Question 3 of 4: Storm Hydrograph Analysis, Unit Hydrograph Development and Application

Question 3: Storm Hydrograph Analysis, Unit Hydrograph Development and Application (33.3 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. 3.1 – a cumulative rainfall/runoff record (Figure 2, read from the source chart) and the direct-runoff hydrograph ordinates (Table 2 below) for an 83 ha basin, storm of April 17, 2003:

Figure 2's cumulative-rainfall curve (dashed) rises from zero near $t=16.0$ hr to a plateau of ≈6.7 cm by roughly $t=17.25$ hr and stays flat thereafter — read directly off the source chart (see check callout below for the reading uncertainty this carries).

3.2 – a 12-hr design storm on a 393 km² watershed, 5 cm excess in the first 6 hr and 15 cm excess in the second 6 hr, constant base flow 100 m³/s, and the watershed's own 6-hr unit hydrograph:

Find. 3.1(a) storm duration and average intensity; 3.1(b) time to peak; 3.1(c) the Φ-index; 3.1(d) the unit hydrograph for the duration found in (a). 3.2 the peak total flow (direct runoff + base flow) and the time at which it occurs.

Approach. 3.1 reads the rainfall mass curve for total depth and duration, integrates the DRH (Table 2) by the trapezoidal rule for total runoff depth, closes a $P=Q+\Phi D$ balance for the Φ-index, then normalizes the DRH by its own runoff depth to obtain the unit hydrograph. 3.2 discretely convolves the two excess-rainfall pulses with the given 6-hr UH and adds the constant base flow.

1. Part 3.1(a) — Storm duration and average intensity. Reading Figure 2's cumulative-rainfall curve, the rise begins at $t\approx16.00$ hr and reaches its flat plateau at $t\approx17.25$ hr, for a total depth of $P\approx6.7$ cm: $$D = 17.25-16.00=\boxed{1.25\ \text{hr}}\ (75\ \text{min}),\qquad i_{avg}=\frac{P}{D}=\frac{6.7}{1.25}=\boxed{5.36\ \text{cm/hr}}$$
2. Part 3.1(b) — Time to peak. Taking the time to peak as the elapsed time from the start of rainfall to the peak of the direct-runoff hydrograph (Table 2 peaks at $Q=3.96$ cms, $t=17.25$ hr): $$T_p = 17.25-16.00=\boxed{1.25\ \text{hr}}$$ (the very short lag reflects the small, 83 ha, fast-responding basin — the peak arrives essentially as soon as the rain itself ends).
3. Part 3.1(c) — Φ-index. Total direct-runoff depth by trapezoidal integration of Table 2 at $\Delta t=900$ s, divided by the 83 ha ($=830{,}000\ \text{m}^2$) basin area: $$V_{runoff}=\sum \tfrac12(Q_i+Q_{i+1})\Delta t=27{,}270\ \text{m}^3,\qquad Q_{runoff}=\frac{27{,}270}{830{,}000}\times100=\boxed{3.29\ \text{cm}}$$ The Φ-index removes a constant loss rate from the storm's average intensity so that the remaining depth equals the observed runoff: $$\Phi=\frac{P-Q_{runoff}}{D}=\frac{6.7-3.29}{1.25}=\boxed{2.73\ \text{cm/hr}}$$
4. Part 3.1(d) — 1.25-hr unit hydrograph. Because Table 2 is already a DIRECT-runoff hydrograph (base flow already removed) generated by 3.29 cm of excess rainfall over the 1.25-hr duration found in (a), the unit hydrograph ordinates follow by simple linear scaling — divide every ordinate by 3.29 cm: $$UH(t)=\frac{Q_{DRH}(t)}{3.29}$$ which gives a UH peaking at $UH=3.96/3.29=\boxed{1.21\ \text{m}^3/\text{s per cm}}$, still at $t=17.25$ hr:

5. Part 3.2 — Convolve the two excess-rainfall pulses with the 6-hr UH. With pulses of 5 cm (hours 0–6) and 15 cm (hours 6–12), the direct-runoff hydrograph at each 6-hr step is $Q_{DRH}(t)=5\cdot UH(t)+15\cdot UH(t-6)$: $$t=18\ \text{hr}:\ 5(85.6)+15(30.9)=428.0+463.5=891.5\ \text{m}^3/\text{s}$$ $$t=24\ \text{hr}:\ 5(41.8)+15(85.6)=209.0+1284.0=\boxed{1493.0\ \text{m}^3/\text{s}}$$ $$t=30\ \text{hr}:\ 5(14.6)+15(41.8)=73.0+627.0=700.0\ \text{m}^3/\text{s}$$ Scanning all eight 6-hr ordinates confirms $t=24$ hr governs the peak. Adding the constant base flow of 100 m³/s: $$Q_{peak,total}=1493.0+100=\boxed{1593\ \text{m}^3/\text{s at }t=24\ \text{hr after the storm began}}$$