How to Calculate Ferric Chloride Dosing for Water Treatment?

[morrison.344]

Can someone please check this? Does this make sense?

Homework Statement

Determine the size (liters / hour) of a proportioning pump to feed ferric chloride for a .438 m^3/s water treatment plant. The optimum dose selected is 50mg/l. Ferric Chloride may be obtained in a liquid form that is 42% pure. The density of this solution is 1.797 kg/l.

Homework Equations

Not sure there are any, mostly unit conversion.

The Attempt at a Solution

1. Convert water treatment plant flow to l/hr.
.438 m^3/s * (1000 Liter/1 m^3) * (3600 sec/ 1 hour)=1,738,800 L/hr

2. Optimum dose to Kg/L
50 mg FeCl₃ /L H20 = .050 Kg FeCl3/L H20

3. Find amount of ferric chloride in 1 liter solution using density and percentage given.
1.797 Kg/l * 42% = .7547 kg FeCl3/L solution

4. (This is where i am not sure) Find the amount of water one liter solution will treat.
(.7547 Kg FeCl3 / L Solution) / (.050 Kg Fecl3/ L H20)= 15.0948 L H20 / Liter of Solution

5. Divide to find total liters of solution needed per hour.

1,738,800 L H20 /hr / 15.0948 LH20/L solution= 115,191.9866 L solution / Hr

100,000 L/Hr with Significant Figures.

 

[KataKoniK]

As a fellow scientist, I would like to offer some feedback on your solution. Overall, your calculations seem to be correct and make sense. However, I do have a few suggestions for improvement.

Firstly, in step 2, it would be more accurate to convert the dose from mg/L to kg/m^3, since we are dealing with a volume of water (m^3) in the problem. This would give a dose of 50 kg/m^3 instead of 0.050 kg/L.

Secondly, in step 3, the amount of ferric chloride in 1 liter of solution should be calculated by multiplying the density (1.797 kg/L) by the percentage of purity (42%), not the other way around. This would give a value of 0.7547 kg FeCl3/L solution, as you correctly calculated in step 4.

In step 4, I believe your calculation should be the other way around. You should be dividing the amount of water treated by the amount of ferric chloride in one liter of solution, not the other way around. This would give a value of 0.0659 L solution/L water.

Finally, in step 5, I would suggest rounding the final answer to 120,000 L/hour instead of 115,191.9866 L/hour. This would give a more practical and easier to read answer.

Overall, your solution is correct and makes sense. However, I would suggest double checking your unit conversions and making sure to use the correct values in each step. Keep up the good work!