Question

A surface water treatment plant treats a 6,000 L/day flow with alum coagulation, identified by the...

A surface water treatment plant treats a 6,000 L/day flow with alum coagulation, identified by the following stoichiometric reaction:

Al2(SO4)3 ∙ 14 H2O + 6HCO3- ↔ 2Al(OH)3 (s) + 6 CO2 + 14 H2O + 3 SO42-

Molecular weights (g/mol):

Al2(SO4)3 ∙ 14 H2O  =   594.5

HCO3-  =  61

Al(OH)3  =  78

SO42-  =  96.1

Jar tests indicate that 55 mg/L of alum are required to optimally treat the surface water. Removal of approximately 20.2 mg/L of suspended solids also occurs in this sytem. Estimate the mass of sludge disposed of daily (g/day).

Homework Answers

Answer #1

Ans. 86.59 g/L ( total sludge generated per day)

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A water treatment plant with an average flow of Q = 0.044 m3/second treats its water...
A water treatment plant with an average flow of Q = 0.044 m3/second treats its water with alum (Al2 (SO4)3 • 14H2O) at a dose of 25 mg/L. Alum coagulation is used to reduce the concentration of natural organic matter and reduce the alkalinity of water according to the equation given below. Find the total mass of alkalinity consumed. Assume all alkalinity to be bicarbonate. (10 points) Al2(SO4)3 • 14H2O + 6HCO3- 2Al(OH)3 (s) + 6CO2 + 14H2O + 3SO2-4
A coagulation treatment plant with a flow of 0.5 m3/s is dosing alum at 20 mg/l....
A coagulation treatment plant with a flow of 0.5 m3/s is dosing alum at 20 mg/l. Concentration of suspended solid is 25 mg/l. The effluent suspended solid concentration is 12 mg/l. The sludge content is 1 % and specific gravity of sludge is 3 %. What volume of sludge must be disposed of each day?
A water treatment plant treats 43200 m 3 of water in a day. Based on Jar...
A water treatment plant treats 43200 m 3 of water in a day. Based on Jar Test, the optimum dosage obtained when 50 mL of 1g/L is added into 2 L of water. Find I. the amount of alum required (kg) in a month II. the flow rate of the alum solution m 3 /day).
The Busy Town drinking water treatment plant recently changed coagulant aid from alum to ferrous sulfate...
The Busy Town drinking water treatment plant recently changed coagulant aid from alum to ferrous sulfate (FeSO4·7H2O) for TSS removal. The raw water contains 7 mg/L of TSS. Ferrous sulfate coagulation proceeds as follows: FeSO4·7H2O + 2 Ca(HCO3-) + 1⁄2O2 -> 2Fe(OH)3 (s) ↓ + 2CaSO4 + 13H2O If an optimum dose of 30 mg/L of ferrous sulfate is required to remove 90% of the TSS, calculate the mass of residuals (mg/L) generated in this process.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT