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7.9: Untitled Page 154

  • Page ID
    18287
  • Chapter 7

    often be discharged into the atmosphere. Determination of these molar flow rates requires the application of Axioms I and II. We begin with the atomic matrix

    Molecular Species  CH

    H

    CO CO

    H O O

    4

    2

    2

    2

    2

    carbon

     1

    0

    1

    1

    0

    0 

    (8)

    hydrogen

    4

    2

    0

    0

    2

    0

    oxygen

     0

    0

    1

    2

    1

    2 

    which leads to the following form of Axiom II:

    R

    CH4

     R

    H2

     

    A  6

    1 0 1 1 0 0

    0

      RCO 

     

    Axiom II:

    N R

    4 2 0 0 2 0

     

    0

    (9)

    JA

    A

     

     RCO

    A  1

    2

    0 0 1 2 1 2

    0

    RH

    2O 

    RO

    2 

    This axiom can be expressed in row reduced echelon form according to

    R

    CH4

     R

    H2

    1 0 0 1 1

    2

      

    0

      RCO 

     

    0 1 0

    2

    3

    4

     

    0

    (10)

     RCO 

    2

    0 0 1

    2

    1

    2

    0

     

    RH

    2O 

    RO

    2 

    Following the development in Sec. 6.4 indicates that the solution for the three non‐pivot global net rates of production is given by R

     R

     R

     2

    CH

    R

    4

    CO2

    H2O

    O2

    R

      2 R

     3R

     4

    H

    CO

    H O

    R

    (11)

    O

    2

    2

    2

    2

    R

      2 R

     R

     2

    CO

    CO

    R

    2

    H2O

    O2

    This completes our analysis of Axiom II, and we move on to the steady state form of Axiom I that is given by

    Material Balances for Complex Systems

    285

    Axiom I:

    c

    dA  R , A  1 , 2 , ..., N

    (12)

    v n

    A

    A

    A

    A

    Application of this result to the control volume illustrated in Figure 7.3

    leads to the general expression

     ( M )  ( M )  ( M )

     R ,

    A  1 , 2 , ... , 7

    (13)

    A 1

    A 2

    A 3

    A

    while the individual species balances take the forms given by CH :

     ( M

    )

      0 004

    .

    M

     R

    (14)

    4

    CH4 1

    1

    CH4

    H :

     ( M )

     0 528

    .

    M

     R

    (15)

    2

    H2 1

    1

    H2

    CO :

     ( M )

     0 383

    .

    M

     R

    (16)

    CO 1

    1

    CO

    CO :

     ( M

    )  ( M

    )

     R

    (17)

    2

    CO2 1

    CO2 3

    CO2

    N :

     ( M )  ( M )  ( M )

     R

    (18)

    2

    N2 1

    N2 2

    N2 3

    N2

    O :

     ( M )  ( M )  ( M )

     R

    (19)

    2

    O2 1

    O2 2

    O2 3

    O2

    H O :

    ( M

    )

     R

    (20)

    2

    H2O 3

    H2O

    At this point we need to simplify the balance equations for carbon dioxide, nitrogen and oxygen. Beginning with carbon dioxide, we obtain CO :

    ( M

    )

     R

     0 055

    .

    M

    (21)

    2

    CO2 3

    CO2

    1

    Moving on to the nitrogen balance given by Eq. 18, we make use of the conditions

    N :

    ( M

    ) 

    0 0

    . 29 M ,

    ( M

    ) 

    0 7

    . 9 M ,

    R

     0

    (21)

    2

    N2 1

    1

    N2 2

    2

    N2

    to express the molar flow rate in the flue gas as

    N :

    ( M

    )

     ( M )  ( M )

     0 0

    . 29 M

     0 . 79 M (22)

    2

    N2 3

    N2 1

    N2 2

    1

    2

    Finally, the use of Eq. 7 leads to the following result for the molar flow rate of nitrogen:

    N :

    ( M

    )

     1 . 94 M

    (23)

    2

    N2 3

    1

    286