(a) A 6-hour unit hydrograph has the following ordinates (time in hr : ordinate in m³/s): 0 : 0; 6 : 200; 12 : 550; 18 : 750; 24…

(a) Resulting flood hydrograph

Incremental rainfall in each 6 h: 3, 5 and 4 cm. The φ index removes 0.3 × 6 = 1.8 cm in each interval, so the effective rainfall is 1.2, 3.2 and 2.2 cm.

The direct runoff is the superposition of the 6-h unit hydrograph, scaled by 1.2, 3.2 and 2.2 and lagged by 0, 6 and 12 h: DRH(t) = 1.2 U(t) + 3.2 U(t − 6) + 2.2 U(t − 12). For example, at 24 h: 1.2 × 650 + 3.2 × 750 + 2.2 × 550 = 780 + 2400 + 1210 = 4390 m³/s.

Adding the base flow of 25 m³/s gives the ordinates (time in h: flow in m³/s): 0: 25; 6: 265; 12: 1325; 18: 3125; 24: 4415; 30: 4415; 36: 3695; 42: 2839; 48: 1949; 54: 1159; 60: 515; 66: 135; 72: 25.

The peak is 4415 m³/s from 24 to 30 h. As a check, the DRH volume is 509 × 10⁶ m³, which equals 6.6 cm of effective rain on the 7710 km² catchment implied by the unit hydrograph.

(b) Catchment water balance

Total area = 15 + 5 + 10 + 18 = 48 km².

(i) Average precipitation = (15 × 1000 + 5 × 800 + 10 × 900 + 18 × 1200)/48 = 49 600/48 = 1033.3 mm.

(ii) Average evaporation = (15 × 500 + 5 × 400 + 10 × 440 + 18 × 650)/48 = 25 600/48 = 533.3 mm.

(iii) With no change in groundwater storage, runoff = P − E. Sub-area runoff: P 500 mm, Q 400 mm, R 460 mm, S 550 mm. Runoff coefficients = runoff/precipitation: P = 0.50, Q = 0.50, R = 460/900 = 0.51, S = 550/1200 = 0.46. For the whole catchment, runoff = 1033.3 − 533.3 = 500 mm, so the coefficient = 500/1033.3 = 0.48.

(c) Muskingum routing

Δt = 4 h, K = 8 h, x = 0.25. C₀ = (0.5Δt − Kx)/(K − Kx + 0.5Δt) = (2 − 2)/8 = 0. C₁ = (0.5Δt + Kx)/8 = 4/8 = 0.5. C₂ = (K − Kx − 0.5Δt)/8 = 4/8 = 0.5. The sum is 1.

O₂ = C₀I₂ + C₁I₁ + C₂O₁, which here is O(t + 4) = 0.5 I(t) + 0.5 O(t).

Outflow (time in h: m³/s): 0: 10; 4: 0.5 × 10 + 0.5 × 10 = 10; 8: 0.5 × 20 + 0.5 × 10 = 15; 12: 0.5 × 30 + 0.5 × 15 = 22.5; 16: 0.5 × 30 + 0.5 × 22.5 = 26.25; 20: 0.5 × 25 + 0.5 × 26.25 = 25.63; 24: 0.5 × 20 + 0.5 × 25.63 = 22.81; 28: 0.5 × 35 + 0.5 × 22.81 = 28.91.

The outflow peak of 26.25 m³/s is lower and later than the inflow peak of 30 m³/s, showing attenuation and lag in the reach.

(d) Water quality indicators

Nitrite is an intermediate of nitrification; its presence shows recent, incomplete oxidation of organic or ammoniacal pollution and it is toxic (methaemoglobinaemia in infants), so it should be very low. Nitrate is the end product of nitrification; it shows older pollution or fertiliser and sewage input, promotes eutrophication and causes blue-baby syndrome; the BIS limit is 45 mg/L. E. coli is the indicator of faecal contamination and of possible pathogens, so it should be absent in 100 mL of drinking water. BOD measures the biodegradable organic load as the oxygen bacteria need in 5 days at 20°C; a high BOD means heavy pollution and oxygen depletion. Dissolved oxygen is needed by aquatic life; low DO (below about 4–5 mg/L) indicates pollution and stress, and DO together with BOD describes the health of a stream.

(e) Settling tank velocity

The settling velocity by Stokes' law is v_s = g(G − 1)d²/(18ν) = 981 × 1.65 × (0.003)²/(18 × 0.01) = 0.081 cm/s. Re = v_s d/ν = 0.024, below 1, so Stokes' law is valid.

For removal of all such particles, the particle must settle the depth H while travelling the length L, so v_s/v = H/L. The flow velocity v = v_s L/H = 0.081 × 80/4 = 1.62 cm/s (about 0.016 m/s), which is the maximum allowed.