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2.9: Untitled Page 21

  • Page ID
    18154
  • Chapter 2

    a   0 2 0 0 

    (2‐36b)

    Pressure is the force per unit area, and we can obtain the fundamental units of pressure by dividing the force by the area to obtain

     units of 

    2

    M L T

    M

     

    M L T

    (2‐37a)

     pressure

     2 L

    1

    1

    2

    2

    LT

    Since division by the units is equivalent to subtraction of the exponents, the 1 4

    array of exponents for pressure can be obtained by

    p  f  a

      1 1 2 0 

    (2‐37b)

    in which we have used p to indicate the array of exponents for pressure.

    EXAMPLE 2.1. Dimensions of the kinematic viscosity

    The kinematic viscosity is defined as the ratio of the viscosity,  , to the density,  , and it is designated by the symbol  . To be explicit, we express the kinematic viscosity as

      

    (1)

    The units of the kinematic viscosity can be obtained from the matrices of exponents for  and  , and from Table 2‐4 we see that these are given by mu   1

    1

    1

    0  ,

    rho   1 3 0 0 

    (2)

    The division indicated in Eq. 1 indicates that we need to subtract these matrices to determine that the units of the kinematic viscosity, i.e., nu  mu  rho

    (3)

    Carrying out the subtraction of the two 1 4 matrices given by Eqs. 2

    leads to

    nu   1 1 1 0    1 3 0 0    0 2 1 0 

    (4)

    This indicates that the units of the kinematic viscosity are given by 2

     units of kinematic

    L

    2

    1

    L T

    (5)

     viscosity

    T

    and one example is given in Table 2‐4 where we see that the kinematic viscosity can be expressed in terms of meters squared per second.

    Units

    29

    EXAMPLE 2.2. Composition of a gas mixture

    We have a gas mixture consisting of 5 kg of methane, 10 kg of ethane, 5

    kg of propane, and 3 kg of butane, and we wish to know the number of moles of each component in the mixture. The number of moles are determined by dividing the mass by the molecular mass of each individual component and these molecular masses are given by

    M

    m

    W ethane

    16.0433 g/mol , M eth

    W ane

    30.07 g/mol

    propane

    MW

    44.097 g/mol ,

    b

    MW utane

    58.124 g/mol

    Here we have two lists or arrays and the result we wish can be obtained by an array division as indicated by Eq. 2‐26. Alternatively the results can be obtained by hand according to

    moles of 

    mass of methane

    5 kg

     

     311.7 mol

    methane

    molecular mass of methane

    16.043 g/mol

    moles of

    mass of ethane

    10 kg

     

     332.6 mol

     ethane 

    molecular mass of ethane

    30.07 g/mol

    moles of

    mass of propane

    5 kg

     

     113.4 mol

    propane 

    molecular mass of propane

    44.097 g/mol

    moles of

    mass of butane

    3 kg

     

     51.6 mol

     butane 

    molecular mass of butane

    58.124 g/mol

    It should be clear that repetitive operations of this type area especially suited to computer assisted calculations.

    2.6 Matrix Operations

    A matrix is a special type of array with well‐defined algebraic laws for equality, addition, subtraction, and multiplication. Matrices are defined as the set of coefficients of a system of linear algebraic equations or as the coefficients of a linear coordinate transformation. A linear algebraic system of three equations is given by

    30