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Rational method peak flow from storm depths

Short answer

Time of concentration by Kirpich: t_c = 25.2 min. Interpolated storm intensity i = 113.8 mm/h. Peak flow by rational method (C=0.7, A=3 km²): Qp = 0.278×0.7×113.8×3 = 66.4 m³/s.

A rainfall storm for the duration of 20 mins, 40 mins, 60 and 80 mins are 40 mm, 70 mm, 90 mm and 100 mm respectively. If the slope of the catchment is 0.01, runoff coefficient is 0.7, maximum length of the travel is 1,100 m and catchment area is 3 km2^2, estimate the peak flow by rational method.

The rational method gives Qp=0.278 C i AQ_p = 0.278\,C\,i\,A (QpQ_p in m³/s, ii in mm/h, AA in km²), where ii is the average intensity of a storm whose duration equals the time of concentration tct_c of the catchment.

Step 1: Time of concentration (Kirpich's formula)

tc=0.0195 L0.77 S−0.385 (min; L in m)=0.0195 (1100)0.77 (0.01)−0.385=25.23 min\begin{aligned} t_c &= 0.0195\,L^{0.77}\,S^{-0.385}\ \text{(min; }L\text{ in m)}\\ &= 0.0195\,(1100)^{0.77}\,(0.01)^{-0.385} = 25.23\ \text{min} \end{aligned}

Step 2: Intensity for each storm duration

Duration (min)Depth (mm)Intensity i = P/t (mm/h)
2040120.0
4070105.0
609090.0
8010075.0

Step 3: Intensity at tct_c

tc=25.23t_c = 25.23 min lies between 20 and 40 min. Interpolating the storm depth linearly:

P=40+25.23−2040−20(70−40)=47.84 mmP = 40 + \frac{25.23 - 20}{40 - 20}(70 - 40) = 47.84\ \text{mm} i=Ptc=47.840.4205=113.8 mm/hi = \frac{P}{t_c} = \frac{47.84}{0.4205} = 113.8\ \text{mm/h}

Step 4: Peak flow

Qp=0.278×0.7×113.8×3=66.4 m3/sQ_p = 0.278 \times 0.7 \times 113.8 \times 3 = 66.4\ \text{m}^3/\text{s}

Answer: Peak flow ≈\approx 66 m³/s (with tct_c = 25.2 min, ii = 114 mm/h). Kirpich's formula was used because no other method for tct_c was specified.

✓ Answer checked by Claude Sonnet 5.5

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