Question

A water treatment plant is to be designed with a treatment capacity of 50 mgd using ferric sulfate as the coagulant with a dose of 35 mg/L as Fe2(SO4)3-9H20. The stock ferric sulfate chemical is supplied as a 60 percent solution of Fe2(SO4)39H20 with a specific gravity of 1.34. Calculate the feed rate for the ferric sulfate solution (in L/min). (Note: 1 mgd = 3,785 m /d, atomic weights: H = 1, C = 12,0 = 16, S = 32, Ca = 40, Fe = 55.8 g/mole)

Answer 1

Firstly, let's convert the treatment capacity from million gallons per day (mgd) to liters per day (L/day).

Given that 1 mgd = 3,785 m^3/day, and 1 m^3 = 1,000 L, we can say that 1 mgd = 3,785,000 L/day.

So, the treatment capacity is 50 mgd * 3,785,000 L/mgd = 189,250,000 L/day.

The dosage of ferric sulfate is given as 35 mg/L. So, the total mass of ferric sulfate required per day is 35 mg/L * 189,250,000 L/day = 6,623,750,000 mg/day, or 6,623.75 kg/day, since 1 kg = 1,000,000 mg.

Now, let's consider the concentration of the stock ferric sulfate solution, which is given as a 60% solution. This means that 60% of the solution by weight is ferric sulfate, and the rest is water. So, the mass of ferric sulfate in 1 L of the solution is 1.34 kg/L * 0.60 = 0.804 kg/L.

Finally, we can calculate the feed rate of the ferric sulfate solution. The feed rate is the total mass of ferric sulfate required per day divided by the mass of ferric sulfate in 1 L of the solution, which gives the volume of solution required per day. Dividing by the number of minutes in a day (24*60 = 1440) gives the volume of solution required per minute.

So, the feed rate is (6,623.75 kg/day) / (0.804 kg/L) = 8238.94 L/day.

To convert this to a per minute rate, we divide by the number of minutes in a day (1440):

Feed rate = 8238.94 L/day / 1440 min/day = 5.72 L/min.

This is the required feed rate for the ferric sulfate solution.