Question

During a 3 h storm event, it was observed that all abstractions other than infiltration are negligible The rainfall was idealized as three 1 h storms of intensities 10 mm/h, 20 mm/h and 10 mm/h respectively and the infiltration was idealized as Horton curve f = 6.8 + 8.7 exp (-t ) (f is in mm/h and t is in h). What is the effective rainfall?

• A. 10.00 mm
• B. 11.33 mm
• C. 12.43 mm
• D. 13.63 mm

Answer 1

The correct option is D 13.63 mm

For the First Hour,
Rainfall = 10 mm
$\text{Infiltration}=\underset{0}{\overset{1}{\int }}f\cdot dt$
$=\underset{0}{\overset{1}{\int }}(6.8+8.7e^{-t})dt$
$=12.299mm$

Effective rainall in first hour
$p_1=10-12.299$
= -ve value
Hence, $p_1=0,$ since effective rainfall cannot be negative.

For the Second Hour,
Effective rainfall,
$p_2=\text{Rainfall - infiltration}$
$=20-\underset{1}{\overset{2}{\int }}(6.8+8.7e^{-t})dt$
$=20-8.823=11.177mm$

In Third Hour,
Effective rainfall
$p_3=10-\underset{2}{\overset{3}{\int }}(6.8+8.7e^{-t})dt$
$=10-7.544=2.456mm$

Total effective rainfall in 3hours
$p=p_1+p_2+p_3$
$=0+11.177+2.456$
$=13.633mm$