Mid Term Review Terry A. Ring CH EN 5253 Design II.
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Transcript of Mid Term Review Terry A. Ring CH EN 5253 Design II.
![Page 1: Mid Term Review Terry A. Ring CH EN 5253 Design II.](https://reader035.fdocuments.in/reader035/viewer/2022062409/56649eae5503460f94bb5460/html5/thumbnails/1.jpg)
Mid Term Review
Terry A. Ring
CH EN 5253
Design II
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Week Day Lecture Topic AssignmentsDue Date
1 12-JanCourse Overview Lecture Review Chapters 7,8,9
14-JanReview of Project EconomicsGeneration of Economics Spread Sheet HW 1
16-JanReview of Project Economics Attainable Rection - HW 2
2 19-JanMLK Holiday
21-JanReview of Reactors - Selectivity & Heat Effects Separation Trains, HW 2 Assigned HW 1
23-JanAttainable Region
3 26-JanSeparations
28-JanDistillation Trains HW-3 Assigned HW 2
30-JanSeparations and Reactors
4 2-FebReactor, Separation and Recycle
4-FebReactor, Separation and Recycle HW-4 Assigned HW 3
6-FebReactor, Separation and Recycle
5 9-FebHeat and Power Integration
11-FebHeat and Power Integration HW-5 Assigned HW 4
13-FebHeat and Power Integration
6 16-FebPresident's Day Holiday
18-FebOptimization on Process Flowsheets HW-6 Assigned HW 5
20-FebEffects of Impurities on Reactors - HX
7 23-FebEffects of Impurites on Separators, Recycle
25-FebPlantwide Control HW-7 Assigned HW 6
27-FebPlantwide Control
8 2-MarReview for Exam
4-MarSequential Batch Processing HW 7
6-MarMid Term Exam Exam
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Reactor Heat Effects
S,S&L Chapter 7
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Managing Heat Effects
• Reaction Run Away– Exothermic
• Reaction Dies– Endothermic
• Preventing Explosions
• Preventing Stalling
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Temperature Effects
• Thermodynamics/Equilibrium
• Kinetics
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Unfavorable Equilibrium
• Increasing Temperature Increases the Rate
• Equilibrium Limits Conversion
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Reactor with Heating or Cooling
Q = UA ΔT
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Best Temperature Path
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Optimum Inlet TemperatureExothermic Rxn
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Inter-stage Cooler
Exothermic Equilibria
Lowers Temp.
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Inter-stage Cold Feed
Exothermic Equilibria
Lowers TempLowers Conversion
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Reaction Selectivity
• Parallel Reactions– A+BR (desired)– AS
• Series Reactions– ABC(desired)D
• Independent Reactions– AB (desired)– CD+E
• Series Parallel Reactions– A+BC+D– A+CE(desired)
• Mixing, Temperature and Pressure Effects
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Rate Selectivity
• Parallel Reactions– A+BR (desired)– A+BS
• Rate Selectivity
• (αD- αU) >1 make CA as large as possible• (βD –βU)>1 make CB as large as possible
• (kD/kU)= (koD/koU)exp[-(EA-D-EA-U)/(RT)]– EA-D > EA-U T– EA-D < EA-U T
)()(A
U
Drr
D/UD
U
D Ck
kS UDU
BC
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Reactor Design to Maximize Desired Product
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Maximize Desired Product
• Series Reactions– AB(desired)CD
• Plug Flow Reactor• Optimum Time in Reactor
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Real Reaction Systems
• More complicated than either – Series Reactions– Parallel Reactions
• Effects of equilibrium must be considered
• Confounding heat effects
• All have Reactor Design Implications
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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
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Attainable Region
S,S&L Chapt. 7
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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
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ProcedureStep 1: Construct a trajectory for a PFR from the feed point,
continuing to complete conversion or chemical equilibriumStep 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.
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Example
BBAB
ABAA
k
kk
k
CkCkCkdt
dC
CkCkCkdt
dC
DA
CdesiredBA
321
2421
4
31
2
2
)(
Reactions
Rate Equations
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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
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PFR Design Equations
BBAB
ABAA
CkCkCkdt
dC
CkCkCkdt
dC
321
2421
x
AA r
dxFV
0
0
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Solve NumericallyRunge-Kutta
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Solve Numerically
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Step 2Plot 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. Tangent line passing (1,0))
Des
ired
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Step 3
Expand the AR as much as possible with straight arcs that represent mixing of reactor effluent and feed stream
PFR
(1-)
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Interpreting points on mixing line
Larger Attainable Region
PFR CA=0. 2187CB=0.00011 CA=0.72
CB=0.00004
CA=1CB=0
(1-)PFR
CA=1CB=0
(1-)
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Mixing of StreamsReactant Bypass
21 )1( ccc Vector Equation, i component is CA, j component is CB
α =fraction of mixture of stream 1in the mixed stream
)1(00011.0000004.0
)1(2187.0172.0
B
A
C
C
Feed mixing fraction: = 0. 64
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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
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CSTR Design Equations
)(
)(
321
2421
BBAB
ABAAAo
CkCkCkC
CkCkCkCC
A
A
r
xFV
0
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Solve numerically at various until complete conversion or
equilibrium is achieved
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CSTR Extends Attainable Region
CSTR
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Plot extends attainable regioni.c. for step 5
CSTR
CA=1CB=0
(1-)
EnlargesAttainableRegion
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Possible Configuration at this point
CSTR
CA=1CB=0
βPFR
β = 0
β = 0
1-α-β
0.38
β = 1α = 0
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Profit ($) = 15000*CB-15*CA2
Attainable Region
-0.00001
0.00001
0.00003
0.00005
0.00007
0.00009
0.00011
0.00013
0.00015
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 1.05
C a, kmol / m3
CSTR
PFR
PFR2
$=0.9
$=2
$=1.5
Optimal point not at highest selectivity
PFRCSTR
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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
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Separation Trains
S, S&L Chapt. 8
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Separation Methods
• Absorption
• Stripping
• Distillation
• Membrane Separations
• Crystallization
• etc
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Use of Separation Units
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Column Sequences
• No. of Columns– Nc=P-1
• P= No. of Products
• No. of Possible Column Sequences– Ns=[2(P-1)]!/[P!(P-1)!]
• P= No. of Products
– P=3, Nc=2, Ns=2– P=4, Nc=3, Ns=5 – P=5, Nc=4, Ns=14– P=6, Nc=5, Ns=42– P=7, Nc=6, Ns=132
No. of Possible Column Sequences Blows up!
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How do I evaluate which is best sequence?
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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
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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 Rmin – V=D (R+1)
• V= Vapor Flow Rate• D= Distillate Flow Rate• R=Recycle Ratio
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Azeotrope Conditions
• Conditions on the Activity Coefficient
• Minimum Boiling, γjL> 1
• Maximum Boiling, γjL< 1
• xj=yj, j=,1,2,…C
Law sRaoult' from Deviations Negative,...2,1,1
Law sRaoult' from Deviations Positive,...2,1,1
)1( 221111
Cj
Cj
PxPxP
Lj
Lj
sLsLT
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Raoult’s Law
satii PxP
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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
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Multi-component Azeotropes
• Residue Curve Map– dxj /dť = dxj /d ln(L) = xj – yj
• Integrate from various starting points
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Defining Conditions for Multi-component Azeotrope
t goes from 0 to 1, ideal to non-ideal to find Azeotrope
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Distillation
• XB, XF and YD form a line for a Distillation Column
• Line can not cross Azeotrope line
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Ethanol/Water Distillation with
BenzeneTo Break Azeotrope
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Pressure Swing to Break Azeotrope
Temp. of Azeotropevs. Pressure
Mole Fraction of Azeotrope
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Reactor-Separation Train-Recycle
Chapt. 7&8
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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
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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
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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
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Compare Recycle Concepts
• Costs
• Benefits
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Feedback effects of Recycle Loop
• Small disturbance on feed
• Large effect on recycle flow rate/composition
• Snowball effect on reactor/separator
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Heat Integration
Chapter 9
Terry Ring
University of Utah
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Costs
• Heat Exchanger Purchase Cost– CP=K(Area)0.6
• Annual Cost– CA=im[ΣCp,i+ ΣCP,A,j]+sFs+(cw)Fcw
• im=return on investment• Fs= Annual Flow of Steam,
– $5.5/ston to $12.1/ston = s
• Fcw=Annual Flow of Cold Water– $0.013/ston = cw
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Lost Work = Lost Money
• Transfer Heat from T1 to T2
• ΔT approach Temp. for Heat Exchanger
• To= Temperature of Environment
• Use 1st and 2nd laws of Thermodynamics
• LW=QToΔT/(T1T2)– ΔT=T1-T2
– To= Environment Temperature
• Q= UAΔTlm
T1
T2
Q
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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
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Terms
• HEN=Heat Exchanger Network
• MER=Maximum Energy Recovery
• Minimum Number of Heat Exchangers
• Threshold Approach Temperature
• Optimum Approach Temperature
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Process
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Minimize UtilitiesFor 4 Streams
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Adjust Hot Stream Temperatures to Give ΔTmin
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Enthalpy Differences for Temperature Intervals
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Pinch Analysis
Minimum Utilities
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Pinch Analysis
ΔTapp
MER values
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How to combine hot with cold?
• At Pinch (temp touching pinch)– Above Pinch Connect
• Cc≥Ch
– Below Pinch Connect• Ch≥Cc
• Not touching Pinch temp.– No requirement for Cc or Ch
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4 Heat ExchangerHEN for Min. Utilities
Cc≥Ch
Ch≥Cc
MER Values
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Stream Splitting
• Two streams created from one
• one heat exchanger on each piece of split stream with couplings
1
1a
1b
1b
1a
1
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Optimization of HEN
• How does approach delta T (ΔTmin) effect the total cost of HEN?
• Q= UA ΔT
• LW=QToΔT/(T1T2)
– More Utility cost
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Costs
• Heat Exchanger Purchase Cost– CP=K(Area)0.6
• Annual Cost– CA=im[ΣCp,i+ ΣCP,A,j]+sFs+(cw)Fcw
• im=return on investment• Fs= Annual Flow of Steam,
– $5.5/ston to $12.1/ston
• Fcw=Annual Flow of Cold Water– $0.013/ston
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Change ΔTmin
CP=K(Area)0.6
Area=Q/(UF ΔTmin)
More Lost Work
LW=QToΔT/(T1T2)
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Optimization of Process Flowsheets
Chapter 24
Terry A. Ring
CHEN 5353
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Degrees of Freedom
• Over Specified Problem – Fitting Data– Nvariables>>Nequations
• Equally Specified Problem – Units in Flow sheet– Nvariables=Nequations
• Under Specified Problem– Optimization– Nvariables<<Nequations
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Optimization
• Number of Decision Variables– ND=Nvariables-Nequations
• Objective Function is optimized with respect to ND
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 Rmin
– Upper and Lower Bounds• Mole fraction is larger than zero and smaller than 1
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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)
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Optimization
• Feasible Region– Unconstrained Optimization
• No constraints– Uni-modal– Multi-modal
– Constrained Optimization• Constraints
– Slack– Binding
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v
v
v
N
i ii=1
i V
N
ij j i Ej=1
N
ij j i Ij=1
MinimizeJ x f xd
Subject to (s.t.) x 0,i 1, ,N
a x b,i 1, ,N
c x d,i 1, ,N
LINEAR PROGRAMING (LP)
equality constraints
inequality constraints
objective function
w.r.t. design variables The ND design variables, d, are adjusted to minimize f{x} while satisfying the constraints
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EXAMPLE LP – GRAPHICAL SOLUTION
A refinery uses two crude oils, with yields as below.
Volumetric Yields Max. Production
Crude #1 Crude #2 (bbl/day)
Gasoline 70 31 6,000
Kerosene 6 9 2,400
Fuel Oil 24 60 12,000The 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?
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EXAMPLE LP –SOLUTION (Cont’d)Step 1. Identify the variables. Let x1 and x2 be the daily production rates of Crude #1 and Crude #2.
Step 2. Select objective function. We need to maximizemaximize profit: 1 2J x 2.00x 1.40x
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 (ND = NV – NE). Here NE = 0.
1 2
1 2
1 2
0.70x 0.31x 6,000
0.06x 0.09x 2,400
0.24x 0.60x 12,000
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EXAMPLE LP –SOLUTION (Cont’d)
Step 5. Compute optimum. a) Inequality constraints define feasible space.
1 20.70x 0.31x 6,000
1 20.06x 0.09x 2,400
1 20.24x 0.60x 12,000Feasible
Space
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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
x1 = 0, x2 = 19,355 bbl/day
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EXAMPLE LP – GRAPHICAL SOLUTION
A refinery uses two crude oils, with yields as below.
Volumetric Yields Max. Production
Crude #1 Crude #2 (bbl/day)
Gasoline 70 31 6,000
Kerosene 6 9 2,400
Fuel Oil 24 60 12,000The 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?
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Dealing with Impurities in Processes and Process Simulators
ChEN 5253 Design IITerry A. Ring
There is not chapter in the book on this subject
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Impurity Effects
• Heat Exchange
• Reactors
• Separation Systems
• Recycle Loops
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Impurities in Heat Exchange
• Impurities effect heat capacity– Lower Cp
• Various options
– Raise Cp
• Increase H2
• Impurities effect the enthalpy of stream– Total heat of condensation is less due to
impurity– Total heat of vaporization is less due to
impurity
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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
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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.
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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?
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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
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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
• Product Flow Control or Feed Flow Control
• Types of control– Single loop PID– Gain Scheduling– Ratio control– Cascade Control– Multi-variable control– Model Based control
(MPC)– Override control
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Distillation Control
• Types of Control– LV control– DV– LB– DB– (L/D)(V/B)– (L/F)(V/F)
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The End