(a) An open drain is to be designed to prevent waterlogging for an area of 576 ha. Given that the drainage coefficient is 0·06 m/day, determine the capacity of the drain required and the dimensions of the trapezoidal section with side slopes 1 : 1 an

Question

(a) An open drain is to be designed to prevent waterlogging for an area of 576 ha. Given that the drainage coefficient is 0·06 m/day, determine the capacity of the drain required and the dimensions of the trapezoidal section with side slopes 1 : 1 and Lacey's f=1·0. Also, compute the slope. (20 marks)
(b) A 0·5 m diameter well fully penetrates an unconfined aquifer whose bottom is 150 m below the undisturbed ground water table. When pumped at a steady rate of 6·0 m³/min, the drawdowns observed in two observation wells at radial distances of 10 m and 50 m are respectively 10 m and 5 m. Determine the drawdown in the well. (15 marks)
(c) What is grit ? Why is it essential to remove the grit ? Why velocity control devices are essential with unaerated horizontal flow grit chambers ? Why are the velocity control devices not required with the aerated grit chambers ? Name any two velocity control devices used with grit chambers. (15 marks)

UPSC Answer Check · Accepted Answer

(a)

Drainage coefficient = 0.06 m/day.
Area = 576 ha = 576 × 10⁴ m² = 5.76 × 10⁶ m².

Capacity required:
Q = area × drainage coefficient = 5.76 × 10⁶ × 0.06 = 3.456 × 10⁵ m³/day
Q = 3.456 × 10⁵ / 86400 = 4.00 m³/s.

Using Lacey’s regime equations with f = 1.0:

V = (Q f² / 140)^(1/6) = (4 × 1 / 140)^(1/6) = (1/35)^(1/6) = 0.5529 m/s.
A = Q / V = 4 / 0.5529 = 7.234 m².
P = 4.75 √Q = 4.75 × 2 = 9.50 m.
R = A / P = 7.234 / 9.50 = 0.7615 m.

For a trapezoidal section with side slopes 1:1, let bed width = b, depth = y.
A = b y + y².
P = b + 2√2 y.

From P: b = 9.50 − 2√2 y.

Substituting in A:
(9.50 − 2√2 y)y + y² = 7.234
9.50 y − (2√2 − 1)y² = 7.234
1.8284 y² − 9.50 y + 7.234 = 0.

Solving:
y = [9.50 − √(9.50² − 4 × 1.8284 × 7.234)] / (2 × 1.8284) = 0.927 m.
The other root gives negative bed width, so it is rejected.

Then:
b = 9.50 − 2.8284 × 0.927 = 6.878 m.

Thus:
Bed width = 6.88 m
Flow depth = 0.927 m
Top width = b + 2y = 6.878 + 1.854 = 8.73 m

Slope by Lacey’s equation:
S = f^(5/3) / (3340 Q^(1/6)) = 1 / (3340 × 4^(1/6))
S = 1 / (3340 × 1.2599) = 0.0002376 = 1 in 4208.

So, capacity = 4.00 m³/s, slope = 2.38 × 10⁻⁴, i.e. 1 in 4208. Valid for a Lacey regime channel; add freeboard separately as per design practice.

(b)

Use the Dupuit–Thiem equation for steady radial flow to a fully penetrating well in an unconfined aquifer:
Q = πK(h₂² − h₁²) / ln(r₂/r₁).

Given:
Q = 6.0 m³/min = 0.10 m³/s.
Aquifer thickness H = 150 m.

At r₁ = 10 m, drawdown s₁ = 10 m, so h₁ = 150 − 10 = 140 m.
At r₂ = 50 m, drawdown s₂ = 5 m, so h₂ = 150 − 5 = 145 m.

ln(r₂/r₁) = ln(50/10) = ln 5 = 1.6094.
h₂² − h₁² = 145² − 140² = 21025 − 19600 = 1425 m².

Thus:
K = Q ln 5 / [π(h₂² − h₁²)]
K = 0.10 × 1.6094 / (π × 1425) = 3.595 × 10⁻⁵ m/s.

Well radius rw = 0.5 / 2 = 0.25 m.

Using the well and the first observation well:
Q = πK(h₁² − hw²) / ln(r₁/rw).

So:
h₁² − hw² = (h₂² − h₁²) × ln(r₁/rw) / ln(r₂/r₁)
= 1425 × ln(10/0.25) / ln 5
= 1425 × ln 40 / 1.6094
= 1425 × 3.6889 / 1.6094 = 3266.14 m².

Therefore:
hw² = 140² − 3266.14 = 19600 − 3266.14 = 16333.86 m²
hw = √16333.86 = 127.804 m.

Drawdown in the well:
sw = H − hw = 150 − 127.804 = 22.196 m ≈ 22.2 m.

Final: drawdown in the well = 22.2 m. Assumptions: steady, homogeneous isotropic unconfined aquifer, no recharge, fully penetrating well, Dupuit assumptions.

(c)
Grit is the inert, dense, abrasive mineral matter present in sewage, such as sand, gravel, cinders, eggshells and bone chips. It generally has specific gravity about 2.5–2.65 and particle size above about 0.2 mm.
Grit must be removed because it:
abrades and wears pumps, valves, pipes and other mechanical equipment;
settles in channels, pipelines, sumps and digesters, reducing capacity and causing blockages;
contaminates sludge, making sludge handling and digestion more difficult;
increases maintenance, cleaning and energy costs;
can damage downstream treatment units.
Velocity control devices are essential with unaerated horizontal-flow grit chambers because the horizontal velocity must remain nearly constant, usually about 0.3 m/s. If velocity is too high, grit does not settle; if too low, organic solids settle with grit, causing septicity and odour. Since sewage flow varies diurnally, a fixed channel section cannot maintain the required velocity. Hence velocity control devices adjust the velocity over the flow range.
Velocity control devices are not required with aerated grit chambers because compressed air creates a spiral rolling motion. The air supply can be adjusted to maintain agitation and velocity independent of flow variation. The rolling motion keeps organic matter in suspension while heavier grit settles at the bottom.
Two velocity control devices used with grit chambers: Sutro proportional weir and Parshall flume.

(a) An open drain is to be designed to prevent waterlogging for an area of 576 ha. Given that the drainage coefficient is 0·06…

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