polykin.transport.diffusion

uptake_convection_sheet

uptake_convection_sheet(Fo: float, Bi: float) -> float

Fractional mass uptake for transient diffusion in a plane sheet subjected to a surface convection boundary condition.

For a plane sheet of thickness \(2a\), with diffusion from both faces, where the concentration is initially \(C_0\) everywhere, and the flux at the surface is:

\[ -D \left.\frac{\partial C}{\partial x} \right|_{x=a}=k(C(a,t)-C_{\infty}) \]

the fractional mass uptake is:

\[ \frac{\bar{C}-C_0}{C_{\infty} -C_0} = 1 - 2 Bi^2 \sum_{n=1}^{\infty}\frac{1}{\beta_n^2[\beta_n^2+Bi(Bi+1)]} \exp (-\beta_n^2 Fo) \]

where \(Fo = D t/a^2\) is the Fourier number, \(Bi = k a/D\) is the Biot number, and \(\beta_n\) are the positive roots of the transcendental equation \(\beta \tan(\beta) = Bi\).

References

• J. Crank, "The mathematics of diffusion", Oxford University Press, 1975, p. 60.

PARAMETER	DESCRIPTION
Fo	Fourier number, \(D t/a^2\). TYPE: float
Bi	Biot number, \(k a/D\). TYPE: float

RETURNS	DESCRIPTION
float	Fractional mass uptake.

See also

• uptake_constc_sheet: related method for constant surface concentration boundary condition.

Examples:

Determine the fractional mass uptake after 500 seconds for a polymer solution film with a thickness of 0.2 mm, a diffusion coefficient of 1e-11 m²/s, and an external mass transfer coefficient of 1e-6 m/s.

>>> from polykin.transport import uptake_convection_sheet
>>> t = 5e2   # s
>>> a = 2e-4  # m
>>> D = 1e-11 # m²/s
>>> k = 1e-6  # m/s
>>> Fo = D*t/a**2
>>> Bi = k*a/D
>>> uptake_convection_sheet(Fo, Bi)
0.3528861780625614

Source code in src/polykin/transport/diffusion.py

def uptake_convection_sheet(Fo: float, Bi: float) -> float:
    r"""Fractional mass uptake for transient diffusion in a plane sheet
    subjected to a surface convection boundary condition. 

    For a plane sheet of thickness $2a$, with diffusion from _both_ faces,
    where the concentration is initially $C_0$ everywhere, and the flux at the
    surface is:

    $$ -D \left.\frac{\partial C}{\partial x} \right|_{x=a}=k(C(a,t)-C_{\infty}) $$

    the fractional mass uptake is:

    $$ \frac{\bar{C}-C_0}{C_{\infty} -C_0} =
        1 - 2 Bi^2 \sum_{n=1}^{\infty}\frac{1}{\beta_n^2[\beta_n^2+Bi(Bi+1)]}
        \exp (-\beta_n^2 Fo) $$

    where $Fo = D t/a^2$ is the Fourier number, $Bi = k a/D$ is the Biot
    number, and $\beta_n$ are the positive roots of the transcendental equation 
    $\beta \tan(\beta) = Bi$.

    **References**

    * J. Crank, "The mathematics of diffusion", Oxford University Press, 1975,
      p. 60.

    Parameters
    ----------
    Fo : float
        Fourier number, $D t/a^2$.
    Bi : float
        Biot number, $k a/D$.

    Returns
    -------
    float
        Fractional mass uptake.

    See also
    --------
    * [`uptake_constc_sheet`](uptake_constc_sheet.md): related method for
      constant surface concentration boundary condition.

    Examples
    --------
    Determine the fractional mass uptake after 500 seconds for a polymer solution
    film with a thickness of 0.2 mm, a diffusion coefficient of 1e-11 m²/s, and
    an external mass transfer coefficient of 1e-6 m/s.
    >>> from polykin.transport import uptake_convection_sheet
    >>> t = 5e2   # s
    >>> a = 2e-4  # m
    >>> D = 1e-11 # m²/s
    >>> k = 1e-6  # m/s
    >>> Fo = D*t/a**2
    >>> Bi = k*a/D
    >>> uptake_convection_sheet(Fo, Bi)
    0.3528861780625614
    """
    if Fo and Bi:
        N = 4  # Number of terms in series expansion (optimal value)
        x = roots_xtanx(Bi, N)
        b2 = x**2
        S = sum(exp(-b2[n]*Fo)/(b2[n]*(b2[n] + Bi*(Bi + 1)))
                for n in range(0, N))
        return 1 - (2*Bi**2)*S
    else:
        return 0