Question

The Rational Method formula for estimating peak runoff is given as \(Q = \frac{CiA}{360}\). For the result \(Q\) to be in cubic meters per second (\(m^3/s\)), what must be the units of rainfall intensity (\(i\)) and catchment area (\(A\))?

• A. \(i\) in \(cm/hr\), \(A\) in \(km^2\)
• B. \(i\) in \(mm/hr\), \(A\) in \(km^2\)
• C. \(i\) in \(mm/hr\), \(A\) in hectares
• D. \(i\) in \(m/hr\), \(A\) in hectares

Hint:

The Rational Method formula appears in different forms depending on the units used. The two most common forms are: 1. \(Q = \frac{CiA}{360}\) for \(Q\) in \(m^3/s\), \(i\) in \(mm/hr\), and \(A\) in hectares. 2. \(Q = \frac{CiA}{3.6}\) for \(Q\) in \(m^3/s\), \(i\) in \(mm/hr\), and \(A\) in \(km^2\). Remembering these two standard forms can help you quickly solve such unit-based problems.

Answer 1

The Correct Option is C

Step 1: Dimensional Check:
We need to determine which units for \(i\) and \(A\) result in a conversion constant of 360 when calculating discharge \(Q\) in \(m^3/s\). The runoff coefficient \(C\) is unitless.
Step 2: Unit Conversion Analysis:
Let's test option (C): \(i\) in \(mm/hr\) and \(A\) in hectares (ha).
- Convert \(i\): \(1 \, mm/hr = \frac{10^{-3} \, m}{3600 \, s}\)
- Convert \(A\): \(1 \, hectare = 10,000 \, m^2\)
Now, multiply the variables: \[ Q = C \times \left( \frac{10^{-3} \, m}{3600 \, s} \right) \times (10,000 \, m^2) \] \[ Q = C \times \frac{10}{3600} \frac{m^3}{s} \] \[ Q = \frac{C \times i \times A}{360} \] This matches the formula provided in the question.
Step 3: Verification of Other Options:
If \(A\) was in \(km^2\) (\(10^6 \, m^2\)) and \(i\) in \(mm/hr\), the numerator would be \(10^3\), leading to \(Q = \frac{CiA}{3.6}\). Since the denominator in the question is 360, the area must be in hectares.
Thus, the correct units are \(mm/hr\) for intensity and hectares for area.