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kok电子竞技:文档简介
1、Hao Ling Department of Petroleum Processing East China University of Science and Technology 64252328, ,Optimization of Chemical Process,Objectives,(1) Learn basic concepts of optimization techniques; (2) Learn optimization modeling of synthesis and planning problems; (3) Learn to use some chemical e
2、ngineering modeling tools。,Assignment and Exams,Assignments (30%) Quiz Homework Energy saving technology of ChE. (submitted before 17th week) Exams (70%) Short conceptual questions; Derivations, proofs, small problems and etc.,Reference,Optimization of chemical processes, Thomas F. Edgar, David M. H
3、immelblau Distillation design and control using Aspen simulation, William L. Luyben 最优化方法施光燕、董加礼,高等教育出kok电子竞技社 Matlab 工具书,Preface,Three key components Cost of energy; Stringent environmental regulations; Product pricing and quality competition. Optimization,How to do optimization job?,Modification in plan
4、t design Modification in operation procedures Automation Optimization other aspects,Ternary distillation column configurations without any thermal coupling.,Thermally coupled ternary distillations.,原料渣油入炉流程图,Problem Formulation,Problem formulation requires identifying the essential elements of a con
5、ceptual or verbal statement of a given application and organizing them into a prescribed mathmatical for The objective function (economic criterion) The process model (constrains),Problem Formulation,Objective function includes in the form of Profit Cost Energy Yield The process model and constrains
6、 describe the interrelationship of the key variables like bridge.,Problem Formulation,Chapter I presents six steps for optimization. Chapter II summarizes the characteristics of process models and explains how to build one. Chapter III treats the most common type of objective function, the cost or r
7、evenue function(收益函数).,Chapter 1,1.1 what optimization is all about? A process can be represented by some equation or perhaps solely by experimental data. The goal of optimization is to find the values of the variables in the process that yield the best value of the performance criterion.,Chapter 1,
8、Optimization is concerned with selecting the best among the entire set by efficient quantitative methods. Critical analysis of the process or design; Insight about what the appropriate performance objectives are; Use of past experience.,Chapter 1,1.2 Why optimize? Improved yields of valuable product
9、s; Reduced energy consumption; Higher processing rates; Long time between shutdown; Reduce maintenance costs; Less equipment wear; Better staff utilization.,Chapter 1,1.3 Scope and hierarchy(等级) of optimization From gross to minute details,A complex combination of plants,Individual plants,Combinatio
10、n of units,Individual pieces of equipment,Subsystems in a piece of equipment,Smaller entities,炼厂典型工艺流程中的加氢单元,Chapter1,Typical industrial company optimization in three levels,Management,Design,Operation,Allocation and scheduling,Individual equipment,Management level,Management decision concerns Proje
11、ct evaluation Product selection Corporate budget Invest. in sales versus research and develop. New plant construction,Uncertainty, risk and the magnitude of objective function is larger than other levels,Design and equipment level,Design and equipment specification includes, A batch process or a con
12、tinuous process? How many reactors? The configurations of plant? How do we arrange the processes? What is the optimum size of a unit or combination of units?,Many large computer programs should be used to arrive at a desirable process flowsheet.,Plant Operation conditions level,Operation control for
13、 a given unit at certain temp., pressures, or flowrates. The allocation of raw materials; Shipping, transportation, distribution of products,Various types affecting the application of opt.,Sales limited by productions; Sales limited by market; Large unit throughputs; High raw material or energy cons
14、umption; Product quality exceeds product specification; Losses of valuable components through waste streams; High labor costs.,More productions, more profits,Reductions in unit manufacturing costs,Small savings are greatly magnified,In the absence of complete optimization we often rely on “incomplet
15、e optimization” Sub-optimization ignores some factors that have an effect, either obvious or indirect, on other systems or processes in the plant.,Suboptimization,Optimization Routing tankers for the distribution of crude and refined products; Sizing and layout of a pipeline; Designing equipment and
16、 an entire plant; Scheduling maintenance and equipment replacement; Operating equipment, such as tubular reactors, columns, and absorbers;,Typical projects for opt.,Evaluating plant data to construct a model of a process; Minimizing inventory charges; Allocating resources or services among several p
17、rocess; Planning and scheduling construction.,Example of optimization,Insulation design (保温设计),Cost of insulation,Cost of lost energy,Minimum Annual cost,Total cost curve,Cost ($/year),Insulation thickness,Example1.1,Example1.2 Optimal operation conditions of a boiler,$/time,Reflux or heat duty,Regi
18、on of feasible operation,Value of product,Cost of heat (high fuel cost),Cost of heat (low fuel cost),R1,R2,Optimal profit,Optimal profit,Constrain,Constrain,Example1.3 Optimization distillation reflux,$/time,Reflux or heat duty,Region of feasible operation,Total profit, low fuel,R1,R2,Optimum,Optimu
19、m,Constrain,Constrain,Total profit, high fuel,Example1.3,Example1.4 Multiplant product distribution,A single product (Y) manufactured at several plant locations. The product needs to be delivered to several customers located at various distances from each plant. Question: How can you arrange the pro
20、duction at various plants? How many factors should we think about?,How much Y must be produced at each of m plants (Y1,Y2,Y3,Y4,Ym)?; How should Ym be allocated to each of n demand point? Transportation Production cost versus capacity curves for each plant (old plant inefficient, new plant more effi
21、cient),The essential features of Opt. problem,Every optimization problem contains three essential categories: At least one objective function to be optimized; Equality constrains (equations); Inequality constraints (inequatilitis).,1.5 The essential features of optimization problems,Example 1.5 Mate
22、rial balance reconciliation,有关部门对32D纯棉纱规定的质量指标为棉结不多于70粒,品质指标不小于2900求最优解。,Example 1.5,根据问题的需要设置变量:设X1 , X2和X3分别为国棉131,229,327的棉花配比。得到数学模型: minZ( 8400 X1 +7500 X2 +6700 X3 ) S.t. 60X1 +65X2 +80 X3 70 3800 X1 +3500 X2 +2500 X3 2900 X1 + X2 + X31, X1 , X2 , X3 0,1.6 General procedure for solving optimiz
23、ation problems,Analyze the process itself so that the process variables and specific characteristics of interest are defined; that is, make a list of all of the variables. Determine the criterion for optimization, and specify the objective function in terms of the variables defined in step 1 togethe
24、r with coefficients. This step provides the performance model(运用模型).,Using mathematical expressions, develop a valid process or equipment model that relates the input-output variables of the process and associated coefficients. Includes both equality and inequality constraints. Use well-know physica
25、l principles (like mass balances, energy balances), empirical relations, implicit concepts and external restrictions. Identify the independent and dependent variables to get the number of degree of freedom.,If the problem formulation is too large in scope: (a) break it up into manageable parts or (b
26、) simplify the objective function and models; Apply suitable optimization technique to the mathematical statement of the problem. Check the answer, and examine the sensitivity of the result to changes in the coefficients in the problem and the assumptions.,Heat exchanger design,Variables in the prob
27、lem: Heat transfer surface, flow rate, number of shell pass, number of tube passes, number and spacing of baffles(隔板), length of the exchanger, diameter of the tubes and shell, the inlet temp., the pressure drop.,T,Feed,CW,Steam,Distillation column,Example1.7 The six steps of optimization for a manu
28、facturing problem,Suppose you are a chemical distributor who wishes to optimize the inventory of a specialty chemical. You expect to sell Q barrels of this chemical over a given year at a fixed price with demand spread evenly over the year. If Q=1000,000 barrels per year, you must decide on a produc
29、tion schedule. Unsold production is kept in in inventory. In order to determine the optimal production schedule, you must quantify those aspects of the problem which are important from a cost viewpoint.,Step 1. How to product the products? One line to produce 1000,000 barrels; Ten runs to produce 10
30、00,000 barrels, 100,000 each line. or N runs to produce 1000,000 barrels, 1000,000/N each line. Let D= 1000,000/N This is an obvious optimization problem. We also need to define two other variables besides D. (1) The number of runs per year (n);,(2) The total number of units produced per year, (Q).
31、We also need to know the cost of operation, which is a cost function and mathematical model. Step2. Let the business costs be split up into two factors: The carrying cost or the cost of inventory; The cost of production. Let D be the number of units produced in one run. Q is assigned to be a known v
32、alue.,Let the cost of inventory be K1D,where the parameter K1 essentially lumps together cost of working capital for the inventory itself and storage costs. The cost per run is assumed to be a linear function of D, given by the following equation: K2 is a setup cost and denotes a fixed cost of produ
33、ction equipment; K3 is an operating cost parameter.,In fact, the realistic assumption will change if the incremental cost(边际成本) of manufacturing could decrease somewhat for large runs, so we have a nonlinear cost function of the form: The total annual manufacturing cost for the product is the sum of
34、 the carrying costs and the production costs, namely,inventory,setup cost,operating cost,The objective function is a function of two variables, D and n. However D and n are directly related, namely Therefore only one independent variable exists for this problem, which we will select to be D. Elimina
35、ting n from the objective function in (c) gives,Several implicit constraints exist: for example, Integer intid quantities of runs. Can D be treated as a continuous variable? As D is the only variable and large, we can treat D as being continuous. After obtaining the optimal D, the practical value fo
36、r D is obtained by rounding up or down. Other constraints exist: D must be positive and etc. Step 4. Not need.,Step 5 Look at the total cost function (d), we can use differential equation of (d) to find the optimal solution, suppose D is a continuous variable. (e),The square root of Q.,Step 6. Sensi
37、tivity of the optimal answer. How much the optimal value of C changes when a variable such as D changes or a coefficient in the objective function changes? Therefore, information concerning the sensitivity of the optimum to changes or variations in a parameter is very important in optimal process de
38、sign. Substitute Dopt from Eq.(f) into the total cost function,Next take partial derivatives of Copt with respect to all variables,Absolute Sensitivity coefficients,Similarly, we can develop expressions for the sensitivity of Dopt,Suppose Then,The most sensitive,Relative sensitivities: Take a look a
39、t : One unit change in Q is quite different from a one unit change in K1. Therefore, in order to put the sensitivities on a more meaningful basis, you should compute the relative sensitivities: for example, the relative sensitivity of Copt to K1 is,Application of the above idea for the other variabl
40、es yields the other relative sensitivities for Copt. Numerical values are Changes in the parameters Q and K3 have the largest relative influence on Copt, significantly more than K1 or K2. The relative sensitivities for Dopt are So that all parameters have the same influence on the optimum value of D
41、 except for K3.,For a problem for which we cannot obtain an analytical solution, you would have to determine sensitivities numerically. You would compute (1) the cost for the base case for a specified value of a parameter; (2) change each parameter separately by some arbitrarily small value, such as plus 1% or 10% then calculate the new cost. This is the same as approximate a differential calculation method.,
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