There are several calculations you will need to understand pertaining to fluoridation, such as feed rate for dry and solution feeders. Lets' take a look at them below.

Percent Fluoride Ion in a Compound

When calculating the percent fluoride ion present in a compound, we need to know the chemical formula for the compound and the atomic weight of each element in the compound. The first step is to calculate the molecular weight of each element in the compound (number of atoms x atomic weight = molecular weight). Then we calculate the percent fluoride in the compound using the following equation:

$LaTeX: \%Fluoride=\frac{MolecularWeightofFluoride}{MolecularWeightofCompound}\times100$

Available fluoride ion concentrations abbreviated as AFI in the calculations that follow.

Example:

Given the following data, calculate the percent fluoride in sodium fluoride (NaF):

Element	No. of Atoms	Atomic Weight	Molecular Weight
Na	1	22.997	22.997
F	1	19.00	19.00
Molecular weight of NaF = 41.997

Now, calculate the percent fluoride in NaF:

$LaTeX: \%Fluoride=\frac{MolecularWeightofFluoride}{MolecularWeightofCompound}\times100$

$LaTeX: \%Fluoride=\frac{19.00}{14.997}\times100=45.2\%$

This would make the AFI equal to 0.452.

Note: The molecular weight of hydrofluorosilicic acid (H₂SiF₆) is 144.08, and the molecular weight of sodium silicofluoride (Na₂SiF₂) is 188.06.

Note: The AFI for the three commonly used fluorides are:

NaF = 0.452 or 45.2%
H₂SiF₆ = 0.792 or 79.2%
Na₂SiF₂ = 0.607 or 60.7%

Fluoride Feed Rate

Adjusting the fluoride level in a water supply to an optimal level is accomplished by adding the proper concentration of a fluoride chemical at a consistent rate. To calculate the fluoride feed rate of any fluoridation feeder in terms of pounds of fluoride to be fed per day, it is necessary to determine:

Dosage
Maximum pumping rate (capacity)
Chemical purity
Available fluoride ion concentration

The fluoride feed rate formula is a general equation used to calculate the concentration of a chemical added to water. It will be used for all fluoride chemicals except sodium fluoride when used in a saturator.

Remember: mg/L is equal to ppm.
Note: Chemical purity is 98.5%, unless specified otherwise.

The fluoride feed rate (the amount of chemical required to raise the fluoride content to the optimal level) can be calculated with the following equations:

Or

Example:

A water plant produces 2000 gpm and the city wants to add 1.1 mg/L of 98.5% purity fluoride. What would the fluoride feed rate be, in lb/day, if sodium silicofluoride was used?

First, convert gpm to MGD:

Now, determine the feed rate, in lb/day:

The fluoride feed rate is 44.2 lb/day. Some feed rates from equipment design data sheets are given in grams per minute. To convert to g/min, do the following conversion:

Now, let's work the same problem, but this time determine the feed rate if hydrofluorosilicic acid was used instead:

Example 2:

A water plant produces 2000 gpm and the city wants to add 1.1 mg/L of 98.5% purity fluoride. What would the fluoride feed rate be, in lb/day, if hydrofluorosilicic acid was used?

Now, let's work it one last time using sodium fluoride (NaF):

Example 3:

A water plant produces 2000 gpm and the city wants to add 1.1 mg/L of 98.5% purity fluoride. What would the fluoride feed rate be, in lb/day, if sodium fluoride was used?

As you can see from the calculations above, for a plant that produces 2000 gpm, hydrofluorosilicic acid (H₂SiF₆) will have the lowest feed rate at 10.7 gallons to obtain 1.1 mg/L of fluoride in the water.

We've seen how to work the lb/day feed rate, but let's take the same problem and determine the kg/day feed rate:

Example 4:

A water plant produces 2000 gpm and the city wants to add 1.1 mg/L of 98.5% purity fluoride. What would the fluoride feed rate be, in kg/day, if sodium fluoride was used?

To convert the capacity from gpm to m³/day do the following conversions:

Now plug in the values to determine the feed rate in kg/day:

Fluoride Feed Rates for Saturators

A sodium fluoride saturator is unique in that the strength of the saturated solution formed is always 18,000 ppm. This is due to the fact that sodium fluoride has a solubility which is practically constant at 4.0 grams per 100 milliliters of water at temperatures generally encountered in water treatment. This means that each liter of solution contains 18,000 milligrams of fluoride ion (40,000 mg/L times the percent available fluoride (45.2%) equals 18,000 mg/L).

This simplifies calculations because it eliminates the need for weighing the chemicals. A meter on the water inlet of the saturator provides this volume, so all we need is the volume of solution added to the water:

or

The fluoride feed rate will have the same units as the capacity. If the capacity is in gpm, the feed rate will be in gpm also. If the capacity is in gpd, the feed rate will be in gpd.

Example:

A water plant produces 1.0 MGD and has less than 0.1 mg/L of natural fluoride. What would the fluoride feed rate be, in gpm, to obtain 1.0 mg/L in the water?

The problem asked for the feed rate in gpm, so let's convert the feed rate now:

Now, let's work the same problem, but this time, let's determine the feed rate in liters per minute:

Example:

A water plant produces 1.0 MGD and has less than 0.1 mg/L of natural fluoride. What would the fluoride feed rate be, in Lpm, to obtain 1.0 mg/L in the water?

First convert the capacity from million gallons per day (MGD) to liters per minute (Lpm):

Now plug in the values to determine the feed rate in Lpm:

Summary

The current enforceable drinking water standard for fluoride if 4.0 mg/L. This is the maximum amount that is allowed in water from public water systems. EPA also has a non-enforceable secondary standard of 2.0 mg/L for fluoride. This is the amount recommended to protect children against the tooth discoloration and/or pitting that can be caused by excess fluoride exposures during the formative period prior to eruption of the teeth. Although water systems are not required to comply with secondary standards, for fluoride, the EPA does require that systems notify customers if the average water levels exceed the secondary standard. Fluoride can occur in drinking water naturally as a result of the geological composition of soils and bedrock. Some areas of the country have high levels of naturally occurring fluoride which can dissolve easily into groundwater as it moves through gaps and pore spaces between rocks. Fluoride can also be added to public drinking water supplies as a public health measure for reducing cavities among the treated population. The decision to add fluoride to drinking water is made on a case by case basis. The three commonly used fluoride chemicals are sodium fluoride, sodium fluorosilicate, and fluorosilicic acid, all of which should meet the American Water Works Associations (AWWA) standards for use in water fluoridation. The AFI for the three commonly used fluorides are:

NaF = 0.452 or 45.2%
H₂SiF₆ = 0.792 or 79.2%
Na₂SiF₂ = 0.607 or 60.7%