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Homework Help: 1-D Bounded, Distributed Diffusion of Contaminant

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm having some difficulty deriving the equation for a concentration of CO2 as a function of length and time. Ultimately I end up with an equation that includes the summation of two error function terms that appear to have incorrect signs.


    A cylinder of infinite length has a mass input, M0 of 1.0 g CO2 added over a width, L, of 5 cm at t = 0. The cylinder has a diameter of 5 cm. Assume a fluid temperature of 25°C.


    The time to reach a concentration of 2 ppm at x = 100 cm for:

    1.) Molecular diffusion in air
    2.) Molecular diffusion in water
    3.) If the fluid was turbulent, how would that affect your answers above? Substitute a value of 1 cm2/s for D as a reflection of uniformly generated (isotropic and homogenous) turbulence in water. How does this time compare to part 2.) above?

    Symmetry (d/dz = d/dy = 0)
    No flow (u = v = w = 0)
    Conservative tracer ( r = 0)

    2. Relevant equations

    At x = 0, the mass is evenly distributed between -L/2 and L/2

    Governing Equation:

    [itex] \frac{\delta c}{\delta t}=D \frac{\delta^{2} c}{\delta t^{2}}[/itex]

    Initial conditions:

    C = 0 for x < -L/2
    C = 0 for x > +L/2
    C = [itex]C_{0}[/itex] for -L/2 < x < +L/2

    Boundary conditions:

    C → 0 as x → ±∞

    3. The attempt at a solution

    Based on a dirac delta function:

    [itex] c \left( x, t \right) = \left[ \frac{M^{0}}{A \sqrt{4 \pi D t}} \right] exp \left[ - \frac{x}{4 D t} \right][/itex]

    Using superposition:

    [itex] dc = \frac{C_{0} A dx_{1}}{A \sqrt{4 \pi D t}} exp \left( - \frac { \left( x - x_{1} \right )^{2} }{4 D t} \right) dt [/itex]

    where A is the cross-sectional area of the cylinder, x1 is the location of the infinitesimal mass dm1

    Integrating the sources (of contaminants) from x1→-L/2 to x1→+L/2:

    [itex] c \left( x, t \right) = \int ^{L/2}_{-L/2} \frac{C_{0} }{ \sqrt{4 \pi D t}} exp \left( - \frac { \left( x - x_{1} \right )^{2} }{4 D t} \right) dx_{1} [/itex]

    [itex] c \left( x, t \right) = \frac{C_{0}}{\sqrt{\pi} } \left[
    \int ^{L/2}_{-\infty} \frac{1}{ \sqrt{4 D t}} exp \left( - \frac { \left( x - x_{1} \right )^{2} }{4 D t} \right) dx_{1} - \int ^{-L/2}_{-\infty} \frac{1}{ \sqrt{4 D t}} exp \left( - \frac { \left( x - x_{1} \right )^{2} }{4 D t} \right) dx_{1} \right] [/itex]

    Via substitution:

    where [itex] \eta = \frac{ \left( x - x_{1} \right) }{ \sqrt{4 D t}} [/itex] and [itex] d \eta = - \frac{dx_{1} }{ \sqrt{4 D t} } [/itex]

    [itex] c \left( x, t \right) = \frac{C_{0}}{\sqrt{\pi} } \left[
    \int ^{\infty}_{\frac{x-L/2}{\sqrt{4 D t}}} - exp \left( - \eta^{2} \right) d \eta - \int ^{\infty}_{\frac{x+L/2}{\sqrt{4 D t}}} - exp \left( - \eta^{2} \right) d \eta \right] [/itex]

    [itex] = \frac{C_{0}}{ 2 } \left[ - \left( 1 - erf \left( \frac{x-L/2}{\sqrt{4 D t}} \right) \right) + \left( 1 - erf \left( \frac{x+L/2}{\sqrt{4 D t}} \right) \right) \right] [/itex]

    [itex] = \frac{C_{0}}{ 2 } \left[ erf \left( \frac{x-L/2}{\sqrt{4 D t}} \right) - erf \left( \frac{x+L/2}{\sqrt{4 D t}} \right) \right] [/itex]

    From what I have seen, this is not correct though. The two error function terms should be switched such that:

    [itex] = \frac{C_{0}}{ 2 } \left[ erf \left( \frac{x+L/2}{\sqrt{4 D t}} \right) - erf \left( \frac{x-L/2}{\sqrt{4 D t}} \right) \right] [/itex]

    I'm wondering if I switched the signs somewhere or I made a mistake in the integration, perhaps the domain.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 24, 2013 #2
    The reference result you gave (with the signs switched) is correct. You must have made a mistake somewhere. It is much simpler to do the integration in a little different way than you have done. When you split it into two integrals, take one integral from -L/2 to 0, and the other integral from 0 to +L/2. The error functions with their correct signs will emerge virtually immediately from this.
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