Question

In Horton's equation fitted to the infiltration data for a soil, the initial infiltration capacity is 10 mm/h; final infiltration capacity is 5 mm/h; and the exponential decay constant is 0.5 /h. Assuming that the infiltration takes place at capacity rates, the total infiltration depth (in mm) from a uniform storm of duration 12 h is \underline{\hspace{2cm}. (round off to one decimal place)}

Hint:

To calculate the total infiltration depth using Horton's equation, remember to integrate the infiltration rate over the time period of interest.

Answer 1

Correct Answer: 69.7

The total infiltration depth \( D \) is calculated using Horton's equation, which gives the infiltration rate as a function of time: \[ f(t) = f_c + (f_0 - f_c) e^{-kt} \] Where: - \( f(t) \) is the infiltration capacity at time \( t \), - \( f_0 \) is the initial infiltration capacity, - \( f_c \) is the final infiltration capacity, - \( k \) is the exponential decay constant, - \( t \) is time in hours. Given values: - Initial infiltration capacity, \( f_0 = 10 \, \text{mm/h} \), - Final infiltration capacity, \( f_c = 5 \, \text{mm/h} \), - Exponential decay constant, \( k = 0.5 \, \text{/h} \), - Duration of storm, \( t = 12 \, \text{h} \). Step 1: Total Infiltration Depth The total infiltration depth \( D \) is the integral of the infiltration rate \( f(t) \) over time from 0 to \( t \), i.e.: \[ D = \int_0^t f(t) \, dt \] Substitute the equation for \( f(t) \): \[ D = \int_0^{12} \left( f_c + (f_0 - f_c) e^{-kt} \right) dt \] This is split into two integrals: \[ D = \int_0^{12} f_c \, dt + \int_0^{12} (f_0 - f_c) e^{-kt} \, dt \] Step 2: Solving the Integrals First, solve the integral of the constant term \( f_c \): \[ \int_0^{12} f_c \, dt = f_c \times 12 = 5 \times 12 = 60 \, \text{mm} \] Now, solve the integral of the exponential term: \[ \int_0^{12} (f_0 - f_c) e^{-kt} \, dt = (f_0 - f_c) \left[ \frac{e^{-kt}}{-k} \right]_0^{12} \] \[ = (10 - 5) \times \left( \frac{e^{-0.5 \times 12} - 1}{-0.5} \right) \] \[ = 5 \times \left( \frac{e^{-6} - 1}{-0.5} \right) \] \[ = 5 \times \left( \frac{0.002478752 - 1}{-0.5} \right) \] \[ = 5 \times \left( \frac{-0.997521248}{-0.5} \right) \] \[ = 5 \times 1.995042496 \approx 9.98 \, \text{mm} \] Step 3: Total Infiltration Depth The total infiltration depth is: \[ D = 60 + 9.98 = 69.98 \, \text{mm} \] Thus, the total infiltration depth is approximately 70.0 mm. \[ \boxed{69.7 \, \text{mm} \, \text{to} \, 70.1 \, \text{mm}} \]