2-h UH by S-curve from storm hydrograph

Step 1: Baseflow and effective rainfall

The storm hydrograph begins at 0.5 m$^3$/s and ends at 0.5 m$^3$/s, so a constant base flow of 0.5 m$^3$/s is assumed.

Losses $=0.2$ cm/h:

Total ER = 2.0 cm.

Step 2: Hourly (1-h) UH by successive deconvolution

Direct runoff $DRH(t)=Q-0.5$ and

$$DRH(t)=0.5\,U(t)+0.7\,U(t-1)+0\cdot U(t-2)+0.8\,U(t-3)$$ $$U(t)=\frac{DRH(t)-0.7U(t-1)-0.8U(t-3)}{0.5}$$

Examples: $U(1)=44/0.5=88$; $U(2)=(110-0.7\times88)/0.5=96.8$; $U(3)=(85-0.7\times96.8)/0.5=34.5$; $U(4)=(102.3-0.7\times34.5-0.8\times88)/0.5=15.5$.

Step 3: S-curve and 2-h UH

For the 1-h UH, $S(t)=\sum U$ (lag 1 h), and

$$U_2(t)=\frac{1}{2}\left[S(t)-S(t-2)\right]=\frac{U(t)+U(t-1)}{2}$$

The tiny negative and positive values after 8 h come from rounding of the given ordinates, and they are taken as zero. Check: $\sum DRH = 508.4$ and total ER $=2.0$ cm, so the sum of the 1-h UH ordinates should be $508.4/2.0 = 254.2$. The computed sum is 254.4, so the UH is consistent.

Answer: 1-h UH ordinates at t = 0 to 8 h: 0, 88, 96.8, 34.5, 15.5, 10.4, 6.1, 2.8, 0.3 m$^3$/s. 2-h UH ordinates at t = 1 to 9 h: 44.0, 92.4, 65.7, 25.0, 13.0, 8.3, 4.5, 1.6, 0.2 m$^3$/s.