(a) A catchment has six rain gauge stations. In a year, the annual rainfall recorded by the rain gauges are as follows: |…

(a)(i) The standard error in the estimation of mean rainfall for an existing set of rain gauges is obtained from the sample mean and sample standard deviation.

Mean rainfall of the six stations: P̄ = (90 + 100 + 200 + 130 + 120 + 150) ÷ 6 = 790 ÷ 6 = 131.667 cm.

Using the sample standard deviation formula: s = √[Σ(P − P̄)² ÷ (n − 1)]

Deviations and squares:

• A: 90 − 131.667 = −41.667; square = 1736.111 cm²
• B: 100 − 131.667 = −31.667; square = 1002.778 cm²
• C: 200 − 131.667 = 68.333; square = 4669.444 cm²
• D: 130 − 131.667 = −1.667; square = 2.778 cm²
• E: 120 − 131.667 = −11.667; square = 136.111 cm²
• F: 150 − 131.667 = 18.333; square = 336.111 cm²

Σ(P − P̄)² = 7883.333 cm² s² = 7883.333 ÷ (6 − 1) = 1576.667 cm² s = √1576.667 = 39.707 cm

Standard error of the mean: SE = s ÷ √n = 39.707 ÷ √6 = 16.210 cm

Coefficient of variation: Cv = 100 × s ÷ P̄ = 100 × 39.707 ÷ 131.667 = 30.157%

Percentage standard error: = Cv ÷ √n = 30.157 ÷ √6 = 12.312%

Final answer: Standard error = 16.21 cm, or 12.31% of the mean.

(a)(ii) The optimum number of rain gauges is found by the hydrologic relation: N = (Cv ÷ E)² where E is the allowable percentage error.

For 22% error: N = (30.157 ÷ 22)² = (1.3708)² = 1.879 Adopt N = 2 gauges. Thus the existing 6 gauges are more than sufficient for 22% error.

For 5% error: N = (30.157 ÷ 5)² = (6.0314)² = 36.378 Adopt N = 37 gauges.

Final answer: Optimum number = 2 gauges for 22% error and 37 gauges for 5% error. This assumes the same coefficient of variation and that the allowable error is expressed as a percentage of the mean rainfall.

(b) For steady radial flow to a fully penetrating well in a confined aquifer, Thiem’s equation is used: s₁ − s₂ = Q × ln(r₂ ÷ r₁) ÷ (2πT) where s is drawdown, r is radial distance, Q is discharge, and T is transmissibility.

Given: Q = 0.04 m³/s r₁ = 20 m, r₂ = 200 m s₁ = 3.5 m, s₂ = 0.5 m s₁ − s₂ = 3.5 − 0.5 = 3.0 m

ln(r₂ ÷ r₁) = ln(200 ÷ 20) = ln 10 = 2.302585

Therefore: T = Q × ln(r₂ ÷ r₁) ÷ [2π(s₁ − s₂)] T = 0.04 × 2.302585 ÷ (2π × 3.0) T = 0.0921034 ÷ 18.84956 T = 0.004886 m²/s

In m²/day: T = 0.004886 × 86400 = 422.2 m²/day

Aquifer thickness b = 40 m. Permeability k = T ÷ b: k = 0.004886 ÷ 40 = 1.2215 × 10⁻⁴ m/s

In cm/s: k = 1.2215 × 10⁻⁴ × 100 = 0.01222 cm/s

In m/day: k = 1.2215 × 10⁻⁴ × 86400 = 10.55 m/day

Final answer: Transmissibility T = 4.886 × 10⁻³ m²/s = 422.2 m²/day. Permeability k = 1.222 × 10⁻⁴ m/s = 0.01222 cm/s = 10.55 m/day. Valid for steady, confined, homogeneous, isotropic aquifer with negligible well loss.

(c) For the sketch, draw time on the horizontal axis and soil moisture on the vertical axis. Mark two horizontal lines: field capacity, FC, above and permanent wilting point, PWP, below.

• A frequent-irrigation curve is a small-amplitude saw-tooth lying high between FC and PWP. Its average moisture content is close to FC.
• An infrequent-irrigation curve is a large-amplitude saw-tooth. It starts near FC, falls close to PWP, and then rises again after irrigation. Its average moisture content is much lower.
• FC and PWP are soil properties and remain unchanged by irrigation frequency. Only the average moisture content and the fluctuation between successive irrigations change.

Frequent irrigation keeps the soil moisture high and reduces crop stress, but excessive frequency may cause waterlogging and poor aeration. Infrequent irrigation allows greater depletion, lowers the average moisture, and may approach PWP, causing moisture stress to the crop.

Calculation: Available water in soil = 50% of root-zone depth = 0.50 × 100 mm = 50 mm

Consumptive use = 3 mm/day

Frequency of irrigation: = available water ÷ consumptive use = 50 ÷ 3 = 16.67 days

Gross depth of water to be applied: = net depth ÷ irrigation efficiency = 50 ÷ 0.80 = 62.5 mm

Final answer: Frequency of irrigation = 16.67 days, approximately 1.8 irrigations per month. Depth of water to be applied = 62.5 mm.

(d) For carbonaceous ultimate BOD, use the stoichiometric oxygen demand formula for an organic compound C_aH_bO_cN_d, assuming nitrogen is released as ammonia: O₂ required = a + b/4 − c/2 − 3d/4 mol O₂ per mol of compound.

For glutamic acid C₅H₁₀N₂O₃: a = 5, b = 10, c = 3, d = 2 O₂ = 5 + 10/4 − 3/2 − 3×2/4 O₂ = 5 + 2.5 − 1.5 − 1.5 = 4.5 mol O₂/mol

Molecular weight of C₅H₁₀N₂O₃: = 5×12 + 10×1 + 2×14 + 3×16 = 146 g/mol

UBOD from 100 mg/L glutamic acid: = 100 × (4.5 × 32) ÷ 146 = 100 × 144 ÷ 146 = 98.63 mg/L

For glucose C₆H₁₂O₆: a = 6, b = 12, c = 6, d = 0 O₂ = 6 + 12/4 − 6/2 = 6 + 3 − 3 = 6 mol O₂/mol

Molecular weight of C₆H₁₂O₆: = 6×12 + 12×1 + 6×16 = 180 g/mol

UBOD from 100 mg/L glucose: = 100 × (6 × 32) ÷ 180 = 100 × 192 ÷ 180 = 106.67 mg/L

Total UBOD of mixture: = 98.63 + 106.67 = 205.30 mg/L

For BOD₅: BOD₅ = UBOD × (1 − e^(−k t)) k = 0.23/d, t = 5 d k t = 0.23 × 5 = 1.15 e^(−1.15) = 0.31664 1 − e^(−1.15) = 0.68336

BOD₅ = 205.30 × 0.68336 = 140.29 mg/L

Final answer: UBOD = 205.30 mg/L. BOD₅ = 140.29 mg/L. This treats ultimate BOD as carbonaceous demand; nitrogenous oxygen demand is not included.

(e) The observed positive tubes are:

• Serial dilution 1.0: 5 positives
• 0.1: 5 positives
• 0.01: 3 positives
• 0.001: 2 positives
• 0.0001: 0 positives

For MPN determination, select three consecutive dilutions such that the highest dilution has at least one positive tube and the next higher dilution has none. The highest positive dilution is 0.001 with 2 positives, and the next dilution 0.0001 has zero positives.

Thus the three consecutive dilutions to be used are: 0.1, 0.01, 0.001 with positive tubes 5, 3, 2.

From the given standard MPN table, the row for positive tubes 5 | 3 | 2 gives: MPN = 140 per 100 mL.

Final answer: MPN = 140 coliforms per 100 mL of river water sample.