(a) (i) A 50 cm well in an unconfined aquifer of saturated thickness 45 m yields 600 lpm under a drawdown of 3 m at the pumping well. What will be the discharge under a drawdown of 6 m ? Consider the radius of influence as 500 m. (ii) What will be th

Question

(a) (i) A 50 cm well in an unconfined aquifer of saturated thickness 45 m yields 600 lpm under a drawdown of 3 m at the pumping well. What will be the discharge under a drawdown of 6 m ? Consider the radius of influence as 500 m. (ii) What will be the discharge in a 30 cm well under a drawdown of 3 m for the unconfined aquifer as mentioned in part (i). (20 marks)

(b) How does the peak hour demand affect the design of a water supply scheme ? Sketch the fluctuation in demand for typical Indian conditions. (15 marks)

(c) (i) The B.O.D. of a sewage incubated for one day at 25°C has been found to be 100 mg/l. What will be the 5 day, 20°C B.O.D. ? Assume K₂₀ = 0·12 at 20°C. Take temperature coefficient, φ = 1·056. (ii) Despite widespread use of B.O.D., it has some limitations. Mention all those limitations. (15 marks)

UPSC Answer Check · Accepted Answer

(a) (i) For steady radial flow to a well in an unconfined aquifer, use the Dupuit–Thiem equation:
Q = π K (H² − h_w²) / ln(R / r_w)
Valid for a homogeneous, isotropic, unconfined aquifer, fully penetrating well, no recharge, and steady flow.

Take the 50 cm well as diameter, so r_w = 0.25 m.
H = 45 m, s₁ = 3 m, h₁ = 45 − 3 = 42 m, R = 500 m.
Q₁ = 600 lpm = 600 × 1.44 = 864 m³/day.

K = Q₁ ln(R / r_w) / [π (H² − h₁²)]
K = 864 ln(500 / 0.25) / [π (45² − 42²)]
K = 864 ln(2000) / (π × 261) = 8.00917 m/day.

For s₂ = 6 m, h₂ = 45 − 6 = 39 m. Same well and R:
Q₂ / Q₁ = (H² − h₂²) / (H² − h₁²)
Q₂ / Q₁ = (45² − 39²) / (45² − 42²)
Q₂ / Q₁ = (2025 − 1521) / (2025 − 1764)
Q₂ / Q₁ = 504 / 261 = 56 / 29.

Q₂ = 600 × 56 / 29 = 33600 / 29 = 1158.62 lpm.
Q₂ ≈ 1158.62 lpm = 1668.41 m³/day.

(a) (ii) For a 30 cm well, r_w = 0.15 m. Drawdown s = 3 m, so H² − h² = 261 m².
Using the same K and R:
Q₃₀ / Q₅₀ = ln(R / 0.25) / ln(R / 0.15)
Q₃₀ / Q₅₀ = ln(500 / 0.25) / ln(500 / 0.15)
Q₃₀ / Q₅₀ = ln(2000) / ln(10000 / 3)
Q₃₀ / Q₅₀ = 7.60090246 / 8.11172808 = 0.93702629.

Q₃₀ = 600 × 0.93702629 = 562.22 lpm.
Q₃₀ ≈ 562.22 lpm = 809.59 m³/day.

(b) Peak hour demand is the maximum water required in any one hour of the maximum-demand day. It affects the design of a water supply scheme because the distribution system must carry this high flow while maintaining adequate residual pressure.
Distribution mains, service connections, valves, meters, and booster pumps are sized for peak hourly flow, plus fire demand where applicable.
Pipe diameters and hydraulic gradients are checked at peak flow; otherwise head loss becomes excessive and pressure at tail ends falls.
Service reservoirs and clear water reservoirs provide balancing storage. Their capacity depends on the cumulative surplus and deficit between inflow and hourly demand; a higher peak increases the required balancing storage.
Pumping mains and pumps may be designed for peak hourly flow if direct pumping is used. If a reservoir is provided, pumps may run at a steadier rate and the reservoir absorbs the peaks.
Source works and treatment units are generally designed for maximum daily demand, while peak hour mainly governs distribution and storage.
Typical Indian peak factor is about 2 to 3 times the average daily demand, and may exceed 3 for small towns, summer days, or festivals. Minimum hourly demand may fall to 0.3–0.6 times average.

```
Demand (× average daily)
3.0 | ╭╮
2.5 | ╱ ╲ ╭╮
2.0 | ╱ ╲ ╱ ╲
1.5 | ╱ ╲ ╱ ╲
1.0 | ╭╯ ╰──╯ ╰╮
0.5 | ╭╯ ╰╮
0.0 +------------------------------- hours
 0 4 8 12 16 20 24
```

The sketch shows two daily peaks: a morning peak about 7–10 h and an evening peak about 17–21 h, with minimum demand at night. This pattern is typical of Indian conditions.

(c) (i) Use first-order BOD kinetics with base-10 rate constant:
BOD_t = L₀ [1 − 10^(−K_T t)]

Temperature correction:
K_T = K₂₀ φ^(T−20)

Given K₂₀ = 0.12 day⁻¹, φ = 1.056, T = 25°C.
K₂₅ = 0.12 × 1.056^(25−20) = 0.12 × 1.056⁵
K₂₅ = 0.12 × 1.313165883 = 0.157579906 day⁻¹.

At 25°C, 1-day BOD = 100 mg/l:
100 = L₀ [1 − 10^(−0.157579906 × 1)]
10^(−0.157579906) = 0.69569694
1 − 0.69569694 = 0.30430306
L₀ = 100 / 0.30430306 = 328.62 mg/l.

For 5-day, 20°C BOD:
BOD₅,₂₀ = L₀ [1 − 10^(−K₂₀ × 5)]
BOD₅,₂₀ = 328.62 [1 − 10^(−0.12 × 5)]
10^(−0.6) = 0.251188643
1 − 10^(−0.6) = 0.748811357
BOD₅,₂₀ = 328.62 × 0.748811357 = 246.07 mg/l.

BOD₅,₂₀ ≈ 246.07 mg/l. Assuming same ultimate BOD and no nitrification.

(c) (ii) Limitations of BOD:
Measures only biodegradable organic matter; ignores inorganic, volatile, refractory, and non-biodegradable organics.
Gives oxygen consumed by microorganisms, not total organic carbon or all pollutants.
Standard 5-day BOD is not ultimate BOD; it represents only a fraction of total carbonaceous demand.
Nitrogenous oxygen demand can interfere; unless nitrification is inhibited, BOD may include ammonia oxidation and overestimate carbonaceous BOD.
Test is slow, requiring 5 days; unsuitable for real-time or online process control.
Results depend on seeding, dilution, temperature, pH, nutrients, dissolved oxygen, and incubation conditions.
Toxic substances, chlorine, heavy metals, and extreme pH inhibit microorganisms, giving falsely low BOD.
Poor reproducibility and precision; requires skilled analyst and careful dilution.
For industrial wastes with variable composition or toxic/refractory organics, BOD correlates poorly with treatability.
Does not distinguish settleable, colloidal, and dissolved biodegradable matter.
Saline, radioactive, or biologically inhibitory wastes may require special adaptation.
COD, TOC, or oxygen uptake rate may be needed as supplementary tests.

(a) (i) A 50 cm well in an unconfined aquifer of saturated thickness 45 m yields 600 lpm under a drawdown of 3 m at the pumping…

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