chemical engineering plant design lecture 01 course
TRANSCRIPT
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Chemical Engineering Plant Design
Lecture 01 Course Introduction & Flowsheets
Instructor: David Courtemanche
CE 408
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Course Introduction
• Welcome to CE 408 Chemical Engineering Process Design
• This is a Capstone Course
• The course will deal with designing a chemical process and will draw from all of your chemical engineering
education
• We will be incorporating material from your entire curriculum• CE 212 Mass and energy balances
• CE 304 Thermodynamics
• CE 329 Reactors and kinetics
• CE 317 Fluid mechanics
• Particularly flow in pipes
• CE 318 Heat and Mass transfer
• CE 407 Separations
• We will be learning some new things, as well• Process simulation
• Evaporation
• Vacuum systems
• Project economics
• Heat exchanger network Pinch Analysis
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Course Introduction
• A chemical Process Design Project is like an onion
• https://www.youtube.com/watch?v=aJQmVZSAqlc
• Yes, at times it will stink and it may make you cry…
• You start with identifying a business proposition
• Need to identify the market and customer requirements
• Identify required product specifications
• CE 404
• You identify ideas for what general process you will pursue
• Sketch out basic process steps
• Block flow diagrams
• Basic mass balances
• Start to identify specifications of required equipment
• Process Flow Diagrams
• Get general estimates of size and cost
• What are energy requirements?
• Begin economic evaluations
• Capital Costs / Raw material cost / energy costs / labor and overhead
• Revenue
• Is this a good investment?
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Course Introduction
• A chemical Process Design Project is like an onion, continued
• This is an iterative process
• We will iterate in that as we get further into the details we’ll see original assumptions that
need to be changed
• We won’t see all of the gate keeping steps
• Design costs money – you don’t go ahead and just drive to a final design
• Evaluate the project economics with a plus/minus accuracy of perhaps 50%
• Use factors to account for all of the supporting equipment, installation costs, etc
• Do we pursue or not?
• More detailed design allows an estimate of perhaps plus/minus accuracy of 10%
• Safety and environmental considerations need to be considered by this point
• Higher detail allows better cost estimates of larger pieces of equipment
• Higher detail allows actual estimates of supporting equipment (you know what it is to a
better degree)
• Do we pursue or not?
• Final design allows an estimate of perhaps plus/minus accuracy of 5%
• Getting to the point where could actually build this plant with the level of design we have
• Decide to build it or not
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Course Introduction
Course website
• http://wwwcourses.sens.buffalo.edu/ce408/
• Username: ce408
• Password: ripE408
Mainpage tab
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Course Introduction
• Syllabus tab
• Tentative list of lectures, with textbook reference and links to additional material
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Course Introduction
• Homework Tab
• Homework will be submitted via assignments in UBLearns
• Each homework problem will be a separate assignment in UBLearns
• This facilitates efficient grading turnaround
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Course Introduction
Linear Algebra for Mass Balances with recycle
• We have the following process
• A (n-octane) is being reacted to B (i-octane)
• Feed is 120 mol/min of pure A
• Reactor yields 40% conversion for reaction 𝐴 → 𝐵• Distillation produces top and bottom products that are 95% pure
• B is lighter (more volatile) than A
• We need to determine all of the quantities indicated by ? in the figure
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Course Introduction
• Let’s unpack this…
• We have 8 unknowns, so we need to find 8 equations…
• Inlet to reactor I
• Component A 120 𝑚𝑜𝑙 + 𝑛𝐵 𝐴 = 𝑛𝐼 𝐴 Note that subscript outside parentheses is component
• Component B 0 𝑚𝑜𝑙 + 𝑛𝐵 𝐵 = 𝑛𝐼 𝐵 Note that subscript inside parentheses is location
• Outlet from Reactor P
• Component A 𝑛𝐼 𝐴 − 𝑛𝑃 𝐴 = 0.4 ∗ 𝑛𝐼 𝐴• 40% of A entering reactor ( 𝑛𝐼 𝐴)is converted
• Component B 𝑛𝑃 𝐵 − 𝑛𝐼 𝐵 = 0.4 ∗ 𝑛𝐼 𝐴• There is one mole of B created for each mole of A converted
• Distillation Column – Distillate D and Bottoms B
• Mass Balance
• Component A 𝑛𝑃 𝐴 = 𝑛𝐷 𝐴 + 𝑛𝐵 𝐴
• Component B 𝑛𝑃 𝐵 = 𝑛𝐷 𝐵 + 𝑛𝐵 𝐵
• 95% Purity
• Distillate is 95% component B 𝑛𝐷 𝐵 = 19 ∗ 𝑛𝐷 𝐴 (95% = 19 * 5%)
• Bottom is 95% component A 𝑛𝐵 𝐴 = 19 ∗ 𝑛𝐵 𝐵 (95% = 19 * 5%)
• Well, that looks easy to solve...
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Course Introduction
• It’s not so bad when using Linear Algebra
• Rearrange the equations to have variables on the left hand side and a constant on the right hand side
• 𝑛𝐼 𝐴 − 𝑛𝐵 𝐴 = 120
• 𝑛𝐼 𝐵 − 𝑛𝐵 𝐵 = 0
• −0.6 ∗ 𝑛𝐼 𝐴 + 𝑛𝑃 𝐴 = 0
• −0.4 ∗ 𝑛𝐼 𝐴 − 𝑛𝐼 𝐵 + 𝑛𝑃 𝐵= 0
• − 𝑛𝑃 𝐴 + 𝑛𝐷 𝐴 + 𝑛𝐵 𝐴 = 0
• − 𝑛𝑃 𝐵 + 𝑛𝐷 𝐵+ 𝑛𝐵 𝐵= 0
• −19 ∗ 𝑛𝐷 𝐴 + 𝑛𝐷 𝐵= 0
• 𝑛𝐵 𝐴 − 19 ∗ 𝑛𝐵 𝐵 = 0
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Course Introduction
• The system of linear equations can be solved by many methods
• Matlab, etc
• Excel
• Excel method
• Fill in the zeros
• This generates the Matrix A in cells A2 to H9
• The array B is in cells J2 to J9
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Course Introduction
• First we need to generate the inverse Matrix A-1
• Select a range of cells of equal range as Matrix A (in this case 8 x 8)
• Type in the expression box the following:
• =MINVERSE (A2:H9) • A2:H9 is the range of our original matrix A – you can type in =MINVERSE( and then select the
range of the Matrix A to do this or manually type in the range
• Simultaneously hit Shift/Control/Enter
• The shaded area below is the inverse Matrix A-1
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Course Introduction
• Next we generate the product of A-1 x B
• Select a range of cells that is the same size as array B
• Type in the expression box the following:
• =MMULT (A11:H18, J2:J9)
• A11:H18 is the range of the inverse matrix, A-1, amd J2:J9 is range of B
• Simultaneously hit Shift/Control/Enter
• The shaded area below is the product of the inverse Matrix A-1 and the array B
• This is the answer to our equations: