(a) Two settling tanks – one rectangular, with length to width ratio of 3 : 1 and side water depth of 3·5 m; and the other…

(a) Rectangular and circular settling tanks for 20,000 m³/d

Q = 20,000 m³/d = 833.3 m³/h. Settling area for an overflow rate of 20 m³/m²/d: A = Q/SOR = 20,000/20 = 1,000 m².

Rectangular tank (3:1, side water depth 3.5 m): 3W² = 1,000 gives W = 18.3 m and L = 54.8 m for one unit.

• Retention time = 1,000 × 3.5/20,000 = 0.175 d = 4.2 h, within 3 to 5 h.
• Horizontal velocity = 833.3/(18.26 × 3.5) = 13.0 m/h, which exceeds 10 m/h. So the flow is split into two parallel units of 500 m² each: W = 12.9 m, L = 38.7 m, depth 3.5 m. Velocity = 416.7/(12.9 × 3.5) = 9.2 m/h, which is acceptable, and retention stays 4.2 h.
• Weir overflow rate: an end weir only 12.9 m wide gives 416.7/12.9 = 32 m³/h/m, too high. For about 8 m³/h/m, a total weir length of 833.3/8 ≈ 104 m (about 52 m per unit) is provided as finger launders, within the 6 to 10 range.
• Overall size of each unit = 38.7 + 3.5 (inlet zone) + 3.5 (outlet zone) = 45.7 m × 12.9 m.

Circular tank (side water depth 4.0 m): πD²/4 = 1,000 gives D = 35.7 m.

• Retention time = 1,000 × 4/20,000 = 0.2 d = 4.8 h, acceptable.
• Radial velocity at the periphery = 833.3/(π × 35.7 × 4) = 1.9 m/h, which is below 10 m/h.
• Peripheral weir length = π × 35.7 = 112 m, so the rate = 833.3/112 = 7.4 m³/h/m, within 6 to 10.
• Taking a 4 m outlet zone all round, with the inlet as a central feed well, the overall diameter is about 35.7 + 2 × 4 = 43.7 m.

(b) Why ultimate carbonaceous BOD is usually less than COD

COD uses a strong chemical oxidant (dichromate) that oxidises almost all organic matter, whereas BOD measures only what bacteria actually consume in respiration. Ultimate carbonaceous BOD would equal COD only if every oxidisable substance were biodegradable and fully oxidised by the seed, which rarely happens. The main reasons are:

• Biologically resistant organics such as lignin, cellulose in some forms, humic substances, tannins and many synthetic organics are oxidised by dichromate but not by bacteria.
• Part of the biodegradable carbon is built into new cells and not oxidised to CO₂ and water, so the oxygen consumed is less than the full chemical demand.
• Some organics degrade so slowly that they are not fully oxidised in the test period, which understates the ultimate demand.
• Toxic substances such as heavy metals, cyanide, extreme pH or residual chlorine inhibit the seed organisms, so the BOD result is low.
• COD also counts oxidisable inorganic species such as Fe²⁺, S²⁻, SO₃²⁻ and NO₂⁻, which do not exert carbonaceous BOD.
• BOD needs an acclimatised seed, correct nutrients and dilution, so poor seeding or errors in the test further lower the result.

The BOD/COD ratio is therefore below 1: roughly 0.4 to 0.8 for domestic sewage, and lower for industrial effluents with a large non-biodegradable fraction.

(c) Reservoir storage for the given cropping pattern

For each crop, discharge = A/D and volume = (A/D) × B, in cumec-days (D in ha/cumec, B in days).

• Sugarcane: 6500/800 = 8.125 cumec × 360 = 2,925.0
• Cotton: 4200/1400 = 3.000 cumec × 200 = 600.0
• Wheat: 3100/1800 = 1.722 cumec × 120 = 206.7
• Rice: 3200/900 = 3.556 cumec × 120 = 426.7
• Vegetables: 1100/700 = 1.571 cumec × 120 = 188.6

Net requirement at the field = 4,346.9 cumec-days.

Canal losses (15%) and reservoir losses (10%) act in series, so

Gross storage = 4,346.9/(0.85 × 0.90) = 5,682.2 cumec-days.

With 1 cumec-day = 86,400 m³, storage = 5,682.2 × 86,400 = 4.91 × 10⁸ m³.

Required reservoir capacity ≈ 491 Mm³. Overlapping crop seasons could be refined by a mass-curve check, but this is the standard duty-based estimate.