What duration D is used in the IDF equation?

Time of concentration, 20 min = 0.3333 h

i from the IDF is in cm/h; what is needed?

Convert to mm/h by multiplying by 10 (149.44 mm/h)

0.25 x 149.44 x 20/360 = 2.08 m^3/s

2080 Bhadra · 5 marks

Peak discharge by rational method with IDF equation

Short answer

Design duration D = 20 min = 0.3333 h. Intensity from IDF: i = 6.93×25^0.189 / (0.3333+0.5)^0.878 = 14.94 cm/h = 149.44 mm/h. Peak discharge by rational formula Qp = C i A / 360 = 0.25×149.44×20 / 360 = 2.08 m³/s.

A catchment area has the time of concentration of 20 minutes and an area of 20 ha. Estimate the peak discharge corresponding to 25 years return period. Take runoff coefficient of 0.25. The intensity duration frequency curve for the storm can be expressed in cm/h by: $i = \frac{KT^x}{(D+a)^n}$ . Take K = 6.93, x = 0.189, a = 0.50 and n = 0.878.

Rational formula: $Q_p=\dfrac{C\,i\,A}{360}$ (Q in m³/s, i in mm/h, A in ha). The design duration of the rain is the time of concentration, D = $t_c$ = 20 min = 0.3333 h.

Intensity from the IDF equation

i=\frac{K\,T^{x}}{(D+a)^{n}}=\frac{6.93\times(25)^{0.189}}{(0.3333+0.50)^{0.878}}

(25)^{0.189}=1.8374,\qquad (0.8333)^{0.878}=0.8521

i=\frac{6.93\times1.8374}{0.8521}=14.944 \text{ cm/h}=149.44 \text{ mm/h}

Peak discharge

Q_p=\frac{0.25\times149.44\times20}{360}=2.08 \text{ m}^3/\text{s}

Answer: peak discharge ≈ 2.08 m³/s (D in hours in the IDF equation, i in cm/h as given).

✓ Answer checked by Claude Sonnet 5.5

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