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Green & Ampt equation
At any time, $t$, the penetration of the infiltrating wetting front will be $Z$. Darcy's law can be stated as follows:
$$ \begin{equation*} q = \frac{dI}{dt} = -K_s * \left[\frac{h_f-(h_s+Z)}{Z}\right] \end{equation*} $$ where $K_s$ is the hydraulic conductivity corresponding to the surface water content, and $I(t)$ is the cumulative infiltration at time $t$, and is equal to $Z*(\theta_s - \theta_0)$.
Using this relation for $I(t)$ to eliminate $Z$ and performing the integration yields,
$$ \begin{equation*} I = K_s*t-(h_f-h_s)*(\theta_s - \theta_0)* log_e \left( 1 - \frac{I}{(h_f-h_s)*(\theta_s-\theta_0)}\right) \end{equation*} $$ $$ \begin{table}
\centering \begin{tabular}{ l l l l }
with & $I(t)$ & infiltration amount & $[cm]$
& $K_s$ & hydr. conductivity & $[cm/h]$ \\ & $h_f$ & wetting front pressure head (negative) & $cm$ \\ & $h_s$ & water pressure at surface (ponding) & $cm$ \\ & $\theta_s$ & moisture content at saturation & $-$ \\ & $\theta_0$ & antecedent moisture & $-$ \\
\end{tabular} \end{table} $$ In order to solve this equation, we need to bring $I(t)$ to one side of the equation:
$$ \begin{equation*} \frac{1}{K_s}*\left[I -(h_f-h_s)*(\theta_s - \theta_0)* log_e \left( 1 - \frac{I}{(h_f-h_s)*(\theta_s-\theta_0)}\right)\right] = t \end{equation*} $$
The R-program to calculate infiltration amounts with Green & Ampt looks like this:
- |Green-Ampt.R
I <- seq(0,100,by=1.0) t0 <- 0.05 ts <- 0.25 hs <- 0.0 # cm hf <- -12.0 # cm Ks <- 8.0 # cm/hour t <- 1/Ks*((I-(hf-hs)*(ts-t0))*log(1-(I/((hf-hs)*(ts-t0))))) # hours plot(t,I,xlim=c(0,6),ylim=c(0,25),xlab="t [hour]", ylab="I in [cm]")