Gas Flow Measurement.pdf

POL Petroleum Open Learning

Gas Flow Measurement Part of the Petroleum Processing Technology Series

OPITO THE OIL & GAS ACADEMY

POL Petroleum Open Learning

Gas Flow Measurement Part of the Petroleum Processing Technology Series

OPITO THE OIL & GAS ACADEMY

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.

Gas Flow Measurements

Petroleum Open Learning

(Part of the Petroleum Processing Technology Series) As a large part of this subject deals with calculations, you will require to be comfortable with maths up to about standard grade level.

Visual Cues

Although some of the equations are fairly complex, all relevant data and information is provided to assist you to solve the problems.



training targets for you to achieve by the end of the unit



test yourself questions to see how much you understand



check yourself answers 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

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 calculations You will also find that a scientific calculator will be useful for this programme.

Contents

Page

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Training Targets

4

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Introduction

5

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Section 1



The Need for Accurate Measurement Units of Measurements

- Gas Flow Measurement Applications and Gas Physics



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Contents (cont’d) *

Section 2 - Fluid Flow Principles



Fluid Flow Principles Fluid Flow Properties



*

Section 3 - Measurement Devices and Methods



Types and Applications Orifice Plate Principles Orifice Plate Flow Calculations

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Section 4 - Orifice Plate Metering Equipment



Types of Plate Sensing Devices Metering Stations Safety Implications



*

Check Yourself - Answers

Page 26

34

Visual Cues training targets for you to achieve by the end of the unit

test yourself questions to see how much you understand

check yourself answers to let you see if you have been thinking along the right lines

46 activities for you to apply your new knowledge

54

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

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Training Targets When 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 Reynold’s number and, given Reynold’s equation, define the terms in it and use it in a calculation. • Describe Bernoulli’s 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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Gas Flow Measurement Oil and Gas Separation

Introduction Introduction

Systems

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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 : Section 1, Gas Flow Measurement Applications and Gas Physics, 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. Section 2, Fluid Flow Principles, presents the concepts of laminar and turbulent fluid flow, develops the Bernoulli and continuity principles to produce a simple flowrate / differential pressure relationship. Section 3, Measurement Devices and Methods, 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. Section 4, Orifice Plate Metering Equipment, 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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Gas Flow Measurement

Section 1 - Applications and Gas Physics The Need for Accurate Measurement

Gas Sales Contracts

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

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

Oil Field Evaluation A 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.

Process Control & Optimisation The 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 antisurge protection.

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Hydrocarbon Taxation In 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.

Fiscal Standards In 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.

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

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Units of Measurement It 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.

Boyle’s Law Figure 1 illustrates the principle behind Boyle’s Law, which describes the relationship between the volume occupied by a given mass, or number of molecules, of gas and its pressure, while the temperature remains constant. 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. Boyle’s Law expresses this formally with the statement that: At constant temperature, the absolute pressure of a given mass of an ideal gas is inversely proportional to the volume.

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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 that atmospheric pressure 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 is bar absolute (bara). • 1.

2.

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

The gas molecules are small spheres, the volumes of which are negligible in comparison to the volume that the gas occupies. 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 Boyle’s Law statement that gas pressure (P) is inversely proportional to its volume (V) can be written as P is proportional to 1v Which means that the pressure is equal to a constant (k) divided by the volume.

Thus : P = kv so 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 : and HENCE

P1 V1 = k P2 V2 = k P1 V1 = P2 V2

which is the mathematical expression of Boyle’s 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 same temperature.

Charles’ Law Charles’ Law describes the relationship between the volume and temperature of an ideal gas, while the pressure is kept constant. As in the case of Boyle’s Law, we can use a cylinder /piston arrangement to demonstrate the principle behind Charles’ Law, as shown in Figure 2. 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.

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 = 3.33 bar 3

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

Figure 3 : Volume / Temperature Graph (Celsius Scale Only)

Figure 4 : Volume / Temperature Graph (Celsius and Kelvin Scale)

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 0°C 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.

Figure 4 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 -273.15°C. Absolute Zero 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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Absolute zero is the lowest point on the absolute temperature scale, which is measured in units called Kelvin (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 the Kelvin and Celsius scales are the same. So a one degree Celsius temperature change is also one Kelvin. (A convention, which is by no means universally applied, is to omit the term degree when using absolute temperature units). O°C, then, is 273.15 K, and 100°C 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 -459.67°F, and a degree on the Rankine scale is the same size as a degree on the Fahrenheit scale. In most practical situations sufficient accuracy is achieved by using 273 as the conversion factor between Celsius and Kelvin, and 460 between Fahrenheit and Rankine. However, where high accuracy is required, such as in fiscal gas flow measurement, the more exact values should be used.

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 We see in Figure 5 that if the absolute temperature is doubled, the gas volume will also be doubled. Charles' Law, then, states that: at constant pressure, the volume occupied by a given mass of gas is proportional to its absolute temperature.

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 new volume or temperature of an ideal gas for a change in which the pressure stays constant.

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Combined Ideal Gas Law Boyle's and Charles' laws combine to give the equation:





P1V1 = P2V2 T T 1 2

We will apply it in the following examples:

Test Yourself 1.1 2 Ltr of an ideal gas at 10˚C and 2 bar is compressed to a volume of 0.5 Ltr. Given that the heat of compression raises its temperature to 25˚C, what will its pressure be?

A 5 Ltr sample of gas at 15˚C and 3 bar a is heated to 45˚C 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





T 1

T2

You will find the answer in Check Yourself 1.1 on page 54.

Molecular Mass

we will ascribe the initial conditions to left side, and the Molecular mass is a physical property of all substances. A comprehensive description of it can be found in 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 = 11.04 Ltr

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 one carbon atom bonded to four hydrogen atoms. Carbon has an atomic mass of 12 atomic mass units (a.m.u.). Hydrogen has an atomic mass of 1 a.m.u. The molecular mass of methane is the sum of the masses of its constituent atoms, which is therefore 12 + (4 x 1) = 16 a.m.u.

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Although it may not be scientifically correct, the terms molecular mass and molecular weight (Mw) are often used interchangeably. The very small size of atoms and molecules makes calculations using their individual masses inconvenient. A more practical approach is to consider the mass of a large number of them, and this involves the concept of the mole. The mole, usually written as mol, or grammemole (g-mol) is defined as the atomic or molecular mass of a substance expressed in grammes. The number of moles of substance is usually assigned the symbol n, and is easily calculated for a given mass (m) of material using the relationship:

n= m Mw

EXAMPLE Calculate the number of kg-mol in 40 kg of methane. We have already seen that the molecular mass of methane is 16. n = m = 40 = 2.5kg-mol Mw 16

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

n= m equation. Hence: Mw



n(kg-mol) Mw







= m(kg) Mw

= m(kg) n(kg-mol)

So, in this case, the unit of molecular mass is kg/kg-mol. It may also be written as kg kg-mol-1 It 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.

Test Yourself 1.2 Calculate the molecular mass of ethane, which is a molecule comprising two carbon and six hydrogen atoms.

Having seen how the molecular mass of pure substances 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 molefractions, 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.

You will find the answer in Check Yourself 1.2 on page 54.

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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 Methane Ethane

Mol. Mass

Mol. Fraction

Mol. mass x Mol.Fraction

16

0.80

12.80

30



0.20

Mixture Mol. Mass

6.00

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.

Test Yourself 1.3 Calculate the molecular mass of a mixture comprising 60% methane, and 40% ethane.

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.

Gas Constants The form of the ideal gas equation we looked at earlier : implies that

PV = a constant 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:

You will find the answer in Check Yourself 1.3 on page 54.

P1V1 = P2V2 T1 T2

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.

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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

EXAMPLE

Because it applies to all ideally behaving gases, R is termed the universal gas constant. 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.

We first find the number of moles of air using:

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



n = m = 1.5 = 0.0517 kg-mol Mw 29

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







P

V kPa m3 bar m3

T

K K

Ltr bar K bar cm3 K oR psia ft3

n

kg-mol kg-mol

g-mol

g-mol

R



1.5 x V = 0.0517 x 0.0831 x 298

0.0831



V = 1.280 = 0.853 m3 1.5

8.130

0.0831 83.1

Ib-mol 10.73



















Test Yourself 1.4



How many kg of methane will, when behaving ideally, occupy a volume of 1 m3 at 2 bar and 20°C?



You will find the answer in Check Yourself 1.4 on page 54.

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Reference Pressures and Temperatures Gas 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 reference conditions for gas volumetric measurements. The values of reference pressures and temperatures may vary between countries and contracts, but the most common ones are

1.

1.013 bar ;

15°C





2.

1.013 bar ;

0°C





3.

14.73 psia ; 60°F

The first one is usually termed Standard Pressure 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 0°C. 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 15°C and 1.013 bar.

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 106 sm3 / d or 106 sft3 / 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 40°C 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.

EXAMPLE

What is the actual volume of 5m3 (st) of an ideal gas at 3 bar and 25°C. We will use

P1V1 = P2V2 T1 T2

and assign the standard conditions to the left side. Hence:

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

and we are required to find V2.

1.013 x 5 288

=



0.01007 X V2 =

V2 = 0.01759 = 0.01007

3 X V2 298 0.01759 1.747 m3

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Non-Ideal Gas Behaviour In our work so far, we have repeatedly used the term 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 compressibility factor. The availability of this factor is due to the large amount of empirical data that has been gleaned from extensive experimental work over many years, particularly in the natural gas industry. The compressibility 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 in Figure 6.

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.

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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 that Z = 1 at low pressure, 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 - 400°C 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 300°C 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 Figure 7. 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.

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EXAMPLE

Activity Use Figure 7 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. 0.88 2. 0.36 3. 0.74

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



a) ignoring non-ideal behaviour

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



b) accounting for non-ideal behaviour.



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

a) Assuming ideal behaviour, we simply apply

PV = nRT



PV = ZnRT V = ZRT P

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

= 0.68 x 0.0831 x 238 70

V = RT P



V = 0.0831 x (-35 + 273) 70







= 19.78 70

V =

0.283 m3

= 13.449 70 V = 0.192 m3

We see that, in this case, failure to consider the non-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%.

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EXAMPLE

Test Yourself 1.5 Using Figure 7 to find the compressibility factor, calculate the volume of 32 kg of methane at 100 bar and -50°C. Estimate the percentage error you would have incurred by assuming the gas to behave ideally.

You will find the answer in Check Yourself 1.5 on page 54. We have already seen that, to perform gas calculations involving changes in conditions from P1, V1 and T1 to P2, V2 and T2 , we use:

P1V1 = P2V2 T1 T2

You will now be aware that, unless the gas is ideal at both sets of conditions, corrections will have to be applied.

We have also seen that the compressibility factor varies with pressure and temperature, so different values of Z are likely to be required for each 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

Find the volume occupied by 3m3 of methane at 65 bar and 20°C, after it has been compressed to 150 bar and chilled to -30°C. 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 0.88 x 293

=

150 x V2 0.61 x 243

1.012 V2 = 0.756

V2 = 0.756 = 0.747 m3 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 V we 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.

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It should now be clear that, if Z is greater than 1, 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.

Figures 8 and 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.

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 molecular mass 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 average molecular mass of the mixture.

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Comparing Figures 8 and 9, it is clear that, except at temperatures well above 100°C, the deviation from ideal gas behaviour is considerably greater for the heavier gas. For example, at 100 bar (10,000 kPa) and 5°C, 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 laws and 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.

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Test Yourself 1.6 1. 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 Check Yourself 1.6 on page 55.

Density

Example

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

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

Density is defined as mass divided by volume, and is usually given the Greek symbol ρ. So

ρ=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,



ρ = PMw = ZRT

1.013 x 16 0.0831 x 288



ρ = 0.678 kg / m3

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

Test Yourself 1.7

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

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

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





m = PMw = ρ V ZRT

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

Ethane (Mw Propane (Mw Butane (Mw

= = =

30) 44) 58)

You will find the answers in Check Yourself 1.7 on page 56.

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Specific Gravity Another method of expressing the density of a material is to use the measurement specific gravity (s.g.). Specific gravity is sometimes referred to as relative density, which is an apt term since it is defined as the density of the substance being evaluated divided by the density of a reference substance. In the case of liquids, the reference material is water; for gases it is air. So for gases we can write:



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



sgg = Mw g Mw a



.˙.

sgmethane = 16 = 0.55 29

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

s.g. = ρg

ρ

a

where the subscripts g and a denote the gas being evaluated and air respectively. To avoid having to account for non-ideal gas behaviour, measurements are usually referred to standard pressure and temperature. Consider the specific gravity of methane:

sgmethane = ρ methane ρ a

As the density (ρ) of both methane and gas are calculated at standard pressure and temperature, we need only use the molecular mass (Mw) for gas and air in the equation.

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Summary of Section 1 Applications in which accurate gas flow measurement is required are :

• • • •

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

Boyle’s 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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Gas Flow Measurement

Section 2 - Fluid Flow Principles 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.

Fluid Flow Characteristics

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Figure 10 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.

The Streamline Concept

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

The construction of streamlines in a diagram is designed to enable a visual image of fluid flow characteristics to be achieved, In simple terms, streamlines are drawn such that adjacent lines represent different fluid flow speeds.

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

Flow Types : Laminar and Turbulent

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 the boundary layer or transition layer, as shown in Figure 11.

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The velocity profile is much flatter in a turbulently flowing fluid, as we see in Figure 12.

Reynold’s Number An indication of whether fluid flow is likely to be laminar or turbulent, or between them, can be obtained by calculating a value called Reynold’s Number (Re), using the following formula: Re = ρDVavg µ Where:

D vavg ρ µ

= Internal Pipe Diameter (m) = 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.

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.

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 Reynold’s numbers.

Viscosity can be considered simply as an indication of a fluid’s resistance to flow. Treacle at temperatures below 10°C, for example, has a much higher viscosity than water. Gases generally have considerably lower viscosities than liquids, but this is partially compensated for in Reynold’s 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 discharge coefficient, as we will see later.

Test Yourself 2.1 Find 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. You will find the answers in Check Yourself 2.1 on page 57.

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Fluid Flow Properties Bernoulli’s Principle Daniel Bernoulli was responsible for considerable advancement of fluid flow theory by developing a principle based on the conservation of energy. The basis for this principle is that the total energy of the fluid remains constant at all points through which it flows. The main assumptions in the development of the theory are that: the fluid is incompressible, frictionless and adiabatic (no heat energy enters or leaves it). The total energy of a flowing fluid is the sum of the following components: Internal Energy (U) can be considered for our purposes, without describing its thermodynamic definition, simply as a function of the fluid temperature. If U is defined as the internal energy per unit mass, the total internal energy of the fluid is Um. Potential Energy is the energy the fluid has by virtue 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.

Pressure Energy can be considered as a form of potential 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 (ρ) is mass (m) divided by volume (V), i.e. þ = m, so V = m and the pressure energy V ρ

therefore equals Pm / ρ. Kinetic Energy is due to the fluid’s 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 2 energy 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 ρ 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 the internal energy will not change and the term Um can be discarded. There will be no significant height difference between the points, so the mgh term can be ignored. Removing these terms and dividing by the mass m to get the energy per unit mass (Em) gives:

Em = P + v2 ρ 2

Multiplying energy per unit mass by density gives pressure. ρ Em = P+ þv2 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

The Continuity Equation

static pressure, and ρv2 is called the dynamic 2 pressure, 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:

The mass flowrate Qm is derived from ρ = M so V m = ρV, and, replacing V with Qv,

ρ Em = PT = P + ρv2 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.

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 in 1 second and the volume of fluid displaced will be 2 x A (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).

Figure 13: Velocity / Cross-sectional Area Relationship Figure 13 represents fluid flowing, at an average velocity v, through a pipe of cross-sectional area A.



Qm =ρ Qv = ρ v A

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

The volumetric flowrate Qv 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: m2 x m/s = m3/s, which is volume per unit time.

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Test Yourself 2.2

Now consider a situation in which the pipeline diameter changes, as in Figure 14.

Water 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.

You will find the answer in Check Yourself 2.2 on page 57.

Figure 14 : Pipeline Diameter Reduction Here we see fluid flowing at average velocity V1, from the section of pipe with cross-section area 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 ρ1 and ρ2 in the wide and narrow sections respectively. The continuity equation, as its name might suggest, is based on the principle that the mass flowrate must be constant through all cross-sections of a flowstream. So the mass flowrate in the wide section (ρ1 A v1) is equal to the mass flowrate in the narrow section (ρ2 a v2) :

ρ1 A v1 = ρ2 a v2 =Qm

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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 π/ 4 x D2 to calculate crosssectional area, although you may be more familiar with π r2



Av =av 1 2

v2 = A v1 = a

π x 0.1502 x 0.5 4 π x 0.0752 4

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

Figure 15 : Pipeline Diameter Reduction Figure 15 is silimar to Figure 14, but with the addition of two pressure gauges, P1 and P2, 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 + ρv12 = P2 + ρv22 2 2

We can also apply the Bernoulli pressure relationship, in which the total pressure is the sum Clearly the pressure P2 must be less than P1 to compensate for v2 being greater than v1 and to obey of the static and dynamic elements, to a situation this equation. involving a changing flowstream diameter.

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Fluid Flow Equation From 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 Bernoulli’s equation: PT = P + ρv2 2 and the mass flowrate expression :

Qm = ρv A

which we rearrange in terms of v :

v = Qm ρA

Substituting this for v in the Bernoulli expression gives:

( )

PT = P+ρ 2

Qm 2 þA

Applying it to the varying area flow depicted in Figure 15, but assuming that the density does not change ( ρ1 = ρ2 = ρ ), we get:

( )

ρ1+ρ Qm 2 = ρ2+ρ 2 ρA 2

( ) Qm ρa

2

Now we have an expression relating the mass flowrate to the change in static pressure, flowstream cross-sectional area, and fluid density; no longer requiring velocities. Crosssectional 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 differential pressure, and it is often called delta p, 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: Qm =

�

(

2∆pρ

A2a2 A2 _ a2

)

While it is possible to measure fluid flow by applying this equation to a changing crosssectional 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 Reynold’s number, the discharge coefficient term will be introduced to compensate for this. Gas flow measurement would be considerably inaccurate 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 the equation will be described in the next section. You will 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 square root of the differential pressure. 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.

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Test Yourself 2.3 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. b. c. d. e. f.

A and a v (note that this is lower case) ρ Qv Qm ∆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 ?

Summary of Section 2 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.

You will find the answers in Check Yourself 2.3 on page 57.

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Gas Flow Measurement

Section 3 - Measurement Devices and Methods Types and Application

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 :

The gas flow measurement devices I will discuss can be described as static, in that they have no moving parts. There is a device which is used for gas flow measurement called a turbine meter. 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 some situations, but the device has not been universally accepted in applications where high accuracy and precision are required. For this reason there will be no further description of it in this book. The static flow measurement devices can be categorised into two groups: velocity head and differential pressure, the names of which give an indication of their operating principle. I will only describe one velocity head device: the Pitot tube. Pitot Tube 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.

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so

Figure 16 Pitot Tube

A diagrammatic sketch of a Pitot tube is shown in Figure 16. 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 static pressure. The pressure at the nozzle is the total pressure: the static pressure plus the dynamic pressure, P +ρv2 2

Pd = ρv2 2 v=

�

2Pd ρ

Note that v is the velocity at the nozzle, not the average velocity of the stream. Before the mass or volumetric flowrate can be evaluated it is essential to 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 providing increased accuracy.

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Test Yourself 3.1 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.

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.

Nozzles Nozzles, an example of which is shown in Figure 17, 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.

You will find the answer in Check Yourself 3.1 on page 58

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Venturi Meter

Dall Tube

In 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 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. Figure 19 shows a typical model.

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 Figure 18. The advantage of low energy loss is often outweighed by their high cost and space required for installation.

Figure 18 Venturi Meter

Figure 19 Dall Tube

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Orifice Plate 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.

Test Yourself 3.2 What is the main difference between the Pitot tube and the other four ∆p flow measurement devices mentioned? You will find the answer in Check Yourself 3.2 on page 58.

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 Figure 20 is an orifice plate configuration which also shows the behaviour of the fluid in terms of streamlines, as it flows through the measurement region. 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 not exactly at the orifice, but is at a distance equal to approximately half the pipe diameter downstream of it, called the vena contracta. 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.

Figure 20 Orifice Plate Meter

This brings us to a description of the location of pressure tappings. Figure 21 on the next page, shows how the pressure varies at locations on the flowpath.

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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 D and D configuration. D is the 2 pipe 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 2 downstream 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

Beta Ratio

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 meter run.

The beta ratio (ß) is defined as d, where d is the D diameter of the orifice and D is the internal diameter of the pipe. ß is an important factor in orifice calculations and it is recommended that it should always be greater than 0.2 and less than 0.7 in natural gas applications.

The objective of these specifications are, as we would expect, to achieve as smooth and symmetrical a flow of fluid through the meter as possible. The presence of bends, valves or other devices within a certain distance of the meter could cause measurement inaccuracy, so specific minimum lengths of straight pipe, both upstream and 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 of the plant makes the minimum straight length unattainable, it is possible to install straightening vanes upstream of the meter, which help to smooth out flow disturbances.

Test Yourself 3.3 1.

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

You will find the answers in Check Yourself 3.3 on page 58.

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Orifice Plate Flow Calculations The ISO 5167 Formula At the end of Section 2 we had the following equation for mass flowrate: Qm = 2∆pþ 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 = CEℇ π d2 2∆p ρ 4 You 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 discharge coefficient, and is a function of Reynold’s number Re and ß. 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 13ß factor in the following formula:

E=

1 - ß4

ℇ, 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 ℇ factor of 1, which means that it can be ignored. For gases, it is obtained from the following formula: ℇ = 1 - ∆p (0.41 +0.35 ß4)

PY

The 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 ℇ, 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 ℇ 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 ρ

We saw in Section 1 that:



þ = 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 ρ, 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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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 square root 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.

Figure 22 demonstrates the problem associated with this square root relationship. Were the relationship 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 greater accuracy.

Orifice Plate Size Selection Reliable flow measurement requires an adequate ∆p 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.

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EXAMPLE

Automated Calculation Methods

An orifice plate meter, with an internal pipe diameter of 200 mm, has been operating satisfactorily with a ß 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 ß 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 ?

Having seen the large number of data necessary to calculate flowrates, you will appreciate that 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.



d ß =D



d = ß D = 0.65 x 200 = 130 mm

To get a beta ratio of 0.45:

d = 0.45 x 200 = 90 mm

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 ß value exactly; hence my use of about 0.45.

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

Test Yourself 3.4 1.

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

2.

What is the factor E called. Given that E = √1-ß4, 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 ℇ, 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 Check Yourself 3.4 on Page 59.

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

Test Yourself 3.5 Given the following: Qm = CEℇ π d2 2∆p ρ 4 E=

1 - ß4

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, 15°C) cubic metres per hour.

þ = PMw ZRT

d = 125mm

D = 257.4 mm

∆p = 100mbar (10000 Pa)

Line pressure P = 24 bar g Line temperature T = 59°C Mw = 22.7 kg / kg-mol Z = 0.937

You will find the answers in Check Yourself 3.5 on page 60.

C = 0.605

ℇ = 0.9987

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Summary of Section 3 The 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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Gas Flow Measurement

Section 4 - Orifice Plate Metering Equipment Types of Plate Plate Geometry Variations Orifice plates are manufactured with various geometrical designs, examples of which are shown in Figure 23, to suit different applications.

Petroleum Open Learning

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 Figure 24. The most common type is the square edge with bevel, and you should note that the plate must be installed with the bevelled edge facing downstream. The only exception to this rule is a plate designed for special applications, which is known as a conical entrance plate. The square edge profile is used in applications where 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.

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.

As you would expect, the location of the orifices and the design of their edges will affect the flow characteristics; so the appropriate factors, mainly C and ℇ, 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.

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Senior Orifice Plate Fitting

Figure 25 is a sketch of the basic design which comprises an orifice plate carrier which is moved by a handle operated rack and pinion arrangement, We have seen that orifice plates are normally fitted between flanges, so a and a chamber situated above the plate. plate changout demands the inconvenience of depressurising and purging the The following description is intended as an explanation of the operating meter run. To alleviate this, the Senior fitting assembly was produced. principle of the system, and it is important that you do not regard it as an operating procedure. 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.

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Sensing Devices 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. Differential Pressure I will describe two differential pressure sensors, one pneumatic and one electrical. The pneumatic one is the torque balance transmitter, and a simplified sketch is shown in Figure 26.

The main feature is a cell containing a liquid-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 capsule depends on the pressure difference between its sides; an increase in flowrate, for example, will reduce the pressure on the downstream side thus causing the diaphragm centre to be displaced to the right. A force bar, which is linked to the capsule, transfers the movement to a flapper/ nozzle arrangement in the signal conditioning system, which is shown simply as a block diagram. The movement is converted to a pneumatic output signal that can be measured directly or, more commonly nowadays, converted to an electrical signal first. The electrical device is the capacitive differential pressure transmitter, a sketch of which is shown in Figure 27, on the next page.

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Gas Density A device which is often used to measure gas density directly uses the vibrating cylinder principle. 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. Figure 28, 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 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.

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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Metering Stations Typical Multi-Stream System Metering 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. Figure 29 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 and density 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 main metering station design requirements that would likely 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 15°Celsius 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 than once every two months.

Safety Implications Most of the safety precautions and procedures that apply to other items of process equipment handling combustible gases are pertinent to metering equipment, and includes: •

isolation, depressurisation, purging and pressurisation

•

checking for leaking fittings

•

monitoring for hazardous gases

You should also be aware that many metering systems are installed in high pressure applications. I will repeat the statement I made when describing the Senior orifice plate assembly, which is that you must always follow the safety precautions and procedures applying to the installation you are working on. Nothing in this book is a substitute for them.

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Summary of Section 4 We looked at some variations of orifice pate design, and at the Senior fitting which is intended as a means of simplifying plate changeouts. The pneumatic torque balance is ofter used to measure differential pressure where electric instrumentation is not employed. Electrical measurement of differential pressure often uses capacitance changes in a sensing cell as the output signal. Gas density can be measured by the principle that the mass of gas in contact with a thin-walled metal cylinder affects its vibrational frequency. The main features of two-stream gas metering station were described, along with some of the fiscal standards it complies with. I finished by drawing your attention to the fact that metering equipment is subjected to specific seafety procedures along with the plant in which it is installed; and you must apply them if you are working on that system.

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Check Yourself - Answers

Check Yourself 1.1

Using







P1V1 = P2V2 T T 1 2

Check Yourself 1.3

Check Yourself 1.5

Molecular mass of mixture

Find n from n = m = 32 = 2 kg-mol Mw 16



= (0.60 x 16) + (0.40 x 30) = 21.6

From Figure 7, at 100 bar and -50°C, Z =0.50

we will assign the initial conditions to the left side, so we need to calculate P2 :

Check Yourself 1.4



PV = ZnRT



2x2 = P2 x 0.5 (10 + 273) (25 + 273)

Start by finding the number of kg-moles of methane using PV = nRT



V = ZnRT P



P2 = 2 x 2 x 298 283 x 0.5

so: 2 x 1 = n x 0.0831 x 293



V = 0.50 x 2 x 0.0831 x 223 100



P2 = 8.42 bar a



V = 0.185 m3

Check Yourself 1.2



n=

2 24.348



n= m Mw

Molecular mass of ethane

0.08214 = m 16





= (2 x 12) + (6 x 1) = 30

= 0.08214 kg-mol

m = 0.08214 x 16 = 1.31 kg

If the compressibility factor had not been used,

V = (2 x 0.0831 x 223) = 0.370 m3 100

which is twice the correct value; ie. an error of 100% !

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For the heavier gas:

Check Yourself 1.6 For both gases we will use PV = ZnRT, but the first point to note is that, since we were given a mass flow rate, we will calculate a molar flow rate using n = m , where m is the mass flow rate. Mw For the lighter gas, n = 50 = 2.632 kg-mol / minute 19 For the heavier gas, n = 50 = 2.174 kg-mol / minute 23 This means that V will be solved as a volumetric flow rate, so we will replace it with the symbol Q. Hence : PQ = ZnRT Q = ZnRT P From Figure 8, Z for the lighter gas is 0.66, and from Figure 9, Z for the heavier gas is 0.51. So, for the lighter gas: Q = 0.66 x 2.632 x 0.0831 x 283 130

Q = 0.314 m3 / minute

Q = 0.51 x 2.174 x 0.0831 x 283 130 3 Q = 0.201 m / minute To replace these actual flow rates at line conditions to standard ones, we use: P1V1 = P2V2 T1 T2 and replace V1 and V2 with Q1 and Q2 respectively. We will assign standard conditions to the left side, so we need to find Q1 for both gas types.



For the lighter gas:

1.013Q1 = 130 x 0.314 228 283



0.003517Q1 = 0.1442



Q1 = 0.1442 = 41.00 m3 (st) / minute 0.003517

For the heavier gas:

1.013Q1 = 130 x 0.201 228 283

Q1 = 0.0923 = 26.25 m3 (st) / minute 0.003517

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Check Yourself 1.7

ρ = PM w ZRT

1. ρ = (1.013 x 30) = 1.270 kg / m3 (0.0831 x 288) 2. ρ = (1.013 x 44) = 1.862 kg / m3 (0.0831 x 288) 3. ρ = (1.013 x 58) = 2.455 kg / m3 (0.0831 x 288)

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Check Yourself 2.1

Check Yourself 2.3

Re = ρDVavg µ

1.

Static pressure and dynamic pressure.

2.

Mass flowrate

3.a. b. c. d. e. f.

A and a : Flowstream cross-sectional area (m3) v (note that this is lower case) : Average fluid velocity (m / s) ρ: Fluid density (kg / m3) Qv : Volumetric flowrate (m3/ s) Qm : Mass flowrate (kg / s) ∆p: Differential pressure i.e. change in static pressure. the SI unit is Pa, but mbar is often used.

ρ = 20 kg / m3 D = 0.125 m vavg = 2 m / s µ = 1.2 X 10-5 Re = 20 x 0.125 x 2 1.2 X 10-5 Re = 416,667; indicating turbulent flow.

Check Yourself 2.2 Qv = A v = (

4.

( )

Divide the mass flowrate by the density Qm ρ 5.

The flow rate is proportional to the square root of the differential pressure.

2 X 3 = 0.0942 m3 / s

)x (0.2) 4

π

Qm = ρ Qv = 1000 x 0.0942 = 94.2 kg / s

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Check Yourself 3.1

Check Yourself 3.2

The total pressure, PT = P + þ v2 2

The pitot measures the ∆p between the static and total pressures. The others measure the static ∆p across the restriction.



PT = 10 mbar P = 9 mbar



ρ v2 = 10 - 9 = 1 mbar = 100 Pa 2

v2 = 2 x 100 = 200 ρ 3

v = 200 = 8.16 m / s 3

Check Yourself 3.3 1.

The vena contracta. It is located at a distance of approximately half the pipe diameter downstream of the plate.

2.

1 and 2.

3.

On the meter flanges.

4.

Fluid friction causes pressure energy loss.

5.

π = d = 130 = 0.52. D 250

It has no units since the length units cancel each other

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Check Yourself 3.4 1.

Re and ß.

2.

E is called the velocity of approach factor. ß= 110 = 0.44 250 E = 1- ß4 = 1 - 0.444 = 0.981

3.

ℇ accounts for gases being compressible and thus experiencing a change in density when flowing through the meter.

4.

Because the resolution is poor at the lower end of a square root scale.

5.

ß=d D

The diameter of the plate to be removed: d = ß D = 0.44 x 250 = 110 mm The diameter of the plate to be installed: d = 0.65 x 250 = 162.5 mm So the nearest available size is 160 mm

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Check Yourself 3.5 E=



ß = d = 125 = 0.486 D 257.4



E=



þ = PMw = 1.013 x 22.7 = 0.961 kg / m3 ZRT 0.0831 x 288

1 - 0.4864 = 0.972 ρ = PMw = 25 x 22.7 = 21.9 kg / m3 ZRT 0.937 x 0.0831 x 332

Qm = CEℇ π d2 4

Q = 4.77 = 4.964 sm3 / s = 17869 sm3 / hr ( 17.87 ksm3 / hr) v 0.961

2∆pρ

Qm= 0.605 x 0.972 x 0.9987 x π x 0.01563 4

b. Qv = Qm Note that we need the volumetric flowrate at standard ρ conditions, so we calculate þ using P = 1.013 bar and T = 288 K, Z = 1

1- ß4

a.

2 x 10000 x 21.9

Qm = 4.77 kg / s

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