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Derive 4-hr UH from triangular direct runoff hydrograph

Short answer

The DRH is a triangle peaked at1 cm/h at4 h with base8 h, produced by4 cm effective rainfall. Divide ordinates by4: the4-h UH is a triangle of base8 h and peak0.25 h⁻¹ at t=4 h.

The direct runoff hydrograph due to an effective rainfall event is given by a triangle such that its base is 8 hours and its height at the midpoint of the base is 1 cm/h. The duration and intensity of the effective rainfall are 4 hours and 1 cm/h, respectively. Derive and sketch a 4 hour unit hydrograph.

A UH is the DRH divided by the depth of effective rainfall that produced it.

Step 1: Depth of effective rainfall

ER=i tr=1 cm/h×4 h=4 cmER = i\,t_r = 1 \text{ cm/h} \times 4 \text{ h} = 4 \text{ cm}

Step 2: Check the DRH volume

The DRH is a triangle: base 8 h, height 1 cm/h. Area = ½ × 8 × 1 = 4 cm. This equals the effective rainfall, so the DRH is consistent (no loss of volume).

Step 3: Ordinates of the 4-h UH

The 4-h UH ordinate = DRH ordinate / 4 (per cm of rain). The DRH rises linearly from 0 at t = 0 to 1 cm/h at t = 4 h, then falls linearly to 0 at t = 8 h.

t (h)012345678
DRH (cm/h)00.250.500.751.000.750.500.250
4-h UH (cm/h per cm)00.06250.1250.18750.250.18750.1250.06250

Sketch of the 4-h UH

 U (per h)
 0.25 |          /\
      |        /    \
      |      /        \
      |    /            \
      |  /                \
    0 +--+--+--+--+--+--+--+-- t (h)
      0  1  2  3  4  5  6  7  8

Answer: the 4-h UH is a triangle of base 8 h and peak 0.25 per hour (0.25 cm/h per cm of effective rain) at t = 4 h. Check: area = ½ × 8 × 0.25 = 1 cm ✓.

✓ Answer checked by Claude Sonnet 5.5

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