MB1 Introduction to Mass Balancing

Mass Balance: Introduction
MB1
Chemical Engineering
www. Chemical Engineering Guy .com

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.

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 solving
Show:
• 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
Restricted Content for Slide Share Users!
Need complete material? Send me an e-mail:
chemical.engineering.guy@gmail.com
Check the video edition of this PPT at my YouTube Channel:
www.youtube.com/ChemEngineeringGuy
Visit me at
www.ChemicalEngineeringGuy.com
Reaction engineering
• 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

Unit Operations
Reactor
Diagram Picture Real-Life Picture

Unit Operations
Distillation Column

Unit Operations
Heat Exchanger

Unit Operations
Centrifugal Pump
Visit me at

Process Variables
• A process variable, process value or process
parameter is the current status of a process
under 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
Visit me at

Transient State
• Transient State: When any of the variables
change with time
Example:
10 liters of a mixture is fed to the pot
After 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
Outlet
(reactions, mixing, and
other operations may occur
here)
Inlet
Inlet
Outlet

System
Outlet
(reactions, mixing, and
other operations may occur
here)
Inlet
Inlet
Outlet
MB Analysis

Outlet
System
(reactions,
mixing, and
other operations
may occur here)
Inlet
Inlet
Outlet
MB
Analysis
Inlet –Outlet + Prod. – Consum. = Accumulation in system

City example
MB
Analysis Immigration = +50
System
Inlet Outlet
(reactions, mixing,
and other
operations may
occur here)
Emmigration = -3
Born = +100
Death = -25
What is the “accumulation” in the city?

City example
MB
Analysis Immigration = +50
System
Inlet Outlet
(reactions, mixing,
and other
operations may
occur here)
Emmigration = -3
Born = +100
Death = -25
What is the “accumulation” in the city? = 50-3+100-25 = 122

Reactor Example of A
MB
Analysis MB in system
System
Inlet Outlet
(reactions, mixing,
and other
operations may
occur here)
Inlet of A = 100 kg
Production = 0 kg
Consumption = 50 kg
Outlet = 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 Need velocity/complete rate: material? g/s or Send kg/me h, an barrel/e-mail:
day
Typical for chemical.• continuous engineering.processes
guy@gmail.com
• TIP: convenient to use time basis (1h, 1 year)
Visit me at
• 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
• Examples
In-Out +P – C = 0
In + P = Out + C
Reactor
Distilation Column
Mixer
20 kg/h
80 kg/h
100 kg/h
100 mol
Total
80 mol mol A 20 mol C
1 mol A
1 mol B

MB Steady State
Exercise 1

MB Steady State
Exercise 1
If you check all MB possibles, you get
the same answer and YOU SHOULD

MB: Batch
• No Inlet/Outlet, closed system
• Production, Consumption and Accumulation
Feeding Production/reaction
Discharge
In-Out + P – C = Accum
P – C = Accum.

MB Batch
Exercise 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 Semibatch
Example 1

MB Semibatch
Example 1
Visit me at

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/w
0.2 H2SO4 w/w
0.3 NaCl w/w
T= 65ºC
P = 1 atm

Diagram Flow Construction
1. Tag each equipment and arrow
2. Assign variables to unknown data/flow/comp.
F = [] kg/h
60 kmol N2 / min
40 kmol O2 / min
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
Airflow ?
60 kmol N2 / min
40 kmol O2 / min

Diagram Flow Construction Example

MB Steady State
Exercises

MB Steady State
Exercises
Visit me at

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

We could assign 2 kg o of P as “Basis”… solving for B, T you get half each
OR
We could assign 1kg to B or T… solving will give you twice P…

We could assign 2 kg o of P as “Basis”… solving for B, T you get half each
OR
We could assign 1kg to B or T… solving will give you twice P…
x10
If we wanted 20 kg of P…
• Just multiply by 20 the flows
• Concentrations stay the same

Multiply
Units
per
time

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!

Exercise 1
• The next process is to be Scaled up from 100
mol/h of Feed (F) to 1250 lbmol/h.

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 = Accum
In-Out = 0
Inlet = Outlet

• 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 Equations
2 Unknown

Exercise 1
Example 4.3-3

Exercise 1
Visit me at

Exercise 1

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
m1
m2
m4
m3
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
If #unknown > # independent equations
• Search more!

Exercise
Visit me at

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

Exercise 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

Exercise 1
• 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 Units
Exercise 1
I labeled them as C1, C2 and P (respectively)

MB in 2 Units
Exercise 1
WHY P? it’s the most simple mass balance…
Recommended ALWAYS to start with the overall mass balance if possible

MB in 2 Units
Exercise 1
Visit me at
Given P, we would like to calculate other flows (if not possible go directly to compositions!)

MB in 2 Units
Exercise 1
Given all Flows (calculated) we only need to find all compositions

MB in 2 Units
Exercise 1
Visit me at
Do Mass Balances in Unit 1, Unit 2… you may get answers from there!

MB in 2 Units
Exercise 1
• 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+ Units
Exercise 1
• Calculate all Flows (E1, E2, R, Et, V, B)
• Calculate all compositions

MB in 2+ Units
Exercise 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?
• Go to www.ChemicalEngineeringGuy.com
• Section: Courses
– Mass Balance Course
• Problems Section
• You will find a problem index there…

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

Bypass
Identify
Bypass

Why Recycle
• For Chemical reactions to increase conversion
(reaction to form product)
• Recover a catalyst ($)
• Dilute a process stream (increasing flow and
decreasing 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/F
2) Conclusion if no Recycle

Bypass/Recycle
Exercise 1
• Draw Diagram… Label flows and write all comp.
1) Calculate  Qw, Pc, Fe, Fc, R/F

Bypass/Recycle
Exercise 1
Visit me at

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!

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 Visit 0.33 me mol at
A/ mol C
– Ratio of www.B:A= ChemicalEngineeringGuy.9/2 : 1 or 4.5 mol B / com
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 Equation
Practice 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 % Conversion
Exercise 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

• 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

Exercise 1
Ni = Nio + βi·ε

Exercise 1
Substitute!
Given data…
Visit me at
From Equation “Stoichiometric Coefficients”

Exercise 2
a) Limiting reactant
b) % excess if any
c) How much is left of the excess?
d) If Xa = 0.50?

Exercise 2
• Limiting reactant
Ei < Ej … Then “i” limits reaction WHY?
A : limiting reactant
B : excess reactant

Exercise 2
• How much is left of the excess?

Exercise 2
• % excess if any

Exercise 2
• If Xa = 0.50?

Exercise 3
• The reaction A+B+3/2 C  D +3E
• Theres is a 100 mol Feed

Exercise 3
• Change “air” composition to
“Oxygen/Nitrogen”

Exercise 3
• We now choose a Basis and calculate all the
moles involved in the flows

Exercise 3
• All the stoichiometric values
from the equation
(BALANCED)
• A+B+3/2 C  D +3E

Exercise 3
• Substitute in the
generic equation
Ni = Nio + βi·ε
βi= + product
- reactant

Exercise 3
• Substitute initial moles
(data is given)

Exercise 3
• A) Calculate the limiting reactant.
– Remember Ei < Ej

Exercise 3
• B) Calculate the % excess of the other 2
reactants (by now, you know its O2 and B)
Visit me at

Exercise 3
• C) What would be the mole flows if Xa = 30%

MB in Chemical Equilibrium
• Chemical Equilibrium: reaction is possible also
reversible
A + B  C
C  A + B
• 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

• 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

Exercise 1
• A) Calculate the composition in equilibrium
• B) Calculate the fractional conversion of limiting reactant

Exercise 1

Exercise 1
Visit me at

Exercise 1
2/3
• 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
• 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

Exercise 1
• C4H4 + ½ O2  C2H4O (1)
• C2H4 + 3 O2  2CO2 + 2H2O (2)
• Express all the species with equation:
• Ni = Nio + β1i·ε1 +β2i·ε2

Exercise 1
• Express all the species
with equation:
• Ni = Nio + β1i·ε1 +β2i·ε2

Yield+Selectivity in Multiple
Reactions
Exercise 2

Reactions
Exercise 2
• We have ε1 and ε2 (RXN1, RXN2)
• β1i and β2i “i” for every species

Reactions
Exercise 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
moles
R: Reacted moles

Reactions
Exercise 2 • Calculating moles of C2H6 @ 50.1%
• From the Yield Information Y = 0.471
Visit me at

Reactions
Exercise 2
• Finally, substitute and solve equations

Reactions
Exercise 2
• With ε1 and ε2

Section 2 - Break
 What we´ve seen
What we´ve seen 
What's left 

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

• 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

Exercise 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

Exercise 1
Do atomic Balances on each specie (C,H,O) N = 3

Exercise 1
Visit me at

Exercise 1
MB/Atomic done!
N1, N2 flows

Exercise 2
• 100 mol of Methane are being burned. There
is a 90% conversion of the limiting reactant.
Calculate all moles @ outlet

Exercise 2

Exercise 2
• Proceed to do Mass Balance of Atomic Species (C,
H, and O)
• We could do balance in species C2 and O2 WHY?

Exercise 2
Visit me at

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

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

Purge
• Solve this type of problems as before, no
difference (it is actually a “Product” line)
• Purge:Feed ratio is a common data given

Purge

Purge: Exercise 1

Purge: Exercise 1
Visit me at

Purge: Exercise 1
• Make a MB in the Reactor (RKT)

Purge: Exercise 1
• Keep doing math… the MB are almost done

MB in Combustion
• Combustion: is a high-temperature
exothermic chemical reaction between a fuel
and 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: Wikipedia
Molecular Weight (average) = 29 g of Air / gmol Air

Air Composition
• Why bother balancing 0.97% of other gases?
• In engineering… Need complete we material? can assume Send me an air
e-mail:
composition chemical.as:
engineering.guy@gmail.com
– 0.79 N2
– 0.21 O2
• We WILL use this assumption Visit me at
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
includingWater 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 Air
0.5 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 %
% Excess Oxygen = [Oxygen feed – Oxygen (theoretical)]/[Oxygen (theoretical)]
• So a 20% oxygen excess for CH4 would be:
20%= (Oxygen feed – 4 mol O2)/ (4 mol O2)
0.2·4 = O -4
O = 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
Visit me at

% 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 Combustion
Exercise 1

MB Combustion
Exercise 1
Visit me at

MB Combustion
Exercise 1
• Table of Composition in Outlet (Stack Gas)

MB Combustion
Exercise 1
• NOTE: Book is
probably wrong
Visit me at
More exercises in problem section!

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!

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

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!

MORE INFORMATION
• Get extra information here!
• FB page:
LOGO AQUI
– www.facebook.com/Chemical.Engineering.Guy
• Contact me by e-mail:
– Chemical.Engineering.Guy@gmail.com
• Directly on the WebPage:
– www.ChemicalEngineeringGuy.com/courses

Bibliography
• Elementary Principles in Chemical Processes. Felder, R;
Rousseau, R. 3rd edition.
• Basic Principles and Calculation in Chemical Engineering.
Himmelblau, D. 7th edition.

Visit me at

MB1 Introduction to Mass Balancing

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Mehr von Chemical Engineering Guy

Mehr von Chemical Engineering Guy (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

MB1 Introduction to Mass Balancing