FUNDAMENTALS OF MATERIAL BALANCE

1
FUNDAMENTALS OF MATERIAL BALANCE

• Chapter 4

2
Chapter 4

• Lecture 6
• 3/12/03

3
Process Classification

• Chemical processes can be classified as batch,
  continuous or semi-batch and as either transient
  or steady state
• Batch process is one in which the feed is charged
  into the system at the beginning of the process,
  and the products are removed all at once some
  time later
• Continuous process is when the inputs and outputs
  flow continuously across the boundaries
  throughout the duration of the process.

4
Process Classification

• Semi-batch process is a process in which its
  inputs are nearly instantaneous but the outputs
  are continuous or vice versa
• If the values of all process variables in a
  process do not change with time, the process is
  said to be operating at steady state. If any
  changes with time, transient or unsteady state
  operation exists

5
Example 1

• CO and steam are fed into tubular reactor at a
  constant rate, and react to form carbon dioxide
  and hydrogen. Products and unused reactants are
  withdrawn at the other end. The reactor
  contains air when the process is started up. The
  temperature of the reactor is also constant.
  Classify the process
• i) initially and
• ii) after a long period of time has elapsed.

6
Material Balance

• The objectives in studying this section are
• 1. Define the system and draw the system
  boundaries for which the material balance is to
  be made
• 2. Explain the difference between an open and a
  closed system
• 3. Write the general material balance equation
  and apply to simple problems
• 4. Write the general balance equation for
  continuous steady-state processes
• 5. Write an integral balance on batch processes

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Open System
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General Balance Equation

• A general balance on a material that enters or
  leaves any process system may be written in the
  following way
• acc. input gen. - output - consumption
• Two types of balances may be written for any
  system
• differential balances and
• integral balances

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General Balance Equation

• Differential balances indicate what is happening
  in a system at an instant of time. Each term is
  a rate and has a unit of quantity unit per time
• Integral balances describe what happens between
  two instant of time. Each term of the equation
  is an amount of the quantity with a corresponding
  unit
• The generation and consumption terms are applied
  only when chemical reaction is involved
• If the balanced quantity is total mass, the
  generation and consumption terms are always zero
  since mass can neither be created nor destroyed

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Balances On Continuous Steady State Processes

• At steady-state there can be no buildup of
  anything in the system, so the accumulation term
  in a general balance must equal zero
• Input generation output consumption
• In addition if there is no reaction,
• Input output

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Example 2

• Write balances on components B and T to calculate
  the unknown flow rates in the output streams

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Integral Balance on Batch Processes

• For any substance participating in a batch
  process, the balance equation becomes
• Initial gen. final consumption
• The input and output terms denote the initial and
  the final amounts of the balanced substance
  rather than the flow rates

13
Example 3

• Two methanol-water mixture are contained in
  separate flasks. The first mixture contains 40
  wt methanol, and the second contains 70
  methanol. If 200 g of the first mixture are
  combined with 150 g of the second, what are the
  mass and composition of the product.

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Chapter 4

• Lecture 6
• 5/12/03

15
Flowchart

• A flowchart is drawn using boxes or other symbols
  to represent the process units and lines with
  arrows to represent inputs and outputs

• The chart must be fully labeled with values of
  known variables at the locations of the streams

OR
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Flowchart

• Assign algebraic symbols to unknown streams and
  write their associated units on the chart
• If a volumetric flow rate of a stream is given,
  convert to mass or molar flow rate since balances
  are not normally written in volumetric quantities

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Flowchart

• Convert all stream quantities to one basis
• Only express quantities in mol or mass
• Translate any information that has no been used
  in labeling the chart into equations
• If you are given that the mass of stream 1 is
  half of stream 2, label these streams as m1 and
  2m1 and not m1 and m2
• If there is three times N2 (in mass) in a stream
  as compared to O2 then, label mass fraction as x
  (g O2/g) and 3x (g N2/g)

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Examples

• An experiment on the growth rate of certain
  organism requires an environment of humid air
  enriched in oxygen. Three input streams are fed
  into an evaporation chamber to produce an output
  stream with the desired composition.
• Liquid water, fed at a rate of 20.0 cm3/min
• Air ( 21 mole O2, balance N2)
• Pure oxygen, with a molar flow rate one-fifth of
  the molar flow rate of stream B.

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Flow Chart Scaling and Basis of Calculation

• Scaling a flowchart is a procedure of changing
  the values of stream flow rates by proportional
  amount while leaving the composition unchanged
• A basis of calculation is an amount or flow rate
  (mass or molar) chosen in the calculation to
  start the material balance all unknown
  variables are determined to be consistent with
  this basis
• Any convenient quantities can be used as a basis,
  and the results can later be scaled to any
  desired values
• It is usually most convenient to use a stream
  amount or flow rate as a basis of calculation and
  if none is specified, choose an amount or flow
  rate of a stream with a known composition as a
  basis

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Example 4

• Scale up the balanced process to a feed of 1000
  kg moles C2H6/hr

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Degree of Freedom Analysis

• Draw and label flow chart
• Count the unknown variables on the flow chart,
  nunknowns
• Count the independent equations relating them,
  nindep eqns
• ndf nunknowns - nindep eqns
• If ndf0, the problem is solvable
• If ndfgt0, the problem is underspecified, need to
  provide more information/equations.
• If ndf0, the problem is overspecified, more
  equations than unknowns, redundant and possibly
  inconsistent information.

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Sources of Equations

• Material balances
• For a non reactive process, no more than nms
  independent equations may be written where nms is
  the number of molecular species (e.g. CH4, O2)
  involved in the process
• Energy balance (Ch. 7-9)
• Process specifications
• Physical properties or laws (e.g. Ideal Gas Law)
• Physical constraints (?yi 1.0)
• Stoichiometric relations (for reactive systems
  only)

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Balancing a Process

• The maximum number of independent equations that
  can be derived by writing balances on a non
  reactive system equals the number of chemical
  species in the streams of each unit (or
  subsystem) added together.
• Write balance equations that involve the fewest
  unknown variables first

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Example 5

• An aqueous solution of NaOH contains 20 NaOH by
  mass. It is desired to produce an 8 NaOH
  solution by diluting a stream of 20 solution
  with a stream of pure water.
• Calculate the ratios (g H2O/g feed solution) and
  (g product solution /g feed solution)
• Determine the feed rates of 20 solution and
  diluting water needed to produce 2310 Ibm/min of
  the 8 solution

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General Procedure for Material Balance
Calculations

• Choose as a basis of calculations an amount or
  flow rate of one of the process streams
• Draw a flowchart of the process. Include all the
  given variables on the chart and label the
  unknown stream variables on the chart
• Write the expressions for the quantities
  requested in problem statement
• Convert all mass and molar unit quantities to one
  basis
• Do the degree of freedom analysis. For any given
  information that has not been used in labeling
  the flowchart, translate it into equations in
  terms of the unknown variables
• If nDF 0, write material balance equations in
  an order such that those involve the fewest
  unknowns are written first
• Solve the equations and calculate the additional
  quantities requested in the problem statement
• Scale the quantities accordingly

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Example 6

• A liquid mixture containing 45.0 benzene (B) and
  55.0 toluene (T) by mass is fed to a
  distillation column. A product stream leaving
  the top of the column (overhead product) contains
  95.0 mole B and a bottom product stream contains
  8.0 of the benzene fed to the column (meaning
  that 92 of B leaves with the overhead product).
  The volumetric flow rate of the feed stream is
  2000 L/h and the SG for mixture is 0.872.
  Determine the mass flow rate of the overhead
  product stream and the mass flow rate and
  composition (mass fraction) of the bottom product
  stream.

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Chapter 4

• Lecture 7
• 8/12/03

28
Balances on Multiple-Unit Processes

• When a multiple-unit process is involved, a
  system is defined as any portion of a process one
  chooses to consider
• A system can be an entire process, an
  interconnected combination of some of the process
  units, a single unit, or a point at which two or
  more process stream come together or split up
• An imaginary boundary is usually drawn around a
  portion of the process on the flowchart to define
  the system on which the balances are written,
  taking as inputs and output all streams crossing
  this boundary
• If several subsystems are required to obtain
  enough equations, choose the boundary which
  intersect streams containing fewest unknown
  variables to solve first

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Example 7

• Calculate the unknown flow rates and compositions
  of streams 1, 2, and 3

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Recycle and Bypass

• Recycle is used in a process when there is a need
  to recover some of the components in the product
  stream
• Bypass is a fraction of a stream, which is
  diverted around one or more units and combined
  with the output stream from the unit
• Recycle an bypass calculations are solved in the
  same manner the flowchart is drawn and labeled,
  then the overall balance and balances around the
  system are used to determine the unknowns

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Recycle and Bypass
32
Recycle and Bypass

• There are several reasons for using
  recycle/bypass in a chemical process
• Recovery and reuse of unconsumed reactants
• Recovery of catalyst
• Dilution of a process
• Control of a process variable
• Circulation of a working fluid

33
Example 8

• An evaporative crystallization process to recover
  crystalline potassium chromate (K2CrO4) from an
  aqueous solution of the salt, 4500 kg/hr which is
  one-third K2CrO4 by mass. The feed is joined by
  a recycle stream containing 36.4 K2CrO4 , and
  the combined stream is fed to an evaporator. The
  concentrate leaving the evaporator contains 49.4
  K2CrO4 this stream is fed into a crystallizer in
  which it is cooled, causing crystals of K2CrO4 to
  come out of solutions and then filtered. The
  filter cake consists of K2CrO4 crystals and a
  solution that contains 36.4 K2CrO4 by mass the
  crystals account for 95 of the total mass of the
  filter cake. The solution that passes thru
  filter, also 36.4 K2CrO4 , is the recycle
  stream.
• Calculate the rate of evaporation, rate of
  production of crystalline, the feed rate to
  evaporator and crystallizer and recycle ratio.
• Suppose that the filtrate is discarded and not
  recycled, calculate the production rate of
  crystals. What is the benefit of recycle?

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Example 8 (cont.)
Water
m2 kg/h
m4kg/h (S)
Crystallizer and Filter
m3kg/h
4500 kg/hr
m1kg/h
Evaporator
33.3 K2CrO4
49.4 K2CrO4
m5kg/h (soln)
36.4 K2CrO4
Filtrate
36.4 K2CrO4
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Chapter 4

• Lecture 8
• 10/12/03

36
Balances on Reactive Systems

• The objectives are to be able to
• Write and balance chemical reaction equations
• Calculate the stoichiometric quantities of
  reactants and products from the given chemical
  equations
• Define excess reactant, limiting reactant,
  conversion, degree of completion, and yield in a
  reaction
• Identify the limiting and excess reactants and
  calculate percent excess reactants, percent
  conversion, percent completion, and yield for a
  chemical reaction with reactants being in
  non-stoichiometric proportions.
• Know the procedures for carrying out material
  balance calculations on reactive systems

37
Stoichiometry

• Stoichiometry is the theory of the proportion in
  which chemical species combine with one another
• Stoichiometric equation of a chemical reaction is
  a statement of the relative number of molecules
  or moles of reactants and products that
  participate in the reaction
• 2SO2 O2 ? 2 SO3
• Stoichiometric coefficients are the numbers that
  precede the formula for each species
• A balanced chemical reaction equation is when the
  number of atoms of each atomic species equals on
  both sides of the equation
• Ratio of the stoichiometric coefficients of two
  molecular species participating a reaction is
  termed stoichiometric and can be used as a
  conversion factor in material balance calculations

38
Example 1
Consider the reaction, C4H8 6O2 -----
4CO2 4H2O Is the equation balanced? What is
the stoichiometric ratio of H2O to O2 if 100
g-mole /min of C4H8 fed into the reactor, and 50
reacts, at what rate water is formed
39
Limiting and Excess Reactants, Fractional
Conversion

• If the ratio of the amounts of two reactants
  equals the stoichiometric ratio obtained from the
  balanced reaction, these reactants are said to be
  present in stoichiometric proportion
• A reactant which is present in less than its
  stoichiometric proportion relative to every other
  reactant is called a limiting reactant and the
  other reactants are excess reactants
• Percentage excess of a reactant is defined as
• excess (n - ns) /ns x 100
• Where n moles of excess reactant
• ns moles correspond to stoichiometric
  proportions
• The fractional conversion of a reactant is the
  ratio
• moles reacted
• f ------------------
• moles fed

40
Example 2

• Acrylonitrile is produced by the reaction of 10
  mole C3H6, 12 NH3 and 78 air. If the
  conversion of the limiting reactant is 30,
  determine the limiting reactant and calculate the
  kg-mole C3H3N produced per kg-mole NH3 fed.
• C3H6 NH3 3/2O2 ? C3H3N 3H2O

41
Multiple Reactions, Yield and Selectivity

• In most chemical processes, reactants can usually
  combine in more than one way and the products
  formed may itself react to form less desirable
  products
• The terms yield and selectivity are used to
  describe the degree to which a desired reaction
  predominates over competing side reactions
• Yield moles of desired product formed
• (based on feed) --------------------------
  -----------------------
• moles of limiting reactant fed
• Yield (based on reactant consumption)
• moles of desired product formed
• -------------------------------------------
  ---------- moles of limiting reactant consumed
• moles of desired product formed
• Selectivity ---------------------------------
  -------------------
• moles of undesired product formed

42
Example 3

• The reactions in a dehydrogenation reactor
• C2H6 ? C2H4 H2
• C2H6 H2 ? 2 CH4
• take place continuously at steady state. The
  molar flow rate of the feed stream is
  100kg.mole/hr, and that of product stream is 140
  kg-moles/hr. The composition of the gases are
  given below. Calculate the fractional conversion
  of ethane, the yields of ethylene based on feed
  and reactant consumption, and the selectivity of
  ethylene relative to methane

43
Balances on Atomic And Molecular Species

• The material balances may also be written for
  atomic species participating in the reaction
  regardless of the molecular species the atoms
  happen to be found
• Balances on atomic species can be written as
• Input output, since atoms can never be
  generated nor consumed in a chemical reaction

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Example 4

• C2H6 -------- C2H4 H2
• Do balances on atomic species to solve for q1
  and q2

45
Product Separation And Recycle

• In the analysis of chemical reactors with product
  separation and recycle stream, the reactant
  conversion can be defined in two ways overall
  conversion and single pass conversion
• Overall conversion
• Reactant input to process - reactant output
  from process
• -----------------------------------------------
  -----------------
• Reactant input to process
• Single pass conversion
• Reactant input to reactor- reactant output from
  reactor
• -----------------------------------------------
  -------------------
• Reactant input to reactor
• Overall and single pass conversions denote the
  reactant conversion with and without recycle
  respectively

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Example 5

• Dehydrogenation of propane
• C3H8 ------ C3H6 H2
• The process is to be designed for 95 overall
  conversion of propane. The reaction products
  are separated into two streams the first, which
  contains H2, C3H6 and 0.555 of the propane that
  leaves the reactor, is taken off as product, and
  the second stream, which contains the balance of
  the unreacted propane and 5 of the propylene in
  the product stream, is recycled to the reactor.
  Calculate the composition of the product, the
  ratio (moles recycled)/(moles fresh feed), and
  the single pass conversion

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Example 5 (cont.)
48
Chapter 4

• Lecture 9
• 12/12/03

49
Combustion Reactions

• Combustion is a rapid reaction of a fuel with
  oxygen
• A major source of oxygen in most combustion
  reactors is air which contains mainly N2 (79)
  and O2 (21).
• Most of the fuel used is either coal, fuel oil or
  gaseous fuels such as natural gas or liquefied
  petroleum gas
• The significance of combustion reactions lies in
  the tremendous quantities of heat released
• The heat is used to produce steam, which is then
  used to drive turbines that generate most of
  world's electrical power

50
Combustion Chemistry

• When a fuel is burned, carbon in the fuels reacts
  to form either CO2 or CO, hydrogen will form H2O
  and sulfur will form SO2
• A combustion reaction in which CO is formed is
  referred to a partial or incomplete combustion
• Composition on wet basis is used to denote the
  component mole fractions of a gas that contains
  water whereas dry basis signifies the mole
  fractions of the same gas without water
• The product gas that leaves the combustion
  reactor is called flue or stack gas

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Combustion Chemistry

• Complete combustion
• C3H8 5O2 --- 3CO2 4H2O
• Incomplete combustion
• C3H8 7/2O2 --- 3CO 4H2O

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Example 6

• a) Convert a stack gas composition on a wet basis
  to its corresponding composition on a dry basis
• N2 60
• CO2 15
• O2 10
• H2O 15
• b) An Orsat analysis yields the following dry
  basis composition
• N2 65
• CO2 14
• O2 10
• CO 11
• The mole fraction of water in the stack gas is
  0.07. Calculate the composition on wet basis

53
Theoretical and Excess Air

• Theoretical air is quantity of air which contains
  amount of oxygen (moles or molar flow rate)
  required for complete combustion of all fuel fed
  to reactor
• Excess air is the amount by which air fed to
  reactor exceeds the theoretical air
• moles airfed - moles airtheoretical
• excess air ----------------------------------
  ----
• moles air theoritical
• The use of excess air has the effect of
  increasing the conversion of the valuable
  reactant (fuel) at the expense of the cost of an
  inexpensive reactant (O2)

54
Theoretical and Excess Air

• Theoretical air required to burn a quantity of
  fuel does not depend on how much the fuel is
  actually burned
• The value of percent excess depends only on the
  theoretical air and the air feed rate, and not on
  how much O2 is consumed or whether combustion is
  complete or partial

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Material Balances on Combustion Reactors

• Procedures for writing and solving material
  balances for a combustion reactor is essentially
  the same as for any other reactive system
• All the components that are present in the system
  but do not involved in the reaction must also be
  included at both inlet and outlet of the reactor
• If the percent excess air is given, the O2
  actually fed may be calculated by multiplying the
  theoretical O2 (determined from reaction
  stoichiometry for complete combustion and fuel
  feed rate) by (1 fractional excess air)
• If several combustion reactions occur
  simultaneously, It is more convenient to write
  atomic balances to calculate unknown stream
  variables

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Example 7

• Ethane is burned with 50 excess air. The
  percentage conversion of the ethane is 90 25
  of the ethane burned reacts to form CO and the
  balance to form CO2. Calculate the composition of
  the flue gas and the ratio of water to dry flue
  gas

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Assignment 3

• Problem 4.4 
• Problem 4.6 
• Problem 4.9.
• Problem 4.18
• Problem 4.29
• Problem 4.43
• Problem 4.51
• Problem 4.64 Due 17/12/03