(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…

(a) Discharge of wells in an unconfined aquifer

For steady radial flow to a fully penetrating well in an unconfined aquifer (Dupuit–Thiem), Q = πK(H² − h_w²)/ln(R/r_w), with H = 45 m and R = 500 m.

(i) For the same well and aquifer, Q ∝ (H² − h_w²). At 3 m drawdown, h_w = 42 m, so H² − h_w² = 2025 − 1764 = 261 m². At 6 m drawdown, h_w = 39 m, so H² − h_w² = 2025 − 1521 = 504 m². Q₂ = 600 × 504/261 = 1159 lpm (about 19.3 L/s, or 1668 m³/day). The discharge does not double, because the extra 3 m of drawdown reduces the saturated thickness.

(ii) With r_w = 0.25 m for the 50 cm well, K = Q ln(R/r_w)/[π(H² − h_w²)] = 864 × ln 2000/(π × 261) = 8.0 m/day, which is the same aquifer. For the 30 cm well, r_w = 0.15 m and the drawdown is 3 m, so Q = πK × 261/ln(500/0.15) = 600 × ln 2000/ln 3333 = 600 × 7.601/8.112 = 562 lpm (about 810 m³/day). The smaller diameter cuts the yield by only about 6%, because Q depends on the logarithm of the radius.

(b) Peak hour demand in water supply design

Peak hour demand is the highest hourly consumption on the day of maximum demand. In India it is taken as 1.5 times the maximum daily demand, which is about 2.7 times the average daily demand (a factor of 3 for small towns and about 2 for large cities).

Effect on design: the distribution network is the part sized for it. The mains, service pipes, valves and booster pumps must carry the peak flow, with fire demand checked as well, while keeping the residual head at the tail ends; otherwise head loss is excessive and pressure falls. Service reservoirs are provided to balance the difference between a steady supply and the variable demand, and a higher peak needs a larger balancing storage. The intake, treatment plant and raw-water mains are sized for the maximum daily demand only, because the reservoirs absorb the hourly peaks.

The sketch of the daily demand curve for Indian towns shows a low demand of about 0.3–0.6 times average at night, a morning peak of about 7 to 10 h, a smaller evening peak of about 17 to 20 h, and lower flow in the afternoon.

(c) BOD of sewage

(i) Using first-order kinetics to base 10, BOD_t = L(1 − 10^(−Kt)). At 25°C, K₂₅ = K₂₀ φ^(T − 20) = 0.12 × 1.056⁵ = 0.1576 per day. From the 1-day BOD of 100 mg/L, L = 100/(1 − 10^(−0.1576)) = 100/0.3043 = 328.6 mg/L. The 5-day BOD at 20°C = 328.6 × (1 − 10^(−0.12 × 5)) = 328.6 × 0.7488 = 246 mg/L. (If the rate constants are taken to base e, the result is about 309 mg/L.)

(ii) Limitations of BOD: it measures only the biodegradable organic matter, and mostly the carbonaceous part within 5 days, not the ultimate demand. Nitrifying bacteria can add oxygen demand unless they are inhibited. The test takes 5 days, so it is useless for process control. It needs a suitable seed, and toxic wastes, heavy metals or extreme pH give falsely low results. Results depend on dilution, temperature and nutrients, and precision is poor. It does not reflect the non-biodegradable load, so COD or TOC is needed as a supplement.