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S-curve: 2-hr hydrograph from 4-hr UH

Short answer

Construct the S-curve by adding the4-h UH ordinates lagged every4 h. The2-h UH is then2[S(t)-S(t-2)]. The resulting2-h UH ordinates (m³/s) at0,2,4,…22 h are: 0,25,100,160,190,170,110,70,30,20,6,0. Peak is 190 m³/s at 8 h.

Derive 2-hrs hydrograph from the following 4 hrs unit hydrograph using s-curve method:

Time (hr)024681012141618202224
Ordinates of 4 hr UH012.562.51301751801409050251330

S-curve method. The S-curve is the hydrograph of a continuous rain of constant intensity. It is built by adding the given 4-h UH repeated every 4 h. A D′-hour UH is then obtained as

UD′(t)=DD′ [S(t)−S(t−D′)]U_{D'}(t) = \frac{D}{D'}\,[S(t) - S(t-D')]

with D = 4 h and D′ = 2 h, so the factor is 4/2 = 2.

Step 1: Construct the S-curve

Ordinates at 2-h intervals: S(t) = U₄(t) + S(t − 4 h), that is, the 4-h UH added to the S-curve lagged by 4 h (2 steps).

Step 2: Lag the S-curve by 2 h and derive the 2-h UH

t (h)4-h UHS(t)S(t−2)2-h UH = 2[S(t)−S(t−2)]
00.00.00.00.0
212.512.50.025.0
462.562.512.5100.0
6130.0142.562.5160.0
8175.0237.5142.5190.0
10180.0322.5237.5170.0
12140.0377.5322.5110.0
1490.0412.5377.570.0
1650.0427.5412.530.0
1825.0437.5427.520.0
2013.0440.5437.56.0
223.0440.5440.50.0

The S-curve reaches a constant equilibrium value of 440.5 m³/s (the sum of alternate ordinates of the 4-h UH), so the construction is consistent.

Check

Σ(4-h UH) = 12.5 + 62.5 + 130 + 175 + 180 + 140 + 90 + 50 + 25 + 13 + 3 = 881 m³/s, and Σ(2-h UH) = 25 + 100 + 160 + 190 + 170 + 110 + 70 + 30 + 20 + 6 = 881 m³/s. The sums are equal, so both UHs carry the same runoff volume of 1 cm ✓.

Answer: 2-h UH ordinates (t = 0, 2, 4, … 22 h): 0, 25, 100, 160, 190, 170, 110, 70, 30, 20, 6, 0 m³/s. Peak = 190 m³/s at t = 8 h.

✓ Answer checked by Claude Sonnet 5.5

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