How do you make drinking water nonaggressive (Larson > 0.5) by physical and not chemical means?
The Larson index indicates the water’s corrosive potential:
The definition is as follows: corrosive index of metals, according to LR (Larson Ratio).
The Larson is equal to: LR = (2 [SO42-] + [Cl-] ) / [HCO3-] with concentrations expressed in moles/liters.
If LR is < 0.2 no probability of corrosion/p>
between 0.2 and 0.4 low probability of corrosion
Between 0.4 an 0.5 slight probability of corrosion
Between 0.5 and 1 Moderate probability of corrosion
> 1 Clear risk of corrosion.
If you want to lower the Larson index, you need to lower sulfates and chlorine or increase hydrogen carbonate.
This is not the index that determines how aggressive the water is. Rather it’s the equilibrium of the water
”Aggressive” water occurs if the saturation index SI = pH - pHs is less than -0.015,
hard, if it is greater than 0.015,
and otherwise at equilibrium.
To answer the question about lowering the Larson index or rather to ensure the water is less aggressive, there really is no physical solution (CO2 degassing via tower or casing if there is too much gas) otherwise I don’t see any other way except adding calcium (with lime, marine or land limestone, calcium carbonate;etc).
There is also the Ryznar index also called the stability index, which is calculated according to a simple formula
Ryznar index = 2 pHs - pH
This index gives a good indication of the water’s tendency to be hard or corrosive.
So you have to calculate the water’s calcium-carbonate balance : Hallopeau Dubin (Equilibriums) or Langelier (LPLwin) to determine whether the water is hard or aggressive.
Aeration will add oxygen to the water and remove CO2 which will increase pH. Higher pHs are less corrosive and more protective.