Gas Flow Measurement.pdf

POLPetroleum Open Learning

OPITO

THE OIL & GAS ACADEMY

Gas Flow MeasurementPart of the

Petroleum Processing Technology Series

Petroleum Open Learning

Designed, Produced and Published by OPITO Ltd., Petroleum Open Learning, Minerva House, Bruntland Road, Portlethen, Aberdeen AB12 4QL

Printed by Astute Print & Design, 44-46 Brechin Road, Forfar, Angus DD8 3JX www.astute.uk.com

OPITO 1993 (rev.2002) ISBN 1 872041 85 X

All rights reserved. No part of this publication may be reproduced, stored in a retrieval or information storage system, transmitted in any form or by any means, mechanical, photocopying, recording or otherwise without the prior permission in writing of the publishers.


Visual Cues trainingtargetsfor you to

achieve by the end of the unit

testyourself questions to see how much you understand

checkyourselfanswers to let you see if you have been thinking along the right lines

activities for you to apply your new knowledge

summaries for you to recap on the major steps in your progress

GasFlowMeasurements(PartofthePetroleumProcessingTechnologySeries)


As a large part of this subject deals with calculations, you will require to be comfortable with maths up to about standard grade level.Although some of the equations are fairly complex, all relevant data and information is provided to assist you to solve the problems.All formulae required for calculations in your examination will be provided for you. However, it is necessary that you are able to recognise the symbols in formulae and allocate the correct units of measurement to each symbol in your calculationsYou will also find that a scientific calculator will be useful for this programme.

Contents Page

* TrainingTargets 4

* Introduction 5

* Section1-GasFlowMeasurementApplicationsandGasPhysics 6 TheNeedforAccurateMeasurement UnitsofMeasurements

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Contents(contd) Page

* Section2-FluidFlowPrinciples 26

FluidFlowPrinciples FluidFlowProperties

* Section3-MeasurementDevicesandMethods 34

TypesandApplications OrificePlatePrinciples OrificePlateFlowCalculations

* Section4-OrificePlateMeteringEquipment 46

TypesofPlate SensingDevices MeteringStations SafetyImplications

* CheckYourself-Answers 54

VisualCuestrainingtargetsfor you to achieve by the end of the unit

testyourself questions to see how much you understand

checkyourselfanswers to let you see if you have been thinking along the right lines

activities for you to apply your new knowledge

summaries for you to recap on the major steps in your progress


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TrainingTargetsWhen you have completed this unit on Gas Flow Measurement, you will be able to:

State the gas laws and perform relevant calculations using the appropriate units of measurements. Define molecular mass, gas density and specific gravity, and perform calculations when given appropriate formulae. Define Reynolds number and, given Reynolds equation, define the terms in it and use it in a calculation. Describe Bernoullis principle and state the types of pressure in a flowing fluid. Describe the main types of gas flow measuring devices. State the relationship between differential pressure and flowrate. Describe the main features of orifice plate meters. Perform a flowrate calculation in which all necessary formulae and data are given. Draw a schematic diagram of a typical multi-stream system which complies with fiscal standards. Describe the main safety implications associated with metering systems.

Tick the box when you have met each target


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Petroleum Open LearningOilandGasSeparationSystemsIntroductionGasFlowMeasurement

IntroductionPetroleum Open Learning

For most of this century there has been a necessity to measure gas flow accurately, both in commercial applications and scientific investigations.

The measurement of gas flow is more complex than that of liquid since gas is more sensitive to physical factors, such as pressure, temperature, composition, etc. It has therefore been subjected to considerable research effort, which has led to a high degree of accuracy now being possible.

In this book we will be mainly concerned with natural gas flow measurement, but the principles are generally applicable to all gas phase matter.

This book comprises four sections :

Section1,GasFlowMeasurementApplicationsandGasPhysics, outlines areas in which the ability to accurately measure gas flow is essential. It then covers the basic physics of gas behaviour, which are essential to an understanding of the measurement and flow calculation methods.

Section2,FluidFlowPrinciples, presents the concepts of laminar and turbulent fluid flow, develops the Bernoulli and continuity principles to produce a simple flowrate / differential pressure relationship.

Section3,MeasurementDevicesandMethods, describes various fluid flow measurement devices which are applicable to gases. It then gives a more detailed treatment of the orifice plate method, since this is the one most widely used. It ends with the ISO 5167 formula, and an explanation of its terms.

Section4,OrificePlateMeteringEquipment, describes : various orifice plate designs, differential pressure and gas density sensing and measuring equipment and a typical gas metering station. It ends by drawing attention to the safety aspects of gas metering systems.

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TheNeedforAccurateMeasurementWe will start by looking at some examples of activities that require accurate measurement of natural gas flow rates.

OilFieldEvaluationA typical field evaluation exercise involves flowing reservoir fluid from an exploration well to a test separator, where the liquid and gas phases are separated. Accurate measurement of the gas and liquid flow rates from the separator is essential to the achievement of a reliable appraisal of the reservoir performance.

ProcessControl&OptimisationThe control and optimisation of gas processes, in both offshore and onshore operations, often requires gas flow rate monitoring as the process variable in automatic control systems.

An example of this is using the gas flow rate through a centrifugal gas compressor as a measured variable to provide automatic flow control / recycling for anti-surge protection.

A comprehensive explanation of surge protection is provided in the Petroleum Gas Compression programme which forms a part of this Petroleum Processing Technology Series.

GasSalesContractsAll natural gas sales contracts are based on accurately measured volumetric flow rates with specific reference pressure and temperature conditions.

HydrocarbonTaxationIn most oil and gas producing countries, governments impose various types of revenue on the production companies. Most of these revenues are applied on a volumetric basis, so flow rates must be measured to a high degree of accuracy.

FiscalStandardsIn the last two examples, in addition to the need for accuracy, there is also the implication of complex legal considerations. This has led to the establishment of a set of fiscal standards, the purpose of which is to achieve consistent levels of high accuracy and reliability in fluid flow measurement.

Petroleum Open LearningGasFlowMeasurement

Section1-ApplicationsandGasPhysics

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UnitsofMeasurementIt is not possible to fully appreciate the methods and procedures of gas flow metering without a basic understanding of the physical behaviour of gases, in particular the relationship between pressure, volume and temperature.

BoylesLawFigure 1 illustrates the principle behind Boyles Law, which describes the relationship between the volume occupied by a given mass, or number of molecules, of gas and its pressure, while the temperatureremainsconstant.

Figure 1 depicts a piston in a cylinder which contains a fixed mass of gas. The highly energetic gas molecules collide with each other and with the cylinder walls and the piston face, resulting in a force being exerted. The property we describe as pressure is defined as the magnitude of that force divided by the area over which it acts.

The force acting on the bottom face of the piston is therefore the pressure multiplied by the cross sectional area of the piston. To prevent the piston being driven out of the cylinder, a force of the same magnitude must be applied downwards, in this case by a weight (W).

The weight (W) balances the force due to the pressure (P), and the gas occupies a volume (V), as shown in Figure 1 (a).

Now consider what happens if we double the force on the piston, a condition we achieve by applying a second weight of the same mass as the first one (we assume that the piston itself is weightless), to exert 2W.

To balance this force the gas must now exert twice the pressure, i.e. 2P. If we measure the new volume we find it to be 1/2 of V1, as shown in Figure 1 (b).

If we double this force to give us four times the original force, Figure 1 (c) shows that a gas pressure of 4P is produced and the volume is reduced to 1/4 of V1.

The relationship between the pressure and volume is now clear. Doubling the pressure halves the volume; quadrupling the pressure reduces the volume to a quarter of its original value. Boyles Law expresses this formally with the statement that: Atconstanttemperature,theabsolutepressureof a givenmass of an ideal gas is inverselyproportionaltothevolume.

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We need to define some of the terms in this statement.

Absolute pressure units must be used instead of gauge units. Most field pressure measurement devices read in gauge units, which means thatatmosphericpressure must be added to the indicated value to obtain the absolute pressure.

NOTE :Unless otherwise stated, all pressure references in this programme will be given as bar, this will infer that the pressure isbarabsolute(bara).

The behaviour of the gas must be Ideal. The kinetic model that was designed to enable the prediction of gas behaviour was based on two main assumptions that pertain to an ideal gas

1. The gas molecules are small spheres, the volumes of which are negligible in comparison to the volume that the gas occupies.

2. There are no attractive or repulsive intermolecular forces, and the behaviour of the molecules when colliding is similar to that of billiard balls, the collisions being elastic.

The Boyles Law statement that gas pressure (P) is inversely proportional to its volume (V) can be written

as P is proportional to 1vWhich means that the pressure is equal to a constant (k) divided by the volume.

Thus : P = kvso k = PV which is the mathematical way of stating that the pressure multiplied by the volume gives a constant value.

Referring to Figure 1 (a) and (b), we can write :

P1 V1 = kand P2 V2 = kHENCE P1 V1 = P2 V2

which is the mathematical expression of Boyles Law.

This equation can be used to calculate a new pressure or volume, where the original pressure and volume and one of the new conditions are given, at the sametemperature.

EXAMPLE

5 Ltr of an ideal gas is contained in a cylinder at 2 bar. A piston then compresses the gas until the volume is reduced to 3 Ltr.

What will the new pressure be, once the temperature has stabilised to its initial value ?

We will use the left side of the equation to represent the initial conditions, and the right the final ones.

Thus : 2 X 5 = P2 X 3

P2 = 2 X 5 = 3.33bar 3

CharlesLawCharles Law describes the relationship between the volume and temperature of an ideal gas, while the pressureiskeptconstant.As in the case of Boyles Law, we can use a cylinder /piston arrangement to demonstrate the principle behind Charles Law, as shown in Figure2.This time, however, we keep the pressure constant by leaving the force on the piston unchanged, and heat the gas in the cylinder. Not surprisingly, we find that the gas volume increases.

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If the volume is measured at various temperatures and the results are plotted on a graph, we obtain a diagram like Figure 3 when we use the Celsius temperature scale.

Figure3:Volume/TemperatureGraph(CelsiusScaleOnly)

We see that there is a simple linear relationship between the volume and temperature (the graph is a straight line). However calculations involving temperatures below 0C are slightly inconvenient due to the presence of negative numbers. This problem is solved by employing a different temperature scale which only has positive values.

Figure4:Volume/TemperatureGraph(CelsiusandKelvinScale)

Figure4 is similar to Figure 3, but with the graph line extrapolated to intersect the Temperature axis. This point is taken as 0 for our other temperature scale, and we see that it corresponds to -273.15C. AbsoluteZero is the term that is commonly applied to this temperature, since it is physically impossible to achieve lower temperatures than it. Absolute zero has been approached experimentally, but has never been quite achieved; and is therefore a theoretical value rather than a practical one. We see in Figure 4 that the gas would occupy no volume at that point; a futile observation, since no substance would be in the gas phase at such a low temperature.

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If the volume is measured at various temperatures and theresults are plotted on a graph, we obtain a diagram likeFigure3when we use the Celsius temperature scale


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Absolute zero is the lowest point on the absolutetemperature scale, which is measured in units calledKelvin(K)in the SI system. In the Imperial system the units are known as Rankine (R).

As evident in Figure 4, the unit step sizes for theKelvin and Celsius scales are the same. So a one degree Celsius temperature change is also one Kelvin. (A convention, which is by no meansuniversally applied, is to omit the term degree when using absolute temperature units). OC, then, is 273.15 K, and 100C is 373.15 K, so to convert from C to K we simply add 273.15 to the C value.

Absolute zero on the Rankine scale is equal to-459.67F, and a degree on the Rankine scale isthe same size as a degree on the Fahrenheit scale.

In most practical situations sufficient accuracy isachieved by using 273 as the conversion factorbetween Celsius and Kelvin, and 460 betweenFahrenheit and Rankine. However, where highaccuracy is required, such as in fiscal gas flowmeasurement, the more exact values should beused.

We see inFigure5that if the absolute temperature is doubled, the gas volume will also be doubled.

Charles' Law, then, states that: atconstantpressure,thevolumeoccupiedbyagivenmassofgasisproportionaltoitsabsolutetemperature.

To derive a mathematical expression of Charles'Law we can employ a similar argument to the one we used for Boyle's Law.

The statement that the volume of gas is proportional to its temperature can be written as :

V = cT

where c is a constant

so c = V T

Referring to Figure 5 (a) and (b) we see that

V1 = c and V2 =c T1 T2

hence : V1=V2 T1 T2

This equation can be used to evaluate the newvolume or temperature of an ideal gas for a change in which the pressurestaysconstant.


CombinedIdealGasLawBoyle's and Charles' laws combine to give the equation:

P1V1 =P2V2 T1 T2We will apply it in the following examples:

A 5 Ltr sample of gas at 15C and 3 bar a is heated to 45C and has its pressure reduced to 1.5 bar. Assuming that it behaves ideally, what will its new volume be?

The first step is to ensure that the pressures and temperatures are in absolute units. The pressures are quoted in bar a, which means that the values are absolute. However we will have to add 273 to the temperatures to convert them from C to K.

Using P1V1=P2V2 T1 T2we will ascribe the initial conditions to left side, and the changed ones to the right.

Thus we need to find V2. 3 x 5 1.5xV2 (15 + 273) (45 + 273)

V2 = 3 x 5 x 318 (1.5 x 288)

V2=11.04Ltr

MolecularMassMolecular mass is a physical property of all substances. A comprehensive description of it can be found in elementary chemistry text books and training manuals, but for our purposes a simple description is sufficient.

All matter consists of atoms. In many substances two or more atoms combine to form molecules. As we wish to keep this description simple, we will accept the atomic mass units (a.m.u.) given in the following text.

Let us consider methane, the lightest alkane hydrocarbon and the main component of natural gas. It is a molecule comprising onecarbonatom bonded to fourhydrogenatoms. Carbon has an atomicmassof12atomicmassunits(a.m.u.). Hydrogen has an atomicmassof1a.m.u. The molecularmass of methane is the sum of the masses of its constituent atoms, which is therefore 12 + (4 x 1) = 16a.m.u.

TestYourself1.12 Ltr of an ideal gas at 10C and 2 bar is compressed to a volume of 0.5 Ltr. Given that the heat of compression raises its temperature to 25C, what will its pressure be?

You will find the answer in CheckYourself1.1on page 54.

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Although it may not be scientifically correct, theterms molecularmass andmolecularweight(Mw)are often used interchangeably.

The very small size of atoms and molecules makes calculations using their individual massesinconvenient. A more practical approach is toconsider the mass of a large number of them, andthis involves the concept of the mole.

The mole, usually written as mol, or gramme-mole(g-mol)is defined as the atomicormolecularmassofasubstanceexpressedingrammes.

The number of moles of substance is usuallyassigned the symboln, and is easily calculated for a given mass (m) of material using the relationship:

n=m MwEXAMPLE

Calculate the number of kg-mol in 40 kg of methane.

We have already seen that the molecular mass ofmethane is 16.

n=m=40=2.5kg-mol Mw16

An appropriate unit of measurement for molecularweight (Mw) is obtained by using the units in the

n=m equation. Hence: Mw n(kg-mol) = m(kg) Mw Mw =m(kg) n(kg-mol)

So, in this case, the unit of molecular mass iskg/kg-mol. It may also be written as kgkg-mol-1It will have the same value when expressed in units of g/g-mol, so molecular mass is one of the few physical quantities for which it is acceptable to omit its unit of measurement.

TestYourself1.2 Calculate the molecular mass of ethane, which is a molecule comprising two carbon and six hydrogen atoms.

You will find the answer in CheckYourself 1.2 on page 54.

Having seen how the molecular mass of puresubstances like methane and ethane are calculated, we will now determine the molecular mass of a mixture of components, such as natural gas. We have already seen that methane is the main constituent of natural gas, but it also contains smaller quantities of heavier hydrocarbons such as ethane, propane and butane. The relative amounts of these can vary considerably between samples of gas, depending on factors such as the reservoir conditions, processing methods, etc. These variations can have very significant effects on the behaviour of gas during handling and measurement of its flow rates.

We will use a simple example of a two-component mixture of methane and ethane.

To perform the calculation, we will obviously need to know the relative quantities of each component. These are expressed as mole-fractions, which simply means the relative number of molecules of each constituent. Let us assume that our mixture has mole-fractions of 80% and 20% for methane and ethane respectively. In other words, 80 of every 100 molecules of the mixture are methane, and 20 are ethane.


The procedure is shown in the following table, and involves adding up the results of multiplying the mole-fraction of each component by its molecular mass.

Component Mol.Mass Mol.Fraction Mol.massx Mol.Fraction

Methane 16 0.80 12.80Ethane 30 0.20 6.00 MixtureMol.Mass 18.80

The mixture in the table could be considered as consisting of molecules with an average mass of 18.80 kg / kg-mol. In the following exercise you will see the effect changing the concentrations of the components has on the mixture molecular mass.

The physical behaviour of the heavier mixture will be considerably different from that of the lighter one.

The procedure for calculating the molecular mass of mixtures of more than two components is exactly the same as for two.

GasConstantsThe form of the ideal gas equation we looked at earlier :

P1V1=P2V2 T1 T2

implies that

PV=aconstant T

In other words, for a given type and mass of an ideal gas, the absolute pressure multiplied by the volume and divided by the absolute temperature will always produce the same answer.

If we call the constant C, we can rewrite the equation as:

PV=CT

A disadvantage of the equation, as it stands, is that C will only be constant for a gas with a given molecular mass. It will have different values for methane and ethane, for example. What we need is a constant that will have the same value regardless of the type of gas under consideration.

TestYourself1.3Calculate the molecular mass of a mixture comprising 60% methane, and 40% ethane.

You will find the answer in CheckYourself1.3on page 54.

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This is achieved by using the number of moles(n)of gas, instead of the mass, in the equation. Thus: PV=nRT

Because it applies to all ideally behaving gases, R is termed the universalgasconstant. Its value will depend on the units by which the other terms in the equation are measured; the following table shows values of R for various combinations of units.

P V T n R kPa m3 K kg-mol 8.130 bar m3 K kg-mol 0.0831 bar Ltr K g-mol 0.0831 bar cm3 K g-mol 83.1 psia ft3 oR Ib-mol 10.73

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EXAMPLE

What is the volume of 1.5 kg of air at 1.5 bar and 25C? (Take the molecularweight of air as 29, and assume ideal behaviour).

We first find the number of moles of air using:

n=m=1.5=0.0517kg-mol Mw29

and use this value in : PV = nRT, having selected the appropriate value of Rfrom the table as 0.0831. Note that this will make our volume units m3. As theseare the most commonly used units for gas measurement calculations, we will use0.0831 for all our calculations where a value of R is required

Inserting these values gives:

1.5 x V = 0.0517 x 0.0831 x 298

V=1.280=0.853m3 1.5 TestYourself1.4 How many kg of methane will, when behaving ideally, occupy a volume of 1 m3 at 2 bar and 20C?

You will find the answer in CheckYourself1.4 on page 54.

This is achieved by using the number of moles(n)of gas, instead of the mass, in the equation. Thus: PV=nRT

Because it applies to all ideally behaving gases, R is termed the universalgasconstant. Its value will depend on the units by which the other terms in the equation are measured; the following table shows values of R for various combinations of units.

P V T n R kPa m3 K kg-mol 8.130 bar m3 K kg-mol 0.0831 bar K g-mol 0.0831 bar cm3 K g-mol 83.1 psia ft3 oR Ib-mol 10.73


ReferencePressuresandTemperaturesGas flow rates are often quoted in volumetric units, such as m3 /minute and ft3 / minute. Having studied the very significant interdependence of pressure, temperature and volume, you will now be aware that it is meaningless to express a volume of gas without stating the pressure and temperature at which it is measured. This has led to the establishment of referenceconditions for gas volumetric measurements.The values of reference pressures and temperatures may vary between countries and contracts, but the most common ones are 1. 1.013bar; 15C 2. 1.013bar; 0C 3. 14.73psia; 60FThe first one is usually termed StandardPressure and Temperature, and the second Normal Pressure and Temperature. However you should be aware that some textbooks use Standard Pressure and Temperature (STP) with a reference temperature of 0C. Clearly, the first two conditions apply to the metric system, and the third to the imperial system.A standard cubic metre, then, is the quantity of gas that has a volume of 1m3 at 15C and 1.013bar.

There are various conventions for expressing reference volumetric and volumetric flow rate units. For example the standard cubic metre may be written as sm3. The oil industry often expresses gas flow rates in millions of volume units per day, which would be 106sm3/d or 106sft3/d. However, the oil industry would normally write these as MMSCMD or MMSCFD, although you may also come across ksm3/hr(1000m3(st)/ hour in place of MMSCMD. Before performing any calculations, you must always ensure that you know which units the quantities you are using are expressed in, and the units that will apply to the result.Gas volumes or volumetric flow rates measured at pressures and temperatures other than reference ones are sometimes called actual volume, or actual volumetric flow rate. 1 m3 at 10 bar and 40C is an actual cubic metre at that pressure and temperature, and would require conversion to be expressed in terms of reference conditions, as the following example shows.

EXAMPLEWhat is the actual volume of 5m3 (st) of an ideal gas at 3 bar and 25C.We will use P1V1=P2V2 T1 T2

and assign the standard conditions to the left side.

Hence:

P1 = 1.013 bar T1 = 15C = 288 K V1 = 5 m3 P2 = 3 bar T2 = 298 K

and we are required to find V2.

1.013 x 5 = 3 X V2 288 298

0.01007 X V2 = 0.01759 V2 =0.01759 = 1.747m3 0.01007

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Non-IdealGasBehaviour

In our work so far, we have repeatedly used theterm ideal to describe the characteristic behaviour of gas. Real gases can deviate significantly from ideal behaviour, since the simple mathematical model from which ideal gas behaviour was predicted is not adequate to describe all gases under all possible conditions. The model has been modified to make prediction more accurate over a wider range of conditions, but it is beyond the scope of this book to present the explanation that these modifications would require.

Although precise gas flow measurement techniques would require use of fairly complex formulae, we will be able to gain an appreciation of how real gases deviate from ideal behaviour, and even to make reasonably accurate calculations, by introducing a term called the compressibilityfactor.

The availability of this factor is due to the largeamount of empirical data that has been gleanedfrom extensive experimental work over many years, particularly in the natural gas industry. Thecompressibility factor is usually given the symbol Z, and is a function of the type of gas, pressure and temperature.

It is customary to present Z in the form of charts,the general form of which is shown inFigure6.

This provides curves that represent values of Z plotted against pressure for various temperatures; so to find the appropriate factor, the point that represents the relevant pressure and temperature is identified and the corresponding value is read from the vertical axis. In Figure 6, for example, we see that if the gas pressure and temperature is P1 and T1 respectively, the compressibility factor is Z1. Note that if the stated temperature lies between two curves, interpolation is necessary.


When Z has been determined, it is inserted in the ideal gas equation thus :

PV=ZnRT

The first point to note is that if Z=1, the gas is ideal ( multiplying by 1 has no effect ). We see in Figure 6 thatZ=1atlowpressure, which includes atmospheric and standard pressure. So, in most cases, ideal gas behaviour can be assumed at these pressures. Points on the chart furthest away from the Z=1 line denote the largest deviations from ideal behaviour, the Z1 in Figure 6 being an example.

In Figure 6, T2 represents a higher temperature than T1, and we see that T2 curve is generally closer to Z = 1 than the T1 curve. This demonstrates the characteristic that higher gas temperatures tend to produce behaviour that is closer to ideal than low ones. However, temperatures above 300 - 400C will have curves that show increasing deviation above the Z = 1 line with increasing pressure.

The general characteristic, then, is for curves of temperatures below about 300C to give decreasing values of Z (increasing deviation from ideal behaviour) as the pressure increases. However, at a certain pressure, a minimum Z value is reached and further pressure increment causes Z to increase until it reaches 1, where the gas is again ideal. Increasing the pressure beyond that point causes Z to become progressively greater than 1, which is again an increasing deviation from ideal behaviour, but in the opposite sense from values of Z that are less than 1. This should become clearer when we look at some examples.

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We will now look at the compressibility factor chart for methane, depicted in Figure7.

The chart covers ranges of pressure and temperature that encompass most processing conditions. The main features of the chart are:

1. At pressures of 10 bar or less the gas virtually behaves ideally.

2. At temperatures below -20oC methane becomes rapidly non-ideal as the pressure rises from 10 to 100 bar.

3. The maximum deviation from ideal behaviour occurs at pressures between 100 and 200 bar, for temperatures below 100oC.

4. Methane virtually behaves ideally between about 100 and 200oC, at pressures below 250 bar.


EXAMPLE

Calculate the volume occupied by 1 kg-mol of methane at 70 bar and 35C,

a) ignoring non-ideal behaviour

b) accounting for non-ideal behaviour.

a) Assuming ideal behaviour, we simply apply

PV = nRT

Since n = 1 kg-mol, we can ignore it. So

V = RT P

V = 0.0831 x (-35 + 273) 70

= 19.78 70 V= 0.283m3

b) Accounting for non-ideal gas behaviour, we find in Figure 7 that Z=0.68 (interpolation between the T = 30C and T = 40C curves was required). So we use this value in :

PV = ZnRT

V = ZRT P

= 0.68 x 0.0831 x 238 70

= 13.449 70

V =0.192m3

We see that, in this case, failure to consider thenon-ideality would have led to an error of

(0.283 - 0.192) x 100 = 47.4%. In other words, 0.192

we would have overestimated the actual volume of the gas by almost 50%.

ActivityUse Figure7 to find values of Z for each of the following sets of conditions :

1. 70 bar ; 20oC

2. 100 bar ; -70oC

3. 150 bar ; 0oC

Your answers should be :

1. 0.882. 0.363. 0.74

The following example will enable you to appreciate the magnitude of the effect that non-ideal gas behaviour can have.

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TestYourself1.5Using Figure7 to find the compressibility factor, calculate the volume of 32 kg of methane at 100 bar and -50C. Estimate the percentage error you would have incurred by assuming the gas to behave ideally.


We have already seen that, to perform gascalculations involving changes in conditions fromP1, V1 and T1 to P2, V2 and T2 , we use:

P1V1 = P2V2 T1 T2

You will now be aware that, unless the gas isideal at both sets of conditions, corrections willhave to be applied.

EXAMPLE

Find the volume occupied by 3m3 of methane at 65 bar and 20C, after it has been compressed to 150 bar and chilled to -30C.Using: P1V1 = P2V2 Z1T1 Z2T2 and assigning the initial conditions to the left side,P1 = 65 bar, V1 = 3 m3, T1 = 293 K and, from Figure 7, Z1 = 0.88. The final conditions are: P2 = 150 bar, T2 = 243 K, at which Z2 = 0.61 ; and V2 is the volume we need to calculate.

65 x 3 = 150 x V2 0.88 x 293 0.61 x 243

1.012 V2 = 0.756

V2 = 0.756 = 0.747m3 1.012

Until now, we have used Z values which were less than 1, which caused the volume occupied by a given mass of gas to be less than that predicted by the ideal gas equation.

Examining PV = ZnRT, from which P = ZnRT Vwe see that the pressure of a given mass of gas at a given volume and temperature would also be less than if it were ideal.

We have also seen that the compressibility factor varies with pressure and temperature, sodifferent values of Z are likely to be required foreach set of conditions; ie. Z1 for P1, V1 and T1,and Z2 for P2, V2 and T2.

The complete equation will be : P1V1=P2V2 Z1T1 Z2T2


It should now be clear that, if Zisgreaterthan1, the volume at a given pressure, or the pressure at a given volume, will be greater than that predicted by the ideal law. You will also observe, when studying Figure 7, that Z is only significantly greater than 1 at pressures and temperatures considerably higher than those we normally encounter.

As stated earlier, natural gas is a mixture of hydrocarbons, so it is worth taking a brief look at the selection of compressibility factors for such mixtures.

When we looked at mixtures earlier, we saw that, in addition to the type of components present, the molecularmass of the mixture is determined by the concentration, or relative amount of each component. A mixture with a high concentration of methane, the lightest hydrocarbon, produces a lower molecular weight mixture than one with a lower methane concentration.

The only difference between finding compressibility factors for gas mixtures and for pure gases, is in the selection of the appropriate chart. Instead of the name of the gas to which the chart applies, a gas mixture chart is identified by the averagemolecularmass of the mixture.

Figures8and 9 are examples of such charts, and apply to mixtures of molecular mass 18.85 and 23.2 kg / kg-mol respectively. Charts are also available for lower, intermediate and higher molecular masses. When applying factors to mixtures with molecular masses that are between values for which charts are available, reasonable accuracy can be achieved by interpolation.

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Comparing Figures8and9, it is clear that, except at temperatures well above 100C, the deviation from ideal gas behaviour is considerably greater for the heavier gas. For example, at 100 bar (10,000 kPa) and 5C, we see that the Z values are 0.47 and 0.68 for the heavier and lighter gases respectively. This is consistent with the fact that its behaviour is closer to ideal when a gas is relatively light.

We have spent some time looking at the gas lawsand the implications of non-ideal behaviour. We did so because they are a very important aspect of gas flow measurement.

We will finish this topic with an exercise in which you will calculate volumetric flow rates instead of simply volumes. This should not present problems, since volumetric flow rate is just volume divided by time and we substitute the symbol Q for V in the equation PV=ZnRT(PQ=ZnRT). This exercise is slightly longer than the ones you have done so far, and it is worth ensuring that you understand how the answers are worked out, especially if you do not get them right first time.


DensityIn fluid flow measurement, and particularly when the fluid is gas, density is an important physical property.

Density is defined as massdividedbyvolume, and is usually given the Greek symbol .

So = m V

Having seen how sensitive gas volume is to pressure and temperature, you will appreciate that density will be similarly affected.

We have already seen that the number of moles,

n = m , and that this can be substituted for Mw

in PV = ZnRT, so that PV =ZRTm MwThis is rearranged to :

m=PMw= VZRT

which is the form that is often used in gas flow calculations.

ExampleFind the density of methane at standard pressure and temperature (1.013 bar and 15C).

= PMw = 1.013 x 16 ZRT 0.0831 x 288

=0.678kg/m3

Note that we ignored Z since the gas is ideal at standard conditions.

TestYourself1.61. For two types of natural gas, with molecular masses of 19 kg / kg-mol and 23 kg / kg-mol, calculate their actual volumetric flow rates in m3 / minute, given the following data:

Mass flow rate = 50 kg / minute Line pressure = 130 bar Line temperature = 10oC

2. Find the standard (15oC, 1.013 bar) volumetric flow rates for both your answers to part 1.

You will find the answers in CheckYourself1.6 on page 55.

TestYourself1.7Find the density, at standard pressure and temperature, of the following natural gas components:

Ethane (Mw = 30) Propane (Mw = 44) Butane (Mw = 58)


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Specific Gravity

Another method of expressing the density of amaterial is to use the measurement specific gravity(s.g.). Specific gravity is sometimes referred to as relativedensity, which is an apt term since it is defined as thedensityofthesubstancebeingevaluateddividedbythedensityofareferencesubstance.

In the case of liquids, the reference material iswater; for gases it is air.

So for gases we can write: s.g. = g awhere the subscripts g and a denote the gas being evaluated and air respectively.

To avoid having to account for non-ideal gasbehaviour, measurements are usually referred tostandardpressureandtemperature.

Consider the specific gravity of methane:

sgmethane=methane aAs the density () of both methane and gas arecalculated at standard pressure and temperature,we need only use the molecular mass (Mw) for gas and air in the equation.

So, in addition to being equal to the density of gas divided by the density of air, specific gravity is equal to the molecularmassofgasdividedbythemolecularmassofair. i.e.:

sgg=Mwg Mwa .. sgmethane=16=0.55 29

Note that sg is a dimensionlessnumber(it has no units).


SummaryofSection1 Applications in which accurate gas flow measurement is required are :

Oil Field Evaluation Process Control and Optimisation Gas Sales Contracts Hydrocarbon Taxation

Boyles Law and Charles Law combine to express the relationship between the pressure, volume and temperature of gases when they behave ideally, with the proviso that absolute temperature and pressure units must be used in the calculations.

The number of moles of a substance is found by dividing its mass by its molecular mass, a procedure that can be applied to mixtures as well as pure substances.

The number of moles (n) can then be used in PV = nRT, where R is defined as the universal gas constant.

The interdependence of these properties demands the use of reference pressure and temperature at which gas volumes are calculated.

Most gases only obey the ideal gas laws at certain pressures and temperatures, so the compressibility factor (Z) is introduced to compensate for non ideal behaviour.

Gas density can be evaluated from the ideal gas equation, and gas specific gravity is defined as its density divided by that of air, both values being referred to standard conditions. Gas specific gravity is also found by dividing the molecular mass of the gas by that of air.

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We will start this section with a general outline of fluid flow principles. You should note that the term fluid applies to gases as well as liquids.

FluidFlowCharacteristicsTheStreamlineConcept

The construction of streamlines in a diagram is designed to enablea visual image of fluid flow characteristics to be achieved, In simpleterms, streamlines are drawn such that adjacent lines representdifferent fluid flow speeds.

FlowTypes:LaminarandTurbulent

Figure10 represents fluid flowing through a pipe, in a condition known as laminar flow. This is characteristic of very gentle flow, in which we see from the velocity profile that the fluid velocity is zero at the pipe wall, and progressively increases to a maximum at a point midway across the pipe.

For calculation purposes, the meanoraveragevelocity is the important value; in a laminar flow situation it would typically be about half the maximum velocity.

Consider now what happens if the flow rate is increased. The laminar profile ismaintained until a certain fluid velocity is reached, at which point eddy currentsstart to appear, indicating a breakdown of the laminar pattern as the layers start to mix, and the onset of turbulent flow.

Turbulence commences near the centre of the pipe, where the velocity is greatest, and spreads towards the pipe wall as the flow rate increases. At the pipe wall a thin layer of laminar flow will survive unless very severe turbulence occurs. A flow pattern exists between the turbulent and laminar regions which is known as theboundarylayer or transitionlayer, as shown in Figure11.

GasFlowMeasurementSection2-FluidFlowPrinciples


The velocity profile is much flatter in a turbulently flowing fluid, as we see in Figure12.

At the boundary layer there is considerable friction between the moving fluid and the static fluid layer, giving rise to the term boundary layer drag. A Telsa pump employs this principle in its design, which is essentially a disc without blades that rotates at a very high speed. The boundary layer provides the friction which allows the disc to impart centrifugal acceleration to the liquid being pumped.

ReynoldsNumberAn indication of whether fluid flow is likely to be laminar or turbulent, or between them, can be obtained by calculating a value called ReynoldsNumber(Re), using the following formula:

Re=DVavg

Where:

D = Internal Pipe Diameter (m) vavg = Average Fluid Velocity (m / s) = Fluid Density (kg / m3) = Fluid Viscosity (kg / m s)

Note that Re is dimensionless, since the units on the right side of the formula cancel each other.

As an approximate guide, values of Re less than 2 000 indicate laminar flow; while values greater than 30 000 indicate turbulence. For intermediate values, the flow would be partially turbulent. Note, however, that this prediction applies to straight sections of pipelines; at elbows, for example, turbulence will occur at lower Reynolds numbers.

Viscosity can be considered simply as an indication of a fluids resistance to flow. Treacle at temperatures below 10C, for example, has a much higher viscosity than water. Gases generally have considerably lower viscosities than liquids, but this is partially compensated for in Reynolds number calculations by their densities also being lower.

Reynolds number is an important factor in flow calculations, and is often incorporated in a quantity called the dischargecoefficient, as we will see later.

TestYourself2.1Find Reynold's number for a process gas, with a viscosity of 1.2 x 10-5 kg / m s and a density of 20 kg / m3, which flows through a 125 mm internal diameter pipeline at an average velocity of 2 m / s.

Predict whether or not the flow is likely to be turbulent.


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FluidFlowPropertiesBernoullisPrincipleDaniel Bernoulli was responsible for considerableadvancement of fluid flow theory by developing aprinciple based on the conservationofenergy.The basis for this principle is that the total energy of the fluid remains constant at all points throughwhich it flows. The main assumptions in thedevelopment of the theory are that: the fluid isincompressible,frictionlessandadiabatic(noheat energy enters or leaves it).

The total energy of a flowing fluid is the sum of the following components:

InternalEnergy(U) can be considered for ourpurposes, without describing its thermodynamicdefinition, simply as a function of the fluidtemperature. If U is defined as the internal energyper unit mass, the total internal energy of the fluid is Um.

PotentialEnergy is the energy the fluid has byvirtue of its position above some reference level. If it is a height h above this reference, its potential energy is mgh, where m is its mass and g is the acceleration due to gravity. Conversely, it is the energy required to propel it to a height h.

PressureEnergy can be considered as a form ofpotential energy in terms of the ability of the fluid to do work, such as driving a piston or impeller. This energy is expressed as the pressure multiplied by the volume (PV). However, we have seen that density () is mass (m) divided by volume (V), i.e.

= m, so V = m and the pressure energy V

therefore equals Pm / .

KineticEnergyis due to the fluids motion, and can be considered as the energy that will be converted to another form, or forms, when it stops moving. It is a function of its mass (m) and average velocity (v)and is calculated from the term mw2 which, along with the potential 2energy expression mgh, you will recognise if you have studied elementary physics.

Adding these terms to express the total fluid energy(E)gives:

E = Um + mgh + mP + mv2 2

Since we will only be concerned with points in a flow line immediately upstream and downstream of flow measuring devices and thus relatively close to each other, we can simplify the expression as follows:

The fluid temperature will be constant, so theinternal energy will not change and the term Um can be discarded.

There will be no significant height differencebetween the points, so the mgh term can beignored.

Removing these terms and dividing by the mass m to get the energy per unit mass (Em) gives:

Em = P + v2 2

Multiplying energy per unit mass by density givespressure.

Em = P+ v2 2

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Here we see that the pressure of a flowing fluid, subjected to the constraints I have described, consists of two components : P is referred to as the

staticpressure, and v2 is called the dynamic 2pressure, because it is associated with the fluid velocity. Clearly, if the fluid is stationary, the total pressure would be P. Using the symbol PT to represent the total pressure:

Em = PT = P + v2 2

We now have an equation with only three physical quantities : pressure, density and velocity. If the fluid density is known, and we measure the total and static pressures, we can calculate the velocity. This is the principle used by the Pitot Tube, which I will describe later.

TheContinuityEquation

Figure13:Velocity/Cross-sectionalAreaRelationship

Figure 13 represents fluid flowing, at an average velocity v, through a pipe of cross-sectional area A.

ThevolumetricflowrateQv is given by:

Qv=Av

You will find that the easiest way to confirm this is to consider the units involved. Using the SI system, A is in m2, and v is in m / s; so multiplying the units gives:

m2xm/s=m3/s, which is volume per unit time.

Alternatively, or additionally, you could consider a particle which is travelling at the average velocity v of the flowstream. If, for example, the velocity is 2 m / s, the particle will move 2 m along the pipe in1 second and the volume of fluid displaced will be2 xA (the volume of a cylinder is calculated by multiplying its cross-sectional area by its length; so the volume of fluid moving along the pipe in one second is, in effect, that of a cylinder of cross-sectional area A and length 2 m).

Themassflowrate Qm is derived from = M so Vm = V, and, replacing V with Qv,

Qm=Qv=vA

We now have two simple formulae relating volumetric and mass flow rates to pipe dimensions and fluid velocities.

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TestYourself2.2Water flows through a 200 mm internal diameter pipeline at an average velocity of 3 m / s. Taking the density of water as 1 000 kg / m3, find the volumetric and mass flow rates.


Now consider a situation in which the pipelinediameter changes, as inFigure14.

Figure14:PipelineDiameterReduction

Here we see fluid flowing at average velocity V1, from the section of pipe with cross-sectionarea A, to the section with area a where its average velocity is v2. To account for compressible fluids which might experience a change in density, we note that the densities are 1 and 2 in the wide and narrow sections respectively.

The continuity equation, as its name might suggest, is based on the principle that the mass flowrate mustbeconstant through all cross-sections of a flowstream. So the mass flowrate in the wide section (1Av1) is equal to the mass flowrate in the narrow section (2av2) :

1 A v1 = 2 a v2 =Qm


EXAMPLE

A liquid flows through a pipeline the diameter of which changes from 150 mm to 75 mm. If the average velocity in the wide section is 0.5 m / s, what will its velocity in the narrow section be ?

Let A and a be the cross-sectional areas of the wide and narrow sections respectively, with v1 the velocity through A and v2 the velocity through a.

Note: We are using / 4 x D2 to calculate cross-sectional area, although you may be more familiar with r2

Av1=av2 x 0.1502 x 0.5 v2 = A v1 = 4 a x 0.0752 4

The s cancel to give: 4 v2=0.01125=2m/s 0.005625

We can also apply the Bernoulli pressure relationship, in which the total pressure is the sum of the static and dynamic elements, to a situation involving a changing flowstream diameter.

Figure 15 : Pipeline Diameter Reduction

Figure15 is silimar to Figure 14, but with the addition of two pressure gauges, P1andP2, which measure the static pressures of the wide and narrow pipe sections respectively.

Consistent with the assumption that energy losses due to friction are negligible, the total pressure PT remains constant; thus we can write:

P1 + v12 = P2 + v22 2 2

Clearly the pressure P2 must be less than P1 to compensate for v2 being greater than v1 and to obey this equation.

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( )

( )

( )( )

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FluidFlowEquationFrom the continuity principle we can express mass and volumetric flow rates in terms of the flowstream dimensions and the fluid densities and velocities. We have been able to calculate flowrates in the preceding example and exercise, but only because we were given the average velocity of the fluid. In practice, accurate measurement of this quantity is difficult due to the susceptibility of the measuring devices to fouling, and other problems; so it is preferable to avoid measuring it directly.

Taking Bernoullis equation:

PT = P + v2 2

and the mass flowrate expression :

Qm = v A

which we rearrange in terms of v :

v = Qm A

Substituting this for v in the Bernoulli expressiongives:

PT = P+ Qm 2 2 A

Applying it to the varying area flow depicted inFigure 15, but assuming that the density does notchange ( 1 = 2 = ), we get:

1+ Qm 2 = 2+ Qm 2 2 A 2 a

Now we have an expression relating the massflowrate to the changeinstaticpressure,flowstream cross-sectional area, and fluid density; no longer requiring velocities. Cross-sectional areas are known, pressures are easily measured and densities can be measured or calculated. This is the principle behind most of the gas measurement devices that will be described in this book.

It is customary to refer to the static pressure change across a measurement device as the differentialpressure, and it is often called deltap, which is written as p. So p = P1 - P2.

When p is substituted for P1 - P2 and the equation is rearranged and simplified, we get:

A2a2 Qm = 2p

A2 _ a2

While it is possible to measure fluid flow by applying this equation to a changing cross- sectional area pipeline configuration as shown in Figures 14 and 15, a high degree of accuracy would not be achieved. In practice, there would be considerable pressure energy loss due to turbulence and friction. As stated when describing Reynolds number, the discharge coefficient term will be introduced to compensate for this.

Gas flow measurement would be considerablyinaccurate from a calculation using the equation as it stands, since compressibility is not accounted for. Again we will see that the equation will be modified by incorporation of a factor to correct this.

These corrections and other modifications to theequation will be described in the next section. Youwill be relieved to know that you are not expected to remember the fluid flow equation, either in this form or when modified. However, you should be able to describe the terms in it, and understand the terms and concepts of the Bernoulli and Continuity principles from which it was developed. One important relationship that you should keep in mind is that the flowrate is proportional to the squarerootofthedifferentialpressure.

Because of the absence of the correction factors, I will not include a calculation exercise at this point; instead, the following exercise will invite you to test your knowledge of the terms and concepts.


TestYourself2.3

1. Which two components does the total pressure of a flowing fluid consist of ?

2. Which quantity is always constant through all cross-sections of a fluid flowstream ?

3. State which quantities are denoted by the following terms and give their SI units

a. A and a b. v (note that this is lower case) c. d. Qv e. Qm f. p

4. If the mass flowrate and the fluid density are known how is the volumetric flowrate calculated ?

5. What is the relationship between the flowrate and the differential pressure ?

You will find the answers inCheckYourself2.3 on page 57.

SummaryofSection2

Fluid flow can be categorised as laminar or turbulent, and Reynold's number can be used to predict which category applies.

Bernoulli's principle of fluid energy conservation can be simplified and expressed in pressure terms, the total pressure being the sum of the static and dynamic components.

Bernoulli's principle, and continuity principle of mass flow conservation, combine to produce a fluid flow formula which does no account for frictional losses.

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TypesandApplicationThe gas flow measurement devices I will discusscan be described as static, in that they have nomoving parts. There is a device which is used forgas flow measurement called a turbinemeter. As its name suggests, it contains a small turbine or fanlike wheel which is rotated by the flowstream, the speed of rotation being a function of the flowrate. The rotational speed of the turbine shaft is converted to an electrical signal which is processed to give an indication of the gas flowrate.

Some models work quite satisfactorily in somesituations, but the device has not been universallyaccepted in applications where high accuracy andprecision are required. For this reason there will be no further description of it in this book.

The static flow measurement devices can becategorised into two groups: velocityhead anddifferentialpressure, the names of which give an indication of their operating principle. I will onlydescribe one velocity head device: the Pitot tube.

PitotTube

I referred to the Pitot tube in the preceding section, when I described pressure as consisting of a static and a dynamic element. This is the device that is normally used to measure the static and dynamic pressure of a flowing gas, from which the velocity is then calculated.

Figure16PitotTube

A diagrammatic sketch of a Pitot tube is shown inFigure16. The main feature is a nozzle (1) pointing in the direction of the flow source and connected to a pressure gauge. A second pressure tapping (2) is subjected to the pressure at the pipe wall, perpendicular to the flow direction, and is thus measuring the staticpressure. The pressure at the nozzle is the total pressure: the static pressure

plus the dynamicpressure, P +v2 2

Clearly, the dynamic pressure is found by subtracting the reading on the static pressure gauge from that on the gauge attached to the nozzle. If we call the dynamic pressure Pd , we can find the velocity from :

Pd = v2 2so v = 2Pd

Note that v is the velocity at the nozzle, not theaverage velocity of the stream. Before the mass orvolumetric flowrate can be evaluated it is essentialto know the relationship between v and the average velocity. We see that, in the example represented by Figure 16, the nozzle is in the centre of the stream, where it measures the maximum velocity. As stated when describing flow patterns, this could be about twice the average velocity if the flow were completely laminar. This, then, is one disadvantage of the Pitot tube; another is that the nozzle is prone to blockage by foreign matter, as would be the case in the majority of natural gas applications. However it should be stated that the device is widely employed with success where the gas is clean and the flow characteristics of the system have been comprehensively determined.

An averaging Pitot tube device is also used in some cases. This has several nozzles spread over a larger area of the flow stream, thereby providingincreased accuracy.

GasFlowMeasurementSection3-MeasurementDevicesandMethods


TestYourself3.1

The pressure gauge attached to a pilot tube nozzle reads 10 m bar g, while the static pressure gauge reads 9 m bar g, and gas density is 3 kg /m3. What is the gas velocity at the location of the nozzle?

Hint: Remember to convert the pressure to Pa (1 m bar - 100 Pa), to get the velocity in m / s. Strictly seaking, you should also use absolute pressure valves, but, since the calculation will involve the difference between two pressures, the same result will be obtained if the gauge values are used.

You will find the answer in CheckYourself3.1 on page 58

We will now look at differential pressure flow measurement devices; their principle being to use the static pressure differential created when the flowstream diameter is reduced.

NozzlesNozzles, an example of which is shown inFigure17, are used in high velocity applications; especially in hostile environments where erosion or corrosion would damage devices such as orifice plates. They produce lower differential pressure at a given flow rate than most other devices.

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VenturiMeterIn the preceding section it was mentioned that energy losses due to friction occur, and will cause a reduction in the total fluid pressure. One objective in the design of a flow measuring device should be to enable the maximum recovery of pressure energy after the fluid has left the measurement location.

The venturi meter is designed to produce a smooth flow, with the minimum turbulence, into and out of the narrow diameter section where the velocity is increased. These features are evident in Figure18.

The advantage of low energy loss is often outweighed by their high cost and space requiredfor installation.

Figure 18 Venturi Meter

DallTube

The dall tube has a similar aerodynamic design to the venturi meter, but is shorter and hence less energy efficient. Its smaller size often makes it a preferred option to the venturi meter. Figure19 shows a typical model.

Figure 19 Dall Tube


TestYourself3.2What is the main difference between the Pitot tube and the other four p flow measurement devices mentioned?


OrificePlate

The simplest, and cheapest, method of creating a restriction in a pipeline is to insert a disc with a hole in it, so that the fluid has to flow through the hole. The concept could hardly be more simple, but it belies the deeper understanding of fluid flow principles that the achievement of accurate and precise flowrate measurements requires.

An orifice plate is mounted between flanged ends in a pipeline, and it is this relative ease of installation and subsequent maintenance that has made orifice plates the most popular gas flow measurement device in commercial applications, especially in the case of natural gas.

It should be noted, however, that the simplicity and low cost are at the expense of the smaller line pressure energy losses enjoyed by devices like the venturi meter. As we would expect, the presence of an orifice plate will create considerably more turbulence than an aerodynamically designed restriction.

We will consider orifice plate design requirements in the next section.

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Orifice Plate Principles

Configuration and Pressure Profile

Figure20 is an orifice plate configuration which also shows the behaviour of the fluid in terms ofstreamlines, as it flows through the measurementregion.

The principle, as we have seen, is that the presence of the orifice reduces the flowstream diameter, and the resulting increase in velocity causes a decrease in static pressure from which the flowrate is calculated.

Note, however, that the maximum velocity is notexactly at the orifice, but is at a distance equal toapproximately half the pipe diameter downstream of it, called thevenacontracta. This effect is mainly due to the inertia of the fluid causing it to continue converging after it passes through the orifice. So the static pressure is slightly lower at the vena contracta than at the orifice, which means that the ideal location for the downstream pressure tapping is at the vena contracta rather than at the orifice plate.

This brings us to a description of the location ofpressure tappings. Figure21 on the next page,shows how the pressure varies at locations on the flowpath.

Figure 20 Orifice Plate Meter


Obviously the best measurement resolution would be obtained from the highest differential pressure(p) between the upstream and downstream regions. In Figure 21, we see that this is between point 1, the corner between the upstream surface of the plate and the pipe wall, and point 2, the vena contracta.

However practical considerations must again prevail; it is easier and cheaper to bore and fit pressure tappings to the flanges instead of the pipe wall, so most industrial installations are fitted with flange tappings. Corner tappings such as would be required to measure the pressure at point 1 are also less convenient than flange ones.

In applications where pipe tappings are used it, is customary to use the DandD configuration. D is the 2pipe internal diameter, and the upstream tapping is located at a distance D from the upstream surface of the plate, while the downstream one is D from the 2downstream surface : at the vena contracta.

The differential pressure, then, measured in the Figure 21 diagram is that between points 3 and 4. We also see that the static component of the line pressure has recovered at point 5, but not to its upstream value. This reflects the pressure energy loss due to turbulence, which it somewhat exaggerated in the diagram. You should also note that a continuous slight pressure reduction is shown to represent the frictional losses that occur in all pipelines.

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Meter Run Specifications

In addition to the orifice plate assembly, the design and configuration of the line in the vicinity of the meter is subjected to certain specifications, particularly in fiscal and other contract situations. The plate assembly and pipeline lengths upstream and downstream of it are referred to as the meterrun.

The objective of these specifications are, as wewould expect, to achieve as smooth and symmetrical a flow of fluid through the meter aspossible. The presence of bends, valves or otherdevices within a certain distance of the meter could cause measurement inaccuracy, so specificminimum lengths of straightpipe, both upstreamand downstream of the plate, are stipulated.

The minimum lengths are quoted as multiples of the pipe diameter, and depend on the types of fitting such as bends, valves, reducers and expanders on the upstream side, and on the diameter of the orifice relative to the pipe diameter. Tables of values are available in international standards publications such as ISO 5167; these range from 5 to 80 times the pipe diameter on the upstream side, and from 4 to 8 on the downstream side.

In some applications where the physical layout ofthe plant makes the minimum straight lengthunattainable, it is possible to install straighteningvanes upstream of the meter, which help to smoothout flow disturbances.

BetaRatio

The beta ratio () is defined as d, where d is the Ddiameter of the orifice and D is the internaldiameter of the pipe. is an important factor inorifice calculations and it is recommended that itshould always be greater than 0.2 and less than0.7 in natural gas applications.

TestYourself3.31. What is the name of the region where the minimum flowstream diameter occurs in an orifice meter, and what is its approximate location?

2. On Figure 21, between which points would the maximum p, and hence best measurement resolution, be obtained?

3. What is the most common location for pressure tappings?

4. Why does Figure 21 show a slight but continuous line pressure drop in the direction of flow, outside the region of the plate?

5. What is the ratio of a meter in which the orifice diameter is 130 mm, and the pipe internal diameter is 250 mm ?



OrificePlateFlowCalculations

TheISO5167Formula

At the end of Section 2 we had the following equation for mass flowrate: Qm = 2p A2a2 A2 - a2

We saw that this equation required to be modified before it could be used for accurate flow calculations, and you will recognise similarities between it and the following flow equation from ISO 5167: Qm = CE d2 2p 4You will be relieved to learn that it is beyond the scope of this book to show how this equation is derived from the preceding one. It looks more simple, but that is because the modifications have been mainly incorporated in the first three terms, which we will now look at.

C is defined as the dischargecoefficient, and is a function of Reynolds number Re and . It can be calculated from a formula, or obtained from tables. ISO 5167 presents discharge coefficient tables for various pressure tapping locations and pipe internal diameters, but if high accuracy is not critical, a value of 0.605 can be used in typical applications.

C is the factor that compensates for the energy losses due to turbulence and friction that were mentioned earlier, and it is interesting to note that a typical value for the venturi tube is 0.98, thus reflecting its more aerodynamically efficient design.

E is called the velocity of approach factor. It has enabled us to eliminate the A and a pipe and orifice cross-sectional area terms since it employs the 13 factor in the following formula: E = 1 - 4

, termed the expansion factor, is important in gas measurement since it accounts for the compressibility, and hence density change, of gases when their pressure changes as they flow through the meter. Liquids, being essentially incompressible, have an factor of 1, which means that it can be ignored. For gases, it is obtained from the following formula:

= 1 - p (0.41 +0.35 4)

PYThe only term that may be unfamiliar to you is Y,which is the specific heat ratio of gases. It is given by: Y = Cp Cv

where Cp is the specific heat at constant pressure, and Cv is the specific heat at constant volume. These values vary between gases, and typical values of Y are 1.4 for air and 1.3 for methane.

Again, you are not expected to remember this formula for , but you should note that its value increases (approaches 1) as the line pressure P increases; which confirms that gases become less compressible as their pressure and density increases. You should also be careful to avoid confusing with Z, the non-ideal gas behavioural factor which was described in an earlier section.

Qv, the volumetric flowrate, is found by dividing the mass flowrate Qm by the gas density at the reference pressure and temperature:

Qv = Qm

We saw in Section 1 that:

= PMw ZRT

So, if the volumetric flowrate is required to be referred to standard conditions, P = 1.013 bar, T = 288 K and Z = 1.

This expression can also be used to obtain a value for , the fluid density upstream of the orifice, in the ISO equation, so non-ideal gas behaviour is accounted for here. Clearly, the P, T, Z and Mw values would be the prevailing ones upstream of the meter. It is fairly common nowadays, however, to measure the gas density directly.

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42

At the end of Section 2, I asked you to take particular note of the relationship between the flowrate Qm and p, in that Qm is proportional to the squareroot of p. This relationship, of course is maintained in the ISO 5167 equation, and has an important implication with regard to the resolution to which flow measurement can be made.

Figure22 demonstrates the problem associatedwith this square root relationship. Were therelationship linear, the scale on the left would apply and the resolution to which it is read is constant over the complete range. With the square root scale, however, we see that high resolution is available on the upper region, but it deteriorates lower down and is extremely poor near the bottom. For this reason a range switching facility is recommended, and stipulated in fiscal systems, so that low flowrate measurements can be measured to greateraccuracy.

Orifice Plate Size Selection

Reliable flow measurement requires an adequatep value. If the flowrate is reduced, the differential pressure across the meter will also be reduced, and, if it falls below a certain value, the reliability of the flowrates calculated from it will decrease. If the flowrate is anticipated to remain near this value for a considerable time, the plate should be replaced with one with a smaller diameter, which will increase the differential pressure.

Conversely, excessively high flow rates will lead to unreliability, as well as creating undue line pressure loss due to excessive friction. In such cases, a larger diameter plate would be fitted.


EXAMPLE

An orifice plate meter, with an internal pipe diameter of 200 mm, has been operating satisfactorily with a ratio of 0.65. The gas flow through the line is to be reduced to a rate which will not produce an adequate p signal. Calculations predict that p would be satisfactory if is reduced to about 0.45. What is the diameter of the plate which is currently fitted, and what is the diameter of the one that will replace it ?

= dD-

d = D = 0.65 x 200 = 130mm

To get a beta ratio of 0.45:

d = 0.45 x 200 = 90mm

You should note that the numbers in this example happen to give round figures for the plate diameter. Orifice plates are only available in certain sizes, so it may not always be possible to achieve a desired value exactly; hence my use of about 0.45.

AutomatedCalculationMethodsHaving seen the large number of data necessary to calculate flowrates, you will appreciatethat longhand methods of performing the calculations are rather tedious and time consuming. Fortunately, recent advances in electronic technology have alleviated this problem. Differential pressure, line pressure, temperature and density measurement signals can be digitised and processed electronically to yield flowrate readouts directly.

We will look at the general layout of such systems in the next section.

TestYourself3.41. Which two factors are used to obtain the discharge co-efficient C?

2. What is the factor E called. Given that E = 1-4, what is its value for a meter in a 250 mm internal diameter pipe, with an orfice diameter of 110 mm ?

3. Why is it , the expansion factor, important when measuring gas flowrates ?

4. Why is it desirable to have a range switching facility on instruments that read gas flow measurements obtained from orifice meters ?

5. It has been decided to increase the gas throughput in a 250 mm internal diameter pipeline which contains an orfice meter operating with a beta ratio of 0.44. The increased rate would create a higher than necessary differential pressure across the plate in addition to causing excessive energy loss due to friction. It is predicted that increasing the beta ration to about 0.65 will alleviate the problem. What is the diameter of the plate to be removed and of the plate which will replace it, given that plates are only available with diameters in multiples of 10 mm?

You will find the answers in CheckYourself3.4 on Page 59.

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The following exercise might look a bit formidable,but is actually not too difficult since you are givenmost of the formulae and the numbers you will need, so it is really a question of inserting them as appropriate.

TestYourself3.5Given the following:

Qm = CE d2 2p 4

E = 1 - 4

= PMw ZRT

d = 125mmD = 257.4 mmp = 100mbar (10000 Pa)

Line pressure P = 24 bar gLine temperature T = 59CMw = 22.7 kg / kg-molZ = 0.937C = 0.605 = 0.9987

a. Evaluate E and p, then calculate Qm, the mass flowrate of the gas in kg/s.

b. Find Qv, the volumetric flowrate in standard (1.013 bar, 15C) cubic metres per hour.

You will find the answers in CheckYourself3.5on page 60.


SummaryofSection3The main features of the following devices were described:

Pitot tube flow nozzles venturi dall tube orifice plate

The Pitot tube is the only velocity head device, the others being dependent on differential pressure measurement.

The orifice plate was identified as the most popluar gas flow measurement device, and the rest of the section was devoted to describing its effects on flowing fluid and how these effects could be used to measure flowrates. The description showed that:

The maximum fluid velocity, and hence minimum static pressure, is at the vena contracta.

Flange-mounted pressure tappings are the most popular type, although they do not tap into the regions of the meter at which the maximum differential pressure occurs.

The ISO 5167 formula is used for orifice and contains terms which correct for fluid frictional energy losses and gas compressibility.

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TypesofPlate

PlateGeometryVariationsOrifice plates are manufactured with variousgeometrical designs, examples of which are shown in Figure23, to suit different applications.

Concentric is the most common type, because the central location of the orifice allows a symmetrical flow pattern.

The segmental type is mainly for slurries, and the eccentric plate is for gas streams which may contain liquid. The hole, being at the bottom of the plate, allows liquid to flow through instead of being trapped, as it would be by a concentric plate. A variation on this is a concentric plate with a small hole near the bottom.

In addition to variations to the location of the orifice on the plate, there are different designs of orifice edge, two of which are shown in Figure24.

The most common type is the square edge withbevel, and you should note that the plate must beinstalled with the bevelled edge facing downstream. The only exception to this rule is aplate designed for special applications, which isknown as a conical entrance plate.

The square edge profile is used in applicationswhere the facility to measure flow in either direction is required. Clearly, its shape will make it slightly less aerodynamically efficient than the bevelled edge plate.

As you would expect, the location of the orificesand the design of their edges will affect the flowcharacteristics; so the appropriate factors, mainlyC and , have to be adjusted accordingly.

It is very important that plates which been damaged are replaced, even if the damage is only slight. Entrained high velocity solid, or even liquid, particles will wear the sharpness off the leading edge; the permissible wear is less than that which is visible to the naked eye. Other damage which can occur is buckling due to excessive temperatures, or high pressure surges.

GasFlowMeasurementSection 4 - Orifice Plate Metering Equipment


SeniorOrificePlateFittingWe have seen that orifice plates are normally fitted between flanges, so a plate changout demands the inconvenience of depressurising and purging the meter run. To alleviate this, the Senior fitting assembly was produced.

Figure25 is a sketch of the basic design which comprises an orifice plate carrier which is moved by a handle operated rack and pinion arrangement, and a chamber situated above the plate.

The following description is intended as an explanation of the operating principle of the system, and it is important that you do not regard it as an operatingprocedure. If your work involves the operation of Senior fittings, then you must follow the procedures pertinent to your installation.

This explanation applies to a unit fitted to a hydrocarbon gas system.

1. The chamber is purged with inert gas and pressurised to the pipeline pressure.

2. The carrier and plate are retracted into the chamber, which is then sealed from the process stream, depressured and purged.

3. The plate is removed from the carrier and chamber, and the replacement plate is inserted.

4. The chamber is sealed and purged with inert gas and pressurised to the pipeline pressure.

5. The plate and carrier assembly is lowered into the meter run.

6. The chamber is safely vented and purged.

In theory, this method can be followed without the gas flow being stopped. In practice, however, many plant operations managers doubt that it is as safe as a proper line isolation, depressurisation and purge procedure. Nonetheless, even if the meter section is isolated, depressurised and purged, the Senior device saves plate changeout time by eliminating the work and problems that can be associated with flanges. 47


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SensingDevices

Line pressure and temperature sensing devices are widely used throughout industry, and the types used in gas flow metering systems are the same as those for other applications. We will therefore confine our attention to the devices that measure the orifice plate differential pressure and the gas density.

Electronic measuring and signal processing devices are becoming increasingly predominant. Advancement in the design of intrinsically safe circuitry has allowed them to replace the pneumatic instrumentation that was traditionally the only viable means of measurement and control in hazardous environments. However, this conversion process is by no means universal, and pneumatic systems are still in use, particularly on older plant installations.

DifferentialPressure

I will describe two differential pressure sensors, one pneumatic and one electrical.

The pneumatic one is the torquebalancetransmitter, and a simplified sketch is shown in Figure26.

The main feature is a cell containing aliquid-filled twin-walled diaphragm capsule, which is subjected to the upstream pressure on one side and the downstream pressure on the other.

The position of the centre of the capsuledepends on the pressure difference between its sides; an increase in flowrate,for example, will reduce the pressure onthe downstream side thus causing thediaphragm centre to be displaced to theright.

A force bar, which is linked to the capsule,transfers the movement to a flapper/ nozzle arrangement in the signal conditioning system, which is shownsimply as a block diagram. The movementis converted to a pneumatic output signalthat can be measured directly or, morecommonly nowadays, converted to anelectrical signal first.

The electrical device is the capacitivedifferential pressure transmitter, a sketchof which is shown in Figure27, on thenext page.


The upstream pressure is transmitted to the isolating diaphragm on one side of the cell, and the downstream pressure to the other side. Movement of the isolating diaphragms is transferred to the sensing diaphragm by silicone oil, and any change in the position of the sensing diaphragm changes the electrical capacitance between it and the capacitor plates. The resultant electrical signal is converted to a current reading in the 4 - 20 mA DC range, which can then be processed to produce the required analogue or digital output.

GasDensityA device which is often used to measure gas density directly uses thevibratingcylinderprinciple. The process gas is passed over the inner and outer surfaces of a thin metal cylinder, which, like all solid objects, has a natural or resonant vibration frequency. However, the vibration frequency of the cylinder is affected by the gas molecules which interact with its surface and vibrate with it. The significant property, here is the mass of gas, the relationship being that increasing the mass will decrease the vibration frequency. This means that, if the frequency is measured, it can be used to evaluate the gas density.

Figure28, on the next page, is a schematic diagram of the system. The cylinder requires an activating signal to make it vibrate; this is achieved by passing current through the cylinder activating coil. The vibration frequency is picked up as an alternating current by the sensing coil, and this signal is amplified and processed to produce a square wave output. This type of signal is ideal for transmission and can be readily converted to an analogue or digital form.

The frequency range of a typical sensing system is:4900 Hz at a gas density of 0 kg / m3 (no gas present) to 3900 Hz at 60 kg / m3. Units are also available which are capable of measuring densities up to 400 kg / m3.

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MeteringStations

TypicalMulti-StreamSystemMetering stations to which fiscal standards apply, and others in applications where maintenance of production rates is essential, comprise two or more meter runs; thus allowing fluid flow to continue while a meter run is shut down for maintenance.

Figure29 is a schematic diagram showing the main components of a two stream metering station with a computer based electronic signal processing system.

We see that each stream has its own computer, and they are connected to the main or station computer. The operator interface with the system is the input / output terminal, which displays live or recorded data and through which entries such as changes to meter factors are made.

The temperature, pressure, density and differential pressure transducers transmit their data to their stream computer where the signals are processed and the computations are done and sent to the station computer.

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An additional feature shown in Figure 29 is a facility to measure the gas specific gravity, upstream of the meter runs. This is an option that is sometimes used so that a more accurate measurement of the density is obtained. The sensing element in the specific gravity transducer is usually a vibrating cylinder, as in the case of the density sensor, but with the facility to incorporate a reference gas.

In addition to mass and volumetric flowrates,measurements of line pressure, temperature anddensity can also be displayed. A fiscal requirement is that flow totalisers are used so that the total mass or volume of gas for a period of time can be measured.

This description has only outlined the basic facilities of an automated metering station. Many installations have additional features such as the ability to transmit data to central monitoring sites.

The following is an example of some of the mainmetering station design requirements that wouldlikely be agreed between the relevant parties, which would include the UK Department of Energy, in a fiscally controlled contract. You should note that this does not include specifications applicable to hardware such as the orifice plates, the meter pipes and the installation of measuring elements. Please regard this as an outline of some of the main clauses in a typical agreement in the UK, and not as a specific contract. If you work on a fiscal metering system you should try to learn as much as possible about the terms of the contract.

1. The gas will be measured in either volume or mass units, depending on the agreement between the interested parties. Volumes will be measured in cubic metres and mass in tonnes. Volume measurement will be referred to the metric standard conditions of 15Celsius and 1.01325 bar.

2. While in most cases gas density will be measured directly using a density transducer, in some instances it may be calculated, by an agreed method, from a knowledge of the composition of the gas together with the measured operating pressure and temperature.

3. Enough meter runs will be provided to ensure that at least one standby meter will be available, at the design production rate. Isolation valves will be provided so that individual meters can be removed from service without shutting down the metering system.

4. All computing functions will be done by a digital microprocessor based flow computer, one of which will be allocated to each meter run.

All the constants and factors which are held in the flow computer will be accessible for inspection in a general display register and it will be possible to modify these values, with authorisation, after overriding some form of security lock.

5. Totalisers on individual meter runs and on station summators will have sufficient digits to prevent cycling occurring more frequently t

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Gas Flow Measurement.pdf