CE322 Basic Hydrology

Jorge A. Ramírez

Infiltration Computations Example

Assume that the time evolution of the infiltration capacity for a given soil is governed by Horton's equation (Note that this equation assumes an infinite water supply at the surface, that is, it assumes saturation conditions at the soil surface).

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For this soil, the asymptotic or final equilibrium infiltration capacity is f_c = 1.25 cm/h; and the initial infiltration capacity is f_o = 8 cm/h. The rate of decay of infiltration capacity parameter is k = 3 h^-1. For the precipitation hyetograph tabulated below, carry out a complete infiltration analysis, including evaluation of cumulative infiltration and rate of production of precipitation excess, s + v.

1. Compute accumulated precipitation volume as a function of time. The incremental volume over each time period of 10 minutes is:

   DP = i Dt

   Time  (min)	Precipitation Intensity, i.  (cm/h)	Cumulative Precipitation, P (cm)	Time  (min)	Precipitation Intensity, i.  (cm/h)	Cumulative Precipitation, P (cm)
   0 - 10	1.5	0.25	40 - 50	4.0	3.583
   10 - 20	3.0	0.75	50 - 60	3.0	4.083
   20 - 30	8.0	2.083	60 - 70	0.8	4.217
   30 - 40	5.0	2.917

2. Compute infiltration capacity using Horton's equation for conditions of unlimited water supply at the surface using equation 1 (Table 1 - Column 2).
3. Compute the accumulated infiltration that would occur under conditions of unlimited water supply at the surface using the following equation 2 (Table 1 - Column 3),

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4. Compare infiltration capacity with precipitation intensity (Figure 1). Observe that during the first 20 minutes of the rainstorm, the infiltration capacity exceeds the precipitation intensity. Thus, during this period, all of the precipitation infiltrates. The actual infiltration rate is (Table 2 - Column 5), 

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5. Because the actual infiltration rate is less than the infiltration capacity during the first 20 minutes, the actual infiltration capacity does not decay as predicted by Horton's equation. This is because, as indicated above, Horton's equation assumes that the supply rate exceeds the infiltration capacity from the start of infiltration. Therefore, we must determine the true infiltration capacity at t = 20 min. To do so, first determine the time t_p by solving the following equation:

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   and then evaluate f_p(t_p) as follows. The left-hand side of equation 4 represents the accumulated volume of actual infiltration, while the right hand side of equation 4 represents the volume of infiltration that would have accumulated up to time t_p if the actual rate of infiltration had been equal to the infiltration capacity.

   At t = 20 min the actual volume of accumulated infiltration is:

   F(t = 20 min) = (1.5 + 3.0) cm/h (10 min/60 min/h) = 0.75 cm. Substituting this value for F(t) in equation 4 and solving for t_p obtain: t_p = 0.107 h = 6.41 min. Finally, the true infiltration capacity at 20 minutes is obtained using equation 1 as f_p(t_p) = 6.15 cm/h = f_op. Alternatively, using equations 1 and 2 to eliminate time and express cumulative infiltration as a function of infiltration capacity obtain the following equation,

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6. The rainfall rate at 20 minutes i = 8 cm/h exceeds the corresponding infiltration capacity f_op = 6.15 cm/h. Therefore, the actual infiltration rate equals the infiltration capacity, and the decay of infiltration capacity follows Horton's equation with an initial infiltration capacity equal to f_op and starting at time t_* = 20 min (Table 1 - Column 5 and Table 2 Column 3). That is (see Figure 2 and Figure 3),

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7. Because the precipitation rate exceeds the infiltration capacity, there is excess precipitation available for runoff and depression storage, s + v (Table 2 - Column 6).

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 Table 1