these quantities depend onthe process going on in the medium.

In addition to function Ein one more function ein is often introduced,representing the internal energy of a unit mass of the considered bodyein = Ein/m, where m is the mass of the body.

We can write Eq. (1.35) for a movable volume of transported mediumenclosed between cross-sections x1(t) and x2(t). The terms of this equation are

d(Ekin + Ein)

dt= d

dt

[∫ x2(t)

x1(t)

(αk

ρv2

2+ ρein

)S dx

],

dQex

dt=

∫ x2(t)

x1(t)πd · qn dx,

dAex

dt= −

∫ x2(t)

x1(t)

∂

∂x(pSv) dx −

∫ x2(t)

x1(t)ρg sin α · v · S dx + Nmech

where qn is the heat flux going through the unit area of the pipeline surfaceper unit time (W m−2); πd · dx is an element of pipeline surface area and d isthe pipeline diameter.

Gathering all terms, we obtain

d

dt

[∫ x2(t)

x1(t)

(αk · ρv2

2+ ρein

)S dx

]=

∫ x2(t)

x1(t)πd · qn dx

−∫ x2(t)

x1(t)

∂

∂x(pSv) dx −

∫ x2(t)

x1(t)ρg sin α · v · S dx + Nmech.

Differentiation of the left-hand side of this equation gives∫ x2(t)

x1(t)

{∂

∂t

[(αkv2

2+ ein

)ρS

]+ ∂

∂x

[(αkv2

2+ ein

)ρvS

]}dx

=∫ x2(t)

x1(t)πd · qn dx −

∫ x2(t)

x1(t)

∂

∂x

(p

ρρvS

)dx

−∫ x2(t)

x1(t)ρvSg

∂z

∂xdx + Nmech


or ∫ x2(t)

x1(t)

{∂

∂t

[(αkv2

2+ ein

)ρS

]+ ∂

∂x

[(αkv2

2+ ein + p

ρ

)ρvS

]}dx

=∫ x2(t)

x1(t)πd · qn dx −

∫ x2(t)

x1(t)ρvSg

∂z

∂xdx + Nmech. (1.36)

If we assume that inside the region [x1(t), x2(t)] the external sources of me-chanical energy are absent, that is Nmech = 0, then it is possible to pass fromintegral equality (1.36) to the corresponding differential equation using, asbefore, the condition that this equation should be true for any volume of thetransported medium, that is the limits of integration x1(t) and x2(t) in (1.36)are to be arbitrarily chosen. Then the sign of the integral can be omitted andthe differential equation is

∂

∂t

[(αkv2

2+ ein

)ρS

]+ ∂

∂x

[(αkv2

2+ ein + p

ρ

)ρvS

]= πd · qn − ρvSg

∂z

∂x. (1.37)

Excluding from Eq. (1.37) the change in kinetic energy with the help of theBernoulli equation with term by term subtraction of Eq. (1.16) from Eq. (1.37)we get one more energy equation

ρS∂

∂t

(αkv2

2

)+ ρvS

∂

∂x

(αkv2

2+

∫dp

ρ+ gz

)= ρvSg · i

called the equation of heat inflow.This equation could be variously written. First, it may be written through

the internal energy ein:

∂

∂t(ein · ρS) + ∂

∂x(ein · ρvS) = πd · qn − p

∂vS

∂x− ρvSg · i

or

ρS

(∂ein

∂t+ v

∂ein

∂x

)= πd · qn − p · ∂vS

∂x− ρvSg · i. (1.38)

This equation proved to be especially convenient for modeling flows ofincompressible or slightly compressible fluids because the derivative ∂(vS)/∂xexpressing the change in fluid volume in the pipeline cross-section is extremelysmall as is the work p · ∂(vS)/∂x of the pressure forces. With this in mindEq. (1.38) may be written in a particularly simple form:

ρdein

dt∼= 4

d· qn − ρvg · i. (1.39)


This means that the rate of internal energy change of the transported mediumis determined by the inflow of external heat through the pipeline surface andheat extraction due to conversion of mechanical energy into heat produced byfriction between the continuum layers.

Second, the equation of heat inflow can be written using the functionJ = ein + p/ρ representing one of the basic thermodynamic functions, enthalpyor heat content, of the transported medium

∂

∂t(ein · ρS) + ∂

∂x

[(ein + p

ρ

)ρvS

]= πd · qn + ρvSg ·

(1

ρg

∂p

∂x− i

)or

∂

∂t(ein · ρS) + ∂

∂x[J · ρvS] = πd · qn + ρvSg ·

(1

ρg

∂p

∂x− i

). (1.40)

If we take into account (as will be shown later) that the expression inparentheses on the right-hand side of this equation is close to zero, since for arelatively light medium, e.g. gas, the hydraulic slope is expressed through thepressure gradient by the formula i = 1/ρg · ∂p/∂x, the equation of heat inflowcan be reduced to a simpler form

∂ρS · ein

∂t+ ∂ρvS · J

∂x∼= πd · qn (1.41)

in which the dissipation of mechanical energy appears to be absent.

Temperature Distribution in Stationary FlowThe equation of heat inflow in the form (1.39) or (1.41) is convenient todetermine the temperature distribution along the pipeline length in stationaryflow of the transported medium.

1. For an incompressible or slightly compressible medium, e.g. dropping liquid:water, oil and oil product, this equation has the form

ρv · dein

dx∼= 4

d· qn − ρvg · i. (1.42)

The internal energy ein depends primarily on the temperature of the fluidT , the derivative dein/ dT giving its specific heat Cv (J kg−1 K−1). If we takeCv = const. then ein = Cv · T + const.

To model the heat flux qn the Newton formula is usually used

qn = −κ · (T − Tex), (1.43)

by which this flow is proportional to the difference between the temperaturesT and Tex in and outside the pipeline, with qn < 0 when T > Tex and qn > 0when T < Tex. The factor κ (W m−2 K−1) in this formula characterizes theoverall heat resistance of the materials through which the heat is transferred


from the pipe to the surrounding medium (anticorrosive and heat insulation,ground, the boundary between ground and air and so on) or the reverse. Thisfactor is called the heat-transfer factor.

The hydraulic gradient i can sometimes be considered constant i = −i0 ≈const., if the dissipation of mechanical energy in the stationary fluid flow inthe pipeline with constant diameter is identical at all cross-sections of thepipeline.

With due regard for all the aforesaid Eq. (1.42) is reduced to the followingordinary differential equation

ρCvv · dT

dx= −4κ

d(T − Tex) + ρvgi0 (1.44)

for temperature T = T(x). From this equation in particular it follows that theheat transfer through the pipeline wall (the first term on the right-hand side)lowers the temperature of the transported medium when T(x) > Tex or raisesit when T(x) < Tex, whereas the dissipation of mechanical energy (the secondterm on the right-hand side) always implies an increase in the temperature ofthe transported medium.

The solution of the differential equation (1.44) with initial conditionT(0) = T0 yields

T(x) − Tex − T⊗T0 − Tex − T⊗

= exp(

− πdκ

CvMx

). (1.45)

Where T⊗ = gi0M/πdκ is a constant having the dimension of temperature;M = ρvS is the mass flow rate of the fluid (M = const.). The formula thusobtained is called the Shuchov formula.

Figure 1.3 illustrates the distribution of temperature T(x) along the pipelinelength x in accordance with Eq. (1.45).

The figure shows that when the initial temperature T0 is greater than(Tex + T⊗), the moving medium cools down, while when T0 is less than(Tex + T⊗), the medium gradually heats up. In all cases with increase in thepipeline length the temperature T → (Tex + T⊗).

In particular from Eq. (1.44) it follows that if the heat insulation of thepipeline is chosen such that at the initial cross-section of the pipeline x = 0

Figure 1.3 Temperaturedistribution along the pipeline length.


the condition of equality to zero of the right-hand side is obeyed

−4κ

d(T0 − Tex) + ρvgi0 = 0

that is the factor κ satisfies the condition

κ = ρvdgi04(T0 − Tex)

= gM · i0πd · (T0 − Tex)

.

And the temperature of the transported medium would remain constant andequal to its initial value over the whole pipeline section. In such a case theheat outgoing from the pipeline would be compensated by the heat extractedby internal friction between the layers. Such an effect is used, for example,in oil transportation along the Trans-Alaska oil pipeline (USA, see the coverpicture). Through good insulation of the pipeline the oil is pumped overwithout preheating despite the fact that in winter the temperature of theenvironment is very low.

From Eq. (1.45) follows the connection between the initial T0 and final TL

temperatures of the transported medium. If in this formula we set x = L,where L is the length of the pipeline section, we obtain

TL − Tex − T⊗T0 − Tex − T⊗

= exp(

−πdκL

CvM

). (1.46)

Expressing now from (1.46) the argument under the exponent and substitutingthe result in Eq. (1.45), we get the expression for the temperature distributionthrough the initial and final values

T(x) − Tex − T⊗T0 − Tex − T⊗

=(

TL − Tex − T⊗T0 − Tex − T⊗

)x/L

. (1.47)

Exercise 1. The initial temperature of crude oil (ρ = 870 kg m−3, Cv =2000 J kg−1 K−1, Q = 2500 m3 h−1), pumping over a pipeline section (d =800 mm, L = 120 km, i0 = 0.002) is 55 ◦C. The temperature of the surroundingmedium is 8 ◦C. The heat insulation of the pipeline is characterized by theheat-transfer factor κ = 2 W m−2 K−1. It is required to find the temperature atthe end of the section.

Solution. Calculate first the temperature T⊗:

T⊗ = gi0M

πdκ= 9.81 · 0.002 · 870 · (2500/3600)

3.14 · 0.8 · 2∼= 2.36 K.

Using Eq. (1.46) we obtain

TL − 8 − 2.36

55 − 8 − 2.36= exp

(− 3.14 · 0.8 · 2 · 120 · 103

2000 · 870 · (2500/3600)

),

from which follows TL∼= 37.5 ◦C.


Exercise 2. By how much would the temperature of the oil (Cv =1950 J kg−1 K−1) be raised due to the heat of internal friction when theoil is transported by an oil pipeline (L = 150 km, d = 500 mm, i0 = 0.004)provided with ideal heat insulation (κ = 0)?

Solution. In this case it is impossible to use at once Eq. (1.45) since κ = 0. Touse Eq. (1.47) one should go to the limit at κ → 0, therefore it would be betterto use Eq. (1.44)

ρCvv · dT

dx= ρvgi0 or Cv · dT

dx= gi0,

from which �T = gi0L/Cv = 9.81 · 0.004 · 150 · 103/1950 ∼= 3 K.

Exercise 3. It is required to obtain the temperature of oil pumping over thepipeline section of length 150 km in cross-sections x = 50, 100 and 125 km,if the temperature at the beginning of the pipeline T0 = 60 ◦C, that at theend TL = 30 ◦C, and that of the environment Tex = 8 ◦C. The extracted heat ofinternal friction may be ignored.

Solution. Using Eq. (1.46), one gets

T(x) − 8

60 − 8=

(30 − 8

60 − 8

)x/L

and T(x) = 8 + 52 · (0.4231)x/150.

Substitution in this formula of successive x = 50, 100 and 125 givesT(50) ∼= 47 ◦C; T(100) ∼= 37.3 ◦C; T(125) ∼= 33.4 ◦C.

2. For stationary flow of a compressible medium, e.g. gas, the equation of heatinflow (1.41) takes the form

ρvSdJ

dx= πd · qn.

In the general case, the gas enthalpy J is a function of pressure and temperatureJ = J(p, T), but for a perfect gas, that is a gas obeying the Clapeyron law p = ρRT ,where R is the gas constant, the enthalpy is a function only of temperatureJ = Cp · T + const., where Cp is the gas specific heat capacity at constantpressure (Cp > Cv; Cp − Cv = R). Regarding Cp = const. and taking as beforeqn = −κ · (T − Tex), we transform the last equation to

CpMdT

dx= −πdκ · (T − Tex)

or

dT

dx= − πdκ

CpM· (T − Tex).

1.8 Mathematical Modeling of One-Dimensional Flows in Pipelines 29

The solution of this differential equation with initial condition T(0) = T0 gives

T(x) − Tex

T0 − Tex= exp

(− πdκ

CpMx

), (1.48)

which is similar to the solution (1.45) for temperature distribution in anincompressible fluid. The difference consists only in that instead of heatcapacity Cv in the solution (1.47) we use heat capacity Cp and the temperatureT⊗ taking into account the heat of internal friction is absent (for methaneCp

∼= 2230 J kg−1 K−1; Cv∼= 1700 J kg−1 K−1).

The temperature TL of the gas at the end of the gas pipeline section is foundfrom

TL − Tex

T0 − Tex= exp

(−πdκL

CpM

)(1.49)

with regard to which the distribution (1.47) takes the form

T(x) − Tex

T0 − Tex=

(TL − Tex

T0 − T

)x/L

(1.50)

allowing us to express the temperature through the initial and finaltemperatures.

Note that for a real gas the enthalpy J = J(p, T) of the medium depends notonly on temperature but also on pressure, so the equation of heat inflow has amore complex form. By the dependence J(p, T) is explained, in particular, theJoule-Thomson effect.

1.8Complete System of Equations for Mathematical Modeling of One-DimensionalFlows in Pipelines

This system consists of the following equations.

1. Continuity equation (1.6)

∂ρS

∂t+ ∂ρvS

∂x= 0;

2. Momentum (motion) equation (1.10)

ρ

(∂v

∂t+ v

∂v

∂x

)= − ∂p

∂x− 4

dτw − ρg sin α(x);

3. Equation of mechanical energy balance (1.15)

∂

∂t

(αkv2

2

)+ v · ∂

∂x

(αkv2

2+ P(ρ) + gz

)= vg · i;


4. Equation of total energy balance (1.37)

∂

∂t

[(αkv2

2+ ein

)ρS

]+ ∂

∂x

[(αkv2

2+ J

)ρvS

]= πd · qn − ρvgS

dz

dx.

The number of unknown functions in this equation is 10: ρ, v, p, S, ein, T, τw,

i, qn, αk, while the number of equations is 4. Therefore there are neededadditional relations to close the system of equations. As closing relations thefollowing relations are commonly used:• equation of state p = p(ρ, T), characterizing the properties of the

transported medium;• equation of pipeline state S = S(p, T) characterizing the deformation ability

of the pipeline;• calorimetric dependences ein = e(p, T) or J = J(p, T);• dependence qn = −κ · (T − Tex) or more complex dependences

representing heat exchange between the transported medium and theenvironment;

• hydraulic dependence τw = τw(ρ, v, v, d, ν, . . .);• dependences αk = f (ρ, v, ν, d, . . .), or i = f (τw),characterizing internal structure of medium flow.

To obtain closing relations a more detailed analysis of flow processes isneeded. It is also necessary to consider mathematical relations describingproperties of the transported medium and the pipeline in which the mediumflows.

The division of mechanics in which properties of a transported mediumsuch as viscosity, elasticity, plasticity and other more complex properties arestudied is called rheology.

31

2Models of Transported Media

Algebraic relations connecting parameters of the transported medium such asdensity, pressure, temperature and so on are called equations of state. Each ofthese relations represents of course a certain schematization of the propertiesof the considered medium and is only a model of a given medium. Let usconsider some models.

2.1Model of a Fluid

By fluid is meant a continuum in which the interaction of the contacting interiorparts at rest is reduced only to the pressing force of pressure. If fluid particlesinteract along the surface element dσ with the unit normal n (Figure 2.1), theforce dFn with which the fluid particles on one side of the element act on thefluid particles on the other side of the element is proportional to the area dσ,is directed along the normal n and has a pressing action on them. Then

dFn = −pn dσ. (2.1)

The magnitude p of this force does not depend on the surface elementorientation and is called pressure.

Thus, p = | dFn/ dσ|. The absence of tangential friction forces in the state ofrest models the fact that the fluid takes the shape of the vessel it fills.

Further classification of fluids is dependent on whether or not tangentialfriction forces are taken into account on exposure to fluid flow. In accordance

Figure 2.1 A scheme of force interactions in a fluid.


32 2 Models of Transported Media

with this there are two models: the model of an ideal fluid and the model of aviscous fluid.

2.2Models of Ideal and Viscous Fluids

In the model of an ideal fluid it is assumed that tangential friction forcesbetween fluid particles separated by an elementary surface are absent, not onlyin the state of rest but also in the state of flow. Such a schematization(or model) of a fluid appears to be very fruitful when the tangentialcomponents of interaction forces, that is friction forces, are far smallerthan their normal components, that is pressure forces. In other cases whenthe friction forces are comparable with or even exceed the pressure forces themodel of an ideal fluid has proved to be inapplicable. Hence expression (2.1)for an ideal fluid is true in the state of rest as well as in the state offlow.

In the model of a viscous fluid tangential stresses resulting in fluid flow aretaken into account. Let for example the fluid layers move as shown in Figure 2.2.

Here u(y) is the velocity distribution in the flow and y is the direction of anormal to the elementary surface dσ.

In the model of a viscous fluid it is accepted that the tangential stress τ

between the layers of the moving fluid is proportional to the velocity differenceof these layers calculated per unit length of the distance between them, namelyto the velocity gradient du/ dy:

τ = µdu

dy. (2.2)

The tangential stress τ is defined as the friction force between the fluid layersdivided by the area of the surface separating these layers. Then the dimensionof the stress τ is

[τ] = force

area= M × L/T2

L2 = M

L × T2 .

In the SI system of units the stress τ is measured by Pa = kg m−1 s−2.

Figure 2.2 Illustration of the definition of the viscousfriction law.

2.2 Models of Ideal and Viscous Fluids 33

The proportionality factor µ in the law of viscous friction (2.2) is called thefactor of dynamic viscosity or simply dynamic viscosity. Its dimension is

[µ] = [τ] × T = M

L × T

In SI units µ is measured in kg m−1s−1 and is expressed through poise P,where 1 P = 0.1 kg m−1 s−1. For example the dynamic viscosity of water isequal to 0.01 P = 0.001 kg m−1 s−1 = 1 cP (centipoise).

The factor of kinematic viscosity or simply kinematic viscosity ν of a fluid isdefined as the ratio µ/ρ, therefore

[ν] =[

µ

ρ

]= M/(L × T)

M/L3 = L2

T.

In SI units ν is measured in m2 s−1 and expressed in stokes St, where1 St = 10−4 m2 s−1. For example, the kinematic viscosity of water is equal to0.01 St = 10−6 m2 s−1 = 1 cSt (centistoke). The kinematic viscosity of benzeneis approximately equal to 0.6 cSt; that of diesel fuel is 4–9 cSt and that oflow-viscous oil 5–15 cSt.

The viscosity of oil and of almost all oil products depends on temperature.As the temperature increases the viscosity decreases, whereas reduction intemperature leads to viscosity enhancement. To calculate the dependence ofthe kinematic viscosity ν on temperature T it different formulas can be usedincluding the Reynolds–Filonov formula

ν(T) = ν0 · e−κ·(T−T0) (2.3)

in which ν0 is the kinematic viscosity of a fluid at temperature T0 and κ (1/ K)is an empirical factor. Equation (2.3) means that the fluid viscosity variesexponentially with temperature.

To use Eq. (2.3) it is necessary to know the factor κ or the viscosity ν1 ofthe same fluid at another temperature T1. Then this factor is found from therelation

κ = ln(ν0/ν1)

(T1 − T0). (2.4)

Exercise. The kinematic viscosity of summer diesel fuel at temperature +20 ◦Cis 5 cSt, whereas at a temperature of 0 ◦C it increases to 8 cSt. Determine theviscosity of the same fuel at temperature +10 ◦C.

Solution. With Eq. (2.4) the factor κ = ln(5/8)/(0 − 20) ∼= 0.0235 is deter-mined. The viscosity is found from Eq. (2.3) ν = 5 · exp[−0.0235 · (10 − 20)] ∼=6.3 cSt.


2.3Model of an Incompressible Fluid

A fluid is called incompressible if its density does not vary when moving, thatis dρ/ dt = 0. If initially all fluid particles had equal densities, that is the fluidwas homogeneous, it remains homogeneous as before, namely ρ = ρ0 = const.while moving.

Of course an incompressible fluid is only a model of a real medium,since as is known absolutely incompressible media do not exist, but whenthe change in fluid density in a certain process can be neglected the modelof an incompressible fluid may be very useful. For example, under normalconditions water density is 1000 kg m−3, benzene ≈ 735–750 kg m−3, dieselfuel ≈ 840 kg m−3, oil ≈ 870–900 kg m−3 etc.

If the fluid density depends only on pressure, that is ρ = ρ(p), the conditionof incompressibility is dρ/ dt = 0 being tantamount to dρ/ dp = 0.

2.4Model of Elastic (Slightly Compressible) Fluid

There are processes which require that account is taken of even a smallvariation in fluid density. In such a case the so-called elastic fluid model is oftenused. In this model the fluid density depends on pressure as follows

ρ(p) = ρ0[1 + β(p − p0)] (2.5)

where β (1/Pa) is the compressibility factor; ρ0 the fluid density at normalpressure p0. The compressibility factor is the inverse of the elastic modulusK (Pa), that is K = 1/β. Then Eq. (2.5) reduces to

ρ(p) = ρ0

(1 + p − p0

K

). (2.6)

Mean values of the elastic modulus for oil and oil products vary in the range1400–1500 MPa, so that K ≈ 1.4 − 1.5 · 109 Pa. It follows that the deviation ofdensity ρ from the normal ρ0 : �ρ = ρ0 · (p − p0)/K is very small for oil andoil products. For example for a fluid with ρ0 = 850 kg m−3 at p − p0 = 5 MPa(≈ 50 atm) the deviation �ρ is 2.8 kg m−3.

2.5Model of a Fluid with Heat Expansion

The expansion of different media on heating and subsequent compression oncooling is taken into account in the fluid model with volume expansion. In the

2.5 Model of a Fluid with Heat Expansion 35

Table 2.1 Factor of volume expansion ξ.

Density ρ, kg m−3 Factor ξ, K−1

720–739 0.001183740–759 0.001118760–779 0.001054780–799 0.000995800–819 0.000937820–839 0.000882840–859 0.000831860–880 0.000782

model to be considered the density ρ is a function of temperature T , so thatρ = ρ(T)

ρ(T) = ρ0[1 + ξ(T0 − T)] (2.7)

in which ξ (1/K) is the factor of volume expansion, ρ0 and T0 are the density andtemperature of the fluid under normal conditions (often T0 = 293 K (20 ◦C);ρ0 = ρ(p0, T0); p0 = pst = 101325 Pa). The values of the factor ξ for oil and oilproducts are given in Table 2.1.

From Eq. (2.6) it follows that on heating, that is at T > T0, the fluid expands,that is ρ < ρ0 whereas at T < T0, ρ > ρ0, that is the fluid is compressed.

Exercise. The density ρ0 of benzene at 20 ◦C is 745 kg m−3. Determine thedensity of the same benzene at 10 ◦C.

Solution. Using Eq. (2.6) and Table 2.1 we haveρ(10 ◦C) = 745 · [1 + 0.00118 · (20 − 10)] = 753.3 kg m−3.Thus the density is increased by 8.3 kg m−3.There is also used a model for fluid expansion with regard to baric and

heat expansion. In such a model the density is a function of pressure andtemperature ρ = ρ(p, T) called the state equation

ρ(p, T) = ρ0

[1 + ξ(T − T0) + p − p0

K

]. (2.8)

Here p0, T0 are the nominal pressure and temperature satisfying the relationρ0 = ρ(p0, T0).

Exercise. The density of benzene ρ0 at 20 ◦C and atmospheric pressurepat ≈ 0.1 MPa is 745 kg m−3. Determine the density of the same benzene attemperature 10 ◦C and pressure 6.5 MPa.

Solution. Using Eq. (2.7) and Table 2.1 we get ρ(p, T) = 745 · [1 + 0.00118 ·(20 − 10) + (6.5 − 0.1) · 106/(1.5 · 109)] = 757 kg m−3. The density is in-creased by 12 kg m−3.


2.6Models of Non-Newtonian Fluids

Fluids modeled by condition (2.2) of viscous friction are called Newtonianviscous fluids in accordance with the name of the law (2.2). The quantityγ = du/ dy having the sense of velocity gradient with dimension s−1, is calledthe shear rate. The linear dependence (2.2) between the tangential frictionstress τ and the shear rate γ is shown in Figure 2.3. This dependence states:‘‘Tangential stresses arising in a medium having been modeled by a Newtonianviscous fluid are proportional to the shear rate of the fluid layers relative toeach other. When the shear rate vanishes, the tangential friction stresses alsodisappear.’’

The dynamic viscosity of fluid µ is represented in this model by the slope of astraight line on the plane (γ, τ) : µ = tan ϕ, where ϕ is the angle of inclinationof the straight line to the abscissa. Many experiments have shown that themodel of a Newtonian viscous fluid well schematizes processes taking placein many real fluids.

And yet there exist dependences of τ on γ (flow curve) that differ significantlyfrom that depicted in Figure 2.3. Such fluids are called non-Newtonian.

As an example of a non-Newtonian fluid is a model of a power Ostwaldfluid (Wilkinson, 1960)

τ = k ·∣∣∣∣ du

dy

∣∣∣∣n−1

· du

dy(2.9)

in which the relation of the tangential friction stresses between fluid layers hasa power nature. In other words the apparent viscosity µ of such a fluid doesnot remain constant as in the model of Newtonian fluid, but depends on thecharacteristics of the flow, namely

µ = k ·∣∣∣∣ du

dy

∣∣∣∣n−1

. (2.10)

In this model k and n are factors. Fluids with n < 1 are called pseudo-plasticfluids. These models are applied to describe the behavior of suspension flows,

Figure 2.3 A model of a Newtonian viscous fluid.

2.6 Models of Non-Newtonian Fluids 37

Figure 2.4 Models of non-Newtonianfluids: 1, pseudo-plastic fluids; 2, dilatantfluids.

that is viscous fluids with suspended small particles. Flow curves of such fluidshave the form of curve 1 in Figure 2.4 (Wilkenson, 1960).

Fluids with n > 1 are called dilatant fluids. Starch glue is an example of fluidwhose behavior is described by the dilatant model. Flow curves of these fluidshave the form of curve 2 in Figure 2.4 (Wilkenson, 1960).

Model of viscous-plastic medium with limit shear stress; model of Shve-dov–Bingham fluid (Wilkenson, 1960). There are fluids in which the stressesbetween the fluid layers are sufficiently well described by the followingrelations

τ = τ0 + µdu

dy, at τ > τ0;

γ = du

dy= 0, at |τ| ≤ τ0; (2.11)

τ = −τ0 + µdu

dy, at τ < −τ0.

These expressions imply that the flow of such a fluid does not begin as long asthe absolute value of the tangential friction stress τ does not exceed a limitingvalue τ0, being the characteristic of the given medium and called the limit shearstress. In this case γ = 0. At |τ| ≥ τ0 and γ �= 0 the medium flows as a viscousfluid.

Real media whose properties are satisfactory modeled by the viscous-plasticBingham fluid are, for example, high-paraffinaceous and solidifying oils, mudsolutions, lacquers, paints and other media.

Flow curves of viscous-plastic media are shown in Figure 2.5.

Figure 2.5 Flow curves of aviscous-plastic fluid.


2.7Models of a Gaseous Continuum

We proceed now to the description of the basic models used for gas flow.First let us consider the properties common to all gases. One such property isthat for all gases in a state of thermodynamic equilibrium there is a relationbetween pressure p, absolute temperature T and density ρ (or specific volumeυ = 1/ρ)

�(p, υ, T) = 0 (2.12)

called the equation of state. The physical nature of this fact is discussed intext-books on statistical physics and thermodynamics. In most models it isalso assumed that when the motion starts the relation (2.12) remains. Thismeans that the establishment of thermodynamic equilibrium happens muchfaster than the non-equilibrium caused by the resulting flow.

The specific form of the dependence (2.12) is set in the course of so-calledcalorimetric measurements, but for the majority of gases this dependencehas one and the same distinctive features. Geometrical representation ofthe dependence (2.12) has the form of a two-dimensional surface in a three-dimensional space of variables (p, υ, T). Figure 2.6 shows isotherms of realgases representing intersections of this surface with planes T = const.

For all gases there exists the so-called critical isotherm, depicted in Figure 2.6by the heavy line, above and below which the properties of the gas are different.If T ≥ Tcr, where Tcr is the critical temperature for a given gas, the gas at anyelevation of pressure remains in the gaseous state. If T < Tcr, then for eachtemperature T there exists a value of pressure p at which the gas begins tochange into the liquid phase, its specific volume being increased from υ′ toυ′′, after which the resulting medium acquires the properties of a liquid.

The point K is called critical point of a given gas, the quantities (Tcr, pcr, υcr)accounting for the individual properties of a gas are constants.

For example, for methane CH4, which is the major constituent of consistsnatural gas, Tcr = 190.55 K and pcr = 4.641 MPa. This means that if the gas

Figure 2.6 Gas isotherms.

2.7 Models of a Gaseous Continuum 39

temperature exceeds 190.55 K, this gas will be changed into the liquid statewithout the need for any pressure increase.

2.7.1Model of a Perfect Gas

If the gas pressure is not too high and the temperature not too low, theisotherms of all gases are similar to each other (see the right-hand part ofFigure 2.6 enclosed in the dotted oval line) and with a high degree of accuracycan be approximated by hyperbolas representing the fact that the pressure p isinversely proportional to the specific volume υ.

Under given conditions the interaction of the molecules of a real gas isindependent of the form of the molecules, that is of the spatial configuration ofthe constituent atoms, and is determined only by their total mass. Figurativelyspeaking, the molecules behave as balls differing in their mass, therefore thenumber of parameters characterizing the gas decreases from three to one,namely the molecular weight µg.

To characterize the thermodynamic state of gases in a given range of pressureand temperature the model of a perfect gas is used. Equation (2.11) of the gasstate then has the simple form

p = RT

υor p = ρRT (2.13)

where R is the only constant in the equation and is called the gas constantR = R0/µg, where R0 is the universal gas constant, equal to 8314 J mol−1 K−1.Thus all gas constants for a perfect gas depend only on the molecular weight.They are: for methane (µg

∼= 16 kg kmol−1) R = 8314/16 ∼= 520 J kg−1 K−1);for oxygen O2 (µg

∼= 32 kg kmol−1) R = 8314/32 ∼= 260 J kg−1 K−1); for carbondioxide CO2 (µg

∼= 44 kg kmol−1) R = 8314/44 ∼= 189 J kg−1 K−1; for air(µg

∼= 29 kg kmol−1) R = 8314/29 ∼= 287 J kg−1 K−1.Equation (2.12) connecting the density, pressure and temperature of a gas is

called the Clapeyron equation.The model of a perfect gas operates sufficiently well over a range of not too

high pressures and moderate temperatures.

2.7.2Model of a Real Gas

Despite the fact that the name of this model contains the word ‘‘real’’,one is dealing only with the next, more general, schematization of a gasmodel. From Figure 2.6 it follows that the hyperbolic dependence (2.13)does not suit observations of gas behavior with increase of pressure anddecrease of temperature. Hence, in processes of gas pipeline transportation


and underground gas storage, where the pressure may be 5–15 MPa the modelof a perfect gas would give improper results were it to be used in calculations.

There is a model of more general form than the model of a perfect gas. Thisis the model of a real gas. One can find the details of such a model for examplein Porshakov et al., 2001. Without dwelling on the details of its derivation wenote that the mathematical form of this equation is represented as follows

p = Z(pr, Tr) · RT

υor p = Z(pr, Tr) · ρRT (2.14)

differing from Eq. (2.13) by the insertion of the dimensionless factor Z(pr, Tr)

called the over-compressibility factor, being a function of two parameters: thereduced pressure pr and the reduced temperature Tr, where

pr = p

pcr, Tr = T

Tcr.

Here pcr and Tcr are the critical pressure and temperature.Hence the model (2.14) takes into account not only the molecular weight

of a gas through the constant R but also its thermodynamic parameters suchas critical pressure and temperature. It is evident that for moderate valuesof pressure and temperature Z ≈ 1 and the model (2.14) transforms into themodel of a perfect gas. For a real gas Z < 1.

Graphs of the function Z(pr, Tr) are shown in Figure 2.7.

Exercise. It is required to determine the over-compressibility factor Z of a gaswith pcr = 4.6 MPa, Tcr = 190 K at p = 7.5 MPa and T = 288 K.

Solution. Let us calculate first the reduced parameters of state: pcr = 1.63;Tcr=1.52. From the plot in Figure 2.7 we determine Z ∼= 0.86.

There are a lot of approximating formulas to calculate the factor Z. In fact thecase is to approximate the equation of state (2.12). However, the properties of areal gas are so complicated that we do not have universal formulas appropriatefor all gases over the whole range of governing parameters. Therefore indifferent cases one should use different approximations. In particular, tosimulate processes in gas-pipelines we can use

Z(pr, Tr) ∼= 1 − 0.4273 · pr · T−3.668r (2.15)

or

Z(pr, Tr) ∼= 1 − 0.0241 · pr

θ, (2.16)

where

θ = 1 − 1.68Tr + 0.78T2r + 0.0107T3

r .

It should be particularly emphasized that Eqs. (2.15) and (2.16) are no morethan approximations of the state equation of a real gas.

2.7 Models of a Gaseous Continuum 41

Figure 2.7 Graphs of Z(pr , Tr) for natural gas.

Exercise. Determine the value of the over-compressibility factor of a gas withpcr = 4.6 MPa, Tcr = 190 K at p = 7.5 MPa and T = 288 K (see the previousexample).

Solution. The reduced parameters of state are:

pr = 7.5

4.6∼= 1.63; Tr = 288

190∼= 1.52.

With formulas (2.15) we obtain

Z = 1 − 0.4273 · 1.63 · 1.52−3.668 ∼= 0.850.

Similar calculation in accordance with Eq. (2.16) gives:

θ = 1 − 1.68 · 1.52 + 0.78 · 1.522 + 0.0107 · 1.523 ∼= 0.2861,

Z = 1 − 0.0241 · 1.63

0.2861∼= 0.863.


It is seen that the error in calculation with Eq. (2.15) is ≈ 2.3%, whereas withEq. (2.16) it is less ≈ 0.8%. But this does not mean that the Eq. (2.16) is moreprecise than Eq. (2.15) because both formulas are only approximations of thestate equation of a real gas.

2.8Model of an Elastic Deformable Pipeline

In the schematization of fluid and gas flow processes in pipelines models of apipeline are also used.

The simplest model of the pipeline is a model of non-deformable pipe, that is ofa cylinder with constant invariable internal diameter d0 and wall thickness δ.The external diameter D = d0 + 2δ of the pipeline in this model remains alsoconstant. The model of a non-deformable pipeline appears to be very usefulwhen researching many technological processes of oil and gas transportation.

However, in some cases, for example in studying the phenomenon ofhydraulic hammer, the model of a non-deformable pipe proves to be inadequateto perceive the point of the phenomenon and it is necessary to use the morecomplicated model of an elastic deformable pipeline.

Experience shows that the volume of the internal space of the pipeline variesinsignificantly with changes in temperature and pressure of the transportedmedium.

In order to account for the volume expansion of a pipeline when thetemperature T deviates from its nominal value T0 one can use

V(T) = V0[1 + 2αp · (T − T0)], (2.17)

where αp is the volume expansion factor of the metal from which the pipelineis produced (for steel αp ≈ 3.3 · 10−5 K−1).

Exercise. How does the volume of the internal space of a steel pipeline sectionwith D = 530 mm, δ = 8 mm, L = 120 km change during even cooling by5 K?

Solution. Using Eq. (2.17) we get:

V(T) − V0 =(

3.14 · 0.5142

4

)· 2 · (−5) · 3.3 · 10−5 · 120 000· ∼= −8.22,

that is the volume decreases by more than by 8 m3.The volume of the pipeline internal space is changed to a greater extent

with variation in the difference between the internal and external pressures.The simplest formula to calculate changes originated by this phenomenon hasbeen suggested by Joukowsky in his work ‘‘On hydraulic hammer in water-supplypipes’’ (1899). Its derivation is illustrated in Figure 2.8.

2.8 Model of an Elastic Deformable Pipeline 43

Figure 2.8 Derivation of the formulafor the cross-section area change in anelastic deformable pipeline.

The equation of equilibrium of the upper half of the pipe shell shown inFigure 2.8 by a heavy line, under the action of a pressure difference (p − p0)

and circumferential stresses σ resulting in the pipe metal has the followingform

(p − p0) · d = σ · 2δ. (2.18)

Hooke’s law of elasticity as applied to the deformed middle filament shown inFigure 2.8 by a dotted line gives the following relation

σ = E · π · (d − d0)

π · d0(2.19)

where E is the Young’s modulus of the pipe material (for steel E ≈ 2 · 105 MPa).Insertion of σ from Eq. (2.18) into Eq. (2.19) and further replacing factor

d by d0 due to the smallness of the wall thickness as compared to the pipediameter gives the dependence of the pipe diameter increment �d = d − d0

on the difference �p = p − p0 of the internal and external pressures

�d = d20

2Eδ· �p. (2.20)

Here d0 may be taken as the internal diameter of the pipe.Remark. When deriving Eq. (2.20) it was assumed that axial stresses in the

pipe are absent. Such a state of the pipe is called a plane-stress-state. But in manycases this assumption is not valid, in particular in steel welded pipelines usedin the oil industry there is a plane-deformable state in which radial expansion ofthe pipe generates axial stresses. In such cases Eq. (2.20) should be replacedby the more general formula

�d = d20(1 − ν2

P)

2Eδ· �p (2.21)

where νP is the Poisson ratio. However, the correction is insignificant sincefor steel pipelines ν2

P ≈ 0.078.


From Eq. (2.20) ensue two useful formulas: one for the increment �S ofthe area of the pipe cross-section and the other for the increment �V of thevolume of a pipeline section with length L

�S = πd30

4Eδ· �p, �V = πd3

0 L

4Eδ· �p (2.22)

and associated formulas

�S = πd30

(1 − ν2

P

)4Eδ

· �p, �V = πd30 L · (

1 − ν2P

)4Eδ

· �p. (2.23)

Exercise. It is required to calculate the increase in the diameter and volumeof a section of a steel pipeline with D = 530 mm, δ = 8 mm, L = 120 km afterbuild up of pressure by 5.0 MPa.

Solution. Using Eq. (2.16) we get

�d = 0.5142

2 · 2 · 1011 · 0.008· 5 · 106 ∼= 0.0004 m = 0.4 mm.

From Eq. (2.17) follows

�V = 3.14 · 0.5143 · 120000

4 · 2 · 1011 · 0.008· 5 · 106 ∼= 40 m3.

For simultaneous deviation of pressure and temperature from their nominalvalues p0 and T0 it is allowable to use the formula

V(T) = V0

[1 + 2αp · (T − T0) + d0

Eδ· (p − p0)

]. (2.24)

There are also more complicated models of pipelines taking into account theviscous-plastic properties of the pipe material. Such models are applicable forpipelines made from synthetic materials, e.g. from plastics.

45

3Structure of Laminar and Turbulent Flows in a Circular Pipe

In the first chapter the flow of a medium transported in a pipeline wasconsidered in the framework of a one-dimensional model, that is a model inwhich the flow is described by characteristics of the fluid velocity, density,pressure, temperature and others averaged over the pipe cross-section. Allcharacteristics of the flow depended only on the longitudinal coordinate xof the cross-section and time t. In such a description additional closurerelations reflecting the relations between the average parameters of the flowwere needed. For example, relation (1.26) connects the mechanical energydissipation with the average velocity of the flow, relation (1.29) expresses thetangential friction stress at the internal surface of a pipeline through averageparameters of the flow and so on.

It was noted above that to get a closure relation it is necessary to scrutinizenot only one-dimensional but also spatial flows taking place in the flow oftransported medium inside the pipeline. Let us consider in greater detailsuch flows with regard to the distribution of the parameters over the pipelinecross-section.

3.1Laminar Flow of a Viscous Fluid in a Circular Pipe

First let us consider the laminar flow of fluid in a circular pipe with radiusr0 (Figure 3.1). Such a flow has only one axial velocity component u(r),dependent on the radial coordinate r equal to the distance from a point underconsideration to the pipe wall.

Next separate inside the flow region a cylinder of arbitrary radius r (r ≤ r0)

and write the balance equation of all the forces acting on the cylinder

(p1 − p2) · πr2 = |τ(r)| · 2πrL

where τ(r) is the tangential friction stress at the lateral surface of the separatedcylinder. From here it follows that the absolute value of the tangential stress τ

between the fluid layers is proportional to the cylinder radius r

|τ(r)| = 1

2· �p

L· r. (3.1)


46 3 Structure of Laminar and Turbulent Flows in a Circular Pipe

Figure 3.1 Calculation of laminar fluid flow.

If we now replace τ by its expression through the velocity gradient du/ dr inaccordance with the law of viscous friction (2.2) and take into account thatτ(r) < 0, we get the differential equation

−µdu

dr= 1

2· �p

L· r (3.2)

for the velocity u(r). This equation should be solved for the boundary conditionu = 0 at r = r0, which is the so-called sticking condition (the velocity at theinternal surface of the pipe vanishes). As a result we obtain

u(r) = umax ·(

1 − r2

r20

); umax = �p · r2

0

4 µL. (3.3)

It is seen that the fluid velocity distribution over the pipe cross-section has aparabolic form with maximal value umax at the pipe axis, that is at r = 0.

The fluid flow rate Q , that is the fluid volume flowing through a unitcross-section of the pipe in unit time, is equal to

Q =∫ r0

0u(r)2πr dr = 2πumax

∫ r0

0r ·

(1 − r2

r20

)dr = 1

2πumaxr

20

or with regard to Eq. (3.3)

Q = r40π�p

8 µL. (3.4)

The relation (3.4) is called the Poiseuille formula. It gives the connectionbetween the flow rate of laminar fluid flow in a circular pipe and the pressuredrop causing the flow.

Let us introduce the mean flow-rate-velocity v which when multiplied by thepipe cross-section area gives the fluid flow rate, that is v = Q/πr2

0 . Then oneobtains very useful formulas

v = r20�p

8 µL= 2 · umax or v = d2�p

32 µL(3.5)

where d = 2r0 is the internal diameter of the pipe.

3.2 Laminar Flow of a Non-Newtonian Power Fluid in a Circular Pipe 47

With the help of these relations it is possible to calculate the tangential stressτw at the internal surface of the pipeline. From Eq. (3.1) follows

|τw| = 1

2· �p

L· r0.

Substitution of �p/L from Eq. (3.5) with regard to Eq. (1.29) yields

|τw| = r0

2· 32µv

d2= Cf · ρv2

2= λ

4· ρv2

2

From here follows Stokes formula (1.30) for the factor λ of hydraulic resistancein laminar flow of a viscous incompressible fluid in a circular pipe

λ = d

4· 32 µv

d2· 8

ρv2= 64

vd/(µ/ρ)= 64

Re. (3.6)

Exercise. Calculate the hydraulic resistance factor of oil λ (ν = 25 cST) inlaminar flow in a circular pipe with diameter 50 mm and flow rate 1 l s−1.

Solution. First determine the mean velocity of the flow

v = Q

πd2/4= 0.001

3.14 · 0.052/4∼= 0.51 m s−1.

Then calculate the Reynolds number

Re = vd

ν= 0.51 · 0.05

25 · 10−6= 1020

Since 1020 < Recr = 2300, the flow is laminar. In accordance with Eq. (3.6)the hydraulic resistance factor is equal to

λ = 64

Re= 64

1020∼= 0.063.

3.2Laminar Flow of a Non-Newtonian Power Fluid in a Circular Pipe

Just as we considered the laminar flow of a Newtonian fluid so we can considerthe laminar flow of a non-Newtonian Ostwald power fluid in a circular pipe (seeChapter 2). The tangential stress τ for this fluid is related to the shear rateγ = du/ dr by the dependence (2.9)

τ = k ·∣∣∣∣ du

dr

∣∣∣∣n−1

· du

dr

so that the equilibrium Eq. (3.2) takes the form

−k ·∣∣∣∣ du

dr

∣∣∣∣n−1

· du

dr= 1

2· �p

L· r


Since du/dr < 0, the last relation transforms to

du

dr= −

(�p

2kL

)1/n

· r1n .

Integration of this equation with the sticking boundary condition u(r0) = 0 atthe pipe wall gives the velocity distribution

u(r) = − n

n + 1·(

�p

2kL

)1/n

·(

rn+1

n − rn+1

n0

). (3.7)

The maximal velocity of the flow umax is achieved at the pipe axis r = 0 as inthe case of a viscous fluid and is equal to

umax = n

n + 1·(

�p

2kL

)1/n

· rn+1

n0 . (3.8)

The flow rate of the fluid Q is calculated by the formula

Q = 2π

∫ r0

0r · u(r) dr

= 2π · n

n + 1·(

�p

2kL

)1/n

·∫ r0

0r ·

(r

n+1n

0 − rn+1

n

)dr

Performing integration one gets after some algebra

Q = πr03n

3n + 1·(

r0�p/L

2k

) 1n

. (3.9)

It is reasonable that at n = 1, k = µ Eq. (3.9) converts to Eq. (3.4) known as thePoiseuille law (Wilkenson, 1960).

Introducing the mean flow-rate-velocity v as in the previous section and thegeneralized Reynolds number Re∗ according to the relations

v = Q

πr02

and Re∗ = v2−n · dn0

k/ρ

where d0 = 2r0, Eq. (3.9) could be written in the habitual form of theDarcy–Weisbach law

�p

L= λ · 1

d0· ρv2

2

where the hydraulic resistance factor λ is

λ =8 ·

(6n + 2

n

)n

Re∗. (3.10)

3.3 Laminar Flow of a Viscous-Plastic Fluid in a Circular Pipe 49

In the same way Eq. (3.10) at n = 1, k = µ transforms to the Stokes Eq. (1.30)given earlier.

To determine the rheological properties of oil and oil products special devicescalled viscometers are often used. The most widespread are capillary viscometers.Operation of all capillary viscometers is based on the determination of the timeof the outflow of a fixed portion of the fluid under test, from the device chamberthrough a narrow cylindrical tube (capillary). This time is calculated with theuse of Eqs. (3.4) and (3.9) replacing in them the pressure gradient �p/L by ρg,where g is the acceleration due to gravity. The flow rate Q in the viscometer ofthe Ostwald power fluid under consideration takes the following form

Q = πr03n

3n + 1·( r0 · ρg

2k

) 1n = πr0

3n

3n + 1·(

r0 · g

2 · k/ρ

) 1n

. (3.11)

Exercise. In order to determine the properties of oil experiments are carriedout on a free outflow of a portion (100 ml) of oil from the viscometer chamber.In the first experiment the outflow flows through a cylindrical capillary withinternal radius 1 mm and in the second experiment through a similar capillarywith internal radius 1.5 mm. In the first case the time of fluid outflow is 1000 s,in the second 180 s. The flow of the oil is modeled by the law of a power fluid.It is required to determine the constants n and k/ρ of the power fluid model.

Solution. From Eq. (3.11) it follows that the ratio of flow rates in bothcases is Q1/Q2 = (r1/r2)

3+1/n. Since the ratio is inversely proportional to thetimes of the fluid outflows, we use the equation to determine n, namely180/1000 = (2/3)3+1/n. The solution of this equation gives n ∼= 0.81.

Using further the results of the first experiment we get

0.0001

1000= 3.14 · 0.81 · 0.0013

3 · 0.81 + 1·(

9.81 · 0.002

4 · k/ρ

)1/0.81

,

from which we get k/ρ ∼= 0.92 · 10−6 m2 s−1.19.

3.3Laminar Flow of a Viscous-Plastic Fluid in a Circular Pipe

Consider the laminar flow of another non-Newtonian fluid – viscous-plasticBingham fluid (Wilkenson, 1960) (see Chapter 2 and Figure 3.2).

The tangential stress τ in this fluid is related to the shear rate γ = du/dr bythe dependence

τ = τ0 + µdu

dr, (du/dr > 0, τ ≥ τ0) (3.12)

where τ0 is the limit shear stress. Due to the existence of such stress the flowdoes not begin immediately after application of the pressure difference to the


Figure 3.2 A scheme of viscous-plastic fluid flow.

pipe ends but only when this difference exceeds the shear strength, namelywhen it obeys the following inequality

�p · πr02 > 2πr0 L · τ0 or

r0 · �p/L

2τ0> 1. (3.13)

If the condition (3.13) is fulfilled then, near the internal surface of the pipe(a < r < r0), the flow of the fluid starts. On approaching the pipe center wherethe shear rate decreases the tangential stresses are also reduced and at somedistance a from the pipe axis, just where τ(a) = τ0, the fluid begins to moveas a rigid core. Therefore the region (0 ≤ r ≤ a) of the flow is called the flowcore.

It is evident that the following relations are valid

τ(r) = τw · r

r0at a ≤ r ≤ r0,

τ(a) = τ0 = τw · a

r0and

a

r0= τ0

τw= 2τ0

r0 · �p/L.

The velocity distribution in the annular region a ≤ r ≤ r0 satisfies the equation

µdu

dr= τ(r) − τ0 = τ0 ·

( r

a− 1

).

Integration of this equation over r from r = r0 to an arbitrary r with regard toboundary condition u(r0) = 0 yields

u(r) = r0τ0

2µ

[(r/r0)

2 − 1

a/r0− 2 · r

r0+ 2

]. (3.14)

In particular the core velocity u(a) is equal to

u(a) = − r0τ0

2µ· (1 − a/r0)

2

a/r0. (3.15)

The flow rate Q of the fluid can be obtained using Eqs. (3.14) and (3.15)

−Q = 2π

∫ r0

ar · u(r) dr + πa2 · u(a).

3.4 Transition of Laminar Flow of a Viscous Fluid to Turbulent Flow 51

Simple rearrangements lead to the following expression

Q = πr04�p/L

8µ

[1 − 4

3·(

2τ0

r0 · �p/L

)+ 1

3·(

2τ0

r0 · �p/L

)4]

. (3.16)

As would be expected this relation at τ0 = 0 becomes the known Poiseuilleformula (3.4) for the flow rate of fluid in a circular pipe (Wilkenson, 1960).

If we introduce two dimensionless parameters Re = v·2r0ν

, the Reynoldsnumber and I = τ0·2r0

µv , the Ilyushin number and take into account the relation2τ0/(r0 · �p/L) = 8I/λRe, the equality (3.16) could be represented as

λ = 64

Re· 1

1 − 4/3 · (8I/λRe) + 1/3 · (8I/λRe)4. (3.17)

If τ0 = 0, then I = 0 and consequently λRe = 64, that is Eq. (3.17) converts tothe known Stokes formula (1.30) for laminar flow of a viscous fluid. In thegeneral case the product λRe depends on the Ilyushin number I. To determinethis product one should resolve Eq. (3.17) with respect to λRe for each value ofthe parameter I (Romanova, 1977).

3.4Transition of Laminar Flow of a Viscous Fluid to Turbulent Flow

With increase in the velocity of viscous fluid flow in a pipe the laminar flowloses hydrodynamic stability and changes into turbulent flow. This flow ischaracterized by fluctuating motions of the fluid, generation and developmentof eddies and intensive mixing. A lot of theoretical investigations performedby many outstanding mathematicians and physicists have been devoted tothe problem of stability of laminar fluid flows and the determination ofthe conditions and criteria of the transition from laminar to turbulent flowregimes.

The criterion of transition of laminar flow into a turbulent one is theReynolds number Re, representing a dimensionless parameter formed by thedimension parameters characterizing the flow and the pipeline: Re = vd/ν, sothat at Re < Recr = 2300 the flow is laminar whereas at Re > Recr = 2300 it isturbulent.

The latter conclusion is true when the critical velocity vcr of the flow isdetermined only by flow parameters such as d, ρ and µ(ν = µ/ρ). In fact, theconditions of transition from laminar to turbulent flow are in many respectsdetermined also by such parameters of the pipeline as the roughness of itsinternal wall surface �. If we accept that vcr = f (d, µ, ρ, �) then, in accordancewith the I-theorem the condition of laminar flow transition to turbulent flowtakes the form

vcr · d

µ/ρ= f

(�

d

)or Recr = f (ε)


where ε is the relative roughness of the pipeline internal wall surface. Thusat Re < Recr(ε) the fluid flow is laminar, while at Re > Recr(ε) the flow isturbulent. The presence of the additional parameter ε besides the Reynoldsnumber means that there is not a single-valued determined boundary of thetransition from laminar to turbulent flow, since the critical Reynolds numberdepends on the degree of preparation of the pipeline and the fluid for the testexperiments. It is known, for example, that in this manner it was possible tolengthen the transition of laminar flow to turbulent flow up to a value of thecritical Reynolds number equal to 6000–12 000.

The hydraulic resistance factor λ in the transition region does not havestable values, it is only characterized by drastic lowering. Some formulasfor λ in this region will be considered further. The factor λ takes a stablevalue in establishing the developed turbulent flow in the pipe, that is atRe > 10 000.

3.5Turbulent Fluid Flow in a Circular Pipe

Consider turbulent flow of a viscous fluid developed in a circular pipe withradius r0 (Figure 3.3). To describe such a flow let us use the average velocityu(r) representing the true time-averaged velocity of fluid particles passingthrough a considered point. It is assumed that the resulting velocities areparallel to the pipe axis and independent of the radial coordinate r, that is ofthe distance between the considered point of the cross-section and the pipeaxis.

Similarly, average tangential stresses τ(r) between the fluid layers movingwith average velocities u(r) are introduced (Figure 3.3). These stresses are alsocalled Reynolds stresses, named in honor of the famous English engineer O.Reynolds who contributed greatly to the development of turbulence theory.The average stress τ(r) represents the ratio of the friction force actingbetween macro-layers of turbulent flow separated by a surface element tothe area of this element. As in laminar flow, turbulent tangential stressesτ(r) are suggested to be proportional to the gradients du/ dr of average flowvelocities

τ(r) = µt · du

dr. (3.18)

However, different from laminar flow, the factor µt in this relation representsnot the intrinsic viscosity of the fluid but the so-called turbulent dynamicviscosity, dependent on the structure and mixing intensity of the fluid layersrather than on fluid properties.

The turbulent dynamic viscosity µt and the turbulent kinematic viscosity νt =µt/ρ of the turbulent flow is caused not by molecular friction of the separatedfluid layers, as in laminar flow, but by large-scale fluctuations and momentum

3.5 Turbulent Fluid Flow in a Circular Pipe 53

Figure 3.3 The calculation of fluid turbulent flow.

transport with eddies from one macro-layer to another. The transport of mo-mentum is perceived as a friction force acting between these layers. When thelayers are moving with equal velocities, that is du/ dr = 0, then the passing ofthe momentum from one layer to another is compensated by equal momentumfrom another layer, therefore τ = 0. If one layer moves faster than another, thatis du/ dr > 0, then there appears between the layers a friction force acceleratingthe backward layer and retarding the leading one. This means that the leadinglayer loses more momentum than it obtains from the backward layer. Thus theintensity of momentum exchange between the layers depends on the regimeof turbulent flow rather than on the viscosity of the fluid characterized by thefactors µ or ν. Hence it follows that the turbulent viscosity µt or νt is not fluidconstant as its molecular analog but depends on parameters of turbulent flow.

Let the law of flow have the following form

1

ρ· τ = νt · du

dr(3.19)

It makes sense to suggest that the turbulent viscosity νt determined by thestructure of turbulent flow be dependent on the parameters of this flow. Inaccordance with the brilliant idea of Karman it can be accepted that

νt = f

(ν,

∣∣∣∣ du

dr

∣∣∣∣ , ∣∣∣∣ d2u

dr2

∣∣∣∣) (3.20)

that is, the turbulent viscosity νt of the flow depends on the molecularviscosity ν of the fluid itself and parameters absolute values of the first u′ andsecond u′′ derivatives of average velocity with respect to the radial coordinate)characterizing the flow.

Since among arguments of the function f there are only two dimensionalindependent variables ν and u′′

[ν] = L

T2 , [|u′′|] = 1

L · T

the dimension of the third argument |u′| could be expressed through thedimensions of the two others as follows (Lurie and Podoba, 1984):

[u′] = [ν]1/3 · [|u′′|]2/3 = 1

T.


Then owing to the �-theorem the number of arguments of the function f canbe reduced to one and the dependence (3.20) takes the following dimensionalform

νt/ν = f

( |u′|3ν · |u′′|2

)⇒ νt = ν · f

( |u|3ν · |u′′|2

). (3.21)

The formula (3.21) shows that the turbulent viscosity νt is equal to themolecular viscosity ν of the fluid multiplied by the dimensionless factorf dependent only on one dimensionless parameter η = |u′|3/(ν · |u′′|2).Experiments testify, however, that in most of the pipe cross-section, thatis in the flow core, except for the narrow wall layer the turbulent viscosity ispractically independent of the molecular viscosity of the fluid. This is becausethe turbulent viscosity is determined by exchange of momentum between fluidlayers due to large-scale eddies rather than by molecular friction. ThereforeEq. (3.21) should have such a structure that the molecular viscosity wouldhave no impact over almost all the cross-section of the flow. The last could beachieved by assuming the function f to be linear, in other words, if f (η) = κ · η,where κ is a constant called the Karman constant. With regard to the acceptedassumption Eq. (3.21) for turbulent viscosity can be rewritten as

νt = ν ·(

κ2 |u′|3ν · |u′′|2

)= κ2 · |u′|3

(u′′)2.

The law of turbulent friction (3.19) is expressed by

1

ρ· τ = κ2 |u′|3

(u′′)2· u′ (3.22)

called the Karman formula (Loizyanskiy, 1987). The last equation representsthe basic equation of the phenomenological, that is resulting from abstractreasoning though adequate for the phenomena under consideration, Karmantheory. The constant κ in Eq. (3.22) has been shown by a plethora of experi-ments to be approximately equal to 0.4 and is a universal constant of the modelin the sense that it is the same for all regimes of turbulent flow of fluids inpipes. The relation (3.22) differs significantly from the analogous relation (2.2)for laminar flows.

Let us determine now the distribution u(r) of average velocities of fluidturbulent flow in a circular pipe. First note that tangential stresses τ(r) as inlaminar flow are linearly distributed over the radius

τ(r) = −1

2· �p

L· r

since the balance of forces acting on an arbitrarily separated fluid cylinder (seeFigures 3.2 and 3.3) is independent of the flow regime (laminar or turbulent).Introduce into the consideration the tangential stress τw caused by the friction


at the pipe walls

τw = τ(r0) = −1

2· �p

L· r0.

Then τ(r) could be expressed through τw

τ(r) = τw · r

r0. (3.23)

Substitution instead of τ(r) from Eq. (3.22) leads to the differential equation

κ2 |u′|3u′

u′′2 = 1

ρτw · r

r0(3.24)

for the velocity distribution u(r) over the radius.The quantity (τw/ρ)1/2 has the dimensions of velocity. It is commonly called

the dynamic velocity and specified by u∗. In fact it has the sense of a frictionstress at the pipe wall since ρu2∗ = |τw|. In addition the dynamic velocity isrelated to the above introduced factor λ of hydraulic resistance

u2∗ = |τw|

ρ= 1

2Cf · v2 = λ

8· v2 ⇒ u∗

v=

√λ

8

where v is the fluid velocity averaged over the cross-section. Since the factorλ is small (λ ≈ 0.01–0.03), the dynamic velocity u∗ ∼= 0.05 · v, from whichfollows that the dynamic velocity is 20–25 times smaller than the average flowvelocity.

In terms of dynamic viscosity the basic equation (3.24) attains a morecompact form

κ2 |u′|3u′

u′′2 = −u2∗ · r

r0, (τw < 0). (3.25)

Solve this equation for a flow in a circular pipe (Figure 3.3) u′ < 0 and u′′ < 0,thus

κ2 |u′3|u′

u′′2 = −κ2 u′4

u′′2 = −u2∗ · r

r0.

From here it ensues that

− u′′

u′2 = κ

u2•·√

r0

ror

d

dr

(1

u′

)= κ

u2∗·√

r0

r.

Repeated integration gives

u(r) = u∗κ

·[√

r

r0+ C1 + C2 · ln

∣∣∣∣√ r

r0− C2

∣∣∣∣] (3.26)

where C1 and C2 are constants of integration.


To determine C1 and C2 one needs two boundary conditions at the internalpipe wall (Lurie and Podoba, 1984). The first condition is evident. It is thesticking condition in accordance with which the velocity uw of fluid particlesshould vanish at the pipe walls

uw = u(r0) = 0. (3.27)

Substituting r = r0 in Eq. (3.26) and taking u = 0 we obtain

0 = u∗κ

· [1 + C1 + C2 · ln |1 − C2|].

Subtracting further the resulting equality term-by-term from Eq. (3.26) we get

u(r) = u∗κ

·[√

r

r0− 1 + C2 · ln

∣∣∣∣√r/r0 − C2

1 − C2

∣∣∣∣] . (3.28)

The second condition is more complicated and was not met in the modelof laminar flow. The problem is that the differential equation (3.25) for theturbulent model has higher order than the analogous differential equation (3.2)for turbulent flow. Therefore to solve it one needs an additional boundarycondition reflecting the interaction of turbulent flow with the pipe walls. Sincethis condition models the fluid flow in a narrow wall layer it has to connectparameters u′

w and u′′w of the turbulent flow at the pipe wall with the molecular

viscosity ν, whose influence is strong in this layer, and the smoothness of theinternal surface of the pipe characterized by its absolute roughness �. In otherwords the missing boundary condition should be expressed by the relation

G(|u′w|, |u′′

w|, ν, �) = 0 (3.29)

in which the subscript ‘‘w’’ indicates that the corresponding derivatives arecalculated at points on the internal pipe surface. The dimension analysis asapplied to Eq. (3.29) allows us to rewrite it in dimensionless form

G

(ν · u

′′2w

|u′w|3 ,

� · |u′′w|

|u′w|

)= 0.

Having resolved the last expression with respect to the first argument, we get

ν · u′′2w

|u′w|3 = g

(� · |u′′

w||u′

w|)

.

For small values of the roughness the right-hand part of this equation couldbe expanded into a Taylor series leaving in it only the first two terms

ν · u′′2w

|u′w|3 = g0 + g1 · � · |u′′

w||u′

w| (3.30)


where g0 and g1 are dimensionless constants. If we now take into account thatat r = r0 in accordance with Eq. (3.25) the relation

κ2 · u′4w

u′′2w

= u2∗

takes place, meaning that, at the pipe walls, |u′′w| = κ · u

′2w/u∗, the condi-

tion (3.30) simplifies to

ν

|u′w|3 · κ2 · u

′4w

u′′2w

= g0 + g1 · �

|u′w| · κ · u

′2w

u∗

and takes the final form (u′w < 0):

ν · u′w = − k · u2∗

1 + a · (u∗ · �/ν)(3.31)

where k = g1/κ2 and a = −g2/κ are dimensionless constants. These constants

are universal like the Karman constant κ, that is, they do not depend on concreteflow and are given once and for all. In other words they are phenomenologicalconstants of the model. A large body of calculation results for different typesof flows correlated to each other gives k = 28 and a = 0.31 (Lurie and Podoba,1984). Hence, the theory of fluid turbulent flow in a circular pipe is based onthree phenomenological constants κ ∼= 0.4; k ∼= 28 and a ∼= 0.31.

In the particular case of an ideal smooth internal surface of the pipe (� ≈ 0),the boundary condition (3.31) reduces to

νu′w = −k · u2

∗. (3.32)

Thus the boundary condition (3.31) permits us to obtain the second integrationconstant C2 in expression (3.28) for the velocity u(r). As a result we have

u′ = u∗κ

·[

1

2√

r · r0+ C2√

r/r0 − C2· 1

2√

r · r0

],

u′w = u′(r0) = u∗

2r0 · κ· 1

1 − C2.

Insertion of the calculated derivative into condition (3.31) yields the equation

− ν · u∗2r0 · κ

· 1

1 − C2= k · u2∗

1 + a · u∗�/ν,

which gives

C2 = 1 + 1 + a · u∗ · �/ν

2kκ · r0u∗/ν


or, in dimensionless form

C2 = 1 + 1 + a · ε · Re · u∗/v

kκ · Re · u∗/v(3.33)

where ε = �/d is the relative roughness; Re = vd/ν the Reynolds number; vthe mean flow velocity; d = 2r0 the pipeline diameter. At the same time notethat for turbulent flows with Re > 104, ε < 10−3, u∗/v ≈ 10−2 the constant C2

is close to one,

C2 ≈ 1 + 1 + 0.31 · 10−3 · 104 · 10−2

28 · 0.4 · 104 · 10−2∼= 1.001. (3.34)

Hence, Eq. (3.28) with constant (3.33) gives the distribution of the averagevelocity of turbulent flow in a circular pipe under the condition that thedynamic velocity u∗ is known. The latter in its turn can be expressed throughthe average flow velocity v and the hydraulic resistance factor λ with Eq. (3.24)as u∗ = v · √

λ/8.Figure 3.4 shows dimensionless turbulent velocity profiles u(r)/umax related

to the maximal value of the fluid velocity at the pipe axis. The lower curvecorresponds to Re = 23 000, the top curve to Re = 3 200 000, the middle one tointermediate values of the Reynolds number (Loitzyanskiy, 1987). The dottedcurve depicts the parabola (3.3) giving the velocity distribution in laminarflow regime. The comparison of laminar and turbulent velocity distributionsshows that the turbulent velocity profile has more plane form and the greater theReynolds number the more the plane become curves. For the laminar flowregime in accordance with Eq. (3.5) umax/v = 2. For the turbulent regime thisratio is far less. In general it depends on the numbers Re, ε and the averageratio is equal to umax/v = 1.15 − 1.25.

The integration constant C2 in Eq. (3.33) can be represented as

C2 = 1 + 1 + a · ε · Re · √λ/8

kκ · Re · √λ/8

. (3.35)

Figure 3.4 Dimensionless averagevelocity profiles in turbulentflows (Loitzyanskiy, 1987).


On the other hand the average velocity is by definition equal to

v = Q

π · r20

= 1

π · r20

·∫ r0

02πr · u(r) dr = 2

r20

·∫ r0

0r · u(r) dr.

Substituting here the distribution (3.28) and taking into account theformula (3.35) for C2 we get the dependence of the hydraulic resistancefactor λ on the Reynolds number, that is in fact on the average velocity v, andon the relative roughness ε of the internal surface of the pipe

v = 2u∗κr2

0

∫ r0

0r

[√r

r0− 1 + C2 · ln

C2 − √r/r0

C2 − 1

]dr

= 2u∗κ

·∫ 1

0η

[√η − 1 + C2 · ln

C2 − √η

C2 − 1

]dη (3.36)

where η = r/r0. Calculating the integral (3.36) with regard to remark (3.34),we obtain√

8

λ= 1

κ·(

lnkκ · Re · √

λ/8

1 + a · ε · Re · √λ/8

− 137

60

). (3.37)

Inserting in Eq. (3.37) the numerical values of the constants k, κ and a we getthe equation for the dependence of λ on the numbers Re and ε

1√λ

= 0.88 · lnRe · √

λ

1 + 0.11 · ε · Re · √λ

− 0.8 (3.38)

called the universal resistance law.

1. If we ignore the effect of roughness, that is we assume0.11 · ε · Re · √

λ 1, e.g. 0.11 · ε · Re · √λ < 0.11 or Re · √

λ < ε−1,then the equation for λ becomes

1√λ

= 0.88 · ln(Re · √λ) − 0.8. (3.39)

Its approximate solution takes the form

λ = 0.31644√

Re(3.40)

mentioned above and called the Blasius formula. The estimation ofallowable Reynolds numbers can be conducted as follows

Re · √λ ∼= 0.56 · Re7/8 <1

ε⇒ Re < 1.93 · ε−8/7 ≈ 2 · ε−1.143,

therefore Eq> (3.39) is true when Re < 2 · ε−1.143.2. At very large Reynolds numbers (Re > 105) the equation for λ acquires

the form1√λ

= 0.88 · ln1

0.11 · ε− 0.8


where λ is independent of Re. We have

1√λ

=(

0.88 · ln1

0.11+ 0.88 · ln

1

ε− 0.8

)−2

⇒

λ ∼=(

0.88 · ln1

ε+ 1.14

)−2

. (3.41)

Exercise 1. It is required to calculate the hydraulic resistance factor λ in theflow of diesel fuel with ν = 4 cSt in a pipeline with d = 500 mm, � = 0.25 mm.The flow rate of the fluid is Q = 1000 m3 h−1.

Solution. Determine the average velocity v of the flow

v = 4Q

πd2= 4 · 1000/3600

3.14 · 0.52= 1.415 m s−1

the Reynolds number

Re = vd

ν= 1.415 · 0.5

4 · 10−6= 176 875

and the relative roughness

ε = �

d= 0.25

500= 0.0005

As a result we get the transcendental equation for λ

1√λ

= 0.88 · ln176 875 · √

λ

1 + 0.11 · 0.0005 · 176 875 · √λ

− 0.8

We look for the solution of this equation by the method of successiveapproximations. Consider the function

�(λ) = 1√λ

− 0.88 · ln176 878

√λ

1 + 9.728 · √λ

+ 0.8

representing the difference in both parts of the resulting equation. We have

λ = 0.02, �(0.02) = −0.279;

λ = 0.019, �(0.019) = −0.086;

λ = 0.018, �(0.018) = 0.123;

λ = 0.0185, �(0.0185) = 0.016;

λ = 0.0186, �(0.0186) = −0.004.

It is seen that λ ∼= 0.0186.

Exercise 2. It is required to calculate the hydraulic resistance factor λ in theflow of benzene with ν = 0.6 cSt in a pipeline with d = 361 mm, � = 0.2 mm.The flow rate of the fluid is Q = 500 m3 h−1.


Solution. Determine the average velocity of the flow

v = 4Q

πd2= 4 · 500/3600

3.14 · 0.3612= 1.358 m s−1,

the Reynolds number

Re = vd

ν= 1.358 · 0.361

0.6 · 10−6= 817 063.

and the relative roughness

ε = �

d= 0.2

361= 0.00055.

As a result we get the transcendental equation for λ

1√λ

= 0.88 · ln817 063 · √

λ

1 + 49.4 · √λ

− 0.8.

We look for the solution of this equation by the method of successiveapproximations. Consider the function

�(λ) = 1√λ

− 0.88 · ln817 063

√λ

1 + 49.4 · √λ

+ 0.8,

representing the difference between both parts of the resulting equation. Wehave

λ = 0.02, �(0.02) = −1.36;

λ = 0.018, �(0.018) = −0.17;

λ = 0.016, �(0.016) = 0.29;

λ = 0.017, �(0.017) = 0.049;

λ = 0.0172, �(0.0172) = 0.003.

It is seen that λ ∼= 0.0172.

Turbulent Flows of Non-Newtonian Fluids

For the turbulent flow of a power fluid (see Section 2.6) the following expressionfor the universal resistance law has been obtained by Dodge and Metzner, 1959

1√λ

= 0.88

n3/4ln

[Re ·

(λ

4

)1−n/2]

− 0.4

n1.2(3.42)

where n is the exponent in the Ostwald rheological law. This equation maybe obtained on the basis of the Karman model (3.25), if in the boundarycondition (3.32) we take the factor k to be dependent on the exponent n, that isk = k(n). As is known k is equal to 28 for a Newtonian viscous fluid.


Romanova (1989) gave the resistance law for a power fluid in the form

1√λ

= 0.88

nln

[k(n)Re ·

(λ

8

)1−n/2]

− 2.83,

where k(n) = 21−n/2 · exp[n · (2.83 − 0.2/n1.2)/0.88

]. In the same publication

the following formulas were suggested as approximate solutions of the aboveequation

λRe1/3 = 0.698 · n − 1.94 · 10−2 at 0.2 ≤ n ≤ 0.5,

λRe1/4 = 0.353 · n − 3.80 · 10−2 at 0.5 < n ≤ 1.25,

λRe1/5 = 0.234 · n − 5.13 · 10−2 at 1.25 < n ≤ 2.0.

In all these formulas the generalized Reynolds number is defined asRe = v2−n · dn/k, where k is the kinematic consistency.

The universal resistance law for a viscous-plastic fluid was suggestedby Potapov (1975)

1√λ

=(

1 − 8He

λ Re2

)·[0.88ln(Re

√λ) − 0.8

]+ 2.76 · 8He

λ Re2(3.43)

where Re = vd/ν is the Reynolds number and He = τ0d2/ρν2 is the Hedstroemnumber.

3.6A Method to Control Hydraulic Resistance by Injection of Anti-Turbulent Additiveinto the Flow

Friction losses are the key reason for electric energy expenditure on fluid andgas pumping along pipelines. They are caused by the forces of internal frictionbetween the layers of the moving fluid. In laminar and turbulent flows thereoccurs the so-called dissipation of mechanical energy of ordered motion andits transition into the energy of chaotic motion of the fluid particles, in otherwords into heat. For turbulent flows this transition has a multi-stage character.The mechanical energy of average motion is transformed first into the energyof large-scale eddies of the turbulent medium, then into the energy of thefluctuation motion of small-scale eddies and finally, due to viscosity forces,into the heat energy of the fluid. Therefore engineers and scientists involvedin the problem of pipelines have for a long time been interested in methodsof governing the turbulent flow structure with the aim of reducing energylosses.

One such method discovered by the English scientist Toms in the late 1940sconsists in injection into the turbulent flow of special high-molecular weightadditives to lower the hydraulic resistance. This effect is called by the name ofthe discoverer, the Toms effect.

3.6 Controlling Hydraulic Resistance by Injection of Anti-Turbulent Additive into the Flow 63

The mechanism of operation of all varieties of anti-turbulent additives isbased on damping turbulent fluctuations near the internal surface of thepipeline by interaction of the long-length molecules of the additive withturbulent eddies generated near the pipeline wall. As a rule this effect isachieved with very small concentrations of additives, measured commonlyin ppm (parts per million of the fluid volume to which the additive isadded).

Owing to the damping of the near-wall turbulence there is a reduction inthe flow hydraulic resistance caused by the pipeline wall. Hence, increasedpumping efficiency is achieved with conservation of the pressure drop orpressure lowering at pumping stations. The decrease in hydraulic resistancereduces expenditure on electric power by 20–60%.

The best known anti-turbulent additives for oil products are CDR producedby the American company Dupon-Conoco and NECCAD-547 produced by theFinnish company Neste. These products are based on hydrocarbons. The firstis equally suitable for benzene and diesel fuel pumping while the second isrecommended primarily for diesel fuels. Both anti-turbulent additives wereput through production tests on pipelines in Russia.

The use of anti-turbulent additives has some specific restriction: duringprolonged action of the additives in turbulent flow they become degraded;their degradation is especially great when passing through pumps. Therefore,when using additives it is necessary to inject a portion of fresh additives intothe flow after each pumping station. It is rational to use anti-turbulent additivesto build up the carrying capacity of certain pipeline sections, first the limitingones.

All anti-turbulent additives reduce the hydraulic resistance factor λ. Tocalculate this factor the universal resistance law (3.37) containing the constantk related to the turbulent flow interaction with pipeline walls was used (seeboundary conditions (3.31) and (3.32)). It was found that k ∼= 28.

The effect of the anti-turbulent additive is that it changes the intensity of the wallturbulence, that is the additive acts on the magnitude of the constant k. Therefore itis reasonable to accept as a model of turbulent flow with anti-turbulent additivea model with variable constant k (Ishmuchamedov et al., 1999) dependent onthe concentration of anti-turbulent additive θ. Thus the universal quantity kwhich was taken earlier as constant would be, in the presence of anti-turbulentadditive, a function of θ, that is k = k(θ). In the absence of anti-turbulentadditive (θ = 0) then k(0) = 28.

The results of tests of anti-turbulent additive CDR have given the dependencek(θ) shown in Table 3.1.

Table 3.1 Dependence k(θ) for CDR.

θ, ppm 20 30 40 50 60 70 80 90k(θ) 61.4 95.1 143 187 249 276 340 380


Table 3.2 Dependence k(θ) for NECCAD-547.

θ, ppm 40 60 100 180k(θ) 50 75 150 340

The dependence k(θ) for anti-turbulent additive NECCAD-547 is shown inTable 3.2.Remark. The data cited in Tables 3.1 and 3.2 could be improved by changing

the anti-turbulent additive composition.

Exercise 1. The pumping of diesel fuel with anti-turbulent additive CDR withθ = 40 ppm is conducted at Reynolds number 40 000. It is required to calculatethe hydraulic resistance factor λ.

Solution. The formula (3.39) at Re = 40 000 and k(40) = 143 taken fromTable 3.1 gives for λ the following transcendental equation

1√λ

= 0.88 · ln(143 · 40 000 · √λ) − 3.745.

Its solution, found by the method of successive approximations, yieldsλ = 0.0153. This value is significantly less than 0.0224 corresponding tothe value of λ in the case of anti-turbulent additive absence in the flow of oilproduct at the same Reynolds number. The effect is about 31.7%.

Exercise 2. The pumping of diesel fuel with anti-turbulent additive NECCAD-547 with θ = 180 ppm is conducted at Reynolds number 40 000. It is requiredto calculate the hydraulic resistance factor λ.

Solution. The formula (3.39) at Re = 40 000 and k(180) = 340 taken fromTable 3.2 gives for λ the following transcendental equation

1√λ

= 0.88 · ln(340 · 40 000 · √λ) − 3.745.

Its solution, found by method of successive approximations, yields λ = 0.0129.This value is significantly less than 0.0224 corresponding to the value of λ inthe case of anti-turbulent additive absence in the flow of oil product at thesame Reynolds number. The effect is about 42.4%.

In order to select the necessary concentration θ of anti-turbulent additiveone can proceed as follows (Ishmuchamedov et al., 1999). From Eq. (3.39) fork(θ) follows the relation

k(θ) = 1

Re · √λ

· e1+3.745·√λ

0.88·√λ (3.44)

which permits the determination of k(θ) for given λ. Then with the help ofTables 3.1 and 3.2 we obtain the concentration θ of the anti-turbulent additive

3.7 Gravity Fluid Flow in a Pipe 65

in the oil product. Multiplying the latter by the total volume of the fluid to bepumped we get the required amount of the anti-turbulent additive.

Exercise 3. It is required at a given pressure resource to increase by 30% thecarrying capacity of an oil-pipeline with D = 377 mm, δ = 8 mm pumpingdiesel fuel with νd = 9 cSt and flow rate 450 m3 h−1. Determine the amountof anti-turbulent additive CDR needed to do this.

Solution. Let us calculate the initial values of the pumping rate v0, Reynoldsnumber Re and the factor of hydraulic resistance λ0:

v0 = 4Q/S = 4 · 450/(3600 · 3.14 · 0.3612) = 1.221 m s−1,

Re0 = v0d/νd = 1.221 · 0.361/(9 · 10−6) = 48 976,

λ0 = 0.0213.

Since the carrying capacity must be increased by 30%, the new values of thepumping rate v and the Reynolds number Re will be

v = 1.3 · v0∼= 1.587 m s−1, ⇒ Re = 1.3 · Re0 = 63 669.

Due to the invariability of the pressure resource it should be

λ0(Re0, 0) · v20 = λ(Re, θ) · v2.

This relation gives a new value of λ

λ = λ0 · (v0/U)2 = 0.0213 · (1/1.3)2 = 0.0126.

The constant k(θ) is determined by Eq. (3.44)

k(θ) = 1

63 669 · √0.0126

· e1+3.745

√0.0126

0.88·√0.0126 ≈ 246.

From Table 3.1 we find that this value of k corresponds to θ = 60 ppm of theanti-turbulent additive.

Exercise 4. It is required to increase by 25% at a given pressure resourcethe carrying capacity of an oil-pipeline section with D = 530 mm, δ = 8 mmpumping diesel fuel with νd = 9 cSt and flow rate 950 m3 h−1. Determine theamount of anti-turbulent additive NECCAD-547 needed to do this.

Answer. About 340 ppm.

3.7Gravity Fluid Flow in a Pipe

The above considered flows of fluid are pertinent to the class of the so-calledenforced (pumped) flows, since the motion of the flow was forced, that is to


overcome the friction force a pressure gradient was needed. However, thereare flows in which the primary driving force could be a component of thegravity force. Such flows are called gravity flows. A variety of gravity flowsare gravity stratified (divided into layers) flows. In these flows the fluid moveswithout completely filling the cross-section, the fluid flows over the lower partof the pipe whereas the upper part of the pipe is filled with vapor and gasesevolved from this fluid.

Considering the stratified gravity flow of a viscous incompressible fluidalong a section of the pipeline with diameter d and roughness � of theinternal wall surface inclined at an angle α to the horizontal (Figure 3.5).Our interest is the dependence of the flow rate Q = v · S on the governingparameters, that is the form of the function Q = f (S, ρ, µ, g, sin α, d, �).We can write the sought dependence in dimensionless form guided bydimensional theory. Among six arguments there are three dimensionallyindependent ones, for example d, ρ, µ. Hence the number of independentarguments in dimensional V would be reduced from six to three (Lurie, 2001)and the function Q = f (S, ρ, µ, g, sin α, d, �) takes the form

Q/S√g sin α · d

= f

(S

d2,

g · sin α

ν2/d3, ε

)(3.45)

where ν = µ/ρ is the kinematic viscosity and ε = �/d is the relative roughnessof the internal surface.

In the theory of gravity fluid flows a parameter Rh, called the hydraulicradius (Leibensone et al., 1934), is often introduced. The hydraulic radius isdefined as the ratio between the area S of a part of pipe cross-section filledwith fluid and the wetted perimeter Ps (Figure 3.6).

Figure 3.5 A scheme of gravity fluid flow in a pipe.

Figure 3.6 Definition of the hydraulic radius.


S = 1

2· r0

2φ − 1

2· r0

2 sin φ = 1

2· r0

2(φ − sin φ);

S

r02

= 1

2(φ − sin φ).

Since

AB = r0 · φ and Ps = r0 · φ

then

Rh = S

PS= 1/2 · r0

2(φ − sin φ)

r0 · φ= r0

2·(

1 − sin φ

φ

)(3.46)

where ϕ is the central angle where is seen the part of cross-section filled withfluid and r0 = d/2 the pipe radius. If the pipe is completely filled with fluid(ϕ = 2π), then Rh = r0/2 = d/4.

Since S/d2 and Rh/d depend only on the degree of the section fillingwith fluid, that is on the angle ϕ, the dependence (3.45) without disturbinggenerality could be written in equivalent form

Q/S√g sin α · Rh

= f1

(S

d2,

g · sin α

ν2/d3, ε

)or

Q = S · CSh · √Rh · sin α (3.47)

where CSh = √g · f1(S/d2,

√gd sin α · d/ν, ε) is the so-called dimensionless

Chezy factor; f1 is a dimensionless factor dependent on the parameters of theflow regime

√gd sin α · d/ν, degree of fluid filling S/d2 and the smoothness

parameter of the internal surface of the pipe ε (Archangelskiy, 1947 andChristianovitch, 1938).

The formula (3.47) could be represented in a form analogous to theDarcy–Weisbach relation for enforced flow

ρg sin α = 2 · (d/Rh)

f 21

· 1

d· ρv2

2(3.48)

when in the last expression we replace the pressure gradient �p/Lresponsible for enforced flow with the rolling down component of thegravity force ρg sin α causing motion in the case of gravity flow. Comparisonof Eq. (3.48) with the Darcy–Weisbach relation shows that the hydraulicresistance factor λ in stratified gravity flow is related to the factor f1 by theequality

λ = 2 · (d/Rh)

f 21


from which follows the expression for the Chezy factor

CSh =√

2g · (d/Rh)

λ. (3.49)

There are many empirical formulas for the Chezy factor for pipes with circularas well as non-circular cross-sections (Leibenson et al., 1934). Avoiding detailedtreatment of these formulas let us only say that in a first approximation itis possible to use Eq. (3.49) replacing d by 4Rh, to give CSh = √

8g/λ, whereλ = λ(Re, ε); Re = 4vRh/ν and Rh the hydraulic radius related to the degree offilling of the pipe cross-section by formulas (3.46).

Pipeline Sections of Gravity Flow

When the pressure in the pipeline section is equal to the saturated vaportension of the transported fluid, then inside the section continuously appearcavities filled with fluid vapor. In this case the flow could be stratified or couldhave a more complicated structure in which portions of fluid alternate withvapor-gas cavities (bubbles). The latter flow regime is called slug flow.

The flow in a section [x1, x2] of the pipeline in which it moves under the actionof the gravity force partially (incompletely) filling the pipeline cross-sectionwhile the remainder is filled with vapor of this fluid, is called gravity flow. Thepressure inside the vapor-gas cavity remains practically invariable and equalto the saturated vapor tension (pressure) pv. In spite of this the difference inpressures between the sections x1 (the beginning of the gravity flow section)and x2 (the end of the gravity flow section) nevertheless exists, it is merely equalto the difference in geometrical heights (z1 − z2) of these sections (Figure 3.7).

Stationary gravity flow can exist only on descending sections of the pipeline.The beginning of the gravity flow section x1 is called the transfer section. Thetransfer section always coincides with the region of the pipeline profile peak.

Figure 3.7 A scheme of pipeline gravity flow.


The line of hydraulic gradient of the gravity flow section passes parallel tothe pipeline profile at the distance pv/ρg over it. Consequently the hydraulicgradient i of the gravity flow section is equal to the slope of the pipeline profileto the horizontal i = tan αp.

The fluid flow rate in the gravity flow section in the stationary flow regimeis equal to the flow rate Q of fluid in the filled sections of the pipeline

Q = v0 S0 = v · S (3.50)

from which may be concluded that the velocity of fluid flow v in the gravityflow section exceeds the velocity v0 of the fluid in the pipeline sections filled bythe fluid, since the area S of the cross-section part of each gravity flow sectionfilled with the fluid is less than the area S0 of the complete cross-section of thepipeline, that is v = v0 · S0/S > v0.

If the flow of fluid in the gravity flow section is stratified, the degree σ = S/S0

of pipeline filling with the fluid depends on the ratio γ = i0/ tan |αp| betweenthe hydraulic gradient i0 = λ0 · 1/d · v2

0/ρg of the pipeline sections completelyfilled with fluid and the absolute value of the gravity flow section slopeαp to the horizontal. This dependence can be obtained from Eqs. (3.47) or(3.48) solving them with respect to S. To calculate the degree σ of pipelinesection filling with fluid it the following approximation formulas have beensuggested (Ishmuchamedov et al., 1999):

1. σ = 1 at γ = i0/ tan |αp| ≥ 1. In this case the pipeline cross-section iscompletely filled with fluid;

2. σ = 1 − 2.98 · 10−2 ·√

2λ0

· (1 − √γ); at 32.32 · λ0 ≤ γ < 1;

3. σ = 9.39 · 10−2 ·√

2γ

λ0+ 0.113; at 4.87 · λ0 ≤ γ < 32.32 · λ0; (3.51)

4. σ = 0.1825 ·(

2γ

λ0

)0.356; at γ < 4.87 · λ0.

Exercise. The oil flow rate (ν = 8.6 cSt) in the gravity flow section of an oil-pipeline (D = 720 mm, δ = 10 mm, � = 0.2 mm) is 900 m3 h−1. The profileof the section is inclined to the horizontal at an angle αp = −1◦. It is requiredto find the degree of pipeline cross-section filling with oil in this section.

Solution. Calculate the pumping rate v0, the Reynolds number Re, the factorof hydraulic resistance λ0 and the hydraulic gradient i0 in the pumping sectionsof the pipeline:

v0 = 4 · 900/(3600 · 3.14 · 0.7002) = 0.650 m s−1; tan 1◦ = 0.0175;

Re = 0.65 · 0.7/(8.6 · 10−6) ∼= 52907; λ0 = 0.0219;

i0 = λ0 · 1/d · v20/(2 · g) = 0.0219 · 1/0.7 · 0.652/(2 · 9.81) ∼= 0.0007.

Determine the parameter γ:

γ = i0/ tan |αp| = 0.0007/0.0175 = 0.04.


Since γ = 0.04 < 4.87 · λ0 = 0.1067, in formulas (3.51) using the fourth case:

σ = 0.1825 · (2γ/λ0)0.356 = 0.1825 · (2 · 0.04/0.0219)0.356 ∼= 0.29,

that is, the considered section of the pipeline is approximately 29% filled.How to determine whether or not there are gravity flow sections in pipeline

sections under consideration. To answer this question we need to build acombined picture of the pipeline profile and the hydraulic gradient line. If theline of the hydraulic gradient is everywhere over the pipeline profile and theamount by which it exceeds it is the quantity pv/ρg, where pv is the saturatedvapor pressure of the fluid, then gravity flow sections in the pipeline are absent.If the line of hydraulic gradient at any point approaches the pipeline profilecloser than pv/ρg or even intersects it, then there exists one or several gravityflow sections in the pipeline (Ishmuchamedov et al., 1999).

Now let us consider Figure 3.8 where a section of pipeline OO1 is depicted.Let us begin to build the line BK2�2K1�1A of the pipeline hydraulic gradient

from the end O1 of the pipeline section under consideration. To do this it issufficient to know the pressure and the hydraulic gradient at the end of thesection. The line of the hydraulic gradient in the segment B K2 lies significantlyover the pipeline profile, therefore cross-sections of the pipeline section arecompletely filled. However, the line of the hydraulic gradient at the pointK2 approaches the pipeline profile up to a distance pv/ρg, thus the point K2

represents the end of the first gravity flow section. Hence, one of the gravityflow sections is found. The line of hydraulic gradient K2�2 at this section isparallel to the pipeline profile.

Continue to build the line of the hydraulic gradient. It leaves the point �2 atan angle whose tangent is equal to the hydraulic gradient, that is in fact parallelto the segment BK2. It turns out that this line at the point K1 approachesthe pipeline profile at the distance pv/ρg for the second time. Consequently,the pressure inside the pipeline again becomes equal to the pressure of thesaturated vapor and in the pipeline there should exist vapor-gas cavities. Thepoint K1 represents the end of the second gravity flow section. Its beginningat the point �1 is a transfer section. It is called a transfer section because it is

Figure 3.8 A scheme to determine the location of a pipeline gravity flow section.


sufficient to deliver the transported fluid to the point �1 so that it reaches thenthe end O1 of the section by itself with the help of the gravity flow. Hence,the second gravity flow section K1�1 is found. The line K1�1 of the hydraulicgradient at this section passes parallel to the pipeline profile at the distancepv/ρg from it.

At the section �1A the line of the hydraulic gradient is parallel to its segmentsBK2 and �2K1 having been built for completely filled pipeline segments.

From Figure 3.8 it follows that the presence of gravity pipeline sections leads toenhancement of the initial hydraulic head H1 (and consequently the pressure p1)at the station and therefore requires higher expenditures of energy for pumpingas compared with a pipeline in which such sections are absent. If the line ofhydraulic gradient beginning from the point K2 were to be lengthened up tothe initial cross-section of the considered pipeline section, it would be possibleto determine the hydraulic head H1 needed to pump fluid with the same flowrate in a pipeline of the same length and with the same diameter but withoutgravity flow sections. It is evident that H1 ≥ H1.

Exercise. Oil (ρ = 870 kg m−3, ν = 8.5 cSt, pv = 0.02 MPa) with flow rate400 m3 h−1 is pumped along an oil-pipeline (L = 140 km; D = 530 mm,δ = 8 mm, � = 0.2 mm). The profile of the section has the form representedin Table 3.3. The pressure at the end of the section is 0.2 MPa. It is requiredto determine the pressure at the beginning of the section.

Solution. Calculate first the hydraulic gradient.

v0 = 4 · 400/(3600 · 3.14 · 0.5142) ∼= 0.536 m s−1;

Re = 0.536 · 0.514/(8.5 · 10−6) = 32 412;

λ0 = 0.11 · (0.2/514 + 68/32 412)0.25 ∼= 0.025;

i0 = 0.025 · 1/0.514 · 0.5362/(2 · 9.81) ∼= 0.00071.

Then determine the head losses in the pipeline section between 120 and140 km. They are h120–140 = i0 · 20 000 = 14.2 m. Therefore the head atthe end of the slope, that is at the cross-section x = 120 km is equal to0.2 · 106/870 · 9.81 + 14.2 ∼= 14.43 m.

Since pv/ρ · g = 20 000/(870 · 9.81) ∼= 2.34 m, the pipeline at the cross-section x = 120 km is still filled with oil. However, the difference in heightat the descending section is 100 m (see the profile of the pipeline), thus itis evident that at some cross-section the pressure of oil will be equal to thesaturated vapor pressure of oil pv, so that a part of the descending pipeline

Table 3.3

x, km 0 80 120 140z, m 100 100 0 0


section will inevitably become a gravity flow section. It is evident that thebeginning of this section coincides with the beginning of the descent atx = 80 km.

The hydraulic gradient at the plain (completely filled) pipeline segmentbetween the section beginning and the 80th kilometer is equal to thehydraulic gradient at the completely filled pipeline segment between 120and 140 km, that is 0.71 m km−1, so that the loss of hydraulic pressure ish0–80

∼= 56.8 m. Therefore the pressure p1 at the beginning of the section isequal to 870 · 9.81 · 56.8 ∼= 484 771 Pa or ≈4.95 atm.

73

4Modeling and Calculation of Stationary Operating Regimesof Oil and Gas Pipelines

In this chapter we consider the calculation of stationary operating regimes ofpipelines for transportation of oil, oil products and gas. The equations obtainedin the first chapter are used as a basis.

4.1A System of Basic Equations for Stationary Flow of an Incompressible Fluid in aPipeline

In stationary flow all parameters of the transported fluid at each cross-sectionof the pipeline remain constant, that is independent of time. Therefore partialderivatives with respect to time ∂()/∂t in the equations of Section 1.8 shouldbe taken equal to zero. Consider successively the basic equations describingthese flows:

1. Continuity equation (1.6) leads to the equation

d

dx(ρvS) = 0

which means that the mass flow rate M of the transported fluid remainsconstant

M = ρvS = const.

If the fluid is incompressible ( dρ/ dt = 0) and homogeneous (ρ = const.)and the pipeline has invariable diameter (S = const.), the velocity of thefluid would be the same at each cross-section of the pipeline

v = const. (4.1)

2. The momentum equation (1.10) gives

ρvdv

dx= − dp

dx− 4

dτw − ρg · sin α(x).


74 4 Modeling and Calculation of Stationary Operating Regimes of Oil and Gas Pipelines

With regard to the condition v = const. the momentum equation issimplified and yields

dp

dx= −4

dτw − ρg · sin α(x).

If we take in the last equation

τw = λ(Re, ε)

4· ρv|v|

2and sin α(x) = dz

dx,

the momentum equation is transformed into

d

dx

(p

ρg+ z

)= −λ(Re, ε) · 1

d· v|v|

2g. (4.2)

Note that the Bernoulli equation (1.19) leads also to Eq. (4.2) if we replacein it the hydraulic gradient i0 by 4|τw|/ρgd = λ(Re, ε) · v2/2gd inaccordance with Eq. (1.23).3. The equation of total energy balance (1.36) for stationary flow has the

form

ρvS · d

dx

(ein + p

ρ+ gz

)= πd · qn

or

ρvdein

dx= 4

d· qn − ρv · d

dx

(p

ρ+ gz

).

If in this equation we take ein = Cv · T + const.,

d

dx

(p

ρ+ gz

)= −λ(Re, ε) · 1

d· v|v|

2and qn = −K · (T − Tex)

the equation of total energy balance takes the form

ρvCvdT

dx= −4K

d(T − Tex) + λ(Re, ε)

1

d· ρ|v|3

2. (4.3)

The system of equations (4.1)–(4.3) with the addition of relations for thehydraulic resistance factor λ(Re, ε) and heat transfer factor K serve as thebasis for calculation of the stationary operating regimes of pipelines tran-sporting incompressible fluids, to which class belong oil and oil products.

4.2 Boundary Conditions. Modeling of the Operation of Pumps and Oil-Pumping Stations 75

4.2Boundary Conditions. Modeling of the Operation of Pumps and Oil-PumpingStations

The Bernoulli equation in algebraic form results from the continuityequation (4.1) and the differential momentum equation (4.2)(

p

ρg+ z

)x=0

−(

p

ρg+ z

)x=L

= λ(Re, ε) · L

d· v2

2g(4.4)

In this relation x = 0 and x = L denote, respectively, the initial and terminalcross-sections of the pipeline section with length L. Thus we have onealgebraic equation relating three parameters of the flow – the pressure p0

at the beginning of the pipeline section, the pressure pL at the end of thepipeline section and the velocity v of the fluid flow. To determine the velocityv or, what is the same, the flow rate of pumping, additional informationon the pressures at both ends of the pipeline is needed. This informationreflecting the interaction of the considered pipeline section with the rest ofthe pipeline is introduced into the mathematical model through boundaryconditions.

In some cases the pressure pL at the end of the pipeline, that is at x = Lcould be taken as given, determined e.g. by the conditions required for fluidpumped into reservoirs through a system of intrabase pipelines. Hence, oneof the boundary conditions can be a simple condition p(L) = pL.

Another boundary condition models the operation of the oil pumpingstation (OPS) located at the beginning of the pipeline section, that is atx = 0. In fluid flow in the pipeline the pressure gradually decreases becausemechanical energy is spent overcoming the force of viscous friction betweenthe fluid layers and is then turned into heat. That is why in a pipeline specialequipment producing pressure is needed. In general such equipment is calleda pump.

4.2.1Pumps

Pumps represent equipment for compulsory fluid movement from a cross-section with lesser head (line of suction) to a cross-section with greater head(line of discharge). Since the elevations of the pump entrance and exit are, asa rule, identical, one can say that these pumps are equipment for forced fluidmovement from a cross-section with lesser head (line of suction) to a cross-sectionwith greater head (line of discharge).

The simplest mathematical model of a pump can be represented as analgebraic equation of the form

�H = pex − pin

ρg= F(Q) (4.5)


characterizing the dependence of the differential head �H produced by thepump on the fluid flow rate Q . For every actual pump the differential head�H appears to be dependent on the fluid flow rate Q called in this case thefeed. The greater the head produced by a pump, the less, as a rule, is its feed.The dependence �H = F(Q) defines the so-called head-discharge (Q − H)characteristic of the pump.

In order to understand the physical basis of this model, let us closelyconsider the operating principle of one of the commonly encountered pumps,namely the centrifugal pump. In centrifugal pumps, used for pumping oil andoil products, the fluid moves from the cross-section with lesser pressure tothat with greater pressure under the action of the centrifugal force producedby the rotation of an impeller with profile blades.

Figure 4.1 shows a scheme of a pump impeller with profile blades.Considering the frame of reference related to the rotating impeller, theimpeller is believed to be immovable whereas the centrifugal force of inertiaρω2r, where ρ is the fluid density, ω the angular velocity of the impellerrotation and r the distance of the fluid particle from the rotation axis, acts onthe fluid filling the pump.

The centrifugal force causes the fluid to move along the impeller blade fromits center to the periphery. This force is capable of overcoming the pressuredrop �p = pex − pin, equal to the pressure difference between the pumpingpressure pex at the impeller periphery and the suction pressure pin at thecenter of the impeller, that is to force the fluid to move from the region of lowpressure to the region of high pressure. It is self-evident that to produce suchforced motion one needs to spend energy for impeller rotation.

For simplicity let us consider an impeller with radially located blades. Thebalance equation of forces acting on the fluid moving along the impeller radiusfrom its center to the periphery can be written as:

ρω2r − dp

dr= ρ · fτ(Q)

where dp/ dr is the radial gradient of pressure opposing the motion and fτ(Q)

the friction force depending on the discharge Q and increasing with Q .Integration of the force balance equation over the radius from 0 to Rim,

where Rim is the radius of the impeller, gives

ρω2Rim2

2− �p = Rimρ · fτ(Q) or �p = ρω2Rim

2

2− Rimρ · fτ(Q).

Division of both sides of this equation by ρg yields

�H = ω2Rim2

2g− Rim

g· fτ(Q). (4.6)

Thus the rotation of the impeller with angular velocity ω can force the fluidto move against the pressure drop �p between the periphery and the central


Figure 4.1 Operating principle of acentrifugal pump.

part of the impeller. The maximal value of the pressure drop which thecentrifugal force is capable of overcoming is equal to ρω2Rim

2/2. This value of�p is achieved at Q = 0 in the absence of a friction force. At Q > 0 Eq. (4.6)determining the (Q − �H) characteristic of the pump, where �H = �p/ρg,is obeyed. The pump feed Q decreases with increase in �p and, conversely,the smaller the pressure drop which the blower has to overcome, the greaterthe pump feed.

Exercise. It is required to determine the maximal differential head developedby a centrifugal pump with radial located impeller blades having radius 0.25 mand rotating at 3000 rpm.

Solution. 3000 rpm corresponds to the angular velocity ω = 2π · 3000/60 =2π · 50 s−1. Then, in accordance with Eq. (4.6) we get

(�H)max = ω2Rb2

2g= 4π2 · 502 · 0.252

2 · 9.81∼= 314.4 m.

The (Q − �H) characteristics of centrifugal pumps operating in stationaryregimes are often approximated by the two-term dependence

�H = a − b · Q2 (4.7)

where the differential head �H is measured in (m) and the flow rate Q in(m3 h−1), therefore the dimension of the factor a is (m) and the factor bis (m/(m3 h−1)2). For example, the main pump HM 1250-260 produced inRussia rated at a nominal feed of 1250 m3 h−1 and nominal head 260 m,has the (Q − �H) characteristic �H = 331 − 0.451 · 10−4 · Q2, main pumpHM 2500-230 with impeller diameter Dim = 430 mm rated at nominal feed2500 m3 h−1 and nominal head 230 m has the (Q − �H) characteristic�H = 280 − 0.792 · 10−5 · Q2 (�H in m, Q in m3 h−1) and so on (Vasil’evet al., 2002).

Figure 4.2 shows the (Q − H) characteristic of the centrifugal pump HM2500-230.

The two upper curves of this figure represent the (Q − H) characteristicsof a pump with accessory impellers (385 and 430 mm, the middle curveshows the dependence of the power consumption N (kW) on the flow rate


Figure 4.2 Characteristics of the centrifugal pump HM 2500-230.

Q and the bottom curve illustrates the dependence of efficiency η(%) onthe flow rate of the transported fluid. In the same figure is also marked theoperating range of the pump, that is the range of flow rates Q of the pump. Inthis range (1800 < Q < 3000 m3 h−1) the efficiency η ≈ 85% and the powerN ≈ 1600 kW of the pump have maximal values.

4.2.2Oil-Pumping Station

Pumps connected in series or parallel provide the basis of oil-pumping stationsintended to produce driving pressure.

The (Q − �H) characteristics of pumps connected in series (Fig. 4.3) aresummarized, the fluid flow rates of the pumps are identical Q1 = Q2 = Q andthe differential heads are given by �H = �H1 + �H2.

If �H1 = a1 − b1 · Q2 is the characteristic of the first pump and �H2 =a2 − b2 · Q2 the characteristic of the second pump, the characteristic of asystem of two pumps connected in series is equal to

�H = (a1 + a2) − (b1 + b2) · Q2. (4.8)


Figure 4.3 Series connectionof pumps.

In parallel connection of pumps (Fig. 4.4) their (Q − �H) characteristics aredifferent. Fluid discharges in pumps are given by Q = Q1 + Q2 but the headsproduced by each pump are identical �H = �H1 = �H2.

If �H = a1 − b1 · Q2 is the characteristic of the first centrifugal pump and�H = a2 − b2 · Q2 that of the second one, the characteristic of the system oftwo pumps connected in parallel is√

(a1 − �H)

b1+

√(a2 − �H)

b2= Q. (4.9)

Exercise 1. The (Q − �H) characteristic of a centrifugal pump with impellerdiameter 440 mm is �H = 331 − 0.451 · 10−4 · Q2. Another pump of the sametype but with impeller diameter 465 mm has the (Q − �H) characteristic�H = 374 − 0.451 · 10−4 · Q2, (�H in m, Q in m3 h−1). What characteristichas a system of two pumps connected in series?

Solution. In accordance with Eq. (4.8) we obtain �H = (331 + 374) − 2 ·0.451 · 10−4 · Q2 = 705 − 0.902 · 10−4 · Q2.

Exercise 2. The (Q − �H) characteristic of a centrifugal pump withimpeller diameter 440 mm is �H = 331 − 0.451 · 10−4 · Q2. Another pumpof the same type but with impeller diameter 465 mm has the (Q − �H)

characteristic �H = 374 − 0.451 · 10−4 · Q2, (�H in m, Q in m3 h−1). It isrequired to find the characteristic of a system of two pumps connected inparallel?

Figure 4.4 Parallel connection of pumps.


Solution. In accordance with Eq. (4.9) we have√(331 − �H)

0.451 · 10−4+

√(374 − �H)

0.451 · 10−4= Q

or√

331 − �H + √374 − �H = 6.716 · 10−3 · Q, where �H < 331 m.

The (Q − �H) characteristic of a pumping station is the total (Q − �H)

characteristic of all pumps operating in the station (connected in series orparallel) minus the (Q − �H) characteristics of the supply communications.The latter is taken as an element connected in series with the pumps of thestation.

Exercise 3. At a pumping station two pumps operate in series withcharacteristics �H = F1(Q) = 331 − 0.451 · 10−4 · Q2 and �H = F2(Q) =374 − 0.385 · 10−4 · Q2. It is also known that the head losses hc in the stationcommunications, that is the pipeline system of the station, are represented bythe dependence

hc = 25 − 0.036 · 10−4 · Q2

(�H, hc in m, Q in m3 h−1). It is required to find the characteristic of thepumping station?

Solution. The characteristic of the pumping station �H = F(Q) is repre-sented by the sum of the characteristics of the pump system minus head lossesin the supply communication F1(Q) + F2(Q) − hc(Q):

�H = F(Q) = 680 − 0.800 · 10−4 · Q2.

Hence, if (Q − �H) the characteristic of the pumping station �H = F(Q) isknown, the boundary condition at the initial cross-section x = 0 of the pipelinesection can be the following condition given by this characteristic

pex − pin

ρg= F(Q) or

p0

ρg= pu

ρg+ ·F(v) (4.10)

that is, a condition similar to condition (4.5), where p0 = pex is the pressureat the initial cross-section of the pipeline section, pu = pin the pressurebefore the oil-pumping station, called the head before pumping station,F(Q) ≡ F(vS) ≡ F(v). When using the two-term dependence of the stationdifferential head �H on the flow rate Q the boundary condition (4.10) at thebeginning of the pipeline section takes the form

p0

ρg= pu

ρg+ a − b · S2(3600)2 · v2 (4.11)

where the velocity v is measured in (m s−1).

4.3 Combined Operation of Linear Pipeline Section and Pumping Station 81

4.3Combined Operation of Linear Pipeline Section and Pumping Station

To calculate the combined operation of a linear pipeline section and thepumping station located at the beginning of the pipeline section the Bernoulliequation (4.4) is used, in which the pressure p0 = p(0) at the initial cross-section of the pipeline section is excluded with the help of boundarycondition (4.11)

(p0

ρg+ z0

)−

(pL

ρg+ zL

)= λ(Re, ε) · L

d· v2

2g,

p0

ρg= pu

ρg+ (a − 1.296 · 107S2b · v2)

where a and b are the approximation factors of the pumping station (Q − �H)

characteristic, the velocity v is measured in (m s−1). After eliminating p0 fromthese equations we obtain

pu

ρg− pL

ρg+ (z0 − zL) + a − 1.296 · 107S2b · v2 = λ(Re, ε) · L

d· v2

2g.

(4.12)

This equation is called the head balance equation. At given values of the headbefore the pumping station pu and pressure at the pipeline section end pL

Eq. (4.12) serves to determine the unknown velocity v of the fluid flow in thepipeline.

To solve Eq. (4.12) it is convenient to rearrange all the terms containingthe unknown velocity v on the right-hand side of the equation leaving on theleft-hand side only the given quantities

pu

ρg− pL

ρg+ (z0 − zL) + a =

(λ · L

d· 1

2g+ 1.296 · 107 S2b

)· v2.

This equation could be solved by the method of successive approximation(iteration method). We can demonstrate it with exercises.

Exercise 1. Two identical pumps connected in series and having iden-tical (Q − H) characteristics �H = 331 − 0.451 · 10−4 · Q2, (�H in m,Q in m3 h−1) are pumping diesel fuel (ρ = 840 kg m−3, ν = 9 cSt)along the pipeline section (D = 530 × 8 mm, L = 120 km, � = 0.2 mm,z0 = 50 m, zL = 100 m). It is required to find the flow rate and pres-sure at the beginning of the section when the pressure pL at the endof the section is 0.3 MPa, the head before the pumping station hu is30 m and it is known that sections of gravity flows are absent in thepipeline.


Solution. Write Eq. (4.11) of the head balance

pu

ρg− pL

ρg+ (z0 − zL) + 2a = (λ · L

d· 1

2g+ 1.296 · 107S2 · 2b)v2.

Insertion of the given data yields

30 − 0.3 · 106

840 · 9.81+ 50 − 100 + 2 · 331 =

[λ · 120 000

0.530 − 2 · 0.008· 1

2 · 9.81

+ 1.296 · 107

(3.14 · 0.5142

4

)2

· 2 · 0.451 · 10−4

]· v2

and

605.6 = v2 · (11 899.2 · λ + 50.3).

If as a first approximation it is accepted that λ(1) = 0.02 then this equationgives v = 1.449 m s−1. We need to verify whether or not the factor λ is correctlytaken. To do this let us determine the first Reynolds number

Re(1) = 1.449 · 0.514(9 · 10−6

) = 82 754.

Then with formula (1.31) we get

λ = 0.11 ·(

0.2

514+ 68

82 754

)0.25 ∼= 0.0205 > λ(1) = 0.02.

It is seen that the obtained value of the hydraulic resistance factor should beenhanced.

As a second approximation we take λ(2) = 0.0205. Then the given equationyields v = 1.434 m s−1, after which it is necessary to verify whether the factorλ is correctly taken. We have

Re(2) = 1.434 · 0.514

(9 · 10−6)∼= 81 897;

λ = 0.11 ·(

0.2

514+ 68

81 897

)0.25 ∼= 0.0206 ≈ λ(2) = 0.0205.

Thus there is good coincidence between the taken and received factor λ.Hence, v ∼= 1.434 m s−1 and

Q = 3.14 · 0.5142

4· 1.434 ∼= 0.2974 m3s−1 or

Q = 0.2974 · 3600 ∼= 1071 m3 h−1.

The pressure p0 at the beginning of the pipeline section is determined withthe formula p0 = ρg · [hu + F(Q)]. As a result we have

4.4 Calculations on the Operation of a Pipeline with Intermediate Oil-Pumping Stations 83

p0 = 840 · 9.81 · [30 + 2 · (

331 − 0.451 · 10−4 · 10712)] ∼= 4.85 · 106 Pa

or 4.85 MPa.

Answer. 1071 m3 h−1; 4.85 MPa.

Exercise 2. The pumping of crude oil (ρ = 870 kg m−3, ν = 25 cSt) is be-ing conducted by two pumps: HM 2500–230 with characteristic �H =251 − 0.812 · 10−5 · Q2 and HM 3600-230 with �H = 273 − 0.125 · 10−4 · Q2

connected in series and rated at feed 1800 m3 h−1. It is known that the(Q − �H) characteristic of the supply communication of the oil-pumpingstation has the form �H = 0.15 · 10−4 · Q2 (here and above �H is in mand Q in m3 h−1). It is required to determine the pumping flow rate underthe condition that the oil-pipeline section (D = 820 × 10 mm, L = 150 km,z0 = 80 m, zL = 120 m, hu = 70 m, hL = 40 m) has an approximately flatcharacter and that sections of gravity flow are absent. Besides it is known thathead losses due to local resistances comprise ≈ 2% of the head losses due tofriction.

Solution. The equation of the head balance is[80 + 70 + (

251 − 0.812 · 10−5 · Q2) + (273 − 0.125 · 10−4 · Q2)

−0.15 · 10−4Q2] − [120 + 40] = 1.02 · λ · 150 000

0.800· v2

2 · 9.81.

After simplification this equation takes the form

514 = v2 · (9748 · λ + 116.6).

This equation is solved by the iteration method.As a first approximation we take λ(1) = 0.02. Then the equation gives

v = 1.284 m s−1. Now verify whether λ has been chosen correctly.

Re = 1.284 · 0.8

(25 · 10−6)= 41 088,

λ = 0.31644√

41 088∼= 0.0222 > λ(1) = 0.02.

As the second approximation we take λ(2) = 0.0222. Then the equation yieldsv = 1.242 m s−1. Now verify whether λ has been chosen correctly

Re = 1.242 · 0.8

(25 · 10−6)= 39 744;

λ = 0.31644√

39744∼= 0.0224 ≈ λ(2) = 0.0222.

Hence v = 1.242 m s−1 which is equivalent to Q ∼= 2246 m3 h−1.

Answer. 2246 m3 h−1.


4.4Calculations on the Operation of a Pipeline with Intermediate Oil-PumpingStations

Consider a pipeline consisting of n successive sections separated by oil-pumping stations (OPS). The transportation of fluid is performed in the so-called pump-to-pump regime. When intermediate fluid dumping and pumpingare absent we can write the Bernoulli equation for each section

[z1 + hn1 + F1(Q)] − [z2 + hn2] = h1–2(Q),

[z2 + hn2 + F2(Q)] − [z3 + hn3] = h2–3(Q),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

[zn + hn,n + Fn(Q)] − [zL + hL] = hn−(n−1)(Q),

(4.13)

where �H = F1(Q), �H = F2(Q), . . . , �H = Fn(Q) are the hydraulic (Q −�H) characteristics of oil-pumping stations; hj−(j−1)(Q) the head losses in thesections between the oil-pumping stations dependent on the pumping flow rateQ ; z1, z2, . . . , zn the elevations of the oil-pumping stations; hu,1, hu,2, . . . , hu,n

the heads before the oil-pumping stations equal to hu,i = pu,i/(ρg); zL, hL =pL/(ρg) the elevation and piezometric head at the pipeline end (x = L),respectively.

Equations (4.13) represent a system of n algebraic equations (according tothe number of sections) with n unknown quantities: flow rate Q and (n − 1)

heads hu,j before the intermediate oil-pumping stations.First consequence (equation of head balance):Term-by-term summation of the equations of the system (4.13) yields the

equation called the balance equation of the heads for the whole pipeline

(hu,1 − hL) +j=n∑j=1

Fj(Q) = (zL − z1) +j=n∑j=1

hj−(j−1)(Q) (4.14)

This equation serves to determine the flow rate Q of the fluid (carrying capacityof the pipeline), with all unknown heads hu,j before the intermediate pumpingstations being excluded.

It should be taken into account that the flow rate Q found from Eq. (4.14)can be realized in the considered pipeline only when the heads hu,j of all theintermediate stations are greater than the minimum allowed value assuringpump operation without cavitation and the pressure in all cross-sections of thepipeline is less than the permissible value defined by the pipeline strength.

Second consequence (equation for heads before the oil-pumping stations):Term-by-term summation of only the first s (s < n) equations of the system

(4.13) yields the equation for the heads hu,s before the s-th intermediatepumping station

hu,s = hu,1 + (z1 − zL) +j=s∑j=1

[Fj(Q) − hj−(j−1)(Q)

]. (4.15)

4.4 Calculations on the Operation of a Pipeline with Intermediate Oil-Pumping Stations 85

The flow rate Q in this equation is assumed to be given, since it can beobtained from Eq. (4.14).

In determining the head losses in the pipeline sections it is necessary toaccount for the possibility of existing transfer points and segments of gravityflow in these sections (see Section 3.7). Let us illustrate the aforesaid with anexercise.

Exercise. An oil-pipeline with length L = 450 km consists of three linearsections the data for which are given in the table below. The head hu.1 beforethe leading oil-pumping station is 50 m and the head hL at the end of thepipeline is 30 m.

At the beginning of each section there is an oil-pumping station with twoidentical pumps connected in series the characteristics of which are given inthe following table

It is required to determine the carrying capacity of the oil-pipeline whenpumping oil (ρ = 900 kg m−3, ν = 30 cSt) through it and the heads of theintermediate oil pumping stations.

Solution. The balance equations of the heads for the pipeline sections are[50 + 50 + 2 · (251 − 0.812 · 10−5 · Q2)

] − [60 + hu,2]

= λ · 150 000

0.704· v2

2 · 9.81,[

60 + hu,2 + 2 · (285 − 0.640 · 10−5Q2)] − [70 + hu,3]

= λ · 180 000

0.704· v2

2 · 9.81,[

70 + hu3 + 2 · (236 − 0.480 · 10−5Q2)] − [180 + 30]

= λ · 120 000

0.704· v2

2 · 9.81.

No. Length, km D, mm δ, mm z0, m zL, m

1. 150 720 8 50 602. 180 720 8 60 703. 120 720 8 70 180

No. Type of pump (Q − �H) characteristic Positive suction head, m

1. II 2500-230 �H = 251 − 0.812 · 10−5Q2 402. II 3600-230 �H = 285 − 0.640 · 10−5Q2 403. II 5000-210 �H = 236 − 0.480 · 10−5Q2 40


Here we assume that, due to the invariability of the pipeline diameter, therate of pumping and the hydraulic resistance factors are identical whenpassing from one section to another; hu,2, hu,3 are the unknown heads of theintermediate stations that are to be determined.

Term-by-term summation of the above cited equations gives

1434 − 3.864 · 10−5 · Q2 ∼= 32 579 · λv2 or

1434 = v2 · (32 579 · λ + 75.8).

This equation (balance of heads for the whole pipeline) is solved by the iterationmethod. We take first λ(1) = 0.02. Then from the above equation we obtainv(1) = 1.404 m s−1. Next we verify the correctness of the factor λ:

Re = 1.404 · 0.704

(30 · 10−6)∼= 32947,

λ = 0.31644√

32947∼= 0.0234 > 0.02.

As the second approximation we take λ(2) = 0.0234. Then the equation yieldsv = 1.308 m s−1. We verify again whether the factor λ is correctly chosen:

Re = 1.308 · 0.704

(30 · 10−6)∼= 30 694;

λ = 0.31644√

30694∼= 0.0239 ≈ 0.0234.

Thus v = 1.308 m s−1 or Q ∼= 1832 m3 h−1.From the first balance equation of heads we determine hu,2

[50 + 50 + 2 · (251 − 0.812 · 10−5 · 18322)] − [60 + hu,2]

= 0.0234 · 150 000

0.704· 1.3082

2 · 9.81and hu,2

∼= 52.7 m.

The second balance equation gives hu,3

[60 + 52.7 + 2 · (285 − 0.640 · 10−5 · 18322)] − [70 + hu,3]

= 0.0234 · 180000

0.704· 1.3082

2 · 9.81and hu,3

∼= 48.0 m.

Both the values for the heads of the intermediate oil-pumping stations complywith the requirement of positive suction head and, therefore, the obtainedpumping regime is realizable.

4.5 Calculations on Pipeline Stationary Operating Regimes in Fluid Pumping with Heating 87

4.5Calculations on Pipeline Stationary Operating Regimes in Fluid Pumping withHeating

To calculate the stationary operating regime of pipelines performing fluidpumping with heating (high-temperature pumping) Eqs. (4.1)–(4.3) are used.

d

dx(ρvS) = 0

d

dx

(p

ρg+ z

)= −λ(Re, ε) · 1

d· v|v|

2g

ρvCvdT

dx= −4K

d(T − TH) + λ(Re, ε) · 1

d· |v|3

2g

(4.16)

The first equation gives ρvS = const., meaning that the mass flow rate isconstant in the stationary operating regime. If we take the pipeline diameterto be invariable d = d0 = const., that is we neglect heat expansion of the pipe,and the fluid density to be insignificantly variable (ρ ≈ const.), that is we alsoignore heat expansion of the fluid, from the constancy of mass flow rate thecondition of pumping rate constancy v ≈ v0 = const. follows.

The latter condition v ≈ v0 = const. allows us to rewrite the system ofdifferential equations (4.16) as

d

dx

(p

ρg+ z

)= −λ(Re, ε) · 1

d0· v0|v0|

2g

ρv0CvdT

dx= −4K

d0(T − TH) + λ(Re, ε) · 1

d0· |v0|3

2g

(4.17)

As initial conditions, that is conditions at the initial cross-section x = 0 of thepipeline, we can take

p(0) = p0; T(0) = T0 (4.18)

signifying that the pressure and temperature at the beginning of the pipelineare known. If it is needed to model the oil-pumping station with given(Q − �H) characteristic �H = a − b · Q2 located at the beginning of thepipeline, the initial conditions at x = 0 should be taken as

p0

ρg= pu

ρg+ a − b · S2

0(3600)2 · v02; T(0) = T0. (4.19)

Here, in accordance with Eq. (4.11) a and b are approximated factors the of(Q − �H) characteristics of OPS; pu the head before the station; S0 = πd2

0/4.In the general case the system of equations (4.16) should be integrated

numerically or solved using one or other simplification suggestions. Oneconsists in ignoring the heat released in the fluid due to the work of internal


friction forces as compared to external heat exchange

λ(Re, ε)1

d0· |v0|3

2g≤ 4K

d0· |T − Tex|.

Then the second equation of system (4.16) can be easily integrated and thesolution of this equation yields (see Eq. (1.44))

T(x) = Tex + (T0 − Tex) · exp

(−πd0 · K

CvM0x

)(4.20)

where M0 = ρv0 S0 is the mass flow rate; Tex the temperature of the externalmedium, taken as constant. It should be noted that the fluid velocity v0 isconstant but unknown.

The first equation of the system (4.16) can be represented as(p0

ρg+ z0

)−

(pL

ρg+ zL

)=

∫ L

0λ(Re, ε) dx · 1

d0· v0

2

2g

This equation differs from the standard form of the Bernoulli equation in thatit takes into account variability of the factor λ with pipeline length. Really, ifT = T(x), then T �= const., the kinematic viscosity of the fluid ν is not constant,it depends on temperature, therefore the Reynolds number Re = v0d0/ν(T) =Re(x) and the factor of hydraulic resistance λ are functions of x. This circum-stance is taken into account in the Bernoulli equation obtained by integrationof λ over the pipeline section length. With regard to Eq. (4.19) we have

pu

ρg− pL

ρg+ (z0 − zL) + a − 1.296 · 107S0

2b · v02

=∫ L

0λ(Re, ε) dx · 1

d0· v0

2

2g

The system of equations

pu

ρg− pL

ρg+ (z0 − zL) + a − 1.296 · 107S0

2b · v02

=∫ L

0λ(Re, ε) dx · 1

d0· v0

2

2g,

T(x) = Tex + (T0 − Tex) · exp(

−πd0 · K

CvM0x

) (4.21)

provides a basis to determine the unknown velocity v0 and consequently theflow rate of fluid pumping with heating. The kinematic viscosity as a functionof temperature ν(T) is assumed to be given, see for example Eq. (2.3).

Analytical SolutionThe system of equations (4.21) has an analytical solution when

ν(T) = νex · e−κ(T−Tex),


where νex is the kinematic viscosity of the fluid at the temperature Tex of theexternal media; κ the dependence factor Eq. (2.3);

λ = 0.31644√

Re

that is the flow regime of the fluid corresponds to the flow region of theso-called hydraulic smooth pipes (Blasius zone). In this case the integral onthe right-hand side of the first equation of the system could be calculated inquadratures∫ L

0λ dx = 0.3164

4√

v0d0·∫ L

0ν1/4 dx = 0.3164

4√

v0d0·∫ L

0ν1/4

ex e− κ4 (T−Tex) dx

= λex ·∫ L

0e− κ

4 (T−Tex) dx, where λex = 0.31644√

v0d0/νex

(4.22)

If we convert to the dimensionless coordinate ξ = x/L. Then∫ L

0λ dx = L ·

∫ 1

0λ dξ = L ·

∫ TL

T0

λdξ

dTdT.

If now in Eq. (1.44) we neglect the heat release, that is we take T⊗ = 0, thenEq. (1.44) gives

ρv0CvdT

dξ= −4KL

d0· (T − Tex) ⇒ dξ

dT= −ρv0Cvd0

4KL· 1

T − Tex.

Thus ∫ L

0λ dx = L ·

∫ TL

T0

λdξ

dTdT = −ρv0Cvd0

4KL· L ·

∫ TL

T0

λ

T − TexdT.

With regard to Eq. (4.22) we have

∫ L

0λ dx = −ρCvv0d0

4KL· L · λex

∫ TL

T0

e− κ4 (T−Tex)

T − TexdT.

The integral on the right-hand side can be transformed to

∫ TL

T0

e− κ4 (T−Tex)

T − TexdT =

∫ − κ4 (TL−Tex)

− κ4 (T0−Tex)

eη

ηdη.

Now gathering together all the results we obtain the following expressionfor head losses in a non-isothermal fluid flow∫ L

0λ dx · 1

d· v0

2

2g= λeff · L

d· v0

2

2g(4.23)


where λeff is the effective factor of hydraulic resistance determined by

λeff = λex · 1

m· [Ei(−k) − Ei(−ke−m)]; λex = 0.3164

4√

v0 d0/νex

;

k = κ

4· (T0 − Tex); m = 4KL

ρCvv0d0= πKd0 · L

CvM0. (4.24)

Here we take into account the equality TL − Tex = (T0 − Tex) · exp(−m)

following from the basic formula (1.44) and as Ei(z) it is denoted the Eulerfunction

Ei(z) =∫ z

−∞eη

ηdη, (4.25)

which is widely encountered in technical applications and for which there arespecial tables. Some values of this function are listed in Table 4.1.

The basic equation (4.21) to determine the velocity v0 of non-isothermal fluidflow with regard to Eqs. (4.22)–(4.24) takes the form

pu

ρg− pL

ρg+ (z0 − zL) + a − 1.296 · 107S0

2b · v02 = λeff · L

d0· v0

2

2g(4.24)

Exercise. Along a practically horizontal oil-pipeline section (D = 720 ×10 mm, L = 120 km) is pumped oil (ρ = 870 kg m−3, Cv = 2000 J kg−1 K−1),ν1 = 5 cSt at T1 = 50 ◦C and ν2 = 40 cSt at T2 = 20 ◦C) with heating. The initialtemperature T0 of the oil is 50 ◦C, the temperature Tex of the environmentis 10 ◦C. The heat-transfer factor K averaged over the pipeline section is3.5 W m−2 K−1). The pumping is carried out with two pumps connected inseries. The characteristic of each pump is �H = 273 − 0.125 · 10−4Q2 (�H inm, Q in m3 h−1, hL = hu). It is required to find the flow rate of pumping andthe temperature of the oil at the section end.

Solution. First determine with Eq. (2.30) the dependence of the oil viscosityon temperature ν(T). We have

ν(T) = 5 · e−κ·(T−50).

Here we take ν(50) = 5 cSt. The second condition ν(20) = 40 cSt yields theequation for the factor κ

40 = 5 · e−κ·(20–50)

from which κ ∼= 0.0693 K−1.

Table 4.1

z −1.0 −0.8 −0.6 −0.4 −0.2 −0.1 −0.05Ei(z) −0.22 −0.31 −0.45 −0.70 −1.22 −1.82 −2.47


Writing Eq. (4.24) for the balance of heads

2 · [273 − 0.125 · 10−4Q2] = λef f · 120 000

0.7· v0

2

2 · 9.81.

The effective factor λeff of the hydraulic resistance taking into account itsvariability with pipeline section length in accordance with Eq. (4.23) is

λeff = 0.31644√

v0 · d0/νex

· 1

m· [

Ei(−k) − Ei(−ke−m)]

where

k = κ

4· (T0 − Tex), m = πKd0 · L

CvM0= 4K · L

ρCvv0d0.

Substituting in the balance equation of heads the expression for the flow rateQ using the velocity v0 of the fluid flow

Q = 3.14 · 0.72

4· v0 · 3600

and taking into account other conditions we get

546 = v02 · (8737.4 · λeff + 47.94) (4.26)

which can be solved by the iteration method.First approximation. Let us take λeff

(1) = 0.02. Then from Eq. (4.26) we findthe velocity of fluid flow v0

(1) = 1.566 m s−1 and verify the correctness of theobtained value. We have

νex = 5 · exp[−0.0693 · (10 − 50)] = 79.95 cSt;

λex = 0.31644√

1.566 · 0.7/(79.95 · 10−6)

∼= 0.029;

k = 1/4 · 0.0693 · (50 − 10) = 0.693;

m = 4 · 3.5 · 120 000

1.566 · 0.7 · 870 · 2000∼= 0.881;

k · exp(−m) = 0.693 · exp(−0.881) ∼= 0.287;

λeff = 0.029 · 1

0.881· [Ei(−0.693) − Ei(−0.287)]

= 0.029 · 1

0.881· [−0.379 − (−0.939)] ∼= 0.0186 < λeff

= 0.02.

Since there is a difference between the taken and calculated values of λeff wemake a second approximation.


Second approximation. Let us take λeff(2) = 0.0186. From Eq. (4.25) we get

the new velocity of fluid flow v0(2) = 1.611 m s−1 and verify its correctness. We

have

κ = 0.0693 K−1; νex = 79.95 cSt;

λex = 0.31644√

1.611 · 0.7/(79.95 · 10−6)

∼= 0.029;

k = 1/4 · 0.0693 · (50 − 10) = 0.693;

m = 4 · 3.5 · 120 000

1.611 · 0.7 · 870 · 2000∼= 0.856;

k · exp(−m) = 0.693 · exp(−0.856) ∼= 0.294;

λeff = 0.029 · 1

0.856· [Ei(−0.693) − Ei(−0.294)]

= 0.029 · 1

0.856· [−0.379 − (−0.921)] ∼= 0.0184 ≈ 0.0186

= λeff(2).

Since the taken and calculated values of the factor λeff show good coincidencethe process of successive approximations ends. Hence, v0

∼= 1.611 m s−1 andQ = 2231 m3 h−1. Hence, the temperature of the oil at the pipeline section endin accordance with Eq. (4.21) is TL = 10 + (50 − 10) · exp(−0.856) ∼= 27 ◦C.

Answer. 2231 m3 h−1, 27 ◦C.

4.6Modeling of Stationary Operating Regimes of Gas-Pipeline Sections

For modeling the stationary flow of a compressible gas in a gas-pipeline thefollowing equations are used:• continuity equation (1.6):

d

dx(ρvS) = 0 ⇒ M = ρvS = const. (4.27)

Since the gas density ρ decreases with pressure drop, from Eq. (4.27) itfollows that the gas velocity v increases from the beginning of the pipelinesection to its end;• momentum equation (1.10):

ρvdv

dx= − dp

dx− 4

dτw − ρg sin α(x).

If the gas velocity v increases, the acceleration v · dv/ dx of the gas is distinctfrom zero. It is evident that the estimation

4.6 Modeling of Stationary Operating Regimes of Gas-Pipeline Sections 93

ρvdv

dx= d

dx(ρv2) � d

dx(p)

is valid when the gas velocity v is small compared to the velocity of soundin a gas c, c = √

γRT ≈ √1.31 · 500 · 300 ∼= 440 m s−1, where γ = Cp/Cv is

the adiabatic index (for methane γ = 1.31). So, for example, when thevelocity of compressed gas with density 50 kg m−3 varies over �v thequantity ρv · �v is 50 · 10 · �v = 500 · �v at gas velocity 10 m s−1, whereasthe pressure variation �p calculated with the Joukowski formula �p = ρc�vis 50 · 440 · �v = 22 000 · �v, that is about 45 times greater. Thus, theacceleration of gas and often the gravity component ρg sin α in the momentumequation may, as a rule, be neglected. Then the momentum equation of the gasexpresses in essence the equality of the driving forces: pressure and friction

dp

dx= −4

dτw = −4

d· Cf

ρv2

2= −λ

1

d· ρv2

2; (4.28)

• balance equation of total energy

ρvS · dJ

dx= πd · qn

This equation ignores the work due to the force of gravity. If in this equationwe use the dependence of enthalpy J on pressure p and temperature T , that iswe take J = J(p, T), and the external heat exchange in the form of Newton’slaw (1.42), then

ρv

[(∂J

∂T

)p

· dT

dx+

(∂J

∂p

)T

dp

dx

]= −4K

d(T − Tex).

Denoting (∂J/∂T)p = Cp as the specific heat of a gas at constant pressure and(∂J/∂p)T = −D∗Cp, where D∗ is the Joule-Thompson factor, the last equationtakes the form

ρvCpdT

dx− ρvCpD∗

dp

dx= −4K

d(T − Tex). (4.29)

Using the expression for gas enthalpy J involving internal energy and otherparameters of state

J = ein + p

ρ= CvT + Z(p, T) · RT = [Cv + Z(p, T)R] · T.

The factors Cp and D∗ can be expressed through the factors Cv and Z(p, T) asfollows

Cp =(

∂J

∂T

)p

= Cv + R ·[

Z + RT

(∂Z

∂T

)p

];

D∗ = − 1

Cp

(∂J

∂p

)T

= −RT

Cp·(

∂Z

∂p

)T

. (4.30)


From this it is seen that the Joule-Thompson effect (cooling of heat-insulatedgas in a gas-pipeline through a pressure drop, (∂Z/∂p)T < 0, see Fig. 2.7)manifests itself only for a real gas when Z �= 1 (as a rule the factorD∗ ≈ 3–5 K MPa−1).

If we neglect the Joule-Thompson effect, the energy balance equation issimplified to

ρvCpdT

dx= −4K

d(T − Tex) (4.31)

and its solution takes the form of Eq. (1.46):

T(x) − Tex

T0 − Tex= exp

(−πd · K

CpMx

). (4.32)

If we define the average temperature Tav of the gas over pipeline section lengthby the expression

Tav = 1

L·∫ L

0T(x) dx

and substitute into it the distribution T(x) from Eq. (4.32), we get

Tav = Tex + T0 − TL

ln(

T0 − Tex

TL − Tex

) (4.33)

where the temperature TL at the end of pipeline section is represented by theexpression

TL − Tex = (T0 − Tex) · exp

(−πd · K · L

CpM

)following from Eq. (4.32).

4.6.1Distribution of Pressure in Stationary Gas Flow in a Gas-Pipeline

This distribution is obtained from Eqs. (4.27) and (4.28) ρvS = M = const.,dp

dx= −λ

1

d· ρv2

2

(4.34)

and the equation of the gas state p = ZρRT . Transformation of the secondequation (4.32) yields

dp

dx= −λ

1

d· ρv2

2= −1

2· λ · ρ2v2S2

d · ρS2= −1

2· λ · M2

(p/ZRT) · (π2d5/16).


If we take• λ = const., see e.g. Eq. (1.34);• T ≈ Tav = const.;• Z(p, T) ≈ Zav = const.the resulting differential equation gives after integration the distribution ofp(x) along the pipeline section length

p

2· dp

dx= − 16

π2· λ · ZcpRTcp · M2

d5

or

p2(x) = p02 − 16λ · ZavRTavM2

π2d5x. (4.35)

Here we used the initial condition p = p0 at x = 0. Equation (4.33) means thatp2(x) decreases linearly with the pipeline section length

p2(x) = p02 − (p0

2 − pL2) · x

L(4.36)

where pL is the pressure at the section end.If we define the average pressure pav over the pipeline section length as

pav = 1

L·∫ L

0p(x) dx

and we insert it into the distribution of p(x) from Eq. (4.36), we get

pav = 2

3

(p0 + pL

2

p0 + pL

). (4.37)

From Eq. (4.36) one gets in particular• pressure pL (in Pa) at the end of the pipeline section with length L for given

mass flow rate M

pL2 = p0

2 − 16λ · ZavRTav · L

π2d5· M2 (4.38)

or

pL =√

p02 − 16λ · ZavRTav · L

π2d5· M2; (4.39)

• mass flow rate M (in kg s−1) for given pressures p0 and pL at the beginningand end of the pipeline section, respectively

M = π

4

√p0

2 − pL2

λ · ZavRTav · L· d5. (4.40)


If we take λ = 0.067(2�/d)0.2 in accordance with Eq. (1.34), the mass flowrate is found to be proportional to d2.6

M = A ·√

p02 − pL

2

L· d2,6

where A is the proportionality factor.The mass flow rate M of gas (kg s−1) can also be measured by the volume flow

rate taken under standard conditions, i.e. at pst = 101 325 Pa; Tst = 293.15 K.To do this it is sufficient to divide M by the density ρst of gas under standardconditions

Qk = M

ρst= ρ

ρst· vS = ρ

ρst· Qv (4.41)

Here ρ and Qv = vS are the density and volume flow rate of gas at the pipelinecross-section, respectively. The quantity Qk (m3 s−1) is the so-called commercialflow rate of gas. In fact the commercial flow rate of gas is the mass flow rateexpressed in volume units under standard conditions, in other words the massflow rate in volume calculus. From Eq. (4.41) in particular follows that thecommercial flow rate Qk of gas is ρ/ρst times greater than the volume flowrate Qv. The key advantage is that in contrast to the volume flow rate thatvaries from one cross-section to another, the commercial flow rate remainsinvariable with the length of the gas-pipeline in stationary flows.

4.6.2Pressure Distribution in a Gas-Pipeline with Great Difference in Elevations

In this case one has to use the following equations instead of Eq. (4.34) ρvS = M = const.,dp

dx= −λ

1

d· ρv2

2− ρg · sin α, α = dz

dx

(4.42)

containing the pipeline profile z(x). If we take T ≈ Tav = const. andZ ≈ Zav = const., we get

dp(x)2

dx= − 2g

ZavRTav

dz

dx· p(x)2 − λ · ZavRTav · M2

(πd2/4)2d(4.43)

to find the function p(x).The solution of this equation with initial condition p(0) = p0 is

p(x)2 = −λ · ZavRTav · M2

(πd2/4)2d·∫ x

0exp

[2g

ZavRTav(z(ς) − z(x))

]dς

+ p02 × exp

[2g

ZavRTav(z0 − z(x))

]. (4.44)


Assuming p(L) = pL in Eq. (4.44), the formula for the mass flow rate of gastakes a form similar to Eq. (4.40)

M = π

4

√p0∗2 − pL

2

λ · ZavRTav · L∗· d5 (4.45)

in which the initial pressure and the length of the pipeline section arechanged to

p0∗ = p0 ·√

exp[

2g

ZavRTav(z0 − zL)

],

L∗ = L ·(

1

L·∫ L

0exp

[2g

ZavRTav(z (ς) − zL)

]dς

) (4.46)

It should be noted that the ratio 2g(z − zL)/(ZavRTav) is usually small when thedifference in pipeline elevations is 100–1000 m (ZavRTav ≈ 150 000 m2 s−1).Therefore Eq. (4.46) could be simplified by expansion of the exponentialfunction in the Taylor series. Accurate to the second term we obtain

p0∗ = p0 ·(

1 + g

ZavRTav(z0 − zL)

),

L∗ = L ·(

1 + 2g

ZavRTav(zav − zL)

) (4.47)

Here zav is the average elevation of the gas-pipeline section

zav = 1

L·∫ L

0z(ς) dς.

Equations (4.47) shows that, even in the case when elevations of the beginningand the end of the pipeline section coincide, that is at z0 = zL, the lengthof the section has to be changed. At zav > zL, it should be enhanced (gas-pipelines running through a mountain pass), whereas at zav < zL it should bediminished (gas-pipelines running along the sea bottom).

4.6.3Calculation of Stationary Operating Regimes of a Gas-Pipeline (General Case)

To calculate the stationary operating regimes of a gas-pipeline the followingsystem of ordinary differential equations is used

dp

dx= −λ

1

d· ρv2

2− ρg · sin α(x), sin α = dz

dx, (v > 0),

ρvCpdT

dx− ρvCpD∗

dp

dx= −4K

d(T − Tex)


When taking into account that ρg dz � dp we have a systemdp

dx= −λ

1

d· ρv2

2,

ρvCpdT

dx= −4K

d(T − Tex) − ρvCpD∗ · λ

1

d· ρv2

2

(4.48)

This system should be supplemented with the following equations

ρvS = M, ρ = p

(ZRT), Z = Z(p, T);

D∗ = −RT(∂Z/∂p)T

Cp; λ = 0.067 ·

(2�

d

)0.2

as well as with initial or initial and boundary conditions.The following problems can be solved:

• the pressure p0 and temperature T0 at the initial cross-section of thegas-pipeline section as well as the mass flow rate M of gas are known. It isrequired to find the pressure pL and temperature TL at the end of thegas-pipeline section;

• the pressure p0 and temperature T0 at the initial cross-section of thegas-pipeline section as well as the pressure pL at the end of the gas-pipelinesection are known. It is required to find the mass flow rate M and thetemperature TL at the end of the gas-pipeline section;

• the pressure p0 and temperature T0 at the initial cross-section of thegas-pipeline section and the pressure pL at the end of the gas-pipelinesection as well as mass flow rate M of gas are known. It is required to findthe diameter d of the gas-pipeline providing this flow rate;

• at the initial cross-section of the gas-pipeline section there is a compressorstation. The pressure pen, flow rate Qen and temperature Ten of gas at theentrance to the compressor station and the pressure pL at the end of thegas-pipeline section are known. It is required to choose the number andtype of gas-pumping aggregates, compression ratio ε = p0/pen and thenumber of revolutions n of the centrifugal blower shafts.

4.6.4Investigation of Thermal Regimes of a Gas-Pipeline Section

To consider the thermal regimes of gas transportation it is convenient to passinto the phase plane of variables (p, O). Dividing the second equation of thesystem (4.48) by the first equation of the same system, we get the differentialequation

dT

dp= D∗(p, T) + 8

q3M

· K

λ· ρ(T − Tex)

Cp(p, T)(4.49)


containing only one unknown function T(p). This equation should be solvedat the segment pL ≤ p ≤ p0 under the condition T(p0) = T0, where p0 and T0

are the pressure and temperature, respectively, of the gas at the beginning ofthe pipeline section; pL the pressure at the section end being unknown and tobe determined; qM = QMS is the specific mass flow rate.

If we neglect the Joule–Thomson effect (D∗ ≈ 0), the right-hand side ofEq. (4.49) would be positive at T > Tex and negative at T < Tex. Hence, in thefirst case the temperature of gas monotone increases with pressure whereas inthe second case it monotone decreases. Mathematically speaking, the straightline T = Tex parallel to the abscissa axis is called the separatrix of Eq. (4.49)because it separates solutions of different types. When the initial temperatureof the gas exceeds the temperature of the environment, the gas cools whenmoving from higher to lower pressure, whereas in the opposite case it heats.

The inclusion of the Joule–Thomson effect changes the pattern of solutionof Eq. (4.49). On the plane (p, T) exists as before a separatrix separating thedifferent types of solutions of Eq. (4.49). However, this separatrix is no longera straight line. It always lies below the straight line T = Tex, i.e. it appears to bein the temperature region below the temperature of the surrounding medium(see Fig. 4.5, curve 4).

If the temperature of the environment is greater than Tex, then dT/ dp > 0.This means that when the pressure falls from p0 at the beginning of thepipeline section to pL at the end of the section, it may become less than the

Figure 4.5 Solutions of differential equation (4.49):1 – monotonic solutions; 2 – solutions with a maximum point;3 – curve dT/ dp = 0 of the temperature maxima; 4 – separatrix ofsolutions of Eq. (4.49).


temperature of the environment, since if T = Tex the derivative dT/ dp > 0 isindependent of the value of the heat-transfer factor (see Fig. 4.5 curve 1).

If the initial parameters of the gas are such that the point (p0, T0) is locatedbelow the separatrix 4 of Eq. (4.49), the derivative dT/ dp could change sign(see Fig. 4.5, curves 2). The latter happens at points on line 3 determined bythe condition of vanishing of the right-hand side of Eq. (4.49), namely

D∗(p, T) + 8

q3M

· K

λ· ρ(T − Tex)

Cp(p, T)= 0.

At points (p, T) of this curve dT/ dp = 0, therefore the temperature of the gas atthese points reaches an extremum, namely a maximum. The gas temperaturefirst increases with decreasing pressure, at the point of intersection with curve 3it reaches a maximum (Tmax < Tex) and then begins to diminish monotonically,remaining as before lower than the temperature of the environment.

The presence of maximum points on the curve T(p) giving temperaturedistribution is related to processes having diverse actions: heating of the gasdue to its mass exchange with the surrounding medium and cooling owing tothe Joule–Thomson effect.

The phenomena under consideration may be used in exploitation of agas pipeline section located in permafrost earth. If we provide an initialtemperature of the gas T0 such that the point (p0, T0) is located below theseparatrix 4 of the solutions of Eq. (4.49), the temperature of the gas willremain below the temperature of the permafrost earth all the way along thepipeline section. This excludes the possibility of warming of the surroundingground and imparts greater stability to the pipeline in the earth.

4.7Modeling of Blower Operation

The motion of gas in a gas-pipeline is determined by compressor stations(GCS) located at the beginning of each pipeline section, or more precisely byblowers accomplishing gas compression. The main purpose of the blowersand of GCS as a whole is to force gas to move from a region of lower pressure(suction region at the GCS entrance) into a region of greater pressure (regionof pumping at the GCS exit). The gas by itself cannot flow against pressure,therefore it is necessary to spend energy for forced flow in this direction.Such forced flow of gas against a pressure force is performed by gas-pumpingaggregates (GPA) consisting of a drive (gas-turbine, electric, gas-motor and soon) producing rotation of an impeller shaft in centrifugal blowers (CFP) orthe reciprocating motion of a piston in piston engines (PP), and by the bloweritself. Displacement of gas from a region of lower pressure into a region ofhigher pressure, in other words gas compression, is accomplished in the blower.It is quite clear that gas in the gas-pipeline moves from a cross-section with

4.7 Modeling of Blower Operation 101

greater pressure to a cross-section with lower pressure overcoming frictionforces.

As a rule the compression station consists of separate plants equippedwith several GPA with blowers connected in parallel in the case of single-step compression or in series in the case of multi-step compression. Onmain gas-pipelines centrifugal blowers are predominantly used following thesame pattern as centrifugal blowers for fluid, see Fig. 4.1. Gas is sucked intothe center of the impeller and thrown by centrifugal rotational force to theperiphery of the impeller in the discharge line. The rotational rate of theimpellers of a large CFP is 4000–15 000 rpm. The gas compression ratiodepends on the type of blower, the rotational speed of its impeller (modernGPA have commonly controlled rotational speed), pressure and temperatureat the discharge line and above all on gas flow rate.

A mathematical model of the centrifugal blower operating in the stationaryregime involves algebraic dependences of the gas compression ratio ε and thedeveloped specific power N/ρe on the gas parameters at the blower suctionline indicated by the subscript e and the number of impeller revolutions n[

ε = p0

pB= ε(ρB, pB, TB, QB, n, . . .),

N = N(ρB, pB, TB, QB, n, . . .)(4.50)

where the dots denote parameters related to structural features of the blower.From dimensional theory (see Chapter 6) it follows that these dependences

may be represented as ε = p0

pB= f

(pB/ρB

n2D2im

,QB/σ

nDim, . . .

),

N

ρB=

(n

n0

)3

· φ

(pB/ρB

n2D2im

,QB/σ

nDim, . . .

) (4.51)

where Dim, σ, n0 are the diameter of the impeller, the area of the suctionbranch pipe cross-section, the nominal number of revolutions of the blowershaft, respectively. If the structural parameters of the blower are taken asinvariable dependences, Eq. (4.50) takes the following form ε = p0

pB= f

(pB/ρB

n2,

QB

n, . . .

),

N

ρB=

(n

n0

)3

· φ

(pB/ρB

n2,

QB

n, . . .

) or

ε = p0

pB= f

(ZBRTB

n2,

QB

n, . . .

),

N

ρB=

(n

n0

)3

· φ

(ZBRTB

n2,

QB

n, . . .

)To determine the universal characteristic of the centrifugal blower the so-calledreduced conditions, denoted by subscripts r, are used. The number of blower


shaft revolutions in such cases is taken equal to a nominal value n0, theproperties of the gas and the conditions at the blower entrance are fixed:R = Rr; T = Tr; Z = Zr. Tests of blowers conducted under these conditionspermit the determination of the functions ε = f

(ZrRrTr

n02

,QB

n0, . . .

),(

N

ρ

)r

= φ

(ZrRrTr

n02

,QB

n0, . . .

) (4.52)

It is evident that the operating characteristics of a blower observed underarbitrary conditions but not under reduced conditions could be found from theuniversal characteristics written as follows

ε = f

(ZBRBTB

n2,

QB

n

)= f

(ZrRrTr

n02

· n02

n2

ZBRBTB

ZrRrTr,

QB · n0/n

n0

),

N

ρB=

(n

n0

)3

·(

N

ρB

)r

=(

n

n0

)3

· φ

(ZrRrTr

n02

· n02

n2

ZBRBTB

ZrRrTr,

QB · n0/n

n0

)It follows that the operation characteristics of each blower under arbitraryconditions are obtained from some universal characteristics of the sameblower, called reduced, through division of the arguments of these characteristicdependences governing the parameters by

(n

n0

)2

· ZrRrTr

ZBRTBand

n

n0,

respectively. This conclusion has a simple geometrical interpretation. If in thespace to build the graphics of the dependences

ε = f [(n/n0)r, (QB)r] and (N/ρB)r = φ[(n/n0)r, (QB)r],

the characteristics of a centrifugal blower operating under arbitrary entranceconditions and at a number of revolutions n distinct from the nominal valuen0 are determined from these graphics by lengthening of the argument axesby factors n/n0 · √

ZrRrTr/ZBRTB and n/n0, respectively. At this the graphic ofthe second dependence is also stretched by the factor (n/n0)

3 in the directionof the function axis.

If we introduce

(n

n0

)r= n

n0·√

ZrRrTr

ZBRTBand (QB)r = QB · n0

n. (4.53)


The characteristics of the centrifugal blower take the universal form ε = f

[(n

n0

)r, (Qe)r

],(

N

ρB

)r= φ

[(n

n0

)r, (Qe)r

] (4.54)

where

N

ρB=

(n

n0

)3

·(

N

ρe

)r.

In Fig. 4.6 are depicted reduced (universal) characteristics of one of thecentrifugal blowers 370-18-1 produced in Russia. The parameters of thisblower are: n0 = 4800 rpm; Tr = 288 K; Zr = 0.9; Rr = 490 J kg−1 K−1.

Figure 4.6 Reduced characteristics of the blower 370-18-1: Tr = 288 K;Zr = 0.9; Rr = 490 J kg−1 K−1.


Exercise. It is required to determine the rotational speed (number ofrevolutions per minute, rpm) of the centrifugal blower shaft 370-18-1 neededto provide transportation of natural gas (µ = 17.95 kg kmol−1, pcr = 4.7 MPa,Tcr = 194 K) with commercial flow rate 22 million m3 day−1 and compressionratio ε = 1, 25. It is known that the pressure and temperature of the gas at thesuction line of the blower are 3.8 MPa and +15 ◦C, respectively.

Solution. Calculate first the parameters of the transported gas

R = 8314

17.95∼= 463.1 J kg−1 K−1),

with Eq. (2.14) we obtain:

ZB = 1 − 0.4273 · 3.8

4.7·(

288

194

)−3.668 ∼= 0.919

ρst = pst

RTst= 101 300

463.1 · 293∼= 0.746 kg m−3

ρB = pB

ZBRTB= 3.8 · 106

0.919 · 463.1 · 288∼= 31.002 kg m−3

QB = Qk · ρst/ρB = 22 · 106

24 · 60· 0.746

31.002∼= 367.6 m3 min−1.

Then determine the reduced parameters of the operating regime of thecentrifugal blower(

n

n0

)r

= n

n0

√ZrRrTr

ZBRTB= n

n0

√0.90 · 490 · 288

0.919 · 463.1 · 288∼= 1.018 · n

n0;

(QB)r = QBn0

n= 367.6 · n0

nm3 min−1.

Since the compression ratio ε is known and is equal to 1.25, it is necessary,using the reduced characteristics of the blower 370-18-1 (see Fig. 4.6), to selecta value of n/n0 such that the point with coordinates QB = 367.6/(n/n0) andε = 1.25 would lie on the characteristic (n/n0)r = 1.018 · n/n0. The solution issought by the iteration method.

1. Let (n/n0)r = 0.9 ⇒ n/n0∼= 0.916;

(QB)r = 367.6/0.916 ∼= 401 m3 min−1 ⇒ ε ∼= 1.2 < 1.25 (see Fig. 4.6),consequently (n/n0)r should be increased.

2. Let (n/n0)r = 1.0 ⇒ n/n0 = 1.0/1.018 ∼= 0.982;(QB)r = 367.6/0.982 ∼= 374 m3 min−1 ⇒ ε ∼= 1.25 (see Fig. 4.6),consequently the solution can be taken as correct.Hence n = 0.982 · n0 = 0.982 · 4800 ∼= 4714 rpm.

Answer. 4714. rpm.


Useful Power of a BlowerLet us show now how we can estimate the useful power needed for gascompression from pressure pB at the blower entrance to pressure p0 at theblower exit.

From the total energy balance equation (1.35) written for a mass of gas goingbetween the entrance cross-section x1 and the exit cross-section x2 of a blowerin the case of stationary flow it follows that[(

αkv2

2+ ein + p

ρ

)· ρvS

]x2

−[(

αkv2

2+ ein + p

ρ

)· ρvS

]x1

= qex + Nus

where qex is the external heat inflow (qex > 0) to the gas or heat outflow fromthe gas (qex < 0); Nus is the useful power of mechanical forces acting on thegas, i.e. the useful power of the blower. Taking into account that the mass flowrate of gas M = ρvS through all gas-pipeline cross-sections in stationary flowremains constant, we have[(

αKv02

2− αKvB

2

2+ (J0 − JB)

)]· M = qex + Nus

where J = ein + p/ρ is the gas enthalpy. Neglecting the difference (αKv02 −

αKvB2)/2 in the kinetic energies of the gas before and after compression

and assuming the process of gas compression in the blower to be adiabatic(qex = 0), we get

Nus∼= M · (J0 − JB) = ρBQB · (J0 − JB) (4.55)

where Q = vS is the volume flow rate of gas (ρBQB = ρ0Q0 = const.).For a perfect gas J = CpT + const. and the following relations are valid

Nus = ρBQB · (J0 − JB) = ρBCpQB · (T0 − TB) = Cp

R

pBQB

TB(T0 − TB)

or

Nus = Cp

Cp − Cv· pBQB

(T0

TB− 1

)= γ

γ − 1· pBQB

(T0

TB− 1

)where γ = Cp/Cv is the adiabatic index (for methane γ ≈ 1.31).

Taking into account that in an adiabatic process the temperature T varies in

accordance with the power law T/Te = (p/pe)γ−1γ , we finally obtain

Nus = γ

γ − 1· pBQB

(ε

γ−1γ − 1

). (4.56)

Here ε = p0/pB is the compression ratio.


A similar formula written as a function of the gas parameters at the exit ofthe blower has the form

Nus = γ

γ − 1· p0Q0

(1 − ε

1−γ

γ

). (4.57)

Here the subscript 0 denotes that the relevant quantity is taken at the exit ofthe blower.

Exercise. The pressure pB before the blower is 3.5 MPa, the compression ratioof a gas with γ = 1.31 is 1.4, the volume flow rate at the entrance QB of theblower is 500 m3 min−1. It is required to determine the useful power Nus spentby the blower for gas compression.

Solution. The useful power is obtained from Eq. (4.48)

Nus = 1.31

1.31 − 1· 3.5 · 106 · 500

60·(

1.41.31−1

1.31 − 1)

∼= 10.2 · 106 W.

Hence the useful power is 10.2 MW.

Answer. 10.2 MW.Finally in this chapter let us consider an exercise on the calculation of a

gas-pipeline section in combination with a compressor station.

Exercise. Natural gas (R = 18.82 kg kmol−1, pcr = 4.75 MPa, Tcr = 195 K) isbeing transported along a 105-km gas-pipeline section (D = 1220 × 12 mm,� = 0.03 mm) with the help of two identical GPA equipped with blowers370-18-1 connected in parallel. It is required to determine the compressionratio of the gas ε and the number of revolutions n of the blower rotors neededto ensure a commercial flow rate of 21 billion m3 year−1 in the gas-pipeline(the number of working days in a year is taken to be 350). It is given thatthe pressure at the end of the pipeline section is 3.8 MPa and at the blowersuction line is 4.7 MPa; the temperature of the gas at the suction line is 12 ◦C;the temperature of the gas after compression is expected to be 30 ◦C; thetemperature of the surrounding ground is 8 ◦C.

Solution. Taking in Eq. (4.33) x = L and Qk = M/ρst, we have

p02 = pL

2 + 16λ · ZavRTavρst2 · L

π2d5· Qk

2.

Now calculate successively the quantities in this relation

Qk = 21 · 109

350 · 24 · 3600∼= 694.4 m3 s−1;

R = R0

µ= 8314

18.82∼= 441.7 J kg−1 K−1;


ρst = pst

RTet= 10 125

441.7 · 293∼= 0.783 kg m−3;

λ = 0.067 ·(

2�

d

)0.2

= 0.067 ·(

2 · 0.03

1196

)0.2 ∼= 0.0093.

Calculate the average gas temperature Tav in the pipeline section

Tav = Tex + T0 − TL

ln(

T0. − Tex

TL − Tex

) = 8 + 30 − 12

ln(

30 − 8

12 − 8

) ∼= 18.6 ◦C = 291.6 K.

The gas over-compressibility factor Z is calculated with Eq. (2.16) taking theaverage pressure pav in a first approximation to be equal to the pressure atthe end of the pipeline section and the temperature to be the temperatureaveraged over the section

Z = 1 − 0.4273 · (3.8/4.75) · (291.6/195)−3.668 ∼= 0.922.

After this we calculate the pressure p0 at the beginning of the pipeline section.We begin with the factor A

A = 16 · λ · ZavRTavρst2 · L

π2d5· Qk

2

= 16 · 0.0093 · 0.922 · 441.7 · 291.6 · 0.7832 · 105 000

3.142 · 1.1965· 694.42

∼= 22.73 · 1012

and then calculate the pressure p0

p0 =√

(3.8 · 106)2 + 22.73 · 1012 ∼= 6.1 · 106 Pa or 6.1 MPa.

The obtained value shows that the average pressure pav in the pipeline section isequal to 2/3 · (4.5 + 6.12/10.6) ∼= 5.34 MPa, which is greater than the expected3.8 MPa. Hence, the calculation should be corrected.

Performing the second approximation for pressure p = pav = 5.34 MPa, weget

Z = 1 − 0.4273 · (5.34/4.75) · (291.6/195)−3.668 ∼= 0.890;

A ∼= 21.94 · 1012 Pa2;

p0 =√

(3.8 · 106)2 + 21.94 · 1012 ∼= 6.0 · 106 Pa or 6.0 MPa.

We see that the obtained value of p0 is practically unchanged. Thus thecompression ratio ε, which should be provided by the blowers 370-18-1, is6.0/4.7 ∼= 1.28.

Once the required compression ratio has been obtained, it is time to calculatethe gas parameters at the suction line of each of the blowers connected


inparallel. We have

ZB = 1 − 0.4273 · 4.7

4.75·(

285

195

)−3.668 ∼= 0.895;

ρB = pB

ZBRTB= 4.7 · 106

0.895 · 441.7 · 285∼= 41.716 kg m−3

.

In a parallel connection of identical blowers the flow rate is distributed equallybetween them, therefore it is

QB = Qk · ρst/ρB = [(21 000/2)/350] · 106

24 · 60· 0.783

41.716∼= 391 m3 min−1.

Determine the reduced parameters of the operating regime of the centrifugalblower:(

n

n0

)r

= n

n0

√ZrRrTr

ZBRTB= n

n0

√0.90 · 490 · 288

0.895 · 441.7 · 285∼= 1.062 · n

n0;

(QB)r = QBn0

n= 391 · n0

nm3 min−1.

As the compression ratio ε has already been found to be equal to 1.28, it isnecessary, using the characteristics of the blower 370-18-1 depicted in Fig. 4.6,to select n/n0 in such a way that a point with coordinates (QB)r = 391/(n/n0)

and ε = 1.28 would lie on the characteristic (n/n0)r = 1.062 · n/n0. Theselection is performed by the iteration method.

1. We take (n/n0)r = 1.0 ⇒ n/n0 = 1.0/1.062 ∼= 0.942;(QB)r = 391/0.942 ∼= 415 m3 min−1 ⇒ ε ∼= 1.25 (see Fig. 4.6), which isless than the required value 1.28. Therefore (n/n0)r should be increased.

2. Let now (n/n0)r = 1.05 ⇒ n/n0 = 1.05/1.062 ∼= 0.989;(QB)r = 391/0.989 ∼= 395 m3 min−1 ⇒ ε ∼= 1.28 (see Fig. 4.6). Hence,the solution is found.As a result we have n = 0.989 · n0 = 0.989 · 4800 ∼= 4750 rpm.

109

5Closed Mathematical Models of One-DimensionalNon-Stationary Flows of Fluid and Gas in a Pipeline

In this chapter are considered the most important models of one-dimensionalnon-stationary flows of fluid and gas in pipelines. The equations obtained inChapter 1 are used as a starting point.

5.1A Model of Non-Stationary Isothermal Flow of a Slightly Compressible Fluidin a Pipeline

At the basis of this model lie the following assumptions:• the variation of fluid density �ρ is much less than its nominal value ρ0, that

is �ρ � ρ0, where �ρ = ρ0 × (p − p0)/K in accordance with Eq. (2.6). Forexample, at ρ0 = 1000 kg m−3, p − p0 = 1.0 MPa (106 Pa ≈ 10 atm),K = 103 MPa (109 Pa), the variation of fluid density �ρ is only 1 kg m−3;

• the variation of pipeline cross-section area �S is much less than itsnominal value S0, that is �S � S0, where �S = πd0

3/4Eδ · (p − p0) or

�S = S0d0/Eδ · (p − p0). For example, at d0 = 500 mm, δ = 10 mm,ρ0 = 103 kg m−3, p − p0 = 107 Pa (≈100 atm), E = 2 · 1011 Pa (pipe steel),the variation of the pipeline diameter �d is 0.06 mm, the variation of thepipeline cross-section area �S is ≈0.5 cm2, whereas S0

∼= 1960 cm2;• the tangential friction stress |τw| at the pipeline walls in accordance with

Eq. (1.28) is determined by the formula |τw| = λ(Re, ε) · ρv2/8 with thefactor λ dependent on the governing parameters Re = vd/ν and ε = �/dgiven in the same form as in stationary flow. Such an assumption is calledthe hypothesis of quasi-stationarity. For example, for λ = 0.02, v = 1.5 m s−1,ρ0 = 1000 kg m−3, then |τw| ∼= 5.6 Pa.

The first equation of the model is continuity equation

∂ρS

∂t+ ∂ρvS

∂x= 0.

With regard to the accepted assumptions this equation could be transformedto the following form

∂ρS

∂t= ρ

∂S

∂t+ S

∂ρ

∂t≈ ρ0

dS

dp

∂p

∂t+ S0

dρ

dp

∂p

∂t=

(ρ0S0d0

Eδ+ ρ0S0

K

)· ∂p

∂t;


110 5 Mathematical Models of 1D Non-Stationary Flows of Fluid and Gas in a Pipeline

∂ρvS

∂x≈ ρ0S0 · ∂v

∂x.

As a result the following equation is obtained(ρ0

K+ ρ0d0

Eδ

)· ∂p

∂t+ ρ0

∂v

∂x= 0.

The factor in parentheses before the derivative of pressure with respect to timehas dimension inverse to the square of the velocity, therefore it can be denotedas 1/c2, where the parameter c, given by

c = 1√ρ0

K+ ρ0d0

Eδ

(5.1)

is called the speed of wave propagation in the pipeline (c ≈ 1000 m s−1). Ifρ0 = 1000 kg m−3, K = 109 Pa, d0 = 500 mm, δ = 10 mm, E = 2 · 1011 Pa,then

c = 1√103

109+ 103 · 0.5

2 · 1011 · 0.01

∼= 895 m s−1.

With regard to the introduced designation the first equation of the model takesthe form

∂p

∂t+ ρ0c2 · ∂v

∂x= 0. (5.2)

The second equation of the model is the momentum equation (1.10)

ρ

(∂v

∂t+ v

∂v

∂x

)= − ∂p

∂x− 4

dθ

τw − ρg sin α(x)

Replacement of τw with the expression containing the average velocity of theflow v yields

ρ

(∂v

∂t+ v

∂v

∂x

)= − ∂p

∂x− λ

1

d0

ρv|v|2

− ρg sin α(x). (5.3)

For slightly compressible fluids, among which are water, oil and oil products,the following simplifying assumptions can be made

∂v

∂t≈ ρ0

∂v

∂t,

ρv∂v

∂x∼= ρ0v

∂v

∂x= ∂

∂x

(ρ0v2

2

),

∂

∂x

(p + ρ0v2

2

)≈ ∂p

∂x.

5.1 A Model of Non-Stationary Isothermal Flow of a Slightly Compressible Fluid in a Pipeline 111

The last approximation is valid because it is easy to verify that �(ρ0v2/2) ��p. Really, at ρ0 ≈ 1000 kg m−3 and v ≈ 1–2 m s−1, �(ρ0v2/2) ≤ 2000 Pa(≈0.04 atm), whereas �p is measured in atmospheres or even tens ofatmospheres. In the general case �(ρ0v2/2) ≈ ρ0v �v while �p ≈ ρ0c �v,hence �(ρ0v2/2)/�p ≈ v/c. Since v � c in pipelines, the ratio �(ρ0v2/2)/�pis negligibly small.

With regard to the given assumptions the momentum equation takes thefollowing form

ρ0∂v

∂t+ ∂p

∂x= −λ(Re, ε) · 1

d0

ρ0v|v|2

− ρ0g sin α(x)

This equation is the second equation of the model.Hence, the mathematical model of slightly compressible fluids is represented

by a system of two differential equations∂p

∂t+ ρ0c2 · ∂v

∂x= 0,

ρ0∂v

∂t+ ∂p

∂x= −λ(Re, ε)

1

d0

ρ0v|v|2

− ρ0g sin α(x)

(5.4)

to determine the two unknown functions p(x, t) and v(x, t) dependent on thecoordinate x and time t.

The system of differential equations (5.4) requires for its solution initial andboundary conditions, being also components of the model under consideration.We will deal with these conditions below.

Virtual MassThe hypothesis of quasi-stationarity in accordance with which the tangentialstress τw at the internal surface of the pipe is represented by the equality τw =λ(Re, ε) · ρ0v|v|/8, asserts in particular that it depends on the instantaneousvalue of the average flow velocity v(x, t) but not on derivatives of the velocitieswith respect to time and coordinate. In fact the following equation holds

4

dτw · vS = ρ0gvS · i0 + ρ0S

d

dt

[(αk − 1)

v2

2

](5.5)

(see Eq. (1.23)), reflecting the transformation of the work of friction forces(4τw · vS/d), and in one-dimensional flow appearing as external forces, into

kinetic energy ρ0S ddt

[(αk − 1) v2

2 ] of the intrinsic motion of the fluid layersrelative to its center of mass and into heat ρ0gS(v · i0) due to the work ofinternal friction forces. The values of the factor αk in this equation vary from4/3 for laminar flow up to 1.02–1.05 for turbulent flow. If we take for αk themean value of this factor, from Eq. (5.5) we get

4

dτw = ρ0(αk − 1) · dv

dt+ ρ0g · i0 (5.6)


indicating that the tangential stress τw contains the term ρ0(αk − 1) · dv/dtproportional to the fluid particle acceleration. The physical nature of this termis hidden in the origin of the additional resistance force ρ0(αk − 1) · v causedby realignment of the internal structure of the flow taking place even whenthere is dissipation of mechanical energy into heat owing to the work of theviscous friction force, that is the term ρ0g · i0, could be neglected. It is evidentthat if v = const. the first term vanishes.

Substitution of Eq. (5.6) into the momentum equation (5.2) yields

ρ0dv

dt= − ∂p

∂x− ρ0(αk − 1)

dv

dt− ρ0gi0 − ρ0g sin α(x)

or

αkρ0dv

dt= − ∂p

∂x− λ(Re, ε) · 1

d0

ρ0v|v|2

− ρ0g sin α(x) (5.7)

Equation (5.7) differs from Eq. (5.3) only in that in the left-hand side entersnot the true fluid density ρ0 but a quantity αkρ0 differing from ρ0 by thefactor αk. The quantity ρ0(αk − 1) may be called the virtual (additional)mass of the fluid. Hence, the inertial properties of a fluid in non-stationaryprocesses are characterized by changing density, to the latter is added acertain quantity dependent on the flow regime. In developed turbulentflows this change is slight (αk ≈ 1.03), but for laminar flow it is greater(αk ≈ 4/3).

5.2A Model of Non-Stationary Gas Flow in a Pipeline

At the basis of this model lie the following assumptions:• the transported media (gas) is compressible, i.e. ρ = ρ(p, T);• the variation of gas-pipeline cross-section area �S can be ignored compared

with the area itself S0, i.e. �S � S0. Therefore S ∼= S0 = πd02/4 = const.;

• the internal energy of the gas is ein = CvT + const.

The system of basic equations is

∂ρ

∂t+ ∂ρv

∂x= 0,

∂ρv

∂t+ ∂

∂x(p + ρv2) = −4τw

d0− ρg sin α,

ρ

(∂ein

∂t+ v

∂ein

∂x

)= 4qn

d0− p

∂v

∂x− ρ · nin

(5.8)

The first equation of this system is the continuity equation reflecting the law ofgas mass conservation in each pipeline cross-section.

5.3 Non-Stationary Flow of a Slightly Compressible Fluid in a Pipeline 113

The second equation of the system is the momentum equation expressingNewton’s second law.

The third equation of the system is the equation of heat inflow followingfrom the laws of total energy conservation of the flow and the variation ofmechanical energy of the transported media.

To the system of equations (5.8) should be added the so-called closingrelations, e.g.

τw = λ(Re, ε)

8· ρv2, nin = −λ(Re, ε) · 1

d0

ρv3

2,

ρ = p

Z(p, T) · RT, ein(T) = Cv · T + const., qn = −κ · (T − Tex).

Hence, if λ(Re, ε) and Z(p, T) are known as functions of their arguments, thesystem of equations (5.8) represents a closed system of three partial differentialequations for three unknown functions p(x, t), v(x, t) and T(x, t) dependenton the coordinate x and time t.

5.3Non-Stationary Flow of a Slightly Compressible Fluid in a Pipeline

Consider the non-stationary flow of a slightly compressible fluid in a pipeline.The basic equations of such flow are represented by the system ofequations (5.4).

5.3.1Wave Equation

Let us consider first the non-stationary flow of a slightly compressible fluid ina horizontal pipeline (α = 0) neglecting for a while terms accounting for thefriction force. Such an assumption is quite allowable for short pipelines andfluids with not too high viscosity. In such a case the system of equations (5.4)takes the form

∂p

∂t+ ρ0c2 · ∂v

∂x= 0,

ρ0∂v

∂t+ ∂p

∂x= 0

or ∂v

∂x= − 1

ρ0c2· ∂p

∂t,

∂v

∂t= − 1

ρ0· ∂p

∂x.

(5.9)


Differentiation of the first equation with respect to t and the second withrespect to x subject to the condition v′′

x,t = v′′t,x yields the equation for p(x, t)

∂2p

∂t2= c2 ∂2p

∂x2. (5.10)

This equation is called the wave equation since it describes the propagation ofwaves and is encountered in different fields of physics.

A similar equation can be obtained for fluid velocity v(x, t)

∂2v

∂t2= c2 ∂2v

∂x2. (5.11)

Equations (5.10) and (5.11) represent partial differential equations whosegeneral solution is expressed by two arbitrary functions

p(x, t) = f1(x − ct) + f2(x + ct) (5.12)

the first of these functions is dependent only on the argument ξ = x − ct andthe second on η = x + ct.

Let us show that Eq. (5.12) gives a solution of Eq. (5.10). We have

∂p

∂t= −cf ′

1ξ + cf ′2η,

∂2p

∂t2= c2f ′′

1ξξ + c2f ′′2ηη = c2(f ′′

1ξξ + f ′′2ηη),

∂p

∂x= f ′

1ξ + f ′2η;

∂2p

∂x2= f ′′

1ξξ + f ′′2ηη.

c2 ∂2p

∂x2= ∂2p

∂t2.

Hence, Eq. (5.12) is a solution of Eq. (5.10) for arbitrary functions f1 and f2.The function f1(x − ct) represents a traveling wave in the positive direction

of the x-axis whereas the function f2(x + ct) represents a traveling wave in thenegative direction of this axis. The magnitude of the velocity propagation ofboth waves, i.e. the propagation velocity of a certain value of function f1 or f2is identical and equal to c.

The form of each of the functions f1 and f2 is determined by the initialconditions for pressure and velocity distributions in the pipeline, that is byp(x, 0) and v(x, 0), as well as by the boundary conditions at the pipeline ends.

The velocity of fluid v(x, t) is determined by the formula

v(x, t) = g1(x − ct) + g2(x + ct).


Functions g1(ξ) and g2(η) are expressed through functions f1(ξ) and f2(η).From Eq. (5.9) follows

g ′1ξ + g ′

2η = 1

ρ0c· [f ′

1ξ − f ′2η],

−c · g ′1ξ + c · g ′

2η = − 1

ρ0[f ′

1ξ + f ′2η],

or {ρ0cg ′

1ξ = f ′1ξ,

ρ0cg ′2η = −f ′

2η.

Integration of the first equation with respect to ξ and the second with respectto η, taking into account that g1 and f1 depend only on ξ and g2 and f2 only onη, yields{

ρ0cg1 = f1 + const.,ρ0cg2 = −f2 + const.

From here follows

ρ0c · v(x, t) = f1(x − ct) − f2(x + ct) = const. (5.13)

With the help of Eqs. (5.12) and (5.13) one can find solutions to differentproblems. We will consider some of them.

5.3.2Propagation of Waves in an Infinite Pipeline

A pipeline is called infinite when it runs in the direction of the x-axis from −∞to +∞. Of course it is only a model of a real pipeline, but it is very useful inthe case of a very long pipeline when boundary effects can be ignored, that iswhen waves reflected from the beginning and the end of a pipeline could beneglected. Suppose also that friction forces are absent.

Addition and subtraction of Eqs. (5.12) and (5.13) yield{p + ρ0c · v = 2 · f1(x − ct) + const.,p − ρ0c · v = 2 · f2(x + ct) + const.

From these expressions it is seen that at those points of the plane (x, t) where(x − ct) remains constant the expression I1 = p + ρ0c · v is also constant and atthose points of the plane (x, t) where (x + ct) remains constant the expressionI2 = p − ρ0c · v is also constant (see Fig. 5.1).

The lines x − ct = const. and x + ct = const. are called characteristics ofthe wave equation and the quantities I1 = p(x, t) + ρ0c · v(x, t) and I2 =p(x, t) − ρ0c · v(x, t) are called the Riemann invariants.

Hence, at each characteristic x − ct = const. with positive slope, dx/dt = +c,the first Riemann invariant I1 is conserved whereas at each characteristic


Figure 5.1 Characteristics on(x, t)-plane.

x + ct = const. with negative slope, dx/dt = −c, the second Riemann invariantI2 is conserved:

at lines x = ct + const. : I1 = p + ρ0cv = const.;

at lines x = −ct + const. : I2 = p − ρ0cv = const.

Problem. Let at an initial instance of time t = 0 in an infinite pipeline(−∞ < x < +∞) there be distributions of pressure p(x, 0) = �(x) and fluidvelocity v(x, 0) = �(x). It is required to determine, what motion appearsin the pipeline at t > 0, i.e. it is required to find the functions p(x, t) andv(x, t) satisfying Eqs. (5.10) and (5.11) and initial conditions p(x, 0) = �(x),v(x, 0) = �(x).

Solution. Consider a plane (x, t) depicted in Fig. 5.1. Let M(x, t) be an arbitrarychosen point of this plane at t > 0. Draw from the point M(x, t) straight lines(characteristics) MA: x − ct = x1 and MB: x + ct = x2.

Since at the characteristic x − ct = x1 the first Riemann invariant I1 isconstant, we can write

pM(x, t) + ρ0c · vM(x, t) = pA(x1, 0) + ρ0c · vA(x1, 0).

At the characteristic x + ct = x2 the second Riemann invariant I2 is constant

pM(x, t) − ρ0c · vM(x, t) = pA(x2, 0) − ρ0c · vA(x2, 0).

Resolving the system of obtained equations relative to pM(x, t) and vM(x, t),we get

pM(x, t) = p(x1, 0) + p(x2, 0)

2+ ρ0c · v(x1, 0) − v(x2, 0)

2,

vM(x, t) = p(x1, 0) − p(x2, 0)

2ρ0c+ v(x1, 0) + v(x2, 0)

2or

pM(x, t) = �(x1) + �(x2)

2+ ρ0c · �(x1) − �(x2)

2,

vM(x, t) = �(x1) − �(x2)

2ρ0c+ �(x1) + �(x2)

2


Substituting in these relations instead of x1 and x2 their expression through xand t, we receive the solution of the problem

pM(x, t) = �(x − ct) + �(x + ct)

2+ ρ0c · �(x − ct) − �(x + ct)

2,

vM(x, t) = �(x − ct) − �(x + ct)

2ρ0c+ �(x − ct) + �(x + ct)

2

(5.14)

Since functions �(x) and �(x) are known, the problem is completely solved.Equalities (5.14) are called the d’Alambert formulas.

5.3.3Propagation of Waves in a Semi-Infinite Pipeline

A pipeline is called semi-infinite when it has the initial cross-section (x = 0)and runs from it in the positive direction of the x-axis (x > 0) to infinity. Itis a model of a pipeline in which the conditions at one of the ends (left end)are taken into account whereas the influence of another end (right end) isneglected.

Problem. Let, at the initial instant of time t = 0 the fluid in a semi-infinite(0 < x < +∞) pipeline be quiescent v(x, 0) = 0 and the pressure constantp(x, 0) = p0. The fluid velocity in the initial cross-section x = 0 at t > 0suddenly begins to change with a law v(0, t) = �(t). It is required to determinethe pattern of fluid flow in the pipeline at t > 0.

Solution. It is required to determine what kind of velocity and pressure wavesbegin to propagate in a semi-infinite pipeline caused by perturbations createdat the left end of the pipeline.

Consider a plane (x, t) at t > 0, x ≥ 0 as depicted in Fig. 5.2. Draw throughthe origin of the coordinates in this plane the characteristic x = ct with positiveslope. At points N of this plane located below the straight line x = ct thed’Alambert formulas (5.14) give the following equations

pN(x, t) = p0 + p0

2+ ρ0c · 0 − 0

2= p0,

vN(x, t) = p0 − p0

2ρ0c+ 0 + 0

2= 0

(5.15)

This result has a simple physical meaning. In the region x > ct of the pipeline,to which the instant of time t has not yet come, a perturbation (signal) ispropagated with velocity c from the initial pipeline cross-section, the fluid is atrest as before, that is its velocity is 0 and the pressure is p0.

Consider now the region x < ct of the pipeline to which at time t have comeperturbations from the initial cross-section of the pipeline. Draw through a


Figure 5.2 Diagram illustratingthe problem on traveling waves.

point M(x, t) of this region two characteristics MC and MB and from the pointC, located at initial cross-section of the pipeline (time t0), the characteristic CA(Fig. 5.2).

With the help of the condition at the characteristic CA consisting in constancyof the second Riemann invariant, and relevant boundary condition we find thepressure in the initial pipeline cross-section at time t0

pC(0, t0) − ρ0c · vC(0, t0) = pA(x1, 0) − ρ0c · vC(x1, 0) = p0 − 0 = p0.

From this follows

pC(0, t0) = p0 + ρ0c · vC(0, t0).

Hence the pressure pC at the initial pipeline cross-section, having been initiallyequal to p0, has varied over ρ0c · vC(0, t0) as a result of the increase in velocityvC(0, t0) in this cross-section.

Since the pressure and velocity of fluid at points C and B are now known,one can find the pressure at the point M(x, t){

pM(x, t) + ρ0c · vM(x, t) = pC + ρ0c · vC = (p0 + ρ0c · v0) + ρ0c · v0,

pM(x, t) − ρ0c · vM(x, t) = pB − ρ0c · vB = p0 − ρ0c · 0 = p0

From this follows{pM(x, t) = p0 + ρ0c · vC(0, t0),

vM(x, t) = vC(0, t0)(5.16)

Note that t0 is determined through the coordinates (x, t) of the point M bythe formula t0 = t − x/c (the characteristic equation for MC is x = c · (t − t0)).Therefore relations (5.16) have the final form

pM(x, t) = p0 + ρ0c · vC

(0, t − x

c

)= p0 + ρ0c · �

(t − x

c

),

vM(x, t) = �(

t − x

c

),

(5.17)

where � is a function determining the velocity change at the initial (x = 0)pipeline cross-section not at the instant of time t but at a later time t − x/c.


It is clear that the quantity x/c is equal to the time in which the perturbation(signal) reaches the considered cross-section x from the initial one.

Formulas (5.17) show that any velocity change at the initial pipeline cross-section propagates rightwards along the pipeline as a traveling wave givingrise to a pressure traveling wave exceeding the initial value p0 by ρ0n · vC.Formulas (5.15) and (5.17) give the complete solution of the consideredproblem.

Exercise. The pumping in quiescent diesel fuel (ρ0 = 840 kg m−3, c =1060 m s−1) in a semi-infinite pipeline has begun with constant flow rateso that v0 = 1.5 m s−1. It is required to determine by how much the pressurehas increased at the pumping cross-section.

Solution. The first formula (5.17) gives p(0, t) − p0 = ρ0c · v(0, t) = ρ0cv0 =840 · 1060 · 1.5 ∼= 1.34 · 106 MPa or approximately 13.6 atm.

Answer. 1.34 MPa.

5.3.4Propagation of Waves in a Bounded Pipeline Section

Consider the problem on wave interaction in a limited pipeline section(0 ≤ x ≤ L) between the initial cross-section x = 0 and the final cross-sectionx = L assuming the absence of friction. In such a problem we need, besidesthe initial conditions p(x, 0) and v(x, 0), boundary conditions reflecting theinteraction of the pipeline section under consideration with equipment locatedat the pipeline ends, that is at x = 0, t > 0 and x = L, t > 0.

Now we should say something about the number of these conditions.Through each point of the left pipeline section boundary x = 0, t > 0 passesonly one characteristic of the wave equation, namely the characteristic ofnegative slope x = −ct + const. Along it the condition p − ρ0c · v = const.should be obeyed. Thus at points on this boundary there is always onealgebraic relation between the quantities p and v. To determine a uniquesolution for p and we need an additional relation, that is one more boundarycondition.

Analogously, through each point of the right boundary x = L, t > 0 of thepipeline section goes only one characteristic with positive slope x = ct + const.Along this characteristic the compatibility condition p + ρ0c · v = const. isobeyed. Thus at these points there should be an additional boundarycondition.

The form of the boundary conditions could be of a great variety, dependenton the type of equipment set at the end cross-sections of the pipeline sectionunder consideration. For example, at the left end of the pipeline can be placeda piston pump providing constant delivery of fluid into the pipeline. Then


the boundary condition at x = 0 would be v(0, t) = v0 = const. at all t > 0. Ifthe right end of the pipeline is open, the boundary condition at the right endwould be p(0, t) = p0 = const. at t > 0. Of course there are also many otherpossible boundary conditions.

Problem. Let a fluid in the pipeline section 0 ≤ x ≤ L be initially (at t = 0)quiescent (v(x, 0) = 0) and the pressure constant (p(x, 0) = p0). At t > 0the fluid starts to be delivered into the pipeline by the law v(0, t) = �(t).The end cross-section x = L of the pipeline is open to the atmosphere,so that the pressure at the cross-section is held constant p(L, t) = p0. Itis required to determine the motion generated in the pipeline at t > 0(Fig. 5.3).

Solution. Consider on the plane (x, t) a strip 0 ≤ x ≤ L, t > 0 correspondingto the variability domain of the problem (Fig. 5.3).

1. The solution in the region 1 enclosed by the triangle OCL, i.e. theregion which the perturbation has not yet reached, is as follows:p(x, t) = p0; v(x, t) = 0.

2. Find now the solution in the region 2 restricted by the triangle OCD(2L/c is the time of the wave double path lengthwise in the pipelinesection).

Figure 5.3 Interaction of waves in the pipeline section.


Determine first the pressure pS at an arbitrary point S of the leftboundary

pS − ρ0c · vS = pQ − ρ0c · vQ = p0 − 0,

pS(0, t) = p0 + ρ0c · vS(t) = p0 + ρ0c · �(t).

Then obtain the pressure and velocity at an arbitrary point M(x, t) of theregion under consideration{

pM + ρ0c · vM = pS + ρ0c · vS = p0 + 2ρ0c · �(tS),

pM − ρ0c · vM = p0 − ρ0c · 0 = p0

from which follows pM(x, t) = p0 + ρ0c · �(

t − x

c

),

vM(x, t) = �(

t − x

c

)3. Find the solution in the region 3 bounded by the triangle CDG (3L/c is

the time of the wave triple path lengthwise in the pipeline section).Determine first the fluid velocity vF at an arbitrary point F of the right

boundary

pF + ρ0c · vF = pS + ρ0c · vS = [p0 + ρ0c · �(tS)] + ρ0c · �(tS)

Since pF = p0 and tS = t − L/c, then vF(L, t) = 2 · �(t − L/c).After this we get the pressure and velocity at the arbitrary point N(x, t) of

the considered regionpN + ρ0c · vN = pR + ρ0c · vR = p0 + 2ρ0c · �

(t − x

c

),

pN − ρ0c · vN = pF − ρ0c · vF = p0 − 2ρ0c · �(

t − x

c

)Hence, pN = p0; vN = 2�(t − x/c).

In a similar manner by the method of characteristics the solution inregions 4, 5 and other regions of the strip under consideration could befound.

5.3.5Method of Characteristics

Let us return to the system of equations (5.4) describing non-stationary flowof a slightly compressible fluid with regard to viscous friction forces

∂p

∂t+ ρ0c2 · ∂v

∂x= 0,

ρ0∂v

∂t+ ∂p

∂x= −λ(Re, ε)

1

d0

ρ0v2

2− ρ0g sin α(x).


Multiplication of the second equation by c and addition of the result to the firstequation yields(

∂p

∂t+ c

∂p

∂x

)+ ρ0c ·

(∂v

∂t+ c

∂v

∂x

)= −λ

ρ0cv|v|2d

− cρ0g sin α.

In a similar manner after subtraction of the second equation multiplied by cfrom the first one we obtain(

∂p

∂t− c

∂p

∂x

)− ρ0c ·

(∂v

∂t− c

∂v

∂x

)= λ

ρ0cv|v|2d

+ cρ0g sin α.

If at the plane (x, t) we consider straight lines determined by the equations

1.dx

dt= c, x − ct = ξ = const.,

2.dx

dt= −c, x + ct = η = const.,

which for the wave equation are called characteristics, it is easy to reveal thatfor any parameter A(x, t)

∂A

∂t+ c · ∂A

∂x=

(dA

dt

)ξ=const.

is valid.This means that the expression on the left-hand side is the derivative of the

function A(x, t) in the direction of the first characteristic (ξ = const.). Similarlyit is true that

∂A

∂t− c · ∂A

∂x=

(dA

dt

)η=const.

i.e. the expression on the left-hand side is the derivative of the function A(x, t)in the direction of the second characteristic (η = const.).

Using now the notion of a directional derivative, one can write the aboveobtained equations as follows(

dp

dt

)ξ=const.

+ ρ0c ·(

dv

dt

)ξ=const.

= −λρ0cv|v|

2d− ρ0cg sin α,(

dp

dt

)η=const.

− ρ0c ·(

dv

dt

)η=const.

= λρ0cv|v|

2d+ ρ0cg sin α,

or d

dt(p + ρ0c · v)ξ=const. = −λ

ρ0cv|v|2d

− ρ0cg sin α,

d

dt(p − ρ0c · v)η=const. = λ

ρ0cv|v|2d

+ ρ0cg sin α.

(5.18)


At λ = 0 and α = 0 the right-hand sides of Eqs. (5.18) vanish. This meansthat along the characteristics of positive slope is conserved the quantityI1 = p0 + ρ0c · v, whereas the quantity I2 = p0 − ρ0c · v is conserved along thecharacteristic of negative slope. This conclusion is consistent with the resultsobtained above for the wave equation.

If λ �= 0, the quantities I1 and I2 are not constants at relevant characteristics.Nevertheless, Eqs. (5.18) may be used to calculate various non-stationary flowsin a pipeline, especially when numerical methods are used.

Let for example at the instant of time tm−1 (in particular at t = 0), thedistributions of pressure p = p(x, tm−1) and velocity v = v(x, tm−1) in thepipeline be known. Show how the values of these parameters at the nextinstant of time tm = tm−1 + �t could be calculated.

Consider on the plane (x, t) a rectangular grid with coordinate step �xand time step �t = �x/c (Fig. 5.4). Through the nodes of the resulting gridlet us draw characteristics x = ct + const. and x = −ct + const. of positiveand negative slope, respectively. Continuous distribution of the soughtfunctions p(x, t) and v(x, t) is replaced with discrete values pk,m = p(xk, tm)

and vk,m = v(xk, tm) of the grid functions at the grid nodes. Suppose that allvalues pk,m−1 and vk,m−1 are known and it is required to find the values pk,m

and vk,m of grid functions at t = tm.Let M(xk, tm) be an arbitrary point on the plane (x, t). Replacing the directional

derivatives in Eqs. (5.18) with finite differences along the characteristics AMand BM we obtain

�(p + ρ0c · v)

�t

∣∣∣∣ξ=const.

= −c · φA,

�(p − ρ0c · v)

�t

∣∣∣∣η=const.

= c · φB,

Figure 5.4 Design diagram of the characteristic method.


where φ = λρ0v|v|

2d + ρ0g · sin α. In addition

�(p + ρ0c · v)∣∣ξ=const. = (pM + ρ0c · vM) − (pA + ρ0c · vA)

�(p − ρ0c · v)∣∣η=const. = (pM − ρ0c · vM) − (pB − ρ0c · vB).

From this follows a system of equations to determine the pressure pM andvelocity vM of the fluid at point M through known values of these parametersat points A and B{

pM + ρ0c · vM = pA + ρ0c · vA − �t · cφA

pM − ρ0c · vM = pB − ρ0c · vB + �t · cφB.

or {pk,m + ρ0c · vk,m = pk−1,m−1 + ρ0c · vk−1,m−1 − �x · φk−1,m−1

pk,m − ρ0c · vk,m = pk+1,m−1 − ρ0c · vk+1,m−1 + �x · φk+1,m−1.

From this system of linear equations we get the pressure pk,m and the velocityvk,m of the flow

pk,m = pk−1,m−1 + pk+1,m−1

2+ ρ0c · vk−1,m−1 − vk+1,m−1

2

+�x

2· (φk+1,m−1 − φk−1,m−1)

vk,m = pk−1,m−1 − pk+1,m−1

2ρ0c+ vk−1,m−1 + vk+1,m−1

2

− �x

2ρ0c· (φk−1,m−1 + φk+1,m−1)

(5.19)

Hence, the recurrent formulas (5.19) give the solution of the formulatedproblem, because they allow one to calculate the pressure and velocity of theflow at the following instant of time tm from known values of these parametersat the preceding instant of time tm−1. Since at the first instant of time we cantake the initial values of the pressure and velocity of the flow at t = 0, thencalculating successively the pressure and velocity with formulas (5.19), we canget flow parameters at an arbitrary instant of time t > 0.

5.3.6Initial, Boundary and Conjugation Conditions

Let us investigate non-stationary fluid flow at a pipeline section 0 ≤ x ≤ Lstarting from a certain instant of time t = 0 taken as the initial time. In orderto know how the non-stationary process progresses it is necessary to haveinformation about the initial and boundary conditions, that is to know the stateof the flow before starting and what is happening at the edges of the pipelinesections, e.g. at the cross-section x = 0 and x = L. The first information iscalled the initial conditions and the second the boundary conditions.


Initial ConditionsThe state of the pipeline section at the initial instant of time can be arbitrary,but often as the initial state is taken the stationary fluid flow existing in thepipeline at the initial instant of time.

Let, for example, for stationary flow, the known fluid flow rate beQ = Q(x, 0) = Q0 and the distribution of head be H(x, 0) = H0 − i0 · x,where H0 = H(0, 0) is the head at the beginning of the pipeline section;i0 = (H0 − Hk)/L, where Hk = H(L, 0) is the head at the end of the pipelinesection. Then as initial conditions (m = 1, tm−1 = t0 = 0) may be accepted

v(x, 0) = Q0

S0= const.; p(x, 0) = ρ0g · [H0 − i0 · x − z(x)]

or

vk,1 = Q0

S0= const.; pk,1 = ρ0g(H0 − i0 · xk − zk)

where k = 1, 2, . . . , N + 1, xk = (k − 1) · �x, x1 = 0, xN+1 = L, zk = z(xk),�x = L/N (N being the number of parts into which the pipeline section isdivided).

Boundary ConditionsThe formulas (5.19) permit one to find p and v at any point on the strip0 < x < L, t > 0 determining the pipeline section except its edges – beginning(x = 0) and ending (x = L).

Only one characteristic of the negative slope dx/dt = −c comes from theintegration domain to a point M(x = 0) of the left pipeline section boundary(Fig. 5.5a). It gives one condition for two unknown quantities p1,m and v1,m

p1,m − ρ0c · v1,m = p2,m−1 − ρ0c · v2,m−1 + �x · φ2,m−1

therefore an additional condition is needed. Such a condition can be analgebraic equation F(p, v) = 0 expressing the relation between pressurepM(0, t) and velocity vM(0, t) at the initial cross-section of the pipeline. Asa rule this condition models the operation of a pumping station and isrepresented by its (Q − �H) characteristic. Thus, boundary conditions atpoints on the left pipeline section boundary may be represented as a system ofequations

x = 0, t > 0 :{

p1,m − ρ0c · v1,m = p2,m−1 − ρ0c · v2,m−1 + �x · ϕ2,m−1

F(p1,m, v1,m) = 0.

Similarly, only one characteristic of the positive slope dx/dt = +c comesfrom the domain of integration to a point E(x = L) of the right pipelinesection boundary (Fig. 5.5b), Therefore the boundary condition at points on


Figure 5.5 Calculation of p and v at boundary cross-sections.

the right pipeline section boundary may be represented as a system of twoequations:

x = L, t > 0 :{pN+1,m + ρ0c · vN+1,m = pN,m−1 + ρ0c · vN,m−1 − �x · ϕN,m−1

G(pN+1,m, vN+1,m) = 0.

The dependence G(p, v) = 0 expresses the relation between the pressurepN(L, t) and velocity vN(L, t) at the end of the pipeline section. It should be notedthat there are also possibilities for more complicated boundary conditions.

Conjugation ConditionsIf the equipment responsible for process non-stationarity is located insidethe pipeline section, e.g. at the cross-section x∗, 0 < x∗ < L, then at thiscross-section there can exist a discontinuity of hydraulic parameters. Such adiscontinuity requires additional conditions called conjugation conditions.

Let take place, for example, at a pipeline cross-section x∗ ejection or injectionof fluid with flow rate q (q < 0 ejection; q > 0 injection). Then such across-section is characterized by continuity of pressure and discontinuityof flow rate or velocity. Let us denote the values of the parameters beforeejection or injection with superscript (−) and after ejection or injection withsuperscript (+). Then at the cross-section x∗ the following conditions shouldbe obeyed:

p+(x∗, t) = p−(x∗, t); v+(x∗, t) − v−(x∗, t) = q

S.

Then to calculate the three unknown parameters pk,m, v+k,m, v−

k,m of non-stationary flow at point x∗ = xk we use the following system of three linearequations

pk,m + ρ0c · v−k,m = pk−1,m−1 + ρ0c · vk−1,m−1 − �x · φk−1,m−1

pk,m − ρ0c · v+k,m = pk+1,m−1 − ρ0c · vk+1,m−1 + �x · φk+1,m−1

v+k,m − v−

k,m = q

S0

with p+k,m = p−

k,m = pk,m.


In the case when a gate valve is located at the cross-section x∗ = xk, itsoperation is modeled by conjugation conditions

v+(x∗, t) = v−(x∗, t)

p−(x∗, t) − p+(x∗, t) = ς(t) · ρ0v2(x∗, t)

2

where ς(t) is the local resistance factor that varies during valve closing oropening. The first condition means the continuity of the flow rate, whereasthe second condition signifies pressure discontinuity at different sides of thevalve. Thus, the model of the valve is represented by the following system ofthree equations

p−

k,m + ρ0c · vk,m = pk−1,m−1 + ρ0c · vk−1,m−1 − �x · φk−1,m−1

p+k,m − ρ0c · vk,m = pk+1,m−1 − ρ0c · vk+1,m−1 + �x · φk+1,m−1

p+k,m − p−

k,m = ς(tm) · ρ0v2k,m/2

for the three parameters p+k,m, p−

k,m, vk,m with v+k,m = v−

k,m = vk,m.

5.3.7Hydraulic Shock in Pipes

In all the above stated it was assumed by default that the functionsρ(x, t), p(x, t) and v(x, t) in the differential equations are differentiable withrespect to time and coordinate and in any case are certainly continuous.Nevertheless, in engineering there are processes in which these functionsvary very quickly with time and in space. An example of such a processis hydraulic shock. The essence of hydraulic shock is that the stationaryflow of fluid in a pipeline is disturbed by the abrupt closing or openingof a gate valve, the switching on or switching off of a pump and so on,resulting in hard braking or acceleration of the fluid and shock compression ofthe fluid particles. The front at which the variation of the hydrodynamicparameters of the fluid takes place has a relatively small extent andpropagates in the form of a pressure wave down-stream and up-stream ofthe fluid.

Similar phenomena occur in the pipeline in other cases when the velocity(flow rate) of the fluid varies in a stepwise manner. The possibility of hydraulicshock should be taken into account in the exploitation of pipelines, since shockpressure can far exceed permissible standards, leading to pipe breakage andan emergency situation.

The explanation of hydraulic shock was given by Joukovski in hisarticle ‘‘On hydraulic shock in water-supply pipes’’ (1899). He connectedthe magnitude of the pressure jump [p] with the properties of fluid


compressibility and the elasticity of the pipe and obtained the followingformula

[p] = ρ0D · [v] (5.20)

where D is the velocity of shock wave propagation in the pipeline (see Eq. (5.2))and [v] the magnitude of the stepwise change in the fluid.

It should be noted that the introduction of stepwise variations (jumps)of hydrodynamic flow parameters is nothing more than a model of thephenomenon under consideration. In fact each such discontinuity has atransition region, though very narrow, from the value of parameter A+to the left of the discontinuity front up to the value A− of the sameparameter to the right of the front. The quantity [A] = A+ − A− is calledthe jump of parameter A at the discontinuity front. To describe thestructure of this transition zone needs as a rule a more complicatedmodel than the given one. Nevertheless, introduction of the system ofequations (5.4) into consideration of discontinuity solutions has proved tobe very fruitful.

Conditions at Discontinuities (Jumps) of Hydrodynamic ParametersNow let us obtain conditions which should satisfy the flow parameters atthe discontinuity front. Let the discontinuity front of the flow parameterspropagate with velocity D = dxo/dt in the positive direction of the x-axis(Fig. 5.6).

The wave of the hydraulic shock is characterized by the parameters of flow(and pipeline cross-section area) that are subjected to discontinuities, or jumps,at their mobile front. However the limiting values of these parameters beforeρ−, v−, p−, S− and after ρ+, v+, p+, S+ the wave front cannot be prescribedarbitrarily. They have to obey the conditions of fluid mass and momentumconservation.

In the time interval �t all fluid particles present at the beginning of thisinterval at a distance (D − v−) · �t before the front pass through the front.Therefore the mass of fluid flowing in the wave in time �t is determined bythe expression ρ−S−(D − v−) · �t.

In the same time interval a fluid mass equal to ρ+S+(D − v+) · �t flows outfrom the wave.

It is evident that the mass of fluid flowing into and out of the wave frontshould be equal

Figure 5.6 Hydraulic shock ina pipeline.


ρ−S−(D − v−) = ρ+S+(D − v+)

or

ρ+v+S+ − Dρ+S+ = ρ−v−S− − Dρ−S.

If we denote a jump of any parameter by [ ] then the last relation mayberewritten as

[ρvS − ρD · S] = 0. (5.21)

Now we can use the theorem that the momentum variation of the fluid massthat has passed through the wave front is equal to the impulse of the pressureforce

ρ+S+(D − v+)�t︸︷︷︸�m

·v+ − ρ−S−(D − v−)�t︸︷︷︸�m

·v− = (p− − p+)S− · �t.

In the right-hand side of this equation the projection of the reaction force ofthe pipeline lateral surface onto the x-axis is taken into account.

We can write the last equation in the following form

(ρ+S+v+2 − ρ+S+v+D) − (ρ−S−v−2 − ρ−S−v−D) = −S−�p

or

[v · (ρvS − Dρ · S)] = −S− · [p].

Then, in accordance with equality (5.21), [ρvS − ρD · S] = 0 the obtainedequation may be simplified to

(ρ−v−S− − Dρ−S−) · [v] = −S− · [p]

or (v−

D− 1

)· ρ−D · [v] = −[p].

With regard to ρ− ≈ ρ0 and the smallness of the ratio v−/D we get theJoukovski formula

[p] = ρ0D · [v]. (5.22)

We can show that the velocity of the hydraulic shock wave D for a slightlycompressible fluid coincides with the velocity of the perturbation propagationin a pipeline with elastic walls. In order to do this we can simplify therelation (5.21) as follows:

ρ0S0[v] − DS0[ρ] − Dρ0[S] = 0


where it is assumed that owing to the smallness of the variations in ρ and Svariations the following approximations can be made

[ρvS] ≈ ρ0S0[v]; [ρS] ≈ ρ0[S] + S0[ρ].

The use of Eqs. (5.22), (2.22) and (2.23)

[v] = − 1

ρ0D· [p]; [ρ] = ρ0

K· [p]; [S] = π · d0

3

4δ · E· [p],

yields

S0

D· [p] − DS0

ρ0

K· [p] − ρ0D

π · d03

4δ · E· [p] = 0.

Since [p] �= 0, we have

1 = D2 ·(

ρ0

K+ ρ0d0

δ · E

).

From here it follows that the absolute value of the hydraulic shock wavevelocity |D| is equal to the velocity of the perturbation propagation in thepipeline (Eq. (5.1))

|D| = 1√ρ0

K+ ρ0d0

δ · E

= c. (5.23)

Exercise 1. Oil (ρ0 = 870 kg m−3, K = 1.5 · 109 Pa) flows with velocityv = 1.0 m s−1 in a steel pipeline (d0 = 0.8 m, δ = 10 mm, E = 2 · 1011 Pa).It is required to determine how much the pressure rises at the pipelinecross-section before the gate valve suddenly closes.

Solution. Calculate first the velocity D of the hydraulic shock wave

D = c = 1√870

1.5 · 109+ 870 · 0.8

0.01 · 2 · 1011

∼= 1038 m s−1.

Then determine with Eq. (5.22) the pressure jump [p] = ρ0D · [v] = 870 · 1038 ·1.0 = 903 060 Pa or approximately 9.21 atm.

Answer. 0.903 MPa.

Exercise 2. It is required to determine how much the pressure risesat the initial cross-section of a steel pipeline (d0 = 0.516 m, δ = 8 mm,E = 2 · 1011 Pa) on abrupt switching on of the pumps providing the benzenefeed (ρ0 = 750 kg m−3, K = 1.3 · 109 Pa) with velocity v = 1.5 m s−1.


Solution. Calculate first the velocity D of the hydraulic shock wave generatedby the benzene pumping into the pipeline

D = c = 1√750

1.3 · 109+ 750 · 0.516

0.008 · 2 · 1011

∼= 1105 m s−1.

Then determine the pressure jump with Eq. (5.22) [p] = ρ0D · [v] = 750 · 1105 ·1.5 = 1 243 125 Pa or approximately 12.7 atm.

Answer. 1.243 MPa.Remark on carrying out calculations. Notice one interesting circumstance

following from Eqs. (5.22) and (5.23). Let, for example, D > 0, i.e. the waveof discontinuity propagates in the positive direction of the x-axis. Then thecharacteristic of negative slope x + ct = const. intersects the discontinuity atpoint I (Fig. 5.7).

Ignoring friction, we have

1. p�− − ρ0c · v�− = pB − ρ0c · vB – the condition at the linear segmentB� of the characteristic BM;

2. p�+ − p�− = ρ0c · (v�+ − v�−) – the condition (5.22) at point I of thediscontinuity;

3. pM − ρ0c · vM = p�+ − ρ0c · v�+ – the condition at the linear segment�+M of the characteristic BM.

Subtraction of the first relation from the last one with regard to the secondrelation yields

pM − ρ0c · vM = pB − ρ0c · vB.

Hence the Riemann invariants (in the considered case I2 = p − ρ0c · v) areconserved at the relevant characteristics, even when these characteristicsintersect the shock front.

Problem 1. On the phases of direct hydraulic shock. It is required to study thenon-stationary flow generated in a pipeline section 0 ≤ x ≤ L by abrupt closingof the gate valve at the right-hand end (x = L) of the pipeline (direct hydraulicshock). It is assumed that the fluid before closing the gate (t = 0) was flowing

Figure 5.7 Interaction of adiscontinuity with a small perturbation.


with constant velocity v = v0 and friction was absent so that the pressure atall pipeline cross-sections was constant p = p0. It is assumed also that at theinitial cross-section (x = 0) the pressure is constant and equal to p0.

Solution. The solution is shown in Fig. 5.8. By convention the solution maybe divided into four phases.

1. The phase of direct shock continues for a time t = L/c. When the fluidflow is stopped, the hydraulic shock wave appears and travels from theend of the pipeline section to its beginning. The wave front brings thefluid to an abrupt stop and the pressure after the jump is raised byρ0c · v0.

2. The phase of the reflected wave runs from t1 = L/c to t2 = 2L/c equal todouble the time taken by the wave path along the pipeline section. Inthis phase the shock wave reflects at the initial cross-section and beginsto move in the opposite direction. The fluid then flows out of thepipeline (v = −v0) and the pressure falls to its initial value p0.

3. In the third phase, lasting from the instant of time t2 = 2L/c tot3 = 3L/c, the fluid continues to flow out of the pipeline with velocityv = −v0 whereas the pressure in the wave reflected from thecross-section x = L is lowered stepwise by ρ0c · v0 becoming lower thanits original value. If the pressure is lowered to the saturated vapor

Figure 5.8 Phases of direct hydraulic shock.


pressure of the fluid, the latter would begin to boil, otherwise at(p0 − ρ0c × v0) > ps the fluid does not vaporize and the pipeline sectionremains completely filled with fluid. The velocity behind the shockfront propagating in the negative direction vanishes.

4. In the fourth phase of the process at 3L/c < t < 4L/c the fluid againbegins to flow into the pipeline (v = v0) and the pressure becomes equalto p = p0. At the instant of time t4 = 4L/c the situation returns to theinitial one, after which the process is periodically repeated.

Problem 2. On the damping of the pressure bow shock at the hydraulic shock wavefront (Charniy, 1975). In the presence of friction the hydraulic shock wave isgradually damped in the pipeline, in particular the magnitude of the pressurejump at the shock front decreases. It is required to determine the intensity ofsuch a fall.

Solution. Let the front of the hydraulic shock travel from the end of thepipeline cross-section x = L to its beginning. Consider two negative slopecharacteristics parallel to the discontinuity along its left- and right-hand sides.Then in accordance with Eq. (5.18) we have

d

dt(p+ − ρ0c · v+)

∣∣∣∣x=−c

= λ+ · 1

d0

ρ0c · v+|v+|2

+ ρ0cg sin α,

d

dt(p− − ρ0c · v−)

∣∣∣∣x=−c

= λ− · 1

d0

ρ0c · v−|v−|2

+ ρ0cg sin α,

p+ − p− = −ρ0c · (v+ − v−).

Subtracting term-by-term from the second equation from the first one andtaking into account the third equation, we get

d

dt(p+ − p−)

∣∣∣∣x=−c

= 1

2· 1

d0

ρ0c

2· [

λ− · v−|v−| − λ+ · v+|v+|]Taking v+ = v0; v− = v+ − (p+ − p−)/ρ0c = v0 − [p]/ρ0c; and [p] = p+ − p−,one obtains the ordinary differential equation for the variation of the pressurejump [p] at the wave front

d[p]

dt

∣∣∣∣x=−c

= − ρ0c

4d0· (λ0v0

2 − λ−v−2) (5.24)

This equation should be integrated under the initial condition [p] = −ρ0c · v0

at t = 0. If λ−(Re−, ε) = λ−(v−) is known, the solution of the problem can beobtained without difficulty.

At λ0v0 ≈ λ−v− = 2a > 0, where a is a constant, the solution has a particu-larly simple form (for laminar flow this result gives an exact solution). We have

d[p]

dt

∣∣∣∣x=−c

= −aρ0c

2d0· (v0 − v−) = − a

2d0· [p] (5.25)


and

[p] = −ρ0cv0 · exp(−a/2d0 · t), (5.26)

that is, the pressure jump at the hydraulic shock wave front decaysexponentially.

From the obtained formula follows, in particular, that in pipelines of largediameter the value of the bow pressure jump at the hydraulic shock wave frontdecays more slowly than in pipelines of smaller diameter.

Problem 3. At the junction of two pipes with different internal diametersd1 and d2 comes, as viewed from the first pipe, the pressure shock wave(incident wave) with pressure amplitude of magnitude [pinc]. There are thengenerated two pressure waves: one is the reflected wave, which travels in theopposite direction with amplitude [prefl]; the other is the transmitted wavewith amplitude [ptrans], which travels through the second pipe. It is requiredto express the amplitudes of the reflected and transmitted waves through theamplitude of the incident wave, if the velocities of wave propagation in the firstand second pipes are c1 and c2 respectively.

Answer. [prefl] = [pinc] · c1d22 − c2d2

1

c1d22 + c2d2

1

; [ptrans] = [pinc] · 2c2d21

c1d22 + c2d2

1

.

Remark. From the solution of the problem it follows that at c1d22 = c2d2

1 ord2 = d1 · √

c2/c1, [pref ] = 0, that is, the reflected wave is not generated at thejunction of the pipes.

Problem 4. It is required to prove that the amplitude of the hydraulic shockwave falling at the closed pipeline end will be doubled when reflecting fromthe pipeline end.

Hint. Use the results of the previous problem.

5.3.8Accounting for Virtual Mass

In Section 5.1 it was indicated that in non-stationary processes the inertialproperties of fluid in a pipeline are characterized by variable density, hence anadditional quantity called virtual mass should be added. The main equations ofnon-stationary flow of a slightly compressible fluid in a pipeline with regard tovirtual mass take the form

∂p

∂t+ ρ0c2 · ∂v

∂x= 0

αkρ0dv

dt= − ∂p

∂x− λ(Re, ε) · 1

d

ρ0v|v|2

− ρ0 g sin α(x)

(5.27)


If we denote ρ∗ = αkρ0 and –c∗ = –c/√

αk, the system of equations (5.27)transforms into

∂p

∂t+ ρ∗c∗2 · ∂v

∂x= 0

ρ∗dv

dt= − ∂p

∂x− λ(Re, ε) · 1

d

ρ0v|v|2

− ρ0g sin α(x)

(5.28)

equivalent to the system of equations (5.4). From this follows that theJoukowski formula (5.20) for the amplitude [p] of the hydraulic shock varies; itwill be enhanced by the factor

√ακ

[p] = ρ∗c∗ · [v] = αkρ0 · c√αk

· [v] = ρ0c√

ακ · [v] (5.29)

The velocity D of hydraulic shock wave propagation will also vary, it will bereduced by a factor

√ακ

|D| = 1√ρ∗K

+ ρ∗d0

δ · E

= c∗ = c√ακ

(5.30)

If we take into account that for turbulent flow αk ≈ 1.03, corrections toJoukovski formula will be small (one of them (5.29) was first obtainedby Leibenson et al., 1934). But for laminar flow αk ≈ 1.33 and these correctionscould be significant.

5.3.9Hydraulic Shock in an Industrial Pipeline Caused by Instantaneous Closing of theGate Valve

In Fig. 5.9 is shown the development of the hydraulic shock in a pipelinesection with diameter D = 1020 mm (δ = 10 mm) and length L = 100 kmupon instantaneous closing of the gate valve at the right-hand edge ofthe pipeline section. The pipeline is transporting crude oil with densityρ0 = 880 kg m−3 and viscosity ν = 20 cSt. At the initial instant of time t = 0the flow in the pipeline is stationary with velocity v(x, 0) = v0 ≈ 1.5 m s−1;the velocity of the hydraulic shock wave propagation c is equal to 870 m s−1.The abrupt closing of the gate valve at the right-hand edge of the pipeline ismodeled by the condition v(L, t) = 0 at t > 0.

Figures 5.9a and 5.9b demonstrate the distributions of the head (z + p/ρg)at 2 and 40 s, respectively, after closing the gate valve.

The initial value of the pressure jump �pf at the wave front is related to theinitial velocity v0 by

�pf = ρ0cv0 = 880 · 870 · 1.5 ∼= 1.15 MPa or ≈ 11.7 atm.Figures 5.10a and 5.10b show the following stages of pressure wave upstream

propagation.


Figure 5.9 Pressure wave propagation: (a) 2 s and (b) 40 s after appearance of the wave.

The velocity v(x, t) of the fluid flow behind the pressure wave front is small,therefore head losses in this region are also small. That is why the head, andconsequently the pressure before the closed gate valve, are always raised. Thehead HL(t) = zL + pL(t)/ρg at the end of the pipeline section, i.e. before thegate valve, can be taken as approximately equal to the head after the pressurewave front, that is to

HL(t) = zL + pL(t)

ρ0g� Hf + [p]

ρ0g,

where zL is the elevation of the section end and Hf the head at the pipeline


Figure 5.10 Pressure wave propagations: (a) 80 s and (b) 100 s after appearance of thewave.

cross-section which the hydraulic wave shock has reached. From this it followsthat for an approximate estimation of the pressure at the pipeline section endone can use the formula

pL(t) � ρ0g(Hf − zL) + [p]

or

pL(t) ≈ ρ0gi0 · ct + [p], (5.31)

where i0 is the hydraulic gradient of the flow in the undisturbed region and tis the time elapsed after closure of the gate valve.


It should be noted that owing to viscous friction the fluid behind the wavefront would not come to a halt immediately but gradually, therefore thepressure shock amplitude [p] reduces monotonically (see formula (5.26)).

Protection of Pipelines from Hydraulic ShocksThe necessity to prevent the destructive force of hydraulic shock in pipelinestransporting heavy dropping liquids (oil, oil products, water and other) ismanifested by the fact that such pipelines, as distinct from gas-pipelines, arenot equipped with cocks that close the pipeline cross-section too rapidly, butare equipped with valves, gates and slowly closing cocks. All of these shouldensure safe fluid braking in the pipeline.

Pumping stations are sometimes equipped with special devices intendedto protect pipelines from hydraulic shock waves. In the suction lines ofpumping stations are, for example, flow dampers of hydraulic shock – specialsafety valves or systems of pressure wave smoothing in case of a sudden pumpingstation switch-off when the pressure before the station begins to build up.These devices operate on the principle of emergency discharge of part of thefluid from the pipeline into a special reservoir to decrease the magnitudeof the pressure and its rate of increase. The safety valves open the fluiddischarge when the pressure exceeds a certain value, the systems of pressurewave smoothing are switched on when the rate of pressure build-up inthe suction line of the pumping station is greater than the permissiblemagnitude.

5.4Non-Isothermal Gas Flow in Gas-Pipelines

Consider now non-stationary and non-isothermal gas flow in gas-pipelines.The main distinction of such flows from the flows considered above is that thegas represents a significantly compressible medium with density dependent notonly on pressure but also on temperature. Thus, to describe these flows it isnecessary not only to use the laws of mass and momentum conservationbut also the law of energy transformation. In other words, as well ascontinuity and momentum equations, the equation of heat inflow shouldbe invoked.

Basic equations for the calculation of non-stationary non-isothermal flows ina gas-pipeline are represented by the system (5.8) giving

∂ρ

∂t+ ∂ρv

∂x= 0

∂ρv

∂t+ ∂

∂x(p + ρv2) = −λ(Re, ε)

1

d0

ρv · |v|2

− ρg sin α

ρ

(∂ein

∂t+ v

∂ein

∂x

)= −4κ

d0(T − Tex) − p

∂v

∂x+ λ(Re, ε)

1

d0

ρv3

2

(5.32)

5.4 Non-Isothermal Gas Flow in Gas-Pipelines 139

in which ρ = p/ZRT is the equation of the gas state and λ(Re, ε) is thehydraulic resistance factor.

The system of equations (5.32) represents partial differential equations forthree unknown functions p(x, t), v(x, t) and T(x, t) dependent on x and t.

These non-linear equations have a complicated structure, therefore thequestion arises: how to solve them? To answer this question it is necessaryto examine the structure of the system (5.32) treating it as well as was donewhen investigating equations (5.4) in the model of non-stationary flow of aslightly compressible fluid. In the course of examining this model we haveseen that on the plane of variables (x, t) there are certain lines (characteristics)along which the system of partial differential equations (5.4) transforms intoan ordinary differential equation providing a relation between unknownfunctions. Through each point M(x, t) of the plane (x, t) go just two (by thenumber of equations) such lines, or strictly speaking

d

dt(p + ρ0cv) = I1(p, v) along the line x = c (or x = ct + const.),

d

dt(p − ρ0cv) = I2(p, v) along the line x = c (or x = −ct + const.).

The presence of two characteristics is the distinctive property of the consideredsystem of equations and allows us to assign the system to a class of hyperbolicdifferential equations and give a constructive method of its solution.

Let us show that the system of equations (5.32) is hyperbolic (the number ofits characteristics is equal to the number of equations, that is three). We lookfor characteristics on the plane (x, t), i.e. lines x = x(t) such that along themEqs. (5.32) give certain differential relations between unknown functions.

Let x = x(t) be a line on the plane (x, t) on which the values of the functionsp(x, t), v(x, t), T(x, t) are known. Then it is possible to write three equationsrelating the derivatives ∂p

∂t ,∂p∂x , ∂v

∂t ,∂v∂x and ∂T

∂t , ∂T∂x of these functions along the

line x = x(t). To do this let us differentiate p, v, T along x = x(t)

dp

dt

∣∣∣∣x(t)

= ∂p

∂t+ ∂p

∂x· dx

dt

dv

dt

∣∣∣∣x(t)

= ∂v

∂t+ ∂v

∂x· dx

dt(5.33)

dT

dt

∣∣∣∣x(t)

= ∂T

∂t+ ∂T

∂x· dx

dt

where dx/dt is the slope of this line (we shall call it a characteristic) tothe t-axis. The slope can be taken as given since the function x = x(t) isknown.

Since the left-hand sides of Eqs. (5.33) are known as well as the slope dx/dtof the line x(t), these equations could be considered as three linear equations todetermine six partial derivatives p, v, T with respect to time and coordinate. If


to these equations we add three equations of the system (5.32), which reducesalso to linear equations with respect to the same derivatives, we get six linearequations for six partial derivatives

(∂ρ

∂p

)T

∂p

∂t+ v

(∂ρ

∂p

)T

∂p

∂x+ ρ

∂v

∂x+

(∂ρ

∂T

)p

∂T

∂t+ v

(∂ρ

∂T

)p

∂T

∂x= 0

v

(∂ρ

∂p

)T

∂p

∂t+

[1 + v2

(∂ρ

∂p

)T

]∂p

∂x+ ρ

∂v

∂t+ 2ρv

∂v

∂x+ v

(∂ρ

∂T

)p

∂T

∂t

+ v2

(∂ρ

∂T

)p

∂T

∂x= J2

p∂v

∂x+ ρCv

∂T

∂t+ ρvCv

∂T

∂x= J3

∂p

∂t+ dx

dt· ∂p

∂x= J4

∂v

∂t+ dx

dt· ∂v

∂x= J5

∂T

∂t+ dx

dt· ∂T

∂x= J6

where J2 = −λρv · |v|/2d0 − ρg sin α; J3 = −4κ(T − Tex)/d0 + λρv3/2d0; J4 =(dp/dt)x(t); J5 = (dv/dt)x(t); J6 = (dT/dt)x(t). The total derivatives of thefunctions p, v and T are taken along the line x(t), on which their valuesare known.

If the principal determinant of the system is non-vanishing, all six partialderivatives as well as the functions p, v, T along the curve x(t) can bedetermined uniquely and independently from each other. If this determinantvanishes and the system of linear equations is compatible, the dependencebetween the values of p, v, T on the curve x(t) exists.

Let us equate the principal determinant of the system of six linear equationsto zero

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(∂ρ

∂p

)T

v

(∂ρ

∂p

)T

0 ρ

(∂ρ

∂T

)p

v

(∂ρ

∂T

)p

v

(∂ρ

∂p

)T

1 + v2

(∂ρ

∂p

)T

ρ 2ρv v

(∂ρ

∂T

)p

v2

(∂ρ

∂T

)p

0 0 0 p ρCv ρvCv

1 dx/dt 0 0 0 00 0 1 dx/dt 0 00 0 0 0 1 dx/dt

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0

and calculate it. To do this, let us multiply the first, third and fifth columns ofthis determinant by dx/dt and then subtract the resulting products from the


second, fourth and sixth columns of the same determinant∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(∂ρ

∂p

)T

(v − x)

(∂ρ

∂p

)T

0 ρ

(∂ρ

∂T

)p

(∂ρ

∂T

)p

(v − x)

v

(∂ρ

∂p

)T

1 + v(v − x)

(∂ρ

∂p

)T

ρ ρ(2v − x) v

(∂ρ

∂T

)p

(∂ρ

∂T

)p

v(v − x)

0 0 0 p ρCv ρCv(v − x)

1 0 0 0 0 0

0 0 1 0 0 0

0 0 0 0 1 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣The obtained determinant can be calculated by the method of determinantdecomposition in terms of the third-column elements. As a result we get thecubic equation

ρCv

(dx

dt− v

)3 (∂ρ

∂p

)T

+[

p

ρ

(∂ρ

∂T

)p

− ρCv

](dx

dt− v

)= 0 (5.34)

with respect to the difference (dx/dt − v). Roots of this equation are evident

1.dx

dt− v = 0 ⇒ dx

dt= v;

2.dx

dt− v = ±

√Cv − p/ρ2 · (∂ρ/∂T)p

Cv

(∂p

∂ρ

)T

.

Since Cv = ∂ein

∂Tand − p

ρ2

(∂ρ

∂T

)p

= ∂

∂T

(p

ρ

)p

, then

Cv − p

ρ2

(∂ρ

∂T

)p

= ∂

∂T

(ein + p

ρ

)p

=(

∂J

∂T

)p

= Cp

where J(p, T) is enthalpy. The expression under the square root is simplified to

dx

dt− v = ±

√Cp

Cv

(∂p

∂ρ

)T

= ±c

where

c2 = Cp

Cv·(

∂p

∂ρ

)T

= γ ·(

∂p

∂ρ

)T

and γ(p, T) = Cp/Cv is the adiabatic index.The quantity c having the dimension of velocity is called the adiabatic velocity

of sound in gas. For a perfect gas (∂p/∂ρ)T = RT , γ = const., c = √γRT. For

example, at γ = 1.31; R = 450 J kg−1 K−1; T = 273 K the velocity of sound isequal to c = √

1.31 · 450 · 273 ∼= 400 m s−1.


The equation (5.34) provides three families of characteristics, two of themhaving velocities (slopes) (dx/dt)1,2 = v ± c and the third dx/dt = v. Sincethe gas velocity v in gas-pipelines as a rule does not exceed 10 m s−1 andc ≈ 400 m s−1, then sometimes we could take dx/dt ∼= ±c. We shall seebelow that the slope of characteristics to the t-axis is simply the velocityof propagation of small perturbations in gas (sound velocity). Therefore theassumption dx/dt ∼= ±c means that this velocity of gas moving in a pipe isapproximately equal to the sound velocity in quiescent gas. In the general caseit is of course not so, especially for gas flowing with high velocity, for examplein propulsive nozzles or for gas flowing from openings, orifices, nozzles andso on. In such cases the difference (dx/dt − c) cannot be ignored.

Since the determinant of the system of the six linear equations underconsideration vanishes at x = v and x = v ± c, for compatibility of this systemthe determinant∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(∂ρ

∂p

)T

v

(∂ρ

∂p

)T

0 ρ

(∂ρ

∂T

)p

0

v

(∂ρ

∂p

)T

1 + v2

(∂ρ

∂p

)T

ρ 2ρv v

(∂ρ

∂T

)p

J2

0 0 0 p ρCv J3

1 x 0 0 0 J4

0 0 1 x 0 J5

0 0 0 0 1 J6

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣obtained from the principle determinant of the system by replacement of thelast column on free terms of the system of equations, should also vanish. Inthe theory of linear equations the condition of the latter determinant vanishingis called the compatibility condition. This condition as applied to our case maybe called the compatibility condition at characteristics.

Characteristic form of equations. Omitting cumbersome calculation of thisdeterminant, we give the final result(

dx

dt− v

)2 (∂ρ

∂p

)T

(J3

ρ− CvJ6

)−

(dx

dt− v

)p

ρ2

(∂ρ

∂p

)T

(ρJ5 − J2)

−[

J3

ρ− CpJ6 + p

ρ2

(∂ρ

∂p

)T

J4

]= 0. (5.35)

Substitution successively in Eq. (5.35) of the values of dx/dt for each of thecharacteristic families yields conditions for the functions p, v, T to be obeyedat these characteristics:

1.dp

dt+ ρc

dv

dt= +c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3 (5.36)

at dx/dt − v = +c or dx/dt = v + c, where J2 = −λρv · |v|/2d0 − ρg sin α,J3 = −4κ(T − T ′)/d0 + λρv3/2d0 and derivatives with respect to time aretaken in the direction of the characteristic dx/dt = v + c.


2.dp

dt− ρc

dv

dt= −c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3 (5.37)

at dx/dt − v = −c or dx/dt = v − c, where derivatives with respect to timeare taken in the direction of the characteristic dx/dt = v − c.

3. CpdT

dt− p

ρ2

(∂ρ

∂p

)T

dp

dt= J3

ρ(5.38)

at dx/dt − v = 0 or dx/dt = v, where derivatives with respect to time aretaken in the direction of the characteristic dx/dt = v.

Method of CharacteristicsFormulas (5.36)–(5.38) give a constructive way to solve the system ofequations (5.41). The method by which this solution is obtained is calledthe method of characteristics.

Basic equations for this method are Eqs. (5.36)–(5.38). The idea of thismethod is that at each point M(xk, tm) of the plane (x, t) relevant to thetime tm ‘‘in the past’’, that is at t < tm, there are just three characteristicsdx/dt = v ± c and dx/dt = v, at which should be satisfied the compatibilityconditions (5.36)–(5.38). Each of these compatibility conditions represents anordinary differential equation which could be integrated over the directionof the respective characteristic. Thus, at the point M(xk, tm), being theintersection point of these characteristics, appear three equations to determinethree unknown quantities p(xk, tm), v(xk, tm) and T(xk, tm). For numericalrealization of the method of characteristics various schemes may be used.Consider one of them.

Let it be required to get the solution of the system of equations (5.33)in the region 0 < x < L, t > 0 of the plane (x, t). Divide this region bya rectangular mesh with straight lines xe = �x · (k − 1), tm = �t · (m − 1),where 1 ≤ κ ≤ N + 1, N = [L/�x]. The time step �t is chosen so that�t = �x/(v + c)max, where vmax and cmax are the maximum possible valuesof the gas and sound velocities respectively (cmax = √

γRTmax; Tmax is themaximum possible value of the gas temperature), m = 1, 2, 3, . . ..

Neglecting the work of the gravity force, the basic system of differentialequations (5.33) may be written in the so-called divergent form (see Eq. (1.36)at αk

∼= 1)

∂ρ

∂t+ ∂ρv

∂x= 0

∂ρv

∂t+ ∂

∂x(p + ρv2) = −λρv · |v|/2d0

∂

∂t

[ρ

(v2

2+ ein

)]+ ∂

∂x

[ρv ·

(v2

2+ ein + p

ρ

)]= −4κ

d0(T − Tex)

(5.39)

The left-hand sides of Eqs. (5.39) are represented by differential operators ofthe form


div{A, B} = ∂A

∂t+ ∂B

∂x

expressing the divergence of a vector with coordinates {A, B} in the space ofvariables (x, t).

Let us first integrate the system of equations (5.39) over the area of the meshcell ABCD (xk ≤ x ≤ xk+1; tm−1 ≤ t ≤ tm) with sides �x and �t (Fig. 5.11)and transform the integrals over the mesh area into integrals over the meshcontour using the formula∫ ∫

ABCD

(∂A

∂t+ ∂B

∂x

)dxdt =

∮ABCD

[A cos(nt) + B cos(nx)]dσ

=∮

ABCD(Adx + Bdt) = (Ak+1/2,m − Ak+1/2,m−1)�x + (Bk+1 − Bk)�t.

Here the quantities with fraction subscripts denote mean values of thecorresponding parameters at the horizontal sides (BC and AD) of the meshcell, whereas the quantities with integer subscripts denote mean values ofparameters at the vertical sides (AB and CD) of the same mesh cell.

Application of this transformation to each equation of the system (5.39)yields the following system of finite difference equations

ρk+1/2,m = ρk+1/2,m−1 − [(ρv)k+1 − (ρv)k] · �t

�x,

ρk+1/2,mvk+1/2,m = ρk+1/2,m−1vk+1/2,m−1 − [(p + ρv2)k+1 − (p + ρv2)k

]· �t

�x− (λρv · |v|/2d0)k+1/2,m−1 · �t,

ρk+1/2,m

(v2

2+ CvT

)k+1/2,m

= ρk+1/2,m

(v2

2+ CvT

)k+1/2,m−1

−{[

∂v ·(

v2

2+ CvT + ρ

∂

)]k+1

Figure 5.11 Mesh cell in the plane ofvariables (x, t).


−[∂v ·

(v2

2+ CvT + ρ

∂

)]k

}× �t

�x− 4κ

d0(T − Tex)k+1/2,m−1 · �t.

(5.40)

Here ρk+1/2,m; vk+1/2,m and Tk+1/2,m are, respectively, the mean values of thedensity, velocity and temperature of the gas in segment (xk, xk+1) of thegas-pipeline at the instant of time t = tm; ρk+1/2,m−1, vk+1/2,m−1 and Tk+1/2,m−1

are the mean values of the same functions at the previous moment of timetm−1 = tm − �t.

The physical meaning of the obtained relations is clear: each of theseequations represents the integral balance of one or other parameter inthe mesh cell. For example, the first relation (5.40) reflects the fact thatthe gas mass ρk+1/2,m · �x in the segment (xk, xk+1) of the pipeline at theinstant of time tm is equal to the gas mass ρk+1/2,m−1 · �x in the samesegment at the previous instant of time added to the mass difference ofthe gas [(ρv)k+1 − (ρv)k] · �t flowing in time �t from the kth segmentinto the (k + 1)th segment and from the (k − 1)th segment into the kthone. The two other relations are interpreted in the same manner withthe only difference being that they deal with momentum and total energy,respectively.

Nevertheless, Eqs. (5.40) are not closed since they include unknownquantities denoted by integer subscripts and representing the transfer ofmass, momentum and energy from one mesh cell into another. The essenceof the method is that the values of these quantities are found from compatibleconditions (5.36)–(5.38) at characteristics.

Let us represent the conditions (5.32)–(5.37) in the form of finite differenceequations

pk,m − pk−1/2,m−1

�t+ (ρc)k−1/2,m−1

vk,m − vk−1/2,m−1

�t

=(

c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3

)k−1/2,m−1

,

pk,m − pk+1/2,m−1

�t− (ρc)k+1/2,m−1

vk,m − vk+1/2,m−1

�t(5.41)

=(

−c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3

)k+1/2,m−1

,

CpTk,m − Tk−1/2,m−1

�t−

[p

ρ2

(∂ρ

∂p

)T

]k−1/2,m−1

· pk,m − pk−1/2,m−1

�t

=(

J3

ρ

)k−1/2,m−1

.


As a result we have three linear equations for three unknown quantitiespk,m, vk,m, Tk,m. First we find quantities pk,m and vk,m from the firsttwo equations, then from the third equation we obtain Tk,m and fromthe relation ρk,m = pk,m/RTk,m we calculate the gas density. Substitutingρk,m, pk,m, vk,m, Tk,m into Eq. (5.40), we get average values of the gas-dynamicparameters in the considered mesh cell at the instant of time tm.

To Eqs. (5.37) and (5.38) should be added initial and boundary conditions.Boundary conditions vary depending on the concrete problem. As an example,they could be taken in the form of two relations between the pressure, flow rateand temperature of the gas at one edge (left) of the pipeline section, reflectingconditions of compressor station operation and one condition at the right edgeof the pipeline section, for example, with a given pressure.

5.5Gas Outflow from a Pipeline in the Case of a Complete Break of the Pipeline

Let us illustrate the use of the method of characteristics as applied to non-stationary processes in a gas-pipeline with an example of the calculation ofgas outflow from a pipeline in the case of its complete break. The dynamicsof gas outflow happen as a rule in two regimes. At first at the cross-section,through which occurs gas outflow, appears a critical regime with the velocityof gas outflow equal to the local velocity of sound (≈380–400 m s−1). Afterthe pressure in the gas-pipeline is lowered by a certain value (for natural gasby a factor of 1.8–1.9 greater than atmospheric pressure) the outflow regimebecomes subsonic and the gas velocity gradually decreases from sound velocityto zero.

The process of gas outflow is not isothermal. The gas temperature, owingto adiabatic expansion and the Joule-Thomson effect falls significantly at thebreak in the cross-section as well as far from it. For example, with a completebreak of a gas-pipeline the gas temperature can be reduced by 80–100 K. Onlyat the final stage of the process is there a gradual restoration of temperaturedue to external heat inflow.

Numerical calculations are carried out on the basis of recurrent formu-las (5.38). The left-hand end of the pipeline (x = 0) is closed, whereasthe right-hand end (x = L) is suddenly opened and remains open to theatmosphere.

Hence, the boundary conditions are as follows.

1. Cross-section x = 0. The first boundary condition is v1,m(0, tm) = 0, sincethe left-hand end of the pipeline is taken to be closed. Hence it followsthat at v = 0 the left integration boundary coincides with one of thecharacteristics of the differential equations (5.36), therefore the secondboundary condition at the cross-section x = 0 (k = 1) is formulated as acondition at the characteristic dx/dt = v = 0

5.5 Gas Outflow from a Pipeline in the Case of a Complete Break of the Pipeline 147

CpT1,m − T1,m−1

�t−

[p

ρ2

(∂ρ

∂p

)T

]1,m−1

· p1,m − p1,m−1

�t=

(J3

ρ

)1,m−1

.

The compatibility condition at the characteristic dx/dt = v − c comingfrom the integration domain to the initial pipeline cross-section (in ourcase at the characteristic dx/dt = 0 − c = −c), provides the third boundarycondition

p1,m − p3/2,m−1

�t− (ρc)3/2,m−1

v1,m − v3/2,m−1

�t

=(

−c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3

)3/2,m−1

.

2. Cross-section x = L. Two characteristics of the system (5.39) come to thiscross-section from the integration domain. These characteristics aredx/dt = v + c and dx/dt = v with positive slope. Thus, the first twoboundary conditions at x = L (k = N + 1) are

pN+1,m − pN+1/2,m−1

�t+ (ρc)N+1/2,m−1

vN+1,m − vN+1/2,m−1

�t

=(

c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3

)N+1/2,m−1

,

CpTN+1,m − TN+1/2,m−1

�t−

[p

ρ2

(∂ρ

∂p

)T

]N+1/2,m−1

· pN+1,m − pN+1/2,m−1

�t=

(J3

ρ

)N+1/2,m−1

.

The third boundary condition at x = L has a different form depending onwhether the gas outflow is subsonic v < c(p, T) or sonic v = c(p, T). If thegas outflow happens with local sound velocity, then dx/dt = v − c = 0 andthe right boundary of the integration domain coincides with the characteri-stic of the system (5.39). Hence, the third boundary condition is nothingbut a condition at this characteristic

pN+1,m − pN+1,m−1

�t− (ρc)N+1,m−1

vN+1,m − vN+1,m−1

�t

=(

−c · J2 + (γ − 1)

p/ρ · (∂ρ/∂p)T· J3

)N+1,m−1

.

In subsonic flow v < c(p, T) only two characteristics of the system (5.39) comefrom the integration domain to the boundary points x = L, therefore one morecondition should be given. Such a condition is pN+1,m = paTM meaning thatthe pressure at the opened pipeline end is equal to the external atmosphericpressure.

Calculations show that the gas is significantly cooled during outflow.Figure 5.12 shows graphs of the gas temperature distribution over a pipeline


L.km

Figure 5.12 Distribution of gas temperature over a pipeline section length: 1, t = 0; 2,t = 5 s; 3, t = 60 s; 4, t = 120 s; 5, t = 200 s; 6, t = 420 s.

section length (D = 1220 mm, L = 5 km) at different instants of time. It isseen that the initial temperature of the gas, equal to 0 ◦C decreases by morethan 100 K. This is explained by gas expansion due to the high outflow velocityand the Joule–Thomson effect. Gas cooling in the pipeline happens becausethe above-mentioned effects are too fast to be compensated by heat inflowfrom the surrounding medium.

Computer modeling shows the effect of gas suction into the pipeline in thefinal stage of the outflow process. Figure 5.13 represents the dependence ofthe gas velocity on time at the cross-section where the pipeline undergoes abreak (D = 1220 mm, L = 1 km) with initial pressure p(x, 0) = 5.5 MPa. Theoscillation process is seen to appear in the final stage of gas outflow and lastsfor about 2 min.

The section AB of the graph characterizes sonic outflow of gas lasting about20 s. The velocity of the outflow gradually decreases as a consequence ofreduction in the pressure and temperature of the gas at the cross-section ofthe break. The section BF of the graph characterizes subsonic outflow of gas.About after a 30th of a second from the outflow beginning in the pipeline anoscillation process occurs in which the gas periodically changes its directionof motion. Once the velocity of the gas becomes equal to zero it continues todecrease and at the section CD of the graph remains negative, testifying thatatmospheric air is being sucking into the pipeline. The same behavior of gas isobserved in the following instants of time. Calculations show that the maximalvelocity of air suction in the oscillation process exceeds 50 m s−1.

Since the mixture of natural gas and air could achieve an explosiveconcentration, the discovered phenomena appears to be a serious hazardto attendants. In particular, it is strongly recommended to take the utmost carein repair-reconditioning operations.

5.6 Mathematical Model of Non-Stationary Gravity Fluid Flow 149

nL, ms-1

t.s

Figure 5.13 Dependence of gas outflow velocity on time.

5.6Mathematical Model of Non-Stationary Gravity Fluid Flow

Such a flow has already been discussed in Section 3.7 but then a stationarygravity flow was considered, that is a flow in which all the hydrodynamicparameters remain constant at each cross-section of the pipeline. In otherwords these parameters were independent of time. Consider now a one-dimensional mathematical model of non-stationary gravity fluid flow. In thismodel there are two parameters governing the flow: S(x, t) the area of thepipeline cross-section and v(x, t) the flow velocity (Fig. 5.14).

The differential equations of the model under consideration are∂ρS

∂t+ ∂ρvS

∂x= 0

∂ρvS

∂t+ ∂ρv2S

∂x= −ρgS cos α · ∂h

∂x− ρgS sin α − ρgS cos α

CSh2Rh

· v|v|.

Figure 5.14 Gravity fluid flow.


There Rh(S) is the hydraulic radius of the flow (see Section 3.7); h(S) isthe depth of the pipeline cross-section filling with fluid; α is the angleof inclination of the pipeline axis to the horizontal; CSh is the Chezyfactor.

The first of these equations (continuity equation) expresses the law ofmass conservation in the fluid flow in a pipeline with cross-section partiallyfilled by fluid. The second equation (momentum equation) is the law ofmomentum change, that is Newton’s Second law: on the left is the derivativeof momentum with respect to time, the quantity ρv2S = ρvS · v representingthe flux of momentum; the term ρgS cos α · ∂h

∂x on the right is the Boussinesqforce acting on the fluid due to its free surface being non-parallel to the pipelineaxis, i.e. due to excess of the fluid level at one cross-section of the pipeline ascompared to the fluid level at another cross-section; the term −ρgS sin α is thecomponent of the gravity force and the term −ρgS cos α · v|v|/(CSh

2Rh) is theforce of the flow resistance due to fluid friction against the pipeline walls.

The system of equations may be rewritten in an equivalent form if we takeinto account that

ρg cos α ·∫ S

0Sdh = ρg cos α ·

∫ S

0S

dh(S)

dSdS

As a result we get

∂ρS

∂t+ ∂ρvS

∂x= 0

∂ρvS

∂t+ ∂

∂x

(ρv2S + ρg cos α ·

∫ S

0Sdh

)= −ρgS sin α − ρgS cos α

CSh2Rh

· v|v|

(5.42)

The system of equations (5.42) belongs to the class of quasi-linear (that is linearwith respect to derivatives) differential equations of the hyperbolic type. Solution ofthese equations can be obtained by specially elaborated methods, which arebeyond the scope of this book. We advise those who are interested in thesemethods to consult the book by Rozhdestvenski and Yanenko (1977), whichamong other things contains voluminous literature devoted to this problem.

For gravity stationary fluid flow (∂/∂t = 0) from the first equation ofthe system (5.42) it follows that ρvS = const. If we take into account thatρ ∼= ρ0 = const., then v · S = const. If we assume in addition that S = const.,which is true everywhere except in small regions close to the transfer sectionswhere gravity flows are formed, from Eq. (5.42) it follows that

0 = −ρgS · sin α − ρgS cos α

CSh2Rh

· v|v| ⇒ v = CSh

√Rh · tan α (5.43)

5.6 Mathematical Model of Non-Stationary Gravity Fluid Flow 151

psat

Figure 5.15 Gravity section in pipeline.

which after some transformation allows one to get the degree of filling withfluid of the gravity flow section, σ, depending on the hydraulic gradient tan α

(see Eq. (3.51)).The point at which the first gravity section in the pipeline begins is called

the transfer section. Since cross-sections after the transfer section are onlypartially filled by fluid, the pressure in this section is constant and equal to thesaturated vapor pressure psat of this fluid. Figure 5.15 illustrates the behaviorof the hydraulic gradient in gravity flow. It is seen from this figure that in thissection the line of the hydraulic gradient goes parallel to the pipeline axis at adistance psat/ρg from it owing to the pressure constancy in the gas cavity overthe fluid-free surface.

An ordinary differential equation serves to calculate the depth h(x) of fluidin the pipe.

dh

dx= − tan α + Q2/(CSh

2RhS2)

1 − (Q2dh/dS)/(g cos α · S3)(5.44)

following from Eq. (5.42) for the case of stationary flow (∂/∂t = 0) withh(x) = h[S(x)]; Rh = Rh(S); Q is the fluid flow rate.

The depth hn of the fluid in the pipe at which the numerator of the fractionon the right-hand side of Eq. (5.44) vanishes is called the normal depth ofgravity flow in the pipe. Fluid flow with a normal depth of flow happens underthe condition hn = const. In this case the relation (5.43) holds.

The depth hcr, at which the nominator of the fraction on the right-hand sideof Eq. (5.44) vanishes is called the critical depth. In cross-sections with such adepth the derivative dh/dx tends to ∞ and the fluid flow varies the level ofpipe filling abruptly. Depending on the relation between the depths hn andhcr different flow regimes are possible. Investigations of these regimes aredescribed in special monographs (Archangelskiy, 1947; Leibenson et al., 1934;Christianovitch, 1938).


5.7Non-Stationary Fluid Flow with Flow Discontinuities in a Pipeline

The preceding classical models of non-stationary flows of a slightlycompressible fluid (see Sections 5.1–5.3) contained an essential restrictionon the absence of phase transition in the fluid. It was tacitly supposed that thefluid under no circumstances passes into the vapor-gas phase, even when thepressure drops to the value of the saturated vapor pressure. However, inthe propagation of a rarefaction wave in the pipeline this condition can beviolated at many pipeline cross-sections and first and foremost at the topsof the pipeline profile. When the pressure in the rarefaction wave reduces tothe saturated vapor pressure the fluid boils, the vapor column breaks and thepipeline cross-section becomes partially filled with vapor. From this point onall further results predicted by classical theory appear to be wrong.

For example, at the disconnection of a pumping station or an aggregateof this station a rarefaction wave is propagated downstream in the pipeline.The pressure in such a wave reduces leading to the formation of voids at thetops of the pipeline profile. These voids are capable of growing and turninginto stationary gravity flow sections or, on the contrary, contracting and evendisappearing altogether. To perform calculation of such processes on the baseof classical theory is of course impossible.

We can give another example: closing the gate valve generates a compressionwave propagating upstream with a rise in the pressure in the wave. This wavehaving been reflected from the open surface of the reservoir or from a vapor-gascavity inside the pipeline initiates a rarefaction wave traveling in the oppositedirection and reduces the pressure in the fluid. This brings into existencetemporal transfer points at some tops of the pipe profile and cavities. Thisleads sometimes to fluid flows with a partially filled pipeline cross-section. Ifthe pressure supply in the pipeline is not high, the reduction in pressure leadsto flow discontinuity and the generation of vapor-gas cavities. So, for example,in laboratory installations it could be observed that the fluid before the gatevalve literally boiled owing to the sharp reduction in pressure. Calculation inthese cases on the basis of classical theory is also impossible.

One more example can be given: Connection of a lateral tap from theoil-pipeline to an intermediate oil tank leads to the propagation of rarefactionwaves up and down stream from the place of tap cutting up- and downstream.These waves are able to break the fluid column at many cross-sections of thepipeline profile and turn the enforced flow into a gravity one characterized bythe presence of vapor-gas cavities and gravity flow sections. Such cases alsodefy calculation in the framework of classical theory.

All the aforesaid is true also for pipelines transporting the so-calledunstable fluids, among which are gas-condensate and a wide fraction oflight hydrocarbons with saturated vapor pressure from 3 to 30 atm. Anysharp pressure reduction gives rise to a plethora of vapor-gas cavities thedisappearance of which in the pipe lead to powerful hydraulic shocks.

5.7 Non-Stationary Fluid Flow with Flow Discontinuities in a Pipeline 153

Profile Hydraulic ShockIt should be noted that the appearance or disappearance of voids in thepipeline is unsafe and is rather dangerous for pipeline integrity. In particular,in Lurie and Polyanskaya (2000), the discovery and investigation of the originof powerful hydraulic waves in the pipeline called profile hydraulic shock aredescribed. For example, each time the gate valve, located before a sectionof pipeline with significant slope, was closed the rarefaction resulting in theregion close to the valve led to a sequence of powerful hydraulic shocks. Theseshocks were gradually damped with time and the flow in the pipeline wasstabilized. Similar phenomena were observed in the pipeline on disconnectionof some pumps or the pumping station as a whole.

The nature of the profile hydraulic shock is as follows. When the gate valve (atx = 0, Fig. 5.16) closes the pressure in the sloping section of pipeline falls andthe fluid column, initially supported by this pressure, begins little by little to slipdown and acquire inverse flow (Figs. 5.17–5.19). The fluid in this inverse flowis accelerated and when vapor-gas voids, having originated before the gate valve,are taken out, there is an abrupt stop of fluid flow and, as a consequence, hy-draulic shock (Fig. 5.18). The power of the shock is then especially great, since

Figure 5.16 Initial stage of the process.

Figure 5.17 Formation of gravity flow in the upward sloping pipeline section.


Figure 5.18 Formation of reverse fluid flow in the upward sloping pipeline section.

Figure 5.19 Profile hydraulic shock.

the inverse flow of the fluid runs against the closed gate valve and, as followsfrom the Joukovski formulas, the amplitude of the hydraulic shock redoubles.

The hydraulic shock wave reflected from the gate valve and accompaniedby fluid stop, propagates downstream. Reaching a temporary transfer section,at which a vapor-gas void has been formed, the wave is reflected from thisvoid and now, in the form of a rarefaction wave, heads back to the gate valve,the fluid column in the ascending section of the pipeline again becomesweightless and begins to slip down to the closed gate valve. Then a secondaryhydraulic shock occurs, after which the process is repeated again and againwith decreasing intensity.

Generalized theory, (Lurie and Polyanskaya, 2000). In accordance with theclassical theory of non-stationary processes the wave processes generating in acompletely filled pipeline at its start-up or stopping, opening or closing of thegate valve or lateral tap and so on, are described by differential equations (5.4)for pressure p(x, t) and velocity v(x, t)

∂p

∂t+ ρ0c2 · ∂v

∂x= 0

ρ0∂v

∂t+ ∂p

∂x= −λ(Re, ε)

1

d0

ρ0v|v|2

− ρ0g sin α(x)

5.7 Non-Stationary Fluid Flow with Flow Discontinuities in a Pipeline 155

These equations represent the laws of mass conservation and variation ofmomentum of fluid particles moving in a pipeline. In the generalized theory itis assumed that in the pipeline there are completely filled enforced (pumped)sections as well as only partially filled gravity sections in which the pressureis equal to the saturated vapor pressure psat. The fluid flow in the sectionsof enforced flow is described by the system of equations (5.4) whereas in thesections of gravity flow it is described by Eqs. (5.42).

Method of calculation. For numerical calculation of oil product flow in com-pletely filled as well as in partially filled pipeline sections there is an elaboratescheme of end-to-end calculation based on the ideas of Godunov (Samarskiy,1977). This scheme involves consideration of the so-called problems on thedisintegration of arbitrary discontinuity in the system of hyperbolic equations.

Results of calculation. Figures 5.16–5.23 demonstrate the results of calcula-tions on successive stages of unloading wave propagation in a 10-km pipelinewith internal diameter d = 516 mm at an abrupt drop in the pumping deliveryof benzene (ρ = 750 kg m−3, psat = 0.7 atm) from 1500 to 200 m3 h−1. It isseen that, as distinct from existing theory, the line of hydraulic gradient at

Figure 5.20 Decay of hydraulic shock wave.

Figure 5.21 Formation of the secondary reverse flow in the upward sloping pipelinesection.


Figure 5.22 The second top of the profile hydraulic shock.

Figure 5.23 Formation of new stationary flow.

no instant of time intersects the pipeline profile, that is, the pressure in thepipeline would never be less than the fluid saturated vapor pressure.

The figures also show how the wave of pressure decrease runs over thepipeline profile top (lower polygonal line of Fig. 5.16), and how, at this top, atemporary transfer point and further gravity flow section with inverse flow offluid is being formed (Fig. 5.17 and Fig. 5.18).

Since the fluid flows in the opposite direction (as if it would run on a closedgate valve), in 24 s after the aggregate disconnection a powerful hydraulicshock is formed in the pipe (Fig. 5.19), in which the pressure is nearly twicethe initial pressure at the station. The amplitude of the hydraulic shock wavegradually decreases (Fig. 5.20) and in a further 10 s in the upward slopingsection of the pipeline slope the inverse flow appears again (Fig. 5.21). After46 s the secondary hydraulic shock occurs (Fig. 5.22), but already with lesseramplitude. In the given calculation there were six such hydraulic shocks. Only4 min after the pumping regime change the stationary regime in the pipelineis achieved (Fig. 5.23), where there exists a single gravity flow section of length500 000 m in the pipeline downward sloping section.

157

6Dimensional Theory

Dimensional theory contains fundamental propositions for representingequations of mathematical regularities modeling different phenomena ininvariant form, that is, in a form independent of the choice of the units of themeasurements. Such representations permit one to compile classes of similarphenomena and to model them on experimental installations.

When setting out the fundamentals of dimensional and similarity theoryand the modeling of phenomena we have followed the methodology of Sedov(1965).

6.1Dimensional and Dimensionless Quantities

A quantitative description of various physical phenomena, including thetransport of fluids and gases in pipelines, is connected with measurements ofthe characteristics of these phenomena, whose numerical values depend on thechoice of measurement units. For example, pipeline diameter can be expressedthrough the numbers 1; 10; 100; 1000; 3.28; 39.4 and so on depending on thatwhat unit of measurement is taken: meter, decimeter, centimeter, millimeter,foot or inch (1 foot ∼= 0.3048 m; 1 inch ∼= 0.0254 m); the length of a pipelinecan be expressed through the numbers 150 000; 150 or 93.21 and so on,depending on what is used as the measurement unit of the length: meter,kilometer or mile (1 mile ∼= 1.6093 km). The same is true for many otherphysical quantities. For example, the volumetric fluid flow rate in a pipelinemay be measured by the numbers 1000, 277.8 or 73.4 depending on whetherthe units of volume and time are taken as cubic meter and hour; liter andsecond; gallon (1 gallon in USA ∼= 3.785 liter) and second. The pressure canalso be measured with different numbers 64, 6.28 or 0.0433 depending on themeasurement units: technical atmosphere (kgf s−1 m−2), millions of pascals(megapascals, MPa) or psi (pound per square inch: 1 force pound ∼= 4.448 N;1 mass pound ∼= 453.6 g, 1 psi ∼= 0.006896 MPa ∼= 0.0703 atm).

It is important to note that the choice of measurement units depends on theresearcher and therefore is, to a great extent, arbitrary. Two different researchers


158 6 Dimensional Theory

describing one and the same phenomenon but at the same time using differentmeasurement units can obtain diverse numerical values of one and the sameparameter.

And yet there exist parameters whose numerical values do not depend on theresearcher, that is, do not depend on measurement units. These parametersare said to be invariant relative to the choice of measurement units. Forexample, the ratio of the pipeline length to the pipeline diameter L/d, theratio of the pressure at a pumping station discharge line to the pressure at apumping station suction line pdl/psl or a more complex combination such asthe Reynolds number Re = vd/ν (v is the average fluid velocity in the pipeline,d is the internal diameter of the pipeline, ν is the fluid kinematic viscosity)are independent of the choice of measurement units. This means that thenumerical values of L/d, pdl/psl and v · d/ν would be unchanged for any choiceof length, pressure, velocity and viscosity units.

Hence, it is possible to give the following definition: quantities whosenumerical values depend on the choice of measurement units are called dimensionalquantities; quantities whose numerical values do not depend on the choice ofmeasurement units are called dimensionless quantities.

6.2Primary (Basic) and Secondary (Derived) Measurement Units

Measurement units, having been introduced empirically by arbitrary condi-tions and propositions, are called primary or basic units. Among these are, inparticular, the units of length, time and mass. In the international SI systemthey are defined as follows.

Meter is a unit of length measurement. In accordance with the definitiontaken at the 11th General Conference on Weights and Measures (1960),1 meter is a length equal to 1650763.73 lengths of the wave emitted by thekrypton (Kr) atom in vacuum at its energy-level transition. The internationalstandard of the meter before 1960 was a bar of platinum–iridium alloy markedon one side of its planes. This bar is kept on deposit at the InternationalBureau of Weights and Measures in Sevre near Paris. At first the meter wasdefined as 10−7 part of a quarter of the Earth’s meridian.

Second is the measurement unit of time. There are recognized the atomic(standard) second reproduced by cesium standards of frequency and time andthe ephemeris second equal to 1/31556925.9747 part of the tropical year.

Kilogram is a unit of mass measurement. The standard kilogram is equalto the mass of the international prototype kept at the International Bureauof Weights and Measures. The prototype of the kilogram is made fromplatinum–iridium alloy in the form of a cylindrical weight. The relative errorof standard copies compared with the original does not exceed 2 · 10−9.

There are also other basic units of measurement, such as coulomb (C) – aunit of electricity quantity (electrical charge), volt (V) – a unit of electrical stress

6.3 Dimensionality of Quantities. Dimensional Formula 159

(voltage, potential); degrees ( ◦C, ◦F, K and other) – a unit of temperature andso on.

Measurement units of other quantities are by definition introduced throughthe basic measurement units. Such units are called secondary or derived units.Among them are the following.

Velocity which is defined as the ratio of length to time, therefore velocityunits can be m s−1, km h−1, mile h−1 and so on.

Acceleration which is defined as the ratio between the velocity increment andtime, hence acceleration units can be m s−2, foot s−2, km s−2, mile h−2 andso on.

Force (weight) which is defined as the product of mass and acceleration,thus force units can be dyne (1 dyn = 1 g cm s−2); newton (1 N = 1 kg m s−2);pound (∼=4.448 kg m s−2) and so on.

Density which is defined as the mass of the medium unit volume, thereforeits units can be kg m−3, g cm−3, t m−3 and so on.

Specific weight which is defined as the weight of the medium unit volume,hence its measurement units can be N m−3, dyn cm−3 and so on.

Pressure which is defined as the ratio of force to unit area, thus force unitscan be: pascal (1 Pa = 1 N m−2 = 1 kg m−1 s−2), pound inch−2 and so on.

Fluid flow rate (e.g. volumetric) which is defined as the fluid volume crossingthe pipeline cross-section area in a unit time, therefore its units can be m3 s−1,m3 h−1, m3 min−1, l s−1 and so on.

Current intensity which is defined as a charge transmitted in a unit time,hence the unit of measurement can be the ampere (1 A = 1 C s−1).

All these and analogous measurement units are derived from the basic units.

6.3Dimensionality of Quantities. Dimensional Formula

Let there be a physical parameter A. Expression of its measurement unitthrough basic units is called the dimension of a given parameter and is denotedby the symbol [A]. This expression written as a formula is called the dimensionalformula.

Denote through L the measurement unit of length, through T themeasurement unit of time and through M the measurement unit of mass.Then the expression of the measurement units of many other quantities canbe written as the following formulas:

(A = v) – velocity: [v] = L

T= L · T−1 = M0 · L · T−1;

(A = a) – acceleration: [a] = L

T2 = L · T−2 = M0 · L · T−2;

(A = F) – force: [F] = M · [a] = M · L · T−2;

(A = ρ) – density: [ρ] = M

L3 = M · L−3 · T0;


(A = γ) – specific weight [γ] = M · L · T−2

L3 = M · L−2 · T−2;

(A = p) – pressure: [p] = [F]/L2 = M · L · T−2

L2 = M · L−1 · T−2;

(A = Q) – volumetric flow rate: [Q ] = L3

T= M0 · L3 · T−1;

(A = G) – mass flow rate: [G] = M

T= M1 · L0 · T−1;

(A = B) – dimensionless parameter [B] = M0 · L0 · T0.

Thus, particular examples show that in all cases the dimensional formula ofparameter A has form of a power monomial

[A] = Mm1 · Lm2 · Tm3 (6.1)

where m1, m2, m3 are certain real positive or negative numbers.What does the dimensional formula mean? It allows one to determine very

simply by how many-fold the numerical value of parameter A would be changed, ongoing from one system of basic measurement units to another one differing from thefirst system only by the scales of the basic units. For example, if the transition fromthe new system of basic measurement units to the old one is accomplished byvariation of the mass unit k1-fold, of the length unit k2-fold, of the time unitk3-fold, the numerical value of the parameter A would vary km1

1 · km22 · km3

3 -fold,that is the new value A′ of this parameter would be determined by the formula

A′ = km11 · km2

2 · km33 · A. (6.2)

Let us explain the afore said by examples.

Example 1. The velocity v in the SI system of basic measurement units (m, s,kg) is 1 m s−1. For a new system of basic measurement units (cm, s, kg) thetransition from this system to the SI system is performed by increasing one ofthe basic units (length) by a factor of 100, the value of the velocity will also beenhanced 100 times

v′ = 10 · 1001 · 1−1 = 100 · v = 100 cm s−1.

Example 2. The pressure p in the SI system of basic units (m, s, kg)is 6 MPa, that is 6 000 000 kg m−1 s−2. For a new system of units (inch,s, pound) the transition from this system to the SI system is carriedout by changing the length scales 100/2.54 = 39.37 times and the massby 1/0.4536 = 2.205 times, hence the value of the pressure will varyby a factor 2.2051 · 39.37−1 · 1−2 = 0.056 and will be 0.056 · 6 000 000 =336 043 (psi).

In Eq. (6.1), obtained empirically, there are only three basic units ofmeasurement because we have considered examples solely from mechanics.However, it is easy to verify that even if the number of physical parameters

6.4 Proof of Dimensional Formula 161

is increased by other parameters from electrical, heat, chemical andother phenomena and the number of basic units grows and exceedsthree, the dimensional formula would not be radically changed. Only thenumber of factors will increase. Hence, it could be assumed, that thedimensional formula in the general case would have the form of a powermonomial

[A] = [a1]m1 .[a2]m2 .[a3]m3 . . . . ..[an]mn (6.3)

where A is a physical parameter, whose dimension is derived from thedimensions of basic quantities denoted by a1, a2, . . . ., an.

6.4Proof of Dimensional Formula

Let us prove the validity of the dimensional formula (6.3) in the general case.This formula means that if we change the scale of a basic measurement unita1 by k1 times, of a2 by k2 times, . . . ., of an by kn times, the numerical valueof the parameter A would be changed by km1

1 · km22 · . . . · kmn

n times, that is thenew value of this parameter A′ would be equal to

A′ = km11 · km2

2 · . . . · kmnn · A.

Let us use this circumstance.Let there be three researchers B, C and D measuring one and the same

physical parameter A but using different systems of basic measurementunits, namely

B: {a1, a2, . . . , an};C:

{a′

1, a′2, . . . , a′

n

}, so that its basic units are related to the basic units of

researcher B by the formulas

a′1 = α1 · a1

a′2 = α2 · a2

. . . . . . . . . . . .

a′n = αn · an

where k1 = α1, k2 = α2, . . . , kn = αn;D:

{a′′

1, a′′2, . . . , a′′

n

}, so that its basic units are related to the basic units of

researcher C by the formulas

a′′1 = β1 · a′

1

a′′2 = β2 · a′

2

. . . . . . . . . . . .

a′′n = βn · a′

n


where k1 = β1, k2 = β2, . . . , kn = βn, and to the basic units of researcher B bythe formulas

a′′1 = β1α1 · a1

a′′2 = β2α2 · a2

. . . . . . . . . . . .

a′′n = βnαn · an

where k1 = β1α1, k2 = β2α2, . . . , kn = βnαn.It is reasonable to suppose that the values of the parameter A measured by

the three researchers would be different: A measured by B, A′ measured by Cand A′′ measured by D.

Let a function F(k1, k2, . . . , kn) show how many-fold the numerical valueof parameter A in one of the system of basic units would be changed whenpassing to another system differing by k1, k2, . . . , kn times from the first onein the scales of the basic units, respectively. Then if we go from the units ofresearcher B to the units of researcher C, we obtain

A′ = F(α1, α2, . . . , αn) · A.

If we then go from the units of researcher C to the units of researcher D,we get

A′′ = F(β1, β2, . . . , βn) · A′

or

A′′ = F(β1, β2, . . . , βn) · F(α1, α2, . . . , αn) · A. (6.4)

On the other hand, if we go at once from the units of researcher B to the unitsof the researcher D, bypassing researcher C, we get

A′′ = F(α1β1, α2β2, . . . , αnβn) · A (6.5)

The result should of course be independent of the transition route fromone researcher to another one, thus it must identically satisfy the followingfunctional equation

F(α1β1, α2β2, . . . , αnβn) = F(α1, α2, . . . , αn) · F(β1, β2, . . . , βn) (6.6)

which ought to be valid for any values of the factors α1, α2, . . . , αn andβ1, β2, . . . , βn.

Find the solution of this equation.Differentiation of both parts of the identity (6.6) with respect to βi, where i

may be equal to 1, 2, 3, . . . , n, yields

αi∂F

∂ξi= ∂F

∂βi· F(α1, α2, . . . , αn), where ξi = αiβi.

6.5 Central Theorem of Dimensional Theory 163

Since this equality also represents an identity, then β1 = 1, β2 = 1, . . . , βn = 1.As a result the following differential equation is obtained

αi · ∂F(α1, α2, . . . , αn)

∂αi= mi · F(α1, α2, . . . , αn) (6.7)

in which mi = ∂F/∂βi at β1 = 1, β2 = 1, . . . , βn = 1.The solution of differential equation (6.7) gives the dependence of the sought

function F on parameter αi. Really, the separation of variables provides

∂

∂αi(ln F) = mi

αi⇒ ln F = mi · ln αi + c ⇒ F = K · α

m1i

where the integration constant K (K = eC) is a function of the remainingparameters αj. Since αi was any one of the arguments of the function F, thelatter should have the following form

F = K0 · αm11 · α

m22 · . . . · αmn

n

where K0 = const. If F(1, 1, . . . , 1) = 1, because the value of the parameter Ais not changed by variation of the basic measurement unit scales, K0 = 1 andthe function will be

F = αm11 · α

m22 · . . . · αmn

n

or redesignating the variables as k1, k2, . . . , kn, one gets

F = km11 · km2

2 · . . . ·, kmnn . (6.8)

Hence, when changing the basic unity a1 by k1 times, the numerical value ofthe parameter A would vary by k1

m1 times, when changing the basic unity a2

by k2 times the numerical value of the parameter A would vary by k2m2 and so

on. Thus, the parameter A has the dimension

[A] = [a1]m1 .[a2]m2 .[a3]m3 . . . . ..[an]mn

which proves the dimensional formula (6.3).

6.5Central Theorem of Dimensional Theory

It is appropriate now to interrupt the description of the theory and to formulatea question concerning an apparent contradiction due to the use of dimensionalquantities.

Let the mathematical form of a certain physical phenomenon be expressed bythe dependence of a parameter A on other parameters a1, a2, . . . , an governingthis phenomenon as follows

A = f (a1, a2, . . . , an). (6.9)


For example, the enhancement of pipeline cross-section area �S whenproducing a positive pressure P depends on the magnitude of this pressure aswell as on the pipeline diameter D, the wall thickness δ and the elastic modulus(Young’s modulus) E of the steel from which the pipeline is made. Hence wehave A = �S, a1 = P, a2 = D, a3 = δ, a4 = E, so that �S = f (P, D, δ, E).

It is evident that the dependence under consideration exists objectively andshould not depend either on the researcher performing the investigation oron the choice of measurement units being used to calculate the values of thefunction, i.e. parameter A, and its arguments (a1, a2, . . . , an).

On the other hand all quantities entering into the dependence (6.9) aredimensional quantities whose numerical values depend on the choice ofmeasurement units and consequently on the researcher.

The question arises as to at which point the dependence (6.9) reflects theobjectively existing physical regularity when numerical values of the function and itsarguments depend on the researcher.

The answer to this question gives the central theorem of dimensional theorycalled the Buckingham �-theorem. This theorem states: ‘‘Each dependencebetween dimensional quantities reflecting an objectively existing physical regularitycould be rewritten in invariant form independent of the choice of measurement units,namely in the form of a dependence between dimensional complexes composed fromgoverning parameters’’.

Now we will prove this theorem, removing the contradiction between theobjective character of any physical regularity and the subjective characterof the choice of measurement units. However, before we do this, weneed first to define dimensionally-dependent and dimensionally-independentquantities.

6.6Dimensionally-Dependent and Dimensionally-Independent Quantities

It is said that a quantity ‘‘a’’ is dimensionally-dependent on the quantitiesa1, a2, . . . , an when its dimension [a] is expressed through the dimensions[a1], [a2], . . . , [an] by the formula

[a] = [a1]m1 [a2]m2 . . . [an]mn (6.10)

namely there exist such numbers m1, m2, . . . , mn that the equality (6.10)is obeyed. If such numbers do not exist, it is said that the quantity a isdimensionally-dependent on the quantities a1, a2, . . . , an.

It is evident that parameters with dimensions of time t, length l and massm are dimensionally-independent of each other. Parameters with dimensionsof velocity v and density ρ are dimensionally-independent. On the other hand,a parameter with the dimension of pressure p is dimensionally-dependent onparameters with dimensions of density ρ and velocity v. It is easy to verify that

6.6 Dimensionally-Dependent and Dimensionally-Independent Quantities 165

[p] = [ρ][v]2

[p] = M1L−1T−2; [ρ] = M1L−3T0; [v2] = M0L2T−2 ⇒M1L−1T−2 = (M1L−3T0)(M0L2T−2).

Hence, the ratio p/(ρv2) is dimensionless since the dimensions of thenumerator and denominator coincide.

There is a general algorithm capable of determining whether one or anotherparameter is dimensionally-dependent or dimensionally-independent of othergiven parameters. For example, let us consider mechanics, in which thereare three basic measurement units: mass (M), length (L), time (T) and it isrequired to clarify whether the parameter P with the dimension of pressureis dimensionally-dependent on parameters µ with the dimension of dynamicviscosity, g with the dimension of acceleration and D with the dimension oflength. Let us show how to do it.

Write the dimensions of all the parameters under consideration

[p] = M1L−1T−2,

[µ] = ML−1T−1,

[g] = LT−2,

[D] = L.

Now let us look for the numbers m1, m2 and m3 that would obey the equality

[p] = [µ]m1 [g]m2 [D]m3 .

Insertion of the considered dimensions in this equality yields

M1L−1T−2 = (ML−1T−1)m1(LT−2)m2(L)m3 .

Equating the exponents of identical basic measurement units in the left- andright-hand sides of the last relation, we get a system of three linear equations

1 = m1,

−1 = −m1 + m2 + m3,

−2 = −m1 − 2m2

to determine the three unknown quantities m1, m2, m3. This system has asingle solution

m1 = 1, m2 = 1

2, m3 = −1

2.

Thus, we obtain

[p] = [µ][g]1/2[D]−1/2 (6.11)

which proves that pressure is a dimensionally-dependent parameter in thesystem of three parameters – viscosity, acceleration, length. From Eq. (6.11)


it follows that the ratio p/(µg1/2D−1/2) is a dimensionless quantity since thedimensions of the numerator and denominator coincide.

In a similar manner many examples of the same type could be considered. Letus focus our attention on the question of the maximal number of dimensionally-dependent quantities for a given set of dimensional parameters {a1, a2, . . . , an}.If this set contains n elements, it is always possible to separate from it asubset containing the maximum possible number k ≤ n of dimensionally-independent parameters.

This assertion follows from the known theorem of algebra that states thatfrom any system of linear equations can be separated the maximum possiblenumber of linearly independent equations, this number being called the rankof the system of equations. If we write dimensional formulas for all theparameters a1, a2, . . . , an,

[a1] = [O1]x1 [O2]x2 . . . .[Ok]xk ,

[a2] = [O1]y1 [O2]y2 . . . .[Ok]yk ,

. . . . . . . . . . . . . . . . . . . . . . . . ..,

[an] = [O1]z1 [O2]z2 . . . .[Ok]zk

where O1, O2, . . . , Ok are symbols of basic measurement units in the givensystem of units and x1, x2, . . . , xk; y1, y2, . . . , yk; z1, z2, . . . , zk are exponents ofthis formula, the question of the dimensional dependence (or independence)of each of them on the other as well as the question of the maximum number ofdimensionally-independent quantities is reduced to solutions of the system oflinear equations

m1x1 + m2y1 + . . . + mnz1 = 0,

m1x2 + m2y2 + . . . + mnz2 = 0,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

m1xk + m2yk + . . . + mnzk = 0

(k equations with n unknowns) with respect to the unknowns m1, m2, . . . , mn

and to determine the rank of the system matrix.In order to separate a subset containing the maximum number of

dimensionally-independent quantities of the set {a1, a2, . . . , an} we proceedas follows. Let us take a quantity a1. If it is a dimensional quantity, the nextquantity a2 is added to it. If a2 has dimension different from the dimension ofa1, the system {a1, a2} would consist of dimensionally-independent quantities.Next a3 is added to the system of quantities {a1, a2} a3. If the dimensionof a3 is expressed through {a1, a2} by the formula (6.10), this quantity isrejected and instead we take a4, which in its turn is tested for independencefrom the quantities {a1, a2}. If the dimension of a3 is not expressed throughthe dimensions of {a1, a2}, the system {a1, a2, a3} represents a system ofdimensionally-independent quantities. In such a way all quantities in the set

6.6 Dimensionally-Dependent and Dimensionally-Independent Quantities 167

{a1, a2, . . . , an} are considered and gradually the subset containing the numberof dimensionally-independent quantities is separated.

Note that for one and the same set several different subsets containingthe maximum number of dimensionally-independent quantities could beseparated. Such subsets by themselves can be different but the number ofelements in them would be equal.

In particular, when the dimensions of all quantities in the set {a1, a2, . . . , an}are expressed through the dimensions M, L, T, this set can have no more thanthree dimensionally-independent parameters.

Exercise. It is required to separate the maximum number of dimensionally-independent quantities among a set of the following parameters: p – pressure;ρ – density; v – velocity; Q – volumetric flow rate; ν – kinematic viscosity;D – diameter; g – acceleration due to gravity.

Solution. Write the dimensional formulas for all given parameters using theSI system of basic measurement units

[p] = kg

m × s2; [ρ] = kg

m3; [v] = m

s; [Q ] = m3

s;

[ν] = m2

s; [g] = m

s2; [D] = m.

Since all the parameters are expressed through mass, length and time, the max-imum number of dimensionally-independent quantities is less than or equal tothree. As the first quantity, we take D. As the second quantity we take g, becauseits dimension contains time and consequently it cannot be expressed throughthe dimensions of D. Finally, as the third dimensionally-independent quantitywe take ρ. It is evident that the dimension of ρ cannot be expressed throughthe dimensions of D and g, since it contains mass. Hence, the set {D, g, ρ}consisting of three parameters represents a basis of the maximum number ofdimensionally-independent quantities in the given set of parameters.

There are of course other possible subsets of the given set consisting alsoof a maximum number of dimensionally-independent quantities. It is easyto verify that, for example, the subsets {p, v, D}, {ρ, Q, D} and so on consistof dimensionally-independent quantities and the number of dimensionally-independent quantities in each of these subsets is a maximum and equal tothree.

Exercises.

1. It is required to separate a basis consisting of the maximum number ofdimensionally-independent quantities among a set of the followingparameters: ρ – density; ω – frequency of revolutions; D – wheeldiameter; Q – volumetric flow rate; g – acceleration due to gravity.

Answer. For example {ρ, ω, D}.


2. It is required to separate a basis consisting of a maximum number ofdimensionally-independent quantities among a set of the followingparameters: v – velocity; S – area of the cross-section occupied by fluid;g – acceleration due to gravity; D – diameter of the pipeline; α – angle ofinclination of the pipeline axis to the horizontal (dimensional quantity);ν – kinematic viscosity, m2 s−1.

Answer. For example {D, v}.

3. It is required to separate a basis consisting of a maximum number ofdimensionally-independent quantities among a set of the followingparameters: p – pressure; v – velocity; θ – temperature; ρ – density;λ – thermal diffusivity factor, W m−1 K−1; c – heat capacity, J kg−1 K−1).

Answer. For example {ρ, v, c, θ}.

6.7Buckingham �-Theorem

Now we go to the proof of the central theorem of dimensional theory, theBuckingham �-theorem, which was partially formulated in Section 6.5.

Let a physical regularity be represented by a function

A = f (a1, a2, . . . , ak, ak+1, . . . , an) (6.12)

of arguments a1, a2, . . . , an among which can be dimensional as well asdimensionless parameters.

Let the maximum number of dimensionally-independent arguments in theset a1, a2, . . . , an be k and without disturbing the generality it can be taken thatthe first k arguments are a1, a2, . . . , ak. Then the remaining (n − k) argumentsof this function ak+1, ak+2, . . . , an would be dimensionally-dependent on thefirst k arguments, that is

[ak+1] = [a1]m1 [a2]m2 . . . [ak]mk ,

[ak+2] = [a1]n1 [a2]n2 . . . [ak]nk ,

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

[an] = [a1]p1 [a2]p2 . . . [ak]pk

where mi, ni, . . . , pi are real numbers. Thus, the relations

�1 = ak+1

am11 am2

2 · · · amk

k

,

�2 = ak+2

an11 an2

2 · · · ank

k

, (6.13)

�n−k = an

ap1

1 ap2

2 · · · apk

k

6.7 Buckingham �-Theorem 169

are dimensionless parameters since the dimensions of the quantities in thenumerators and denominators of these fractions are identical.

The dimension of the quantity A should also be expressed through thedimensions of the arguments a1, a2, . . . , ak. If it is not expressed throughthese dimensions, it would also not be expressed through the dimensions ofall the quantities a1, a2, . . . , an. Hence, there should exist a relation

[A] = [a1]q1 [a2]q2 . . . [ak]qk

meaning that the ratio

� = A

aq1

1 aq2

2 · · · aqk

k

(6.14)

has to be dimensionless.Let us revert to the dependence (6.12). This dependence may be rewritten as

A

aq1

1 aq2

2 · · · aqk

k

= f

(a1, a2, . . . , ak,

ak+1

am11 am2

2 · · · amk

k

, . . . ,an

api

1 ap2

2 · · · apk

k

)

where f represents a function resulting from f after redefinition of itsarguments. This dependence with (6.13) and (6.14) can be represented asfollows

� = f (a1, a2, . . . , ak, �1, �2, . . . , �n−k). (6.15)

If we now arbitrarily and independently from each other vary the numericalvalues of the arguments a1, a2, . . . , ak by going from one system of basicmeasurement units to another one, the numerical values of the parameters �

and �1, �2, . . . , �n−k would not be changed because they are dimensionlessquantities. From here it follows that the function � cannot depend on itsfirst k arguments a1, a2, . . . , ak, since the dependence (6.15) would have thefollowing simple form

� = f (�1, �2, . . . , �n−k). (6.16)

Thus, we have shown that any physical dependence between dimensionalquantities of the type (6.12) will be invariant, that is, independent of thechoice of measurement units, of the form (6.16) between dimensionlesscomplexes made up from arguments of the dependence under consideration.The number of these complexes would be less than the number of argumentsof the initial dependence by a number equal to the maximum number ofdimensionally-independent quantities among these arguments.

Exercise 1. The augmentation of the cross-section area �S of a steel pipelinewhen setting up in it excess pressure P depends on the value of this pressure,the diameter D of the pipeline, the thickness δ of the pipeline wall and theelastic modulus (Young’s modulus) E of the steel from which the pipe is made.


Using the �-theorem, it is required to write this dependence in dimensionlessform and to clarify how many dimensionless parameters determine it.

Solution. The dependence to be sought can be written in a general formas �S = f (P, D, δ, E). The dimensions of the arguments of this dependencein the SI system are: [P] = kg m−1 s−2; [D] = m; [δ] = m; [E] = kg m−1 s−2.Among these there are only two dimensionally-independent quantities, forexample D and E. Consequently, the number of arguments in the dependenceunder consideration may be reduced to two. Thus we have:

� = S/D2; �1 = P/E; �2 = δ/D;

� = f (�1, �2) or �S = D2 · f (P/E, δ/D).

Hence, the dimensional analysis has shown that in the considered dependencethere are only two dimensionless complexes P/E and δ/D instead of fourdimensional arguments.

Moreover, if additionally we invoke the reasoning that the variation ofpipeline cross-section area �S should be proportional to P/E (the function, f ,could be expanded in a Taylor series in the vicinity of the point �1 = 0 leavingin the expansion only the first term, because �S = f (0, �2) = 0 at P = 0 andthe ratio P/E is very small: P ≈ 2–7 · 106 �a; E ∼= 2 · 1011 �a, and inverselyproportional to δ/D (δ ≈ 5–10 mm; D ≈ 300–100 mm), the dependence understudy could be written as

�S = D2 · P

E· 1

δ/D· f0 = f0 · D3 · P

δ · E

where f0 is a certain constant. Hence, in the dependence under study there isonly one constant left to be obtained. Theoretical investigation shows that theconstant f0 is equal to π/4 ∼= 0.785.

Exercise 2. It is known that the laminar flow of a viscous incompressiblefluid in a circular pipe loses stability and becomes turbulent. It is requiredto investigate the dependence of the critical flow velocity vcr at which thistransition happens, taking the critical velocity as a function of three parameters:pipeline diameter d, dynamic viscosity µ and density ρ of the fluid. In otherwords it is required to investigate the dependence vcr = f (d, µ, ρ).

Solution. In the given case A = vcr, a1 = d; a2 = µ; a3 = ρ; n = 3. The di-mensions of the parameters in the SI system are: [vcr] = m s−1; [d] = m;[µ] = kg m−1 s−1; [ρ] = kg m−3. From this it follows that all three argumentsof the function are dimensionally-independent, i.e. k = n = 3, and the numberof arguments in the dimensionless writing of the function under study maybe reduced by three, i.e. to 0.

6.7 Buckingham �-Theorem 171

The single dimensionless complex � can be written as � = vcr · d/(µ/ρ).The sought dependence then takes the especially simple form

�cr = vcr · d

(µ/ρ)= const. (6.17)

The ratio µ/ρ is called the kinematic viscosity of the fluid and is denotedby ν ([ν] = L2/T). In the SI system the unit of kinematic viscosity isstokes (St); 1 St = 10−4 m2 s−1; the viscosity of water is ≈0.01 St = 1 cSt(centistokes) = 10−6 m2 s−1.

The dimensionless parameter �cr determining the transition of laminarflow to turbulent flow is called the critical Reynolds number and is denoted byRecr. The theory and experiments have shown that Recr

∼= 2300. At Re < Recr

the flow is laminar whereas at Re > Recr it is turbulent.

Exercise 3. A ball of mass m and diameter D is dropped in a viscous fluid withdensity ρ and viscosity ν under the action of gravity (gravity acceleration – g)with constant velocity v. The dependence of this velocity on the governingparameters: v = f (m, D, ρ, ν, g) is to be investigated. Using the �-theorem itis required to write the sought dependence in dimensional form.

Solution. Dimensions of the parameters (n = 5), in this problem (A = v, a1 =m, a2 = D, a3 = ρ, a4 = ν, a5 = g), in the SI system are [v] = m s−1; [m] = kg;[D] = m; [ρ] = kg m−3; [ν] = m2 s−1; [g] = m s−2. The maximum number ofdimensionally-independent parameters among the arguments is equal tothree (as such parameters can be taken for example ρ, D and g), the numberof arguments in the sought dimensionless writing may be reduced to two(5 − 3 = 2).

Setting up dimensionless complexes �, �1 and �2

� = v

g1/2D1/2, �1 = m

ρ · D3, �2 = ν

g1/2D3/2,

the sought dependence can be represented as follows

� = f (�1, �2)

or

v = √gD · f

(m/ρD3, ν/

√gD3

).

It is seen that the dependence contains in fact not five dimensional arguments,but only two dimensionless complexes.

Exercises.

1. Using the �-theorem, it is required to write in dimensional form thedependence of the resistance force F experienced by a submarine


moving in water (density ρ, kinematic viscosity ν) with velocity v, if weaccept that this force depends also on the diameter of the submarinecross-section D and the submarine length L, that is F = f (v, ρ, ν, D, L).

Answer. F/(ρv2D2) = f (vD/ν, D/L).

2. Using the �-theorem, it is required to write in dimensional form thedependence of the oscillation period T of a mathematical pendulum(massive point on non-stretchable line), if the mass of the latter is m,the length of the line is L and the acceleration due to gravity is g, that isT = f (m, L, g).

Answer. T/√

L/g = const.

3. Using the �-theorem, it is required to write in dimensional form thedependence of the time T of complete outflow of fluid (density ρ andviscosity ν) from a tank car with diameter D and length L. The outflowhappens under gravity (acceleration due to gravity g) through the drainsystem with flow area s located at the tank bottom. The sought functionis T = f (ρ, ν, g, D, L, s).

Answer. T/√

D/g = f (√

gD3/2/ν, L/D, s/D2).

4. Using the �-theorem, it is required to write in dimensional form thesame dependence as in the previous exercise with the single distinctionthat the outflow happens not only under the action of gravity but alsounder the action of positive pressure �p created inside the tank. Thesought function is T = f (�p, ρ, ν, g, D, L, s).

Answer. T/√

D/g = f (√

gD3/2/ν, �p/(ρgD), L/D, s/D2).

5. Using the �-theorem, it is required to write in dimensional form thedependence of the volumetric flow rate Q of a fluid with density ρ andviscosity ν in an inclined pipeline (inclination angle α) havingcross-section area S0, if this flow represents voluntary (gravity) flow.

The flow happens under the action of the gravity force projectiong sin α, the area of the pipeline cross-section filled with fluid is S < S0.The sought function is Q = f (g sin α, S, S0, ρ, ν).

Answer. Q/[S5/40 · (g sin α)1/2] = f (S/S0, (g sin α)1/2 · S3/4

0 /ν).

173

7Physical Modeling of Phenomena

The main advantage of the dimensional theory is that it opens up possibilitiesto use the similarity laws of physical phenomena and allows modeling of thesephenomena through replacing them in nature by similar phenomena on areduced or enlarged scale under experimental conditions.

7.1Similarity of Phenomena and the Principle of Modeling

In order to elucidate the essence of modeling, let us consider several examples.Assume that it is required to calculate the radius of a circle inscribed in a

triangle whose sides are very large, for example 1, 2 and 3 km. This problemmay be easily solved by simple algebraic calculation. However, if it wasnecessary to perform this measurement experimentally, one could proceedas follows: on a sheet of paper draw a triangle on a reduced scale similar tothe given triangle, for example a triangle with sides 10, 20 and 30 cm. Thesimilarity factor of this triangle to the full-scale one would be equal to 10 000.Inscribing a circle in the depicted triangle, it is easy to measure its radius. Thenthe obtained number multiplied by the similarity factor, i.e. 10 000, yields thesought radius of the circle inscribed in the full-scale triangle. Of course in thecase under consideration we are dealing with a simple geometric similarity andgeometric modeling but it clarifies the essence of modeling in the general case.

Let it be required to determine experimentally the dump time of a tank ofcomplex geometrical form and very large size. In order to solve this problem itis decided to make a copy of the tank on a reduced scale, fill it with some modelliquid and then measure the dump time. The question arises as to whetherit is enough to provide merely geometric similarity of full-scale and modeltanks and to use in experiments the same liquid or is it necessary to replacethe liquid with another one with specially selected properties. Moreover, is itnecessary to determine the ratio between the measured dump time and thatactually taking place in the full-scale object. The answers to these questionsshould give the theory of simulation (modeling).

Finally, one more example from engineering practice. It is required to clarifywhether the river dam will withstand the dynamic head of flooding water.


174 7 Physical Modeling of Phenomena

For this purpose a reduced size copy of the dam is made, mounted in anexperimental channel in a hydrological laboratory. It is evident that if the damwere made from the same material as in the natural conditions, i.e. fromferroconcrete, the dam would withstand any water head. It is necessary todetermine the velocity of water in the experimental channel in order to modelthe river head. When reducing the linear sizes of the considered phenomenon,it is insufficient to provide geometric similarity. It is necessary especially tochange the scales of many other parameters of the phenomenon.

Similar problems take place in different fields of engineering: hydraulicengineering, aviation, transport, the storage of oil and gas and so on.

Definition. Two phenomena are called similar when, from the given parame-ters of one phenomenon, analogous parameters of another phenomenon aredetermined by simple recalculation of the same kind as for transition from onesystem of measurement units to another. Each of such phenomena is called amodel of another phenomenon from this set of phenomena.

7.2Similarity Criteria

Let us determine the necessary and sufficient conditions of two phenomenato be similar. Such conditions are called similarity criteria.

Let a phenomenon be such that a certain physical quantity A is determinedby a set of physical parameters a1, a2, . . . , an, so that

A = f (a1, a2, . . . , an). (7.1)

The model under consideration consists of the dependence of an analogousphysical quantity A′ on the same physical parameters the numerical values ofwhich a′

1, a′2, . . . , a′

n differ from those determining the quantity A. Thus, wehave

A′ = f(a′

1, a′2, . . . , a′

n

). (7.2)

In accordance with the I-theorem both dependences (7.1) and (7.2) may berewritten in dimensionless form as follows

� = f(�1, �2, . . . , �n−k

),

(7.3)�′ = f

(�′

1, �′2, . . . , �′

n−k

),

where k is the number of dimensionally-independent parameters among thequantities a1, a2, . . . , an.

Relations (7.3) show that if the parameters a′1, a′

2, . . . , a′n are chosen such

that the following conditions are obeyed

�′1 = �1, �′

2 = �2, . . . , �′n = �n. (7.4)

7.3 Modeling of Viscous Fluid Flow in a Pipe 175

Then the following condition would also be satisfied

�′ = �. (7.5)

The value of the parameter A could be found by simple recalculation of theparameter A′ through

A = A′ · am11 am2

2 · · · amk

k

a′m11 a′m2

2 · · · a′mk

k

(7.6)

and the considered phenomena would, by definition, be similar.Thus, necessary and sufficient conditions of two phenomena to be

similar are equalities of the dimensionless complexes determining thesephenomena, namely conditions (7.4). Hence, the dimensionless parameters�1, �2, . . . , �n−k are the sought similarity criteria.

7.3Modeling of Viscous Fluid Flow in a Pipe

As an example of two similar phenomena, consider the modeling of thestationary flow of a viscous incompressible fluid in a model pipe with reducedsize as compared to the full-scale one.

In accordance with the results given in Section 1.6 this dependence is

�p

L

d· ρv2

2

= λ(Re, ε).

Denote through ρ′, v′, d′, µ′, �′, L′, �p′ the values of the hydrodynamicparameters relating to the flow in the model pipe. The same parameterswithout the superscript ‘‘prime’’ refer to the phenomena to be modeled. Then

�p′

L′

d′ · ρ′v′2

2

= λ(Re′, ε′).

The parameters of the model pipe and the flow regime in it are chosen to obeythe relations

Re = Re′, ε = ε′ (7.7)

or

vd

ν= v′d′

ν′ ,�

d= �′

d′ .

In so doing we ensure the equality

�p′

L′

d′ · ρ′v′2

2

= �p

L

d· ρv2

2


or

�p = �p′ · L

L′ · d′

d· ρ

ρ′ ·( ν

ν′)2

. (7.8)

If we now take the ratio d′/d, showing how many times smaller are the sizesof the model pipe than the sizes of the full-scale pipe (factor of geometricsimilarity), and the ratio ν′/ν of the viscosities of fluids flowing in full-scaleand model pipes, respectively, we could determine the fluid flow velocity androughness of the pipe walls of the experimental (model) installation needed toachieve the similarity

v′ = v · d

d′ · ν′

ν, �′ = � · d′

d. (7.9)

Formulas (7.9) state that the similarity in the case under consideration isafforded by fulfilling two conditions: equality of the Reynolds numbers andequality of the wall relative roughness. Consequently, the Reynolds numberand relative roughness serve as similarity criteria for the problem on the flowof a viscous fluid in a pipe.

7.4Modeling Gravity Fluid Flow

This type of flow has already been considered in Section 3.7. We now set a ques-tion on the physical modeling of this process. Let the flow rate Q of the fluid in apipe inclined to the horizontal at an angle α be given. It is required to determinethe depth h of the fluid filling the pipe cross-section, that is, the function

h = f (Q, d, g sin α, ν, �).

Rewrite the dependence to be sought in dimensionless form using the �-theorem. Among five arguments of this function there are two dimensionally-independent, for example ν, d, therefore in dimensionless form the numberof independent arguments will be reduced from five to three.

h

d= f

((Q/d2) · d

ν,

√gd sin α · d

ν,�

d

).

It is more convenient in this dependence to use the ratio of the first argumentto the second one instead of the second argument. As a result we have

h

d= f1

((Q/d2) · d

ν,

(Q/d2)√gd sin α

,�

d

).

In the first argument it is easy to recognize the Reynolds number Re of the flowcalculated from the velocity v = Q/d2 and the kinematic viscosity ν = µ/ρ,so that the first similarity criteria would be �1 = Re. The second argument

7.4 Modeling Gravity Fluid Flow 177

is called the Froude number, Fr. The Froude number is, in general, equal tov2/gd, thus in our case we are dealing with a Froude number calculated fromthe velocity Q/d2 and the component of the acceleration due togravity g sin α.Nevertheless, the second similarity criterion can be taken as �2 = Fr. Finally,the third argument of this dependence �3 = ε is the relative roughness of thepipe’s internal wall surface. Hence, the dependence under study is

h

d= f (Re, Fr, ε). (7.10)

To model this phenomenon it is necessary to provide the following similarityconditions

Re′ = Re ⇒ Q ′

ν′d′ = Q

νd⇒ Q ′ = Q ·

(ν′

ν

)·(

d′

d

),

Fr′ = Fr ⇒ Q ′2

g sin α′ · d′5 = Q2

g sin α · d5,

⇒ Q ′ = Q ·√

sin α′

sin α·(

d′

d

)5/2

, (7.11)

ε′ = ε ⇒ �′

d′ = �

d⇒ �′ = � · d′

d.

If these conditions are obeyed, the depth h to which the pipe is filled can becalculated using the formula

h′

d′ = h

d⇒ h = h′ · d

d′ . (7.12)

In order for the first two conditions to be consistent, the following equalitiesshould be true(

ν′

ν

)·(

d′

d

)=

√sin α′

sin α·(

d′

d

)5/2

or

sin α′ = sin α ·(

ν′

ν

)2

·(

d

d′

)3

. (7.13)

If in the model the same liquid as in the full-scale pipe is used, then a differentslope of the pipe should be used in the model, it must be chosen in accordancewith relation (7.13). If α′ = α, then the viscosity of the fluid used in the modelshould be different, namely ν′ = ν · (d′/d)3/2.

For example, let the linear size of the experimental pipe be 1/5th thatof the full-scale pipe, that is, d′/d = 1/5, and the viscosity ν′ of the modelliquid be five times less than the viscosity of the natural liquid. Thenthe conditions for the experiment should be: Q ′ = 1/5 · 1/5 · Q = 0.04 · Q ;sin α′ = 1/25 · 53 · sin α = 5 · sin α; �′ = 0.2 · �, that is, the fluid flow rate inthe experiment has to be reduced by a factor of 25, the internal surface of the


pipe should be so polished that the absolute roughness is reduced by five timesas compared to the full-scale pipe and the sine of the inclination angle of themodel pipe should be increased by five times. Thus, the depth h of the flow inthe full-scale pipe has to be five times greater than the model depth h′, that is,h = 5 · h′.

In experiments it is not always possible to obey all the required similarityconditions, therefore only the main conditions are satisfied. If, for example, ina given case a flow of diesel fuel with viscosity 3 cSt is to be investigated, thenbenzene with viscosity ∼=0.6 cSt could be taken for the model. Reduction ofthe flow rate and enhancement of the pipe slope (at small values of α) presentno special problems. It is far more complicated to fulfill the last similaritycondition and so sometimes it is neglected.

7.5Modeling the Fluid Outflow from a Tank

Let there be a railway tank with a boiler diameter D and length L or, in general,a tank of arbitrary geometric form intended to transport oil with density ρ andkinematic viscosity ν. The tank is provided with bottom drain equipment withthe area of the flowing cross-section S. It is required to design an experimentalinstallation to model the process and to investigate the dependence of the timeT of the oil outflow on the parameters of the liquid and tank.

The dependence under study T = f (D, L, ρ, ν, g, S) in dimensionless formis written as

T√D/g

= f

(L

D,

S

D2,

√gD · D

ν

).

Therefore the similarity criteria are:

�1 = L

D, �2 = S

D2the criteria of geometric similarity;

�3 = g1/2 · D3/2

νthe criterion of dynamic similarity.

At the fulfillment of the first two criteria at which the model tank shouldbe similar to the full-scale one, the phenomena would be similar under thecondition

g1/2 · D′3/2

ν′ = g1/2 · D3/2

ν,

where the superscripts ‘‘prime’’ refer to the model parameters. The last relationshows that the sufficient condition for the dynamic similarity of the model

7.6 Similarity Criteria for the Operation of Centrifugal Pumps 179

tank to the full-scale one is the condition

ν′ = ν ·(

D′

D

)3/2

.

If this condition is fulfilled, then T ′/√

D′/g = T/√

D/g or

T = T ′ ·√

D

D′ . (7.14)

From this follows, in particular, that for modeling of oil outflow on theinstallation with sizes distinct from the full-scale tank, it is necessary touse a liquid with velocity also differing from the oil under study. Let theexperimental tank be a cylindrical tank 10 times smaller than the full-scaleone, that is D/D′ = 10. Then ν′ ∼= 0.0316 · ν, i.e. the kinematic viscosity of theliquid used in the experiments, should be 32 times lower than the viscosityof the oil being modeled. If this oil has sufficiently high viscosity then it iseasy to select such a liquid, otherwise the problem is complicated. The timeT of oil outflow is determined by the time T ′ measured in the experiment. Inaccordance with the formula (7.14) it is T ∼= 3.162 · T ′.

7.6Similarity Criteria for the Operation of Centrifugal Pumps

As already described in Section 4.2.1 pumps are equipment to make liquidsflow against a pressure force, that is, in the direction from lesser pressure psuc

at the suction line to the greater pressure pd at the line of discharge underpressure. Of course, it is possible only by the work of external energy sources(mechanical, thermal, electrical and so on).

Centrifugal pumps represent a variety of pump in which the centrifugalforce, acting on fluid particles rotating in the impeller, makes the liquid flowagainst the pressure. The propulsive device of the impeller or, as it is called,the pump drive, can be an internal combustion engine, vapor-gas turbine orany other source of rotational moment (see Fig. 4.1).

The volumetric fluid flow rate Q (or, as it is called, the feed) depends onthe pressure drop �p = pd − psuc which the liquid should overcome at its flowfrom the impeller center to the periphery. The greater the pressure drop thelesser is the flow rate of the liquid.

If the area σ of the outlet branch pipe at the discharge line is given by thepump design, the fluid velocity Q/σ is a function of the impeller diameterDim (centrifugal pumps may have accessory impellers), the angular velocity ω

of the impeller rotation (centrifugal pumps may be supplied with equipmentto change ω), the density ρ and viscosity ν of the fluid to be pumped, sothat

Q

σ= f (�p, Dim, ω, ρ, ν). (7.15)


The dependence (7.15) is called the (Q − p) characteristic of the cen-trifugal pump and is to a large extent determined by its constructivepeculiarities.

Among the arguments of the function (7.15) there are three dimensionally-independent parameters, for example Dim, ω, ρ, therefore its dimensionlessform could be reduced to

Q/σ

ωDim= f

(�p

ρω2Dim2 ,

ν

ωDim · Dim

).

The last argument represents none other than a quantity inversely proportionalto the Reynolds number Re. This parameter reflects the effect of viscosity onthe characteristic of the pump operation.

If we introduce into consideration the quantity �H = �p/ρg (the differentialhead of the pump) and solve Eq. (7.15) with respect to this head, we get theso-called (Q − �H) characteristic of the pump

�H = ω2Dim2 · F

(Q

ωDim,ωDim

2

ν

). (7.16)

Here the constants σ and g are taken into account by the form of the function F.It is interesting to note that the density ρ of the liquid to be pumped does

not enter into the dependence (7.16), i.e. the form of the latter is true for thepump operated by liquid of any density. As for the influence of the viscosityν on the form of (Q − �H) characteristic of centrifugal pumps, it is small.Reynolds numbers ωDim

2/ν have rather large values (≈106) due to the highrotation velocity of the pump impeller (ω ≈ 300 s−1, ω = 2π · n; n ≈ 3000 rpm;Dim ≈ 0.2–0.7 m; v ≈ 1–10 · 10−6 m2s−1), therefore in practice their variationonly slightly affects the (Q − �H) characteristic of the centrifugal pump andthus the influence of the pumping fluid viscosity is also small. The latterconclusion is of course true only up to certain limits.

If we ignore the influence of pumping fluid viscosity on the (Q − �H)

characteristic of the centrifugal pump, then it can be represented in the simpleform

�H = ω2Dim2 · F

(Q

ωDκ

). (7.17)

From the derived formulas some practically important conclusions follow:• If a centrifugal pump operating with angular velocity ω0 or revolutions per

minute n0 has a characteristic �H = F∗(Q), the same pump working withvaried rotation frequency ω or revolutions per minute n, has thecharacteristic

�H =(

ω

ω0

)2

· F∗(ω0

ω· Q

); (7.18)

• If a centrifugal pump, operating with an impeller of diameter Dκ0 has thecharacteristic �H = F∗(Q), the same pump operating with another

7.6 Similarity Criteria for the Operation of Centrifugal Pumps 181

impeller of diameter Dim has the characteristic

�H =(

Dim

Dim0

)2

· F∗(

Dim0

Dim· Q

); (7.19)

• If a centrifugal pump, operating with rotational velocity ω0 and impellerdiameter Dim0 has the characteristic �H = F∗(Q), the same pumpoperating with varied rotational frequency ω and impeller diameter Dim hasthe characteristic

�H =(

ωDim

ω0Dim0

)2

· F∗(

ω0Dim0

ωDim· Q

). (7.20)

Rules (7.18)–(7.20) allow us to change the (Q − �H) characteristic ofcentrifugal pumps by changing the rotation speed or/and the impellerdiameter.

In many cases, as has been said, the (Q − �H) characteristics of centrifugalpumps are represented in the form of a parabola

�H = F∗(Q) = a − b · Q2 (7.21)

where a and b are approximation factors. If we now change the impellerdiameter from Dim0 to Dim and the rotation frequency from ω0 to ω, the(Q − �H) characteristic of the same pump takes the form

�H = a · ω2Dim2

ω20D2

im0

− b · Q2, (7.22)

that is the parabola graphic in the plane (Q, �H) undergoes a displacementalong the H-axis by

a ·(

1 − ω2Dim2

ω20Dim0

2

).

Exercise. The diameter of a centrifugal pump impeller is 490 mm and theimpeller velocity is 3200 rpm. The pump has the following characteristic

�H = 331 − 0.451 · 10−4 · Q2

(�H in m, Q in m3 h−1). It is required to determine the characteristic of thesame pump if we reduce the impeller diameter to 480 mm and the number ofrevolutions per minute to 3000.

Solution. In accordance with Eq. (7.22) the sought characteristic has thefollowing form

�H = 331 ·(

3000 · 480

3200 · 490

)2

− 0.451 · 10−4 · Q2

= 279 − 0.451 · 10−4 · Q2.


Exercises.1. How does the (Q − �H) characteristic of a centrifugal pump vary if the

frequency speed of the impeller is increased from 3000 to 3200 rpm?The given characteristic of the pump is: �H = 360 − 0.42 · 10−4 · Q2,where �H is measured in m, Q in m3 h−1.

Answer. �H = 410 − 0.42 · 10−4 · Q2.

2. How does the (Q − �H) characteristic of a centrifugal pump vary if thefrequency speed of the impeller is increased from 3000 to 3200 rpm?The given characteristic of the pump is: �H = 360 − 0.42 · 10−4 · Q1.75,where �H is measured in m, Q in m3 h−1.

Answer. �H = 410 − 0.375 · 10−4 · Q1.75.

3. How does the (Q − �H) characteristic of a centrifugal pump vary if theimpeller diameter is reduced from 480 to 470 mm. The givencharacteristic of the pump is: �H = 360 − 0.42 · 10−4 · Q1.75, where�H is measured in m, Q in m3 h−1.

Answer. �H = 345 − 0.436 · 10−4 · Q1.75.

183

8Dimensionality and Similarity in Mathematical Modelingof Processes

In the previous chapter we have seen that dimensionless similarity criteria�1, �2, . . . , �n−k appear in the considered problems after fixation of aset of governing parameters a1, a2, . . . , an by way of heuristic reasoning.Hence, it is by no means necessary to know which equations satisfy theseparameters and with which physical laws they are connected. It is enoughto know only the dimensions of these parameters in order to set updimensionless similarity criteria by which the class of considered phenomenais characterized.

8.1Origination of Similarity Criteria in the Equations of a Mathematical Model

A somewhat different approach takes place in the mathematical modeling ofphenomena in construction systems of algebraic or differential equations withinitial and boundary conditions in which we would like to see an adequatemodel of the considered phenomena or process. Dimensionless similarityparameters originate in these models in a strictly specified way of bringingmodel equations to a dimensionless form.

Let us illustrate the afore said by a simple example. It is known thatone-dimensional oscillations of a pont weight with mass m on an elasticspring around an equilibrium position (x = 0) are described by an ordinarydifferential equation

md2x

dt2= −k · x (8.1)

where x(t) is the linear dependence of the weight coordinate on time and k isthe restoring force factor. The initial, at t = 0, position and velocity x(0) = v0

of the weight are also given, that is the mathematical model of the process.Let us bring the model equation to dimensionless form. First we introduce

dimensionless variables

x = x

Land ¯x = x

v0, (L �= 0, v0 �= 0).


184 8 Dimensionality and Similarity in Mathematical Modeling of Processes

Then the dimensionless time t = t/(L/v0) is determined. In new variables thedifferential equation transforms to

m · L · d2x

(L/v0)2 · dt2= −k · (L · x)

or

d2x

dt2= − kL2

mv20

· x. (8.2)

The initial conditions take the form x(0) = 1; ¯x = 1.Hence, the oscillation of different weights with different mass on springs

with different elasticity caused by different initial conditions is in factdescribed by a mathematical model containing only one dimensionless parameter� = kL2/mv2

0. If in two situations this parameter appears to be identical, thesolutions of Eq. (8.2) are indistinguishable and consequently we have to dealwith similar situations.

It is easy to verify that the solution of the Eq. (8.2) is

x(t) = cos(√

�t)

+ 1√�

· sin(√

�t)

.

Returning to dimensional quantities, we obtain

x(t) = L · cos

(√k

m· t

)+

√m

k· v0 · cos

(√k

m· t

).

As is known√

k/m = ω is the frequency of harmonic oscillations of the weightand the amplitude h is the square root of the sum of the squared factors of thesine and cosine

h =√

L2 + mv20

k= L ·

√1 + 1

�.

8.2One-Dimensional Non-Stationary Flow of a Slightly Compressible Fluidin a Pipeline

The theory of one-dimensional non-stationary flows of a slightly compressiblefluid in a pipeline was considered in detail in Sections 5.1–5.3. The mainequations modeling such flows are

∂p

∂t+ c2 ∂(ρv)

∂x= 0

ρ

(∂v

∂t+ v

∂v

∂x

)= − ∂p

∂x− λ(Re, ε)

d· ρv2

2

(8.3)

8.2 One-Dimensional Non-Stationary Flow of a Slightly Compressible Fluid in a Pipeline 185

in which the quantity

c = 1√ρ0

K+ ρ0d

δ · E

(8.4)

is called the velocity of the pressure wave propagations in pipeline.Consider a certain problem on the calculation of non-stationary flow of a

slightly compressible fluid in pipeline taking Eqs. (8.3) as the mathematicalmodel of this flow.

Problem. Let there be at the pipeline section 0 ≤ x ≤ L stationary fluidflow with velocity v0. However, starting from some time a valve locatedat the end cross-section of the pipeline x = L begins to vary its openinglevel with frequency ω. It is required to reveal the similarity criteria of thisphenomenon.

Solution. Introduce the following dimensionless variables marked by thehorizontal bar at the top

t = 1

ω· t, x = L · x, v = v0 · v, ρ = ρ0 · ρ, p = p0 · p.

Equations (8.3) in the new variables take the formp0

1/ω· ∂ p

∂ t+ c2 · ρ0v0

L· ∂(ρv)

∂ x= 0

ρ0ρ

(v0

1/ω· ∂ v

∂ t+ v2

0

L· v

∂v

∂ x

)= −p0

L· ∂ p

∂ x− λ(Re, ε)

d· ρ0v2

0 · ρv2

2

If now we take p0 = ρ0v0c, the system of equations simplifies toωL

c· ∂ p

∂ t+ ∂(ρv)

∂ x= 0

ωL

c· ∂ v

∂ t+ v0

c· v

∂ v

∂ x= − ∂ p

∂ x− λ(Re, ε) · L

d· v0

c· ρv2

2

(8.5)

From Eqs. (8.5) it is seen that there are three dimensionless criteria governingthe class of problems under consideration and differing from each other onlyby the numerical values of the parameters entering in these equations

�1 = v0

c; �2 = ωL

c; �3 = λL

d.

The first of these criteria is called the Mach number. The ratio of the firstcriterion to the second one �1/�2 = v0/(ωL) is called the Strouchal numberSt = v0/(ωL).

The Mach number of the fluid or gas flow in a main pipeline is, as a rule,very small (for example, for fluid flow M ≈ 0.001; for gas flow M ≈ 0.03),


therefore it often turns out that the second term in the left-hand side ofthe last equation (8.5) can be neglected in comparison with the first one.If the flow velocity v0 is comparable to the sound velocity c, as it may be,for example, at the gas outflow from a pipeline through a short nozzle,the second term would become almost the primary term and could not beignored.

The Strouchal number characterizes the degree of process non-stationarity.If this number is great (St = v0/ωL � 1, that is, the characteristic time ofthe process 1/ω is large compared to L/v0 the time of fluid particle passagethrough the pipeline), the non-stationarity degree is small and the process isclose to stationary. If the Strouchal number is small (that is the frequencyof the process ω is large compared to v0/L) then the non-stationarity of theprocess cannot be neglected.

The third criterion λL/d does not have a special name. It characterizes themagnitude of the resistance to the fluid friction in the pipeline.

8.3Gravity Fluid Flow in a Pipeline

Consider now problems connected with thegravity flow of incompressible fluidin a pipeline (see Sections 3.7 and 7.3). Earlier we were dealing with stationarygravity fluid flow, that is with flow in which all the hydrodynamic parametersat each pipeline cross-section remained constant, now we will consider thegeneral case of non-stationary gravity flow in a descending pipeline sectioncharacterized by the slope angle to the horizontal α (α < 0). The differentialequations describing such flow are

∂ρS

∂t+ ∂ρvS

∂x= 0

∂ρvS

∂t+ ∂

∂x

(ρv2S + ρg cos α ·

∫ S

0S dh

)= −ρgS sin α − ρgS cos α

CSh2Rh

· v|v|

Here S(x, t) is the area of the pipeline cross-section filled by fluid; Rh(S) thehydraulic radius of the flow; h(S) the depth of fluid in the pipeline cross-section;CSh the Chezy factor (see Section 3.7).

Let v0 be the fluid velocity at the inlet to the descending section of thepipeline and ρ ∼= ρ0 = const. Let us introduce the following dimensionlessparameters

x = x

d; t = t · v0

d; v = v

v0; S = S

S0= S

(πd2/4);

Rh = Rh

d= 0.25 · (1 − sin φ/φ) (see Eq. (3.46)); CSh = CSh√

g.

8.4 Pipeline Transportation of Oil Products. Batching 187

With new variables the system of equations describing gravity fluid flowtransforms to

ρ0S0v0

d·(

∂S

∂ t+ ∂ vS

∂ x

)= 0

ρ0v02S0

d·[

∂ vS

∂ t+ ∂

∂ x

(v2S + gd cos α

v02

∫ S

0S dh

)]

= −ρ0gS0S sin α−ρ0g cos α · S0v02

gd· S · v|v|

RhCSh

or, after simplification∂S

∂ t+ ∂ vS

∂ x= 0

∂ vS

∂ t+ ∂

∂ x

(v2S + gd cos α

v02

∫ S

0S dh

)= −gd sin α

v02

S · − v|v| · S cos α

RhCSh

.

(8.6)

It is seen that, in the system of equations (8.6), besides the dimensionlessangle α (�1 = α) there is one more dimensionless complex �2 = gd/v0

2 orv0/

√gd. We already met the latter one in Section 7.4; it is the so-called Froude

number

Fr = v0√gd

. (8.7)

Thus, the Froude similarity criterion (Froude number) is obtained in themathematical problem on gravity fluid flow in a pipeline, which wasearlier derived on the basis of general reasoning on the similarity of suchflows.

8.4Pipeline Transportation of Oil Products. Batching

At the present time light oil products (benzenes, kerosenes, diesel fuels andothers) are pumped by the batching method (Ishmuchamedov et al., 1999).In oil treatment the factory plants simultaneously produce a plethora of oilproducts, mostly the so-called light oil products and, above all, motor oils.As a result of further compounding (mixing of two or several oil products toprepare fuels with given properties) there are obtained different sorts of oilproducts ready for use. It is evident that the building of a separate pipelinefor each of produced oil products would be unprofitable, therefore most of theoil products are transported by one and the same pipeline, pumping them inseries one after another (batching).


8.4.1Principle of Oil Product Batching by Direct Contact

The essence of batching by direct contact consists in the different oil productsbeing combined in separate batches, each of several thousand or even tenthousand tons, which are pumped into the pipeline in series, one afteranother, and transported to the user. In this way each batch displaces theprevious one and is, in turn, displaced by the following batch. It is as if theoil-pipeline along its full length were filled by different oil products arrangedin a chain and contacting with each other at the places where one batch comesto an end and another begins. Thus, the key advantage in the batching of oilproducts is that different sorts of oil products are pumped not along differentpipes but along one and the same pipe.

However, despite all the advantages of batching it has one significantdisadvantage consisting in the formation of a mixture of different oil productsby their mutual displacement in the pipeline. Although the mixing of similaroil products, for example benzenes of different sorts or diesel fuels of differentsorts, threatens little the quality of the resulting oil product, because oilproducts relating to one group of fuels are more compatible than oil productsrelating to different groups, the mixing of dissimilar oil products, for examplebenzenes, kerosenes and diesel fuels threatens significantly the quality of theoil products. Nevertheless, batching with direct contact of the oil productshas received wide recognition because the quantity of mixture forming in thecontact zones of batches moving in series is relativey small compared to thelarge volume of transported fuels and the whole mixture can be decomposedinto the initial oil products, preserving the quality of the latter.

The mixture formed in the contact zone of oil products is caused by physicalprocesses inherent to fluid flow in the pipeline and to displacment of one oilproduct by another. If contacting oil products displaced each other as rigidbars with plane interface boundaries, mixing in the contacting zone wouldof course be absent. However, fluid oil products are not rigid bodies andmutual displacement happens nonuniformly over the pipeline cross-section.The velocities of fluid particles at different points of the pipeline cross-sectionare distinct. At the pipeline wall they vanish whereas at the pipeline axis theyachieve a maximum value. Thus the displacement of one oil product by anotheroccurs more at the pipeline center than at the pipeline wall. At each instanceof time the wedge behind the moving oil product becomes as if penetratedinto the leading fluid, the penetration happening more when the profile of theaverage velocity is more stretched along the pipeline axis. There takes placeso-called convection (or convective diffusion) of the impurity of one oil productinto another one owing to, and together with, fluid layers transferring relativeto each other.

Nonuniformity of the fluid average velocity distribution at the pipeline cross-section is not the only reason for mixture formation of oil products in thezone of their contact. Light oil products are, as a rule, pumped in a turbulent


flow regime in which the fluid particles do not move parallel the pipeline wallsbut execute chaotic turbulent motion, as can be seen in smoke jets gushingfrom heat plants. In turbulent flows there exists intensive mixing of differentparticles over the pipeline cross-section caused by velocity fluctuations and thechaotic motion of particles. This process is called turbulent diffusion. It mixesover the pipeline cross-section the edge of the displacing fluid as well as therest of the fluid to be displaced, causing their more or less uniform distributionin each pipeline cross-section.

Hence, the mixing process of oil products displacing and to be displacedhappens in accordance with the following scheme: the edge of the oil productmoving behind penetrates the oil product moving in front while processes ofturbulent diffusion mix the penetrated impurity over the pipeline cross-section.Since the concentration of the displacing oil product at the pipeline axis isgreater than at the wall, transport of the displacing oil product into the regionoccupied by oil product to be displaced occurs. Conversely, back transport ofthe oil product to be displaced into the region of the displacing oil productalso occurs. These processes are inseparable. They operate permanently andsimultaneously over the course of the displacement time, determining thusthe intensity of the longitudinal mixing, the volume and length of the resultingmixture.

8.4.2Modeling of Mixture Formation in Oil Product Batching

Light oil products possess the following property: if a volume V1 of the first oilproduct is mixed with a volume V2 of the second oil product, the volume Vc ofthe resulting mixture is, to a high degree of accuracy, equal to the sum of thevolumes of the components Vc = V1 + V2. Therefore, the additivity propertyof fluid volume on mixing of its components is taken as a main assumption inthe construction of the model of mixture formation in a pipeline.

If we denote through ρ1 and ρ2 the densities of the contacting oil products,the volume concentrations θ1 and θ2 of the oil products may be expressedthrough these quantities and the density ρc of the mixture. In accordance withthe mass conservation law we have

ρ1V1 + ρ2V2 = ρcVc.

From this it follows that

ρ1V1

Vc+ ρ2

V2

Vc= ρc ⇒ ρ1θ1 + ρ2θ2 = ρc.

Taking further that θ2 = 1 − θ1 or θ1 = 1 − θ2 we obtain the following formulas

θ1 = ρc − ρ2

ρ1 − ρ2, θ2 = ρc − ρ1

ρ2 − ρ1. (8.8)


We now develop the model of mixture formation in the fluid flow in pipeline.The mixture of oil products in the contact zone can be characterized by theconcentration θ(x, t) of one of the oil products, for example the displacing one,in one-dimensional fluid flow in a pipeline. The cases θ �= 1 or θ �= 0 have adirect relationship to the mixture. In the mixture region 0 < θ < 1. The caseθ = 1 corresponds to the region of displacing oil product, whereas the caseθ = 0 corresponds to the oil product ahead that is to be displaced.

In order to derive a mathematical model of mixture formation it is necessaryto reveal the mass exchange regularities of oil products in the mixture region,that is to specify the relation between the volumetric flow rate q(x, t) ofthe displacing oil product (the volumetric flow rate of the oil product to bedisplaced would clearly be equal to v0S − q(x, t)) and the parameters of theconcentration distribution θ(x, t) in the flow.

Figure 8.1 represents a scheme of mass exchange in an arbitrary cross-sectionof the mixture region. The total fluid flow rate through the cross-section xin the moving frame of reference is equal to zero, but the transfers of fluid(mixture of oil products) from left to right and from right to left are non-zero;they are equal in magnitude but opposite in sign. The flow of the mixturethrough the cross-section x from left to right with flow rate w1 happens mainlyin the central part of the pipeline, while the flow of the mixture in the reversedirection from right to left with flow rate w2 = −w1 occurs chiefly close to theinternal surface of the pipeline.

The flow rate w = w1 = −w2 is determined by the profile of the velocity u(r)averaged over the interval 0 ≤ r ≤ r∗

w = 2π ·∫ r∗

0r · [u(r) − v0] dr. (8.9)

If we take the velocity profile u(r) equal to the logarithmic profile in turbulentflow (Loitzyanskiy, 1987)

u(r) − umax

u∗= 1

κ· ln

(1 − r

r0

)(8.10)

from Eqs. (8.9) and (8.10) ensue the following relations

v0 = umax − 4.08 · u∗, r∗ = 0.805 · r0, w = 1.26 · u∗ · S. (8.11)

Here κ ∼= 0.4 is the Karman constant and u∗ is the dynamic velocity, seeSection 3.5. The latter is expressed through the tangential frictional stress |τw|at the pipeline wall

|τw| = ρ · u2∗, u∗ =

√|τw|ρ

.


Figure 8.1 Scheme of mass exchange in the mixture region.

Since |τw| = λ/8 · ρv02, where λ is the hydraulic resistance factor, the dynamic

velocity is related to this factor by

u∗ =√

λ

8· v0.

Insertion of the expression for u∗ in Eq. (8.11) yields the connection betweenthe mass exchange flow rate w and the pumping flow rate Q = v0S

w = 1.26 ·√

λ

8· v0S = 0.446 · √

λ · Q. (8.12)

From this formula it follows that the quantity exchange flow rates are relativelynot large. So, for example, at λ = 0.022 the quantity w = 0.066 v0S, which isonly 6.6% of the pumping flow rate.

Counter flows of fluid transfer both the first and second oil productthrough the cross-section x of the moving frame of reference, but the averageconcentrations θ′ and θ′′ in the transfer flows are different. Thus, the flow rateq(x, t) of the displacing oil product through the cross-section x is given by

q(x, t) = w · θ′ − w · θ′′ = w · (θ′ − θ′′).

In the first flux (from left to right) the concentration θ′ is equal to theconcentration of the displacing oil product averaged over the cross-section atsome distance l1 behind the cross-section x. In the second flux (from rightto left) the concentration θ′′ is equal to the concentration of the displacing oilproduct at some distance l2 ahead of the cross-section x. Lengths l1 and l2 canbe called mixing lengths, since they are equal to the lengths over which theturbulent diffusion mixes the invading impurity over the pipeline cross-section.

Accurate to small quantities of the highest order it could be written asθ′ = θ(x − l1, t) ∼= θ(x, t) − l1 · ∂θ

∂x+ . . . ,

θ′′ = θ(x + l2, t) ∼= θ(x, t) + l2 · ∂θ

∂x+ . . .

(8.13)


and

q(x, t) = w · θ′ − w · θ′′ = w · (θ′ − θ′′) = −w · (l1 + l2) · ∂θ

∂x.

Substitution of w from (8.12) in the latter relation yields

q(x, t) = −0.446 · √λ · (l1 + l2) · ∂θ

∂x· v0S

or

q(x, t) = −K · ∂θ

∂x· S (8.14)

where

K = 0.446 · √λ · (l1 + l2) · v0. (8.15)

The relation (8.14) expressing the proportionality of the volumetric flow rateq(x, t) of the displacing oil product gradient ∂θ/∂x to its concentration iscalled the law of longitudinal mixing and the factor K (m2 s−1) the effective factorof the longitudinal mixing. The minus sign in Eq. (8.14) shows that the fluxof each oil product is directed from the higher concentration to the lowerone, that is opposite to the concentration gradient of the oil product underconsideration.

There are many theoretical and experimental formulas for the factor K oflongitudinal mixing (Ishmuchamedov et al., 1999). We consider one of them,namely the formula derived by Taylor when investigating the dispersion of animpurity in turbulent fluid flow in a pipe

K = 1.785 · √λ · v0d. (8.16)

Comparing this formula with Eq. (8.15), we find for the sum of the mixinglengths (l1 + l2) the value 4d. This result appears to be true for Reynoldsnumbers higher than 3 · 104.

8.4.3Equation of Longitudinal Mixing

To derive a mathematical model describing the process of mixture formationin a pipeline for displacement of one fluid by another one we use thevolume balance equation for each component. Since the sum of the volumeconcentrations θ1 and θ2 of oil products is equal to 1, it is enough to write onlyone balance equation for one of the components, for example the displacingcomponent, taking θ1 = θ(x, t), θ2 = 1 − θ(x, t).


Figure 8.2 Derivation of the volumebalance equation.

Consider a fluid volume (see Section 1.2) enclosed between cross-sectionsx1(t) and x2(t), Fig. 8.2.

The first (displacing) oil product occupies only a part of this volume V1

which can be written as

V1 =∫ x2(t)

x1(t)θ(x, t) · S dx.

If the exchange of oil products through the cross-sections x1(t) and x2(t)were absent, the quantity V1 would be constant, but in reality it is not. Thequantity V1 varies owing to mutual penetration of the oil products into eachother and this variation is determined by the difference in transfer flow ratesq(x1, t) − q(x2, t). Thus

dV1

dt= d

dt

(∫ x2(t)

x1(t)θ(x, t) · S dx

)= q(x1, t) − q(x2, t). (8.17)

If we take into account the following identities:

d

dt

(∫ x2(t)

x1(t)θ(x, t) · S dx

)=

∫ x2(t)

x1(t)

(∂θS

∂t+ ∂v0θS

∂x

)dx, see Eq. (1.4),

q(x1, t) − q(x2, t) = −∫ x2(t)

x1(t)

∂q

∂xdx,

the relation (8.17) may be written as∫ x2(t)

x1(t)

(∂θS

∂t+ ∂v0θS

∂x

)dx = −

∫ x2(t)

x1(t)

∂q

∂xdx.

If, in addition, we recall that the fluid volume under consideration is arbitrarilychosen, that is the integration limits x1(t) and x2(t) are arbitrarily chosen, thenfrom the latter integral equation follows the differential equation

∂θS

∂t+ ∂v0θS

∂x= − ∂q

∂x

expressing the volume balance of the first oil product in the mixture.Insertion instead of q, using its expression through the concentration

gradient q(x, t) = −K · S · ∂θ/∂x (see Eq. (8.14)) with regard to the conditionsS = const. and v0 = const. yields the differential equation for the concentration


θ(x, t) of the displacing oil product in the mixture

∂θ

∂t+ v0

∂θ

∂x= −K · ∂2θ

∂x2(8.18)

Equation (8.18) represents the differential equation of longitudinal mixingreferring to the class of heat conduction equations (Ishmuchamedov et al., 1999).

If we introduce the dimensionless variables x = x/L and t = t/(L/v0),Eq. (8.18) can be rewritten in dimensionless form

1

(L/v0)· ∂θ

∂ t+ v0

L· ∂θ

∂ x= K

L2· ∂2θ

∂ x2

or

∂θ

∂ t+ ∂θ

∂ x= Pe−1 · ∂2θ

∂ x2(8.19)

where the dimensionless parameter Pe = v0L/K called the Peclet number is themain characteristic of quantity of mixture forming in the course of oil productpumping and the similarity criterion in problems on longitudinal mixing offluids by their displacement in the pipe.

8.4.4Self-Similar Solutions

The reasoning of dimensional theory can sometimes bring very importantresults, being capable of finding solutions of differential equations reflectingthe most important peculiarities of processes and being in some sense limitingfor solutions, taking into account the minor peculiarities and details ofprocesses. To such solutions belong the so-called self-similar solutions. Let usdemonstrate these with the example of Eq. (8.18).

In problems on the displacement of one oil product by another we canconsider the dimensional parameters x, t, K, v0, L. However, the number ofthese parameters could be reduced without serious consequences for thephysics of the processes. If for example we change to a frame of referencex∗ = x − v0t moving with average velocity v0 of the pumping fluid, thenEq. (8.18) would be simplified to

∂θ

∂t= K · ∂2θ

∂x∗2(8.20)

If the boundary conditions at the pipeline section edges have become, afterframe of reference transformation, movable x∗ = −v0t and x∗ = L − v0t, butfar from the region occupied by the mixture they can be replaced by conditionsat ‘‘infinitiy’’ (±∞), that is θ → 0 when x → +∞ and θ → 1 when x → −∞


and the initial conditions at t = 0 we can take as instantaneous changing of oilproducts, namely θ(x∗, 0) = 1 at x∗ < 0 and θ(x∗, 0) = 0 at x∗ > 0 or

θ(x∗, 0) ={

1, x∗ < 0,

0, x∗ > 0.(8.21)

Then, in the problem under consideration, there remain only threedimensional quantities x, t, K . Since the solution of the problem is thedimensional function θ, depending on three dimensionally-independentparameters from which may be built only one dimensionless combination then

ξ = x∗√Kt

(8.22)

in accordance with the �-theorem the solution should depend on one variable.Thus, the solution of the problem (8.20) and (8.21) has to be sought in theform θ(x∗, t) = θ(ξ).

Such a solution is called self-similar because it is as if it transforms itselfin space in a similar way, namely the solution of the problem at an arbitraryinstant of time t may be obtained from the solution of the problem at anyprevious instant of time t1 by stretching the distribution θ(x∗, t1) along thex∗-axis with factor

√K · t/t1, because it is well known that the multiplication of

the function argument by a certain number brings extension or compressionof the function graph along the abscissa axis.

Substitution of θ(x∗, t) = θ(x∗/√

Kt) in Eq. (8.20) leads to an ordinarydifferential equation of the second order

−ξ

2· dθ

dξ= d2θ

dξ2

from which follows

θ(ξ) = A ·∫ ξ

0e−α2/4

dα + B

where A and B are constants of integration.Using the boundary conditions θ → 0 at ξ → +∞ and θ → 1 at ξ → −∞

gives A = 1/(2√

π), B = 1/2. In deriving the latter we use the known equality∫ ∞

0e−α2

dα =√

π

2.

The solution of the problem is

θ(ξ) = 1

2

(1 − 2√

π·∫ ξ/2

0e−α2

dα

)


or in dimensional variables

θ(x∗, t) = 1

2

(1 − 2√

π

∫ x∗/√

4Kt

0e−α2

dα

). (8.23)

Graphs of the concentration distribution θ(x•, t) in the moving frame ofreference are shown in Fig. 8.3. The heavy line depicts the initial distributionwhereas other lines show the concentration distribution at successivelyincreased instants of time.

A remark on the imperfection of the model. It can be seen at once that theobtained solution has a defect. This is that the mixture, which initially wasabsent, at just the next instant of time would propagate through the wholepipeline length. In fact such a case could not occur in practice and the obtainedparadox is the result of imperfectness of the model. The dispersion model oflongitudinal mixing, as in general all models of this kind, represents only acertain schematization, in the given case it is the formation of a mixture inthe contact zone of the displacing fluid and the fluid to be displaced. But theresult of such a schematization does not give a particularly bad result. Thefunction θ(x, t) tends very quickly to zero at x → ∞ and to one at x → −∞.In the main domain of variability this function approximates the concentra-tion distribution in the mixture zone well. Therefore, the dispersion modelof longitudinal mixing based on Eq. (8.20) has received wide application incalculations of mixtures forming in the batching of oil products.

Mixture volume. The volume Vc of a mixture of pumping oil products formingin a pipeline at the instant of time t calculated by Eq. (8.23) in the range ofconcentrations 0.01 < θ < 0.99 is determined by the expression

Vc∼= 6.58 · πd2

4· √

K · t. (8.24)

A remark on the quantity of mixture volume. Taylor obtained Eq. (8.16) for thefactor K of longitudinal mixing under the assumption that the densitiesand viscosities of contacting fluids are close to each other. Therefore,strictly speaking, this equation is not suitable for the case when thedensities and viscosities of contacting fluids differ significantly from eachother. Such situations occur in the pumping of oil products, for example

Figure 8.3 Self-similar distribution of concentration.


when pumping benzenes (ρ ≈ 730–750 kg m−3, ν ≈ 0.6 cSt) with diesel fuels(ρ ≈ 830–850 kg m−3, ν ≈ 4–9 cSt). It is evident that for these fluids thefactors of hydraulic resistance λ are also different.

The formula for the mixture volume Vc as applied to the case underconsideration can be improved if we calculate the mixture volume as thearithmetical mean of two volumes: the first calculated by Eq. (8.24) withthe factor K1 = 1.785 · √

λ1 · v0d and the second by Eq. (8.24) with the factorK2 = 1.785 · √

λ2 · v0d. It is equivalent to formula (8.24) if we take

K = 0.446 ·(

4√

λ1 + 4√

λ2

)2 · v0d. (8.25)

Exercise. It is required to calculate the length and the volume of themixture region in a symmetric concentration range 0.01 < θ < 0.99 whenbatching benzene (νB = 0.6 cSt) and diesel fuel (νD = 6 cSt) in an oil-pipeline(D = 530 × 8 mm, � = 0.15 mm, L = 700 km) when the transportation of oilproducts occurs with flow rate Q = 1000 m3 h−1.

Solution. Determine the average transportation velocity v0 of the oil products

v0 = 4Q

πd2= 4 · 1000/3600

3.14 · (0.530 − 2 · 0.008)2∼= 1.34 m s−1.

Calculate the Reynolds numbers

ReB = v0d

νB= 1.34 · 0.514

0.6 · 10−6∼= 1 147 933,

ReD = v0d

νD= 1.34 · 0.514

6 · 10−6∼= 114 793.

Calculate the hydraulic resistance factors λB and λD

λB = 0.11 ·(

0.15

514+ 68

1 147 933

)0.25 ∼= 0.015,

λD = 0.11 ·(

0.15

514+ 68

114 793

)0.25 ∼= 0.019.

Calculate with Eq. (8.25) the factor K of longitudinal mixing

K = 0.446 ·(

4√

0.015 + 4√

0.019)2 · 1.34 · 0.514 ∼= 0.222 m2 s−1.

With Eq. (8.24) calculate the mixture volume Vc

Vc∼= 6.58 · πd2

4· √

K · t = 6.58 · πd2

4·√

K · L

v0

= 6.58 · 3.14 · 0.5142

4·√

0.222 · 700 000

1.34∼= 465 m3.


The length of the mixture region lc is

lc = Vc

πd2/4= 465

3.14 · 0.5142/4∼= 2242 m or 2.242 km.

Answer: 2242 m; 465 m3.

It should be noted that non-self-similar solutions of the considered problem,that is solutions taking into account the finite extent of the pipeline and theconditions at its edges, differ slightly from the self-similar solution. The facilityto formulate the problem, making it self-similar and yet without changing itsmost important features, characterizes the high skill of the scientist. That iswhy self-similar solutions play such an important role in the different fields ofscience (Sedov, 1965).

199

References

Archangelskiy V.A., (1947) Calculation ofNon-Stationary Flow in Open WaterCourses, Academy of Sciences USSR,Moscow (in Russian).

Charniy I.A., (1975) Non-Stationary Motionof Real Fluid in Pipes, 2nd edn., Nedra,Moscow (in Russian).

Christianowitch S.A., (1938)Non-Stationary Flow in Channels andRivers, Collected articles on Some Problemsin Continuum Mechanics, Academy ofSciences USSR, Moscow (in Russian).

Dodge D.W., Metzner A.B., (1958)Turbulent Flow of Non-NewtonianSystems, AIChE J., 2, 189–204 .

Ginsburg I.P., (1958) AppliedHydro-Gas-Dynamics, LGU, Leningrad(in Russian).

Ishmuchamedov I.T., Isaev S.L., LurieM.V., Makarov C.P., (1999) PipelineTransportation of Oil Products, Oil andGas, Moscow (in Russian).

Leibenson L.S., Vilker D.S., Shumilov P.P.,Yablonskiy V.S., (1934) Hydraulics, 2ndedn., Gosgorgeolnephteizdat, Moscow,Leningrad (in Russian).

Loitzyanskiy L.G., (1987) Mechanics of Fluidand Gas, Nauka, Moscow (in Russian).

Lurie M.V., (2001) Technique of ScientificResearches. Dimensionality, Similarityand Simulation of Phenomena inProblems of Oil Transportation and Oil-and Gas-Storage, Oil and Gas, Moscow(in Russian).

Lurie M.V., Polyanskaya L.V., (2000) AboutOne Dangerous Source of HydraulicShock Waves in Oil and Oil Products, OilFacilities, No. 8 (in Russian).

Lurie M.V., Podoba N.A., (1984)Modification of Karman Theory for

Turbulent Shear Flows, Papers of theAcademy of Science of the USSR, 279(3),(in Russian).

Vasil’ev G.G., Korobkov G.E., Lurie M.V.et al., (2002) Pipeline Transportation ofOil, Vol.1, S.M. Veinstock, Nedra (inRussian).

Polyanscaya L.V., (1965) Investigation ofnon-stationary processes in changingoperation regime with centrifugalpumps, Kand. Sci. Thesis, Gubkin Oil &Gas Institute, Moscow (in Russian).

Porshakov Yu.P., Kosachenko A.N.,Nikishin V.I., (2001) Power of PipelineGas Transportation, Oil and Gas, Moscow(in Russian).

Potapov A.G., (1975) Hydraulic ResistanceFactor in Turbulent Flow of Viscous-PlasticFluids, Volgogradnipinepht, Volgograd,No. 23 (in Russian).

Rozhdestvenskiy B.L., Yanenko N.N.,(1977) Systems of Quasi-Linear DifferentialEquations, Nauka, Moscow (in Russian).

Romanova N.A., (1985) Laminar andTurbulent Flows in Pipes and Channelswith Moving Walls, Kand. Sci. Thesis,Gubkin Oil & Gas Institute, Moscow (inRussian).

Samarskiy A.A., (1977) Introduction to theTheory of Difference Schemes, Nauka,Moscow (in Russian).

Sedov L.I., (1965) Methods of Similarity andDimensionality in Mechanics, Nauka,Moscow (in Russian).

Tihonov A.N., Samarskiy A.A., (1966)Equations of Mathematical Physics,Nauka, Moscow (in Russian).

Wilkenson U.L., (1960) Non-NewtonianFluids, Pergamon Press, London,NewYork.


201

AppendicesAppendix A. Increment and Differential of a Function

In some sections of this book the formula for the increment �f of thedifferentiable f (x)

f (x + �x) − f (x) ≈ df

dx

∣∣∣∣x

· �x

where �x is the increment of the argument. Let us explain this formula.Figure A.1 shows a graph of a differentiable and, consequently, contin-

uous function y = f (x). The difference f (x + �x) − f (x) of the values ofthis function at two points x and (x + �x) is called the increment of thefunction �f .

If at the point M on the abscissa x we draw a tangent to the plot of thefunction, the increment of the function can be interpreted as the sum of thetwo segments AB and BC, Fig. A2.

Since

df

dx= lim

�x→0

f (x + �x) − f (x)

�x

this means that

df

dx

∣∣∣∣x

− f (x + �x) − f (x)

�x= ε → 0 at �x → 0

that is, ε is an infinitesimal quantity. From this it follows that the increment�f = (x + �x) − f (x) of a differentiable function f (x) can be represented as

f (x + �x) − f (x) = df

dx

∣∣∣∣x

· �x︸︷︷︸AB

+ ε · �x︸︷︷︸BC

.

Since the value of the derivative at the point x is equal to the slope of thefunction plot at this point to the horizontal, it is evident that the first term onthe right-hand side of the last formula is nothing but the length of the segmentAB. If df / dx �= 0, it appears to be a quantity of the same order as �x.


202 Appendices

Figure A.1 Increment �f and differentialdf of a function.

The second term (ε · �x) is depicted in Fig. A.1 by segment BC. This term isa smaller quantity than the first term, because ε → 0 at �x → 0. This meansthat the smaller �x then the greater the accuracy when neglecting the secondterm. Thus, there is an approximation formula

f (x + �x) − f (x) ≈ df

dx

∣∣∣∣x

· �x.

It is commonly said that this formula is true up to an infinitesimal quantity ofhigher order.

The principal linear (in �x ≡ dx) part dfdx

∣∣∣x· �x of the function increment

�f is called the function differential df

df = df

dx

∣∣∣∣x

· dx; �f = df + ε · dx.

The differential of function df is plotted by segment AA, Fig. A.1.

Appendix B. Proof of the Formula

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx =

∫ x2(t)

x1(t)

∂

∂t[A(x, t) · S(x, t)] dx

+ A(x, t) · S(x, t)|x2(t) · dx2

dt− A(x, t) · S(x, t)|x1(t) · dx1

dtto calculate the total derivative with respect to time of an integral with integrationlimits determining the motion of a fluid (individual) volume of a continuum in thepipeline.

Proof. Consider two successive locations at the time instances t and t + �tof one and the same fluid volume in a pipeline, Fig. B.1.

In accordance with general definition of a derivative as a limit of the ratio ofa function increment to an argument increment when the latter tends to zero

Appendix B. Proof of the Formula 203

Figure B.1 Derivation of the formula of time differentiation of an integral with timedependent limits.

we have

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx

= lim�t→0

1

�t

[∫ x2(t+�t)

x1(t+�t)A(x, t + �t) · S(x, t + �t) dx

−∫ x2(t)


].

The integrals in square brackets can be represented as follows (see Fig. B.1)∫ x2(t+�t)

x1(t+�t)A(x, t + �t) · S(x, t + �t) dx

=∫ x2(t)

x1(t+�t)A(x, t + �t) · S(x, t + �t) dx

+∫ x2(t+�t)

x2(t)A(x, t + �t) · S(x, t + �t) dx

∫ x2(t)


=∫ x1(t+�t)

x1(t)A(x, t) · S(x, t) dx +

∫ x2(t+�t)

x1(t+�t)A(x, t) · S(x, t) dx.

Consequently∫ x2(t+�t)

x1(t+�t)A(x, t + �t) · S(x, t + �t) dx −

∫ x2(t)


=∫ x2(t)

x1(t+�t)[A(x, t + �t) · S(x, t + �t) − A(x, t) · S(x, t)] dx

+∫ x2(t+�t)

x2(t)A(x, t + �t) · S(x, t + �t) dx

−∫ x1(t+�t)

x1(t)A(x, t) · S(x, t) dx.

204 Appendices

Up to infinitesimal quantities of higher order this can be written as∫ x2(t)

x1(t+�t)[A(x, t + �t) · S(x, t + �t) − A(x, t) · S(x, t)] dx

∼=∫ x2(t)

x1(t+�t)

∂(A · S)

∂t· �t · dx

∫ x2(t+�t)

x2(t)A(x, t + �t) · S(x, t + �t) dx ∼= A(x2, t) · S(x2, t) · dx2

dt· �t,

∫ x1(t+�t)

x1(t)A(x, t) · S(x, t) dx ∼= A(x1, t) · S(x1, t) · dx1

dt· �t.

Dividing the sum of the latter three expressions by �t and passing to the limitat �t → 0 we get the formula

d

dt

∫ x2(t)

x1(t)A(x, t) · S(x, t) dx =

∫ x2(t)

x1(t)

∂

∂t[A(x, t) · S(x, t)] dx

+ A(x, t) · S(x, t)|x2(t) · dx2

dt− A(x, t) · S(x, t)|x1(t) · dx1

dt.

When going to the limit we took into account the continuity prop-erty of functions x1(t) and x2(t), namely lim�t→0 x1(t + �t) = x1(t) andlim�t→0 x2(t + �t) = x2(t).

205

Author Index

aArchangelskiy V.A. 67, 151

cCharniy I.A. 133Christianowitch S.A. 67, 151

dDodge D.W. 62

gGinsburg I.P. 21

iIshmuchamedov I.T. 63, 65, 187,

192, 194

jJoukowski N.E. 42, 93

lLeibenson L.S. 20, 66, 135, 151Loitzianskiy L.G. 58, 59, 190Lurie M.V. 19, 54, 56, 57, 66, 153,

154,

mMetzner A.B. 61

pPodoba N.A. 53, 56, 57Polyanskaya L.V. 153, 154Porshakov B.P. 40Potapov A.G. 62

rRomanova N.A. 51, 62Rozhdestvenskiy B.L. 150

sSamarskiy A.A. 114, 155Sedov L.I. 3, 157, 198

tTichonov A.N. 114

vVasil’ev G.G. 77

wWilkinson U.L. 36, 48, 49, 51

yYanenko N.N. 150


207

Subject Index

aabsolute roughness 19acceleration 4adiabatic

–expansion of gas 146–index 93–velocity of sound in gas 141

air density 4Altshuler formula 21analytical solution 88angular velocity of impeller rotation

77anti-turbulent additive 62apparent viscosity 36area of pipeline cross-section 7average

–physical parameter 4–tangential stress 52–velocity 52

bbalance equation of forces 76barotropic

–gas 15–fluid 15–medium 13

basic measurement units, seeprimary measurement units

batching 187Bernoulli equation 15Blasius formula 21boundary condition 46Boussinesq force 150bringing of model equations to

dimensionless form 183Buckingham �-theorem 164

ccalorimetric dependences 30capillary viscometers 49Cauchy problem, see initial value problemcentral theorem of dimensional theory

163centrifugal

–blower 98–force 76–pump 76

change of total energy 22characteristic equation 118characteristics of wave equation 119Chezy factor 67circular pipe 45Clapeyron law 28Clapeyron equation 39closed mathematical model 2closed mathematical model of

one-dimensional–non-stationary flows of fluid and

gas in pipelines 109closing relations 30combined operation of linear pipeline

section and–pumping station 81

commercial flow rate of gas 96compatibility condition 119

–at characteristics 142compressibility 3

–factor 34compressible medium 28compression wave 152concentration of anti-turbulent additive

63compressor 16


208 Subject Index

concrete mathematical model 2conditions at discontinuities (jumps) of

hydrodynamic–parameters 128

conditions of transition from laminar toturbulent flow 51

conjugation conditions 124conservation law of transported medium

mass 7continuity equation 7continuum 3convective diffusion 188Coriolis factor 6correctness of the model 2criteria of

–dynamic similarity 178–geometric similarity 178–transition from laminar to turbulent

flow 51critical

–depth 151–isotherm 38–point of gas 38–pressure 40–Reynolds number 52–temperature 40–velocity 51

dD’alambert formulas 117damping of pressure bow shock at

wave front of hydraulic shock 133Darcy-Weisbach

–law 48–relation 67

density of–internal energy 6–medium 5

–transported 7dependences characterizing internal

structure of medium flow 30derived measurement units,

see secondary measurement unitsdescending sections of pipeline 68developed turbulent flow 52differential

–equation of longitudinal mixing194

–head 17–of pump 180

differentiation of integral quantityregarding fluid volume 7

dilatant fluid 37

dimensional–formula 159–quantities 158

dimensionality of quantities 159dimensionless quantities 157dimensional theory 157dimensional-dependent quantities 164dimensional-independent quantities 164direct hydraulic shock 131discharge line 17discontinuity of

–flow rate 126–velocity 126

dissipation of mechanical energy 15divergence of vector 144divergent form of differential equations

143drag force 4dynamic pressure 8dynamic

–velocity 55–viscosity 33

eeffective factor of longitudinal mixing

192efficiency of the model 2elastic

–modulus 34–spring 1

elasticity 3elementary surface 32energy equation 24energy of material point system 6enforced flow 67enthalpy 25equation for

–head before oil-pumping stations84

–variation of pressure jump 133equation of

–fluid motion 9–heat inflow 24–internal motion kinetic energy change

17–mechanical energy balance 11–medium state 29–pipeline state 30–state 30–total energy 29

Euclidian geometry 2Euler function 90Eulerian derivative, see partial derivative

Subject Index 209

external–force 9–inflow of heat 22–sources of mechanical energy

24

ffactor of

–dynamic viscosity 33–kinematic viscosity 33–volume expansion 35

feed 76filling degree of gravity flow 151first law of thermodynamics 22flow

–in hydraulic smooth pipe 21–core 50–curve 36

fluid particle of transported medium7

fluid volume 193force of dry friction 2force of viscous friction 2formation of oil product mixtures in

contact zones 188friction

–factor 20–force 13

Froude number 177fundamental laws of continuum

physics 4Funning factor 20

ggas

–compression 100–compressor station (GCS) 100–constant 28–enthalpy 105–isotherm 38–outflow from a pipeline 146–specific capacity at constant

pressure 28gas-pumping aggregates (GPA) 98gate valve 127

–closing 153generalized

–Reynolds number 48–theory of hydraulic shock 154

geometric–head 14–modeling 173–similarity 174

governing factors 3gravity

–acceleration 4–flow 66

–of incompressible fluid in a pipeline186

–force 9–stratified flow 66

hhead 75

–balance equation 81–before pump 81–before pumping station 80

head-discharge (Q − H) characteristic ofpump 76

heat–conduction equation 194–conductivity 3–energy inflow 23–exchange between transported medium

and environment 30–flux 23

heat-transfer factor 26height of pipeline axis above sea level 5Hedstroem number 62homogeneous fluid 15homogeneous incompressible fluid 15Hooke law of elasticity 43hydraulic

–dependence 30–gradient 15–hammer 42–losses 18–radius 66–resistance factor 20–resistance in laminar flow of viscous

incompressible–fluid in circular pipe 47

–shock 127–in industrial pipeline caused by

instantaneous–closing of gate valve 135–wave velocity 130

–smooth pipe 21–smooth surface 21

hydrodynamic stability 51hyperbolic differential equations 139hypothesis of quasi-stationarity 109

iIlyushin number 51impeller 76

210 Subject Index

incident wave 134incompressible

–fluid 8–medium 18

independent dimensionlesscombinations 4

individual–derivative 11–particle of continuum 6

inertial properties of fluid in non-stationary processes 112

initial and boundary conditions124

initial value (Cauchy) problem 2input of external energy 16integral characteristics of fluid

volume 5integral with variable integration

limits 13interaction of waves in pipeline

section 120intermittency factor 21internal energy 23

–of fluid volume 6international system of

measurement units SI158

invariant form 4–of equations 153

isotherm of real gas 38iteration method 81

jJoukowski formula 93Joule–Thompson

–effect 94–factor 93

jump of parameter at discontinuityfront 128

kKarman

–constant 54–formula 54–model 61

kinematic–consistency 62–viscosity 19

kinetic energy 11–of fluid volume 6–of internal motion relative mass

centre 12–of particle mass centre 12

kinetic head 14

lLagrangian derivative, see individual

derivativelaminar flow

–of non-Newtonian Ostwald power fluidin circular pipe 47

–of viscous fluid 45–of viscous-plastic fluid in circular pipe

49–regime 12

law of–longitudinal mixing 192–momentum change 9–total energy conservation 22

limit shear stress 37line of hydraulic gradient 16local derivative with respect to time 6longitudinal mixing 189loss of head in Darcy-Veisbach form 20

mMach number 185mass 4

–conservation law 7mass flow rate 5

–of fluid volume 5–of material point system 5

mathematical model 1–of centrifugal blower operating in

stationary regime 101–of mixture formation 190–of non-stationary gravity fluid flow

149–of pump 75–of slightly compressible fluid 109

mathematical modeling 1mean asperity height of pipeline internal

surface 19mean flow rate velocity 46measurement units 157mechanical energy

–balance equation 11–energy change law 14

medium macroscopic volume 3mesh cell 144method of

–characteristics 121–successive approximation, see iteration

methodmixture volume 146model 1model of

–fluid baric and heat expansion 34

Subject Index 211

–continuum 3–elastic deformable pipeline 42–elastic fluid 34–fluid 5

–with heat expansion 34–gaseous continuum 31–ideal fluid 32–homogeneous fluid 34–incompressible fluid 34–material point 1–medium 31–non-deformable pipeline 42–non-Newtonian fluid 36–non-stationary gas flow in pipeline

112–non-stationary isothermal flow of

slightly compressible–fluid in pipeline 109–perfect gas 39

–power Ostwald fluid 36–real fluid 5–real gas 5–Shvedov-Bingham fluid 37–valve 127–viscous fluid 32

modeling of–blower operation 100–fluid outflow from tank 178–gravity fluid flow 176–mixture formation in oil product

batching 188–stationary flow of compressible gas

in gas-pipeline 92–stationary flow of viscous

incompressible fluid in a pipe 175molecular weight 39momentum

–equation 29–of fluid volume 6–of material point system 6

motion equation 29movable

–fluid volume of continuum 5–individual fluid volume 5

multi-step compression 101mutual displacement of oil products

188

nnear-wall turbulence 63negative slope characteristic 131Newton formula 25

Newton-Leibniz formula 7Newtonian viscous fluid 36non-isothermal fluid flow 89non-Newtonian fluid 36non-stationary

–fluid flow with flow discontinuities152

–non-isothermal gas flow ingas-pipelines 138

normal depth 151

ooil

–product batching by direct contact188

–pumping station 75one-dimensional

–flow 1–model 45–non-stationary flow of slightly

compressible fluid–in pipeline 184

–theory 12operation of pipeline with intermediate

oil-pumping stations 84origination of similarity criteria in

equations of mathematical–model 183

oscillation of small load 1Ostwald rheological law 61overcompressibility factor 40

–for natural gas 40

pparabolic (Q − H)-characteristic of

centrifugal pump 181parallel connection of pumps 79parameter of pipeline internal surface

smoothness 21partial derivative 10Peclet number 194perfect gas 28phase of

–direct hydraulic shock 131–reflected wave 132

phenomenological Karman theory 54physical modeling 173piezometric head 14pipeline

–profile 14–transportation of oil products 187

piston 100–engine 100

212 Subject Index

plane-deformable state 43plane-stress-state 43point mass 1Poiseuille

–formula 46–law 48

Poisson ratio 43positive slope characteristic 115power of

–external mechanical devices 13–gravity force 13–internal friction forces 13–pressure force 13

pressure 5–jump 128–wave 127

primary measurement units 158problem on

–disintegration of arbitrarydiscontinuity 155

–oil and gas transportation 4–traveling waves 118–wave interaction in limited pipeline

section 119profile

–blades 76–hydraulic shock 153

propagation of waves in–bounded pipeline section 119–infinite pipeline 115–semi-infinite pipeline 117

protection of pipeline from hydraulicshock 138

pseudo-plastic fluid 36pump 16

–efficiency 17–power 17

pump-to-pump regime 84pumped flow, see enforced flowpumping

–plant 16–pressure 76

pumps connected in series 78

qquasi-linear differential equations of

hyperbolic type 150

rrarefaction wave 152rate of

–elementary volume change 13–internal energy change 25

reaction force 9rectangular mesh 143reduced

–pressure 40–temperature 40–universal characteristic of centrifugal

blower 103reflected wave at the joint of two pipe

sections 134relative roughness 66

–of pipeline inner surface 19relativistic effects of the relativity theory 2restoring force 1Reynolds

–number 19–stresses 52

Reynolds-Filonov formula 33rheological properties 49rheology 30Riemann invariants 115root-mean-square (rms) value 12rule of integral quantity differentiation with

respect to time 6

ssafety valve 138schematization of

–initial conditions 2–one-dimensional flows of fluids and

gases in pipelines 4–phenomenon 2

second Newtonian law 2secondary measurement units 158separatrix 99shear rate 36shear stress 5shock front 131Shuchov formula 26similarity 173

–criteria 174–of operation of centrifugal pumps

179–factor 173–theory 157

slightly compressible fluid 24slug flow 68saturated vapor tension (pressure) 68self-similar

–distribution of concentration 196–solutions 194

small perturbations 142space coordinate measured along

pipeline 4

Subject Index 213

specific–mechanical energy dissipation 19–volume 38

speed of wave propagation in pipeline110

square flow 21stability of laminar fluid flow 51stationary

–flow 8–of barotropic medium 15

–operation regime of–high-temperature pumping

87–gas-pipeline together with

compressor station 98steady motion 4stepwise variations (jumps) of

hydrodynamic flow parameter128

sticking condition 46Stokes formula 20string elasticity factor 1Strouchal number 185suction

–line 16–pressure 76

surface element 31system matrix rank 166system of

–finite difference equations 144–measurement units 4

systems of pressure wave smoothing138

ttangential

–friction tension 18–stress 32

temperature 5thermodynamic equilibrium 23time 4Toms effect 62total

–derivative with respect to time6

–energy balance equation 22–head 14–kinetic energy 17

transfer section 68transition of laminar flow into

turbulent flow 52transmitted wave at the joint of two

pipe sections 134

traveling wave 114turbulent

–diffusion 189–dynamic viscosity 52–flow

–in circular pipe 52–kinematic viscosity 52–of non-Newtonian fluid 61–of power fluid 61–regime 12

–velocity profile 58two-term (Q − �H)-characteristic of

centrifugal pump 78

uundamped periodic oscillations 2undeformable pipeline. 5unit normal 31universal

–characteristic of centrifugal blower103

–gas constant 39–resistance law 59

unstable fluid 152useful power

–of blower 105–needed for gas 105

vvapor-gas

–cavities 68–phase 152

velocity–gradient 32–of medium 5–of perturbation propagation in

pipeline 129–of propagation of small

perturbations in gas (sound velocity)142

–of shock wave propagation inpipeline 128

–of transported medium 8–step-wise change 128

virtual mass of fluid 112viscosity 3viscous-plastic Bingham fluid 37void 152volume

–balance equation 192–expansion factor of metal pipeline

42volumetric flow rate 157

214 Subject Index

wwave equation 113wetted perimeter 66work of

–external forces 9–internal forces 13

–pressure and internal friction, see workof internal forces

yYoung modulus 43