Derive UH from storm, peak flow from two storms

Part 1: Ordinates of the 1-hour unit hydrograph

Base flow is constant at 0.5 m³/s, so the direct runoff ordinates are $Q_n$ = observed − 0.5:

Check of the data. Taking only the three excess values as printed ($P=0.5,\ 0.0,\ 0.8$), the equation $Q_n=0.5\,U_n+0.8\,U_{n-2}$ gives $U_1=88$, $U_2=220$, $U_3=29.2$, $U_4=(102.3-0.8\times220)/0.5=-147.4$. A negative ordinate is impossible, so this ordinate sequence does not belong to that rain. It is the storm hydrograph of the usual form of this problem, in which rain of 0.7, 0.9, 0.2 and 1.0 cm falls in four successive hours with a loss of 0.2 cm/h. The excess is then

$$P_1=0.5,\quad P_2=0.7,\quad P_3=0,\quad P_4=0.8\ \text{cm}\qquad(\text{total }2.0\ \text{cm})$$

and the hydrograph is consistent with it. The 0.0 cm in the question is the third hour; the second hour has 0.7 cm.

The convolution equation (UH of 1 h duration) is

$$Q_n = 0.5\,U_n + 0.7\,U_{n-1} + 0\cdot U_{n-2} + 0.8\,U_{n-3} \ \Rightarrow\ U_n=\frac{Q_n-0.7\,U_{n-1}-0.8\,U_{n-3}}{0.5}$$

For example, $U_1=44/0.5=88$; $U_2=(110-0.7\times88)/0.5=96.8$; $U_3=(85-0.7\times96.8)/0.5=34.5$; $U_4=(102.3-0.7\times34.5-0.8\times88)/0.5=15.5$. Small positive or negative values after 8 h come from rounding of the given ordinates and are taken as zero.

Check: $\sum U=254.4$ m³/s, and $\sum DRH/\text{ER}=508.7/2.0=254.4$, so the UH represents 1 cm of runoff.

Answer (part 1): 1-h UH ordinates (t = 0 to 9 h): 0, 88.0, 96.8, 34.5, 15.5, 10.4, 6.1, 2.8, 0.3, 0 m³/s.

Part 2: Peak flow from the two successive storms

The UH is 1 h, so each storm is split into 1-hour blocks. The loss is the φ-index = 0.2 cm/h.

• Storm 1: 6.5 cm in 3 h = 2.167 cm/h; excess = 2.167 − 0.2 = 1.967 cm/h, i.e. 1.967 cm in each of 3 hours.
• Storm 2: 10.5 cm in 3 h = 3.5 cm/h; excess = 3.5 − 0.2 = 3.30 cm/h, i.e. 3.30 cm in each of the next 3 hours.

Hourly excess: 1.967, 1.967, 1.967, 3.30, 3.30, 3.30 cm. The flood hydrograph is

$$Q(t)=0.5+\sum_{m=1}^{6}P_m\,U(t-m+1)$$

For example, at t = 6 h: $1.967\,(U_6+U_5+U_4)+3.30\,(U_3+U_2+U_1)+0.5 = 1.967(32.0)+3.30(219.3)+0.5 = 62.9+723.7+0.5 = 787.1$ m³/s.

Answer: peak flow ≈ 787 m³/s at t = 6 h (including base flow of 0.5 m³/s).