Infiltration is a complex process; however, numerical solution of the Richards equation is expensive. MIKE SHE includes two additional UZ models that require surface infiltration to be calculated: the 2-layer Water Balance (2LWB) method and the Gravity Flow (GF) method. In both of these methods, infiltration is calculated based on the infiltration capacity of the soil and ponding occurs if the rainfall rate exceeds the infiltration capacity. This method neglects the very important water adsorption processes that occur in dry soils, which means that the amount of infiltration is underestimated in dry soils and when the water table is relatively deep.
The Green and Ampt (GA) method is a widely used, physics based approximation to the Richards solution for calculating infiltration. As an infiltration model, the GA method can be used to calculate the infiltration in both the 2LWB and the GF UZ methods.
The GA method results in a high infiltration rate to dry soils. This is followed by ponding on the ground surface as the upper zone becomes saturated. This is followed by a slowly decreasing infiltration rate that is ultimately limited by the saturated hydraulic conductivity.
The Green and Ampt method assumes:
· an infinite thickness of soil
· no water table, capillary fringe, or lower soil boundary
· uniform soil characteristics with depth
· uniform water content with depth at t=0
· insignificant depth of ponded water
The necessary input parameters are:
· saturated hydraulic conductivity
· initial water content
· saturated water content
· rainfall rate, w
· effective soil suction ahead of wetting front
The GA method generates three key outputs:
· f(t), the infiltration rate at time t
· F(t), the total infiltration at time t
· the time until ponding after the start of the time step
The GA method distinguishes between two main cases:
Case 1: Rainfall rate less than the saturated hydraulic conductivity
In this case, no ponding occurs and the infiltration rate equals the rainfall rate and the total infiltration equals the rainfall rate times the time step length. Thus,
f(t) = w
F(t) = w t
Case 2: Rainfall rate greater than saturated hydraulic conductivity.
Initially, rainfall will be absorbed into the soil and ponding will occur when the top of the unsaturated zone becomes saturated. However, the initial infiltration rate will be greater than the saturated hydraulic conductivity because the soil will adsorb water ahead of the wetting front. Thus, ponding will occur after the start of the timestep. Prior to the start of ponding Case 1 applies. After ponding starts, the infiltration rate starts to decrease, and the infiltration time is a function of the total amount of infiltration. This must be solved implicitly to find the amount of infiltration corresponding to the time step.
Since the GA method assumes that the infiltration occurs as a sharp front plug flow, the depth of infiltration can be easily calculated by dividing the original deficit by the amount of infiltration.
Actually, there are a couple of special cases in addition to the two main cases.
Case 3: Initial ponding
The GA method assumes that there is no ponding on the ground surface. In fact, the infiltration rate depends on the degree of ponding but the ponding level is generally neglected. This is a reasonable assumption since the ponding due to rainfall is generally negligible. However, in MIKE SHE there is often ponding at the beginning of the time step due to flooding. In this case, the we can assume that the top of the soil is saturated and the infiltration rate will equal Ks.
Case 4: Rainfall + Ponded storage depleted during the time step
In some cases, there may be a small amount of initial ponded storage in the cell. This should be infiltrated as per Case 3, but once this has been depleted, ponded storage infiltration will cease if the rainfall rate is less than Ks. If the rainfall rate is greater than this, then water will continue to pond and infiltrate at the rate equal to Ks. However, if the rainfall rate is less than Ks, then the infiltration rate should continue at the rainfall rate.
Case 5: Insufficient UZ storage
The total amount of infiltration cannot exceed the amount of available UZ storage, so there must be a check to see that this amount is not exceeded. Excess must be added to OL.