mb1 introduction to mass balancing

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Mass Balance: Introduction

Chemical Engineering

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

Section 1

• Some basic theory in engineering and process engineering

• Easy but Important to understand!

• Learn by hard!

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

Process

• A series of actions or steps taken in order to achieve a particular end.

Process

• A series of actions or steps taken in order to achieve a particular end.

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

Chemical Process Example

Chemical Reaction!

Types of Diagrams

• Block Diagram

• Flow Diagram

• P&ID Diagram

Block Diagram

• Simple diagram that “shows at a glance” the process

• Most used for MB solvingShow:• Flows• Unit Operations• Some extra data

Flow Diagram

• Recommended for “general” information

P&ID• Pipe and Instrumentation Diagram

• Formal “language” for diagram

• Type of valves• Piping information• Op. Unit detail• Automation units

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!

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

Unit Operations

Distillation Column

Unit Operations

Heat Exchanger

Unit Operations

Centrifugal Pump

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.

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

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

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

Flow Example

System State

• Steady State

– Typical for continuous processes

• Transient State

– aka unsteady-state/non-steady state

– Typical for batch, semibatch processes

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

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.

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)

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

Types of Systems

• Open • Closed • Isolated

Type of Systems

• Isobaric system: Constant Pressure

• Isothermal system: Constant temperature

• Isochoric system: Constant Volume

• Adiabatic system: No heat exchange Q=0

End of section 1

• Please make sure you understood all material before advancing to the next section

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!

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

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.

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

System(reactions, mixing, and

other operations may occur here)

Outlet

MB Analysis

Outlet

MB Analysis

Inlet –Outlet + Prod. – Consum. = Accumulation in system

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?

City example

occur here)

Inlet Outlet

MB Analysis Immigration = +50

Emmigration = -3Born = +100Death = -25

What is the “accumulation” in the city? = 50-3+100-25 = 122

Reactor Example of A

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

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

Types of Mass Balances

• Differential

• Integral

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%

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

Types of process

• Steady State (Continuous)

• Transient (Batch)

• Transient-Steady (Semi-batch)

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

20 kg/h

80 kg/h

100 kg/h

100 molTotal

20 mol C80 mol mol A

1 mol A

1 mol B

MB Steady StateExercise 1

If you check all MB possibles, you get the same answer and YOU SHOULD

MB Steady StateExercise 1

MB: Batch

• No Inlet/Outlet, closed system

• Production, Consumption and Accumulation

Feeding Production/reactionDischarge

In-Out + P – C = AccumP – C = Accum.

MB BatchExercise 1

NOTE: We do not need the MB of Water (WHY??)MB of water proves 1=1 or 0=0 WHY??

MB: Semibatch

• More complex, include continuous and batch

• Accumulation always is included

• In, Out, P, C may vary

• See next example

MB SemibatchExample 1

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

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

Diagram Flow Construction Example

MB Steady State Exercises

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

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…

If we wanted 20 kg of P…• Just multiply by 20 the flows• Concentrations stay the same

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…

Multiply Units per

By time (x60 min hr)

By mass(x2.2 lb kg)

• Conclusion all processes are balanced!

Which Process is Balanced?

• NO, MB don’t match

• Yes

• Impossible to know (reaction?)

• Not enough data!

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.

Scale-up & Basis of Calculation Exercise 1

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

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

MB: Non-reactive systemsExercise 1

Example 4.3-3

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)

Is it Possible to solve?

Variables that are explicitly given

Equations from the MB (N=2)

Physical restrictions

Only MATH can STOP you!

Variable-Equation counting

• Don’t just do problems! Analyze

• Many times you have NO sufficient data to solve the problem

100 kg/h

Impossible to solve for m2, m3, m4

• 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

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)

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

Variable-Equation countingExercise

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

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

• 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

• Conclusion:

– Following the “methodology” helps!

– The difficult part is the assumption of equations

– MB are easy if data is known

Need more Exercises & Problems?

• Go to www.ChemicalEngineeringGuy.com

• Section: Courses

– Mass Balance Course

• Problems Section

• You will find a problem index there…

MB in 2 Units

• In chemical processes, its normal to have 2+ Operation Units

MB in 2 Units

• In chemical processes, its normal to have 2+ Operation Units

Global Balance:Inlet = Sweet gas + Acid gas

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)

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

MB in 2 Units

• Many equations may be redundant/dependent

• Example of redundant equations:

MB in 2 UnitsExercise 1

I labeled them as C1, C2 and P (respectively)

WHY P? it’s the most simple mass balance… Recommended ALWAYS to start with the overall mass balance if possible

Given P, we would like to calculate other flows (if not possible go directly to compositions!)

Given all Flows (calculated) we only need to find all compositions

Do Mass Balances in Unit 1, Unit 2… you may get answers from there!

• Data Conclusion:– Start doing Overall Mass Balance– Try to get all the flows first– Calculate compositions by doing MB in units

MB in 2+ Units

• Complexity increases

• MB: Global, unit, group of units

MB in 2+ UnitsExercise 1

• Calculate all Flows (E1, E2, R, Et, V, B)

• Calculate all compositions

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

Need more problems?

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

Recycle

Identify Recycle

Recycle

Identify Recycle

Bypass

Identify Bypass

Bypass

Identify Bypass

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)

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)

Bypass/Recycle Exercise 1

1) Qw, Pc, Fe, Fc, R/F2) Conclusion if no Recycle

• Draw Diagram… Label flows and write all comp.

1) Calculate Qw, Pc, Fe, Fc, R/F

1) Calculate Qw, Pc, Fe, Fc,

2) R/F 5630/4500 = 1.25

If no recycle: you will loss all the crystals in solution!

Section 2 Break #1

What we´ve seen

What's left

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

MB in Reactive systems

• Reaction basics kinetics

• Stoichiometry

• We will use now “Production – Consumption” term

• NO Accumulation in the system (continuous)

Stoichiometry for MB reactive systems

• Stoichiometry Coefficient

• Stoichiometric Ratio (A/B)

• Balancing Equations

• Limiting reactant

• Excess reactant

• % Conversion

• Extent of reaction ε

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

• 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

• 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?

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

Stoichiometry: Balance Equation

Stoichiometry: Balance EquationPractice Problems

Balance the next equations to master “Combustion” Equations

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)

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

Stoichiometry Limiting/Excess Exercise 1

Stoichiometry Limiting/Excess Exercise 2

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

Stoichiometry % ConversionExercise 1

You actually don’t need B,C moles nor P

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

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

Extent of reaction εExercise 1

Ni = Nio + βi·ε

Substitute!

Given data…

From Equation “Stoichiometric Coefficients”

a) Limiting reactantb) % excess if anyc) How much is left of the excess?d) If Xa = 0.50?

• Limiting reactant

Ei < Ej … Then “i” limits reaction WHY?A : limiting reactantB : excess reactant

• How much is left of the excess?

• % excess if any

• If Xa = 0.50?

• The reaction A+B+3/2 C D +3E

• Theres is a 100 mol Feed

• Change “air” composition to “Oxygen/Nitrogen”

• We now choose a Basis and calculate all the moles involved in the flows

• All the stoichiometric values from the equation (BALANCED)

• A+B+3/2 C D +3E

• Substitute in the generic equation

Ni = Nio + βi·ε

βi= + product- reactant

• Substitute initial moles (data is given)

• A) Calculate the limiting reactant.

– Remember Ei < Ej

• B) Calculate the % excess of the other 2 reactants (by now, you know its O2 and B)

• C) What would be the mole flows if Xa = 30%

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

• 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

• 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

MB in Chemical EquilibriumExercise 1

• A) Calculate the composition in equilibrium

• B) Calculate the fractional conversion of limiting reactant

• A)

• B) Fractional conversion of CO

– fCO = (1-yco) = 1- 0.333 = 0.667

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 -$$

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

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

• 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

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

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

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

MB in systems with Multiple RXNExercise 1

• Express all the species with equation:

• Ni = Nio + β1i·ε1 +β2i·ε2

Yield+Selectivity in Multiple ReactionsExercise 2

• We have ε1 and ε2 (RXN1, RXN2)

• β1i and β2i “i” for every species

• 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

• Calculating moles of C2H6 @ 50.1%

• From the Yield Information Y = 0.471

• Finally, substitute and solve equations

• With ε1 and ε2

Section 2 - Break

What we´ve seen

What's left

What we´ve seen

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

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

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

Do atomic Balances on each specie (C,H,O) N = 3

MB/Atomic done!

N1, N2 flows

• 100 mol of Methane are being burned. There is a 90% conversion of the limiting reactant. Calculate all moles @ outlet

• Proceed to do Mass Balance of Atomic Species (C, H, and O)

• We could do balance in species C2 and O2 WHY?

Conversion: Global vs. Single-Pass

• Global conversion “Xa global” or “Xa process”

• Single-Pass Conversion “Xa in Unit”

• Global:

– Process Inlet

– Process Outlet

• Single-Pass

– Unit Inlet

– Unit Outlet

• It must be specified

• If not, suppose it is Operation Unit conversion

• 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

• Solve this type of problems as before, no difference (it is actually a “Product” line)

• Purge:Feed ratio is a common data given

Purge: Exercise 1

• Make a MB in the Reactor (RKT)

Purge: Exercise 1

• Keep doing math… the MB are almost done

Purge: Exercise 1

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!

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)

MB in Combustion

• General idea:

• Fuel + Oxygen CO2 + H2O + Heat

• Ex: CH4 + O2 CO2 + H2O (not balanced)

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)

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

Air Composition

• Nitrogen

• Oxygen

• Noble gases (Ar, Kr, Ne, He)

• CO2, Methane, H2, H2O

Air Composition (Table)

Source: WikipediaMolecular Weight (average) = 29 g of Air / gmol Air

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!

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

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

Dry & Wet Base: Exercise 1

Now go backwards… From Dry basis to mass of Water

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)

Theoretical Oxygen

• Oxygen needed to perform a 100% combustion

• For CH4 + 4·O2 CO2 + 2H2O

• Theoretic Oxygen is 4 mol of O2

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

% 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

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

% 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

Air & Oxygen excess: Exercise 1

• NOTE: the definition implies 100% conversion! So there should be no difference

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!

MB CombustionExercise 1

• Table of Composition in Outlet (Stack Gas)

• NOTE: Book is probably wrong

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