(a) What is a flood hydrograph ? The monthly discharge (Q) data of a river at a gauging station are given below : | Month | Q…

(a) A flood hydrograph is a graphical plot of discharge (or stage) of a river at a gauging station against time during a flood event. It generally shows the rising limb, crest/peak, and recession limb. The given monthly data give an annual flow hydrograph rather than a single-storm flood hydrograph, but the same plotting method applies.

Plot months January to December on the x-axis and discharge Q in m³/s on the y-axis. The points to be joined by straight lines are:

• Jan: 30 m³/s
• Feb: 40 m³/s
• Mar: 50 m³/s
• Apr: 60 m³/s
• May: 80 m³/s
• Jun: 90 m³/s
• Jul: 100 m³/s
• Aug: 120 m³/s
• Sep: 110 m³/s
• Oct: 90 m³/s
• Nov: 80 m³/s
• Dec: 70 m³/s

Joining these points gives the required hydrograph. From the graph, the highest ordinate occurs in August and the lowest in January.

Peak flow = 120 m³/s (August) Minimum flow = 30 m³/s (January)

(b)(i) Evaporation is the process by which water changes from liquid state to vapour state from free water surfaces, soil surfaces, or wet surfaces. It depends on vapour-pressure deficit, temperature, wind speed, relative humidity, solar radiation, and atmospheric pressure. It is expressed as depth of water lost per unit time, usually mm/day. For evaporation to continue, both energy and available water must be present.

(b)(ii) Transpiration is the loss of water vapour from living plant tissues, mainly through stomata in leaves. Water absorbed by roots moves through the plant and escapes as vapour. It depends on plant type, stomatal opening, soil moisture, temperature, wind, and humidity. It is essentially evaporation from plant surfaces, but it is biologically controlled.

(b)(iii) Evapotranspiration is the combined loss of water to the atmosphere by evaporation from water and soil surfaces and by transpiration from vegetation. Potential evapotranspiration (PET) is the maximum possible loss when water supply is not limiting. Actual evapotranspiration (AET) is the actual loss under existing soil-moisture conditions. It is important in irrigation planning, water-balance studies, and catchment yield estimation.

(b)(iv) Infiltration is the process by which water enters the soil surface from rainfall or irrigation. The maximum rate at which soil can absorb water is called infiltration capacity. It depends on soil texture, structure, initial moisture content, vegetation, compaction, and rainfall intensity. Infiltration rate generally decreases with time and is often modelled by Horton’s equation: f = f_c + (f_0 − f_c)e^(−kt). Water in excess of infiltration capacity becomes surface runoff.

(c) The mean rainfall is first found using the arithmetic-mean method. Sum of rainfall = 72.3 + 86.4 + 94.2 + 103.8 + 71.4 = 428.1 cm Mean x̄ = 428.1/5 = 85.62 cm

For the required number of raingauges, the coefficient of variation method is used: N = (Cv/ε)² where Cv = coefficient of variation in percent and ε = permissible percentage error = 5%.

Deviations and squares:

• 72.3 − 85.62 = −13.32; square = 177.4224
• 86.4 − 85.62 = 0.78; square = 0.6084
• 94.2 − 85.62 = 8.58; square = 73.6164
• 103.8 − 85.62 = 18.18; square = 330.5124
• 71.4 − 85.62 = −14.22; square = 202.2084

Σ(x − x̄)² = 784.368 cm² Sample standard deviation s = √[784.368/(5 − 1)] = √196.092 = 14.003 cm Cv = (s/x̄) × 100 = (14.003/85.62) × 100 = 16.355%

Required total number of raingauges: N = (16.355/5)² = (3.271)² = 10.70 ≈ 11 Existing raingauges = 5 Additional raingauges needed = 11 − 5 = 6

Additional number of raingauges needed = 6

(d) At the mixing point, use the mass-balance principle for ultimate CBOD. River flow Q_r = 8 m³/s, upstream ultimate CBOD L_r = 10 mg/L Wastewater flow Q_w = 2 m³/s, ultimate CBOD L_w = 1000 mg/L Mixed flow Q_m = 8 + 2 = 10 m³/s

Ultimate CBOD just after mixing, L_0: L_0 = (Q_r L_r + Q_w L_w)/Q_m L_0 = (8 × 10 + 2 × 1000)/10 L_0 = (80 + 2000)/10 = 2080/10 = 208 mg/L

For first-order CBOD decay, k = 0.1/day. 5-day CBOD at 0 km: CBOD₅ = L_0(1 − e^(−k × 5)) = 208(1 − e^(−0.5)) = 208(1 − 0.60653) = 208 × 0.39347 = 81.84 mg/L

At 20 km downstream, travel time t = distance/velocity = 20/10 = 2 days. Remaining ultimate CBOD: L_20 = L_0 e^(−k t) = 208 e^(−0.1 × 2) = 208 e^(−0.2) = 208 × 0.81873 = 170.30 mg/L

5-day CBOD at 20 km: CBOD₅,20 = L_20(1 − e^(−0.5)) = 170.30 × 0.39347 = 67.01 mg/L

At 0 km: ultimate CBOD = 208 mg/L; 5-day CBOD = 81.84 mg/L At 20 km: ultimate CBOD = 170.30 mg/L; 5-day CBOD = 67.01 mg/L The calculation assumes complete mixing at the outfall, plug flow downstream, first-order decay with base e, and no additional pollution load between the two sections.

(e) Per capita demand is the average quantity of water required per person per day for domestic, commercial, industrial, public, and firefighting uses, including losses and waste. It is usually expressed as litres per capita per day (Lpcd). It is determined by dividing the total annual water consumption of a town by its population and by 365 days. It can also be estimated as the sum of domestic, commercial, industrial, public, fire, and unaccounted-for demands. It varies with climate, standard of living, habits, cost of water, pressure in the distribution system, and efficiency of water management.

Given average daily demand = 135 Lpcd.

(e)(i) Maximum daily demand Using 1.8 times the average daily demand: = 1.8 × 135 = 243 Lpcd

(e)(ii) Maximum weekly demand Using 1.48 times the average daily demand: = 1.48 × 135 = 199.8 Lpcd ≈ 200 Lpcd

(e)(iii) Maximum monthly demand Using 1.28 times the average daily demand: = 1.28 × 135 = 172.8 Lpcd ≈ 173 Lpcd

(e)(iv) Maximum hourly demand Average hourly demand on the maximum day = maximum daily demand/24 = 243/24 = 10.125 L per capita per hour. Maximum hourly demand = 1.5 × 10.125 = 15.19 L per capita per hour This is equivalent to 15.19 × 24 = 364.5 Lpcd if expressed as a daily volume.