Question

The parameter in Horton's infiltration equation $\left[f\right(t)=f_c+(f_0-f_c\left)e^{-kt}\right]$ are given as, $f_0=7.62cm/hour$ , $f_c=1.34cm/hour$, and $k=-4.182/hour$. For assumed continuous ponding the cumulative infiltration at the end of 2 hours is

• A. 2.68 cm
• B. 1.50 cm
• C. 1.34 cm
• D. 4.18 cm

Answer 1

The correct option is D 4.18 cm

Horton's infiltration equation,
$f_t=f_c+(f_o-f_c)e^{-kt}$
$f_t=1.34+(7.62-1.34)e^{-4.182t}$
$f_t=1.34+6.28e^{-4.182t}$
Cumulative infiltration at the end of 2 hours
=$\underset{0}{\overset{2}{\int }}{f}_{t}dt$
=$\underset{0}{\overset{2}{\int }}(1.34+6.28e^{-4.182t})dt$
= 4.18 cm