(a) (i) Define unit hydrograph. Explain two basic assumptions made in the derivation of a unit hydrograph. What are the…

(a)(i) A unit hydrograph is the direct-runoff hydrograph resulting from one unit depth (usually 1 cm) of effective rainfall occurring uniformly over the catchment at a uniform rate during a specified duration D. It is applicable only to that catchment and that duration.

Two basic assumptions:

• Linear response: direct runoff is proportional to effective rainfall. Thus R cm of rainfall excess gives ordinates R times those of the unit hydrograph.
• Time invariance and uniform distribution: the same effective rainfall pattern produces the same time distribution, merely shifted in time; rainfall excess is uniformly distributed over the catchment and uniformly intense during the duration.

Applications: flood forecasting, design of spillways and culverts, extension of unit hydrographs to different durations, derivation of synthetic unit hydrographs, and storm-drainage design.

Limitations: it is not valid for very large catchments, snow-melt runoff, non-uniform rainfall, or catchments with large storage; linearity fails for extreme storms; adequate rainfall-runoff records are needed.

(a)(ii) By the method of superposition, a 12 h unit hydrograph is obtained by adding three 4 h unit hydrographs lagged by 4 h and 8 h, then dividing by 3. Thus y(t) = [u(t) + u(t−4) + u(t−8)]/3.

• 0 h: y = 0
• 4 h: y = (30+0+0)/3 = 10 m³/s
• 8 h: y = (120+30+0)/3 = 50 m³/s
• 12 h: y = (200+120+30)/3 = 350/3 = 116.67 m³/s
• 16 h: y = (250+200+120)/3 = 190 m³/s
• 20 h: y = (210+250+200)/3 = 220 m³/s
• 24 h: y = (130+210+250)/3 = 590/3 = 196.67 m³/s
• 28 h: y = (75+130+210)/3 = 415/3 = 138.33 m³/s
• 32 h: y = (50+75+130)/3 = 85 m³/s
• 36 h: y = (20+50+75)/3 = 145/3 = 48.33 m³/s
• 40 h: y = (10+20+50)/3 = 80/3 = 26.67 m³/s
• 44 h: y = (0+10+20)/3 = 10 m³/s
• 48 h: y = (0+0+10)/3 = 10/3 = 3.33 m³/s
• 52 h: y = 0

Final 12 h UH ordinates: 0, 10, 50, 350/3, 190, 220, 590/3, 415/3, 85, 145/3, 80/3, 10, 10/3, 0 m³/s at 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52 h respectively.

(b)(i) For Indian conditions, a separate system is generally better for new and planned areas. India receives intense monsoon rainfall in short spells, while dry-weather sewage flow is relatively small. A combined system must be designed for storm flow, making sewers and treatment units very large and costly. During storms, combined sewers overflow and discharge dilute but polluted sewage into rivers, causing serious pollution.

A separate system keeps sanitary sewage free of storm water, so the treatment plant receives a more uniform flow and stronger sewage; storm water can be managed by rainwater harvesting, recharge, or separate drains. This suits water-scarce Indian cities where treated sewage can be reused and storm water can recharge groundwater. It also reduces urban flooding if storm drains are properly designed.

However, in old congested Indian city cores with narrow streets and existing combined sewers, complete separation may be impractical and costly. There, combined or partially separate systems with interception and treatment of overflows may be preferred. Overall, for new Indian towns and planned areas, the separate system is the better choice.

(b)(ii) Break point chlorination test is a test in which increasing doses of chlorine are added to water samples and the residual chlorine is measured after a fixed contact time. Initially chlorine is consumed by reducing matter and ammonia; chloramines form. As the dose increases, the residual may fall again as chloramines are destroyed. The break point is the dose at which ammonia and other chlorine-demanding substances are fully oxidized and free residual chlorine begins to increase almost linearly.

It is needed to determine the minimum chlorine dose that gives a stable free chlorine residual for effective disinfection, avoids chloramine taste and odour, removes ammonia, and controls algae and slime growth.

(c) Wastewater flow: Q = 100000 × 150 × 0.80 L/d = 12,000,000 L/d = 12,000 m³/d.

BOD removed: S₀ = 150 mg/L = 0.150 kg/m³, S = 5 mg/L = 0.005 kg/m³. ΔS = 0.145 kg/m³. BOD removed per day = QΔS = 12000 × 0.145 = 1740 kg/d.

MLVSS X = 3000 mg/L = 3 kg/m³. Using the activated-sludge MCRT relation: 1/θc = YQΔS/(VX) − kd So V = θc YQΔS / [X(1 + kdθc)]. V = 10 × 0.5 × 12000 × 0.145 / [3(1 + 0.05 × 10)] = 8700 / (3 × 1.5) = 8700/4.5 = 1933.33 m³.

Check HRT = V/Q = 1933.33/12000 = 0.1611 d = 3.87 h ≈ 4 h.

MLVSS in reactor = VX = 1933.33 × 3 = 5800 kg. Sludge waste mass per day = VX/θc = 5800/10 = 580 kg/d. Also net production = YQΔS − kd VX = 0.5 × 1740 − 0.05 × 5800 = 870 − 290 = 580 kg/d.

Underflow concentration = 10000 mg/L = 10 kg/m³. Waste sludge volume = 580/10 = 58 m³/d.

Final: Reactor volume ≈ 1933.33 m³; sludge waste mass ≈ 580 kg/d; waste sludge volume ≈ 58 m³/d. If HRT = 4 h is imposed exactly, V = 2000 m³ and the corresponding waste values are 600 kg/d and 60 m³/d.