(a) (i) What is Lacey's Silt Factor ? What is its value for standard silt ? Design an earthen canal using Lacey's silt theory…

(a)(i) Lacey's silt factor (f) expresses how the size of the bed material controls the regime (non-silting, non-scouring) dimensions of a channel in alluvium:

f = 1.76 √d_mm, where d_mm is the mean diameter of the bed material in millimetres.

Standard silt has f = 1.0, corresponding to d = (1/1.76)² ≈ 0.32 mm. Coarser material gives f > 1 and finer material f < 1.

Design data: Q = 15 m³/s, f = 1.0, side slope ½ H : 1 V.

1. Regime velocity: V = (Qf²/140)^(1/6) = (15/140)^(1/6) = 0.689 m/s

2. Area: A = Q/V = 15/0.689 = 21.77 m²

3. Hydraulic mean depth: R = (5/2)(V²/f) = 2.5 × 0.689² = 1.19 m

4. Wetted perimeter: P = 4.75√Q = 4.75√15 = 18.40 m

5. Bed slope: S = f^(5/3)/(3340 Q^(1/6)) = 1/(3340 × 15^(1/6)) = 1/(3340 × 1.570) = 1 in 5245

6. Section dimensions. Each ½ : 1 side slope has a length D√(1 + ¼) = 1.118D, so:

• P = B + 2.236D
• A = BD + 0.5D²

Eliminating B: A = PD − 1.736D², i.e. 1.736D² − 18.40D + 21.77 = 0.

Solving: D = 1.36 m and B = 18.40 − 2.236 × 1.357 = 15.36 m

Check: A = 15.36 × 1.357 + 0.5 × 1.357² = 21.77 m² and R = 21.77/18.40 = 1.18 m, consistent with the regime values.

Adopt: bed width 15.4 m, full supply depth 1.36 m, side slopes ½ : 1, bed slope 1 in 5250, with freeboard and banks as per the canal class.

(a)(ii) Spillways. A spillway is the safety valve of a dam. It passes surplus flood water from the reservoir safely to the river downstream, so that the reservoir never rises above its maximum water level and the dam is not overtopped. It must carry the design flood, and it ends in an energy dissipator (stilling basin, ski-jump bucket or roller bucket) that protects the river bed at the toe.

Spillways at the named dams:

• Bhakra (concrete gravity): controlled overflow (ogee) spillway on the dam body, with four radial crest gates.
• Ramganga, Kalagarh (earth and rock-fill): gated chute spillway built separately from the embankment, with five radial gates, because water cannot be passed over an earth dam.
• Hirakud (composite earth, concrete and masonry): gated ogee overflow spillways in the concrete and masonry sections, with crest gates and under-sluices.
• Rihand (concrete gravity): gated ogee overflow spillway on the dam body.

Forces on a mass-concrete (gravity) dam:

• self-weight;
• water pressure, from the reservoir and the tailwater;
• uplift pressure;
• silt pressure;
• wave pressure;
• ice pressure (in cold regions);
• earthquake forces, both horizontal and vertical, including hydrodynamic pressure of the reservoir.

1. Water pressure. This is the main destabilising force. On a vertical upstream face the pressure rises linearly from zero at the surface to γ_w h at the base, so the resultant is P = ½ γ_w h² per metre run, acting horizontally at h/3 above the base. If the upstream face is battered, the weight of water on the slope adds a vertical, stabilising component. Tailwater pressure acts on the downstream face in the opposite direction.

2. Uplift pressure. Water seeping through the foundation and through joints in the body pushes upward on the base and reduces the effective weight of the dam.

• Without drains, the uplift varies linearly from γ_w h at the heel to γ_w h′ (tailwater head) at the toe, and acts on the full area.
• With a drainage gallery, the intensity at the line of drains is reduced to the tailwater pressure plus one-third of the difference between reservoir and tailwater pressures, which cuts the total uplift sharply.

Uplift lowers the resistance to both sliding and overturning, which is why grout curtains and drainage holes are provided.

(b) Given 100 persons, peak flow 240 L/min and L/B = 2.5.

1. Surface area for sedimentation. 240 L/min is 24 units of 10 L/min: Area = 24 × 0.92 = 22.08 m²

2. Volumes required:

• Sedimentation: 22.08 × 0.3 = 6.62 m³
• Sludge digestion: 0.032 × 100 = 3.20 m³
• Digested-sludge storage for one year: 0.0002 × 100 × 365 = 7.30 m³
• Total liquid capacity = 17.12 m³

3. Plan dimensions: 2.5B² = 22.08, so B = 2.97 m and L = 7.43 m. Provide L = 7.5 m and B = 3.0 m (area 22.5 m²).

4. Depth:

• Liquid depth = 17.12/22.5 = 0.76 m, made up of 0.30 m for sedimentation, 0.14 m for digestion and 0.32 m for sludge storage.
• Adding 0.3 m freeboard gives 1.06 m. Provide 1.1 m overall depth.

Result: a rectangular septic tank 7.5 m × 3.0 m × 1.1 m overall (including 0.3 m freeboard). It should have inlet and outlet tees or baffles dipping below the scum layer, a vent pipe, and be desludged once a year. The effluent goes to a soak pit or dispersion trench.

(c) Data: d = 0.6 mm, φ = 0.85, G = 2.67; bed L = 0.65 m, η = 0.45; expanded η_e = 0.75; ρ = 998.2 kg/m³, μ = 1.002 × 10⁻³ N·s/m².

1. Terminal settling velocity of the grains

v_s = [4g(G − 1)d/(3C_D)]^½, with C_D = 24/Re + 3/√Re + 0.34 and Re = φρv_s d/μ.

Trial v_s = 0.10 m/s gives Re = 0.85 × 998.2 × 0.10 × 0.0006/1.002 × 10⁻³ = 50.8, C_D = 1.23 and v_s = 0.103 m/s. Iterating to convergence gives Re = 53.0 and C_D = 1.205:

v_s = [4 × 9.81 × 1.67 × 0.0006/(3 × 1.205)]^½ = 0.104 m/s

Since 1 < Re < 10⁴, the flow is in the transition range and Stokes' law cannot be used.

2. Backwash velocity for η_e = 0.75

Using the expansion relation η_e = (v_b/v_s)^0.22, so v_b = v_s η_e^(1/0.22) = v_s η_e^4.545:

v_b = 0.1043 × 0.75^4.545 = 0.1043 × 0.2705 = 0.028 m/s (about 28 mm/s or 1.7 m/min)

3. Expanded depth

The volume of sand is unchanged, so L(1 − η) = L_e(1 − η_e):

L_e = 0.65 × (1 − 0.45)/(1 − 0.75) = 1.43 m

4. Head loss through the expanded bed

Once the bed is fluidised, the head loss equals the submerged weight of the grains per unit area:

h_L = L(1 − η)(G − 1) = 0.65 × 0.55 × 1.67 = 0.60 m

The same value follows from the expanded bed: 1.43 × 0.25 × 1.67 = 0.60 m. The bed expands by 120%, the head loss stays constant at about 0.6 m once fluidised, and the wash-water troughs must be set above the 1.43 m expanded level so that no sand is lost.