(a) A box culvert has an expected life of 10 years. (i) If the acceptable risk of at least one event exceeding the culvert…

(a)(i)

Let T be the design return period in years and n = 10 years the expected life. For independent annual exceedance events, the risk of at least one event exceeding the culvert capacity is R = 1 − (1 − 1/T)ⁿ. Given R = 0.05 and n = 10: (1 − 1/T)¹⁰ = 0.95. So 1 − 1/T = 0.95^(1/10) = exp(ln 0.95 / 10) = exp(−0.051293/10) = 0.994884. Thus 1/T = 1 − 0.994884 = 0.005116. T = 1/0.005116 = 195.46 years. Design return period T ≈ 195.46 years, say 195.5 years.

(a)(ii)

For T = 195.46 years, the annual non-exceedance probability is 1 − 1/T = 0.994884. The probability that capacity is not exceeded for 50 years is P(no exceedance in 50 years) = (1 − 1/T)⁵⁰ = (0.95^(1/10))⁵⁰ = 0.95⁵. 0.95⁵ = 0.7737809375. Chance of not exceeding capacity in 50 years = 77.38%. Chance of at least one exceedance = 22.62%.

(b)

Sketch (vertical section, not to scale):

For steady radial flow to a fully penetrating well in a confined homogeneous aquifer, the Thiem equation is s₁ − s₂ = Q/(2πT) · ln(r₂/r₁). Therefore T = Q · ln(r₂/r₁) / [2π(s₁ − s₂)]. Given Q = 0.10 m³/s, r₁ = 50 m, r₂ = 150 m, s₁ = 3 m, s₂ = 1.2 m, b = 50 m. s₁ − s₂ = 3 − 1.2 = 1.8 m. r₂/r₁ = 150/50 = 3. T = 0.10 × ln 3 / (2π × 1.8) = 0.10 × 1.098612 / 11.309733 = 9.714 × 10⁻³ m²/s. Hydraulic conductivity K = T/b: K = 9.714 × 10⁻³ / 50 = 1.943 × 10⁻⁴ m/s. In m/day: K = 1.943 × 10⁻⁴ × 86400 = 16.79 m/day. K ≈ 1.94 × 10⁻⁴ m/s ≈ 16.8 m/day; T ≈ 9.71 × 10⁻³ m²/s. Condition of validity: steady state, confined aquifer, fully penetrating well, homogeneous and isotropic aquifer, constant thickness, and Darcy flow.

(c)

Five adverse effects of reservoir sedimentation:

1. Loss of live storage capacity, reducing water supply, irrigation, hydropower generation, and flood moderation.
2. Upstream aggradation and rise in backwater levels, causing flooding, waterlogging, and river-bed rise.
3. Delta formation at the reservoir head and deposition in navigation channels, obstructing navigation and increasing dredging.
4. Downstream channel degradation, bank erosion, and ecological disturbance due to sediment-starved releases.
5. Abrasion and erosion of turbines, pumps, gates, valves, and outlet works, increasing maintenance and reducing efficiency.

Reservoir sedimentation can be reduced by:

• Catchment-area treatment: soil conservation, contour bunding, check dams, afforestation, vegetated buffers, and stream-bank stabilisation.
• Sediment traps and detention basins upstream, plus bypass channels or tunnels to divert sediment-laden flows.
• Reservoir operation measures: sluicing, density-current venting, drawdown flushing, and low-level outlet operation.
• Dredging, hydro-suction, siphoning, and mechanical removal of deposited sediment.
• Proper design with allocated dead storage for sediment, multiple outlets, and sediment-pass-through facilities. Integrated watershed management plus sediment routing/flushing is the most sustainable reduction strategy.

(d)

Using the standard base-10 BOD equation used in Indian practice: BOD_t = L₀[1 − 10^(−k t)], where L₀ is ultimate BOD. Temperature correction: k_T = k₂₀ θ^(T−20), with θ = 1.047. Given k₂₀ = 0.1/day at 20°C. At 30°C: k₃₀ = 0.1 × 1.047¹⁰ = 0.1583/day. At 30°C for 1 day, BOD₁ = 120 mg/L: 120 = L₀[1 − 10^(−0.1583 × 1)]. 10^(−0.1583) = 0.6946. 1 − 0.6946 = 0.3054. L₀ = 120/0.3054 = 392.9 mg/L. Now BOD₅ at 20°C: BOD₅ = 392.9[1 − 10^(−0.1 × 5)] = 392.9[1 − 10^(−0.5)] = 392.9[1 − 0.3162] = 392.9 × 0.6838 = 268.6 mg/L. Unoxidized BOD after 20 days at 20°C: Remaining = L₀ × 10^(−0.1 × 20) = L₀ × 10^(−2) = 0.01 L₀. Percent unoxidized = (0.01 L₀/L₀) × 100 = 1.00%. 5-day BOD at 20°C ≈ 268.6 mg/L; unoxidized BOD after 20 days = 1.00%, i.e. about 99.00% oxidised.

(e)

Composting of municipal solid waste is the controlled aerobic biological decomposition of organic matter by bacteria, fungi, and actinomycetes into a stable humus-like material. The organic fraction is first segregated to remove plastics, metals, glass, and inerts. It is shredded and mixed with bulking agents to adjust moisture, porosity, and C/N ratio. In the presence of oxygen, microorganisms oxidise organic matter: organic matter + O₂ → CO₂ + H₂O + heat + compost. The process passes through mesophilic, thermophilic (about 55–65°C), cooling, and maturation phases. Aeration is provided by turning windrows, forced aeration in static piles, or in-vessel reactors. The thermophilic phase destroys pathogens and weed seeds, while the maturation phase stabilises the product.

Important design considerations of aerobic composting:

• C/N ratio: optimum about 25–30:1. High carbon slows decomposition; low carbon causes ammonia loss and odour.
• Moisture content: maintain about 50–60%. Too dry stops microbial activity; too wet causes anaerobic conditions.
• Aeration and oxygen: maintain oxygen above about 5–10%. Aeration controls temperature, moisture, and odour and is achieved by turning, forced air, or proper porosity.
• Temperature: keep 55–65°C for effective pathogen destruction; avoid exceeding about 70°C, which kills beneficial microbes.
• Particle size and porosity: 25–75 mm is generally suitable; bulking agents such as wood chips improve air flow.
• pH: maintain about 6.5–8.0; lime may be added if the waste is too acidic.
• Retention time: active composting for about 3–6 weeks, followed by maturation for 2–4 weeks or more.
• Pile/reactor design: windrow height, width, base slope, drainage, leachate collection, and turning equipment must be designed for the waste quantity and climate.
• Environmental control: manage leachate, odour, vectors, runoff, and gas emissions; provide buffer zones and proper siting.