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5.1: Untitled Page 91

  • Page ID
    18224
  • Chapter 5

    m

    m

     (

    ) v A

    o

    1

    CH

    o o

    o

    4

    (10)

     

    3

    133 . 7 kg/m 15 . 24 m/s

    2

    0 . 2027 m   413 kg/s

    from which we determine the constant molar flow rate to be

    ( m

    )

    CH

    o

    413 kg/s

    4

    3

    M

    M

     25 . 74 10 mol/s (11)

    o

    1

    MW

    16 . 043 g/mol

    CH4

    In order to use Eq. 5‐3 to estimate the density of a pure gas, we divide both sides by VRT and multiply both sides by MW to obtain n MW

    p MW

     

    (5‐13)

    V

    R T

    For an ideal gas mixture, one uses the definitions of the total mass density of a mixture (Eq. 5‐1b) and total pressure (Eq. 5‐4) along with Eq. 5‐2 to obtain A N

    A N

    A N

    n MW

    p MW

    p MW

    A

    A

    A

    A

     

    (5‐14)

    A

    V

    R T

    R T

    A  1

    A  1

    A  1

    Here we have used MW to represent the molar average molecular mass defined by

    A N

    A N

    1

    MW

    p MW

    y

    MW

    (5‐15)

    A

    A

    A

    A

    p A 1

    A  1

    These results are applicable when molecule‐molecule interactions are negligible, and this occurs for many gases under ambient conditions. At low temperatures and high pressures gases depart from ideal gas behavior. Under those conditions one should use more accurate equations of state such as those presented in a standard course on thermodynamics (Sandler, 2006).

    EXAMPLE 5.2. Molecular mass of air

    The air we breathe has a composition that depends on position. Air pollution sources abound and these sources add minute amounts of chemicals to the atmosphere. Combustion of fuels in cars and power plants are a source for sulfur dioxide, oxides of nitrogen, and carbon monoxide. Chemical industries add pollutants such as ammonia, chlorine, and even hydrogen cyanide to the atmosphere. In some locations, there

    Two‐Phase Systems & Equilibrium Stages 163

    are minute amounts of other gases such as argon, helium, and radon.

    Standard dry air, for the purpose of combustion computations (see Sec. 7.2), is assumed to be a mixture of 79% by volume of nitrogen and 21% by volume of oxygen. In this example, we want to determine the molar average molecular mass of standard dry air.

    For an ideal gas, the volume percentage is also the molar percentage.

    Thus, the volume percentages of nitrogen and oxygen can be simply translated to mole fractions according to

    79% by volume of nitrogen  y

     0 . 79

    (1)

    N2

    21% by volume of oxygen  y

     0 . 21

    (2)

    O2

    We can use Eq. 5‐15 to compute the molar average molecular mass of standard dry air according to

    MW

    MW

    y

    MW

    y

    MW

     28 . 85 g/mol

    air

    N2

    N2

    O2

    O2

    Here we have used the subscript “air” to represent standard dry air as described by Eqs. 1 and 2.

    5.2 Liquid Properties and Liquid Mixtures

    When performing material balances for liquid systems, one must have access to reliable liquid properties. Unlike gases, the densities of liquids are weak functions of pressure and temperature, i.e., large changes in pressure and temperature result in small changes in the density. The changes in liquid density due to changes in pressure are determined by the coefficient of compressibility which is defined by

     coefficient of 

    1    

      

    (5‐16)

    compressibility

     

    p T

    Changes in liquid density due to changes in temperature are determined by the coefficient of thermal expansion which is defined by

    coefficient of thermal

    1    

      

     

    (5‐17)

     expansion

     

    T p

    The coefficient of thermal expansion,  , is defined with a negative sign since the density of most liquids decreases with increasing temperature. Using a negative

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