Download presentation
Presentation is loading. Please wait.
Published byBertram Fox Modified over 8 years ago
1
Mid Term Review Terry A. Ring CH EN 5253 Design II
2
WeekDayLecture TopicAssignments 111-JanCourse Overview LectureReview Chapters 7,8,9 T&S 13-JanReview of Project EconomicsGeneration of Economics Spread Sheet HW 1 15-JanReview of Project Economics 218-JanMLK HolidayChapter 15 T&S 20-JanReview of Reactors - Selectivity & Heat EffectsHW 2 Assigned - Reactor 22-JanAttainable RegionAttainable Region Article 325-JanSeparationsChapter 17 T&S, Distillation Sequences Article 27-JanDistillation TrainsHW-3 Assigned-Separations 29-JanSeparations and Reactors 41-FebReactor, Separation and RecycleReactor-Separator-Recycle Networks Article 3-FebReactor, Separation and RecycleHW-4 Assigned Reactor/Separation and Recycle 5-FebReactor, Separation and Recycle 58-FebHeat and Power IntegrationChapter 3, 19 T&S 10-FebHeat and Power IntegrationHW-5 Assigned Heat Integration 12-FebHeat and Power Integration 615-FebPresident's Day Holiday 17-FebOptimization on Process FlowsheetsHW-6 Assigned Impurities and Optimization 19-FebEffects of Impurities on Reactors - HX 722-FebEffects of Impurites on Separators, Recycle 23-FebPlantwide ControlHW-7 Assigned Plant-wide Control 26-FebPlantwide ControlChapter 5 T&S 829-FebSequential Batch ProcessingHW-8 Assigned Sequential Batch Processing 2-MarSequential Batch Processing 4-MarSequential Batch Processing 97-MarReview for Exam 9-MarSequential Batch Processing 11-MarMid Term Exam
3
Equipment Costing Review 0.6 th rule Factored Estimate of Major Equipment Installation Factor Peripheral equipment, foundations, piping, etc Biggest Mistakes –No Off Sites - OSBL No Electrical Substation No Natural Gas Connection, etc. –Neglecting Major Equipment
4
Measures of Profitability Return on Investment (ROI) –ROI=annual earnings/capital investment –ROI > cost of capital (commercial interest rate, i) Payback period (PBP) Annualized Cost (C A ) Venture profit (VP) Investor’s Rate of Return (IRR) Discounted Cash Flow Rate of Return (DCFR) Hard work to get this all together Same
5
Definitions of Profitability Measures
6
Depreciation Straight-line Depreciation –Equipment Lifetime – Plant = 12 yr –Constant % each year so that plant is totally written off at end of life time 8% of Total Depreciable Capital, C TDC –What would be the depreciation % for 20 yr lifetime? Other types of depreciation –Accelerated Cost Recovery System (ARCS) –Modified Accelerated Cost Recovery System (MARCS) for Taxes –Declining-Balance Method (DB) –Double Declining-Balance Method (DDB) –Others
7
Taxes Property Taxes –Based upon the value of the property Used in Cost of Manufacturing Severance Tax = 12.5% of extracted mineral’s net value State Taxes –Very between states, UT = 6.2% Federal Income Tax for Business –Typically taken to be 34% of gross earnings
8
Total Production Cost, C Cost of Manufacturing minus general expenses C=COM+General Expenses General Expenses (see slide 6) –Selling expenses, R&D, Admin. (top management), Management Incentive Package
9
Working Capital, C WC Typically 15 % C TCI More Accurate Working Capital Calculation C WC =Cash Reserves+Inventory+accounts receivable- accounts payable Cash reserves - 30 days of raw materials, utilities, operations, Maintenance, operating overhead, property taxes, insurance and depreciation 8.33% of COM Inventories = 7 days of products at sales prices Accounts receivable - 30 days at sales price 8.33% of annual sales Accounts payable – 30 days of feedstocks at purchase price, 8.33% of annual feed stock costs
10
Reactor Heat Effects S,S&L Chapter 7
11
Managing Heat Effects Reaction Run Away –Exothermic Reaction Dies –Endothermic Preventing Explosions Preventing Stalling
12
Temperature Effects Thermodynamics/Equilibrium Kinetics
13
Unfavorable Equilibrium Increasing Temperature Increases the Rate Equilibrium Limits Conversion
14
Reactor with Heating or Cooling Q = UA ΔT
15
Best Temperature Path
16
Optimum Inlet Temperature Exothermic Rxn
17
Inter-stage Cooler Exothermic Equilibria Lowers Temp.
18
Inter-stage Cold Feed Exothermic Equilibria Lowers Temp Lowers Conversion
19
Reaction Selectivity Parallel Reactions –A+B R (desired) –A S Series Reactions –A B C(desired) D Independent Reactions –A B (desired) –C D+E Series Parallel Reactions –A+B C+D –A+C E(desired) Mixing, Temperature and Pressure Effects
20
Rate Selectivity Parallel Reactions –A+B R (desired) –A+B S Rate Selectivity (α D - α U ) >1 make C A as large as possible (β D –β U )>1 make C B as large as possible (k D /k U )= (k oD /k oU )exp[-(E A-D -E A-U )/(RT)] –E A-D > E A-U T –E A-D < E A-U T
21
Reactor Design to Maximize Desired Product
22
Maximize Desired Product Series Reactions –A B(desired) C D Plug Flow Reactor Optimum Time in Reactor
23
Engineering Tricks Reactor types –Multiple Reactors Mixtures of Reactors –Bypass –Recycle after Separation Split Feed Points/ Multiple Feed Points Diluents Temperature Management Sorted Out with Attainable Region Analysis
24
Attainable Region S,S&L Chapt. 7
25
Attainable Region Graphical method that is used to determine the entire space feasible concentrations Useful for identifying reactor configurations that will yield the optimal products
26
Procedure Step 1: Construct a trajectory for a PFR from the feed point, continuing to complete conversion or chemical equilibrium Step 2: When the PFR bounds a convex region, this constitutes a candidate AR. The procedure terminates if the rate vectors outside the candidate AR do not point back into it. Step 3: The PFR trajectory is expanded by linear arcs, representing mixing between the PFR effluent and the feed stream, extending the candidate AR. Step 4: Construct a CSTR trajectory to see if the AR can be extended. Place linear arcs, which represent mixing, on the CSTR trajectory to ensure the trajectory remains convex. Step 5: A PFR trajectory is drawn from the position where the mixing line meets the CSTR trajectory. If the PFR trajectory is convex, it extends the previous AR to form a expanded AR. Then return to step 2. Otherwise, repeat the procedure from Step 3.
27
Example Reactions Rate Equations
28
Step 1 Begin by constructing a trajectory for a PFR from the feed point, continuing to the complete conversion of A or chemical equilibrium Solve the PFR design equations numerically –Use the feed conditions as initial conditions to the o.d.e. –Adjust integration range, (residence time), until complete conversion or to equilibrium
29
PFR Design Equations
30
Solve Numerically Runge-Kutta
31
Solve Numerically
32
Step 2 Plot the PFR trajectory from the previous results. Check to see if rate vectors outside AR point back into it (e.g. Look for non-convex regions on the curve)
33
Step 3 Expand the AR as much as possible with straight arcs that represent mixing of reactor effluent and feed stream
34
Interpreting points on mixing line Larger Attainable region PFR C A =0. 2187 C B =0.00011 C A =0.72 C B =0.0004 C A =1 C B =0 (1- ) PFR C A =1 C B =0 (1- ) Desired Mixing point C A =1 C B =0
35
Mixing of Streams Reactant Bypass Vector Equation, i component is C A, j component is C B α =fraction of mixture of stream 1in the mixed stream Feed mixing fraction: = 0. 64
36
Step 4 If a mixing arc extends the attainable region on a PFR trajectory, check to see if a CSTR trajectory can extend the attainable region For CSTR trajectories that extend the attainable region, add mixing arcs to concave regions to ensure the attainable region remains convex Solve CSTR multiple NLE numerically –Vary until all feed is consumed or equilibrium is reached
37
CSTR Design Equations
38
Solve numerically at various until complete conversion or equilibrium is achieved
39
Plot extensions to attainable region i.c. for step 5 CSTR C A =1 C B =0 (1- )
40
Possible Configuration at this point CSTR C A =1 C B =0 (1- ) CSTRPRF
41
But you can do better than even this! Add PFR after mix point
42
Step 5 A PFR trajectory is drawn from the position where the mixing line meets the CSTR trajectory. If this PFR trajectory is convex, it extends the previous AR to form an expanded candidate AR. Then return to Step 2. Otherwise repeat Step 3
43
Solve PFR equations with modified initial conditions New feed point Vary integration range
44
Attainable Region
45
Keep track of feed points Initial feed point occurs at far right on AR Mixing lines connect two feed points Connect reactors and mixers with feed points to get network
46
Reactor configuration for highest selectivity CSTR C A =1 C B =0 C A =0.38 C B =0.0001 C A =0.185 C B =0.000124 Reactor series occur when multiple feed points exist PFR
47
Go back to calculations for optimal reactor sizing
48
Other factors to consider Annualized, operating, and capital costs might favor designs that don’t give the highest selectivity If objective function (e.g. $ = f{C A } + f{C B }) can be expressed in terms of the axis variable, a family of objective contours can be plotted on top of the AR –The point where a contour becomes tangent to the AR is the optimum Temperature effects –Changing temperature will change the AR –Need energy balance for non-isothermal reactions Make sure to keep track of temperature
49
Profit ($) = 15000*C B -15*C A 2 Optimal point not at highest selectivity PFR CSTR
50
Conclusions Need to know feed conditions AR graphical method is 2-D and limited to 2 independent species Systems with rate expressions involving more than 2 species need to be reduced –Atom balances are used to reduce independent species –Independent species = #molecular species - #atomic species If independent species < 2, AR can be used by Principle of Reaction Invariants
51
Separation Trains S, S&L Chapt. 8
52
Separation Methods Absorption Stripping Distillation Membrane Separations Crystallization etc
53
Use of Separation Units
54
Column Sequences No. of Columns –N c =P-1 P= No. of Products No. of Possible Column Sequences –N s =[2(P-1)]!/[P!(P-1)!] P= No. of Products –P=3, N c =2, N s =2 –P=4, N c =3, N s =5 –P=5, N c =4, N s =14 –P=6, N c =5, N s =42 –P=7, N c =6, N s =132 No. of Possible Column Sequences Blows up!
55
How do I evaluate which is best sequence?
56
Marginal Vapor Rate Marginal Annualized Cost~ Marginal Vapor Rate Marginal Annualized Cost proportional to –Reboiler Duty (Operating Cost) –Reboiler Area (Capital Cost) –Condenser Duty (Operating Cost) –Condenser Area (Capital Cost) –Diameter of Column (Capital Cost) Vapor Rate is proportional to all of the above
57
Selecting Multiple Column Separation Trains Minimum Cost for Separation Train will occur when you have a –Minimum of Total Vapor Flow Rate for all columns –R= 1.2 R min –V=D (R+1) V= Vapor Flow Rate D= Distillate Flow Rate R=Recycle Ratio
58
Azeotrope Conditions Conditions on the Activity Coefficient Minimum Boiling, γ j L > 1 Maximum Boiling, γ j L < 1 x j =y j, j=,1,2,…C
59
Raoult’s Law
60
Importance of Physical Property Data Set In all cases –Need sophisticated liquid phase model to accurately predict the activity coefficient for the liquid. For High Pressure Cases Only –Also need sophisticated (non-ideal) gas phase fugacity model
61
Multi-component Azeotropes Residue Curve Map –dx j /dť = dx j /d ln(L) = x j – y j Integrate from various starting points
62
Defining Conditions for Multi- component Azeotrope t goes from 0 to 1, ideal to non-ideal to find Azeotrope
63
Distillation X B, X F and Y D form a line for a Distillation Column Line can not cross Azeotrope line
64
Ethanol/Water Distillation with Benzene To Break Azeotrope
65
Pressure Swing to Break Azeotrope Temp. of Azeotrope vs. Pressure Mole Fraction of Azeotrope
66
Reactor-Separation Train- Recycle Chapt. 7&8
67
Trade-off between Reactor and Separator Factors –Reactor Conversion of limiting reactant Effects cost and size of Separation Train –Reactor Temperature and mode of operation (adiabatic, isothermal, etc.) Effect utility costs for separation and reaction Effect impurities from side reactions –High Reactor Pressure for Le Chatlier cases (less moles of product) Higher cost for recycle compression
68
Trade-off between Reactor and Separator Factors, cont. –Use of excess of one or more reactant to increase equilibrium conversion and/or reaction rate Increases cost of separation train –Use of diluents in adiabatic reactor to control temperature in reactor Increases cost of separations train –Use of purge to avoid difficult separation. Decreases the cost of separations Loss of reactants – increase cost of reactants May increased cost of reactor, depending on the purge-to- recycle ratio
69
Factors that effect recycle/purge Factor –Excess reactants Increases recycle flow Increases separation costs –Concentration of impurities to be purged Effects the recycle-to-purge ratio –Reactor outlet temperature and pressure Increase cost of utilities in separation Increase cost to recycle - compressor
70
Compare Recycle Concepts Costs Benefits
71
Feedback effects of Recycle Loop Small disturbance on feed Large effect on recycle flow rate/composition Snowball effect on reactor/separator
72
Heat Integration Chapter 9 Terry Ring University of Utah
73
Costs Heat Exchanger Purchase Cost – C P =K(Area) 0.6 Annual Cost –C A =i m [ΣC p,i + ΣC P,A,j ]+sF s +(cw)F cw i m =return on investment F s = Annual Flow of Steam, –$5.5/ston to $12.1/ston = s F cw =Annual Flow of Cold Water –$0.013/ston = cw
74
Lost Work = Lost Money Transfer Heat from T 1 to T 2 ΔT approach Temp. for Heat Exchanger T o = Temperature of Environment Use 1 st and 2 nd laws of Thermodynamics LW=QT o ΔT/(T 1 T 2 ) –ΔT=T 1 -T 2 –T o = Environment Temperature Q= UAΔT lm T1T1 T2T2 Q
75
Heat Integration Make list of HX Instead of using utilities can you use another stream to heat/cool any streams? How much of this can you do without causing operational problems? Can you use air to cool? –Air is a low cost coolant. Less utilities = smaller cost of operations
76
Terms HEN=Heat Exchanger Network MER=Maximum Energy Recovery Minimum Number of Heat Exchangers Threshold Approach Temperature Optimum Approach Temperature
77
Process
78
Minimize Utilities For 4 Streams
79
Adjust Hot Stream Temperatures to Give ΔT min
80
Enthalpy Differences for Temperature Intervals
81
Pinch Analysis Minimum Utilities
82
Pinch Analysis ΔT app MER values
83
How to combine hot with cold? At Pinch (temp touching pinch) –Above Pinch Connect C c ≥C h –Below Pinch Connect C h ≥C c Not touching Pinch temp. –No requirement for C c or C h
84
4 Heat Exchanger HEN for Min. Utilities C c ≥C h C h ≥C c MER Values
85
Stream Splitting Two streams created from one one heat exchanger on each piece of split stream with couplings 1 1a 1b 1a 1
86
Optimization of HEN How does approach delta T (ΔT min ) effect the total cost of HEN? Q= UA ΔT LW=QT o ΔT/(T 1 T 2 ) –More Utility cost
87
Costs Heat Exchanger Purchase Cost – C P =K(Area) 0.6 Annual Cost –C A =i m [ΣC p,i + ΣC P,A,j ]+sF s +(cw)F cw i m =return on investment F s = Annual Flow of Steam, –$5.5/ston to $12.1/ston F cw =Annual Flow of Cold Water –$0.013/ston
89
Change ΔT min C P =K(Area) 0.6 Area=Q/(UF ΔT min ) More Lost Work LW=QToΔT/(T1T2)
90
Optimization of Process Flowsheets Chapter 24 Terry A. Ring CHEN 5353
91
Degrees of Freedom Over Specified Problem –Fitting Data –N variables >>N equations Equally Specified Problem –Units in Flow sheet –N variables =N equations Under Specified Problem –Optimization –N variables <<N equations
92
Optimization Number of Decision Variables –N D =N variables -N equations Objective Function is optimized with respect to N D Variables –Minimize Cost –Maximize Investor Rate of Return Subject To Constraints –Equality Constraints Mole fractions add to 1 –Inequality Constraints Reflux ratio is larger than R min –Upper and Lower Bounds Mole fraction is larger than zero and smaller than 1
93
PRACTICAL ASPECTS Design variables, need to be identified and kept free for manipulation by optimizer –e.g., in a distillation column, reflux ratio specification and distillate flow specification are degrees of freedom, rather than their actual values themselves Design variables should be selected AFTER ensuring that the objective function is sensitive to their values –e.g., the capital cost of a given column may be insensitive to the column feed temperature Do not use discrete-valued variables in gradient-based optimization as they lead to discontinuities in f(d)
94
Optimization Feasible Region –Unconstrained Optimization No constraints –Uni-modal –Multi-modal –Constrained Optimization Constraints –Slack –Binding
95
LINEAR PROGRAMING (LP) equality constraints inequality constraints objective function w.r.t. design variables The N D design variables, d, are adjusted to minimize f{x} while satisfying the constraints
96
EXAMPLE LP – GRAPHICAL SOLUTION A refinery uses two crude oils, with yields as below. Volumetric YieldsMax. Production Crude #1Crude #2(bbl/day) Gasoline70316,000 Kerosene692,400 Fuel Oil246012,000 The profit on processing each crude is: $2/bbl for Crude #1 and $1.4/bbl for Crude #2. a)What is the optimum daily processing rate for each grade? b)What is the optimum if 6,000 bbl/day of gasoline is needed?
97
EXAMPLE LP –SOLUTION (Cont’d) Step 1. Identify the variables. Let x 1 and x 2 be the daily production rates of Crude #1 and Crude #2. maximize Step 2. Select objective function. We need to maximize profit: Step 3. Develop models for process and constraints. Only constraints on the three products are given: Step 4. Simplification of model and objective function. Equality constraints are used to reduce the number of independent variables (N D = N V – N E ). Here N E = 0.
98
EXAMPLE LP –SOLUTION (Cont’d) Step 5. Compute optimum. a)Inequality constraints define feasible space. Feasible Space
99
EXAMPLE LP –SOLUTION (Cont’d) Step 5. Compute optimum. b)Constant J contours are positioned to find optimum. J = 10,000 J = 20,000 J = 27,097 x 1 = 0, x 2 = 19,355 bbl/day
100
EXAMPLE LP – GRAPHICAL SOLUTION A refinery uses two crude oils, with yields as below. Volumetric YieldsMax. Production Crude #1Crude #2(bbl/day) Gasoline70316,000 Kerosene692,400 Fuel Oil246012,000 The profit on processing each crude is: $2/bbl for Crude #1 and $1.4/bbl for Crude #2. a)What is the optimum daily processing rate for each grade? b)What is the optimum if 6,000 bbl/day of gasoline is needed?
101
Dealing with Impurities in Processes and Process Simulators ChEN 5253 Design II Terry A. Ring There is not chapter in the book on this subject
102
Impurity Effects Heat Exchange Reactors Separation Systems Recycle Loops
103
Impurities in Heat Exchange Impurities effect heat capacity –Lower C p Various options –Raise C p Increase H 2 Impurities effect the enthalpy of stream –Total heat of condensation is less due to impurity –Total heat of vaporization is less due to impurity
104
Impurities in Separation Trains Non-condensable Impurities –Build up in Distillation column – Big Trouble!! Condensable Impurities –Cause some products to be less pure May not meet product specifications Can not sell this product – Big Trouble!! –Rework cost –Waste it –Sell for lower price
105
Processes are tested for Impurity Tolerance Add light and heavy impurities to feed –Low concentration All impurities add to 0.1 % of feed (may need to increase Tolerance in Simulation) –Medium concentration All impurities add to 1% of feed –High concentration All impurities add to 10% of feed Find out where impurities end up in process Find out if process falls apart due to impurities –What purges are required to return process to function.
106
Impurities in Separation Trains It is important to know where the impurites will accumulate in the train Which products will be polluted by which impurities –Is that acceptable for sale of product?
107
Purging Impurities Find the point in the process where the impurities have the highest concentration –Put Purge here Put a purge in almost all recycle loops
108
Plant-wide Control T&S Chapter 5, SS&L Chapter 20 Terry A. Ring University of Utah
109
10 - Design and ControlDESIGN AND ANALYSIS II - (c) Daniel R. Lewin 109 PROCESS DESIGN STAGES AND TOOLS
110
Instrumentation and Control Objectives
111
Control Valves Safety Aspects –Fail Open, Fail Closed Types of Control Valves –Linear –Non-linear –Restricted open range (gain scheduled controller)
112
Pressure Control Loop Pressure Control Valve – fail closed –Is fail closed best? Pressure Transducer – PT –Pieziolelectric, diaphragm displacement Pressure Indicator Controller – PIC –PID, maybe gain scheduling if valve is non-linear –Gain scheduling for pressure relief above a given pressure. Alarm High Alarm Low
113
Level Control Loop Level Control Valve – fail open Level Transducer – LT –DP cell, guided radar, conductivity Level Indicator Controller – LIC –PID, maybe gain scheduling if valve is non-linear Alarm High Alarm Low Pump may dead head –Interlock or gain schedule valve to never go to zero pump
114
Flow Control Centrifugal Pump Flow Control Valve – Fail ?? Flow Transducer –Orifice, venture, v-cone, many others Flow Indicator Controller – FIC PID, maybe gain scheduling if valve is non-linear Note, check valve to prevent back flow and by- pass loop to prevent dead-heading pump VFD can be costly Missing Alarms on Pressure Indicator (PI) Flow Ratio
115
Ratio Control Ratio Control Valve – FFV – fail Closed –May have two control valves – one for each stream 2 Flow Transducers –Orifice, venture, v-cone, many others Flow Ratio Controller –PID with calculator for total flow(F T =F T1 +F T2 ) to be held constant –CV1 SP=Ratio*SP LC –CV2 SP=(1-Ratio)*SP LC Add Pressure Relief
116
Temperature Control w Utility Utility Control Valve – TV – Fail ?? Temperature Sensing Element (TE) and Temperature Transducer (TT) or Temperature Indicator (TI) Temperature Indicator Controller –PID, maybe gain scheduling if valve is non-linear
117
Vaporizer Control Flow Control on Liquid and Level Control in HX Pressure relief valve???
118
CSTR Contol System Utility Temperature Control –TT and FT feed signals to Flow control – Cascade Control –Valve Fail??? Pressure Control –PID, gain scheduling = on/off –Valve Fail ??? Level Control PID, gain scheduling if non-linear Valve Valve Fail?? Feed Flow Rate Control –This needs ratio control!! –Valves Fail ??
119
PID Tuning – Lamda Tuning P only P=1 Step Test – CV +/- 10% Observe increase/decrease in PV –Τ d and τ (63.2%) and K p =%PV/%CV(output) K c =(K p ) -1 *(T r )/(λ+T d ) T r =τ, λ=0.5 to 4 (typically 3) times Max(T d, T r ) Cascade Control inner loop 5x to 10x faster than outer loop
120
Lamda Tuning Results Max(T d, T r ) = 4 s
121
These are all simple 1 input, 1 output controllers with alarms for safety Plant Wide Control is Different –Active Control System Objectives Meet production rate and product quality targets Keep system in Safe Operating Range for equipment, catalysts and materials of construction Minimize Energy Utilized Minimize Process Variability Meet environmental regulations –Air quality –Water discharge quality Profit Optimizer –Provide inputs for Active Control System Separate Safety Control System separate sensors, valves, controllers, power system –Identify Unsafe conditions –Take Action to Safely Bring the Plant/System Down to a Safe Point of Operation or Shut Down
122
Procedure for Plant Wide Control I. Top-down analysis to identify degrees of freedom and primary controlled variables (look for self- regulating variables) II. Bottom-up analysis to determine secondary controlled variables and structure of control system e.g. pairings (CVs with Sensors).
123
Degrees of Freedom (DoF) Analysis DoF (= Control Valves) DoF=N variables -N externally_defined -N equations Practice DoF on Smaller Problems
124
Heat Exchanger Network Flow Rate Set Floating Flow Rates for C1 & C2
125
Heat Exchanger Network Control DoF = N v -N Def -N eq =15-4-9=2 ( # Control Valves ) N v =15 (9Ts,3 flows, 3 Hx Qs) N Def =4 (F 1,T o, Θ o, Θ 1 ) Equations for Each Exchanger N eq =3x3=9 –Q 1 =F 1 C p1 (T o -T 1 ) –Q 1 =F 3 C p3 (Θ 4 - Θ 3 ) –Q 1 = U 1 A 1 ΔT LM With By-pass, DoF = N v -N Def -N eq =17-4-10=3 –Θ 3 =(1-φ) Θ 0 + Θ’ 3 N eq =10, N v =17(& φ,Θ’ 3 ) – Hot Stream Streams 2 and 3 Cold
126
Additional Control Valve on Feed is Possible which is used for Production Rate Control Control Valves ???? Degrees of Freedom Analysis DoF= N v -N Def - N eq Variables, N v = 4 N T +13, N Def =2 (Feed Flow and Composition) Equations = 4 N T +6, DoF=5 (# Control Valves) Distillation Control with Total Condenser
127
Distillation Control What measurements? What should be the pairings of measurements to control valves? P D controlled with CV Q C L R controlled by CV B L D controlled by –CV L if so what does CV D control?? –CV D if so what does CV L control?? What does CV Q R and CV B control?
128
Distillation Control Material Balance Control 4 Control Schemes Inferred Composition Analysis = Temp. of Stage –Top CA or Bottom CA
129
With Composition Analysis LV control DV control L/F V control D/F V control
130
Distillation Control Types of Control –LV control –DV –LB –DB –(L/D)(V/B) –(L/F)(V/F)
131
Key Concerns for Plant-wide Control System Establish Control Objectives –Safe Operation – Process Within its Constraints –Meets Environmental Constraints –Control Production Rate Feed Flow Controls Product (on-demand) Flow Controls –Control Product Quality Determine Degrees of Freedom for Control System –Position Control Valves Establish Energy Management System Recycle Loop Flows Fixed – watch out for snowball effects Vapor and Liquid Inventories Fixed Improve Dynamic Controllability
132
Acrylic Process A B Objectives –Production Rate of B high & constant –Conversion in reactor highest possible –Constant Composition for B
133
Where to set the production rate? Should it be at the inlet or outlet? Very important! Determines structure of remaining inventory (level) control system Set production rate at (dynamic) bottleneck Link between Top-down and Bottom-up parts of Plant-wide Control System
134
On-demand Product or Feed Flow Control System 20.13 FC on Product B Feed of A controlled as needed by reactor 20.14 FC on Feed A Flow Level Controls Product B Flow Rate Reactor Temp is controlled with Cascade Controller –To meet Composition Requirements Note: –Trim steam heater for Feed A for accurate temperature control at reactor Note: Flash Vessel will have a spray head for cold B to condense all possible Vapors Vapor Typically is fed above Liquid level Demister on outlet of Vessel
135
Plantwide Control Design Determine all manipulated variable. Number of manipulated variables MAY be equal to the number of control valves. Determine how production rate is set: (a) upstream process; (b) downstream process; (c) free to vary. Determine the best way to control the production rate: (a) valve selection; (b) setpoint on temperature, recycle, etc. can sometimes be used to control production rate. Decide how to control product quality. –The closed loop system for product quality control should have adequately small time constant, time delays, and sufficiently large gains. –Consider interaction between different loops and decide if multivariable control is needed. –Make sure that flows are not too small to achieve the objective (i.e. it is difficult to control condenser level using small flow rate of distillate in the column with high internal flows). Determine and stabilize unstable units. Design inventory control loops. Develop component balance control loops (control of makeup streams of reactants, makeup gaseous streams to maintain pressure, liquid makeup to control level, etc.). Be careful with recycle loops: They introduce feedback (positive or negative), which makes it difficult to analyze the consequences of the particular control design on the overall performance. For instance, it was found that unless one flow somewhere in the recycle loop is fixed, the recycle flow might grow to very high rate when disturbance occurs or when throughput is increased (snowball effect). Heat integration saves cost but can make plantwide control more difficult because integrating loops also “spread” disturbances. Use cascade control to reduce the effect of disturbances. Using steady state simulations to study controllability of the designed system in face of disturbances. For example, perturb the feed flow rate or composition and study how flows and other process variables change to compensate for the disturbances. If small disturbance requires large changes in flow rate or other manipulated variables to compensate for its effect, the control system must be redesigned. Using steady state simulations, determine the ranges of acceptable disturbances, which can be taken care of while maintaining the production goals. Use dynamic simulation to determine dynamic response of the designed system. Redesign as needed. For instance, redesign to eliminate inverse response, if at all possible. Use dynamic simulation to find the range of disturbances, which can be compensated for using the designed control system. Use dynamic model to determine the range of stability of the closed-loop system when different model parameters change. Use the remaining degrees of freedom (manipulated variables) for steady state optimization (maximize the profit, etc.) and/or to improve controllability of the plant (disturbance rejection, flexibility in changing operation point or product mix, etc). Ensure safe and environmentally sound operation. Startup shutdown, safety and emergency handling control.
136
Vinyl Chloride Process Reactor 1 FeCl 3 solid Catalyst Conversion is >90% –C 2 H 4 + Cl 2 C 2 H 2 Cl 2 Fired Heater for Pyrolysis Reaction –Conversion is 60% –C 2 H 2 Cl 2 C 2 H 3 Cl + HCl »Product –
137
Vinyl Chloride Process
138
Vinyl Chloride Process w Control System
139
Approach to Optimizer DoF Analysis for Optimizer –Levels are not important in Optimization Constraint Analysis Optimization Itself
140
Example of self-optimizing structure Recycle process: J = V (min. boil-up => min. energy) N m = 5 N 0y = 2 Levels DoF=N ss = 5 - 2 = 3 Given feed rate F 0 and column pressure: 1 2 3 4 5 Constraints: Maximize reactor volume (residence time and thus conversion) Product spec: x B > 0.98
141
Plant-Wide Controllability Control Architecture –DoF analysis Dynamic Analysis No. of valves –DoF analysis Steady State Analysis No. of valves – No. of liquid level loops Types of control –Single loop –Gain Scheduling –Ratio control –Cascade Control –Multi-variable control –Model Based control (MPC) –Override control
142
The End
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.