KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY, KUMASI INSTITUTE OF DISTANCE LEARNING CHE 451 Chemical Process Design & Economics (Credits: 4)
Benjamin Afotey, PhD Chemical Engineering Department August, 2014
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Course Introduction Design is the synthesis of ideas to achieve a desired goal (product). The designer starts with an idea and proceeds to develop several alternative designs that he evaluates and finally settles on the one that satisfies his objective (goal). The search for alternative designs is important to the economic viability of the design project. CHE 451 is a fourth year core course offered in first semester in BSc. Chemical Engineering.
Course Overview Methodology of the Design Process: Constraints on a Design Problem; Fixed/Rigid constraints, Less rigid constraints, The design process; Design objectives, Data collection, Generation of possible designs,Selection, Chemical manufacturing processes, Continuous and batch processes, Organization of a chemical engineering design project. Codes and Standards, Design Factors, Variable & Mathematical Representation of Design Problems: Codes and standards, Design factors, Systems of units, Mathematical representation of the design problem, Selection of design variables. Optimization and Batch Production Process: Introduction, Simple models, Multiple variable systems, Methods of analysis, Other optimization methods, Batch production process. Process Synthesis: Introduction, Raw materials and chemical reactions, Summary of process design heuristics, Heuristics in equipment design. Process Simulation: Introduction, Process simulator, Types of process simulation, Units operation solvers, Uncertainty and sensitivity issues. Flow sheeting, Piping and Instrumentation: Introduction, Piping & instrumentation diagrams, Valve selection, Pumps, Classification of pumps, Factors to consider in pump selection, Centrifugal pumps, Effective characteristics curves, Design parameters of centrifugal pumps, Operating point, Choice of rotational speed,. Process Economics: Cost estimation, Cash flow for industrial operation, Factors affecting investment and production costs, Capital investment, Estimation of capital investment, Types of capital cost estimates, cost indices, Methods of estimating capital investment, Turnover ratio, Estimation of total product cost, Break-even point. Process Economics: Depreciation and profitability analysis, Service life, Salvage value, Present value, Methods for determining depreciation, Profitability standards, Basis for evaluating project profitability, Mathematical methods for profitability evaluation, Rate of return on investment, Discounted cash flow, Capitalized cost, Pay-back time, Sensitivity analysis.
Course Objective The following is the main course Objective 1. To enable students understand and acknowledge the importance of economic analysis in chemical process design, by way of efficiently maximizing profit.
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Course Outline The course outline is divided into eight units. Each of the eight units is broken into subtopics. Each unit addresses one or more of the course objectives. Unit 1: Methodology of the Design Process Unit 2: Codes and Standards, Design Factors, Variable &Mathematical Representation of Design Problems Unit 3: Optimization and Batch Production Process Unit 4: Process Synthesis Unit 5: Process Simulation Unit 6: Flow Sheeting, Piping and Instrumentation Unit 7: Process Economics: Cost Estimation Unit 8: Process Economics: Depreciation and Profitability Analysis
Grading Continuous Assessment: 30% End of Semester Examination: 70%
Reading List/Recommendation Textbooks/Websites/CDs 1. Plant Design and Economics for Chemical Engineers: By Peters and Timmerhaus 2. Coulson & Richardson’s Chemical Engineering Design: By Sinnot, Volume 6
Course Writer Dr. Benjamin Afotey received his BSc. Degree in Chemical Engineering in 2000 at Kwame Nkrumah University of Science and Technology, Kumasi. He received his MSc. and PhD Degrees at the University of Texas, Arlington, and U.S.A in 2003 and 2008 respectively. He worked with the Texas Commission on Environmental Quality, U.S.A between 2008 and 2009 and joined the Chemical Engineering Department in 2010.
Acknowledgement I wish to thank the Almighty God for His guidance throughout the write up. I would also like to thank my colleagues who provided encouragement of any kind. Finally, I acknowledge the effort of my Teaching Assistant who contributed in a way to the successful completion of the material.
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TABLE OF CONTENT 1.0 METHODOLOGY OF THE DESIGN PROCESS SESSION 1-1 : 1-1.1 Introduction 1-1.2.Constraint on a Design Problem 1-1.2.1 Fixed/Rigid constraint 1-1.2.2 Less rigid constraint 1-1.3 The Design Process 1-1.3.1 The design objective 1-1.3.2 Data collection 1-1.3.3 Generation of possible designs 1-1.3.4 Selection 1-1.4 Chemical Manufacturing Processes 1-1.5 Continuous and Batch Processes 1-1.5.1 Choice of continuous verses Batch processes 1-1.5.2 Organization of a chemical engineering design project 2.0 CODES AND STANDARDS, DESIGN FACTORS, VARIABLE & MATHEMATICAL REPRESENTATION OF DESIGN PROBLEMS SESSION 2-1 2-1.1 Codes and Standards 2-1.2 Factors of Safety 2-1.3 System of Units 2-1.4 Mathematical Representation of the Design Problem 2-1.5 Selection of Design Variables 3.0 OPTIMIZATION AND BATCH PRODUCTION PROCESS SEESION 3-1 3-1.1 Introduction 3-1.2 Simple Methods 3-1.3 Multiple Variable Systems 3-1.4 Methods of Analysis 3-1.5 Other Optimization Methods 3-1.6 Batch Production Process 4-0 PROCESS SYNTHESIS SESSION 4-1 4-1.1Introduction 4-1.2 Raw Materials and Chemical Reactions 4-1.3 Summary of Process Design Heuristics 4-1.4 Heuristics in Equipment Design 5.0 PROCESS SIMULATION SESSION 5-1 5-1.1 Introduction 5-1.2 Process Simulator 5-1.3 Types of Process Simulation 5-1.4 Unit Operation Solvers 5-1.5 Uncertainty and Sensitivity Issues iv
1 1 1 1 1 1 2 3 4 4 5 5 7 7 7 11 11 11 12 13 13 18 20 20 20 21 22 23 24 24 26 26 26 26 34 34 38 38 38 39 39 40 40
6.0 FLOW SHEETING, PIPING AND INSTRUMENTATION SESSION 6-1 6-1.1 Introduction 6-1.2 Piping and Instrumentation Diagrams 6-1.3 Valve Selection 6-1.3.1 Gate valves 6-1.3.2 Globe valves SESSION 6-2 6-2.1 Introduction of Pumps 6-2.2 Classification of Pumps 6-2.3 Factors to Consider in Pump Selection 6-2.4 Centrifugal Pumps 6-2.4.1 Effective characteristics curves 6-2.4.2 Design parameters of centrifugal pumps 6-2.4.3 Operating point 6-2.4.4 Q/H Curve versus Technical choices 6-2.4.5 Choice of rotation speed 6-2.4.6 Suction Conditions: Concept of NPSH 7.0 PROCESS ECONOMICS: COST ESIMATION SESSION 7-1 7-1.1 Cost Estimation 7-1.2 Cash Flow for Industrial Operations 7-1.3 Factors affecting Investment and Production Costs 7-1.4 Capital Investment 7-1.5 Estimation of Capital Investment 7-1.5.1 Introduction 7-1.5.2 Types of capital cost estimates 7-1.5.3 Cost indexes 7-1.5.4 Methods for estimating capital investment 7-1.5.4.1 Power factor applied to plant-capacity ratio 7-1.5.4.2 Detailed item estimate 7-1.5.4.3 Other Methods for Estimating Equipment or Capital Investment 7-1.5.5 Turn-over ratio SESSION 7-2 7-2.1 Estimation of Total Product Cost 7-2.2 Break-even Point 8.0 PROCESS ECONOMICS: DEPRECIATION AND PROFITABILITY ANALYSIS SESSION 8-1 8-1.1 Depreciation 8-1.2 Service Life 8-1.3 Salvage Value 8-1.4 Present Value 8-1.5 Methods for Determining Depreciation 8-1.5.1 Straight-line method 8-1.5.2 Declining – balance ( or fixed percentage) method 8-1.5.3 Other methods for determining depreciation v
42 42 42 43 44 45 46 46 46 46 47 48 48 49 49 51 52 53 57 57 57 58 59 59 61 61 61 62 63 63 66 67 68 68 68 70 75 75 75 76 76 76 77 77 77 78
SESSION 8-2 8-2.1 Profitability Analysis 8-2.2 Profitability Standards 8-2.3 Basis for Evaluating Project Profitability 8-2.4 Mathematical Methods for Profitability Evaluation 8-2.4.1 Rate of return on investment 8-2.4.2 Discounted cash flow 8-2.4.3 Capitalized cost 8-2.4.4 Payout period (or Pay-back time) 8-2.4.5 Sensitivity Analysis
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LIST OF FIGURES Figure 1-1.1 Design Constraints Figure 1-1.2 The Design Process Figure 1-1.3 Anatomy of a Chemical Process Figure 1-1.4 The Structure of a Chemical Engineering Project Figure 1-1.5 Project Organisation Figure 3-1.1 Effect of Constraints on the Optimum of a Function Figure 3-1.2 Yield as a Function of Reactor Temperature and Pressure Figure 6-1.1 Flow-sheet of Simplified Nitric Acid Production Process Figure 6-1.2 Polymer Production Diagram Figure 6-1.3 Commonly used Valves Figure 6-2.1 Approximate Range of Operation for the Three Main Types of Pump Figure 6-2.2 Basic Curves Characterizing a Centrifugal Pump Figure 6-2.3 Curve Characteristic of the System Figure 6-2.4 Variation of Operating Point by means of a Valve Figure 6-2.5 Variation in Specific Speed versus the Type of Impeller used Figure 6-2.6 Different Types of Characteristic Curves Figure 6-2.7 Variation in the Operating Point versus the Rotation Speed Figure 7-1.1 Cash Flow for an Overall Industrial Operation Figure 7-2.1 Cost Involved in Total Product Cost for a Typical Chemical Process Plant Figure 7-2.2 Break-even Chart for Chemical Processing Plant Figure 7-2.3: Project Cash Flow Diagram
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2 3 5 8 9 23 24 42 43 45 48 49 50 50 51 52 53 58 70 71 72
LIST OF TABLES Table 6-2.1 Main Types of Pumps Table 7-1.1: Cost indexes as Annual Averages Table 7-1.2 Typical Exponents for Equipment Cost vs. Capacity
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UNIT
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METHODOLOGY OF THE DESIGN PROCESS Introduction This unit discusses the methodology of the design process and its application to the design of chemical manufacturing processes. Learning Objectives After reading this unit, you should be able to: 1. Explain chemical process design 2. Distinguish between fixed and less rigid constraints 3. Know what constitutes the design process 4. List the basic components of a chemical manufacturing process SESSION 1-1 In this session we shall discuss what a chemical process design is, the constraints of the design problem, the components of the chemical manufacturing process and the difference between a continuous and a batch process.
1-1.1 Introduction: Design is the synthesis of ideas to achieve a desired goal (product). The designer starts with an idea and proceeds to develop several alternative designs that he evaluates and finally settles on the one that satisfies his objective (goal). The search for alternatives: this step becomes necessary because the designer will be constrained by several factors.
1-1.2 Constraints on a Design Problem 1-1.2.1 Fixed/Rigid constraints: these are constraints the designer must live with outside his influence. E.g. physical laws, government regulations and standards. The fixed constraints define the outer boundary of all possible designs.
1-1.2.2 Less rigid constraints: these are constraints the designer can manipulate inorder to arrive at the best design. E.g. materrials of construction, time. These are interrnal constraints over which the designer has some control. In summarry we have the following diagramatic sketch.
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Figure 1-1.1: Design Constraints 1-1.3 The Design Process The design process can be shown in schematic form in Figure 1-1.2.
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Figure 1-1.2: The Design Process
The diagram shows the design process as an iterative procedure because as the design proceeds the designer will be looking for information and ideas to refine the design.
1-1.3.1 The Design objective In the particular case of a chemical process plant, the objective/goal is to satisfy the public need for a product. In large commercial organizations, this need is identified by the sales/ marketing department. Before starting to work, the designer should obtain complete information/background on the need for the product and its application areas/uses.
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1-1.3.2 Data collection To proceed with the design, the designer must assemble all the relevant facts and data required. For process design, the information should include process alternatives, equipment performances, physical property data. In large design companies, they have “in house” manuals containing all the process “know how” on which the design is based and preferred methods and data for the frequently used design procedures.
1-1.3.3 Generation of possible designs At this stage the designer must come up with all possible solutions for analysis, evaluation and selection. To do this, he must rely on his own experience or that of others, using tried or tested methods. Chemical engineering projects can be divided into 3 types: 1. Modification, additions to existing plant often undertaken by the plant design group. 2. New production capacity to meet growing sales demand, and the sale of established processes by contractors. Repetition of existing designs, with only minor design changes. 3. New processes, developed from laboratory research, through pilot plant, to a commercial process. Here most of the unit operations and process equipment will use established designs.
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1-1.3.4 Selection The selection process can follow the following screening stages:
Possible designs (credible) – within the external constraints
Plausible designs (feasible)- within the internal constraints
Probable designs – likely candidates
Best designs (optimum) – judged the best solution to the problem
To select the best design from the probable designs, detailed design work and costing will be necessary.
1-1.4 Chemical Manufacturing Processes The basic components of a typical chemical process can be shown using the block diagram below.
Figure 1-1.3: Anatomy of a Chemical Process
Each block represents a stage in the overall process for producing a product from the raw materials. Each stage is a collection of equipment required to accomplish a defined task.
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Stage 1: Raw Material Storage Unless the raw materials are supplied as intermediate products from a neighboring plant, storage space is needed to hold several days or months supply. Types of storage required will depend on the nature of the raw materials, and the methods of delivery.
Stage 2: Feed Preparation Some purification of the raw materials will be necessary to render them in a form required for feed to the reaction stage.
Stage 3: Reactor In the reactor the raw materials are brought together under conditions that promote the production of the desired product. However by-products and unwanted compounds/ impurities will also be formed.
Stage 4: Product Separation After the reactor, the products and by-products are separated from any unreacted material. If in sufficient quantity, the unreacted material will be recycled to either the reactor directly or to the feed purification and preparation stage.
Stage 5: Purification Before sale, the main product is purified to meet product specification. If the by-product is also produced in sufficiently large quantities, it must also be purified for sale.
Stage 6: Product Storage Provision for product packaging and transport will be required. Besides some inventory of finished product must be held to match production with sales
Ancillary Process In addition to the main process stages/ units, provision will have to be made for the supply of utility services, process water, cooling water, compressed air, steam. Facilities are required for maintenance, firefighting, offices, laboratories and accommodation. 6
1-1.5 Continuous and Batch Processes Continuous processes are designed to operate 24 hours a day, 7 days a week, throughout the year (365 days). However down time is allowed for maintenance and process catalyst regeneration. Plant attainment: this is the percentage of the available hours in a year that the plant operates. This is usually 90-95%.
1-1.5.1 Choice of continuous versus batch production The choice will not be clear - cut, however one can use as a guide the following rules: Continuous 1. Production rate greater than 5*106 kg/h (5000tonnes/h) 2. Single product 3. No severe fouling 4. Good catalyst life 5. Proven process design 6. Established market 7. Cost can be reduced 8. Less labor Batch 1. Production rate less than 5*106 kg/h (5000tonnes/h) 2. A range of products or product specifications 3. Severe fouling 4. Short catalyst life 5. New product 6. Uncertain design
1-1.5.2 Organization of a chemical engineering design project The design work required in the engineering of a chemical manufacturing process plant can be divided into two broad phases: 7
Phase 1.Process Design This covers the steps from the initial selection of the process to be used, through to the issuing of the process flow-sheets; and includes the selection, specification, and chemical engineering design of equipment. In any organization, this phase is handled by the process design group composed of chemical engineers. The group is also responsible for the preparation of piping and instrumentation diagrams. Organization of a project group is shown in Figure 1-1.4 below.
Figure 1-1.4: The Structure of a Chemical Engineering Project
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Phase 2. The detailed mechanical design of equipment; the structural, civil and electrical design; specification and design of ancillary services. Other specialist groups will be responsible for cost estimation, and the purchase and procurement of equipment and materials.
The sequence of steps in the design, construction and start up of a chemical process plant is shown diagrammatically in Figure 1-1.5. .
Figure 1-1.5: Project Organization Project manager; a chemical engineer by training is responsible for the coordination of the project.
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SELF ASSESSMENT 1-1 (a) Distinguish between Fixed and less rigid constraints (b) List the components of a basic chemical manufacturing process and in just two sentences explain the importance of each in the manufacturing process. (c) In a tabular form, list six differences between a continuous process and a batch process.
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UNIT
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CODES AND STANDARDS, DESIGN FACTORS, VARIABLE & MATHEMATICAL REPRESENTATION OF DESIGN PROBLEMS Introduction This unit discusses the codes & standards, design factors and variables and mathematical representation of the design problems. Learning Objectives After reading this unit, you should be able to: 1. Distinguish between codes and standards 2. Understand the importance of design factors as a margin of safety in meeting design specifications 3. Appreciate the importance of mathematical representation of the design problem SESSION 2-1 In this session we shall study the difference between codes and standards, design factors and their relation to equipment safety and the mathematical representation of the design problem.
2-1.1 Codes and Standards The terms CODE and STANDARD are used interchangeably, though CODE should be reserved for a code of practices. That is a recommended design or operating procedure. STANDARD on the other hand refers to preferred sizes, eg. pipes, composition etc. In modern engineering practice we have standards and codes that cover various functions e.g 1. Materials, properties and composition 2. Testing procedures for performance, composition and quality 3. Preferred sizes, eg tubes, plates, sections 4. Design methods, inspection, fabrication 5. Codes of practice, for plant operation and safety All developed countries have national organizations responsible for the issue and maintenance of standards for the manufacturing industries and for the protection of consumers.
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In the U.K:; British Standards Institution In the U.S; National Bureau of Standards They are responsible for coordinating information on standards. Standards are issued by the Federal State and various commercial organizations. The major ones of interest to chemical engineers are: American National Standards Institute (ANSI) American Petroleum Institute (API) American Society for Testing Materials (ASTM) American Society of Mechanical Engineers (ASME)
International Organization for Standardisation (ISO) coordinates the publication of International Standards.
In Ghana, there is no national standards organization to coordinate local standards for industries. However there are national standard organizations with standards for the protection of consumers eg. Standards Boards, Food and Drug Administration (FDA), and the Environmental Protection Agency (EPA). Equipment manufactures work together to produce standardized designs and size ranges for commonly used items; electric motors, pumps, pipes, and pipe fittings.
2-1.2 Factors of Safety (Design Factors) Design is an inexact act; errors and uncertainties can arise in the design data available and in the approximations necessary in the design calculations. To meet design specifications, factors are included to give a margin of safety in the design so that the equipment will not fail to perform satisfactorily, and that it will operate safely. e.g. in a mechanical and structural design, the magnitude of design factors used to allow for uncertainties in material properties, design methods fabrication and operating loads are well established. In process design, design factors are used to give tolerances in the design e.g. process stream average flows calculated from material balances are often increased by a factor of 10%, to give some flexibility in process operation. This factor then sets the maximum flows for equipment, instrumentation and piping design. 12
2-1.3 System of Units Chemical engineering uses a diversity of units from American and British engineering Systems, CGS (grain, centimeter, second) MKS( kilogram, meter, seconds) English and American – pound mass (lb), foot, second or hours, pound force. If working in S. I units is preferred, data expressed in the American and British engineering systems can be converted to S.I units. Conversions factors are available in the literature.
2-1.4 Mathematical Representation of the Design Problem A process unit e.g. distillation unit in a chemical process plant performs some operation on the inlet material stream to produce the desired outlet stream.
Inlet stream
Process Unit
Outlet stream
In the design of such a unit, the design calculations model the operation of the unit. Thus the flow of materials is replaced by flow of information into the unit and flow of derived information out of the unit.
Input Information
Calculation method
Output Information
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Information flows are the values of the variables which are involved in the design. Example:
Input stream
Output stream
variables
variables
Flow rate
same as the input variables.
Composition Temperature Pressure Enthalpy
Composition, temperature, pressure are called intensive variables, i.e. independent of quantity of material flow (flow rate)(Nv).
The constraints on the design will place restrictions on the possible values that these variables can take; for example the values of some of the variables will be fixed directly by process specification. The values of other variables will be determined by design relationships arising from constraints.
Some of the design relationships will be in the form of explicit mathematical equations (design equations): such as those arising from material and energy balances, thermodynamic relationships, and equipment performance parameters.
Other relationships will be less precise such as those arising from the use of standards and preferred sizes and safety considerations (Nr).
The difference between the number of variables in the design and the number of design relationships is called the number of degrees of freedom.
If Nv – the number of possible variables in a design problem Nr – the number of design relationships Nd = Nv – Nr Nd = number of degrees of freedom
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Case 1. If Nv=Nr implies Nd = 0; this implies that there is only one unique solution to the problem. The problem is not a true design problem and no optimization is possible.
Case 2 Nv is less than Nr, Nd is less than 0; this implies that the problem is over defined, and only the trivial solution is possible.
Case 3 Nv >Nr, Nd > 0; implies there is an infinite number of possible solutions. However for a practical problem, there will be only a limited number of feasible solutions. Nd represents the number of variables which the designer must assign values to solve the problem.
EXAMPLE 2-1.1: Consider a single phase stream (liquid/vapour) containing C components.
Input
Process Unit
Output
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Note (1) The sum of the mass/mol fractions equal 1. (2) The enthalpy is a function of stream composition, temperature and pressure. Therefore, Degrees of Freedom, Nd = Nv – Nr =(C+4) – (2) = C+2 Specifying C+2 variables completely defines the stream.
EXAMPLE 2-1.2: Flash distillation / Equilibrium distillation In this process unit, a feed is passed into a still (fractionating column), where part is vaporized, and the vapour remaining in contact with the liquid. The mixture of vapor and liquid leaves the still and is separated so that the vapour is in equilibrium with the liquid.
Where, F= Stream flow rate, P= pressure, T=temperature, xi= concentration of component i, q = heat input. Surfixes, 1= inlet; 2=outlet vapor; 3 =outlet liquid
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Note: 1. given the temperature and pressure, the concentration of any component in the vapor phase can be obtained from the concentration in the liquid phase ( v-l-e data) 2. An equilibrium separation implies that the outlet streams and the still are the same pressure and temperature. This implies that
P2 = P
(1)
T2 = T
(3)
P3 = P
(2)
T3 = T
(4)
Gives 4 equations. Degrees of freedom Nd (No of degrees of freedom)=Nv - Nr= (3C + 9) - (2C + 5) = C + 4 Though the total degrees of freedom calculated is (C+4), some of the variables will be fixed by the process conditions and will not be free for the designer to select as design variables. For example: the flash distillation unit will normally be one unit in a process system and the feed composition and feed conditions will be fixed by the upstream process. Hence defining the feed, fixes (C+2) variables and the designer is left with, (C+4)-(C+2) = 2
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2-1.5 Selection of Design Variables To solve a design problem, the designer has to decide which variables are to be chosen as design variables, i.e the ones he can manipulate to produce the best design. This choice is crucial to enable the simplification of the calculations. Example: Flash distillation problem in the previous example. For a binary mixture, C=2; this implies Nd=C+4=6 If the feed stream flow, composition, temperature, pressure are fixed by upstream conditions, then the number of design variables is Nd=6-(C+2)=6-4=2. This implies, the designer is free to select 2 variables from the remaining variables to proceed with the calculation of the outlet stream composition and flows.
Scenerio 1
Suppose you select the still pressure implies for a binary system, vapour-liquid equilibrium (V-L-e) relationship is determined, and one outlet stream flow rate, then the outlet compositions can be calculated by the simultaneous solution of mass-balance and (v-l-e) relations.
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Scenerio 2
If you select the still pressure and the liquid outlet stream composition, then the simultaneous solution of the mass balance and the v-l-e relationship will not be necessary. Following the procedure below, one can calculate the stream compositions.
1. Specify P determines the v-l-e curve from experimental data 2. Knowing the outlet liquid composition, the outlet vapor composition can be calculated from the v-l-e. 3. Knowing the feed and outlet compositions, and the feed flow rate, the outlet stream flows can be calculated from a material balance. 4. An enthalpy balance gives the heat input required.
SELF ASSESSMENT 2-1 (a) List 3 American organizations responsible for coordinating information on standards. (b) List 3 national standard organizations with standards for the protection of consumers.
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UNIT OPTIMIZATION AND BATCH PRODUCTION PROCESS Introduction This unit discusses batch production processes and the importance of optimization. Learning Objectives After reading this unit, you should: 1. Appreciate the importance of optimization in plant design 2. Understand the process of batch production
SESSION 3-1 In this session we shall discuss the steps involved in optimizing the design of a chemical process plant. Further, we shall
3-1.1 Introduction Optimizing the design of a chemical process plant is a foreboding task. This can be achieved by subdividing the plant into subunits and optimizing each subunit. However this does not result in the optimal design of the whole plant because the optimization of each subunit is at the expense of the other. The general procedure for optimizing process units and equipment design: Step1: the first step is to clearly define the objective. i.e the criteria to be used for measuring the performance of the system. For a chemical process plant, the overall objective is to maximize profit. This overall objective can be broken down into sub objectives such as; minimize operating cost, minimize capital investment, maximize yield of the product, reduce labour requirements, reduce maintenance, operate safely. Step 2: the second step is to determine the objective function; the system of equations and other relationships, which relate the objectives with the variables to be manipulated to optimize the function. Step 3: this step is to find values of the variables that give the optimum value of objective function i.e maximum or minimum. The best technique to be used will depend on the complexity of the system and the type of mathematical model used to represent it.
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3-1.2 Simple Models If the objective function can be expressed as a function of one variable (single degree of freedom), the function can be differentiated or plotted to find the maximum or minimum. This situation arises only for exceptional cases. In most practical situations, the numbers of variables exceed the number of relationships. Example of a simple model: Determine the optimum proportions for a closed cylindrical container. D
L
The surface area A in terms of the dimensions is: A DL 2 D 2 DL D 2 2 4 the objective function is f D, L D L
D2 2
(1)
for a given volume (V) we have, V
D2 4V L L 4 D2
(2)
exp res sin g the objective function in terms of one variable D, we have
f(D)
4V D 2 D 2
differenti ating ; f ' (D)
4V D D
(3)
finding the optimum D, we set the above equation to zero 4V 4V 4V D 0 D3 D D 4V 4V substituti ng D in (2) above L 21
2
3
1
3
4V
1
3
For a cylindrical container, the minimum surface area to enclose a given volume is obtained when the length (height) is made equal to the diameter. In practice, when the cost is the objective function; L=2D; this is because the cost must include that of forming the vessel, making the joints in addition to the cost of the material.
3-1.3 Multiple Variable Systems The general optimization problem can be represented mathematically as:
f f (v1 , v 2 , v3 ,.......v n ) where f objective function and v1 , v 2 , v3 ,.......v n are the variables In a design situation, there will be constraints on the possible values of the objective function due to constraints on the variables. -
Equality constraints are expressed by equations of the form
m m (v1 , v 2 , v3 ,.......v n ) 0 -
Inequality constraints are expressed by equations of the form p p (v1 , v 2 , v3 ,.......v n ) Pp
Optimization of the problem involves finding values for the variables v1 , v 2 , v3 ,.......v n that will optimize the objective function ( i.e give maximum or minimum values within the constraints).
3-1.4 Methods of Analysis: a. Analytical method: objective functions can be expressed as a mathematical function; use the methods of calculations to find maximum or minimum values. -
For practical situations where the values of the variables are subject to constraints, the optimum of the constrained objective function will not necessarily occur where the partial derivatives of the objective function are zero.
e.g.
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Figure 3-1.1: Effect of Constraints on the Optimum of a Function
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The method of Lagrange’s undetermined multipliers is a useful analytical technique for dealing with problems that have equality constraints (fixed design values).
b. Search Methods: Relationships between variables and constraints that arise in practical design problems are such that analytical methods are not feasible; hence the use of search methods. For single variable problems where the objective function is unimodal, the simplest approach is to calculate the value of the objective function at uniformly spaced values of the variable until a maximum or minimum is obtained.
Figure 3-1.2 Yield as a Function of Reactor Temperature and Pressure 23
3-1.5 Other Optimization Methods - Linear programming; a technique used when the objective function and constraints can be expressed as a linear function of the variables. - Dynamic programming; used for the optimization of large systems.
3-1.6 Batch Production Process -
Productive period: this is the period when product is being produced
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Nonproductive period; this is the period when the product is discharged and equipment prepared for the next batch.
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Total batch time: productive period + nonproductive period
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The rate of production is determined by the total batch time as follows:
Batches per year = 8760 x Plant attainment Total batch time (batch cycle time)
Annual production rate = (quantities produced per batch)x(batches per year)
Cost per unit of production= annual cost of production Annual production rate
SELF ASSESSMENT 3-1 A rectangular tank with a square base is constructed from 5 mm steel plates. If the capacity required is eight cubic meters. Determine the optimum dimensions if the tank has a closed top.
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UNIT PROCESS SYNTHESIS Introduction This unit discusses the synthesis of chemical processes. Learning Objectives After reading this unit, you should be able to: 1. Explain process synthesis 2. Define and explain heuristics
SESSION 4-1 In this session we shall discuss process synthesis and the importance of heuristics in chemical process optimization.
4-1.1 Introduction: process synthesis aims at the optimization of the logical structure of a chemical process; specifically the sequence of steps; reaction, separation (distillation, extraction etc), the source and destination of recycle streams. The logical structure of a chemical process: Given the following; -
Raw materials, required products, allowed byproducts, a set of unit operations for consideration, cost factors for materials and unit operations required to generate and rank in order of preference and feasible chemical plant flow sheets.
Approach 1. Combinatorial algorithms are used to find all feasible flow sheets contained in a toolkit of raw materials and operating steps. The flow sheets are then reviewed and optimized based on performance, economic and safety criteria. Approach 2. Heuristic rules based on experiences that are used for the selection and positioning of processing operations as flow sheets are assembled.
4-1.2 Raw Materials and Chemical Reactions Heuristic 1: select raw materials and chemical reactions to avoid or reduce the handling and storage of hazardous and toxic chemicals. e.x. Manufacture of ethylene glycol
C 2 H 4 1 O2 C 2 H 4 O 2
(1)
C 2 H 4 O H 2 O HOCH 2 CH 2 OH 25
(2)
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Both reactions are extremely exothermic, therefore they need to be controlled carefully. Such processes are designed with two reaction steps with storage of the intermediate, to enable continuous production, even when maintenance problems shut down the first reaction operation.
Alternative to the 2-step example process: 1. Use chlorine and caustic in a single reaction step, to avoid the intermediate:
CH 2 CH 2 Cl 2 2 NaOH (aq) HOCH 2 CH 2 OH 2 NaCl
2. Use the 2-step reaction with the following modifications: -
As ethylene-oxide is formed, react it with carbon dioxide to form ethylene carbonate, a much less active intermediate that can be stored safely. This can then be hydrolysed to form the required ethylene glycol product
C 2 H 4 1 O2 C 2 H 4 O 2 C 2 H 4 O CO2 C3 H 4 O3 C3 H 4 O3 H 2 O HOCH 2 CH 2 OH CO2
Heuristic 2: Distribution of Chemicals Use an excess of one chemical reactant in a reaction to completely consume a second valuable, toxic or hazardous reactant. e.x. use an excess of ethylene in the production of Dichloroethane.
Cl C2
C2H4Cl2 + C2H4
2H4
C2H4 26
Heuristic 3: when nearly pure products are required, eliminate the inert species before the reaction operations, when the separations are easily accomplished, or when the catalyst is adversely affected by the inert.
Heuristic 4: introduce liquid or vapour purge streams to provide exit for species that; -
Enter the process as impurities in the feed
-
Produced by irreversible side reactions
e.x. Ammonia, NH3 synthesis loop
27
Heuristic 5: Do not purge valuable species or species that are toxic and hazardous, even in small concentrations -
Add separators to recover valuable species
-
Add reactors to eliminate toxic and hazardous species
e.g. catalytic converter in car exhaust
Heuristic 6: For competing series or parallel reactions, adjust temperature, pressure, and catalyst to obtain high yields of desired products. In the initial distribution of chemicals, assume that these conditions can be satisfied; obtain kinetic data, and check this assumption before developing a base-case design. e.x. Manufacture of alkyl-chloride
k1 CH 2 CHCH 3 Cl 2 CH 2 CHCH 2 Cl HCl
Cl 2 k2
k3
CH 3 CHClCH 2 Cl
CHCl CHCH 2 Cl HCl
Dichloropropane
dichloropropene
This is a series/parallel reaction; -
for each reaction, obtain H R , K o , E
28
R
For each reaction, obtain kinetic data and examine the dependency of reaction rate on temperature; implies k k o e
E
RT
Since for multiple reactions, high temperature favors the reaction of higher activation energy and vice versa.
Heuristic 7: for reversible reactions, consider conducting them in a separation device capable of removing products, and hence driving the reaction to the right. e.g. manufacture of ethyl-acetate (ethyl ethanoate) using reactive distillation Conventionally, this will call for the reaction: EtOH + HOAc
EtOAc +H2O
followed by separation of products using a sequence of separation of towers, using a reactive distillation:
29
Heuristic 8: Separations : separate liquid mixtures using distillation, stripping towers and liquid-liquid extractors.
Heuristic 9: attempt to condense vapour mixtures with cooling water, then use heuristic 8.
30
Heuristic 10: Heat Transfer in reactors: to remove highly exothermic heat of reaction, consider the use of excess reactant, an inert diliuent.
Heuristic 11: For less exothermic heat of reaction, circulate reactor fluid to an external cooler, or use a jacketed vessel or cooling coils.
31
Heuristic 12: Pumping and Compression: To increase the pressure of a stream, pump a liquid rather than compress a gas: i.e condense a vapor as long as refrigeration ( and compression) is not needed before pumping.
Instead of:
Compressors have large capital cost and consume a lot of power. 32
4-1.3 Summary of Process Design Heuristics The discussion focused on the following: -
understanding the importance of selecting reaction paths that do not involve toxic or hazardous chemicals.
-
Be able to distribute the chemicals in a process flowsheet, to account for the presence of inert species, to purge species that would otherwise build up to unacceptable concentrations, to achieve a high selectivity of the desired products.
-
Apply heuristics in in selecting separation processes to separate liquids, vapours, and vapour-liquid mixtures.
-
Understand the advantages of pumping a liquid rather than compressing a vapor.
4-1.4 Heuristics in Equipment Design 1. Equipment Size Need information on the required throughput to determine vessel size -
General guidelines for vessel size o Height: 2-10 m o L/D: 2-5m
-
Towers/ Columns o Height: 2-50m o L/D: 2-30m
Note: Do not specify units outside these ranges
2. Heat Exchangers Several kinds are used -
Area: 10-1000m2
-
For shell and tube (tubular) o Tube diameter: 1-2 cm o Tube length: 2-6m o Shell diameter:0.3-1m
33
-
Plate and frame Newer technology- thin, gasketed plates separate hot and cold fluids. Advantage: more compact and very efficient
3. Heat transfer considerations -
Minimum temperature approach: for fluids 10oC and for refrigerants, 5oC
-
Cooling water in: 30oC; Cooling water out: 45oC
-
Equipment heat transfer(overall) coefficient in decreasing order of magnitude Reboiler>Condenser>Liquid-to-Liquid>gas-to-gas
-
Heat Exchangers:Δtlm < 100oC
4. Towers/ Columns Tower operating pressure is usually determined by the temperature of condensing medium or maximum allowable reboiler temperature
Sequencing multiple towers: typically
-
Do easy separation first and leave difficult ones for last
-
If relative volatilities of all species are close , remove one-by-one from overhead
-
When volatilities are close but feed concentrations vary, remove high concentrations first.
Distillation operating conditions
-
Economically reflux ratio is typically 1.2-1.5 times Lmin
-
Optimum number of theoretical trays trays is typically 2 times Nmin
-
Find Nmin from Fenske-Underwood equation
Tower design
-
Tray spacing are typically 20-24”
-
Pressure drop typically 0.1psi per tray
-
Tray efficiencies: 60-90% for gas absorption and stripping
34
5. Process Conditions: General Guidelines (a) High Pressure To achieve a high pressure in a process: -
For a liquid use a pump; for a vapour first condense it and pump it into an evaporatior
-
For a gas compression; P/Po<3
Note that , compressors are high cost items: large capital cost and high power consumption
Advantages of high pressure o Smaller volumes o Greater concentration o Change thermodynamics
Lower pressures processing conditions o 1-10 bar is preferable o Higher pressure or vacuum requires special equipment o Unless there is a compelling reason, operate at 1-10 bar pressure
(b) High Temperature
Operate at high temperature to achieve the following: o High reaction rates o Change equilibrium of reaction or VLE
Achieve a high temperature in a process by the following means: o Furnaces(using fuel gas, oil)- very high temperatures o Heat exchangers(using steam, thermal fluids, molten salt)
Steam o High pressure steam: 40 – 50bar (approx. 260oC) o Medium pressure steam: 20 bar(approx. 200oC) o Low pressure steam: 5- 10 bar(approx. 150oC)
35
(c) Low Temperature
Operate at low temperature to achieve the following: o Slow reaction rate o Prevent thermal sensitive species from degradation o Change thermodynamics
Achieve a high temperature in a process by the following means: o Using a heat exchanger with the following fluids( cold steam)
Cooling tower water
Chilled water
Active refrigeration
SELF ASSESMENT 4-1 In the manufacture of ethylene glycol, a two step reaction process is used; the first step is the formation of the intermediate ethylene oxide which can be stored, followed by its hydrolysis to form the ethylene glycol. Since both reactions are extremely exothermic and explosive, show with balanced chemical equations a safer two step approach after the initial production of ethylene oxide.
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5
UNIT PROCESS SIMULATION Introduction This unit discusses the simulation of a chemical process. Learning Objectives After reading this unit, you should be able to: 1. Explain what process simulation is. 2. Identify the types of process simulations SESSION 5-1 In this session we shall discuss the importance of process simulations, the various types of simulations available and the importance of mathematical models in process simulation.
5-1.1 Introduction -
Process simulation is the act of representing some aspects of the real world by numbers or symbols that may be easily manipulated to facilitate their study.
-
The important step in process simulation is the representation of that aspect of the real world to be studied in terms of a mathematical model.
-
With respect to chemical engineering, the real world is a chemical process described by a process flow sheet. Therefore process simulation is used to solve problems related to the process flow sheet, i.e process design, process analysis, process control etc.
-
The mathematical model that simulates an aspect of the process needs an appropriate method of solution.
-
A process simulation is a computer program developed to solve the model and study its behavior by manipulating the model parameters.
37
5-1.2 Process Simulator Computational package that enables predictions of the process behavior Inputs for the package are: Basic engineering relationships such as: -
Mass and energy balances
-
Phase and chemical equilibrium
-
Flowsheet topography
Uses are: -
For design of new plants
-
Increase profitability in existing plants
-
Run many possible cases
-
Sensitivity studies, and optimization
Accuracy depends on the use of -
Reliable thermodynamic and property data
-
Realistic operating conditions
-
Rigorous equipment models
5-1.3 Types of Process Simulation There are several commercial packages. E.g Aspen Plus, Chemcad, Hysis Pro/II. Common features of all these are: 1. Database of hundreds – thousands( a large number) of compounds 2. A parameter library to compute /estimate properties of these compounds 3. Flowsheet builder- graphical interface that enables units to be defined and connected. 4. Thermo Solver; computes phase equilibrium and thermodynamic properties given the database 5. Unit operations Solver: short cut and rigorous solvers for equipment 6. Overall flowsheet solver; mathematical convergence of the flowsheet
38
5-1.4 Unit Operations Solvers; e.g Reactors - specifies the reaction stoichiometry, temperature, pressure and conversion - when the reaction is equilibrium limited, specifies the stoichiometry , the approach to equilibrium and equilibrium constants - for the reaction involved, specifies the kinetic expression, reactor type and enables the reactor size calculation. - distillation columns - Uses the Feuske, Underwood methods to provide a good initial guess for subsequent calculations - for rigorous plate-by-plate approach, it solves simultaneously, material and energy balance equations, and VLE relations for each plate.
5-1.5. Uncertainty and Sensitivity Issues It is important to be able to quantify the uncertainty of results: -
Determine the probability of accuracy of results
-
Determine what part of results obtained is most likely to be incorrect, and estimate error range
-
Sensitivity issues; in cost estimation and profitability studies, estimate the sensitivity of the results to variations in capital cost, operating cost etc.
Other causes of uncertainty of results using a simulator. -
Thermo model used
-
Physical properties data
-
Convergence tolerance
-
Simulation method ( simulator) Because there are disturbing variations between different process simulators using the same models.
39
-
What can we do? o Estimate uncertainties by performing simulations using a range of parameters, different models. o Determine the sensitivity of results o Use statistical methods to design experiments over ranges of parameters and the results will provide confidence limits o Ultimately, final designs should be tested on pilot studies.
40
6
UNIT FLOW SHEETING, PIPING AND INSTRUMENTATION
Introduction This unit discusses flow sheeting, piping and instrumentation in chemical process design. Learning Objectives After reading this unit, you should be able to: 1. Appreciate the importance of flow sheet in chemical process design 2. Understand why instrumentation in plant design is critical 3. Acknowledge the processes involved in valves and pumps selection SESSION 6-1 In this session we shall study the essence of flow sheets in process design. We shall also look at piping and instrumentation diagram and the types of valves used in a process plant design.
6-1.1 Introduction Industrial equipment are always arranged and interconnected in a certain fashion. The process flow sheet gives a pictorial representation of the various equipment selected to carry out the process. The process flow sheet always convey information on the operating conditions, stream flow-rates and composition. The information included in a typical flow sheet is either essential or optional. An example flow sheet is shown in Figure 6-1.1
Figure 6-1.1: Flow-sheet of Simplified Nitric Acid Production Process
41
The essential information included in the flow-sheet are; 1. Stream composition, either:
The flow-rate of each individual component, kg/h, which is preferred, or
The stream composition as a weight fraction.
2. Total stream flow-rate, kg/h. 3. Stream temperature, degrees Celsius preferred. 4. Nominal operating pressure (the required operating pressure).
Optional information 1. Molar percentages composition. 2. Physical property data, mean values for the stream, such as:
Density
Viscosity
3. Stream name, a brief, one or two-word, description of the nature of the stream 4. Stream enthalpy, kJ/h. Figure 1 below shows the information included in a polymer production process.
6-1.2 Piping and Instrumentation Diagram (P&ID) The piping and instrumentation diagram shows the engineering details of the equipment, instruments, piping, valves and fittings; and their arrangement.
Figure 6-1.2: Polymer Production Diagram 42
In the polymer production flow sheet above, all the following information should be included in the P&ID:
1. All process equipment identified by an equipment number. The equipment should be drawn roughly in proportion, and the location of nozzles shown. 2. All pipes, identified by a line number. The pipe size and material of construction should be shown. The material may be included as part of the line identification number. 3. All valves control and block valves, with an identification number. The type and size should be shown. The type may be shown by the symbol used for the valve or included in the code used for the valve number. 4. Ancillary fittings that are part of the piping system, such as inline sight-glasses, strainers and steam traps; with an identification number. 5. Pumps, identified by a suitable code number. 6. All control loops and instruments, with an identification number. For simple processes, the utility (service) lines can be shown on the P and I diagram. For complex processes, separate diagrams should be used to show the service lines, so the information can be shown clearly, without cluttering up the diagram. The service connections to each unit should, however, be shown on the P and I diagram.
6-1.3 Valve Selection Control of flow in lines and provision for isolation of equipment when needed are accomplished with valves. Depending on primary function, valves are classified as: 1. Shut-off valves (block valves), whose purpose is to close off the flow. 2. Control valves, both manual and automatic, used to regulate flow. Figure 6-3 shows the valves commonly encountered in industry.
43
Figure 6-1.3: Commonly used Valves The main types of valves used are: Gate
Figure 6-1.3a
Plug
Figure 6-1.3b
Ball
Figure 6-1.3c
Globe
Figure 6-1.3d
Diaphragm
Figure 6-1.3e
6-1.3.1 Gate valves In gate valves, the flow is straight through and is regulated by raising or lowering the gate. The majority of valves in the plant are of this type. In the wide open position they cause little pressure drop.
44
6-1.3.2 Globe valves In globe valves, the flow changes direction and results in appreciable friction even in the wide open position. This kind of valve is essential when tight shutoff is needed, particularly of gas flow.
The ball and plug valves are also frequently used for the purpose of flow shutoff. Butterfly valves are often used for the control of gas and vapor flows. Automatic control valves are basically globe valves with special trim design.
SESSION 6-2: In this session we shall discuss the classification of pumps and factors to consider in pump selection.
6-2.1 Introduction of Pumps They are mechanical devices used to increase the energy of a liquid stream flowing in a closed conduit or pipe. This energy may be used to increase the velocity (move the fluid), the pressure or the elevation of the fluid.
6-2.2 Classification of Pumps Pumps are generally classified into two main categories namely:
Centrifugal pumps
Positive displacement pumps
45
Table 6-2.1: Main Types of Pumps Category Centrifugal
Types Single-stage
Structure Volute Diffuser
Multi-stage
Regenerative Vertical Hellico-centrifugal Axial flow
Positive
Rotary
Gear
displacement
Screw Vane Reciprocating
Piston Diaphragm Plunger
There are many subgroups of pumps as indicated in Table 6-2.1.
6-2.3 Factors to Consider in Pump Selection
Capacity (flow rate in m3/h)
The pressure head that is generated by the pump
The type of liquid pumped (its viscosity and vapour pressure under inlet conditions)
An initial selection is generally made based on the first two criteria mentioned, i.e. the capacity and the pressure generated. The centrifugal pump is often the only possible choice for high capacities whereas positive displacement pumps are better suited to generating high pressure heads. Other criteria such as the viscosity of the liquid can modify this initial choice. A positive displacement pump is generally recommended to pump liquids with a viscosity higher than 2000 cP. Figure 6-2.1shows the approximate range of operation generally covered by the main types of pump mentioned above.
46
Figure 6-2.1: Approximate Range of Operation for the Three Main Types of Pump
6-2.4 Centrifugal Pumps Basically consists of an impeller equipped with radial vanes rotating inside a shell called the pump casing. It works by the transfer of centrifugal force of the rotating impeller into kinetic energy of the liquid. This energy is then converted into pressure when the fluid velocity decreases.
6-2.4.1 Effective characteristic curves A centrifugal pump is characterized by 4 basic curves, all of which are expressed versus the flow rate as shown in Figure 6-2.2. A. The head generated B. The mechanical/hydraulic conversion efficiency C. The mechanical power input consumed at the shaft D. The pump suction capacity or NPSH.
47
Figure 6-2.2: Basic Curves Characterizing a Centrifugal Pump.
6-2.4.2 Design parameters of centrifugal pumps -
The rotation speed
-
The number of impellers
-
The impeller diameter
-
The impeller design
6-2.4.3 Operating point The pressure head HA required by the installation is represented by the system curve versus flow rate, Q. It is the sum of the static and dynamic heads of the installation as shown in Figure 6-2.3. The static heads are independent of the flow rate and include differences in height and pressure between the unit inlet and outlet. The dynamic heads correspond to pressure drops and are proportional to the square of the flow rate.
48
Figure 6-2.3: Curve Characteristic of the System.
A centrifugal pump adjusts itself on an operating point B, corresponding to the intersection between the Q/H curve of the pump and the H A curve of the system (Figure 6-2.4). A variation in the operating point (and therefore in the flow rate and the head) can be obtained by a physical modification in the pump, but also by modifying its speed or the system curve, usually by means of a valve.
Figure 6-2.4: Variation of Operating Point by means of a Valve.
49
6-2.4.4 Q/H Curve versus Technical Choices 1. Basic choices; concept of specific speed The number Nq, called the “specific speed”, allows all centrifugal pumps to be compared with one another. It is calculated from the following expression:
Nq
N Q 60H
3
4
With: N = rotational speed in rpm Q= flow rate the best efficiency in m3/h (through one eye if double-suction impeller) H= head in m generated at the best efficiency (for one stage) For the same specific speed, the hydraulic designs will be similar on varying scales. The choice of Nq is a major parameter in impeller hydraulic design (Figure 6-2.5). The specific speed also considerably influences the best efficiency achievable by a pump. If high pressure is required, a compromise will have to be found between a reasonable number of stages and an acceptable number efficiency.
Figure 6-2.5: Variation in Specific Speed versus the Type of Impeller used.
2. Choice of Design: Concept of Q/H Curve Slope For a given hydraulic choice, the different design parameters, in particular the number of vanes can modify the curve shape downward, flat or bell-shaped (Figure 6-2.6). These curves should be considered according to the requirements of the resisting network. Bellshaped curves in particular, should be avoided for pumps that have to work in parallel (risk of instability). 50
Figure 6-2.6: Different Types of Characteristic Curves
6-2.4.5 Choice of Rotation Speed The rotation speed is a dominant parameter for the characteristic curve of a centrifugal pump. Figure 6-2.7 shows how the operating point varies with speed. The flow rate varies linearly with speed:
Q2 Q1
N2 N1
The head generated varies with the square of the speed: N H 2 H 1 2 N1
2
As a result, the power absorbed varies with the cube of the speed N P2 P1 2 N1
3
51
Figure 6-2.7: Variation in the Operating Point versus the Rotation Speed.
6-2.4.6 Suction Conditions: Concept of NPSH Vapour pressure For a given temperature, each liquid has a specific boiling pressure, called the vapour pressure Tv. if the pressure at one point in the liquid becomes less than T v, the liquid vaporizes instantly. Cavitation The lowest static pressure inside a centrifugal pump is located at the impeller inlet. If vaporization begins at this point, the liquid will be repressurized nearby downstream. The bubbles formed condense by collapsing suddenly, most often near a wall. This very noisy phenomenon is called cavitation. The head generated by the pump and the absorbed power then drop, the vibrations and noise increase and erosion can be observed, mainly in the impeller, in the form of characteristic pits. If the pump is kept working under these conditions, permanent damage may occur. Required NPSH To prevent cavitation, the total liquid pressure at the inlet must be such that no vaporization can occur. The minimum value depends on the pump design Available NPSH It is the pressure available to force a given flow through the suction piping into the pump. This is a function of the system. It is easily calculated with the formula below.
52
=
ℎ
+ −
ℎ
ℎ
−
ℎ ℎ
(
ℎ
) ℎ
EXAMPLE 6-1.1: Pressure drop calculation A pipeline connecting two tanks contains four standard elbows, a plug valve that is fully open and a gate valve that is half open. The line is commercial steel pipe, 25mm internal diameter, length120m. The properties of the fluid are: viscosity 0.99mNm-2s, density 998kg/m3.Calculate the total pressure drop due to friction when the flow rate is 3500 kg/h. SOLUTION 6-1.1 Cross-sectional area of pipe = Fluid velocity, u =
25 10 3 4
2
0.491 10 3 m 2
3500 1 1 1.98m / s 3 3600 0.491 10 998
998 1.98 25 10 3
Reynolds number, Re =
9.9 10 4
49,900 5 10 4 Absolute roughness commercial steel pipe = 0.046mm Relative roughness =
0.046 0.0018 , round to 0.002 25 10 3
From friction factor chart, f = 0.0032 Fitting/valve
Number of velocity Equivalent heads, K
pipe diameters
Entry
0.5
25
Elbows
(0.8×4)
(40×4)
Globe valve, open
6.0
300
Gate valve, ½ open 4.0
200
Exit
1.0
50
Total
14.7
735 53
Method 1, velocity heads A velocity head
u2 1.98 2 0.20m of liquid, 2 g 2 9.8
Head loss = 0.20 × 14.7 = 2.94 m As pressure = 2.94 × 998 × 9.8 = 28,754 N/m2 Friction loss in pipe, Pf 8 0.0032
120 1.98 2 998 2 25 10 3
= 240,388 N/m2 Total pressure = 28,754 + 240,388 = 269,142 N/m2 = 270 kN/m2
Method 2, equivalent pipe diameters Extra length of pipe to allow for miscellaneous losses = 735 × 25 × 10-3 = 18.4 m So total length for ΔP calculation = 120 + 18.4 m
Pf 8 0.0032
138 1.98 2 998 2 25 10 3
= 277 kN/m2 Note: the two methods will not give the same results. The method of velocity heads is the more fundamentally correct approach. But the use of equivalent diameters is easier to apply and sufficiently accurate for use in design calculations.
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SELF ASSESSMENT 6-1 A process fluid is pumped from the bottom of one distillation column to another, using a centrifugal pump. The line is standard commercial steel pipe 75 mm internal diameter. From the column to the pump inlet the line is 25 m long and contains six standard elbows and a fully open gate valve. From the pump outlet to the second column the line is 250 m long and contains ten standard elbows, four gate valves (operated fully open) and a flowcontrol valve. The fluid level in the first column is 4 m above the pump inlet. The feed point of the second column is 6 m above the pump inlet. The operating pressure in the first column is 1.05 bara and that of the second column 0.3 barg. Determine the operating point on the pump characteristic curve when the flow is such that the pressure drop across the control valve is 35 kN/m2. The physical properties of the fluid are: density 875 kg/m3, viscosity 1.46 mNm-2s. Pump characteristic Flow-rate, m3/h 0.0 18.2 27.3 36.3 45.4 54.5 63.6 Head, m of liquid 32.0 31.4 30.8 29.0 26.5 23.2 18.3
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7
UNIT PROCESS ECONOMICS: COST ESIMATION Introduction This unit discusses the estimation of capital investment, total product cost and the determination of breakeven point.
Learning Objectives After reading this unit, you should be able to: 1. Define total capital investment 2. Apply the Marshall & Swift cost index and capacity ratio to estimate equipment cost 3. Distinguish between fixed capital and working capital 4. Define turnover ratio SESSION 7-1 In this session we shall discuss the estimation of capital investment. Further we shall discuss fixed capital investment, working capital, cost indices and turnover ratio.
7-1.1 Cost Estimation An acceptable plant design must present a process that is capable of operating under conditions which will yield a profit. Since net profit equals total income minus all expenses, it is essential that the chemical engineer be aware of the many different types of cost involved in manufacturing process. Capital must be allocated for direct plant expenses, such as those for raw materials, labor and equipment. Besides direct expenses, many other indirect expenses are incurred and these must be included if a complete analysis of the total cost is to be obtained. Examples of these indirect expenses are administrative salaries, product-distribution cost, and cost for interplant communications.
A capital investment is required for any industrial process, and determination of the necessary investment is an important part of a plant-design project. The total investment for any process consist of fixed-capital investment for physical equipment and facilities in the plant plus working capital which must be available to pay salaries, keep raw materials and products on hand and handle other special items requiring a direct cash outlay.
56
Thus in an analysis of cost in industrial processes, capital investment costs, manufacturing costs, and general expenses including income taxes must be taken into consideration.
7-1.2 Cash Flow for Industrial Operations Figure 7-1.1 shows the concept of cash flow for an overall industrial operation.
Figure 7-1.1: Cash Flow for an Overall Industrial Operation.
57
7-1.3 Factors affecting Investment and Production Costs
Sources of equipment
Price fluctuations
Company policies
Operating time and rate of production
Governmental policies
7-1.4 Capital Investment Before an industrial plant can be put into operation, a large sum of money must be supplied to purchase and install the necessary machinery and equipment. Land and service facilities must be obtained, and the plant must be erected complete with all piping, controls, and services. In addition it is necessary to have money available for the payment of expenses involved in the plant operation.
-
Fixed capital investment The capital needed to supply the needed manufacturing and plant facilities.
-
Working capital The capital necessary for the operation of the plant
-
Total capital investment = Fixed capital investment + Working capital
-
Fixed capital investment= Manufacturing fixed capital investment + Nonmanufacturing fixed capital Investment.
-
Manufacturing fixed capital investment: the fixed capital necessary for the installed process equipment with all auxiliaries that are needed for complete process operation. These include expenses for piping, instruments, insulations, foundations, and site preparation.
58
-
Non- manufacturing fixed capital investment: the fixed capital required for construction overhead and for all plant components that are not directly related to the process operation. These plant components include the land, processing buildings, administrative and other offices, warehouses, laboratories, transportation, shipping and receiving facilities, utilities and waste disposal facilities, shops and other permanent parts of the plant.
-
Construction overhead cost: consists of field-office and supervision expenses, home-office expenses, engineering expenses, miscellaneous construction costs, contactor’s fees and contingencies. In most cases, construction overhead is proportioned between manufacturing and non-manufacturing fixed-capital investment.
-
Working capital: the working capital for an industrial plant consist of the total amount of money invested in
o Raw materials and supplies carried in stock o Finished products in stock and semi-finished products in process of being manufactured o Accounts receivable o Cash kept on hand for monthly payment of operating expenses , such as salaries, wages, and raw materials purchases o Accounts payables o Taxes payables
The ratio of working capital to total capital investment varies with different companies. Most chemical plants use an initial working capital of 10 to 20 percent of the total capital investment. It is 50% or more for companies producing products of seasonal demand because of the large inventories which must be maintained for appreciable period of time.
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7-1.5 Estimation of Capital Investment 7-1.5.1 Introduction Of the many factors which contribute to poor estimates of capital investments, the most significant one is usually traceable to sizeable omissions of equipment, services, or auxiliary facilities rather than to gross errors in costing. A check list of items covering a new facility is an invaluable aid in making a complete estimation of the fixed capital investment. A typical list of items for estimating fixed capital investment is: Direct cost 1. Purchased equipment; all equipments listed on a flow sheet 2. Purchased-equipment installation 3. Instrumentation and controls 4. Piping 5. Electrical equipment and materials 6. Buildings(including services) 7. Yard improvements 8. Service facilities 9. land Indirect cost 1. Engineering and supervision 2. Construction expenses 3. Contractor’s fees 4. Contingencies
7-1.5.2 Types of capital cost estimates An estimate of the capital investment for a process may vary from a predesign estimate based on little information except the size of the proposed project to a detailed estimate prepared from complete drawings and specifications. The following five categories represent the accuracy range and designation normally used for design purposes: 1. Order-of-magnitude estimate(ration estimate) based on similar previous cost data: probable accuracy of estimate over ±30 percent 60
2. Study estimate (factored estimate) based on knowledge of major items of equipment; probable accuracy of estimate up to ±30 percent
3. Preliminary estimate (budget authorization estimate; scope estimate) based on sufficient data to permit the estimate to be budgeted; probable accuracy of estimate within ±20 percent.
4. Definitive estimate(project control estimate) based on almost complete data but before completion of drawings and specifications; probable accuracy of estimate within ±10 percent
5. Detailed estimate (contractor’s estimate) based on complete engineering drawings, specifications and site surveys; probable accuracy of estimate within ±5 percent. 7-1.5.3 Cost indexes Most cost data which are available for immediate use in a preliminary or predesign estimate are based on conditions at some time in the past. Because prices may change considerably with time due to changes in economic conditions, some methods must be used for updating cost data applicable at a past date to costs that are representative of conditions at a later time. This is done using a cost index. A cost index is hence an index value for a given point in time showing the cost of an equipment or plant at that time relative to a certain base time.
index value at present time Pr esent cost Original cost index value at time original cost was obtained Cost indexes can be used to give a general estimate, but no index can take into account all factors, such as special technological advancements or local conditions. The common indexes permit fairly accurate estimates if the time period involved is less than 10 years.
61
Some indexes are used for estimating equipment cost, others apply specifically to labor, construction, materials etc. The most common of these indexes are the Marshall and Swift all-industry and process-industry equipment indexes, the Engineering New-Record construction index, the Nelson refinery construction index and the Chemical Engineering plant cost index. This is shown in Table 7-1.1.
Table 7-1.1: Cost indexes as Annual Averages
7-1.5.4 Methods for Estimating Capital Investment 7-1.5.4.1 Power factor applied to plant-capacity ratio:
C n CFeR
n
Where: R = size (or capacity) ratio of equipment=
capacity of new equipment or plant capacity of old equipment or plant 62
Fe = equipment cost index ratio=
index value in present time index value at the time original cost was obtained
n = equipment size (or capacity) exponent C = purchased cost of old equipment Cn = purchased cost of new equipment Table 7-1.2 shows typical exponents for equipment cost vs. capacity
Table 7-1.2 Typical Exponents for Equipment Cost vs. Capacity
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EXAMPLE 7-1.1 If a process plant was erected in Kumasi at a fixed capital investment of $436,000 in 1970, determine what the capital investment will be in 1975 for a similar process plant located near Dansoman in Accra with twice the process capacity but with an equal number of process equipment.
SOLUTION 7-1.1
C n CFeR
n
Where: R = size ( or capacity) ratio of equipment Fe = equipment cost index ratio n = equipment size exponent C = purchased cost of equipment in 1970 Cn = cost of equipment in 1975 From Table 7-1.1 M & S index, 1970 =303, M & S index 1975 = 444,
From Table 7-1.2 Equipment size exponent, n = 0.6 C n 436,000
444 0 .6 2 $968,378 303
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7-1.5.4.2 Detailed item estimate EXAMPLE 7-1.2 Initial design work was done for a chemical plant to revamp the process in order to recover valuable product from an effluent gas stream. The gas will be scrubbed with a solvent in a packed column and the recovered product and solvent separated by distillation. The solvent will then be cooled and recycled. The major items of equipment required and their purchased costs in cedis are: Absorption column Column purchased cost = 19,800 Cost of column packing = 6,786 Recovery column Column purchased cost = 45,000 Cost of 30 sieve trays = 8,670 Reboiler Cost of reboiler = 7,600 Condenser Purchased cost = 4,800 Recycle solvent cooler Cost of cooler = 2,550 Storage tank Cost of tank = 7,790 The relevant component factors for this processing plant are; f1: equipment erection = 0.40 f2: piping = 0.70 f3: Instrumentation = 0.2 f4: electrical = 0.10 f10: design and engineering = 0.30 f12: contingencies = 0.20 Estimate the total capital investment for the project, if the working capital can be taken as 10 % of the fixed capital cost.
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SOLUTION 7-1.2 Total capital investment, TCI = Fixed capital investment, FCI × Working capital, WC Total purchased cost of equipment, PCE Absorption column = 19,800 + 6,786 = 26,586 Recovery column = 45,000 + 8,670 = 53,670 Reboiler = 7,600 Condenser = 4,800 Recycle solvent cooler = 2,550 Solvent and product storage tank = 7,790 Total purchased cost of major equipment = ₵ 102,996 Equipment erection = 102,996 (0.4) = ₵ 41,198 Piping cost = 103,996 (0.70) = ₵ 72,097 Instrumentation cost = 102,996 (0.2) = ₵ 20,599 Electrical = 102,996 (0.10) = ₵ 10,300 Total physical plant cost (Direct cost) = 10,300 + 20,599 + 72,097 + 41,198 + 102,996 = ₵ 247,190 Indirect cost Design and engineering, f10 = 0.30 (247,190) = ₵ 74,157 Contingencies = 0.20 (247,190) = ₵ 49,436 Total indirect cost = ₵ 123,595 Fixed capital investment (Direct cost + Indirect cost) = 247,190 + 123,595 = ₵ 370,785 Working capital = 0.10 (370,785) = ₵ 37,079 Total capital investment = 370,785 + 37,079 = ₵ 407,864 7-1.5.4.3 Other Methods for Estimating Equipment or Capital Investment (i) Unit cost Estimate (ii) Percentage of delivered-equipment Cost (iii) Lang factors for approximation of capital Investment
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7-1.5.5.Turnover Ratios: A rapid evaluation method suitable for order-of-magnitude estimates is known as the turnover ratio method. Turnover ratio is defined as the ratio of gross annual sales to the fixed capital investment
Turnoverratio
gross annual sales fixed capital investiment
Where the product of the annual production rate and the average selling price of the commodities produced is the gross annual sales figure.
Capital ratio or investment ratio
1 Turnover ratio
Turnover ratio ranges between 0.2 to 5. For chemical industries, as a very rough rule of thumb, the ratio can be approximated to 1.
SESSION 7-2: In this session we shall study the estimation of total product cost and determination of breakeven point.
7-2.1 Estimation of Total Product Cost •
Manufacturing cost: all expenses directly connected with the manufacturing operation or the physical equipment of a process plant itself.
•
There are 3 classifications of these expenses: – Direct production cost: expenses directly associated with the manufacturing operation. E.g. expenses for raw materials ( including transportation, unloading)
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– Fixed charges: expenses that remains practically constant from year to year. Do not vary widely with changes in production rate. E.g. property taxes, insurance. – Plant overhead cost: are for medical and hospital services; general plant maintenance and overhead; safety services; social security
General Expenses: – Administrative expenses – Distribution and marketing expenses – Research and development – Financing expenses: extra costs involved in procuring the money necessary for the capital investment. – Gross earnings expenses
•
Total Product Cost= Manufacturing cost + General Expenses.
•
Gross earnings (Gross Profit)= The total income – the total product cost.
•
Net annual earnings = Gross annual earnings – income taxes.
Figure 7-2.1 shows the components constituting the estimation of total product cost
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Figure 7-2.1: Cost Involved in Total Product Cost for a Typical Chemical Process Plant
7-2.2 Break-even Point (a) Break-even point occurs (or is the percentage of plant capacity) when the total annual product cost equals the total annual sales. i.e: Total Annual Product Cost = Total Annual Sales (total income from all products sold).This is shown by Figure 7-2.2.
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Figure 7-2.2 Break-even Chart for Chemical Processing Plant
(b) In a different context, Break-even point is the time reached at which all the investment has been paid off and the plant begins to make profit as shown in Figure 7-2.3.
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Figure 7-2.3: Project Cash Flow Diagram •
A-B The investment required to design the plant.
•
B-C The heavy flow of capital to build the plant, and provide funds for start-up.
•
C-D The cash-flow curve turns up at C, as the process comes on stream and income is generated from sales. The net cash flow is now positive but the cumulative amount remains negative until the investment is paid off, at point D.
•
Point D is known as the break-even point and the time to reach the break-even point is called the pay-back time. In a different context, the term "break-even point" is used for the percentage of plant capacity at which the income equals the cost for production.
•
D-E In this region the cumulative cash flow is positive. The project is earning a return on the investment.
•
E-F Toward the end of project life the rate of cash flow may tend to fall off, due to increased operating costs and falling sale volume and price, and the slope of the curve changes.
•
The point F gives the final cumulative net cash flow at the end of the project life.
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EXAMPLE 7-2.1 The annual direct production costs for a plant operating at 70 % capacity is $ 280,000 while the sum of the annual fixed charges, overhead costs and general expenses is $ 200,000. What is the break-even point in units of production per year if the total annual sales are $ 560,000 and the product sells at $ 40 per unit? (ii) What are the annual gross earnings and net profits for this plant at 100 % capacity in the year 1974 when corporate income taxes required 22 %, normal tax on the total gross earnings plus 26 % surtax on gross earnings above $ 25,000?
SOLUTION 7-2.1 Units sold per annum × selling price per unit = total annual sales Units sold = $ 560,000 per yr. / $ 40 per unit = 14,000 units Total product cost per unit = $ 280,000/14,000 = $ 20 per unit At break-even, Total product cost per annum = Total annual sales $ 200,000 + 20x = 40x Where x is the units required to break-even x = 10000 units Therefore at break-even, 10000 units will be produced. (ii) Gross annual earnings = total annual sales – total annual product cost = 14,000 units × $ 40 per unit/0.70 – ($ 200,000 + (14,000 units × $ 20per unit/0.7)) = $ 200,000 Net annual earnings = $ 200,000 – [0.22(200,000) + 0.26(175,000)] = $ 110,500
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SELF ASSESSMENT 7-1 The total capital investment for a chemical plant is $1 million, and the working capital is $100,000. If the plant can produce an average of 8000 kg of final product per day during a 365-day year, what selling price in dollars per kilogram of product would be necessary to give a turnover ratio of l.0? SELF ASSESSMENT 7-2 The total capital investment for a conventional chemical plant is $1,500,000, and the plant produces 3 million kg of product annually. The selling price of the product is $0.82/kg. Working capital amounts to 15 percent of the total capital investment. The investment is from company funds, and no interest is charged. Raw-materials costs for the product are $0.09/kg, labor $0.08/kg, utilities $0.05/kg, and packaging $0.008/kg. Distribution costs are 5 percent of the total product cost. Estimate the following: (a) Manufacturing cost per kilogram of product. (b) Total product cost per year. (c) Profit per kilogram of product before taxes. (d) Profit per kilogram of product after taxes (use current rate)
SELF ASSESSMENT 7-3 What is the breakeven point (units/year) for a process plant making 2,000,000 units per year of products with direct production cost of $1million per year? The selling price per unit of product is $1.2. Assume 100% capacity production and other fixed charges amounts to $700,000.
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8
UNIT PROCESS ECONOMICS: DEPRECIATION AND PROFITABILITY ANALYSIS Introduction This unit discusses depreciation and the various methods for determining depreciation. Further, it discusses profitability analysis of process plants.. Learning Objectives After reading this unit, you should be able to: 1. Define depreciation and distinguish between physical and functional depreciation. 2. Understand the different methods for estimating depreciation 3. Solve depreciation problems 4. Explain profitability analysis and its importance in process economics. SESSION 8-1 In this session we shall discuss depreciation, the different types of estimating depreciation, service life, salvage value and present value or book value.
8-1.1 Depreciation An analysis of costs and profits for any business operation requires recognition of the fact that physical assets decrease in value with age. This decrease in value may be due to physical deterioration, technological advances, economic changes, and other factors which ultimately will cause retirement of the property. The reduction in value due to any of these causes is a measure of the depreciation. Types of Depreciation 1. Physical depreciation: is the measure of the decrease in value due to changes in the physical aspect of the property. Wear and tear, corrosion, accidents and deterioration due to age. 2. Functional depreciation: depreciation due to all other causes. One common type of functional depreciation is obsolescence. This is caused by technological advances or developments that make an existing property obsolete. Though the property does not has suffered no physical change, its economic serviceability is reduced because it is inferior to improved types of similar assets that have been made available through advancements in technology.
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8-1.2 Service life The period during which the use of a property is economically feasible. Both physical and functional depreciation are considered in determining service life and the term is synonymous with economic or useful life. In estimating the probable service life, it is assumed that a reasonable amount of maintenance and repairs will be carried out at the expense of the owner.
8-1.3 Salvage Value It is the net amount of money obtainable from the sale of used property over and above any charges involved in removal and sales. If a property is capable of further service, its salvage value may be high. However, other factors such as location of the property, existing price levels, market supply and demand, and difficulty of dismantling, may have an effect. If the property cannot be disposed of as a useful unit, it can often be dismantled and sold as junk to be used again as a manufacturing raw material. The profit obtainable from this type of disposal is scrap or junk value.
8-1.4 Present Value It is the value of an asset in its condition at the time of valuation. Types of present values are: 1. Book value or unamortized cost: the difference between the original cost of a property, and all the depreciation charges made to date. 2. Market value: the price which could be obtained for an asset if it were placed on sale in the open market. 3. The cost necessary to replace an existing property at any given time with one at least equally capable of rendering the same service.
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8-1.5 Methods for determining Depreciation 8-1.5.1 Straight line method With this method, it is assumed that the value of the property decreases linearly with time. Equal amount are charged for depreciation each year throughout the entire service life of the property.
d
V Vs n
where d annual depreciation, $/year V original value of the property at start of the service life period, completely installed and ready for use, dollars. Vs salvage value of property at end of service life, dollars n service life, years
The book value ( or asset value)of the equipment at any time during the service life may be determined from:
Va V ad where Va book value, dollars a number of years in actual use 8-1.5.2 Declining – balance ( or Fixed percentage) method When this method is used, the annual depreciation cost is a fixed percentage of the property value at the beginning of the particular year. The fixed percentage factor remains constant throughout the entire service life of the property, while the annual cost for depreciation is different each year. Under these conditions, the depreciation cost for the first year of the property’s life is Vf. where f = the fixed-percentage factor. At the end of the first year,
Asset value Va V (1 f )
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At the end of the second year, Asset value Va V (1 f ) 2
At the end of the a year, Asset value Va V (1 f ) a
At the end of the n year ( i.e. at the end of service life) Asset value Va V (1 f ) n Vn
Therefore, 1
V n f 1 s . V
8-1.5.3 Other methods for determining depreciation: (i) Sum-of-the-Years-Digits Method and (ii) Sinking-Fund Method.
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EXAMPLE 8-1.1 The original cost of a piece of equipment is $ 30,000 completely installed and ready to use. The salvage value of the equipment is estimated to be $ 5000 at the end of a service life estimated to be 20 years. (i) Determine the book value of the equipment at the end of 10 years using (a) straight line method (b) textbook declining balance method. (ii) Using the textbook declining method: (a) What will be the first year depreciation? (b) What will be the depreciation at the end of the 6th year?
SOLUTION 8-1-1 Original cost, v = $ 30,000 Salvage value, vs = $ 5000 Service life, n = 20 yrs. Depreciation, d by the straight line method:
d
v vs n
30,000 5,000 $1,250 20
Book value at the end of 10 yrs. v10 = v – ad = 30,000 – 10(1,250) = $17,500 (i) b. textbook declining balance method First year depreciation, d = v × f 1
1
5,000 20 v n f 1 s 1 0.0857 v 30,000
d = 30,000(0.0857) = $2,571 Asset value after 10 years v10 = v (1-f)10 = 30,000 (1- 0.0857)10 = $ 12,246 78
(ii) a. first year depreciation d=v×f = $30,000 × 0.0857 = $2,571 b. depreciation at the end of 6th year = $30,000 - $17,525 = $12,475
EXAMPLE 8-1.2 A piece of equipment having a negligible salvage and scrap value is estimated to have a service life of 10 yrs. The original cost of the equipment was $ 40,000. Determine the following (i) the depreciation charge for the fifth year if double declining balance depreciation method is used.
SOLUTION 8-1.2 d
v vs n
40,000 0 $4,000 10
f = 2(4,000/40,000) = 0.2 The asset value at the end of year 5 is given by: v5 = 40,000 (1 – 0.2)5 v5 = $13,107 The asset value at the end of year 4 is given by: v4 = 40,000 (1 – 0.2)4 v4 = $16,384 The depreciation for year 5 is: d = v 4 – v5 d = $16,384 - $13,107 d = $2,277 79
EXAMPLE 8-1.3 Consider an equipment costing $20,000 and with a service life estimated at 5 years. If the scrap value is $2,000, find the depreciation cost for the 1st , 2nd and 3rd years using the sum-of-the-years-digits method.
SOLUTION 8-1.3 Yearly depreciation factor = Year 1 depreciation factor
5 0.33 15
Annual depreciation factor = yearly depreciation factor × total depreciable value at the start of the service life Total depreciable value = $20,000 - $2,000 = $18,000 Year 1 depreciation
5 $18,000 $6,000 15
Depreciation factor for second year
4 15
Second year depreciation cost
4 $18,000 $4,800 15
Third year depreciation factor
3 15
Third year depreciation cost
3 $18,000 $3,600 15
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SESSION 8-2: In this session, we will discuss profitability analysis, indicating the basis for evaluating profitability. Further, we will discuss the mathematical methods for profitability evaluation.
8-2.1 Profitability Analysis Profitability is generally a term for the measure of the amount of profit that can be obtained from a given situation. Before capital is invested in a project or enterprise, it is necessary to know how much profit can be obtained and whether or not it might be more advantageous to invest the capital in another form of enterprise. Thus, the determination and analysis of profits obtainable from the investment of capital and the choice of the best investment amongst various alternatives are major goals of an economic analysis. The design engineer, however, usually deals with investments which are expected to yield a tangible profit.
8-2.2 Profitability Standards In the process of making an investment decision, the profits anticipated from the investment of funds should be considered in terms of a minimum profitability standard. The profitability standard, which can normally be expressed on a direct numerical basis, must be weighed against the overall judgment evaluation for the project in making the final decision as to whether or not the project should be undertaken.
The judgment evaluation must be based on the recognition that a quantified profitability standard can serve only as a guide. Thus it must be noted that the profit evaluation is based on a prediction of future results so that assumptions are necessarily included. Many intangible factors, such as future changes in demand or prices, possibility of operational failure, or premature obsolescence, cannot be quantitized. It is in areas of this type that judgment becomes critical in making a final investment decision. A primary factor in the judgment decision is the consideration of possible alternatives. For example, the alternative to continuing the operation of an existing plant may be to replace it with a more efficient plant, to discontinue the operation entirely, or to make modifications in the existing plant. 81
8-2.3 Basis for Evaluating Project Profitability Total profit alone cannot be used as the deciding profitability factor in determining if an investment should be made. The profit goal of a company is to maximize income above the cost of the capital which must be invested to generate the income. If the goal were merely to maximize profits, any investment would be accepted which would give a profit, no matter how low the return or how great the cost.
EXAMPLE 8-2.1 Assume two equally sound investments A and B could be made. Investment A: Capital needed=$100,000; Profit yielded=$10,000/year Investment B: Capital needed=$1 million; Profit yielded=$25,000/year Which of the investments is worthy of undertaking?
SOLUTION 8-2.1 Investment B gives a greater yearly profit than A but annual rate of return on the second investment is only ($25,000/$1,000, 000) x100=2.5% While annual rate of return on $100,000 is 10%.
Because reliable bonds and other conservative investments would yield annual rates of return in the range of 6 to 9%, the $1 million investment in this example would not be very attractive; the 10% return on the $100,000capital would make this investment worthy of careful consideration.
Therefore, the rate of return rather than the total amount of profit is the important profitability factor in determining if the investment should be made.
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8-2.4 Mathematical Methods for Profitability Evaluation The most commonly used methods for profitability evaluation are: 1. Rate of return on investment 2. Discounted cash flow rate of return based on full-life performance 3. Capitalized cost 4. Payout period 5. Net present worth Note: No single method is best for all cases.
8-2.4.1 Rate of return on investment ROR
Profit 100% Total initial investment Years
Profit= income – expenses
Total investment determination; consider both fixed capital and working capital
To determine profit; estimates must be made of direct production cost, fixed charges including depreciation, plant overhead costs, general expenses
Profits may be expressed on a before-tax or after-tax basis but the condition must be indicated.
NOTE :
Pay-back time is the reciprocal of ROR
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EXAMPLE 8-2.2 A proposed manufacturing plant requires an initial fixed capital investment of $900,000 and $100,000 of working capital. It is estimated that the annual income will be $800,000 and the annual expenses including depreciation will be $520,000 before income taxes. A minimum annual return of 15% before income taxes is required before the investment will be worthwhile. Income taxes amount to 48% of all pre-tax profits. Determine the following: a. The annual percent return on the total initial investment before income taxes. b. The annual percent return on the total initial investment after income taxes
SOLUTION 8-2.2 (a ) Total Capital Investment 900,000 100,000 1,000,000 Annual Profit before tax 800,000 520,000 280 ,000 Pr ofit 100% Total initial investment year 280,000 Annual % ROR 100% 1,000,000 1 % ROR 28% Annual Percentage ROR
(b) Annual Profit after tax 0.52 280,000 145,600 145,600 100% 1,000,000 1 % ROR 14.56%
Annual % ROR
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8-2.4.2 Discounted cash flow The method of approach for a profitability evaluation by discounted cash flow takes into account the time value of money and it is based on the amount of the investment that is unreturned at the end of each year during the estimated life of the project. A trial and error procedure is used to establish a rate of return which can be applied to a yearly cash flow so that the original investment is reduced to zero (or to salvage and land value plus working-capital investment) during the project life. Thus, the rate of return by this method is equivalent to the maximum interest rate ( normally after taxes) at which money could be borrowed to finance the project under conditions where the net cash flow to the project over its life would be just sufficient to pay all principal and interest accumulated on the outstanding principal.
This particular rate is called the “discounted cash flow rate of return” (DCFRR) and it is a measure of the maximum rate that the project could pay and still breakeven by the end of the project life. n t
NFW 0 n n 1 (1 r )
The Total NPW (orNPV) where NPW net present worth NPV net present value NFW net future worth
r discounted rate (interest rate) t life of project n nth year
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EXAMPLE 8-2.3 Initial fixed capital investment = $100,000 Working capital investment = $10,000 Salvage value at end of service-life = $10,000 The predicted after-tax cash flow to project based on total income minus all costs except depreciation in dollars is shown in the table below. Estimate the discounted cash flow rate of return (DCFRR). Year
Cash flow into the project,$
0
110,000
1
30,000
2
31,000
3
36,000
4
40,000
5
43,000
SOLUTION 8-2.3 The discounted factor is given by: 1
1 i n Where i = discounted cash flow rate of return n = year Sample calculation of Present value For i = 0.10 Present value in year 1 = $30,000 Present value in year 2 = $31,000
1 $27,273 1 0.1 1
1 0.12
$25,620
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The DCFRR is the interest rate at which the initial capital investment is equal to the sum of the present values. The DCFRR is estimated by a trial and error approach. The table of values is shown below. Year
Cash
Present
flow,$
value
Present
Present
Present
at value at value at value at
i=0.10
i=0.15
i=0.20
i=0.1763
1
30,000
27,273
26,087
25,000
25,504
2
31,000
25,620
23,440
21,528
22,404
3
36,000
27,047
23,671
20,833
22,118
4
40,000
27,321
22,870
19,290
20,892
5
43,000
26,700
24,585
17,281
19,093
Total
$133,961
$120,653
$103,932 $110,011
From the table, the discounted cash flow rate of return is estimated to be 17.63%.
8-2.4.3 Capitalized cost The capitalized cost profitability concept is used for comparing alternatives which exists as possible investment choices within a simple overall project.
For example, if a decision based on profitability analysis were to be made as to whether stainless-steel or mild-steel should be used in a chemical reactor as one part of a chemical plant, capitalized cost comparison will be a useful and appropriate approach.
Capitalized cost related to investment represents the amount of money that must be available initially to purchase the equipment and simultaneously provide sufficient funds for interest accumulation to permit perpetual replacement of the equipment.
If only one portion of an overall process to accomplish a set objective is involved and operating costs do not vary, then the alternative giving the least capitalized cost would be 87
the desirable economic choice.
Basic equation for capitalized cost is:
K Cv
CR C R (1 i ) n Vs (1 i ) n 1 (1 i ) n 1
where K Capitalized cost C R Re placement cost C v Original cost of equipment Vs Salvage value at end of estimated useful life n Estimated useful life of equipment i int erest rate (1 i ) n capitalised cost factor (1 i ) n 1
EXAMPLE 8-2.4 A reactor, which will contain corrosive liquids, has been designed. If the reactor is made of mild-steel, the initial installed cost will be $5000 and the useful life period will be 3 years. Since stainless- steel is highly resistant to the corrosive action of the liquids, stainless- steel as the material of construction has been proposed as an alternative to mildsteel. The stainless-steel reactor will have an initial installed cost of $15,000. The scrap value at the end of the useful life would be zero for either type of reactor, and both could be replaced at the cost equal to the original price. On the basis of equal capitalized cost for both types of reactors, what should be the useful life period for the stainless-steel reactor if money is worth 4% compounded annually?
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SOLUTION 8-2.4
Capitalized cost for mild steel is : CR 5000 K Cv 5000 n (1 i ) 1 (1 0.04) 3 1 K $45,030 therefore, capitalized cost for stainless steel reactor must also be $45, 030. For stainless steel reactor, CR 15000 45,030 C v 15000 n (1 i ) 1 (1 0.04) n 1 n 10.4 years Thus, the useful-life period of a stainless-steel should be 10.4 years for the two types of reactors to have equal capitalized cost.
If the stainless-steel reactor would have a useful life of more than 10.4 years, it would be the recommended choice, while the mild-steel reactor would be recommended if the useful-life using the stainless-steel were less than 10.4 years.
8-2.4.4 Payout period (or Pay-back time) Payout period or pay-back time is the minimum length of time theoretically necessary to recover the original capital investment in the form of cash flow to the project based on total income minus all costs except depreciation. In other words, it is the length of time to reach a cumulative cash flow of zero. Generally for this method, original capital investment means only the original, depreciable, fixed-capital investment, and interest effects are neglected. Thus
Payout period in years
depreciable fixed capital investment avg profit/yr avg depreciation/yr
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Pay-back time (or payout period), is a useful criterion for judging projects that have a short life, or when the capital is only available for a short time. It is often used to judge small improvement projects on operating plants. Typically, a pay-back time of 2 to 5 years would be expected from such projects. Pay-back time as a criterion of investment performance does not, by definition, consider the performance of the project after the pay-back period.
8-2.5 Sensitivity Analysis The economic analysis of a project can only be based on the best estimates that can be made of the investment and the cash flows. The actual cash flows achieved in any year would be affected by any changes in raw material costs and other operating costs and will be very dependent on the sales volume and price. A sensitivity analysis, therefore, is a way of examining the effects of uncertainties in the forecasts on the viability of a project.
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SELF ASSESSMENT 8-1 The original value of a piece of equipment is $22,000, completely installed and ready for use. Its salvage value is estimated to be $2000 at the end of a service life estimated to be 10 years. Determine the asset ( or book) value of the equipment at the end of 5 years using: (a) Straight –line method (b)Text book declining-balance method
SELF ASSESSMENT 8-2 A process modification to a small process unit has the following cash flow profile: Year 0 1 2 3 4 5
Cash Flow $ -91,093 20,000 40,000 40,000 40,000 30,000
Calculate the discounted cash flow rate of return (DCFRR) of the project. HINT: Use discount rates from 10% to 30%
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