A municipal water treatment plant is designed to produce 20 million gallons per

Question

A municipal water treatment plant is designed to produce 20 million gallons per day (MGD) of potable water. The plant treats surface water using coagulation followed by sedimentation, filtration, and chlorination. Although the water is hard, it is not presently softened. The water is coagulated using a 30 mg/L dosage of alum (aluminum sulfate, Al2(SO4)3•14.3H2O, MW = 600). This chemical reacts in water to form Al(OH)3 as follows:

Al2(SO4)3 + 6HCO3- 2Al(OH)3(s) + 3SO4-2 + 6CO2

1. Some consumers might be concerned about adding aluminum to their drinking water. If the pH of the finished water (following filtration, chlorination, and final pH adjustment) is 8.0, and if the solubility of Al+3 is governed by the solubility of Al(OH)3, how much dissolved Al+3 is expected to remain in solution? Assume pKs = 32.9 for freshly precipitated Al(OH)3(s).

How many pounds of alum will be used per day when the plant is operated at the design flow rate using an alum dosage of 30 mg/L? (Hint: there is a very useful conversion factor you can use to solve this problem, and future problems of this nature: (mg/L) x MGD x 8.34 = lbs/day. The basis for this conversion factor is that 1 mg/L = 1 ppm = 1 gal/MG and that 1 gal of water weighs 8.34 lbs. In making this calculation, be sure to express the flow rate in MGD, not gal/day, or your answer will be a million times too high. If in doubt, check your answer using routine conversion factors.)

The treatment plant has two identical circular sedimentation basins operated in parallel, with each one treating half of the flow (10 MGD). Each basin has a hydraulic retention time of 3 hours and is 18 feet deep. What is the diameter of the basins? (Hint: there are 7.48 gallons in 1 ft3.)

Explanation / Answer

Solution:

1.Dissolved aluminium:

Alum = Al2(SO4)3.18H2O.

Molecular weights of Aluminium Al = 27

Sulfur S = 32

Oxygen O = 16

Hydrogen H = 2

Molecular weight of alum = (2*27)+(32+(4*16))*3+18*((2*2)+16) = 666

Molecular weight of Flocculant Al(OH)3. = (27)+3*(16+1) = 78

Molecular weight of aluminum Al+3 = 27.

Dosage of alum = 30 mg/l.

Total water per day = 20 million gallons (or) 20 MG.

Water supply = 20MGD (or) 75.8 Million litres per day

Dosage of alum = 30*75.8 = 2274 kg/day.

Dissolved flocculant per day = (78*2274)/(666) = 266.32 kg per day.

Dissolved Aluminum per day = (27*266.32)/(78) = 92.189 kg per day.

2.Pounds of alum used per day:

Alum dosage = 30 mg/l (or) 30 ppm (or) 30 gal/MG

Alum dosage per day = 30 * 20 = 600 gallons per day

1 gallons = 8.345 pounds

Alum dasage per day = 600*8.345 = 5007 pounds.

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