chapter 4a - chemical process optimization (overview)

8/18/2019 Chapter 4a - Chemical Process Optimization (Overview)

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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


Optimization classification and the approach (Part I)02/03/2016

10Individual Assesement (5%)

Presentation based on the group assignment07/03/2016


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)

chapter 4a - chemical process optimization (overview)

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