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EP426Chemical Process Design and Optimization
Chapter 4
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Teaching plan (Wk8 to Wk14)
8 Chapter 4Chemical Process Optimization.
Optimization overview.22/02/2016
Chapter 4Chemical Process Optimization.
Optimization application on Chemical processes.24/02/2016
9 Chapter 4Chemical Process Optimization.
Optimization application on Chemical processes.29/02/2016
Chapter 4Chemical Process Optimization.
Optimization classification and the approach (Part I)02/03/2016
10Individual Assesement (5%)
Presentation based on the group assignment07/03/2016
Chapter 4Chemical Process Optimization.
Optimization classification and the approach (Part II) 09/03/2016
11 Chapter 5Heat & Energy Integration.
Overview of process integration and the applicaton14/03/2016
Chapter 5Heat & Energy Integration.
HENs analysis (Part I) - Composite Curves and Problem16/03/2016
12 Test 1 (10%) 21/03/2016
Chapter 5 Heat & Energy Integration.HENs analysis (Part II) - Area & Unit targeting
23/03/2016
13 Chapter 5Heat & Energy Integration.
HENs analysis (Part III) - Pinch design28/03/2016
Chapter 5Heat & Energy Integration.
HENs analysis (Part IV) - Maximum Recovery design.30/03/2016
14
Revision and Tutorial
Group Report Submission (10%)
04/04/2016
Due: 5:00 PM
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Student attainment
CLO4: Determine optimal solution for a chemicalprocess using Linear Programming.
Note:
Teaching method - Lecture & Group ProjectAssessment - Test, Final Exam and report presentation.
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Chapter 4: Topics
1. Optimization overview.
2. Optimization application on Chemical processes.
3. Basic elements in the optimization; ObjectiveFunction, Parameters, and Constrains.
4. Optimization classification and the approach ofLinear Programming method.
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EP426Chemical Process Design and Optimization
Chapter 4a - Chemical Process Optimization.
OVERVIEW
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Objective
1. To differentiate the type of optimisation
2. To formulate and solve a linear program (LP) byusing
• Graphical Method
• Simplex Method
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Revision
= 2 − 12 + 20
Q. Find the minimum value of function f(x) defined by the
equation
Solution
=
2 − 12 + 5
We need to find x = ? for min value of f(x) , when df/dx = 0
= 4 − 12
0 = 4 − 12
=12
4
= 3
Thus, the minimum value is f(3)
3 = 2 3 − 12 3 + 20
3 = 2
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Graphical Solution (Alternative)
x f(X)
0 20
0.5 14.5
1 10
1.5 6.5
2 4
2.5 2.5
3 2
Step 1: Prepare dataset of x and f(x)
= 2 − 12 + 20
Step 2: Plot f(x) vs x
From the graph,
the minimum
value is f(3) = 2
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Exercise 1
Given = − 3 + 0.6
Determine the optimal solution of and the corresponding
value of , analytically and graphically.
Answer: 1.5 = -1.65
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Summary
• F(x) is the objective of the case study
•
X is the design variable
• The case study can be constraint or un-constrainscenario.
= − 3 + 0.6
∈
0 ≤ ≤ 1
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Standard Optimisation Format
Min : = − 3 + 0.6
Subject to : ∈
0 ≤ ≤ 1
(equality constraint)
(bound constraint)
(objective function)
(constraint)
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OBJECTIVE FUNCTION
Candidates for the measure of goodness of adesign, f(x), where x is a design variable(s) thatapproximate profitability measures:
Example of objective function
• ROI – Return of Investment (max)• VP – Venture Profit (max)
• PBP – Payback period (min)
• CA - Annualized Cost (min)
or more rigorous measuresNPV – Net present value (max)
IRR – Investors rate of return (max)
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CONSTRAINTS
•
In process simulators, most of the equality constraints,c{x} = 0, are the model equations relating to M&Ebalances.
• These are not stated explicitly, but are invoked as each unitoperation is installed on the flowsheet.
• Some equality constraints are due to performancespecifications
e.g., 95% recovery of species i in the distillate flow:
D
xi
B
F
zi
xiD - 0.95ziF = 0
For in-equality:More/less than 95% recovery of species i in
the distillate flow
Thus,
xiD - 0.95ziF≥
0 xiD - 0.95ziF≤
0or
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v
v
v
N
i ii=1
i V
N
ij j i E j=1
N
ij j i I j=1
Minimize J 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
design variablesThe ND design variables, d, are
adjusted to minimize f{x} while
satisfying the constraints
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Example 1 : Two-dimensionalobjective function
Determine the maximum and the corresponding value of x1and x2 for this function GRAPHICALLY.
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+ 2 ≤ 6
+ ≤ 3
≥ 0
≥ 0
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= − ; = 1
= − ; = 0
= − ; = 2
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= − ; = 5
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Exercise 2 : Two-dimensionalobjective function
Determine the maximum and the corresponding value of xand y for this function GRAPHICALLY.
+ 0.2 ≤ 6
+ 0.1
− ≤ 3
, ≥ 0
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Homework
A batch distillation facility has a bank of columns of Type1 and another bank of Type 2. Type 1 columns are
available for processing 6,000 hr/week, while Type 2
columns are available 10,000 hr/week. It is desired to use
these columns to manufacture two different products, A
and B. Distillation time to produce 100 gal of product A is2 hr in Type 1 columns and 1 hr in Type 2 columns.
Distillation time to produce 100 gal of product B is 1 hr in
Type 1 columns and 4 hr in Type 2 columns. The net
profit is $5.00 per gal for product A and $ per gal for product B.
Determine the production plan to maximizes the net profit
in $ per week.
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Type of Optimisation
1. Parametric optimization focuses on adjustingoperating (decision) variables in order to improve theobjective function.
Example:
• Adjusting the T and P at which a reactor operates.
• Adjusting the surface area of a heat exchanger.• Number of trays for a distillation column.
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Type of Optimisation
2. Topological optimization focuses on adjusting thelayout or topology of the flowsheet in order toimprove the objective function.
Example:
• Changing the order in which a separation sequenceis implemented.
• Looking at the effect of adding a heat recoveryexchanger.
• Changing a utility (CW to refrigerated fluid).
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To be continuedLinear Programming Method (Application)