mb1 introduction to mass balancing
TRANSCRIPT
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Mass Balance: Introduction
MB1
Chemical Engineering
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Content
• Section 1
– Theory “Basics of Process/Chemical Engineering”
– System, Process, Unit Operation, Streams, Flows, Variables, Flow Diagrams, MB
• Section 2
– MB basics
– Theory + Problems + Exercises
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Section 1
• Some basic theory in engineering and process engineering
• Easy but Important to understand!
• Learn by hard!
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Theory: Basic Definitions
• Process vs. Chemical Process• Unit Operation• Stream and Flows• Process Variables• Flow Diagram• Block Diagram• System
– Open System– Closed System– Isolated System
• Mass Balance
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Process
• A series of actions or steps taken in order to achieve a particular end.
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Process
• A series of actions or steps taken in order to achieve a particular end.
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Chemical Process
• …chemical process is a method or means of somehow changing one or more chemicals or chemical compounds. It can occur by itself or be caused by an outside force, and involves a chemical reaction of some sort…
From almighty Wikipedia
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Chemical Process Example
Chemical Reaction!
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Types of Diagrams
• Block Diagram
• Flow Diagram
• P&ID Diagram
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Block Diagram
• Simple diagram that “shows at a glance” the process
• Most used for MB solvingShow:• Flows• Unit Operations• Some extra data
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Flow Diagram
• Recommended for “general” information
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P&ID• Pipe and Instrumentation Diagram
• Formal “language” for diagram
• Type of valves• Piping information• Op. Unit detail• Automation units
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Unit Operations
• Basic Process Step in chemical engineering
• Unit operations involve:
– physical change or chemical transformation
• Examples:
– separation, crystallization, evaporation, filtration, polymerization, isomerization, and other reactions
From almighty Wikipedia!
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Unit Operations
• Reactors
• Condensers
• Evaporators
• H-Exchangers
• Crystallizers
• Filters
• Cyclone
• Separators
• Mixers
• Distillation C.
• Absorption Tw.
• Adsorption Tw.
• Pumps
• Compressors
• Storage Tank
Mass operation
Heat operation Momentum operation
Reaction engineering
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Unit Operations
Reactor
Diagram Picture Real-Life Picture
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Unit Operations
Distillation Column
Diagram Picture Real-Life Picture
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Unit Operations
Heat Exchanger
Diagram Picture Real-Life Picture
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Unit Operations
Centrifugal Pump
Diagram Picture Real-Life Picture
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Process Variables
• A process variable, process value or processparameter is the current status of a processunder control.
• Example:
– Temperature of a furnace, process variable.
– Desired temperature is known as the set-point.
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Process Variables
• Typical P.V.:– Pressure
– Volume
– Temperature
– Density
– Height (Level)
– Concentration
– Flow
– % Conversion
TIP: Get to know how these PV are measured in engineering
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Flow (mass, mole, volume)
• Mass flow [kg/s] = mass flow per unit time
• Mole flow [mol/s] = mole flow per unit time
• Volume flow [m3/s] = volume flow per unit time
• How much quantity of (mass/mole/volume) is “flowing” per unit time
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Flow (mass, mole, volume)
• Mass flow examples:– 10 kg/min
– 1 lb/s
– 350 kTon/year
– 3 mg per day
• Mole Flow examples:– 1 mol H2O per min
– 3.02 kmol/h
– 125 lbmol/s
• Volume Flow examples:
– 10 m3/s
– 1 liter per minute
– 4.5 ml/day
– 18,000 gal/year
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Flow Example
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System State
• Steady State
– Typical for continuous processes
• Transient State
– aka unsteady-state/non-steady state
– Typical for batch, semibatch processes
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System State
• Steady State: when all values of all variables in a process do not change in time
Operation of plant: 365 days, 24h
After one year, concentrations, flows and other P.V. Should be the same value
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Transient State
• Transient State: When any of the variables change with time
Example:
10 liters of a mixture is fed to the potAfter 5 hours, the mixture left in the pot is about 2 liters.
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Other Process Classification
• Batch process: Process is being fed once, then time elapses and then the process products are removed.
• Continuous process: Input and Outputs flow/react continuously throughout the duration of the process
• Semibatch process: Combination of Batch-Continuous (charging a reactor)
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Types of Systems
• Open System: inlet/outlet of materials to system
• Closed System: no inlet/outlet of materials
• Isolated System: no inlet/outlet of materials; no inlet/outlet of energy to the system
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Types of Systems
• Open • Closed • Isolated
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Type of Systems
• Isobaric system: Constant Pressure
• Isothermal system: Constant temperature
• Isochoric system: Constant Volume
• Adiabatic system: No heat exchange Q=0
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End of section 1
• Please make sure you understood all material before advancing to the next section
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Section 2
• Now lets actually study Mass Balancing!
• Theory about mass balance
• Exercises about mass balance
– Mastering MB implies lot of exercise
• BE PATIENT!
• Read, re-read, calculate, recalculate!
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Section 2
• Mass Balance (Principle and Equation)
• MB in Steady, Unsteady and Semibatch
– Diagram Construction
– Scale-up and Basis of Calculation
– Variable-Equation counting (Solve?)
– General Methodology
• MB in Steady state: 1 unit, no reaction
• Recycle + Bypass
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Section 2 (cont.)
• MB with 1 chemical reaction– Chemistry review: Stoichiometry, limiting reactant,
excess reactant, equation balancing, conversion, selectivity, yield and extent of reaction ε.
• MB in Chemical Equilibrium (1 reaction)• MB with 2+ chemical reactions• MB of Atomic and Molecular Species• Purge• MB in Combustion processes
– Excess air, theoretical oxygen, air composition, etc.
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Mass Balance Principle
• Mass can neither be created nor destroyed, only transformed
• Energy-Mass relationship: E=mc2
– Not used in this course; no nuclear reactions!
• We will not analyze ENERGY in this course
– Energy+Mass balances are analyzed in the Energy Balance course
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Mass Balance Principle
System(reactions, mixing, and
other operations may occur here)
Inlet
Inlet
Outlet
Outlet
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Mass Balance Principle
System(reactions, mixing, and
other operations may occur here)
Inlet
Inlet
Outlet
Outlet
MB Analysis
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Mass Balance Principle
System(reactions, mixing, and
other operations may occur here)
Inlet
Inlet
Outlet
Outlet
MB Analysis
Inlet –Outlet + Prod. – Consum. = Accumulation in system
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City example
System(reactions, mixing,
and other operations may
occur here)
Inlet Outlet
MB Analysis Immigration = +50
Emmigration = -3Born = +100Death = -25
What is the “accumulation” in the city?
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City example
System(reactions, mixing,
and other operations may
occur here)
Inlet Outlet
MB Analysis Immigration = +50
Emmigration = -3Born = +100Death = -25
What is the “accumulation” in the city? = 50-3+100-25 = 122
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Reactor Example of A
System(reactions, mixing,
and other operations may
occur here)
Inlet Outlet
MB Analysis MB in system
Inlet of A = 100 kgProduction = 0 kgConsumption = 50 kgOutlet = 25 kg
What is the “accumulation” in the system? = 100+0-50-25 = 25 kg accumulated in system
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The Mass Balance Equation
• General Mass Balance Equation– Inlet – Outlet + Production – Consumption = Accumulation
• It can also be done on a specie:
• Mass balance of species “i” – Inleti – Outleti + Productioni – Consumptionii = Accumulationi
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Types of Mass Balances
• Differential
• Integral
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Differential MB
• Analysis in an “instant” of time
• Use of velocity/rate: g/s or kg/h, barrel/day
• Typical for continuous processes
• TIP: convenient to use time basis (1h, 1 year)
• We WILL use this a lot in the course 80%-90%
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Integral MB
• It occurs between two instants
• We use quantities rather than velocities (kg, ton, lb)
• Time elapsed: Tf-Ti
• Typical for batch processes
• Feed, reaction, discharge; repeat
• We ONLY use this in transient process about 10-20% of the course
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Types of process
• Steady State (Continuous)
• Transient (Batch)
• Transient-Steady (Semi-batch)
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MB: Steady State
• NO accumulation term (no differential equations XD)
• Inlet – Outlet + Production – Consumption = 0
• ExamplesIn-Out +P – C = 0In + P = Out + C
Reactor
Distilation Column
Mixer
20 kg/h
80 kg/h
100 kg/h
100 molTotal
20 mol C80 mol mol A
1 mol A
1 mol B
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MB Steady StateExercise 1
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If you check all MB possibles, you get the same answer and YOU SHOULD
MB Steady StateExercise 1
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MB: Batch
• No Inlet/Outlet, closed system
• Production, Consumption and Accumulation
Feeding Production/reactionDischarge
In-Out + P – C = AccumP – C = Accum.
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MB BatchExercise 1
NOTE: We do not need the MB of Water (WHY??)MB of water proves 1=1 or 0=0 WHY??
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MB: Semibatch
• More complex, include continuous and batch
• Accumulation always is included
• In, Out, P, C may vary
• See next example
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MB SemibatchExample 1
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MB SemibatchExample 1
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MB SemibatchExample 1
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Diagram Flow Construction
• Look for information Organize it and Draw!
• Unit operation are blocks or “box”
• Flows are “arrows”
• No matter size/direction of arrows
F = 100 kg/h
0.5 Water w/w0.2 H2SO4 w/w0.3 NaCl w/w
T= 65ºCP = 1 atm
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Diagram Flow Construction
1. Tag each equipment and arrow
2. Assign variables to unknown data/flow/comp.
3. Avoid excess of variables
– “C” is ½ of the inlet “F” ½·F = C
– E is equal to the flow of the distillate “D” F = D
4. Use same units to avoid error/conversions
F = [] kg/h
60 kmol N2 / min40 kmol O2 / min
Airflow ?
60 kmol N2 / min40 kmol O2 / min
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Diagram Flow Construction Example
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MB Steady State Exercises
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MB Steady State Exercises
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MB Steady State Exercises
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Scale-up & Basis of Calculation
• Review the next Diagrams…
• There is a “Basis of Calculation” for each.
– This basis is a number assigned to simplify the problem
B flow = T flow
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Scale-up & Basis of CalculationWe could assign 2 kg o of P as “Basis”… solving for B, T you get half eachORWe could assign 1kg to B or T… solving will give you twice P…
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Scale-up & Basis of Calculation
If we wanted 20 kg of P…• Just multiply by 20 the flows• Concentrations stay the same
x10
We could assign 2 kg o of P as “Basis”… solving for B, T you get half eachORWe could assign 1kg to B or T… solving will give you twice P…
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Scale-up & Basis of Calculation
Multiply Units per
time
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Scale-up & Basis of Calculation
By time (x60 min hr)
By mass(x2.2 lb kg)
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Scale-up & Basis of Calculation
• Conclusion all processes are balanced!
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Which Process is Balanced?
• NO, MB don’t match
• Yes
• Impossible to know (reaction?)
• Not enough data!
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Scale-up & Basis of Calculation Exercise 1
• The next process is to be Scaled up from 100 mol/h of Feed (F) to 1250 lbmol/h.
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Scale-up & Basis of Calculation Exercise 1
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MB: Non-reactive systems
• There is no reaction… therefore Production and Consumption = 0
• It is steady state so Accum = 0
In-Out + P – C = AccumIn-Out = 0Inlet = Outlet
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MB: Non-reactive systems
• For N substances… you can ONLY write N equations:
• If you write N+1 equations (1 balance for global system and N balances for substances)
– You will prove 1 = 1 or 0 = 0
3 Equations2 Unknown
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MB: Non-reactive systemsExercise 1
Example 4.3-3
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MB: Non-reactive systemsExercise 1
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MB: Non-reactive systemsExercise 1
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Is it Possible to solve?The problem shown: 10 lbmol of Feed (0.5 of A) enter and then is separated in two streams. D = 5 lbmol and C is not known. D has 20% of A in its composition and 40% of A is in C
All the variables in the problem (known and unknown)
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Is it Possible to solve?
Variables that are explicitly given
Equations from the MB (N=2)
Physical restrictions
Only MATH can STOP you!
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Variable-Equation counting
• Don’t just do problems! Analyze
• Many times you have NO sufficient data to solve the problem
m1
m2
m4
m3
100 kg/h
Impossible to solve for m2, m3, m4
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Variable-Equation counting
• Relating Variable-Equations:
1. MB: global, species … up to N equations
2. Energy Balance: (not included in this course)
3. Problem “text”:
• Product is ½ feed
• 30 kg of A in stream B
• L/B ratio is 0.55
• Final amount of F in C is the same as A
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Variable-Equation counting
4. Physical properties, laws and data
– Density (relates volume to mass, moles)
– Gas laws (ideal gas law, SKR, van der Waals)
– Phases between substances (x, y, P, T) see CH6
5. Physical restriction
– Compound fraction (1 = xa+xb+xc…xz)
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Systems that can be solved
• When unknown variables = independent equations
• The hard part is to find the independent equations
• Try to decrease the # of unknown variables
• Search more!
If #unknown > # independent equations
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Variable-Equation countingExercise
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General MB Problem Solving
1. Draw and label the block diagram
2. Choose a convenient Basis of calculation
3. Tag all variables in Diagram (Flow, conc., T, P)
4. Account all variables & equations (DOF)
5. If solvable, change all volume data to mass/mole
6. When data is given, change all to either m/n
7. Translate all text to equations
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General MB Problem Solving
8. Write Equation of MB (N equations fo N substances). Order from simple to solve to difficult.
9. Do math! Solve for all equations and variables
10. Be sure to scale-up the basis of calculus according to the problem statements
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MB: Non-reactive systemsExercise 1
• A distillation column is being fed with a stream of 45% Benzene and the balance Toluene. The Product “D” flow contains 95% of benzene. An 8% of the fed benzene is being produced in the bottoms “B”.
• If Feed “F” flow is 2000 kg/h, determine:
– A) Flow D
– B) Mass flow of Benzene and Toluene in Bottoms
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MB: Non-reactive systemsExercise 1
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MB: Non-reactive systemsExercise 1
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MB: Non-reactive systemsExercise 1
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MB: Non-reactive systemsExercise 1
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• Conclusion:
– Following the “methodology” helps!
– The difficult part is the assumption of equations
– MB are easy if data is known
MB: Non-reactive systemsExercise 1
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Need more Exercises & Problems?
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• Section: Courses
– Mass Balance Course
• Problems Section
• You will find a problem index there…
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MB in 2 Units
• In chemical processes, its normal to have 2+ Operation Units
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MB in 2 Units
• In chemical processes, its normal to have 2+ Operation Units
Global Balance:Inlet = Sweet gas + Acid gas
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MB in 2 Units
• We could create “ sub-systems” inside the Global System
Subsystems:• Distillation 1• Distillation 2• Reservoir 1• Reservoir 2
Each has a new MB (inlet = outlet)
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MB in 2 Units
• Subsystem = any portion of a chemical process
– Unit Operation
– Two Unit Operation
– 1 Point connecting flows (mixer)
– 1 Point dividing flows (splitter)
• Applying different sub-system we could find more independent equation to solve
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MB in 2 Units
• Many equations may be redundant/dependent
• Example of redundant equations:
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MB in 2 UnitsExercise 1
I labeled them as C1, C2 and P (respectively)
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MB in 2 UnitsExercise 1
WHY P? it’s the most simple mass balance… Recommended ALWAYS to start with the overall mass balance if possible
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MB in 2 UnitsExercise 1
Given P, we would like to calculate other flows (if not possible go directly to compositions!)
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MB in 2 UnitsExercise 1
Given all Flows (calculated) we only need to find all compositions
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MB in 2 UnitsExercise 1
Do Mass Balances in Unit 1, Unit 2… you may get answers from there!
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MB in 2 UnitsExercise 1
• Data Conclusion:– Start doing Overall Mass Balance– Try to get all the flows first– Calculate compositions by doing MB in units
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MB in 2+ Units
• Complexity increases
• MB: Global, unit, group of units
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MB in 2+ UnitsExercise 1
• Calculate all Flows (E1, E2, R, Et, V, B)
• Calculate all compositions
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MB in 2+ UnitsExercise 1
• From the previous problem, it is impossible to calculate all flows
• We need more data– Explicit data (composition values, flow falues)
– Relationship
– Ratios of flows
– % of mass of flows
– Many other data which could be written mathematically
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Need more problems?
• Go to www.ChemicalEngineeringGuy.com
• Section: Courses
– Mass Balance Course
• Problems Section
• You will find a problem index there…
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Recycle + Bypass
Recycle: product stream of a unit being returned to a previous stream/unit
Bypass: a fraction of the feed is diverted around the unit and combined with the output stream
BACKWARDS FORWARD
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Recycle
Identify Recycle
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Recycle
Identify Recycle
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Bypass
Identify Bypass
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Bypass
Identify Bypass
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Why Recycle
• For Chemical reactions to increase conversion(reaction to form product)
• Recover a catalyst ($)
• Dilute a process stream (increasing flow anddecreasing composition)
• Control of process variables (T, P, level, etc)
• Circulation of a working fluid (refrigeration
refrigerant; plant generation steam)
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Why Bypass
• Increase properties of product by adding raw material (inlet material)
• Operation Unit does not work as desired
• Control Process variables (T,P,level)
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Bypass/Recycle Exercise 1
1) Qw, Pc, Fe, Fc, R/F2) Conclusion if no Recycle
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Bypass/Recycle Exercise 1
• Draw Diagram… Label flows and write all comp.
1) Calculate Qw, Pc, Fe, Fc, R/F
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Bypass/Recycle Exercise 1
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Bypass/Recycle Exercise 1
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Bypass/Recycle Exercise 1
1) Calculate Qw, Pc, Fe, Fc,
2) R/F 5630/4500 = 1.25
If no recycle: you will loss all the crystals in solution!
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• Section: Courses
– Mass Balance Course
• Problems Section
• You will find a problem index there…
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Section 2 Break #1
What we´ve seen
What's left
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MB in Reactive systems
• We've seen so far MB in non-reactive systems:
– MB for 1 unit
– MB for 2 and 2+ unit
– MB with recycle and bypass
• Now we must Balance Problems involving REACTIONS
• We need to cover some theory first
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MB in Reactive systems
• Reaction basics kinetics
• Stoichiometry
• We will use now “Production – Consumption” term
• NO Accumulation in the system (continuous)
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Stoichiometry for MB reactive systems
• Stoichiometry Coefficient
• Stoichiometric Ratio (A/B)
• Balancing Equations
• Limiting reactant
• Excess reactant
• % Conversion
• Extent of reaction ε
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Stoichiometry Coefficient/Ratio
• Coefficient: “number” applied to a mole in a reaction
– For: C3H8 + 9/2·O2 3CO2 + 3H2O
– Coefficient of Ch3H8 = 1
– Coefficient of O2 = 9/2 or 4.5
– Coefficient of H2O = 3
– Coefficient of CO2 = 3
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Stoichiometry Coefficient/Ratio
• Ratio: relationship between two species (may be product:product; product:reactant or reactant:reactant)– For: C3H8 + 9/2·O2 3CO2 + 3H2O
– Ratio of A:B
– Ratio of A:C
– Ratio of B:A
– Ratio of C:C
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Stoichiometry Coefficient/Ratio
• Ratio: relationship between two species (may be product:product; product:reactant or reactant:reactant)– For: C3H8 + 9/2·O2 3CO2 + 3H2O
– Ratio of A:B = 1:9/2 or 0.22 mol A/molB
– Ratio of A:C = 1:3 or 0.33 mol A/ mol C
– Ratio of B:A= 9/2 : 1 or 4.5 mol B / mol A
– Ratio of C:C = 1:1 WHY?
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Stoichiometry: Balancing Equations
• From mass balance concept: mass is not created nor destroyed… Same applies for atoms
• Example: Propane
• C3H8+O2 CO2 +H2O
– Balance C: C3H8 + O2 3CO2 + H2O
– Balance H: C3H8 + O2 3CO2 + 3H2O
– Balance O: C3H8 + 9/2·O2 3CO2 + 3H2O
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Stoichiometry: Balance Equation
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Stoichiometry: Balance EquationPractice Problems
Balance the next equations to master “Combustion” Equations
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Stoichiometry Limiting Reactant
• If A + B C
• We need 1 mol of A and 1 mol of B to react
• If A is little less than 1 mol.. We wont be able to achieve 1 mol of C, B will not react completely
• A is said to be the limiting reactant (it limits the reaction; B could further react to form C)
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Stoichiometry Excess Reactant
• From A + B C
• B is in “excess” since there is still B able to react but no A
• If A + B + 2F + 2 H C
– There is only one limiting reactant
– There is 3 excess reactants
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Stoichiometry Limiting/Excess Exercise 1
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Stoichiometry Limiting/Excess Exercise 2
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Stoichiometry % Conversion
• If A + 2B C
• Not all A reacts…
• % of conversion : amount of reacted A vs feed A…
• XA denotes conversion of A … min = 0; max 1
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Stoichiometry % ConversionExercise 1
You actually don’t need B,C moles nor P
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Extent of reaction ε
• If A + B 2C
• This applies always:
• Ni = actual moles of species “i”
• Nio= initial moles of species “i” fed
• ε = extent of reaction
• βi = “reactant/product coefficient of species “i””• - for reactants
• + for products
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Extent of reaction ε
• Then at any moment…
• Ni = Nio + βi·ε
• “Actual moles of I is equal to, Moles fed of I +/- de coefficient of I in the reaction times the extent of reaction”
Hard to get… better go to examples
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Extent of reaction εExercise 1
Ni = Nio + βi·ε
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Extent of reaction εExercise 1
Substitute!
Given data…
From Equation “Stoichiometric Coefficients”
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Extent of reaction εExercise 2
a) Limiting reactantb) % excess if anyc) How much is left of the excess?d) If Xa = 0.50?
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Extent of reaction εExercise 2
• Limiting reactant
Ei < Ej … Then “i” limits reaction WHY?A : limiting reactantB : excess reactant
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Extent of reaction εExercise 2
• How much is left of the excess?
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Extent of reaction εExercise 2
• % excess if any
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Extent of reaction εExercise 2
• If Xa = 0.50?
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Extent of reaction εExercise 3
• The reaction A+B+3/2 C D +3E
• Theres is a 100 mol Feed
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Extent of reaction εExercise 3
• Change “air” composition to “Oxygen/Nitrogen”
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Extent of reaction εExercise 3
• We now choose a Basis and calculate all the moles involved in the flows
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Extent of reaction εExercise 3
• All the stoichiometric values from the equation (BALANCED)
• A+B+3/2 C D +3E
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Extent of reaction εExercise 3
• Substitute in the generic equation
Ni = Nio + βi·ε
βi= + product- reactant
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Extent of reaction εExercise 3
• Substitute initial moles (data is given)
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Extent of reaction εExercise 3
• A) Calculate the limiting reactant.
– Remember Ei < Ej
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Extent of reaction εExercise 3
• B) Calculate the % excess of the other 2 reactants (by now, you know its O2 and B)
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Extent of reaction εExercise 3
• C) What would be the mole flows if Xa = 30%
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MB in Chemical Equilibrium
• Chemical Equilibrium: reaction is possible also reversible
• In general, not all reactions are reversible
• If you are nterested in this type of reactions…– Watch:
• Thermodynamics course• Equilibrium Thermodynamics course• @ www.ChemicalEngineeringGuy.com
A + B CC A + B
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MB in Chemical Equilibrium
• We will use Equilibrium constants! @T = constant
• Imagine: CO + H2O <-> CO2 + H2• K eq= [Products]^y/[Reactants]^x = [CO2][H2] / [CO][H2O]
– [X] = typical nomenclature for “molar concentration”
• Lets use another concentration… – yA = mol of A / total moles
• The equilibrium concentration is now:– K eq = (yCO2·yH2)/(yCO2·yH2O)
• @T = 1105 K … K = 1.0
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MB in Chemical Equilibrium
• K eq: Equilibrium constant
• K eq @ T (it is Temperature dependent)
• Want to learn more about chemical equilibirum?
– Review Chemistry Classes in “Chemical Equiblirium”
– Review Equilibrium Termodynamics Course
– @ www.ChemicalEngineeringGuy.com
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MB in Chemical EquilibriumExercise 1
• A) Calculate the composition in equilibrium
• B) Calculate the fractional conversion of limiting reactant
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MB in Chemical EquilibriumExercise 1
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MB in Chemical EquilibriumExercise 1
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MB in Chemical EquilibriumExercise 1
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MB in Chemical EquilibriumExercise 1
• A)
• B) Fractional conversion of CO
– fCO = (1-yco) = 1- 0.333 = 0.667
2/3
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MB in systems with Multiple Reactions
• Theory
– Yield real vs. theoretical
– Selectivity desired vs. non-desired
– Multiple Reactions 2+ reactions at same place
• Some reactants will react to give our desired product(s)
• Some product/reactant may form side reactions to give non-desired product(s)
• This decreases %Conversion which means -$$
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MB in systems with Multiple
• Example:
– C2H6 C2H4 + H2 (1)
– C2H6 + H2 2·CH4 (2)
– C2H6+C2H6 C3H6 + CH4 (3)
• We would like to produce C2H4
• CH4, C3H6 are expensive to separate
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MB in systems with Multiple
• Example:
– C2H6 C2H4 + H2 (1)
– C2H6 + H2 2·CH4 (2)
– C2H6+C2H6 C3H6 + CH4 (3)
• We would like to produce C2H4
• CH4, C3H6 are expensive to separate
• Conclude why is it difficult to produce C2H6 only
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Yield
• Yield: Describes how our desired reaction performs
• Special attention in “limiting reactant had reacted completely”
• As Yield of A increases, more moles of A are being produced
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Selectivity
• Selectivity: how the reaction arrange to produce our desired component
• Generally as SA/B= selectivity of A over B
• As SA/B increases, more moles of A are being produced
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MB in systems with Multiple RXN
• Theory
– Yield real vs. theoretical
– Selectivity desired vs. non-desired
– Multiple Reactions 2+ reactions at same place
• We can use Extent of reaction (ε)
– We need one E.ofR (ε) for every reaction!
– ε1 and ε2 if there are two reactions
– βi1, βj1 ; βi2, βj2
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MB in systems with Multiple RXNExercise 1
• C4H4 + ½ O2 C2H4O (1)
• C2H4 + 3 O2 2CO2 + 2H2O (2)
• Express all the species with equation:• Ni = Nio + β1i·ε1 +β2i·ε2
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MB in systems with Multiple RXNExercise 1
• Express all the species with equation:
• Ni = Nio + β1i·ε1 +β2i·ε2
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Yield+Selectivity in Multiple ReactionsExercise 2
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Yield+Selectivity in Multiple ReactionsExercise 2
• We have ε1 and ε2 (RXN1, RXN2)
• β1i and β2i “i” for every species
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Yield+Selectivity in Multiple ReactionsExercise 2
• Just substitute values to Equation
– Ni = Nio + β1i·ε1 +β2i·ε2
From the Conversion given in the text… Xe = 50.1%
NR: Non-reacted molesR: Reacted moles
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Yield+Selectivity in Multiple ReactionsExercise 2
• Calculating moles of C2H6 @ 50.1%
• From the Yield Information Y = 0.471
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Yield+Selectivity in Multiple ReactionsExercise 2
• Finally, substitute and solve equations
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Yield+Selectivity in Multiple ReactionsExercise 2
• With ε1 and ε2
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Need more Exercises & Problems?
• Go to www.ChemicalEngineeringGuy.com
• Section: Courses
– Mass Balance Course
• Problems Section
• You will find a problem index there…
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Section 2 - Break
What we´ve seen
What's left
What we´ve seen
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MB atomic species vs. molecular
• We could either do the mass balance:
– Molecules (H2O, CO2, N2, CH3·OH)
– Atoms (H, C, N, C)
• Some times is more efficient to do atomic species MB
• In Molecular MB: N = molecules
• In Atomic spcies MB: N = atomic species
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MB atomic species vs. molecular
• Advantage of atoms:– Atoms don’t create nor destroy!
– No production, no consumption
– Inlet = Outlet (Since Accumulation = 0 )
• NOTE: take care in diatomic elements:– HONClBrIF
– H2, O2, N2, Cl2, Br2, I2, F2
– H2 ->Hydrogen gas or even just “Hydrogen”
– H -> Hydrogen Atom
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MB atomic species vs. molecularExercise 1
Ethane is going to be dehydrated to form ethene. The reactor is fed 100 mol of C2H6.
The outlet has 40 gmol of H2
A) Calculate n1, n2
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MB atomic species vs. molecularExercise 1
Do atomic Balances on each specie (C,H,O) N = 3
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MB atomic species vs. molecularExercise 1
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MB atomic species vs. molecularExercise 1
MB/Atomic done!
N1, N2 flows
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MB atomic species vs. molecularExercise 2
• 100 mol of Methane are being burned. There is a 90% conversion of the limiting reactant. Calculate all moles @ outlet
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MB atomic species vs. molecularExercise 2
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MB atomic species vs. molecularExercise 2
• Proceed to do Mass Balance of Atomic Species (C, H, and O)
• We could do balance in species C2 and O2 WHY?
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MB atomic species vs. molecularExercise 2
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Conversion: Global vs. Single-Pass
• Global conversion “Xa global” or “Xa process”
• Single-Pass Conversion “Xa in Unit”
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Conversion: Global vs. Single-Pass
• Global:
– Process Inlet
– Process Outlet
• Single-Pass
– Unit Inlet
– Unit Outlet
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Conversion: Global vs. Single-Pass
• It must be specified
• If not, suppose it is Operation Unit conversion
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Purge
• We are producing non-desired products in the system
• We are recycling some sort of stream with this non-desired product– We need to purge them…
– accum. of substance = 0
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Purge
• Solve this type of problems as before, no difference (it is actually a “Product” line)
• Purge:Feed ratio is a common data given
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Purge
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Purge: Exercise 1
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Purge: Exercise 1
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Purge: Exercise 1
• Make a MB in the Reactor (RKT)
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Purge: Exercise 1
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Purge: Exercise 1
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Purge: Exercise 1
• Keep doing math… the MB are almost done
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Purge: Exercise 1
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Purge: Exercise 1
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Purge: Exercise 1
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Need more Exercises & Problems?
• Go to www.ChemicalEngineeringGuy.com
• Section: Courses
– Mass Balance Course
• Problems Section
• You will find a problem index there…
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MB in Combustion
• Combustion: is a high-temperature exothermic chemical reaction between a fueland an oxidant, usually atmospheric oxygen, that produces oxidized, often gaseous products
• Why burning? To get heat energy electrical energy electricity!
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MB in Combustion
• Importance of Combustion– Heat energy (not analyzed in this course)
– Reactants• Fuel
• Oxygen
• Inerts (don’t react)
– Products• CO2
• CO
• H2O
• SO2
• Inerts (just flow out)
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MB in Combustion
• General idea:
• Fuel + Oxygen CO2 + H2O + Heat
• Ex: CH4 + O2 CO2 + H2O (not balanced)
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MB in Combustion
• Balancing equation is SUPER important!
• Failing wrong mass balance
• Tips for balancing (order)
– Balance all Carbon atoms
– Balance all Hydrogen Atoms
– Balance all Oxygen Atoms (don’t hesitate to use fractions in moles)
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MB in Combustion
• Example: Propane
• C3H8+O2 CO2 +H2O
– Balance C: C3H8 + O2 3CO2 + H2O
– Balance H: C3H8 + O2 3CO2 + 3H2O
– Balance O: C3H8 + 9/2·O2 3CO2 + 3H2O
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Air Composition
• Nitrogen
• Oxygen
• Noble gases (Ar, Kr, Ne, He)
• CO2, Methane, H2, H2O
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Air Composition (Table)
Source: WikipediaMolecular Weight (average) = 29 g of Air / gmol Air
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Air Composition
• Why bother balancing 0.97% of other gases?
• In engineering… we can assume air composition as:
– 0.79 N2
– 0.21 O2
• We WILL use this assumption in all Combustion problems!
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N2 – O2 relationship
• 1 mol of air contains:
– 0.21 mol of O2
– 0.79 mol of N2
• Relationship of O2/N2
– 0.79 mol N2 per 0.21 mol O2
– 0.79/0.21 = 3.76 mole of N2 per mol O2
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Wet vs. Dry Base
• Wet base composition: composition of a flow including Water as a component.
• Dry Base composition: composition of a flow NOT including Water as a component.
Example: Calculate Dry and Wet compoisition of Air if: 1 mol O2, 1 mol N2, 1 mol H2O
0.33 O2; 0.33 N2, 0.33 H2O Wet composition of Air0.5 O2; 0.5 N2; H2O not included Dry composition of Air
View more exercises HERE
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Dry & Wet Base: Exercise 1
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Dry & Wet Base: Exercise 2
Now go backwards… From Dry basis to mass of Water
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Dry & Wet Base: Exercise 2
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Theoretical Oxygen, Excess Oxygen
• Many times, there is an excess of oxygen to avoid incomplete combustion…
– CH4 + 4·O2 CO2 + 2H2O but may also react:
– CH4 + 3/2·O2 CO + 2H2O
• Using excess:
– CH4 + 5·O2 CO2 + 2H2O (not probably to form CO)
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Theoretical Oxygen
• Oxygen needed to perform a 100% combustion
• For CH4 + 4·O2 CO2 + 2H2O
• Theoretic Oxygen is 4 mol of O2
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Excess Oxygen
• Excess oxygen: oxygen not reacted after a 100% combustion
• CH4 + 5·O2 CO2 + 2H2O
• But we need only 4 mol O2 = 1 mol wont react
• Excess oxygen: 1 mol
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% of Excess Oxygen
• Express the excess in %
• So a 20% oxygen excess for CH4 would be:
% Excess Oxygen = [Oxygen feed – Oxygen (theoretical)]/[Oxygen (theoretical)]
20%= (Oxygen feed – 4 mol O2)/ (4 mol O2)
0.2·4 = O -4O = 4+0.8 = 4.8 moles of O2
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Theoretical Air, Excess Air
• Since we use air to get oxygen in the reaction
– We need to calculate air flows from Oxygen
• Theoretical Air: air needed to get the theoretical oxygen contained
• Excess Air: excess air fed to get higher combustion rates
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% Excess Air
• Similar to % Excess oxygen:
• Be aware of the NITROGEN content of “air”
• Nitrogen is 3.76x in quantity than O2
• Will definitively change compositions
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Air & Oxygen excess: Exercise 1
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Air & Oxygen excess: Exercise 2
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Air & Oxygen excess: Exercise 2
• NOTE: the definition implies 100% conversion! So there should be no difference
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Air & Oxygen excess: Exercise 2
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MB Combustion Tips
• Never forget N2 (inert) in MB equations
• Review CO vs. CO2 balances
• Any side reactions of S or N to SOx or NOx?
• If Oxygen in excess, review Oxygen balance in the outlet stack gas
• Remember the relationship 3.76 mol N2 per mol O2
• Balance equation and double check balance!
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MB CombustionExercise 1
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MB CombustionExercise 1
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MB CombustionExercise 1
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MB CombustionExercise 1
• Table of Composition in Outlet (Stack Gas)
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MB CombustionExercise 1
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MB CombustionExercise 1
• NOTE: Book is probably wrong
More exercises in problem section!
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End of Section 2
• WE are done with Section 2
• You know the basics of MB
– 1,2 or + units
– No Reactive + Reactive systems
– With purge, recycle and bypass
– Combustion
More exercise in problem section!
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Problems & Exercises
• All pair problems of Elementary Principles in Chemical Processes. Felder, R; Rousseau, R. 3rd edition CH4 are solved in the section of “Solved Problems” in my webpage
• Visit www.ChemicalEngineeringGuy.com
• Remember: practice makes the master
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End of MB1: Introduction• You should be now able to perform MB of process
• Hopefully:
– You are able to draw a Diagram from a “text” problem
– You are able to apply the methodology of MB
– You are now able to do a DOF analysis
– You can differentiate between: non-reactive MB, reactive MB, 2+ Unit MB, Combustion MB, Purge, Recycle, Bypass…
– You practice a lot of problems!• Theory alone will not help
– This is the basis of chemical engineering! Learn it well!
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MORE INFORMATION
• Get extra information here!
• FB page:
– www.facebook.com/Chemical.Engineering.Guy
• Contact me by e-mail:
• Directly on the WebPage:
– www.ChemicalEngineeringGuy.com/courses
LOGO AQUI
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Bibliography
• Elementary Principles in Chemical Processes. Felder, R; Rousseau, R. 3rd edition.
• Basic Principles and Calculation in Chemical Engineering. Himmelblau, D. 7th edition.
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