Skip to main content

Rational method peak flow, Melamchi storm

Short answer

Time of concentration by Kirpich formula is 84.2 min. Rainfall intensity for this duration is 50.0 mm/h from the storm hyetograph. Peak flow by rational method Qp = 0.278 × 0.6 × 50.0 × 100 = 834 m³/s.

Consider a rainfall storm in Melamchi during a recent flood event spanned for a period of 160 minutes. Observed records showed that during the first 30 minutes, 50 minutes, 84 minutes and 100 minutes the rainfall depths were 25 mm, 40 mm, 70 mm and 80 mm respectively. If the catchment area of the watershed is 100 km2^2, length of the longest flow path is 6 km, general slope of the catchment is 0.013 and runoff coefficient is 0.6, estimate the peak flow using the Rational method.

Rational formula: Qp=0.278 C i AQ_p = 0.278\,C\,i\,A (Q in m³/s, i in mm/h, A in km²), where ii is the intensity for a duration equal to the time of concentration.

Step 1: Time of concentration (Kirpich)

tc=0.0195 L0.77S−0.385=0.0195 (6000)0.77(0.013)−0.385=84.2 min\begin{aligned} t_c &= 0.0195\,L^{0.77}S^{-0.385} \\ &= 0.0195\,(6000)^{0.77}(0.013)^{-0.385} = 84.2 \text{ min} \end{aligned}

Step 2: Rainfall intensity for duration tct_c

Cumulative rainfall from the start of the storm:

Time (min)0305084100
Depth (mm)025407080

The depth in the first 84.2 min, by linear interpolation between 50 and 84 min (this is also the most intense 84.2-min window of the storm):

P=40+84.2−5084−50(70−40)=70.2 mmP = 40 + \frac{84.2-50}{84-50}(70-40) = 70.2 \text{ mm} i=70.284.2×60=50.0 mm/hi = \frac{70.2}{84.2}\times 60 = 50.0 \text{ mm/h}

Step 3: Peak flow

Qp=0.278×0.6×50.0×100=834 m3/s\begin{aligned} Q_p &= 0.278\times 0.6\times 50.0 \times 100 \\ &= 834 \text{ m}^3/\text{s} \end{aligned}

Answer: peak flow ≈ 834 m³/s (note: a 100 km² catchment is large for the rational method, so this is a rough estimate).

✓ Answer checked by Claude Sonnet 5.5

All Engineering Hydrology chapter 6 questions →