(a) What is the difference between a flood hydrograph and a mass curve ? Using the following data of a river at a dam site…

(a) Hydrograph and mass curve; plotting the river flow (10 marks)

Difference. A hydrograph is a plot of discharge (flow rate, in m³/s) against time; its ordinates are rates of flow, and it shows the peak, the minimum and the timing of the flow. A mass curve is a plot of the cumulative volume of flow against time, so it is the integral of the hydrograph. Its slope at any time is the discharge, and it is used to find the storage a reservoir needs (Rippl's method), while the hydrograph is used to find flood peaks and the design flood.

Plot. The monthly discharges are plotted against the months, January to December, with the discharge (m³/s) on the vertical axis and the months on the horizontal axis, and the points joined by a smooth line.

Reading the hydrograph. The flow rises from 110 m³/s in January to a peak in August, and then falls steadily to December.

• The maximum discharge is 210 m³/s, in August.
• The minimum discharge is 100 m³/s, in December.

The mean flow is 1720/12 = 143.3 m³/s.

(b) Rain-gauge network: standard error and optimum number of gauges (10 marks)

Statistics. Mean rainfall x̄ = (82.7 + 102.9 + 180.3 + 110.3 + 95.8 + 136.7)/6 = 708.7/6 = 118.1 cm.

Standard deviation σ = √[Σ(x − x̄)²/(n − 1)] = 35.4 cm

Coefficient of variation C_v = 100σ/x̄ = 100 × 35.4/118.1 = 29.9%

(b)(i) Standard error of the estimated mean. With N = 6 gauges,

ε = C_v/√N = 29.9/√6 = 12.2%

(b)(ii) Optimum number of gauges for a 10% error.

N = (C_v/e)² = (29.9/10)² = 8.97 ≈ 9 gauges

So 3 more gauges are needed in addition to the existing 6.

(c) Permeability and transmissivity from a pumping test (10 marks)

The aquifer is unconfined and the flow is steady, so the Thiem equation for an unconfined aquifer applies:

Q = πK(h₂² − h₁²)/ln(r₂/r₁)

Data: Q = 5400 l/min = 0.09 m³/s. The saturated thickness before pumping is 25 m, and the well is assumed to reach the impervious base. Drawdowns of 1.11 m at r₁ = 30 m and 0.53 m at r₂ = 90 m give the water depths h₁ = 25 − 1.11 = 23.89 m and h₂ = 25 − 0.53 = 24.47 m.

h₂² − h₁² = 598.8 − 570.7 = 28.05 m²

K = Q ln(r₂/r₁)/[π(h₂² − h₁²)] = 0.09 × ln 3/(π × 28.05) = 0.09 × 1.0986/88.12 = 1.12 × 10⁻³ m/s

Coefficient of permeability K ≈ 1.12 × 10⁻³ m/s ≈ 97 m/day.

Transmissivity T = K × b = 1.12 × 10⁻³ × 25 = 0.0281 m²/s

T ≈ 0.028 m²/s ≈ 2420 m²/day.

(d) Design of water supply units for demand variations (10 marks)

Demand variations. The average demand is 15 000 × 270 = 4 050 000 l/day = 4050 m³/day = 168.75 m³/h. The maximum daily demand is 1.8 × average = 7290 m³/day = 303.75 m³/h (0.0844 m³/s). The maximum hourly demand is 1.5 × the maximum day's average hourly demand = 2.7 × average hourly = 455.6 m³/h (126.6 l/s).

• (i) Filter units. They are designed for the maximum daily demand, because the plant must meet the demand of the peak day, and it works at a steady rate through the day. Design flow = 7290 m³/day (303.75 m³/h). At an assumed filtration rate of 5 m/h, the filter area is 303.75/5 ≈ 61 m², plus a stand-by unit and 3–5% for backwash water.
• (ii) Pumping units. They are designed for the maximum daily demand, spread over the pumping hours. In 24 hours of pumping, the capacity is 303.75 m³/h (84.4 l/s); if pumping runs for 16 hours, it is 7290/16 = 455.6 m³/h (126.6 l/s). The hourly fluctuation is met from the storage, not by the pumps.
• (iii) Distribution mains. They are designed for the maximum hourly demand of the maximum day, because the mains have to carry the peak flow when the demand is greatest (with a check for the fire demand). Design flow = 455.6 m³/h = 126.6 l/s.
• (iv) Sedimentation tank. It is designed for the maximum daily demand, since the detention time must be provided at the highest flow of the day. With flow 303.75 m³/h and an assumed detention time of 2.5 h, the volume is 303.75 × 2.5 ≈ 760 m³.
• (v) Clear water reservoir. It is designed for the hourly variation, because it balances the constant rate of supply from the plant against the varying demand, and holds the fire and emergency reserve. As a rule of thumb, take about one-third of the maximum daily demand: 7290/3 ≈ 2430 m³.

(e) Theoretical oxygen demand of glycine (10 marks)

Glycine, CH₂(NH₂)COOH, is C₂H₅NO₂ with molar mass 2 × 12 + 5 × 1 + 14 + 2 × 16 = 75 g/mol.

Step 1, carbon oxidised to CO₂ and nitrogen released as NH₃.

C₂H₅NO₂ + 1.5 O₂ → 2 CO₂ + H₂O + NH₃

Step 2, ammonia to nitrite.

NH₃ + 1.5 O₂ → NO₂⁻ + H⁺ + H₂O

Step 3, nitrite to nitrate.

NO₂⁻ + 0.5 O₂ → NO₃⁻

Sum. C₂H₅NO₂ + 3.5 O₂ → 2 CO₂ + 2 H₂O + NO₃⁻ + H⁺

The oxygen required is 1.5 + 1.5 + 0.5 = 3.5 mol O₂ per mol of glycine.

ThOD = 3.5 × 32 = 112 g O₂ per mole of glycine. Per gram of glycine, ThOD = 112/75 = 1.49 g O₂/g glycine, made up of 0.64 g/g for the carbonaceous stage (Step 1) and 0.85 g/g for the nitrogenous stages (Steps 2 and 3).