Written from the practitioner's perspective, this book is designed as a companion for engineers who are working in the field and faced with various problems related to pressure vessels and stacks, such as: modification, retrofitting existing pressure vessels or stacks to either enhance process capability, lift, move or replace damaged equipment. This makes the book a valuable guide for new engineers who need to develop a feel for these types of operations or more experienced engineers who wish to acquire more useful tips, this handy manual provides the readers with rules of thumbs and tips to mitigate or remediate problems which can occur on a daily bases.
Because of their size, complexity, or hazardous contents, pressure vessels and stacks require the highest level of expertise in determining their fitness for service after these operations. Care must be taken in installation / removal of the vessel to avoid damage to the shell. Damage to the shell can result in catastrophic failure and possible injury to personnel. The book will cover topics such as: lifting and tailing devices; an overview of rigging equipment; safety consideration; inspection and repair tips; methods to avoid dynamic resonance in pressure vessels and stacks; wind loads and how to apply them for various applications and assessment guidelines for column internals, tables and pressure vessel calculations, and code formulas.
The examples in the book are actual field applications based on 40+ years of experience from various parts of the world and are written from a view to enhance field operations. In many parts of the world, often in remote locations, these methods were applied to repair pressure vessels and stacks. These problems will still continue to happen, so there is a need to know how to address them. This book is to present assessments and techniques and methods for the repair of pressure vessels and stacks for field applications. Also the book is to be a repair manual for easy use for mechanical engineers, civil-structural engineers, plant operators, maintenance engineers, plant engineers and inspectors, materials specialists, consultants, and academicians.
- Lifting and tailing devices
- An overview of rigging equipment
- Inspection and repair tips
- Guidelines for column internals
- Tables and pressure vessel calculations, and code formulas
|Product dimensions:||6.20(w) x 9.10(h) x 0.80(d)|
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Pressure Vessel and Stacks Field Repair Manual
By A. Keith Escoe
Butterworth-HeinemannCopyright © 2008 Elsevier Inc.
All right reserved.
Chapter OneSystems of Units
This chapter presents two systems of units so that you can follow the examples ahead. These two systems of units are the metric SI and what is termed by the American Society of Mechanical Engineers (ASME) as the U.S. Customary system of units, namely in the ASME Section II Part D. This system is also termed the American Engineering System (AES) by the U.S. government. I mentioned the latter term in my book Piping and Pipelines Assessment Guide, in how to use the two systems of units. In this book, we will discuss briefly the other variants of the metric SI system, but it is the prevailing metric system of units. Likewise, we will concentrate on the U.S. Customary system versus the British Imperial system. Even though the latter two are similar, there are some differences.
This book is about engineering and discusses how to engineer with each system. It is not of interest to get into a historical discussion about how the system of units evolved, as there are many sources available if you have this interest. There are strong emotions associated with using each system, but this book is not interested in the polemics of using one system versus the other. The other reason for this discussion is that I have worked extensively in each system and have noticed the level of apprehension and intimidation among those using U.S. Customary units toward the metric SI system. This apprehension is totally unnecessary and is without warrant, as the metric SI is used in almost every country of the world except the United States, where it has made headway in medicine and the pure sciences. After reading this chapter, you will not need to convert from one system to the other in the discussions that follow; this text is for users of each system of units.
If you have used only the U.S. Customary system of units, the younger you are, the more likely you will be in the future to encounter the metric SI in practice. If you work outside the United States, then chances are certain that you will have to work in this system of units in one form or another. With more and more foreign projects and foreign engineers coming to the United States, the more likely the event of your using metric SI. Instead of resisting metric SI, consider it as a new friend, which it has been to me. In the metric SI, there are no fractions to worry about, like adding 3/32 to 11/64! The thought of not having to work with fractions is addictive in itself.
The metric SI is an absolute system of units, meaning that it does not depend on where the measure is made. The measurements can be made at any location. For example, the meter has the same (or absolute) length regardless of where the measurement is taken—here on earth or elsewhere. The unit of force is a derived unit. The metric SI system has been called the meter, kilogram, and second system, or MKS. These three units are primary units. In this system the Newton is the amount of force needed to give 1Kg mass an acceleration of 1m/sec2. Thus, Newton's second law is the crux of the system.
To derive force from mass, you have to use Newton's second law:
F = M*A, Newtons (1.1)
The unit of mass is kilogram (Kg) and acceleration is m/sec2. To perform the conversion, you use
F = (g/gc) M(Kg) (1.2)
In the metric SI system, you use
g = 9.807 m/sec2 and gc = 1.0 Kg - m/N - sec2
Thus, the force required giving 1Kg of mass an acceleration of 1m/sec2 is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.3)
Yes, that's right: It takes 9.807N (Newtons) to accelerate 1Kg (kilogram) of mass 1m/sec2—almost 10 times. This is a number to remember. See the note later in this section.
Regarding the U.S. Customary system, the same discussion is presented in my book Piping and Pipelines Assessment Guide, pp. 498–500, as follows:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
From Newton's second law, we have the following:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.4)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.5)
As you can see, in the U.S. Customary system, mass is a derived unit, with the primary units being force, pound, and second. Some authors refer to it as the FPS system. This is a gravitational system, where force is a primary unit. Since most experiments involve a direct measurement of force, engineers prefer a gravitational system of units as opposed to an absolute system. Often the units g and gc are rounded to 32.2ft/sec2.
As you can see in equations 1.4 and 1.5, the terms lbf and lbm are used interchangeably. In the U.S. Customary system of units, lbf and lbm have the same magnitude (value). Pound (mass), lbm, and pound (force), lbf, have identical numerical values. Thus, 1 pound mass is equal to 1 pound force; hence, it is not uncommon to use the term pound, or lb, interchangeably. This usage has unfortunately caused confusion. Force is not mass, and this is hard to understand using the U.S. Customary system of units, where the same term is used for both mass and force. In locations without gravity, such as outer space, weight is meaningless. The official unit of mass in the U.S. Customary system of units is the slug. A pound is the force required to accelerate 1 slug of mass at 1ft/sec2. Since the acceleration of gravity in the U.S. Customary system is 32.2ft/sec2, it follows that the weight of one slug is 32.2 pounds, commonly referred to as 32.3 lbm. The comparison of the slug and the pound makes it clear why the size of the pound is more practical for commerce. With the current scientific work, it is undesirable to have the weight of an object as a standard because the value of g does vary at different locations on Earth. It is much better to have a standard in terms of mass. The standard kilogram is the mass reference for scientific work. This book is for industrial practice by practicing engineers, inspectors, maintenance engineers, plant and pipeline personnel, rigging engineers, and others that work in industry. It is not intended for scientific work. The value of the gravitational constant does not vary enough to affect most engineering applications. The slug is rarely used outside of textbooks, which has contributed to the confusion between the pound mass and the pound force. When expressing mass in pounds, it is necessary to recognize that we are actually expressing "weight," which is a measure of the gravitational force on a body. When used in this manner, the weight is that of a mass when it is subjected to an acceleration of 1g. In academia, where the study of dynamics involves forces, masses, and accelerations, it is important that mass be expressed in slugs, that is, m = W/g, where g is approximately equal to 32.174 ft/sec2. These points are arguments for the use of the metric SI system of units. Particularly in the study of dynamics, the SI system is much easier. The slug is defined as
1 Slug = 32.174 lbm = lbf - sec2/ft (1.6)
gc = 1 = 32.174 lbm - ft/lbf - sec2 (1.7)
Like text books in academia, the slug is rarely used in industrial circles, but where it is used remember that 1 slug 32.2lbm. This will come up briefly in Chapter 3 where the ASME STS-1 uses the slug as mass. However, ASME is in the process of making the SI system the preferred system of units. In locations without gravity, such as outer space, weight is meaningless. If two bodies were to collide in outer space, the results would be due to their differences in mass and velocity. The body with the greater mass would win out.
Note: Because lbf and lbm have the same unit and both are often referred to as pounds, it is a common mistake for users of the U.S. Customary system to forget to convert kilogram mass to Newton's force, or vice versa. When using the metric SI, don't forget the conversion factor of 9.807 derived earlier. Repeating again, kilograms are not Newtons. With the metric SI, this phenomenon does not exist, as 1 kilogram is 9.807 Newtons, so mass and weight cannot be confused.
GETTING FAMILIAR WITH METRIC SI UNITS
Civil-structural engineers prefer to work in units of force in designing foundations. With the U.S. Customary system, this is obvious: A pound is a pound. In the metric SI system, you must make a conversion. In the metric SI, KiloNewtons (KN) are used for foundations. Often I have heard the question "How do I convert kilograms to KiloNewtons?" The answer is simple. Suppose you have a pressure vessel that is a large reactor that is to go into a new refinery. This reactor weighs 1,000,000 kilograms. This is converted to force as follows:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.8)
This means 1 million kilograms almost equals 10 million Newtons. So the civil-structural engineers would design for 9807 KiloNewtons (KN).
In Europe it is quite common to see lifting devices, such as small cranes in automobile shops, rated in KN. I saw a lifting crane in an automobile shop in Germany marked as 20KN. This marking means that the crane could safely lift
20 KN = 20 000N
Using Eq. 1.3, we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
So the crane is rated at roughly 2039Kg. It would not be wise to ask how many pounds this is, as often many in the European Union are as emotional about the metric SI as some Americans are about the U.S. Customary system. In secret, you can calculate
2039.4 Kg 4496.1 lbm
Most people using the U.S. Customary system would then say the measurement is "4496.1 pounds."
If you are beginning to use the metric SI for the first time, it is quicker to learn the system by carrying all calculations solely in metric. This will enable you to become familiar with the system more quickly and obtain a "feel" for the answer.
OTHER IMPORTANT METRIC SI UNITS USED IN MECHANICS
The basic units—area, section modulus, and moment of inertia—are mm2, mm3, and mm4, respectively.
The density of steel is 0.283lbm/in3. In the SI metric system, this measurement converts to approximately
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Bending Moments and Torque
Because moment of force (bending moment) and torque are equal to a force times a distance (moment arm or lever arm), their SI unit is N*m. The Joule (J = N*m), which is a special name for the SI unit of energy and work, should not be used as a name for the unit of moment of force or of torque. Typically, the moment of torque is written as N m, with a space between the N and m or as N*m.
The Joule is equal to a 1N*m, but is reserved for a unit of energy and can have more than one application, as discussed later. When we get into thermal stresses and heat transfer, it is confusing to use Joule as a bending moment of torque and as a thermal unit. We will spend more time later on the proper use of metric units.
UNITS FOR STRESS AND PRESSURE
The term for pressure and stress is 1 Newton per square meter, which is named in honor of the famous mathematician, physicist, and philosopher Blaise Pascal. Since the area of a meter is rather large, 1,000 pascals is a kilopascal (KPa), and 1 million pascals is 1 megapascal (MPa). Simply written, we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
A megapascal, or MPa, is most commonly used for stress. It can also be used for pressure, but the numbers may remain small for small values of pressure. Typically, the kilopascal is used for pressure. The bar has been used for pressure often in the past, but the bar is not an SI unit. Although it may be accepted in the SI, it is discouraged. Now
1 kilopascal = 0.01 bar = 0.001 megapascal
In U.S. Customary units, these metric units are
1 megapascal = 145 psi 10 bars 1 bar = 14.5 psi 1 kilopascal = 0.145 psi
A WARNING ABOUT COMBINING METRIC SI UNITS
When you are using the metric SI system of units, it is wise to remember that many units are named in terms of a magnitude of 10, e.g., kilo or mega as a prefix. When you are performing computations, it is advised to reduce these terms to their most basic set of units. For example, if you have a cylinder that is 609.6mm (24") ID that contains 1000KPa of pressure that is 24mm thick, the hoop stress is
σ = PD/2t
Entering the equation as
σ = (1575) KPa (609.4) mm/2(24)mm
can lead to mistakes, since the stress term is in MPa and the pressure is in KPa. The best way to avoid mistakes is to write the equation as follows:
1575 KPa (0.001) = 1.575 MPa = 1.575 N/mm2
Excerpted from Pressure Vessel and Stacks Field Repair Manual by A. Keith Escoe Copyright © 2008 by Elsevier Inc.. Excerpted by permission of Butterworth-Heinemann. All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
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Table of Contents1. System of Units
2. Handy Pressure Vessel Formulas
3. Dynamic Response of Pressure Vessel and Stacks
4. Wind Loadings on Pressure Vessels and Stacks
5. Assessment of Pressure Vessel Internals
6. Safety Considerations for Lifting and Rigging
7. Lifting and Tail Attachments
8. Assessing Weld Attachments
9. Rigging Devices