Example Assume that you are assigned to design sedimentation tank for the water treatment plant of a community that consume 40m^3 water per hour.. Initially you proposed that the depth of tank is 3 meter and the detention I time is 2 hour What should be: A. the surface loading of the tank B. the surface area of the tank C. average sludge accumulation in the tank per month (assuming the raw water contains ; a suspended load of 150mg/l which is to be reduced to 10mg/l by sedimentation processes) D. Sludge capacity

. The surface loading of the tank is the rate at which water is added to the tank, which is given as .B. The surface area of the tank can be calculated using the formula: Surface Area = Volume / Depth. Given that the volume is and the depth is 3 meters, the surface area of the tank is approximately .C. The average sludge accumulation in the tank per month can be calculated using the formula: Sludge Accumulation = (Influent Concentration - Effluent Concentration) x Flow Rate x Time. Given that the influent concentration is , the effluent concentration is , the flow rate is , and the time is 2 hours, the sludge accumulation is approximately .D. The sludge capacity of the tank can be calculated using the formula: Sludge Capacity = Surface Area x Depth. Given that the surface area is and the depth is 3 meters, the sludge capacity of the tank is approximately .