2.2 Chemical Feed, Dosage & Dilution Calculations

Water and wastewater treatment processes rely on precise chemical addition to achieve regulatory compliance and protect public health. Certified operators add coagulants (alum, ferric chloride, polyaluminum chloride) to destabilize colloidal turbidity, lime or sodium hydroxide for pH and corrosion control, sodium hypochlorite or chlorine gas for disinfection, and polymers for sludge conditioning.

Under-dosing chemicals leads to regulatory non-compliance, such as exceeding finished water turbidity Maximum Contaminant Levels (MCLs) under 25 Pa. Code Chapter 109 or exceeding total phosphorus limits under Chesapeake Bay tributary strategies. Over-dosing creates chemical waste, increases operating costs, generates excessive sludge volumes, and may create hazardous secondary water quality violations such as elevated trihalomethanes (THMs) or aluminum carryover.

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The Universal Pounds Formula

The most critical mathematical tool in water and wastewater operations is the Pounds Formula (often referred to on certification examinations as the chemical feed equation).

The Fundamental Equation

$\text{Chemical Feed (lbs/day)} = \text{Flow (MGD)} \times \text{Dosage (mg/L)} \times 8.34\text{ lbs/gal}$

Physical Derivation of the Constant 8.34

Understanding where $8.34$ originates eliminates confusion on exam day:

1. One liter of pure water has a mass of $1,000\text{ grams}$, which equals $1,000,000\text{ milligrams}$.
2. Therefore, a concentration of $1\text{ milligram per liter (mg/L)}$ represents 1 part of chemical per $1,000,000$ parts of water by weight—identically equal to $1\text{ part per million (ppm)}$.
3. One gallon of water weighs $8.34\text{ pounds}$.
4. Consequently, one million gallons of water weighs: $1,000,000\text{ gallons} \times 8.34\text{ lbs/gal} = 8,340,000\text{ pounds}$
5. If one million gallons of water weighs $8,340,000\text{ pounds}$, adding $1\text{ part per million}$ (1 pound of chemical per million pounds of water) requires: $\frac{1}{1,000,000} \times 8,340,000\text{ lbs} = \mathbf{8.34\text{ pounds}}$

Thus, dosing water at $1.0\text{ mg/L}$ requires feeding exactly $8.34\text{ pounds}$ of pure chemical per million gallons treated.

Algebraic Rearrangements of the Pounds Formula

Depending on the operational question, an operator may know the daily chemical usage and need to calculate the delivered dosage or verify plant throughput:

• Solving for Chemical Dosage (mg/L): $\text{Dosage (mg/L)} = \frac{\text{Chemical Fed (lbs/day)}}{\text{Flow (MGD)} \times 8.34\text{ lbs/gal}}$

• Solving for Plant Flow (MGD): $\text{Flow (MGD)} = \frac{\text{Chemical Fed (lbs/day)}}{\text{Dosage (mg/L)} \times 8.34\text{ lbs/gal}}$

• Solving for Total Volume Treated in a Batch Tank (Million Gallons): $\text{Volume (MG)} = \frac{\text{Pounds of Chemical Added}}{\text{Target Dosage (mg/L)} \times 8.34\text{ lbs/gal}}$

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Chemical Purity and Active Ingredient Adjustments

The basic Pounds Formula calculates the mass of 100% active pure chemical required. However, almost all commercial chemicals delivered to treatment plants are mixtures, solutions, or hydrated compounds containing inactive ingredients, carrier water, or inert binders.

The Golden Rule of Chemical Purity

Core Rule: If a commercial product contains less than $100%$ active chemical, an operator must feed a GREATER total mass of commercial product to deliver the required active dose. Therefore, you must DIVIDE by the decimal purity.

$\text{Commercial Chemical Required (lbs/day)} = \frac{\text{Active Chemical (lbs/day)}}{\text{Purity (decimal)}} = \frac{\text{Flow (MGD)} \times \text{Dosage (mg/L)} \times 8.34}{\text{Decimal Purity}}$

Common Commercial Chemical Purities

• Calcium Hypochlorite (HTH): Granular chlorine compound, typically $65%$ available chlorine ($0.65$ decimal purity).
• Hydrated Lime: Powdered calcium hydroxide, typically $90%$ to $93%$ pure ($0.90$ to $0.93$ decimal purity).
• Dry Alum: Filter alum, approximately $17%$ available active $Al_2O_3$.
• Gas Chlorine ($Cl_2$): $100%$ pure active element ($1.0$ decimal purity).

If an operator mistakenly multiplies by the purity fraction rather than dividing, the feeder delivers less chemical than needed, causing a severe under-dose.

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Liquid Chemical Feed Calculations

Many facilities utilize liquid chemical solutions rather than dry powders due to ease of handling and automated feed systems. Calculating liquid feed rates requires incorporating Specific Gravity (SG).

Understanding Specific Gravity

Specific gravity is the ratio of the density of a liquid solution to the density of pure water ($8.34\text{ lbs/gal}$ at $60^\circ\text{F}$):

$\text{Density of Liquid Solution (lbs/gal)} = 8.34\text{ lbs/gal} \times \text{Specific Gravity (SG)}$

Once solution density is established, the mass of active chemical per gallon of liquid is calculated:

$\text{Active Mass per Gallon (lbs active/gal)} = \text{Density (lbs/gal)} \times \frac{\% \text{ Active Strength}}{100}$ $\text{Active Mass per Gallon} = (8.34 \times \text{SG}) \times (\text{Decimal Active Fraction})$

Liquid Feed Rate in Gallons per Day (gpd)

Once the active mass per gallon is established, the daily liquid feed volume is determined by dividing the required active chemical poundage by the active pounds per gallon:

$\text{Liquid Feed Rate (gpd)} = \frac{\text{Active Chemical Required (lbs/day)}}{\text{Active Mass per Gallon (lbs active/gal)}}$

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Metering Pump Calibration and Draw-Down Testing

Positive displacement metering pumps (diaphragm or peristaltic) inject liquid chemicals directly into pressurized mains or rapid mix basins. Because chemical metering pumps are rated in milliliters per minute ($\text{mL/min}$), gallons per hour ($\text{gph}$), or gallons per day ($\text{gpd}$), operators must calibrate pump discharge rates using a graduated calibration cylinder (draw-down tube).

The Standard Calibration Conversion Factor: 2.629

Converting daily volume in gallons to continuous flow in milliliters per minute:

1. There are $3,785.4\text{ mL}$ in one gallon.
2. There are $1,440\text{ minutes}$ in one day ($24\text{ hr} \times 60\text{ min}$).
3. Therefore, the continuous flow equivalent of $1.0\text{ gallon per day}$ is: $\frac{3,785.4\text{ mL/gal}}{1,440\text{ min/day}} = \mathbf{2.6288\text{ mL/min per gpd}} \approx \mathbf{2.629\text{ mL/min per gpd}}$

$\text{Feed Pump Rate (mL/min)} = \text{Feed Rate (gpd)} \times 2.629\text{ mL/min per gpd}$

The Volumetric Draw-Down Calibration Test

To verify that a pump is delivering the intended chemical rate, an operator performs a draw-down test:

1. Isolate the chemical feed pump from the bulk supply tank and open the valve to the graduated calibration column.
2. Record the starting volume on the cylinder scale in milliliters.
3. Run the pump for a precisely timed interval (typically $1.0\text{ minute}$ or $2.0\text{ minutes}$).
4. Record the final volume and determine the milliliters withdrawn:

$\text{Actual Pump Output (mL/min)} = \frac{\text{Volume Withdrawn (mL)}}{\text{Elapsed Test Time (minutes)}}$

If the measured rate deviates from the setpoint, the operator adjusts the pump speed (frequency) or stroke length.

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Solution Dilution and Batch Mixing ($C_1 V_1 = C_2 V_2$)

Operators frequently prepare day tanks, dilute concentrated coagulants, or mix polymer solutions. The principle of mass conservation dictates that the mass of active chemical before dilution equals the mass of active chemical after dilution:

$C_1 \times V_1 = C_2 \times V_2$

Where:

• $C_1$ = Initial concentration of stock chemical (in $%$ or $\text{mg/L}$)
• $V_1$ = Volume of stock chemical required
• $C_2$ = Target final concentration of blended solution (in $%$ or $\text{mg/L}$)
• $V_2$ = Total final volume of blended solution desired

Calculating Added Dilution Water

After solving for the required volume of concentrated stock ($V_1$), the volume of carrier water needed to complete the batch is calculated by subtraction:

$V_{\text{water}} = V_2 - V_1$

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Step-by-Step Worked Mathematical Examples

Example 1: Liquid Sodium Hypochlorite Feed Rate (gpd and mL/min)

A water filtration plant treats $4.0\text{ MGD}$ with a target finished chlorine dosage of $2.5\text{ mg/L}$. The plant feeds commercial sodium hypochlorite solution having a specific gravity of $1.21$ and containing $12.5%$ available chlorine by weight. Calculate the required chemical feed rate in gallons per day (gpd) and the metering pump calibration setting in milliliters per minute (mL/min).

• Step 1: Calculate active chlorine demand in pounds per day. $\text{Active Chlorine (lbs/day)} = 4.0\text{ MGD} \times 2.5\text{ mg/L} \times 8.34\text{ lbs/gal} = 83.4\text{ lbs/day}$

• Step 2: Calculate total weight of one gallon of hypochlorite solution. $\text{Solution Weight} = 8.34\text{ lbs/gal} \times 1.21\text{ SG} = 10.0914\text{ lbs/gal}$

• Step 3: Calculate pounds of active chlorine per gallon of solution. $\text{Active Chlorine per Gallon} = 10.0914\text{ lbs/gal} \times 0.125 = 1.2614\text{ lbs Cl}_2\text{/gal}$

• Step 4: Calculate the required liquid feed rate in gallons per day (gpd). $\text{Feed Rate (gpd)} = \frac{83.4\text{ lbs Cl}_2\text{/day}}{1.2614\text{ lbs Cl}_2\text{/gal}} = 66.12\text{ gpd}$

• Step 5: Convert the feed rate to milliliters per minute (mL/min). $\text{Feed Rate (mL/min)} = 66.12\text{ gpd} \times 2.6288 = 173.8\text{ mL/min}$ The operator should adjust the chemical feed pump to discharge $174\text{ mL/min}$ on the calibration column.

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Example 2: Dry Chemical Feeder Adjustment for Product Purity

A municipal wastewater facility must dose ferric chloride to secondary effluent for total phosphorus precipitation to comply with Chesapeake Bay watershed permit limits. The plant flow is $1.8\text{ MGD}$ and jar testing dictates an active ferric dose of $18.0\text{ mg/L}$. The delivered dry ferric chloride material has a verified purity of $82%$. How many pounds per day of commercial dry chemical must the gravimetric dry feeder feed?

• Step 1: Calculate pure active chemical demand. $\text{Active Chemical (lbs/day)} = 1.8\text{ MGD} \times 18.0\text{ mg/L} \times 8.34\text{ lbs/gal} = 270.22\text{ lbs/day}$

• Step 2: Adjust for product purity. $\text{Commercial Feed Rate (lbs/day)} = \frac{270.22\text{ lbs/day}}{0.82} = 329.54\text{ lbs/day}$ The dry chemical feeder must be set to deliver approximately $330\text{ lbs/day}$ of bulk product to provide the required $270.2\text{ lbs/day}$ of active ferric.

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Example 3: Chemical Day Tank Preparation ($C_1 V_1 = C_2 V_2$)

An operator needs to prepare a $250\text{-gallon}$ batch of $2.0%$ potassium permanganate solution in a day tank to treat seasonal iron and manganese in a reservoir supply. The plant storage room holds bulk liquid permanganate solution concentrated at $10.0%$. How many gallons of the $10.0%$ stock solution and how many gallons of tap water are required?

• Step 1: Identify the known dilution parameters.

  • $C_1 = 10.0%$
  • $V_1 = ?$
  • $C_2 = 2.0%$
  • $V_2 = 250\text{ gallons}$

• Step 2: Solve for the required stock volume ($V_1$). $V_1 = \frac{C_2 \times V_2}{C_1} = \frac{2.0\% \times 250\text{ gallons}}{10.0\%} = \frac{500}{10.0} = 50\text{ gallons}$

• Step 3: Calculate the volume of dilution water. $V_{\text{water}} = V_2 - V_1 = 250\text{ gallons} - 50\text{ gallons} = 200\text{ gallons}$ The operator must add $50\text{ gallons}$ of $10.0%$ stock permanganate into the day tank and add $200\text{ gallons}$ of water to reach the $250\text{-gallon}$ mark.

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Common Operational Pitfalls and Exam Traps

• Multiplying by Purity Instead of Dividing: Multiplying active poundage by $0.85$ results in feeding less chemical, when in reality lower purity requires feeding more bulk chemical. Remember: commercial feed rate is always larger than active chemical demand.
• Omitting Specific Gravity in Liquid Feed Calculations: Assuming a liquid chemical weighs $8.34\text{ lbs/gal}$ introduces a major error when handling dense solutions. Sodium hypochlorite ($SG \approx 1.20$), liquid alum ($SG \approx 1.33$), and caustic soda ($SG \approx 1.53$) are substantially denser than water.
• Confusing Trade Percent with Weight Percent: Sodium hypochlorite is often described in "trade percent available chlorine" ($12.5%$ to $15%$), which equals weight percent multiplied by specific gravity. Exam problems will state whether the percentage is by weight or trade; adhere strictly to the given specification.