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An integral relation in soil physics

  1. May 27, 2014 #1
    Hi every one,

    Here is my question: In soil physics, knowing the relation between suction head, h, and the soil water content, S, one can derive the hydraulic conductivity, K, of that soil using a formula like:

    1znpn6f.png

    (ignore the superscripts "cap")
    where in my problem, τ=0.5, κ=1, β=2.

    Now what if we need the reverse procedure? If I have the relation between h and K, and want to calculate S, how it is possible?


    Thanks in advance,
    Mohammad
     
  2. jcsd
  3. May 27, 2014 #2

    SteamKing

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    Iteration is one way.

    You know K on the LHS of the equation, and you have h as a function of x, along with the constants tau, kappa, and beta. The integral in the denominator seems to be a fixed quantity, but the value of the integral in the numerator depends on the value of S for the upper limit.

    The engineering approach, would be to assume different values of the parameter S and calculate the LHS and the RHS of the equation, or re-write the equation as LHS - RHS = 0. Then you can iterate until a particular value of S makes this relation arbitrarily close to zero.

    You could probably put together a spreadsheet to calculate the value of the expression for various input values of S.
     
  4. May 27, 2014 #3
    The issue here is in the reverse procedure we don't have any knowledge about the relation between h and x (or in fact h and S), so the integrals could not be calculated. How about this?
     
  5. May 27, 2014 #4

    pasmith

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    If you know [itex]h : [0,1] \to \mathbb{R}[/itex], [itex]\beta[/itex], [itex]\kappa[/itex], [itex]\tau[/itex] and [itex]K[/itex], then you can calculate [tex]
    I(S) = \left(\frac{\int_0^S h^{-\kappa}\,dx}{\int_0^1 h^{-\kappa}\,dx}\right)^{\beta}[/tex] for any [itex]0 \leq S \leq 1[/itex] and from that you can solve [tex]
    K = S^\beta I(S)[/tex] numerically for [itex]S[/itex]. (I'm assuming here that only values of [itex]S[/itex] between 0 and 1 inclusive make physical sense.)

    [itex]h[/itex] is a function of a real variable. If you can't determine [itex]h(x)[/itex] for every [itex]x[/itex] in the domain then there is no sense in which you know [itex]h[/itex]. Thus you can't calculate [itex]I(S)[/itex] for any [itex]S[/itex] other than the trivial [itex]I(0) = 0[/itex] (if [itex]\beta > 0[/itex]) and [itex]I(1) = 1[/itex]. If you can't calculate [itex]I(S)[/itex] you have no hope of solving [itex]K = S^{\tau}I(S)[/itex] for [itex]S[/itex].
     
  6. May 27, 2014 #5


    In [itex]h(x)[/itex], x is a dummy variable of integration (in calculating K , when the known relation [itex]h(S)[/itex] is available, S is replaced by x in the integrands. So [itex]h(x)[/itex] or [itex]h(S)[/itex] is exactly what we are looking for.
     
  7. Jun 11, 2014 #6
    I've gone a bit further to get this: (assuming τ=0.5, κ=1, β=2)

    [itex]\frac{\mathrm{d} }{\mathrm{d} S} \int_{0}^{S}\frac{1}{h(x))}dx=\int_{0}^{1}\frac{1}{h(x))}dx \cdot \frac{\mathrm{d} }{\mathrm{d} S} \left ( \sqrt{\frac{K_{r}}{\sqrt{S}}} \right ) [/itex]

    [itex]\Rightarrow

    \frac{1}{h(S))}=\int_{0}^{1}\frac{1}{h(x))}dx \cdot \frac{\sqrt{S}\frac{dK}{dh}\cdot \frac{dh}{dS}-\frac{K}{2\sqrt{S}}}{2\sqrt{\frac{K}{\sqrt{S}}}} [/itex]


    now knowing the K and [itex]\frac{dK}{dh} [/itex], the last equation seems to be an ODE but in this stage I don't know what to do with the [itex]\int_{0}^{1}\frac{1}{h(x))}dx[/itex] ?

    Can I in someway handle that?
     
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