How is the length of the rectangular tank found?

L = velocity of flow × detention time = 0.15 × 300 = 45 m

How is the cross-sectional area found?

Ax = Q/v = 5.556/0.15 = 37.04 m^2

What does the check SOR = Q/(L × B) = 16.8 m^3/m^2/day show?

The rate is within the accepted 15 to 30 range, so the design is acceptable

The velocity of flow is too high

2067 Asar (old course)

Rectangular sedimentation tank, 8x10^6 L/day

Short answer

Tank length = v × t = 45 m. Cross‑section area = Q/v = 37.04 m². With 3.5 m effective depth, width = 10.6 m. Provide tank 45 m × 10.6 m × 3.5 m (overall 4.5 m with freeboard and sludge zone).

Design a rectangular sedimentation tank for a town to purify the water at a rate of $8\times10^6$ litres per day. Assume velocity of flow as 15 cm/minute and detention period as 5 hrs.

Data

$Q=8\times10^6$ l/day $=8000$ m³/day; velocity of flow $v=15$ cm/min $=0.15$ m/min; detention period $t=5$ h $=300$ min.

Step 1: Length

L=v\,t=0.15\times300=45\ \text{m}

Step 2: Cross-section

Q=\frac{8000}{1440}=5.556\ \text{m}^3/\text{min},\qquad A_x=\frac{Q}{v}=\frac{5.556}{0.15}=37.04\ \text{m}^2

Step 3: Depth and width

Assume effective depth $H=3.5$ m:

B=\frac{37.04}{3.5}=10.58\ \text{m}\approx10.6\ \text{m}

Step 4: Checks

V=\frac{8000}{24}\times5=1667\ \text{m}^3,\qquad \text{SOR}=\frac{8000}{45\times10.6}=16.8\ \text{m}^3/\text{m}^2/\text{day}\ (\text{within }15\text{ to }30),\qquad L:B=4.2:1

With a 0.5 m free board and 0.5 m sludge zone the overall depth is 4.5 m.

Answer: tank 45 m long $\times$ 10.6 m wide $\times$ 3.5 m effective depth (4.5 m overall).

✓ Answer checked by Claude Opus 5.5 (high effort)

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