(a) A sewer having a 60 cm diameter (D) is required to flow at 0·4 depth (i.e., a proportional depth d/D = 0·4) on a grade…

(a) Principle. Two flows are equally self-cleansing when the drag (tractive force) they exert on the invert per unit area is the same: γ r s (part-full) = γ R S (full). Because the hydraulic mean depth r at 0.4 depth is smaller than R at full depth, the part-full sewer must be laid steeper: s = S·R/r.

1. Full-bore flow at the reference velocity (Manning's formula)

• D = 0.60 m and N = 0.015
• A = π(0.6)²/4 = 0.2827 m², P = π × 0.6 = 1.885 m, R = D/4 = 0.150 m
• V = (1/N) R^(2/3) S^(1/2), so 0.80 = (1/0.015)(0.150)^(2/3) S^(1/2) = 66.67 × 0.2823 × S^(1/2)
• S^(1/2) = 0.0425, so S = 0.001807, i.e. 1 in 553
• Q = 0.2827 × 0.80 = 0.226 m³/s

2. Hydraulic elements at d/D = 0.4

• Central angle of the wetted arc: cos(α/2) = 1 − 2d/D = 0.2, so α/2 = 78.46° and α = 156.93° (2.739 rad)
• Area ratio: a/A = (α − sin α)/2π = (2.739 − 0.392)/6.283 = 0.3735, so a = 0.1056 m²
• Perimeter ratio: p/P = α/360° = 0.4359, so p = 0.822 m
• Hydraulic mean depth ratio: r/R = 0.3735/0.4359 = 0.857, so r = 0.1285 m

3. Grade for equal self-cleansing

• s = S × R/r = 0.001807 × 0.150/0.1285 = 0.002109, i.e. 1 in 474
• Check: tractive force γrs = 9810 × 0.1285 × 0.002109 = 2.66 N/m², the same as γRS for full-bore flow at 1 in 553.

4. Velocities and discharges at the required grade of 1 in 474

• At 0.4 depth: v = (1/0.015)(0.1285)^(2/3)(0.002109)^(1/2) = 66.67 × 0.2546 × 0.04592 = 0.78 m/s, and q = 0.1056 × 0.78 = 0.082 m³/s (82 L/s)
• At full depth: V = (1/0.015)(0.150)^(2/3)(0.002109)^(1/2) = 66.67 × 0.2823 × 0.04592 = 0.864 m/s, and Q = 0.2827 × 0.864 = 0.244 m³/s (244 L/s)

The ratios v/V = 0.90 and q/Q = 0.34 agree with the partial-flow curves for constant N. So a 60 cm sewer laid at 1 in 474 and running 0.4 full (0.78 m/s, 82 L/s) is as self-cleansing as the same sewer running full at 0.8 m/s on 1 in 553. Note that at 0.4 depth the velocity is almost the reference value, yet a flatter grade would not be enough, because the drag on deposits depends on r·s and not on velocity alone.

(b)

(i) Aquitard. A saturated bed of low permeability, such as sandy or silty clay. It stores appreciable water but transmits it too slowly to supply a well. It does leak water vertically to the aquifers above or below it, which are then called leaky aquifers. It differs from an aquiclude (e.g. clay), which holds water but transmits practically none, and from an aquifuge (e.g. unfractured granite), which neither holds nor transmits water.

(ii) Specific retention (S_r). The volume of water a saturated soil or rock retains against gravity drainage, as a fraction of its total volume. The water is held as films on grains and in fine pores by molecular attraction and surface tension. Porosity n = S_y + S_r, where S_y is the specific yield. Clays have high S_r and low S_y, and gravels the reverse.

(iii) Storage coefficient (storativity, S). The volume of water an aquifer releases from storage, or takes into storage, per unit surface area per unit change in head. It is dimensionless.

• Confined aquifer: water is released by expansion of the water and compression of the aquifer skeleton, S = γ_w b(α + nβ), where b is the thickness, α the aquifer compressibility and β the water compressibility. Typical values are 10⁻⁵ to 10⁻³.
• Unconfined aquifer: the pores actually drain, so S ≈ specific yield, about 0.01 to 0.3.

(iv) Coefficient of permeability (hydraulic conductivity, K). The discharge per unit cross-sectional area under unit hydraulic gradient, i.e. the Darcy velocity when i = 1, in m/s or m/day. It depends on both the medium and the fluid, K = kγ/μ, where k is the intrinsic permeability of the medium. It falls by orders of magnitude from gravel (about 10⁻² m/s) through sand and silt to clay (below 10⁻⁹ m/s).

(v) Darcy's law. For laminar flow through a saturated porous medium, the discharge velocity is proportional to the hydraulic gradient:

V = K i and Q = K i A, where i = −dh/dl

Here V is the apparent velocity over the gross area; the actual seepage velocity through the pores is V/n. The law is valid for laminar flow, with a particle Reynolds number up to about 1 (at most about 10), and it underlies well hydraulics, seepage analysis and pumping tests.

(c)(i) Geometry from the figure: total height 30 m (3 m freeboard and 27 m water depth); top width 6 m; the downstream face is vertical for the top 10 m and then slopes at 0.75 H : 1 V over the lower 20 m, giving a base B = 6 + 15 = 21 m. There is no tailwater.

Assumptions: the question gives no uplift data, so full uplift varying linearly from γ_w h = 10 × 27 = 270 kN/m² at the heel to zero at the toe (no drainage gallery) is taken, and the no-uplift case is also given. q = 1400 kN/m² is the shear strength of the dam–foundation joint, used in the shear-friction factor. γ_c = 24 kN/m³.

Forces per metre length (lever arms from the toe):

• W₂ (rectangle 6 × 30 m): 6 × 30 × 24 = 4320 kN ↓, arm 15 + 3 = 18 m, resisting moment 77,760 kN·m
• W₁ (triangle 15 × 20 m): ½ × 15 × 20 × 24 = 3600 kN ↓, arm ⅔ × 15 = 10 m, resisting moment 36,000 kN·m
• Water pressure P: ½ × 10 × 27² = 3645 kN →, arm 27/3 = 9 m, overturning moment 32,805 kN·m
• Uplift U: ½ × 270 × 21 = 2835 kN ↑, arm ⅔ × 21 = 14 m, overturning moment 39,690 kN·m

ΣV = 4320 + 3600 − 2835 = 5085 kN and ΣH = 3645 kN

I. Factor of safety against sliding

• Friction alone: FS = μΣV/ΣH = 0.70 × 5085/3645 = 0.98. This is below 1, so friction alone cannot resist sliding.
• Shear-friction factor: SFF = (μΣV + Bq)/ΣH = (3559.5 + 21 × 1400)/3645 = 32,959.5/3645 = 9.04. This exceeds the usual minimum of 3 to 5, so the dam is safe against sliding once the shear strength of the joint is counted.

II. Factor of safety against overturning (about the toe)

• FS = ΣM_R/ΣM_O = 113,760/(32,805 + 39,690) = 113,760/72,495 = 1.57, above the required 1.5, so the dam is safe.
• The resultant cuts the base at (113,760 − 72,495)/5085 = 8.12 m from the toe, so e = 10.5 − 8.12 = 2.38 m, which is less than B/6 = 3.5 m. The resultant lies in the middle third and there is no tension at the heel.

If uplift is ignored (ΣV = 7920 kN): friction FS = 1.52, SFF = 9.59 and overturning FS = 113,760/32,805 = 3.47. The comparison shows how strongly uplift governs stability, and why drainage galleries and a grout curtain are provided.

(c)(ii) Delta (Δ) is the total depth of water a crop needs over its whole base period, for example about 120 cm for sugarcane. Duty (D) is the area that one cumec of water, flowing continuously for the base period B, can irrigate (ha/cumec). The two are related by:

D = 8.64 B/Δ ha/cumec (B in days, Δ in metres)

The kor period is the critical period of the first watering after sowing, when the demand rate is highest, and the kor depth is the depth applied in it. The outlet capacity is therefore fixed by kor watering, not by the average over the season.

Calculation:

• D = 8.64 × 30/0.17 = 1524.7 ha/cumec
• Q = area/duty = 2200/1524.7 = 1.44 m³/s

The outlet must deliver about 1.44 cumecs.