Material Balance Calculations

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Material Balance Calculations

• Department of Chemical Engineering

Based on Notes from Prof. M. Ioannidis
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Outline

• Single Unit in the Absence of
• Chemical Reactions
• Multiple Units in the Absence of
• Chemical Reactions
• Multiple Units in the Presence of
• Chemical Reactions

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1. Single Unit Analysis

• An everyday example
• The Process Brewing Coffee (technical term
  leaching)
• The Machinery Coffeemaker (technical term
  solid-liquid contactor)

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Process Description
A water (W)
B CS, CG
C CS, W
D coffee solubles (CS), coffee grounds (CG) and
water (W)
() Batch process
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Stream Description

• We know everything about a stream if we know its
  mass (or mass flow rate for continuous processes)
  and its composition.
• n variables needed to describe a stream with n
  components. What may these variables be?

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Stream Description (contd)

• Stream D (three components CS, CG, W) needs 3
  variables to describe it.
• The 3 mass fractions xCS, xCG and xW are a poor
  choice, because xCS xCG xW 1 (i.e., they
  are not independent). Thus, at least one variable
  must be a mass (total or component mass).

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Stream Description (contd)

• To describe stream D, we could use either
• the total mass D and any two mass fractions (say,
  xCS and xCG), or
• the mass of each component DCS, DCG, DW, or
• any other combination, except xCS, xCG and xW !
• For example, upon finding DCS, DCG, DW, we also
  know the mass fractions
• xCS DCS/(DCS DCG DW), etc.

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Process Description (contd)
CW, CCS

• Verify that if you knew AW, BCG, BCS, CW, CCS,
  DCS, DCG and DW, youd know everything there is
  to know about the mass and composition of all
  four streams.

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Material Balance Equations

• To determine unknowns you must solve an equal
  number of independent equations that relate them.
  Around a process unit involving n components we
  can write at most n independent material balance
  equations
• Total balance plus n-1 component balances, or
• n component balances.

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Material Balance Equations (contd)

• Suppose we write
• Water balance, AW CW DW
• Coffee grounds balance, BCG DCG
• Coffee solubles balance, BCS DCS
  CCS
• then,
• Total balance (not independent) A B C D
• If we are to solve this problem, we must have
  no more than 3 unknowns (because we can write at
  most 3 independent equations). The remaining 5
  variables must be obtained from data. Data are
  frequently interpreted as extra (auxiliary)
  equations.

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Data and Auxiliary Equations

• The following equations are not material
  balances, they are auxiliary equations coming
  from data (as they may be given in a problem
  statement)
• One kg of W is used to brew coffee
• A AW 1 kg,
• Coffee contains 1 CS
• BCS/(BCSBCG) 0.01,
• Coffee extract contains 0.4 CS
• CCS/(CCSCW) 0.004,
• Waste product contains 80 CG and 19.6 W
• DCG/(DCGDW DCS) 0.8,
• DW/(DCGDW DCS) 0.196.

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Linear_Algebra_at_Work

• Thus far, we have established 7 independent
  simultaneous linear equations in 7 unknowns (AW
  is directly available)
• (0)BCG (0)BCS (0)CCS (1)CW (0)DCG
  (0)DCS (1)DW 1
• (1)BCG (0)BCS (0)CCS (0)CW - (1)DCG
  (0)DCS (0)DW
  0
• (0)BCG (1)BCS - (1)CCS (0)CW (0)DCG -
  (1)DCS (0)DW
  0
• - (0.01)BCG (0.99)BCS (0)CCS (0)CW
  (0)DCG (0)DCS (0)DW 0
• (0)BCG (0)BCS (0.996)CCS - (0.004)CW
  (0)DCG (0)DCS (0)DW 0
• (0)BCG (0)BCS (0)CCS (1)CW (0.2)DCG -
  (0.8)DCS - (0.8)DW 0
• (0)BCG (0)BCS (0)CCS (1)CW - (0.196)DCG -
  (0.196)DCS (0.804)DW 0

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Linear_Algebra_at_Work (contd)

• Alternatively we write , MX b, where
• Solution X M-1b
• Excel and Mathcad can both solve linear systems
  easily...

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Advantages of Matrix Solution

• Analytical clarity
• Ability to investigate what-if scenarios
• Convenient treatment of processes involving many
  streams and many components

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2. Multiple Unit Analysis

• Two-distillation column process to separate
  benzene (B), toluene (T) and xylene (X).
• First column produces overhead product containing
  mostly B.
• Second column produces overhead product
  containing mostly T and bottom product containing
  mostly X.
• All chemicals (B,T,X) are present in all streams.

1
2
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Our Methodology
A
D

• Pretend that you have no data.

F

• Then give a unique name to each stream.

1
2
C
E
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Our Methodology (contd)

• Recall that each stream has 3 components ? each
  stream is fully described by 3 variables.
• Lets use component mass flows to describe them.
• We have a total of 15 unknowns (remember, we
  pretend we have no data!)

AB, AT, AX
DB, DT, DX
FB, FT, FX
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)

• Recall that around each unit we can write as many
  independent mass balances as the number of
  components involved, that is 3 balances.
• For unit 1
• 1 FB AB CB (B-balance)
• 2 FT AT CT (T-balance)
• 3 FX AX CX (X-balance)

AB, AT, AX
DB, DT, DX
FB, FT, FX
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)

• For unit 2
• 4 CB DB EB (B-balance)
• 5 CT DT ET (T-balance)
• 6 CX DX EX (X-balance)
• Total of 6 independent mass balances
• Anything more (e.g., overall balance for B)
• FB AB DB EB
• is redundant (to see this add equations 1
  and 4!)

AB, AT, AX
DB, DT, DX
FB, FT, FX
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)

• Observe that without data we cannot proceed,
  because we have 6 equations and 15 unknowns!
• Data can be translated into auxiliary equations
  we need 9 such equations and we want them to be
  independent!
• Suppose they give us the composition of stream
  A...

AB, AT, AX
DB, DT, DX
FB, FT, FX
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)
AB, AT, AX (4B, 91T, 5X)

• Knowledge of stream A composition allows us to
  write no more than 2 auxiliary equations, e.g.,
• 7 AB/(ABATAX) 0.04
• 8 AT/(ABATAX) 0.91
• The following would not be independent (why?)
• AX/(ABATAX) 0.05
• Generalize knowing the composition of a stream
  of n components affords us n-1 auxiliary
  equations.

DB, DT, DX
FB, FT, FX
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)
AB, AT, AX (4B, 91T, 5X)
DB, DT, DX (4.3B, 91.2T, 4.5X)

• Knowledge of the composition of streams F and D
  would give 4 additional auxiliary equations
• 9 DB/(DBDTDX) 0.043
• 10 DT/(DBDTDX) 0.912
• 11 FB/(FBFTFX) 0.35
• 12 FT/(FBFTFX) 0.50

FB, FT, FX (35B, 50T, 15X)
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)
AB, AT, AX (4B, 91T, 5X)
DB, DT, DX (4.3B, 91.2T, 4.5X)

• A basis provides one more auxiliary equation,
  e.g.
• 13 FBFTFX 100
• The last two auxiliary equations may come from
  knowing that stream E contains 10 of B in the
  feed and 93.3 of X in the feed
• 14 EB 0.1FB
• 15 EX 0.933FX

FB, FT, FX (35B, 50T, 15X)
1
2
CB, CT, CX
EB, ET, EX
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Our Methodology (contd)

• Think Why do we prefer to work with component
  mass flows as our stream variables, e.g., CB, CT
  and CX for stream C, and not with total mass flow
  and mass fractions (e.g., C, xCB and xCT) ?

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Our Methodology (contd)
AB, AT, AX (4B, 91T, 5X)
DB, DT, DX (4.3B, 91.2T, 4.5X)

• Suppose we used C, xCB and xCT to describe stream
  C. Then, the mass balances for, say, unit 2
  would be
• CxCB DB EB (for B)
• CxCT DT ET (for T)
• C(1-xCT-xCB) DX EX (for X)
• This formulation would result in non-linear
  equations which are more difficult to solve!

FB, FT, FX (35B, 50T, 15X)
1
2
C, xCT, xCB
EB, ET, EX
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3. Single Unit Balance with Reaction
A (NH3)
4NH3 5O2 ? 4NO 6H2O
B (Air O2, N2)
C (O2, N2, NO, NH3, H2O)
NOTE Since no data are available at this stage,
we must assume that all reactants and products
are present in the effluent stream (8 stream
variables)
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Mass balances using the extent of reaction...

• If X is the number of ammonia moles that
  reacted
• Ammonia CNH3 ANH3 - X
• Nitrogen monoxide CNO X
• Oxygen CO2 BO2 - (5/4)X
• Nitrogen CN2 BN2
• Water CH2O (6/4)X
• To the 8 stream variables we must add the extent
  of reaction, i.e., we have 9 unknowns and 5
  equations
• from mass balances (one for each chemical
  species)

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Auxiliary equations from data...

• In addition to the 8 stream variables, the extent
  of reaction is a also an unknown, i.e., we have 9
  unknowns and only 5 equations from mass balances
  (one for each of the 5 chemical species). We
  need 4 auxiliary equations before we can solve
  this problem. These may be
• One from the basis e.g., 100 mol of NH3 feed
• One from the composition of stream B (why not
  two?)
• One from knowledge of NH3 fractional conversion
• One from knowledge of the percent excess air