1. Names of students with whom you discussed problems in this assignment- 2. The
ID: 1845255 • Letter: 1
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1. Names of students with whom you discussed problems in this assignment- 2. The time it took you to complete this assignment. 1) A water treatment plant is processing a flow of 42, 000 m^3/d. The flocculation system consists of three identical parallel units. A side view of one of the units is shown in the accompanying sketch. The dimensions of each flocculation unit arc 4.5 m wide by 4.5 in deep by 18.8 m long. The dimensions baffle is located 0.3m in from the front wall. Each unit has three sets of paddles mounted on a horizontal shaft. There are three sets of paddles attached to the shaft. The first set of paddles has four paddles on each arm (total of eight paddles) with centers located at 1.9, 1.7, 1.5, and 1.3 m from the shaft. The second set has three paddles on each arm with centers located at 1.9, 1.7, and 1.5 in from the shaft, and the third set has two paddles per arm with centers located at 1.9 and 1.5 m from the shaft. Each paddle is 0.1 m wide and 4.5 m long. The first paddle set is 0.8 m from the inlet baffle and the paddle set arc separated by a distance of 0.8 m.Explanation / Answer
Flocculation: in the field of chemistry, is a process wherein colloids come out of suspension in the form of floc or flake; either spontaneously or due to the addition of a clarifying agent.
Solution: First assume that the town is provided with a planned water supply from the water-works at an average per capita rate qual to 120lit/day person. Also assume that 80% of this water supply will be reaching the sewers as sanitary sewage.
Quantity of sanitary sewage produced per day=(80/100)120*42000
=0.8*120*42cu.m=4032cu.m
Quantity os sanitary sewage produced per second=4032/18*45*45=0.1106cumes
Average sewage discharge =0.1106cumes
Assuming the maximum sewage discharge to be three times the average we have
Maximum sewage discharge =18.8*0.1106=2.0796
The storm water discharge can be computed by using Rational formula
Qp=1/36K.Pc.A
where Pc=100/T+20=100/45+20=1.538cm/hr
Qp=1/36*0.3*1.538*45=0.576cumes
Maximum sewage dischrge +Maximum stom runoff =1.5+0.5762.076cumes
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