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 $40m^{3}/hour$.<br /><br />B. The surface area of the tank can be calculated using the formula: Surface Area = Volume / Depth. Given that the volume is $40m^{3}$ and the depth is 3 meters, the surface area of the tank is approximately $13.33m^{2}$.<br /><br />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 $150mg/l$, the effluent concentration is $10mg/l$, the flow rate is $40m^{3}/hour$, and the time is 2 hours, the sludge accumulation is approximately $0.28m^{3}/month$.<br /><br />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 $13.33m^{2}$ and the depth is 3 meters, the sludge capacity of the tank is approximately $40m^{3}$.