Chemical equation and calculation for remineralization system

Dear all,

I am designing a remineralization system(Lime and CO2 gas injection to BWRO permeate ). So, our company received consultant from a vendor.

Below chemical equation and calculation had calculated from a  vendor.

However, we don't know how to calculate below data "As such required: 39.22, 46.64, 85.86", "Actual alkalinity produced: 47.567, 87.5675".

If you know well chemical equations and this calculation, please let me know how to calculate.

Thank you.   

Dose Calculation for Lime + CO2

                                                    Ca(OH)2 + 2CO2 = Ca(HCO3)2
                                                          74           88            162
As such required:                         39.22      46.64        85.86
As CaCO3:                                        53           53              53
Dosages as Such provided:           40           50          92.045
Actual alkalinity produced:                        47.567      87.5675
 

Taxonomy

• Alkaline
• Industrial Water Treatment
• Brackish Water
• Chemical Analysis
• Industrial Use
• Reverse Osmosis
• Membrane Filtration
• Produced Water From Oil & Gas Industry
• Chemical Engineering
• Hydrogeology & Hydrochemistry

2 Answers

1. You decide that you want an alkalinity of 53 mg/L of calcium carbonate.

   You add the molar masses of Ca = 40 g/mole, C = 12 g/mole and three times O = 16 g/mole to find that 1 mole of CaCO3 or calcium carbonate weighs 100 g.

   You decide that you need 0.53 mole of calcium carbonate in 1000 L.

   With 1 mole of Ca(OH)2 and 2 moles of CO2 you can make 1 mole of Ca(HCO3)2, or 1 mole of CaCO3 alkalinity.

   While also knowing that 1 mole of H weighs 1 g, you now calculate the weights of 1 mole Ca(OH)2, 2 moles of CO2 and 1 mole of Ca(HCO3)2, which are 74, 88, and 162 g, respectively.

   You only need 0.53 mole, or 0.53*74 = 39.22 g of Ca(OH)2, 0.53*88 = 46.46 g of CO2, to make 0.53*162 = 85.86 g of Ca(HCO3)2, equivalent to the alkalinity of 53 g of CaCO3.

   So you weigh, as a matter of convenience, 40 g instead of 39.22 g of Ca(OH)2 and you add 50 g of CO2.

   If 50 g of CO2 react instead of 46.46 g, then you produce (50/46.46)*85.86 = 92.045 g of Ca(HCO3)2, but then you have not enough Ca(OH)2.

   So, 40 g of Ca(OH)2 shall react, then you produce (40/39.22)*85.86 = 87.568 g of Ca(HCO3)2 and some CO2 is left.

   In attached excel file you find the calculations also.

   2 Comments

   1. i have an additional question. According to above equation and calculation, vendor had calculated Free CO2(3.36ppm). Please let me how to calculate free CO2(3.36ppm), if you can.

      1 Comment reply

      1. When you add 50 ppm of CO2 and you only need 46.64 ppm, then the difference of 'Free' CO2 equals 50 - 46.64 = 3.36 ppm.

   2. Thanks!!

      1 Comment reply

      1. You are welcome. If you want to use a free, quick and simple model to calculate concentrations of 'mineralized solutions', you may like to download 'Chemical Equilibrium Diagrams' of the Royal Technical University of Sweden (KTU). 

         https://www.kth.se/che/medusa/chemeq-1.369367

2. Please give the detail analysis of permeate. This will enable me to make calculation. Most practical approach is to carry out the experiment in laboratory and then extrapolate to plant scale. For calculation the purity of lime must be known.  RO permeate is slightly acidic

   1 Comment

   1. Tank you for interest. Please refer to below RO permeate quality.

      Calcium, Ca - 0.01mg/L

      Magnesium, Mg - 0.04 mg/L

      Sodium, Na - 2.68 mg/L

      Potassium, K - 0.16 mg/L

      Ammonium, NH4 - 0.22 mg/L

      Bicarbonate, HCO3 - 0.08 mg/L

      Sulphates, SO4 - 0.02 mg/L

      Chlorides, Cl - 4.28 mg/L

      Nitrate, NO3 - 0.07 mg/L

      Total Dissolved Solids - 7.56 mg/L