Civil Engineering > GATE 2026 SET-1 > Water Treatment
The maximum demand at a water purification plant has been estimated as 12 million litres per day. For the raw supplies, a rectangular sedimentation tank is to be designed with mechanical sludge removal arrangement. Consider depth of the tank as 4 m, detention period as 6 hours, and velocity of flow as 0.003 m/s.

The width (in m) of the detention tank is _________ (rounded off to two decimal places).

Correct : 11.57 m

Convert flow rate:
Q = 12 million litres/day = 12 × 106 / 86400 = 138.89 litres/s = 0.13889 m3/s
Volume of tank:
Detention period T = 6 hours = 21600 s
Volume = Q × T = 0.13889 × 21600 = 3000 m3
Length of tank:
L = velocity × T = 0.003 × 21600 = 64.8 m
Width of tank:
Volume = L × W × D
3000 = 64.8 × W × 4
W = 3000 / 259.2 = 11.57 m
This can also be verified using the cross-sectional area: A = Q/v = 0.13889/0.003 = 46.30 m² = W × D = W × 4 → W = 46.30/4 = 11.57 m
Correct answer: 11.57 m

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Related Topics

rectangular sedimentation tank width GATE 2026 GATE Civil 2026 Set-1 Q65 sedimentation tank design 12 MLD GATE water treatment sedimentation tank width GATE detention period velocity sedimentation tank GATE water purification tank width depth GATE civil

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