PE114-11-07 Process Plant Troubleshooting_1.pdf

QAF011 Rev. 04 Jan. 12, 06

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Haward Technology Middle East

Process Plant Troubleshooting& Engineering Problem Solving

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Table of Contents

Troubleshooting Definition, Potential Sources Engineering Problem Solving Course Approach Components of Plant Problem

Solving Limitations to Plant Problem

Solving Sources of Historical Data Daily Monitoring System

Guidelines Setting Trigger Points Disciplined Learned Problem

Solving Approach Step 1 to Step 6 - Considerations Risk Analysis - HAZOP - MSDS Troubleshooting Manual:

Sample problems

Applied Economics Valuation Principles and Methods Other Views Valuation Principle

and Methods Compressor - Compressor

Problems - Simplified Approach

Section 1

Section 2

Table of Contents

Distillation, Plates, Tray Stability Guidelines for Problems Solving

Temperature, Pressure, LevelMeasurements , Verification

Sample Exercise Kinetics, Flow,Mechanical, Design

Fluids Overview -Basic principles Fluids Overview- Head definition Equivalent piping Lengths Commercial correlations Practical Exercises – Hand outs Two Phase Flow / Theory and

Applications Practical Exercises – Hand outs Process Control – Introduction;

PID Controllers, Feedback, Feed

Forward and Cascade Controls Advanced Control ; Multi - loop Controllers; Process Control &

Optimization; On Line Optimization; Process

Analyzers; Distillation Multiple Control ;

Volume Control;

Section 3

Table of Contents

Condenser Control; PracticalConsiderations; Advanced

Control Project Drawbacks

Heat Transfer Overview Troubleshooting Techniques/

Applications Practical Exercises – Hand outs Distillation Column Packing Practical Exercises

Hazards Demonstration QRA

“Ishikawa” diagramsExercises

MSDS Needed Information, Is it Good

Enough? Incomplete? Accidents FLIXBOROUGH ACCIDENT Lessons Learned, General

Information

Section 4

Section 5

Section 1

Troubleshooting Definition, Potential Sources Engineering Problem Solving Course Approach Components of Plant Problem

Solving Limitations to Plant Problem Solving Sources of Historical Data Daily Monitoring System Guidelines Setting Trigger Points Disciplined Learned Problem

Solving Approach Step 1 to Step 6 - Considerations Risk Analysis - HAZOP – MSDS Troubleshooting Manual:

Sample problems

This document is the property of the course instructor and/or Haward Technology Middle East. No part of this publication may be reproduced,stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise,without the prior written permission of Haward Technology Middle East


PIONEERS IN TECHNOLOGY TRANSFER

Process Plant Troubleshooting& Engineering Problem Solving

Section 1 Haward Technology Middle East 2

Process Plant Troubleshooting &Engineering Problem Solving

Section 1




Course ProgramDay 1 : November - 2007

08:00 – 08:30 Troubleshooting08:30 – 09:00 Definition, Potential Sources09:00 – 10:00 Engineering Problem Solving

Seminar ApproachComponents of Plant Problem SolvingLimitations to Plant Problem Solving

10:00 – 10:15 Break10:15 – 11:45 Sources of Historical Data

Daily Monitoring System GuidelinesSetting Trigger Points

11:45 -12:00 Break12:00 -13:00 Lunch13:00 -14:30 Disciplined Learned Problem Solving Approach

Step 1 to Step 6 - Considerations14:30 -15:30 Risk Analysis - HAZOP - MSDS15:30 – 15:45 Break15:45 – 16:30 Troubleshooting Manual : Sample problems



Definition of Troubleshooting

What is the definition of troubleshooting from theengineer’s point of view ?



Troubleshooting Sources

Usually troubleshooting is related to an “anomaly”detected that deviates from the “normal” performanceof the unit.

The normal performance of the unit is usually definedinside a design parameters given by the Projectexecutors or by the Licensor’s who built the unit orsupply the technology.



Troubleshooting SourcesUnder this scenario the personnel in charge of executingthe troubleshooting exercise is usually assuming that thedesign in place is correct and the performance of the unithave deviated from the “normal” performance of the unitfor other reasons that need to be determined.

Under this scenario the troubleshooting is thereforefollowing the concept that the deviation from thetargeted performance of the unit could be originatedfrom technical /operational/ maintenance problems(equipment failure, operational mistakes, incorrectmaintenance practices, over design conditions, etc..)but not from design problems.



Samples of TroubleshootingPotential Sources

TroubleshootingSources

Hiding Design

Problems

Stress Mechanical

Failures

Instrumentation and

Control

Environmental

Violations

Project Drawbacks

Over / undercapacity

Operational

& Maintenance

Mistakes

Utilities

Failures

Optimization

Projects

Safety and healthViolations

Open discussion

Start Up/ Shut downProcedures



Engineering Problem Solving

Sometimes this narrow definition bring to the table thefollowing statement :

The “problem went away, but it came back”

This is often a syndrome using process engineering andproblem solving skills in modern process plants.



What you Can Expect

THE NEED:

A recent survey by the Department of Labor listed“problem solving skills” and “vocational-technicalskills” in the top 10 skills that employers wish theiremployees had.

The lack of either of these skills often causes the“problem went away, but it came back” syndrome.

The lack of “looking outside the box” is another factorthat avoids to find the best solution most the times andbrings the “problem went away, but it came back”syndrome.



Course Approach

The following areas will be covered:

Some of the essential problem solving skills and otherexternal requirements are:

Daily Monitoring System Disciplined Problem Solving Approach Determining Optimum Technical Depth Outside the box thinking Total Back Up from management An adequate training program Communication and political skills

The above skills will be discussed and potential guidelinesare going to be presented to evaluate successfulimplementation of each of these.



Course Approach

The following areas will be covered (Cont.):

Vocational-technical skills will be enhanced by helpfulhints and practical knowledge that expand the problemsolver’s academic training.

Practical exercises together with a brief requiredtheory will be presented for the first two daysfollowing the course guide book.

Three real cases not presented in the course guidebook will be assigned to different groups to apply someprinciples presented in the guide book to determinetheir level of acceptance .



Course Approach

The following areas will be covered (Cont..):

Different groups will be formed to execute technical ,

financial and safety / health (HAZOP) evaluations to

determine potential solutions for each practical daily

case (s).

Determine the best possible solution through a

comparative evaluation of the groups presented

solutions.

Open discussion



Course Interchange Mode

Interactive - Ask questions.

Problem working - Techniques must be practiced.

The manual contains more material than it is possibleto cover during the five days course time frame.

The manual has the main target to provide guidancefor the course and present solving skills.



Course Outline

Limitations to Plant Problem

Solving

Successful Plant Problem

Solving

Examples of Plant Problem

Solving

Application to Prime Movers

Application to Plate

Processes

Application to Kinetically

Limited Processes

Application to Unsteady

State

Verification of Data

Utilization of Manual

Computations

Three real cases evaluation

to determine principles

applicability and drawbacks



Components Of Plant Problem Solving

A SYSTEMATIC APPROACH TO PROBLEM SOLVING

A GOOD PROCESS UNDERSTANDING OF EQUIPMENT

A GOOD UNDERSTANDING OF SPECIFIC PROCESSTECHNOLOGY

BE WILLING TO GET OUTSIDE THE “ BOX”



Limitations To Plant Problem Solving

Is problem solving really engineering?

• A definition of engineering from Webster saysengineering is “The science of making practicalapplication of knowledge in any field”

Why is this an important question?

• Our paradigm or viewpoint will often determineour enthusiasm in approaching our work.



Limitations to Plant Problem Solving

Often engineers may conclude that problem solving isnot truly engineering because of the following:

• Engineering is defined in such narrow terms thatonly “design work” appears to be engineering.

• Intuition and “gut feel” have replaced thoroughanalysis as a preferred tool for problem solving.

• Considerations of “optimum technical depth” arenot well understood.

These paradigms lead to inadequate engineeringanalysis of plant problems.



Plant Problem Solving Considerations

Modern day process plants are large and complex.

The problem is usually more complicated than firstdescribed.

Conflicting data is always present.

Modern day plants have a high degree of variableinteraction.

Besides a high degree of variable interaction, there isalso a high degree of interaction between the variousengineering disciplines.



Plant Problem Solving Considerations

System dynamics involve long holdup times.

Inadequate Application of Engineering Principles

Lack of a Methodical Approach

Failure to see the whole picture

Over dependence on history

Internal Competition among departments

Internal / External politics



At base level the pressure at the level instrument willbe less than the same pressure in the drum as follows:

(P2-P1)/62.4 + (36-0)/64.4 = 0P1-P2 = 34.9 psfthis is equivalent to 0.5 feet in measurement of level

this ignores the friction loss in the line and nozzle

Connecting the level instrument in theprocess line as shown will result in themeasured level reading being 0.5 feetlower than actual.

This is based on Bernoulli’s theoremdP/D + dV2/2g + dZ = 0

wheredP = difference in pressureD = density of liquiddV2 = difference in liquid velocities squaredg = gravitational constantdZ = difference in liquid height

fluid is waterat 6fps velocity

P1 and P2are at essentially

the same level

P1

P2

LIC

FIGURE 1-2EXAMPLE OF IMPROPER LEVEL

INSTRUMENTATION



The most commonly mistuned loops in auditedrefineries were level controllers, where the

operating objective is usually to keep the flowsteady and accept level swings, but where tuners

often prefer the opposite.



Real World Applications

Do any of these sound familiar?

How many of these are present in the exampleproblem?



Proposed Problem Solving Steps

ProblemDetection

DesignProblem

OtherReasons

DetermineRoot of the Problem

HypothesisDevelopment

TechnicalFeasibility

FinancialFeasibility

LegalInvestigation

Recommendation

Evaluation

Implementation

HAZOP

MSDS

CLIENTAPPROVAL

RESULTSEVALUATION

TechnicalResults

Financialresults

Negative

Positive

Negative Results

Licensor

Negative Results

ManagementPresentation



Successful Plant Problem Solving

A definition will be helpful:

• Engineering problem solving is defined as theapplication of engineering principles to allowfinding, defining, and solving plant operatingproblems in an expedient and complete fashion.

Note that many problem solving courses do notadequately cover the concept of finding and definingproblems.



Successful Plant Problem Solving

Successful plant problem solving requires thefollowing:

• A daily monitoring system.

• A disciplined (not intuitive), learned (not inherited)engineering problem solving approach.

• The ability to distinguish between problemsrequiring technical problem solving and those onlyrequiring an expedient answer. This is referred tolater as optimum technical depth.



Daily Monitoring System Guidelines

Develop a managerial objective/data source system forproblem finding.

Develop process models that incorporate severalvariables into a single theoretical variable.

Recognize that there is a difference between usingstatistics to control a process and using statistics tofind problems.

• Control - Be 99% confident before making a change.

• Problem Finding - is rarely as high as 99%.



Table 2-1Sources of Historical Data

MANAGERIAL OBJECTIVE

Maximize Max. VariableMinimize Maximize Finding Trend RetentionsRoutine Work Hidden Problems Spotting Volume Key

Computer Data X XStorage

Computer or Hand Plots X X

Delta Data Plots X X X

Communication with XHourly Forces

Visual Observation of XField Equipment



Figure 2-1Essential Variable % Of Theory Vs Time

% of Theory

Time80

85

90

95

100



Daily Monitoring System Guidelines

Develop a “trigger point” for each important variable.

Trigger points can be set for variables based on thefollowing:

• Theory/Laboratory or Pilot Plant Demonstrated

• Plant Demonstrated

• Licensor (s) Demonstrated or Guaranteed

• Personnel with Extensive experience on site

• Benchmarking



Daily Monitoring System Guidelines Pick 6 to 10 essential variables and plot (computer or

hand) them on a continuous daily basis using delta plotsand theoretically determined target values.

Compare these data plots to well established “triggerpoints”.

Obtain operator comments daily and follow-up on anyunusual observations.

Visually observe equipment in the field on a weekly basis.

Store the essential variable plots so that this informationcan be easily accessed.

Continuous interchange of information with otherdepartments that are involved in production or relatedactivities .



Setting Trigger Points

CONCEPT

Should be based on statistics when feasible.

However, criteria for declaring problem is differentfrom criteria for taking control action.

Positive deviations must always be considered.

Always keep on mind that not all the variables arebeing taken into consideration and their synergy isnot always detectable or easy to determine.



Setting Trigger Points

CRITERIA FOR

Should be a function of seriousness ofconsequence.

A secondary consideration is location on learningcurve. For example, in a new process operatingwith tighter trigger points can escalate thelearning curve.

Experience of personnel on site could help ordistortion the learning process curve.



Disciplined Learned Problem solving Approach

Disciplined Learned Problem Solving is an approach that allowsboth determining if the problem really occurred and specifyingthe problem in quantitative terms. The approach discussed herediffers significantly from traditional problem solving training asfollows:

It emphasizes using techniques for verifying that theproblem really occurred.

It emphasizes the need to use engineering principles informulating a hypothesis to explain the problem.

It emphasizes that a problem solution must not create newproblems. That is any hypothesis must be confirmed with aplant test or by making “directionally correct changes”.



Disciplined Learned Problem Solving Approach

Step 1: Verify that the problem actually occurred.

Communications in an operating environment are almostalways 2nd or 3rd hand and often highly “garbled”.

Step 2: Write out an accurate specification of the problem.Answers to the following questions may be helpful:

• What happened?

• When did it happen?

• Where did it happen?

• What was the magnitude of the problem?

• What else happened at the same time or shortlybefore?



Disciplined Learned Problem Solving Approach

Step 3: Develop a theoretically sound working hypothesisthat explains as many specifications of the problemas possible.

Step 4: Provide a mechanism to test the hypothesis.Calculations are a valid means to test a hypothesis.

Step 5: Recommend remedial action to eliminate theproblem without creating another problem.

Step 6: Follow up recommendation results and determinehow close/far are from the technical and financialevaluation done.

Find out accuracy (positive/negative) of recommendations andreasons of deviations.



Considerations for Step 2

Step 2: Write out an accurate specification of theproblem. Answers to the following questions maybe helpful:

What happened?

When did it happen?

Where did it happen?

What was the magnitude of the problem?

What else happened at the same time orshortly before?




The preparation of a written statement of the problem by theone who knows best is the key to correctly initiating the process.The problem statement provides:

A means to communicate directly to verify that theproblem actually occurred as described.

A means to uncover data gaps.

A means to clarify the data and problem.

A simple tool to allow communications between differentlayers of management and the problem solver.

A tool to allow the problem solver to assess the severityand solution difficulty of the problem.

QUESTION : WHO WRITES THE DOCUMENT ?



Problem Specification Example SHORT TITLE OF PROBLEM ____________________

DESCRIPTION OF EVENT (make sure that step 2 is utilized to provide acomplete problem description) ____________________

HOW WAS PROBLEM DISCOVERED (was it by data plotting, operatordiscussion….) ____________________

PRELIMINARY PROBLEM ASSESSMENT

• COST OF PROBLEM (HIGH, MODERATE, LOW) _______

• IS IT AN OPERATING OR TECHNICAL PROBLEM _______

• IS THERE AN OBVIOUS IMMEDIATE FIX _______

• IF YES WHAT IS PROBABILITY OF SUCCESS _______

• IF NO WHAT AMOUNT OF EFFORT IS INVOLVEDIN PROVIDING A FIX? _______

• ARE YOU ACTIVELY WORKING ON THIS PROBLEM _______



Considerations For Step 3

Step 3: Develop a theoretically sound working hypothesis thatexplains as many specifications of the problem aspossible.

A theoretically sound hypothesis (Step 3) is rarely developedby unstructured “brainstorming”, but is almost always based onengineering principles such as:

Unit operations and/or design calculations.

Unsteady state accumulation calculations.

Mass and energy balances.

Other problem solving techniques based on scientificprinciples.




Step 4: Provide a mechanism to test the hypothesis.Calculations are a valid means to test ahypothesis.

The testing of the proposed hypothesis (Step 4) should beconsidered a success if it proves conclusively that theproposed hypothesis is either right or wrong.

Hypothesis testing can consist of:

• Fundamentally sound calculations.

• Plant test of new operating conditions.

• Increased data collection frequency and/or new data.

• Temporary “mechanical fix”.




Step 5: Recommend remedial action to eliminate the problemwithout creating another problem.

Once a proposed hypothesis has been demonstrated to be true, theproblem must now be eradicated (Step 5). The four keys to Step 5(Recommend remedial action to eliminate the problem withoutcreating another problem) are as follows:

Conduct a thorough potential problem analysis. This analysisshould include safety aspects also.

Make sure that the problem solution is the simplest one that willwork (keep it simple) KIS Theory.

Make allowances for “follow-up”. Remember the importance of communications and working in

harmony with all the other departments , specially the client.



Considerations for Step 6Step 6: Follow up recommendation results and determine how

close/far are from the technical and financialevaluation done.

The implementation of the recommendations is usuallydone with the participation and approval of otherdepartments.

The required HAZOP and pre-established follow system todetermine the success/failure of the recommendationneeds to be presented and accepted by the client whichusually is the production department.

The recollection of the data to prove that therecommendation has been successful and match thefinancial/technical evaluations done should be done underthe team concept.




In such a case that the data collected after theinstallation of the recommendation done proves thatit does not work as expected, then, a detail evaluationof the engineering principles used must be donetogether with revision of all other aspects of therecommendation and its implementation.

If otherwise the data shows that the recommendationis as expected, a financial evaluation is required todetermine the economical benefits and present then tomanagement for their information.



Optimum Technical Depth

The ability to compromise between expediency andthoroughness can be referred to as the optimum technical

depth. While this is a difficult area to quantify, there are

several helpful guidelines.

The confidence level that the problem solution is correctis directly proportional to the technical depth involved inthe problem solving.

The required confidence level in an industrialenvironment is much lower than in an academic orresearch environment.




The required confidence level is directly proportionalto the cost of the solution and the execution time ofthe solution, but inversely related to the cost of theproblem.

These concepts are illustrated in Figure 2-2.

Unfortunately, the very expensive problems oftenrequire a detailed technical analysis. Rather thandoing this technical analysis, the problem solver oftensubmits to the temptation to “try something”. He thenfinds himself/herself spending some of his limitedamount of time implementing the “something” ratherthan doing a technical analysis.



Figure 2-2Confidence Level Versus Solution Cost

Cost of Solution and/or Time to implement

SolutionConfidenceLevelRequired

moderate

high

ParametersAre problemcost




Optimum TechnicalDepth

Another View

Pure IntuitionNo Calculations

Large Main FrameComputer



Directionally Correct Hypothesis

This approach assumes that if one can make low cost(in either time or money) changes that have a 75%confidence level, the result of this move will by itselfeither prove or disprove the working hypothesis.

It should be noted that this concept still requires atechnical analysis and a theoretically correcthypothesis. The required confidence level that thesolution is correct is just reduced to allow low costchanges.



Risk Analysis



Risk Evaluation Methods 1

A typical FMEA incorporates some method to evaluatethe risk associated with the potential problemsidentified

Risk Priority Numbers

To use the Risk Priority Number (RPN) method theteam must:

• Rate the severity of each effect of failure

• Rate the likelihood of occurrence for each cause offailure




• Rate the likelihood of prior detection for eachcause of failure

• Calculate the RPN by the product of the threeratings:

• RPN = Severity x Occurrence x Detection

• The RPN can then be used to compare issues withinthe analysis and to prioritize problems forcorrective action

• The higher the RPN the higher the priority




Risk Priority Number rating scales usually range from 1to 5 or 10, with the higher number representing thehigher risk.

For example, on a ten point Occurrence scale, 10indicates that the failure is very likely to occur and ismuch worse than 1, which indicates that the failure isvery unlikely to occur.

For Detection, the scale is reversed since somethingvery likely to be detected would be 1 and somethingvery unlikely to be detected would be 10.




The specific rating descriptions and criteria are definedby the organization or the analysis team to fit theproducts or processes that are being analyzed.

Typically, if the RPN falls within a pre-determinedrange, corrective action is required to reduce the risk(i.e. to reduce the likelihood of occurrence, increasethe likelihood of prior detection or, if possible, reducethe severity of the failure effect.



Risk Evaluation Method 2

An example rating table for severity follows:

Generic Five Point Severity Scale

Safety-related catastrophicfailures

Very High or Catastrophic5

Loss of functionHigh4

Gradual performance degradationModerate or Significant3

Operable at reducedperformance

Low or Minor2

Minor nuisanceVery Low or None1

CriteriaDescriptionRating




Criticality Analysis

Quantitative Criticality Analysis Method

• Define the basic unreliability for each item at agiven operating time

• Identify the portion of the item’s unreliability thatcan be attributed to each potential failure mode

• Rate the probability of loss (or severity) that willresult from each failure mode that may occur

• Calculate the criticality for each potential failuremode by the product of the three factors:




• Mode Criticality = Item Unreliability x Mode Ratioof Unreliability x Probability of Loss (or Severity)

• Calculate the criticality for each item by obtainingthe sum of the criticalities for each failure modeidentified for the item

• Item Criticality = Sum of Mode Criticalities




Criticality Analysis

Qualitative Criticality Analysis Method• Rate the severity of the potential effects of failure

• Rate the likelihood of occurrence for each potential failure mode

• Compare failure modes via a Criticality Matrix, which identifiesseverity on the horizontal axis and occurrence on the vertical axis

Not VeryCritical

Occurrence

Severity

ExtremelyCritical



Risk Management Methods

There are many established procedures that we can useto manage the risk involved with the safe and reliableoperation of oil and chemical plants.

Risk ManagementPlanning

Process Safety Management (PSM)

PHA (Hazan)

Failure Mode Effect & Criticality Analysis(FMECA)

Root Cause failure Analysis (RCFA)

Safetyorientated

Reliabilityorientated

Job Safety Analysis (JSA)

Hazop

Risk ManagementPlanning

Process Safety Management (PSM)

PHA (Hazan)

Failure Mode Effect & Criticality Analysis(FMECA)

Root Cause failure Analysis (RCFA)

Safetyorientated

Reliabilityorientated

Job Safety Analysis (JSA)

Hazop



Hazop Study



Hazop Study

A Hazop study identifies hazards and operabilityproblems by identifying how the plant might deviatefrom the design intent.

If a solution to a problem becomes apparent, it isrecorded as part of the Hazop result but the primeobjective for the Hazop is problem identification.

Hazop studies are normally conducted during thedesign phase, especially when new technology isinvolved but can be used at almost any phase of aplant's life.



Hazop Study

Hazop is based on the principle that several expertswith different backgrounds can interact and identifymore problems when working together than whenworking separately and then combining their results.

The most common form Hazop study employs guidewords to test the consequences of parametersdeviating from design.



Hazop Study

Objectives of a Hazop study may include:

• Check the safety of a design

• Check the maintainability and operability of adesign

• Decide whether and where to build

• Develop a list of questions to ask a supplier

• Check operating & safety procedures

• Improve the safety of an existing facility

• Verify that safety instrumentation is reacting tobest parameters



Consequences to be Considered

It is also important to define what specificconsequences are to be considered.

The following lists many of the consequences normallyevaluated:

• Employee safety

• Loss of plant or equipment

• Loss of production

• Liability

• Insurability

• Public safety & impact on neighborhood

• Environmental impacts



Consequences to be Considered The Hazop team must be chosen from experienced people

preferably with knowledge of a similar facility who willlikely be involved with the operation of the plant.

The Team Leader should be chosen for his ability to get theteam to focus on making the analysis rather than his abilityto solve problems.

Issues identified can be resolved after the Hazop.

Depending on objects the following team assignment issuggested:

Safety Engineer (Team Leader)Design EngineerProcess EngineerPlanning Engineer

Operations SupervisorInstrument EngineerChemistMaintenance Supervisor



Hazop Guide Words

Simple words are used to qualify or quantify theintention in order to guide and stimulate thebrainstorming process and so discover deviations.

Examples are shown in the following table:

Maintenance

Two Phase

High Pressure

No Flow

DeviationDeviation

OperationOther Than

One PhaseAs Well As

PressureMore

FlowNo

ParameterParameterGuide WordGuide Word



Hazop Team Review Process

NOT SURE

Any Hazard’sor Operability

Problems?

Need MoreInformation

RecordConsequences &

CausesSuggest

Remedies

Divide System intoStudy Nodes

Select a Node

Apply All GuideWords in Turn

YES NO

YES NO



Successful Hazop Criteria

The success or failure of the Hazop depends onseveral factors:

• The completeness and accuracy of drawings andother data used as a basis for the study.

• The technical skills and insights of the team.

• The ability of the team to use the approach as anaid to their Imagination in visualizing deviations,causes and consequences.

• The ability of the team to prioritize andconcentrate on the more serious hazards that areidentified.



MSDS Sheets



MSDS Sheets

A Material Safety Data Sheet (MSDS) is a technicalbulletin containing detailed information about ahazardous substance.

OSHA requires that manufacturers prepare a MSDS foreach chemical that it sells.

The MSDS contains more extensive information than isconveyed on the label.

The MSDS must accompany each chemical it ships thefirst time that the chemical is shipped to thatrecipient.



MSDS Sheets

The following minimum information must be provided in

the MSDS:

1. The identity of the product as used on the containerlabel.

2. The chemical and common name for all ingredientspresent in concentrations greater than 1% and 0.1%for a cancer causing substance (carcinogen).

3. The physical and chemical properties of thehazardous components.



MSDS Sheets

4. The physical and health hazards, including signs andsymptoms of exposure and/or prior and/or existingconditions that can warn against exposure.

5. Primary routes of entry into the body.

6. Any known exposure limits.

7. Whether the hazardous substance is a carcinogen.

8. Precautions for safe handling and use.



MSDS Sheets

9. Procedure for spill or leak cleanup.

10. Control measures

11. Emergency first-aid procedures.

12. The date of preparation

13. The name, address, and telephone number of thecompany or responsible employee distributing theMSDS.



EXAMPLES OF PLANT PROBLEM SOLVING



Examples of Plant Problem Solving

In an industrial environment where the emphasis is usuallyon increased productivity, doubts about the validity ofthis technique will always be present. Typical questionsare:

Does this technique really work?

On what kind of problems can it be used?

Is it really possible in an industrial environment to useengineering calculations as opposed to intuitiveproblem solving?

Let’s find the answer from the licensor, they knowbetter.



Troubleshooting Manual : First problem



Reactor Temperature Runaway

At 0200 hours on April 2, one of the six continuouspolymerization reactors experienced a temperaturerunaway. That is the reactor temperature roseexponentially from 150oF to 175oF in a 30-minute period..When the reactor in question reached 175oF the reactionwas terminated by injection of a quench agent. All theother reactors were operating normally.




The temperature control system on the reactor was suchthat an increase in temperature caused an immediateincrease in the cooling water supply system. It was knownthat a small increase in catalyst rate occurred right beforethe temperature began increasing. However in the past,catalyst rate increases of this magnitude only resulted in aslight temperature increase. Following this slight increase,the reactor temperature very quickly returned to normalas the cooling water control system responded.



Polymerizationreactor

coolingwater return

cooling water supply

pumparound pump

Data Values at Midnight

Temperatures

Cooling WaterIn 90Out 120

Pumparound LiquidIn 150Out 142

Flow Rates, pph

Cooling Water 195000

Reactor Slurry Pumparound 1,440,000

Other InformationThe valve on the cooling wateris 95% open

Technology Information

Reaction Heat Generated = K e ˜(-11000/T)

Where K is a constant that containing monomerConcentration, catalyst concentration, reactorvolume and heat of reaction

K = 3.9( 10ˆ14)T is in Rankin

The specific heat of the reaction fluid – 0.5 B TU/lb-F

Figure 3-1Reactor Schematic



The problem solver is faced with 3 questions:

• What should be done to return the reactor back toservice?

• What caused the episode?

• What can be done to prevent it from recurring inthe future?

Problem solving focuses on the last 2 questions.




Step 1 – Verify that the problem actually occurred.

Did all temperature points show an increase?

Could the phase rule be used to confirm the hightemperature?




Step 2- Write out an accurate statement of what problemyou are trying to solve.

“Determine why temperature control was lost on April2. This loss of control occurred at about 0200 followinga very small increase in the reactor temperature causedby a slight increase in catalyst flow. This loss of controloccurred on only one of six reactors all operating at thesame charge rate on the same feedstock. There was nomechanical or utility failure on the reactor in question.The weather turned slightly warmer on March 30. Oncethe reactor temperature began increasing it roseexponentially from 150 F to 175 F in an extended period(30 minutes).




Once the cause has been determined, developrecommendations to prevent this problem fromrecurring.”




Step 3- Develop a theoretically sound working hypothesis that

explains the problem.

Several possible hypotheses could be proposed and theproblem statement could eliminate all but one.

Hypothesis Why it can be eliminated

Recirculation Pump Stopped “no mechanical failure”

Pump-around Exchanger Plugged “no mechanical failure”

Cooling Water Supply Lost “no utility failure”

Catalyst activated by feedstock “only single reactor”

Heat Generated>Heat Removal not eliminated

Capability




Step 4 – Provide a mechanism to test the hypothesis.

Calculations can be used to test the hypothesis.

The hypothesis is summarized below:

dQg / dT > dQr / dT

The manual shows these calculations in detail.




A summary of the calculations is as follows:

dQg/dT = (K*11000/T2)*e(-11000/T)

dQr/dT = U*A

dQg/dT = 170000 BTU/hr-F

dQr/dT = U*A = 144000 BTU/hr-F

Since dQg/dT > dQr/dT, Hypothesis is valid





Step 5 - Recommend remedial action to eliminate theproblem without creating another problem.

Determine the minimum “UA” value.

Begin daily monitoring and plotting of this variable.

Remove the exchanger from service when the “UA”value drops below the minimum.




Items to consider:

How would you estimate a minimum “UA” value?

What is the advantage of calculating a “UA” value asopposed to just watching the position of the coolantcontrol valve on a daily basis?

What kind of problems can you visualize inimplementing this problem solution?



MANUAL SECOND PROBLEM



Rotary Filter Screen Tear

A process plant using a rotary filter was plagued bydowntimes caused by tears of the screen cloth on one outof 3 rotary filters. Whenever the screen cloth would tear,solids would enter the liquid stream causing a shutdown ofcritical equipment and a resulting shutdown of the plant.After each screen cloth tear, the screen cloth and filterdrum were carefully examined and the followingobservations were made:




The metal drum would be scratched.

Solids would be present between the cloth and thedrum.

The cloth would be torn in a circumferential mannerwith most of the tears occurring in the middle 60 to70% of the cloth.



FIGURE 3-2ROTARY FILTER SCHEMATIC

baffle

Blowback gas

solidsSlurry to filter

Blowback gas fromCompressor KO drum



While there was no doubt that the problem occurred,there was serious question if the problem was worsethan in the past.

Key people in different parts of the organization hadalready taken very strong positions as follows:

• Mechanical—The hard solids deposits are cutting thecloth.

• Research – There is liquid in the blowback gas and itis cutting the cloth.




• Technical – Something inside the filter is rubbingagainst the cloth cutting it and scratching the metaldrum.

There were huge financial losses for each screenfailure. In other words “Do something quick!!!”




Step 1 – Verify that the problem actually occurred.

A review of mechanical records indicated the following:

Time period MTBF, Days Tear type

Past data 43 horizontal

Current data (all runs) 16 circumfer

Current data (ex short runs) 25 circumfer




A further review of what changed between the pastand current data revealed that the filtrationtemperature on this filter was increased from 130F to170F. Returning to 130F the previous processcondition was not a satisfactory solution to theproblem.




Step 2- Write out an accurate statement of what problemyou are trying to solve.

“Determine the cause for the significant change in screentearing frequency that occurred on only one filter at thesame time as the filtration temperature was increased. Inaddition, to a reduction in mean time between screenfailures, the nature of the screen failure changed.Previous failures were fatigue failure caused by the clothbeing weakened during flexing while being held in placeby the tension rods. The current failure is a catastrophiccircumferential failure. The current failure is alsocharacterized by scratch marks on the metal drum.”




Step 3- Develop a theoretically sound working hypothesisthat explains the problem.

Since the new failure mode appears to be related to theincrease in filtration temperature, the followinghypotheses were developed:

• The screen cloth is decomposing at the highertemperatures.

• The baffle (see Figure 3-2) is expanding due tothermal growth and bowing into the filter cloth andmetal drum. See Figure 3-3.




• The rotating drum is deforming at the highertemperatures causing poor distribution of blow backgas. The poor distribution causes an increase inblowback gas in the middle of the drum that thenblows the filter cloth into the baffle causing thecloth to tear.




Figure 3-3Hypothetical Baffle Deformation

TOP VIEW

Original Baffle Position

Deformed Baffle Position



Step 4 – Provide a mechanism to test the hypothesis.

This hypothesis can be tested by calculations ofthermal growth of the baffle as shown in the manualon page 35.

These calculations show that a filtration temperature90Fo above ambient temperature would causesufficient baffle growth for the baffle to contact therotating filter drum.

The baffle could then bow either into the rotatingdrum or away from it.




Two alternatives were available for further testing this

hypothesis:

The filtration temperature could be reduced to130Fo. This alternative would require 215 days oflower temperature operation to provide a 90%confidence level that the hypothesis was correct.

A mechanical constraint could be provided tocause the baffle to always bow away from thedrum.




Step 5 - Recommend remedial action to eliminate the

problem without creating another problem.

The alternative technique of mechanicallyconstraining the baffle so that it always bows awayfrom the drum provides both a testing procedure anda permanent solution.

What is missing from this problem solving analysis?




MANUAL THIRD PROBLEM



Kinetic SystemsBernoulli’s theory is the key to understanding these systems.

P/ + (v2)/2gc + z = -w – lw

Where:

P = The pressure difference between twopoints.

= The fluid density.(v2) = The difference in velocity squared

between two points.gc = The gravitational constant.z = The difference in elevation between two

points.w = The amount of work added by the prime

mover.lw = The frictional loss in the piping system.



Displacement Systems The term “Displacement Systems” refers to the prime

movers which displace a fixed amount of fluid (liquid orgas) essentially independent of the differential pressureacross the pump or compressor. Typical equipment itemsthat fit this category are reciprocating pumps/compressorsand rotary pumps/compressors.

The important concepts to understanding this class ofprime movers are as follows:

• Energy is imparted by displacement of a fixed volumeof fluid.

• The mass flow rate will depend on fluid suctionconditions and physical dimensions of the equipment.

• Figure 4-4 shows a typical flow pattern for areciprocating pump/compressor.



Figure 4-4 Reciprocating Flow Path In DoubleActing Equipment



Dynamic Systems“Head Curve” Considerations

The head curve is developed by the equipment supplierand is provided as part of the equipment purchase.

Operation to the left of the stability limit will result inflow instabilities as flow surges forward and thenbackwards through the prime mover. The stability isusually well defined for compressors and blowers.However for pumps, it is usually 25 to 40% of the BEP(best efficiency point).



Dynamic Systems“Head Curve” Considerations

While the head curve is usually developed using wateror air, it is valid for any fluid if the correct units areutilized for flow and head. These units are definedlater.

As shown in Figure 4-2, the horsepower requirementsnormally peak at the “end of the curve” (maximumflow rate). The driver for the prime mover may or maynot be provided with “end of the curve” protection.



Figure 4-2 Characteristic Centrifugal Pump orCompressor Curve

Head- Feet

StabilityLimit

BHP

Flow / Volume / Minute



Important Definitions

NPSHR = Net Positive Head Required. This is the headin feet required to overcome the pressure loss betweenthe pump suction flange and pump impeller eye. Thepump supplier will specify this. A typical NPSHR vs.flow rate curve is shown in Figure 4-3. Note that thiscurve is usually developed with water, but it is validfor any fluid.




NPSHA = Net Positive Head Available. This is thedifference in feet of head between the actual pressureat the pump suction flange and the vapor pressure ofthe liquid being pumped. If the liquid has been storedunder a nitrogen, air or inert gas blanket, somequestion may arise regarding the actual vapor pressureof the liquid. The most conservative approach is toassume that the vapor pressure is equivalent to thepressure in the storage vessel.



NPSH

FLOW RATEVOL/TIME

FIGURE 4-3 NPSH REQUIRED




Cavitation = A condition that occurs if NPSHR > NPSHA.If this situation occurs, some of the liquid beingpumped will vaporize between the pump suction flangeand the pump impeller. This will cause the pump tooperate off the head curve and damage may occur tothe impeller.




Horsepower load point – This is a unique feature of apositive displacement compressor. It is the point on aplot of horsepower versus suction pressure where therequired fluid horsepower is at a maximum. To one sideof this point, increasing mass flow increases therequired horsepower. To the other side of this point,decreasing compression ratio decreases the requiredhorsepower.



Volumetric Efficiency - The actual volume of fluid

displaced relative to the dimensions of the cylinder of a

reciprocating pump/compressor or rotating pocket of a

rotary pump/compressor. For a liquid, this efficiency

approaches 100%. However for a gas, it is approximately

70%. The differences are due to the compressibility of

gases.

Leakage – This is an additional loss in volumetric

efficiency caused by leakage through clearances.




Clearance – This is the part of the cylinder in areciprocating pump/compressor that is not displacedcompletely by the piston. Figure 4-5 shows a typicalsketch of a “clearance pocket”. Because of thecompressibility of gases, this becomes more importantfor gases than liquids.




Figure 4-5 Clearance Pocket ReciprocatingCompressor

Simplified Sketch

ClearancePocket

Cylinder

PistonAnd Rod

Gas at Discharge Pressure is trapped in the clearance pocket andexpands into the cylinder as the cylinder pressure is reduced. Alarger clearance pocket results in more gas expanding into thecylinder.



Calculation Considerations

Compressor calculations are more complicated thanpump calculations because gases are compressible.

Polytropic/Adiabatic compression must be used toevaluate head. DO NOT USE DIFFERENTIAL PRESSURE.

The determination of the amount of NPSHA is morecomplicated for reciprocating type pumps than forcentrifugal or rotary pumps.

Section 2

Applied Economics Valuation Principles and

Methods Other Views Valuation Principle

and Methods Compressor - Compressor

Problems - Simplified Approach Distillation, Plates, Tray

Stability Guidelines for Problems Solving

Temperature, Pressure, LevelMeasurements , Verification

Sample Exercise Kinetics, Flow,Mechanical, Design



Section 2

Process Operation Troubleshooting

&

Problem Solving

DAY 2



Course ProgramDay 2 : November 2007

08:00 – 08:30 Applied Economics08:30 – 09:00 Valuation Principles and Methods09:00 – 10:00 Other Views Valuation Principle and Methods10:00 – 10:15 Break

10:15 – 11:45 Compressor - Compressor Problems - Simplified ApproachPumps - Pumps Problems - Simplified Approach

11:45 -13:00 Lunch13:00 -14:30 Distillation, Plates, Tray Stability14:30-15:30 Guidelines for Problems Solving Temperature, Pressure,

Level Measurements , Verification15:30 – 15:45 Break15:45 – 16:30 Sample Exercise Kinetics, Flow, Mechanical, design



Applied Economics

In a free enterprise system, the primary motive forindividuals investing in a business is to make a profit.

Business decisions are also influenced by laws beingenforced by regulatory agencies and Govermentrulings depending on the Country.



In the area of energy conservation the followingactivities could reduce the amount of energyconsumed per unit of production:

• Changes in start-up, shutdown, and operatingprocedures.

• Changes in plant equipment could bring ways tocut energy usage.

Applied Economics



Applied Economics

Changes in start-up, shutdown, and operating procedures

These changes did not require any additionalinvestment of money in the form of equipment ormaterials.

The only costs incurred are the man-hours plantpeople expended in investigating, analyzing, andchanging the operating procedures.

If the people involved are the supervisors andsuperintendents the man-hour costs are minimized.These people should be continually upgrading theiroperations as one of their job duties.



Changes in plant equipment could bring ways to cutenergy usage:

When new equipment is required, some existingequipment may have to be scrapped.

All these decisions require additional investment in theplant.

Unless this investment can be justified from the profitviewpoint, it should not be done.

Applied Economics



Profit

Profit is the excess of revenues of products over theircost. It is also the compensation to investors for theassumption of the risk in the business enterprise.

Money returned to you comes from the profitsgenerated by the business.

Management should have economic guidelines toevaluate energy savings proposals.

Applied Economics- Definitions



Net Back

The definition of net back depends upon your company'saccounting procedures.

It may be defined as the total price of products sold toyour customers less all the transportation costs todeliver the product from the plant to the customers.

Another definition is: Instead of subtracting out all thetransportation costs, use the part of the transportationcosts that your company pays that exceeds the cost oftransportation from the nearest competitor to thecustomer.

Applied Economics - Definitions



Net Back (Cont...)

Net back is also considered as the total price ofproducts sold to your customers less all selling andtransportation costs.

At the plant site, the operating people have nocontrol over transportation and selling costs, so foryour plant's economic decisions, the last definitionseems best.




Depreciation.

Depreciation is the reduction in value of physicalassets (i.e. plant equipment due to physicaldeterioration, technological advances, economicchanges, etc.) that leads to retirement of the physicalasset.

For tax purposes, depreciation is different from truephysical deterioration in determining if the additionalequipment can be purchased and installed for energysavings and be attractive to management.




Let us assume you estimate the company must buy$100,000 in plant installed equipment for your energysaving idea. The company must use its money to makethe installation. It has converted capital as money tocapital as equipment.

When this equipment is operated, it deteriorates fromuse.

The money deposited in a savings bank stays thesame, but the money represented by investments inequipment (car) disappears as the equipment is usedand ages.




Depreciation

For tax purposes, the Internal Revenue Servicerecognizes depreciation as a cost. The IRS has setguidelines on the life of the capital equipment.Various accounting methods distribute cost over theofficial life. Note that under inflationary, economicconditions, the replacement cost of equipment ismuch higher than the depreciation recovered.




If the Internal Revenue makes a company depreciate aspecific piece of equipment over a ten year period,after this time there is no penalty for keep using itbecause the accounting sheets will no longer show theequipment, but the company can continue to use it.

If your energy saving proposal requires the removal ofequipment from the plant that still has say five moreyears of depreciation on it, how does the accountanthandle this?




Depreciation (Cont..)

An example will illustrate the procedure:

Assume the equipment originally cost $10,000, $8,000 ofdepreciation had been taken, and this equipment was soldfor $1,500.

The book value was $10,000 - $8,000 = $2,000.

It was sold for $1,500 or a $500 loss on the books.

Thus, the Internal Revenue Service allows the company toconsider this a loss in sales revenue.

If the company sold the equipment for $2,500, the gain was$500 and the sales revenues would be increased by $500.

Applied Economics -Definitions



Investment Tax Credit

The federal government may attempt to stimulateeconomic activity by permitting tax deduction equalto some percentage of a plant's new investment inequipment.

In some cases a proposed processing unit is noteconomically attractive because it does not generatesufficient profits.

If the federal government allows less tax money to goto the government, more money is retained by thecompany. The proposed venture may now beattractive.

Presently, an investment tax of 10% is allowed.




Fixed Costs

Fixed costs are defined as those costs which do notdepend on the production rate of the processing unit.

For example, a fixed cost of $1,000,000 each yearmeans this cost has the same value whether theprocess produced 50% of its yearly operating capacity,or 100% of its yearly operating capacity.

Examples of fixed costs are rents, property taxes,insurance, maintenance labour, repair parts, andoperating labour.




Variable Costs

Variable costs are manufacturing costs that varydirectly with volume of production. Examples arechemical materials used in the process and theutilities used. Utilities include fuels, electricity,steam, and cooling water.

Although costs are usually considered either fixed orvariable, sometimes a fixed cost could have someelements of a variable cost, and vice versa.




For example operating labour is generally considereda fixed cost. At 75%, 85% or 100% of operatingcapacity, the processing unit requires the samenumber of operators and supervisors. Maybe at 40% ofcapacity, the company can operate the unit ten daysand shut down for four days without affecting otheroperations. Operating labour costs have been reducedby an incremental drop of 4/14 x 100 or 29%.

In the above case the Operating costs take on avariable cost aspect.




Cash Flow

Cash flow is the difference between actual cash thatcomes into the plant and the actual cash that leaves.

This cash is primarily in the form of checks.

Cash generated from selling product is returned to theplant.

Cash expended for paying wages, fringe benefits,utilities, taxes, raw materials, operating supplies,etc. leaves the plant.




Discounted Cash Flow (D.C.F.)

An investment is usually evaluated by the discountcash flow method when payments are made toconstruct the facility at the beginning of a periodfollowed by varying returns over the life of theproject.




It takes into account the time value of money. Forexample, if a company had the following twoprocesses with a life of two years to consider:


Process 1 Process 2

Investment $1,000,000 $1,000,000

1st year 900,000 100,000

2nd year 200,000 1,000,000Cash Generated

TOTAL $1,100,000 $1,100,000



Return on Investment (R.O.I.)

This is the ratio of the yearly profits averaged overthe life of the investment to the original investment.

The original investment includes working capital. Inthe example under "Discounted Cash Flow" let usassume each process made $50,000/1,000,000 x 100 =5% each year.

Over the life of the processes, they generated$100,000 in profit and recovered the $1,000,000 inthe investment before the processes becametechnically obsolete and were torn down.




When we inspect the two methods of determiningwhether to invest in a process, we realize that theR.O.I. showed both processes equally attractive.

Since the D.C.F. method included the time value ofmoney, it proved Process 1 to be more attractive.

Although both methods are generally included formanagement to decide what to do, the D.C.F. methodis more significant.




The Concept of Investment Equivalence to Save Money

Plant people need an easy way to cut out energysaving ideas so that valuable manpower will only beexpended on economically reasonable ideas.

Management can give to plant people the dollarvalues that can be spent to buy and install equipmentthat will save a unit of each type of energy.

Applied Economics



For example, management says you can invest up to$800 to save a continuous kilowatt demand ofelectricity by removing a pump as a result of revisingthe piping system.

The cost of the change is estimated at $6,000.

You can invest up to (20 KWHr/hr)(800) or $16,000.

Thus, you readily conclude your idea is viable andshould be presented to management for action.

Applied Economics



The Concept of Investment Equivalence to Save Money(Cont..)

When management studies the idea, they will performmore precise calculations.

For example, the depreciated value of the equipmentbeing disposed of may be $5,000. This is a loss.

Thus the total cost is not $6,000, but $6,000 + $5,000 or$11,000.

There is also a loss in production during the period ofremoving old equipment and adding the new piping.

Applied Economics



The Concept of Investment Equivalence to Save Money(Cont..)

When the production unit is operating at the maximumeconomic rate, management may postpone any changesuntil demand drops off and any loss in production can berecovered.

Since these type of decisions are the responsibility ofmanagement, the concept of having guidelines for theoperating people to initiate ideas that have a goodchance to be accepted is very important.

Applied Economics



Valuation

Principles and Methods



The value of any asset, tangible or intangible,commercial, industrial or financial, is the present valueof the expected stream of cash flows from this asset atthe cost of capital

CF = CF1 / ( 1+ k) + CF2 / ( 1+ k)2 + ………… + CFn / ( 1+ k)n

CF = cash flow available to the investors (shareholders andlenders) who financed the asset.

k = Is the opportunity cost of capital. This is the return thatinvestors can get from alternative investments in the samerisk class as the asset.

n = The expected economic life of the asset.

Other Views – Valuation Principles and Methods



The basic discounted cash flow (DCF) approach:

Simple example :

An investor has decide to purchase a flat(apartment) withrenting purposes. A rent forecast has been prepared inEuros for the next three (3) years. The annual rent isexpected to be paid at the end of each year:

Year 1 Year 2 Year 3

Rent 10000 10500 11000

Resale Value 180000

Forecast in Euros




The investor is expecting a minimum return of 8%, reflectingthe rate that he could expect to earn from an alternativeinvestment .

How much is the investor should pay for this apartment today ?

In the same way, the 10,500 Euros that the tenant would payyou at the end of the second year is worth today :

10,000 Euros expected one year from now are worth for the investor:

10,000 / (1+0.08) = 9,259 Euros

10,500 / (1+0.08)2 =9,002 Euros




Financially a Euro in the bank is worth more than aEuro which you will receive in a year´s time .

If you have the Euro today you could use it to makemore money in a year´s time.

A bird in the hand is worth two in the bush.

Discounting is a method of translating future valuesinto present values in order to enable apples – to –apples comparisons.




Following previous simple example, the 11,000 Euros(third year) due from renting the flat is worth today :

11,000 / (1+0.02)3 = 8,732 Euros




The investor would like to sell the flat right after theend of the third year contract when the tenant isleaving the flat .

The resale value presented at the beginning of 180,000Euros is equal to :

180,000 / (1+0.08)3 = 142,890 Euros




The value of the flat in today´s money is 142,890Euros

The total resale value of the apartment today will bethen :

£ 9259 +£ 9002 +£ 8732 = Apartment rent today’s value£ 142, 8909 sales price today

£ 169,883 will be the total resale price today




If the investor pay £ 169,883 for the flat today andthe forecast is right on target that means that thisinvestment will provide a return of the 8% meetingthe initial requirements.

Now if the investor paid lower than £ 169,883, thenthe return obviously will be higher than 8%.

On the contrary, if the investor paid more than £169,883 the rate of return will be lower than 8%.

These above statements are the bases forunderstanding valuation of assets.




Studies done in England has proven that the high rate of failures

of mergers and aquisitions all over the world failed because:

The acquires pay too much

Acquirers are over optimistic in their assumptions and

estimations of the potential synergies.

If the investor was completely unrealistic about the

rent of the flat,it should be quite difficult to obtain

the expected return from your investment.

The acquirers are too optimistic in the return that it could

get from and alternative investment, which is the perfect

recipe for over value any asset.




Investment Decision Techniques

Using the flat example presented before :

• If the flat acquirer found that the flat could be saleat £150,000 today.

• The acquirer knows that the flat is worth £169,883taking into account the same assumptions for rentand terminal value over the next three years.

• Paying £150,000 today for the flat today it willgenerate a surplus of £19,883 in todays´money.

• This surplus of £19,883 is called the Net PresentValue (NPV) of the investment project.





Using the flat example presented before (Cont..):

NPV is the difference between the present value ofexpected cash flows and present value of initialinvestment.

NPV = CF1 / ( 1 + k) + CF2 / ( 1 + k)2 + ……..+ CFn / ( 1 + k)n - Io




Following this above correlation, some simple rules can

be established:

• If a project has a positive NPV, it creates valueand should be accepted.

• If a project has a negative NPV, it destroys valueand should be rejected.

NPV incorporates the magnitude, the timing and the risk

of expected future cash flow.





The most accurate investment decision criterion isthe NPV.

In some cases it is preferred to use IRR (internalrate of return) which seems more appealing.

IRR is the rate which makes the present value ofexpected cash flows equal to the present value ofinitial investment.

I0 = CF1 / (1+ IRR) + CF2 / (1+IRR)2 +…….+ CFn / (1+IRR)n




IRR is the rate at which NPV is equal to zero (0) ; NPV = 0

Using this formula it is found that the IRR of the real stateinvestment is 12.85% which is significantly higher thanyour minimum required rate of 8%. This investment underthe assumptions made is attractive.

CF1 / (1+IRR) + CF2 / (1+IRR)2 +…….+CFn / (1+IRR)n –I0 = 0





The IRR gives the correct accept-reject decision ifused carefully but it gives incorrect rankings for:

• Mutually exclusive projects, specially if theydiffer in scale (amount of initial investment) oreconomic life.

• When cash flows change signs, reinvestmentassumptions for cash flow that is toounrealistic.

The conclusion is do not use IRR use NPV instead.

The investors are in the business of maximizing valuecreation not in the business of maximizing rate ofreturn.




Value Creation

The economic value of a business depends on itsability to generate future cash flows.

A good measure of a business value is the presentvalue of the expected stream of cash flows.

The main tasks of management is to increase theseexpected cash flows to create value. The main tasksare:

• Acting on sales (growth of sales and its duration)

• Maximizing the operating margin

• Managing taxes

• Minimizing the investment in net working capital

• Rationalizing the investment in fixed capital




Value Drivers

Promote growth

• Increase business with current customers

• Pursue high growth segments within broadermarkets

• Expand global presence

• Pursue complementary alliances and acquisitions

Improve margins

• Focus on restructuring, efficiency, productivityand cost control




Lower working capital

• Increase inventory turns

• Focus on collection processes

• Get best conditions from suppliers

Optimize asset utilization

• Lower capital expenditures

• Improve turnover ratios




Reduce effective taxes

• Lower tax rates, by using all tax benefits providedby law

• Use international status to benefits from best taxprovisions

Optimize cost of capital

• Reduce cost of various financing means

• Do not deviate from optimal capital structure




Other Views–ValuationPrinciples and Methods

SampleSample ofof ProjectProject ValuationValuation







Sample of Project Valuation

Simplified Model

Exercise



COMPRESSORS



For critical compressors, use suction and dischargetemperatures to monitor compressor performance.Equation (4-8) can be used on a daily basis to confirmthe performance of a critical centrifugal or positivedisplacement compressor. If the compressionexponent () begins to increase, it is an indicationthat the compressor is becoming less efficient.

Compressor Problem Solving Approaches



A similar approach can be utilized for monitoringsteam turbines that are often used as drivers on largepumps and compressors. The efficiency of a steamturbine can be determined by comparing the actualchange in enthalpy with that predicted assuming anisentropic (constant entropy) expansion. For criticalturbines, the efficiency can be calculated andmonitored on a daily basis. This will allow spottingmechanical problems before they become so severethat an immediate repair is required.




For centrifugal compressors, use the compressorperformance curve to analyze problems. To do thisaccurately will require the following:

• Make sure that all field instruments have beencalibrated before taking any data.

• Develop the kinetic head as described in equation(4-6).

• Calculate the ACFM as accurately as possible. Ifnecessary, adjust the metered flow rate fordifferences in pressure, temperature and molecularweight between the meter calibration and actualconditions.




Make sure that you know the gas composition since itcan effect:

• Molecular weight and flow rate in lbs/hr. ,• Calculated head,• Temperature difference between the suction and

discharge.

If problem solving involves a plant test on acentrifugal compressor, beware of increasing thespeed at constant volumetric flow. Since the surgepoint normally increases with speed, the compressorcould go into surge if the volumetric flow rate ismaintained constant.




Remember that internal restrictions in a compressorcan cause the actual pressures to be different fromthe measured pressures. This could cause thecompression exponent calculated from inlet andoutlet temperatures to be higher than anticipated.

The most likely cause for loss of capacity for areciprocating compressor is leaking valves. Thisproblem can be detected by using equation (4-8).Chronic valve malfunction is often due to liquid orsolids entrainment.




Compressor ProblemsSimplified Approach



Compressors

Factors to consider:

• Flow rate

• Head or pressure

• Temperature limitations

• Method of sealing

• Method of lubrication

• Power consumption

• Serviceability

• Cost



The combination of the most adverse conditions occurringsimultaneously during the process must be determinedwhen selecting a compressor:

Lowest barometric pressure

Lowest intake pressure

Maximum intake temperature

Highest ratio or specific heat (k values)

Lowest specific gravity

Maximum intake volume

Maximum discharge pressure

Compressors



Choice is made on economics and mechanical factors

Centrifugal compressors is the most common choice.

It produces the most power per unit weight and volume.

Less expensive per unit of power output

Ideal for high flow rate, medium head situations

But

Overall efficiency is lower than a reciprocal and fuelconsumption is higher than a reciprocal compressor

More complex in terms of specifications and changes inoperating specifications

General Range of Application of Compressors



The theoretical amount of energy needed to compress agiven amount of a gas between specified suction anddischarge conditions is independent of the compressorunit.

The basic thermodynamic equation is written:

∆H = ∫ Vdp = - Wtheor

based on the assumptions: process is reversible (S1 = S2)and adiabatic (Q= 0)

Wact = (Wtheor) / E ; E= overall efficiency (isentropic andmechanical efficiency)

Compressors



The performance of the centrifugal compressor, at

speeds other than design, is such that

The capacity will vary directly as the speed, Qproportional to N

The head developed as the square of the speed, Hproportional N2

And the power as the cube of the speed, Powerproportional N3

As the speed deviates from design values, the error of

these rules increases

Performance Laws



Estimating Performance

Using charts from GPSA handbook it is possible to makean estimate of compressor performance.

Head Volume flow Discharge temperature Number of compressor wheels Horsepower Efficiency



Centrifugal compressors handleInlet flow ranges going from 800m3 / hour up to 340,000 m3 / hour.

Single stage (single wheel)800 – 250,000 m3 / hour

Multi stage (multi- wheel)800 – 340,000 m3 / hour

In the higher values of flow inletsIs a overlap between theCentrifugal compressor andAxial CompressorIn this case for a technicaldecision other factors needto be taken into account :economics, service,Operational requirements, etc.































1) For adiabatic Calculation:

a) Headadiab = [ T1ZaR / (k-1/k)(MW)] [(P2/P1)(k-1/k) -1 ]

(P2/P1) compression ratio

lb/lbmolKg/kmolMW Molecular Weight

K ratio of heat capacities Cp/Cv

1545 ft.lbf/lbmol.R1.99 Btu / lbmol.R

8.314 KJ/kmol K;

848 kg.m / kmol.K

R gas law constant

Za average compressibilityfactor (Z1+Z2)/2

RKT1 suction temperature

ft – lbf /lbm ; Btu/lbmkJ/ kg ; mHead adiabatic

EnglishMetric

Calculation Procedures,(Mollier chart is not available)



b) Compression Power:

kW = (Mass of gas per time) (Head per unit of mass)/ (E)(Energy Conversion Factor)

Compression Power kW = w1 (kg/hr) X Hadiab

ηad X 3 600 000 [J / (kW.h)]c) Brake Power, includes bearing and seal losses

Brake Power = Compression Power + Losses• Losses = 40 kW for 40 small machines

= 80 kW for large machines (8000 kW)• Losses = (Compression Power)0.9• Losses (UOP criteria) = 0.1 * Compression Power

Calculation Procedures(Adiabatic Calculation, Cont.)



d) Brake Power, includes

bearing and seal losses

Mechanical Losses WL

WL = FL (N/1000)2

WL = Mechanical Losses kW

FL = loss factor

N = Compressor speed, rpm15.417

8.312.5

4.458.7

2.386

1.284.25

0.693.1

0.372.1

0.201.5

FLQ, m3/s




e) Theoretical discharge Temperature:

∆Tideal = T1 [(P2/P1)(k-1)/k - 1]

T2 =T1 + ∆Tideal (theoretical discharge temperature)

Actual Theoretical discharge Temperature:

∆Tactual = T1 [(P2/P1)(k-1)/k - 1] / ηad

T2 =T1 + ∆Tactual (actual discharge temperature)

Note: For centrifugal compressors ηad could be estimated as0.7 to 0.75

For reciprocating 0.7 to 0.75 for high speed units and0.8 to 0.85 for low speed




Polytropic efficiency, ηp, is independent of gas properties orwheel performance with respect to pressure ratio

Polytropic efficiency : n/(n-1) = [k/(k-1)] ηp

Calculations: Headp = [ T1ZaR / (k-1/k)(MW)] [(P2/P1)

(n-1/n) -1 ] Compression Power kW = w1 (kg/hr) X Headp

ηp X 3 600 000 [J / (kW.h)] Brake Power = Compression Power + Losses Polytropic and adiabatic head are related by Headp = Headadiab * ηad / ηp

Calculation Procedures(Polytropic Calculation)



Number of Impellers and Speed

Using a fairly conservative tip speed of 250 m/s (820 ft/s) themaximum head which can be imparted to the gas across oneimpeller is approximately 3200 m.

• Number of impellers = Head m / 3200 m per impeller

Impeller Diameter

• d = (q/0.050 u)0.5

q inlet flow rate ( m3/s - ft3 /s)

u impeller tip speed

Compressor Speed

• N = 60 u /d`π

u impeller tip speed

d impeller diameter

Calculation Procedures



With the given inlet conditions the enthalpy can beobtained from the Mollier chart, Point 1

From Point 1 follow the line of constant entropy tothe required discharge pressure P2, locating theadiabatic discharge state point.

With these two point, calculate the differentialadiabatic enthalpy

∆Had = H2ad – H1,

To find the discharge enthalpy:

H2 = ∆Had / ηad + H1

Calculation Procedures Mollier Chart Method



∆Had

∆H

1

2

2ad

2

Calculation ProceduresMollier Chart Method



The actual discharge temperature can now beobtained from the Mollier chart

Compression Power kW = w1 (kg/hr) X Head pηp X 3 600 000 [J / (kW.h)]

Brake Power = Compression Power + Losses

Polytropic and adiabatic head are related byHeadp = Head adiab * ηad / ηp

Calculation ProceduresMollier Chart Method



Centrifugal Compressor Characteristics:

A motor-driven centrifugal compressor will operate at aconstant speed. The constant-speed feature, along withseveral other intrinsic characteristics, result in a lack offlexibility in the centrifugal machine.

The most infamous restrictive characteristic is surge.

Below a certain capacity or suction volume, a centrifugalcompressor will enter an operating range where itsperformance becomes unstable.

The minimum suction volume required to prevent acompressor from surging is different for each machine.

Compressor vendors supply a performance curve with eachmachine.

Understanding Compressor Surge



When the volume that the compressor is pumping falls tothe left of the surge line, the rotor (i.e., that part of thecompressor that is spinning) begins to slide back and forthacross its radial bearings.

The end of the compressor shaft hits the thrust bearing(i.e., the component that constrains the axial movement ofthe rotor).

Each movement of the shaft against the thrust bearing iscalled a surge. Depending on the speed of the compressorand its mechanical strength, it can withstand 100-2000surges before the thrust bearing is damaged and thecompressor self destructs.

In general, the slower the speed of a compressor, the moreinsensitive it is to surging. A speed of 3000 rpm is quiteslow for a centrifugal machine, while 10,000 rpm isextremely fast.

What is Surge?



A centrifugal compressor backed-up on its curve to the left ofthe surge line will self-destruct

Compressor Surge

PE114-11-07 Process Plant Troubleshooting_1.pdf

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