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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 ReactionsMultiple Units in the Absence of
Chemical Reactions
Multiple Units in the Presence ofChemical 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)
D: coffee solubles (CS), coffee grounds
(CG) and water (W)
B: CS, CG C: CS, W
(*) Batch process
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Stream Description
We know everything about a stream if we
know: its mass (or mass flow rate forcontinuous processes) and its composition.
n variables needed to describe a stream withn components. What may these variables
be?
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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, atleast one variable must be a mass (total or
component mass).
Stream Description (contd)
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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, exceptxCS, xCG and xW !
For example, upon finding DCS, DCG, DW, we
also know the mass fractions:
xCS = DCS/(DCS + DCG + DW), etc.
Stream Description (contd)
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CW, CCS
AW
DCS, DCG, DW
BCG, BCS
Process Description (contd)
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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To determine unknowns you must solve an
equal number ofindependentequations that
relate them. Around a process unit
involving n components we can write at
most n independentmaterial balance
equations: Total balanceplus n-1 component balances, or
n component balances.
Material Balance Equations
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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.
Material Balance Equations (contd)
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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/(BCS+BCG) = 0.01,
Coffee extract contains 0.4% CS:
CCS/(CCS+CW) = 0.004,
Waste product contains 80% CG and 19.6% W:
DCG/(DCG+DW + DCS) = 0.8,
DW/(DCG+DW + DCS) = 0.196.
Data and Auxiliary Equations
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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
Linear_Algebra@Work
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Alternatively we write , [M]{X} = {b}, where
Solution: {X} = M-1{b}
Excel and Mathcad can both solve linear systems easily...
Linear_Algebra@Work (contd)
M
1
0
0.01
0
0
0
0
0
1
0.99
0
0
0
0
0
1
0
0.99
0
0
1
0
0
0
0.004
0
0
0
1
0
0
0
0.
0.19
0
0
1
0
0
0.
0.19
1
0
0
0
0
0.
0. 04
b
1
0
0
0
0
0
0
X
S
S
W
S
W
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Analytical clarity
Ability to investigate what-if scenarios
Convenient treatment of processes
involving many streams and many
components
Advantages of Matrix Solution
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2. Multiple Unit Analysis
1 2
Two-distillation column
process to separate benzene
(B), toluene (T) and xylene
(X).
First column producesoverhead 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.
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Pretend that you haveno data.
Our Methodology
1 2 Then give a unique
name to each stream.
F
A
C
D
E
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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 of15unknowns (remember,
we pretend we have no
data!)
Our Methodology (contd)
1 2
FB, FT, FX
AB, AT, AX
CB, CT, CX
DB, DT, DX
EB, ET, EX
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Our Methodology (contd)
1 2
FB, FT, FX
AB, AT, AX
CB, CT, CX
DB, DT, DX
EB, ET, EX
Recall that around each
unit we can write as many
independent mass balances
as the number ofcomponents 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)
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Our Methodology (contd)
1 2
FB, FT, FX
AB, AT, AX
CB, CT, CX
DB, DT, DX
EB, ET, EX
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]!)
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Our Methodology (contd)
1 2
FB, FT, FX
AB, AT, AX
CB, CT, CX
DB, DT, DX
EB, ET, EX
Observe that without data
we cannot proceed,
because we have 6
equations and1
5unknowns!
Data can be translated into
auxiliary equations; we
need9
such equations andwe want them to be
independent!
Suppose they give us the
composition of stream A...
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Our Methodology (contd)
1 2
FB, FT, FX
AB, AT, AX
(4%B, 91%T, 5%X)
CB, CT, CX
DB, DT, DX
EB, ET, EX
Knowledge of stream A
composition allows us to write
no more than 2 auxiliary
equations, e.g.,
[7] AB/(AB+AT+AX) = 0.04
[8] AT/(AB+AT+AX) = 0.91
The following would not be
independent (why?)
AX/(AB+AT+AX) = 0.05 Generalize: knowing the
composition of a stream ofn
components affords us n-1
auxiliary equations.
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Our Methodology (contd)
1 2
FB, FT, FX
(35%B, 50%T, 15%X)
AB, AT, AX
(4%B, 91%T, 5%X)
CB, CT, CX
DB, DT, DX
(4.3%B, 91.2%T, 4.5%X)
EB, ET, EX
A basisprovides one more
auxiliary equation, e.g.:
[13] FB+F
T+F
X= 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 thefeed:
[14] EB = 0.1FB
[15] EX = 0.933FX
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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 andCX 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)
1 2
FB, FT, FX
(35%B, 50%T, 15%X)
AB, AT, AX
(4%B, 91%T, 5%X)
C, xCT, xCB
DB, DT, DX
(4.3%B, 91.2%T, 4.5%X)
EB, ET, EX
Suppose we used C, xCB and xCTto 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-linearequations
which are more difficult to
solve!
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3. Single Unit Balance with
Reaction
4NH3 + 5O2 4NO + 6H2O
A (NH3)
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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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)XTo 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)
Mass balances using the extent of
reaction...
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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 5chemical 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
Auxiliary equations from data...