Chapt4 Balances

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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...