+ ( f o. (t) = f c. f c. )e kt (1)

Transcription

1 CIVE Basic Hydrology Jorge A. Ramírez 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). + ( f o )e t () For this soil, the asymptotic or final equilibrium infiltration capacity is f c =. cm/h; and the initial infiltration capacity is f o = cm/h. The rate of decay of infiltration capacity parameter is = h -. For the precipitation hyetograph tabulated below (Table ), carry out a complete infiltration analysis, including evaluation of cumulative infiltration and rate oroduction orecipitation excess, s + v. Table : hyetograph ) Compute infiltration capacity using Horton's equation for conditions of unlimited water supply at the surface using equation (Table - Column ). ) Graph hyetograph (from Table ) and infiltration capacity function (equation, from Table - Column ). See Figure. Rates (cm/min) Figure Rate

2 ) Compare infiltration capacity with precipitation intensity (Figure ). Observe that during the first minutes of the storm, the infiltration capacity exceeds the precipitation intensity. Thus, during the first minutes of the storm, all of the precipitation infiltrates. The actual infiltration rate is (Table - Column ), f (t) = min[i(t), (t)] = i(t), (t), i(t) (t) otherwise () ) Compute the accumulated infiltration that would occur under conditions of unlimited water supply at the surface using the following equation (Table - Column ), F p t + f o ( e t ) () ) Compute accumulated precipitation volume (see Table ) as a function of time. t n P(t) = i(t)dt ΔP j () The incremental volume over each time period of minutes is: Table : Intensity, i., P DP = i Dt j= Intensity, i., P ) Because i(t) <= (t) for t * <= minutes, all the rainfall infiltrates and the actual volume of infiltration at t * = minutes is equal to the volume of rainfall up to that time. Thus, F(t = t * ) = P(t = t * ) ) Because the actual infiltration rate is less than the infiltration capacity during the first 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 = min. To do so, first determine the time t p by solving the following equation:

3 t * F(t * ) = i(t)dt and then evaluate (t p ) as follows. = f c t p + f o ( e t p ) () At t * = min the actual volume of accumulated infiltration is (use equation ): F(t = min) = (. +.) cm/h ( min/ min/h) =. cm. Substituting this value for F(t * ) in equation and solving for t p obtain: t p =. h =. min. Finally, the true infiltration capacity at t * = minutes is obtained using equation as (t p ) =. cm/h = f op. Alternatively, use equations and to eliminate time and express cumulative infiltration as a function of infiltration capacity and obtain the following equation (see Figure ), F p (t) = ln( ) + f o f o () (cm/min) Figure ) The rainfall rate at t * = minutes i = cm/h exceeds the corresponding infiltration capacity f op =. 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 * = min (Table - Column and Table Column ). That is (see Figure and Figure ), + ( f op )e (t t * ) ()

4 Rates (cm/min) Rate Figure ) Because for t >= minutes the precipitation rate exceeds the infiltration capacity, there is excess precipitation available for runoff and depression storage, s + v (Table - Column ). σ + v = i(t) f (t) () Table - Eq. - Eq. - Eq.,, F, P

5 Table - Eq. - Eq. - Eq. - Eq., Intensity, i Rate, f(t) Excess s + v >