4.1: It's a Pound Formula!
 Page ID
 7135
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)It's pretty easy to calculate how many pounds of chlorine are needed to provide a certain dosage if we are using 100% gas chlorine. Most operators in the water industry have put to memory the “Pound Formula” by multiplying the flow or volume of water in MG or MGD by 8.34 pounds per gallon and then by the dosage in ppm, the pounds of chlorine needed is calculated.
\[MG \times 8.34\, lbs/gallon \times ppm = lbs \label{eq1}\]
or
\[\dfrac{MG}{Day} \times 8.34\, lbs/gallon \times ppm = lbs/day \label{eq2}\]
However, in many treatment plants and at treatment sites within distribution systems, the use of gas chlorine is in decline, unless the plant is of considerable size. The reduction in chlorine gas usage is primarily due to safety concerns and other forms of chlorine being more costcompetitive. For example, groundwater wells are commonly disinfected with solid (calcium hypochlorite) or liquid (sodium hypochlorite) forms of chlorine. In addition, other chemicals such as Alum, ferric chloride, sodium hydroxide are used in varying concentration strengths at treatment plants in addition to chlorine. Most often these chemicals are not in the pure 100% form.
When solving dosage problems with chemicals of different strength the following two statements are helpful in remembering whether you need to multiply or divide by the percent concentration.
“If you are solving for pounds you divide by the percent concentration.”
“If pounds are given you multiply by the percent concentration.”
Therefore, if you are calculating for the amount of pounds needed you divide by the decimal equivalent of the percent concentration. You need more of the chemical since it is not 100% and dividing by a number less than one yields a larger number. If the amount of pounds is known, then by multiplying by the decimal equivalent of the percent concentration you will calculate how much of that chemical is available in the total pounds of the substance. Multiplying by a number less than one yields a smaller number.
Once you understand the concept behind the problem it makes solving them easier. Think of it this way... it takes much more 10% ferric chloride in the coagulation process than let's say ferric chloride at 75% strength. The same is true if you are using calcium hypochlorite as opposed to gas chlorine, because gas is a greater strength than calcium hypochlorite. Similarly, if you have 100 pounds of 65% calcium hypochlorite, you don’t have 100 pounds of chlorine. Only 65% or 65 pounds of the substance is actually chlorine. As opposed to 100 pounds of gas chlorine which is 100 pounds of available chlorine. When using chemicals of different strengths, the pound formula can be looked at like this:
% concentration ___________ MG x 8.34 lbs/gallon x ppm = lbs
% concentration ___________ MG/Day x 8.34 lbs/gallon x ppm = lbs/day
Placing the decimal equivalent of the percent concentration of the chemical being used under the left side of the equation will allow the appropriate amount of chemical needed to be calculated. In addition, if MG or ppm are the unknowns, the percent would be multiplied by the weight of the chemical in pounds. Examples are provided later in this section.
The last chemical dosage concept we need to look at is when the chemical being used is in the form of a liquid. Since the Pound Formula is after all measuring chemicals in “pounds,” then the chemical needs to be expressed as pounds. In Section number 3, we learned about specific gravity and how it affects the weight of a substance. You will need to use that information when presented with a pound formula question where the chemical used is a liquid.
Finally, a general understanding of chlorine dosage is needed. The concept is straightforward. The reason drinking water is disinfected is to prevent pathogenic organisms from contaminating the supply causing illness to the population drinking the water. The amount of disinfectant added is not always what is measured later by a water utility operator. Typically the amount of disinfectant measured after the original dosage occurs is lower than the dosage measured which is referred to as the residual. What happens to the disinfectant? The chlorine that “disappears” is the chlorine demand. It is the amount of chlorine that is inactivating the pathogens. Once the demand is satisfied, the remaining chlorine is termed the residual. This formula is provided below.
\[\text{Dosage} = \text{residual} + \text{demand}\]
Example \(\PageIndex{1}\)
How many pounds of chlorine are needed to dose 2 MG of water to a dosage of 3.25 ppm?
Solution
We use Equation \ref{eq1} to solve this.
\[\begin{align*} MG \times 8.34\, lbs/gallon \times ppm &= lbs \\[4pt] 2\, MG \times 8.34\, lbs/gal \times 3.25 ppm &= 54.21\, lbs \end{align*}\]
In the example above, it is a straightforward chemical dosage problem.
Example \(\PageIndex{2}\)
How many pounds of 10% Alum are needed to dose a treatment flow of 5 MGD to a dosage of 10 ppm?
Solution
We use Equation \ref{eq2} to solve this
(5 MG/Day x 8.34 lbs/gallon x 10 ppm)/0.1 = lbs/day
417/0.1 = 4,170 lbs/day
It takes 417 lbs of Alum to dose 5 MGD to 10 ppm. However, in this problem, the Alum being used is only a 10% concentration. Therefore, you need to divide by 10% (or 0.1) to calculate the total amount of this form of Alum that is needed.
Example \(\PageIndex{3}\)
How many gallons of 15% strength sodium hypochlorite are needed to dose a well flowing 1,500 gpm to a dosage of 1.75 ppm? (Assume the sodium hypochlorite has a specific gravity of 1.42)
Solution
First, the flow rate needs to be converted to MGD.
1,500 gpm x 1440 = 2,160,000 GPD or 2.16 MGD
(2.16 MG/Day x 8.34 lbs/gallon x 1.75 ppm)/0.15 = 210 lbs/day
210 lbs of 15% sodium hypochlorite are needed to dose 1,500 gpm to 1.75 ppm. Now the 210 lbs need to be converted to gallons.
210 lbs/day x gallons/(8.34 lbs x 1.42) = 17.75 gal/day
Exercises
Solve the following chemical dosage problems. Be sure to account for the differences in chemical percent concentrations.

How many gallons of water can be treated with 100 pounds of 65% High Test Hypochlorite (HTH) to a dosage of 2.55 mg/L?

An operator added 165 pounds of 25% ferric chloride to a treatment flow of 10.5 MGD. What was the corresponding dosage?

How many pounds of 12.5% sodium hypochlorite are needed to dose a well with a flow rate of 1,000 gpm to a dosage of 1.75 ppm? (Assume the well runs 17 hours a day)

In the above problem, how many gallons of chemical are needed per hour? (Assume the SG is 1.4)

A treatment operator has set a chemical pump to dose 75 gallons of NaOH (sodium hydroxide) per day for a flow rate of 2.25 MGD. What is the corresponding dosage? (Assume the SG is 1.65)

11,250 feet of 18” diameter main line needs to be dosed to 50 ppm. Answer the following questions.

How many gallons of 12.5% (SG = 1.44) sodium hypochlorite are needed?

How many pounds of 65% HTH are needed?

Assuming the following costs, which one is least expensive?
 Sodium hypochlorite = $2.45 per gallon
 HTH = $1.65 per pound


A water treatment operator adjusted the amount of 20% Alum dosage from 85 mg/L to 70 mg/L. Based on a treatment flow of 10 MGD, what is the cost savings if 100% pure Alum costs $2.50 per pound?

A water utility produced 11,275 AF of water last year. The entire amount was dosed at an average rate of 1.5 ppm. If the chemical of choice was 65% HTH at a per pound cost of $1.85, what was the annual budget?

Ferric chloride is used as the coagulant of choice at a 5.75 MGD rated capacity treatment plant. If the plant operated at the rated capacity for 75% of the year and operated at 60% of rated capacity for 25% of the year, how many pounds of the coagulant was needed to maintain a dosage of 45 mg/L?

A water softening treatment process uses 25% NaOH during 20% of the year and 50% NaOH for 80% of the year. Assuming a constant flow rate of 1,100 gpm and a dosage of 70 mg/L, what is the annual budget if the 25% NaOH (SG = 1.18) costs $0.95 per gallon and the 50% NaOH (SG = 1.53) costs $1.70 per gallon?

An operator added 275 gallons of 12.5% (SG=1.32) into 2,550 ft of 12” diameter pipe. After 24 hours, the residual was measured at 10.25 ppm. What was the demand?