Question

The intensity-duration relationship for a rainfall on a rectangular plot ABCD of area 7 ha (\(1 \, \text{ha} = 10^4 \, \text{m}^2\)) can be modelled by the following equation:
\[ I = \frac{25}{(t+10)} \]
where \(I\) is rainfall intensity (in cm/h), and \(t\) is the duration (in minutes) of rainfall. The average runoff coefficient over the area is 0.60. The time of entry (in minutes) to the outfall from the corners A, B, C, and D is 10, 20, 15 and 25, respectively.

The design flowrate (in \(m^3/h\)) of the storm-sewer at the outfall is _____ (in integer).

Hint:

Use the longest time of entry among A, B, C, D as the design duration \(t\), then apply \(Q = CIA\) after making the units of \(I\) and \(A\) consistent.

Answer 1

Correct Answer: 300

Step 1: Set up with the metric Rational formula.
A common alternate form of the Rational formula lets you work directly in hectares and mm/h, without hand converting the area to $m^2$: $Q = \dfrac{C I A}{360}$, where $C$ is the runoff coefficient, $I$ is intensity in mm/h, $A$ is the area in hectares, and $Q$ comes out in $m^3/s$.

Step 2: Pick the design duration.
The outfall receives flow from all of plot ABCD only once water from the slowest corner has arrived. The entry times from A, B, C, D are 10, 20, 15, 25 minutes, so the full catchment is contributing only after 25 minutes, the largest of the four. So the design duration is $t = 25$ min.

Step 3: Compute the design intensity in mm/h.
$I = \dfrac{25}{t+10} = \dfrac{25}{35} = 0.7143$ cm/h $= 7.143$ mm/h.

Step 4: Substitute into the metric formula.
With $C = 0.60$ and $A = 7$ ha:
$Q = \dfrac{0.60 \times 7.143 \times 7}{360} = \dfrac{30.0}{360} = 0.08333 \ m^3/s$

Step 5: Convert to $m^3/h$.
There are 3600 seconds in an hour:
$Q = 0.08333 \times 3600 = 300 \ m^3/h$
This route and the direct unit conversion route both land on the same number, which checks the arithmetic.
\[ \boxed{Q = 300 \ m^3/h} \]