| Amount of raw water |
: A m³
(amount of water accumulated in tank and auxiliary facilities) |
| Target concentration |
: B mg/L |
| Sodium hyphochlorite |
: C% |
| Condition: chlorine consumption |
: Zero |
ϑCalculation Formula
Amount of injected sodium hypochlorite = 6.94 x A x B x 12/C mL
When the amount of water is 1m³ and target concentration is 1 mg/L
1 m³ = 1,000 L1 mg/L target
1 mg/L x 1,000 L = 1,000 mg = 1 g (amount of effective chlorine)
As sodium hypochlorite = concentration 12%, then 1 g x 100 ./. 12 = 8.33 g
As the specific gravity when the concentration of sodium hypochlorite is 12%,
then the volume becomes 6.94 mL
[Extra Information]
On our home page bulletin board, we were asked the following question:
"How much chlorine should be introduced as a countermeasure for Legionnaire's
disease to make the concentration of chlorine in a 10m³ tank 10 mg/L?"
As the amount of introduced chlorine in the above question is as follows:
10 m³ = 10,000 L10 mg/ target
the amount of required chlorine (not amount of reagent but effective component)
is as follows:
10 mg/L x 10,000 L = 100,000 mg = 100g
Here, let's assume that the reagent to introduce is sodium hypochlorite of
concentration 12%. Then, the required amount becomes as follows:
100 g x 100 ./. 12 = 833.3 g
As the specific gravity of the 12% concentration sodium hypochlorite is 1.2, and
then the volume becomes 694 mL.
The amount of water is not only the volume of the tank itself; an amount of water
for the filtering unit, inside the piping and entire other systems is required.