The initial infiltration capacity (fo) of a catchment is estimated as 5.5 mm/hr, the time constant as 0.45/hour, and the capacity (fc) as 0.5 mm/hr. Use Horton’s equation to find;
a) The value (ft) at t = 15 mins, 30 mins, 45 mins, 1 hr, and 3 hrs.
b) The total volume of infiltration over the 3 hrs period. Assume continuously ponded conditions.
To solve this problem, we will use Horton's equation. Horton's equation describes the rate of infiltration into the soil as a function of time. It is given by the equation:
I(t) = fo - (fo - fc) * e^(-kt)
Where:
- I(t) is the infiltration rate at time t.
- fo is the initial infiltration capacity.
- fc is the final infiltration capacity.
- k is the time decay constant.
- t is the time in hours.
a) To find the infiltration rate at different time intervals, we can substitute the given values into Horton's equation:
At t = 15 mins:
Convert 15 minutes to hours: 15 mins ÷ 60 = 0.25 hours
I(0.25) = fo - (fo - fc) * e^(-kt)
At t = 30 mins:
Convert 30 minutes to hours: 30 mins ÷ 60 = 0.5 hours
I(0.5) = fo - (fo - fc) * e^(-kt)
At t = 45 mins:
Convert 45 minutes to hours: 45 mins ÷ 60 = 0.75 hours
I(0.75) = fo - (fo - fc) * e^(-kt)
At t = 1 hr:
I(1) = fo - (fo - fc) * e^(-kt)
At t = 3 hrs:
I(3) = fo - (fo - fc) * e^(-kt)
Substitute the given values: fo = 5.5 mm/hr, fc = 0.5 mm/hr, k = 0.45/hr
Now, we will calculate the values of I(t) at each time interval using the equation.
b) To find the total volume of infiltration over the 3-hour period, we need to integrate the infiltration rate over time. The volume of infiltration (V) is given by the equation:
V = ∫[fo - (fo - fc) * e^(-kt)] dt
We will integrate the infiltration rate equation with respect to time from 0 to 3 hours.
V = ∫[fo - (fo - fc) * e^(-kt)] dt from 0 to 3
By evaluating this integral, we will find the total volume of infiltration over the 3-hour period.
Now, let's substitute the given values into the equations and calculate the requested values.