Page 263 - Advanced Mine Ventilation
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240                                               Advanced Mine Ventilation

            Thakur [1] has shown mathematically that

                          1
              M t   6    D  2  1
                 ¼ p ffiffiffi  2  t 2                                       (14.16)
             M N     p  a
         where M t is the volume of gas diffused in time t; M N is the Langmuir volume, V m ,
         discussed earlier, and t is the time.
                             1
            A plot of  M  against t / 2 will yield a straight line. The gradient of the straight line is
                   M N
                       1
                       2

                 6   D
                p    2 .
                  p  a
         equal to ffiffiffi
            The time taken for a piece of coal to desorb (1 e 1/e) or 63.21% of gas is called its
         “sorption time” or s. This expresses the rate of desorption in mining parlance better
                                      2
         than the absolute value of D or (D/a ).
            Modifying Eq. (14.16), we can write
                                                    1
                                M  =  (  1−  1  )  =  6 ⎛  ⎞ D  2  τ 1 2  (14.17)
                               M ∞       e   π ⎝  ⎜  a ⎠  2 ⎟
            Rearranging and solving for s, we get

                 3:49   10  2
             s ¼       2                                               (14.18)
                   ðD=a Þ
            Another way to determine s is to solve the equation.


                               n
              M             t
                 ¼ 1   exp                                             (14.19)
             M N            s
            To illustrate and compare Eqs. (14.17) and (14.19), a typical gas desorption curve
         shown in Fig. 14.9 was plotted both ways and results were compared.
            The plot of curve in Fig. 14.9 using Eq. (14.17) results in a straight line with an
         equation:

              M             1
                 ¼ 0.000436 s 2                                        (14.20)
             M N
                        2
            This gives (D/a ) equal to 1.6589   10   8  1  and s ¼ 24.35 days.
                                             s
            Eq. (14.19) was rewritten as
                 0       1
                      1
                 B       C
             ln ln B     C  ¼ nlnt   nln s                             (14.21)
                 @     M t  A
                   1
                      M N
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