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allsm [11]
3 years ago
13

A process engineer performed jar tests for a water in order to determine the optimal pH and dose using alum. A test was conducte

d by first dosing each jar with the same alum dose of 10 mg/L and varying pH in each jar from 5.0 to 7.5 with an increment of 0.5 unit of pH. After the first test, he/she plotted the results of remaining turbidity versus pH and found the optimal pH was 6.25. He/she continuously perform a second set of jar tests by holding the optimal pH of 6.25 constant and varying alum doses from 10 to 15 mg/L. Here are the results:
Results of Jar Tests for raw water at optimal pH of 6.25
Turbidity of raw water = 15 NTU
Alkalinity of raw water = 5 mg/L expressed as CaCO3.
Alum Dose 10 11 12 13 14 15
(mg/L)
Turbidity 5.0 4.6 4.5 3.0 5.0 6.0
Remaining
(NTU)
Determine:
1) Plot turbidity versus dose and find the optimal dose of alum (mg/L) with the water lowest remaining turbidity.
2) The theoretical amount of alkalinity consumed at the optimal dose expressed as CaCO3, mg/L.
3) Compared with the theoretical alkalinity from the above results, is the raw water alkalinity sufficient for the coagulation? If not, what kinds of chemical do you need in order to enhance the alkalinity?

Engineering
1 answer:
IRINA_888 [86]3 years ago
3 0

Answer:

1) 13 mg/liters

2)  72.22 mg/lit

3)  The Alkanity in raw water is not sufficient enough and the kinds of chemicals needed to enhance its acidity are

  • CaO
  • KOH
  • Na2CO3
  • NaOH
  • CO2
  • NaHCO3

Explanation:

1) plot of turbidity versus dose and optimal dose of Alum ( mg/L )

Optimal dose of Alum = 13 mg/liters from the graph attached below

2) Theoretical amount of alkalinity consumed at the optimal dose can be calculated as follows

Alkanity is due to HCO^-_{3}

given optimal dose of Alum = 13 mg/liters for question 1

I mole of alum = 2 moles of AL(OH)3

666 grams of alum = 2*27 = 54 grams of AL(OH)3

hence 1 mole of AL^{+3}  = (13/54 ) mMole / lit

The  moles of HCO3 = 6 * \frac{13}{54}   because 1 mole of Alum reacts with 6 moles of HC03

[HCO3] as CaCO3 = 6 * (13/54) * 50

                               = 72.22 mg/lit (theoretical amount of alkalinity consumed)

3) The Alkanity in raw water is not sufficient enough and the kinds of chemicals needed to enhance its acidity are

  • CaO
  • KOH
  • Na2CO3
  • NaOH
  • CO2
  • NaHCO3

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