Concept: Rational method and time of concentration.
The storm-sewer at the outfall collects runoff from the whole rectangular plot ABCD. In the Rational Method, the design rainfall duration used to find the intensity \(I\) must equal the time of concentration \(t_c\), the time needed for runoff from the farthest corner to reach the outfall. Using a shorter duration would mean part of the catchment has not yet started sending water to the outfall, so the whole-area design flow would come out wrong.
Step 1: Find the governing time of concentration.
The time of entry to the outfall from corners A, B, C and D is 10, 20, 15 and 25 minutes. The corner that takes the longest to drain is the one that fixes \(t_c\), since only after that time has the entire plot started contributing water to the outfall.
\[ t_c = \max(10, 20, 15, 25) = 25 \text{ min} \]
Step 2: Find the design rainfall intensity.
Put \(t = t_c = 25\) min into the given intensity-duration equation:
\[ I = \frac{25}{t+10} = \frac{25}{25+10} = \frac{25}{35} = 0.7143 \text{ cm/h} \]
Step 3: Convert the intensity to metres per hour.
The area is in \(m^2\), so the intensity must be changed to m/h to get \(Q\) directly in \(m^3/h\). Since \(1 \text{ cm} = 0.01 \text{ m}\):
\[ I = 0.7143 \times 0.01 = 0.0071429 \text{ m/h} \]
Step 4: Apply the Rational formula.
The Rational formula for peak design discharge is \(Q = C \, I \, A\), where \(C\) is the runoff coefficient, \(I\) is the rainfall intensity, and \(A\) is the catchment area. Here \(A = 7 \text{ ha} = 7 \times 10^4 \text{ m}^2 = 70000 \text{ m}^2\) and \(C = 0.60\).
\[ Q = 0.60 \times 0.0071429 \times 70000 \]
\[ Q = 0.60 \times 500 = 300 \text{ m}^3/\text{h} \]
Final Answer:
The design flowrate of the storm-sewer at the outfall is 300 m3/h.
\[ \boxed{Q = 300 \text{ m}^3/\text{h}} \]