13.2 The Pounds Formula, Chemical Feed Rates & Solution Concentration Calculations

13.2 The Pounds Formula, Chemical Feed Rates & Solution Concentration Calculations

Accurate chemical dosing is vital for both regulatory compliance and plant economic efficiency. Underdosing coagulants, disinfectants, or pH adjusters risks pathogen breakthrough, permit violations, or corrosive finished water. Overdosing wastes operating funds, accelerates equipment scaling, and can generate hazardous concentrations of disinfection byproducts (DBPs) such as trihalomethanes (THMs) and haloacetic acids (HAAs). Water and wastewater operators must be completely proficient in calculating chemical feed rates for gaseous, dry, and liquid chemicals across variable plant flows.

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1. The Fundamental Pounds Equation (Chemical Dosage Formula)

The Pounds Equation is the most widely applied mathematical formula in water and wastewater operations. It determines the mass of pure chemical required per day to achieve a target concentration in a given volume of treated water:

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

Dimensional Analysis of the 8.34 Factor

Many operators wonder why multiplying Million Gallons per Day by milligrams per liter directly produces pounds per day. The dimensional derivation proves this relationship:

1. One liter of pure water has a mass of 1,000 grams, or $1,000,000\text{ milligrams}$. Therefore, $1\text{ milligram per liter (mg/L)}$ represents 1 part per million (ppm) by mass: $$1\text{ mg/L} = \frac{1\text{ lb of chemical}}{1,000,000\text{ lbs of water}}$$
2. One gallon of water weighs $8.34\text{ lbs}$. One million gallons of water weighs: $$1,000,000\text{ gallons} \times 8.34\text{ lbs/gal} = 8,340,000\text{ lbs of water}$$
3. Combining the terms for a flow of 1.0 MGD treated at 1.0 mg/L: $$\text{lbs/day} = \left(1.0\text{ MGD} \times \frac{8,340,000\text{ lbs water}}{1\text{ MG}}\right) \times \left(\frac{1\text{ lb chemical}}{1,000,000\text{ lbs water}}\right) = 8.34\text{ lbs chemical/day}$$

The constant $8.34$ incorporates the density of water and the metric-to-English dimensional cancellation.

Algebraic Rearrangements (Davidson's Pie / Box Method)

Operators frequently utilize the Davidson Pie Method as a visual calculation aid. A circle is divided horizontally; the top half contains Chemical Mass (lbs/day), and the bottom half is divided into three wedges: Flow (MGD), Dosage (mg/L), and the constant 8.34.

By covering the desired variable, the required mathematical operation is revealed:

1. Solving for Feed Rate: $$\text{Feed Rate (lbs/day)} = \text{Flow (MGD)} \times \text{Dosage (mg/L)} \times 8.34$$
2. Solving for Dosage: $$\text{Dosage (mg/L)} = \frac{\text{Feed Rate (lbs/day)}}{\text{Flow (MGD)} \times 8.34}$$
3. Solving for Flow Capacity: $$\text{Flow (MGD)} = \frac{\text{Feed Rate (lbs/day)}}{\text{Dosage (mg/L)} \times 8.34}$$

Batch Dosing Variation: When calculating the chemical mass needed to treat a static, standing volume (e.g., disinfected storage tank, clarifier, or pipeline) rather than a flowing stream, substitute Tank Volume in Million Gallons (MG) for Flow (MGD): $\text{Chemical Required (lbs)} = \text{Volume (MG)} \times \text{Dosage (mg/L)} \times 8.34\text{ lbs/gal}$

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2. Chemical Purity and Active Ingredient Corrections

The fundamental pounds formula calculates the weight of 100% pure active chemical required. However, many commercial water treatment chemicals are delivered in impure solid forms, aqueous blends, or hydration complexes:

• Calcium Hypochlorite (HTH): Solid granules or tablets containing approximately $65\%\text{ available chlorine}$ ($0.65$ decimal purity).
• Dry Alum (Aluminum Sulfate): Typically contains $17\%\text{ available } Al_2O_3$, with bulk dry product commercial purity rated at approximately $85\%\text{ to } 90\%$.
• Quicklime ($CaO$): Approximately $90\%\text{ available } CaO$; Hydrated Lime ($Ca(OH)_2$): Approximately $95\%$.

Because the commercial product is not pure active chemical, the operator must feed a larger total weight of the commercial product to deliver the required active pounds:

$$\text{Commercial Chemical Feed Rate (lbs/day)} = \frac{\text{Pure Chemical Calculated (lbs/day)}}{\text{Decimal Purity}}$$

Common Sense Verification: Dividing by a decimal fraction (a number less than 1.0) always produces a result larger than the starting number. If your calculated commercial feed rate is smaller than your pure active requirement, you mistakenly multiplied instead of dividing.

Worked Example: Commercial Dry Chemical Feed

Problem Statement: A water treatment plant treats 4.2 MGD. The coagulation jar test establishes an optimum alum dosage of 22 mg/L. The bulk dry alum delivered has an active chemical purity of 85% (0.85). Calculate the daily feed rate of commercial alum required in pounds per day.

Step 1: Calculate the pure alum requirement using the pounds formula.

$$\text{Pure Alum (lbs/day)} = 4.2\text{ MGD} \times 22\text{ mg/L} \times 8.34\text{ lbs/gal} = 770.616\text{ lbs/day}$$

Step 2: Correct for the 85% chemical purity.

$$\text{Commercial Alum (lbs/day)} = \frac{770.616\text{ lbs/day}}{0.85} = 906.607\text{ lbs/day} \approx 907\text{ lbs/day}$$

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3. Liquid Chemical Feed Calculations & Specific Gravity

Many municipal treatment facilities utilize liquid chemicals rather than dry powders or gaseous chlorine to eliminate dust hazards and gas leak risks. Common liquid chemicals include:

• Sodium Hypochlorite ($NaOCl$): Delivered as a liquid solution at concentrations between $10\%$ and $15\%$ available chlorine.
• Liquid Alum: Typically supplied as a $48\%\text{ to } 50\%$ aqueous solution.
• Ferric Chloride ($FeCl_3$): Supplied as a $35\%\text{ to } 42\%$ solution.
• Sodium Hydroxide (Caustic Soda, $NaOH$): Commonly delivered as a $25\%\text{ or } 50\%$ solution.

Specific Gravity ($SG$)

Liquid chemical solutions are denser than plain water. The Specific Gravity ($SG$) is the ratio of the liquid chemical's density to the density of water ($8.34\text{ lbs/gal}$):

$$\text{Weight of 1 Gallon of Chemical Solution (lbs/gal)} = \text{Specific Gravity} \times 8.34\text{ lbs/gal}$$

Active Ingredient Mass per Gallon

To find the actual weight of the active therapeutic chemical contained in each gallon of bulk liquid:

$$\text{Active Chemical Weight (lbs/gal)} = \text{Specific Gravity} \times 8.34\text{ lbs/gal} \times \text{Decimal Concentration}$$

Liquid Feed Rate Formulas

Once the pure active pounds per day are calculated from the pounds formula, the required liquid volume feed rate is:

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

Chemical metering pumps (diaphragm or peristaltic pumps) are calibrated in milliliters per minute (mL/min) or gallons per hour (gph):

$$\text{Liquid Feed Rate (gph)} = \frac{\text{Feed Rate (gal/day)}}{24\text{ hr/day}}$$ $$\text{Liquid Feed Rate (mL/min)} = \frac{\text{Feed Rate (gal/day)} \times 3,785\text{ mL/gallon}}{1,440\text{ minutes/day}} = \text{Feed Rate (gal/day)} \times 2.6285$$

Worked Example: Sodium Hypochlorite Metering Pump Sizing

Problem Statement: An operator must disinfect a finished water flow of 3.0 MGD with a target free chlorine dosage of 2.4 mg/L. The utility feeds commercial liquid sodium hypochlorite containing 12.5% available chlorine with a specific gravity of 1.20. Determine:

1. The required active chlorine in lbs/day.
2. The weight of active chlorine per gallon of solution.
3. The liquid feed rate in gallons per day (gpd).
4. The metering pump calibration rate in milliliters per minute (mL/min).

Step 1: Calculate active chlorine required per day.

$$\text{Active } Cl_2\text{ (lbs/day)} = 3.0\text{ MGD} \times 2.4\text{ mg/L} \times 8.34\text{ lbs/gal} = 60.048\text{ lbs/day}$$

Step 2: Determine total weight of one gallon and active chlorine content.

$$\text{Weight of 1 gallon of solution} = 1.20 \times 8.34\text{ lbs/gal} = 10.008\text{ lbs/gal}$$ $$\text{Active } Cl_2\text{ per gallon} = 10.008\text{ lbs/gal} \times 0.125 = 1.251\text{ lbs } Cl_2/\text{gal}$$

Step 3: Calculate liquid feed rate in gallons per day.

$$\text{Liquid Feed Rate (gal/day)} = \frac{60.048\text{ lbs } Cl_2/\text{day}}{1.251\text{ lbs } Cl_2/\text{gal}} = 48.00\text{ gal/day}$$

Step 4: Convert feed rate to mL/min for calibration.

$$\text{Feed Rate (mL/min)} = \frac{48.00\text{ gal/day} \times 3,785\text{ mL/gal}}{1,440\text{ min/day}} = \frac{181,680}{1,440} = 126.17\text{ mL/min} \approx 126.2\text{ mL/min}$$

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4. Dry Chemical Feeder Calibration & Catch Testing

Volumetric and gravimetric dry chemical feeders (screw, belt, or oscillating hopper feeders) must be physically calibrated by performing a catch-and-weigh test at various feeder dial settings. An operator catches the dry chemical discharged into a tare-weighed container for an exact time period and weighs the sample.

Catch-and-Weigh Equations

$$\text{Feed Rate (grams/minute)} = \frac{\text{Grams Caught}}{\text{Catch Time (minutes)}}$$

To convert grams per minute to operational field units ($1\text{ pound} = 453.6\text{ grams} \approx 454\text{ grams}$):

$$\text{Feed Rate (lbs/day)} = \frac{\text{Grams/min} \times 1,440\text{ min/day}}{453.6\text{ grams/lb}}$$ $$\text{Feed Rate (lbs/hour)} = \frac{\text{Feed Rate (lbs/day)}}{24\text{ hr/day}} = \frac{\text{Grams/min} \times 60\text{ min/hr}}{453.6\text{ grams/lb}}$$

Worked Example: Dry Feeder Calibration

Problem Statement: An operator tests a gravimetric dry lime feeder set at 45% stroke. A clean plastic catch bucket weighs 320 grams empty. After catching dry lime for exactly 4.0 minutes, the bucket and lime weigh 880 grams. What is the feeder output in pounds per day?

Step 1: Determine the net weight of lime caught.

$$\text{Net Lime Caught} = 880\text{ g} - 320\text{ g} = 560\text{ grams}$$

Step 2: Determine delivery rate in grams per minute.

$$\text{Delivery Rate} = \frac{560\text{ grams}}{4.0\text{ minutes}} = 140.0\text{ grams/min}$$

Step 3: Convert grams per minute to pounds per day.

$$\text{Feed Rate (lbs/day)} = \frac{140.0\text{ g/min} \times 1,440\text{ min/day}}{453.6\text{ g/lb}} = \frac{201,600}{453.6} = 444.44\text{ lbs/day} \approx 444\text{ lbs/day}$$

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5. Chlorine Demand, Residual, and Dosage Mass Balance

Chlorine added to water or wastewater does not remain entirely available as a free disinfectant. A portion is consumed immediately by inorganic reducing agents (ferrous iron $Fe^{2+}$, manganese $Mn^{2+}$, hydrogen sulfide $H_2S$, nitrite $NO_2^-$) and organic compounds. The basic disinfection mass balance relationship is:

$$\text{Chlorine Dosage} = \text{Chlorine Demand} + \text{Chlorine Residual}$$

Rearranging to determine demand or residual:

$$\text{Chlorine Demand} = \text{Chlorine Dosage} - \text{Chlorine Residual}$$ $$\text{Chlorine Residual} = \text{Chlorine Dosage} - \text{Chlorine Demand}$$

Where:

• Dosage (mg/L): Total concentration of chlorine applied to the water by the feed system.
• Demand (mg/L): The amount of chlorine consumed by chemical reactions and microbiological destruction during a defined contact period.
• Residual (mg/L): The concentration of active chlorine remaining in the water at the end of the contact time. Chlorine residual is further divided into Free Residual (hypochlorous acid $HOCl$ and hypochlorite ion $OCl^-$) and Combined Residual (chloramines formed with ammonia).

Worked Example: Determining Chlorine Demand

Problem Statement: A surface water plant treats 5.0 MGD. The gas chlorinator feeds 160 pounds of chlorine gas per day. Effluent sampling after 45 minutes of contact chamber detention reveals a total chlorine residual of 0.9 mg/L. What is the chlorine demand of the water in mg/L?

Step 1: Calculate the applied chlorine dosage using the rearranged pounds formula.

$$\text{Dosage (mg/L)} = \frac{\text{Feed Rate (lbs/day)}}{\text{Flow (MGD)} \times 8.34\text{ lbs/gal}} = \frac{160\text{ lbs/day}}{5.0\text{ MGD} \times 8.34\text{ lbs/gal}} = \frac{160}{41.7} = 3.837\text{ mg/L} \approx 3.84\text{ mg/L}$$

Step 2: Solve for chlorine demand.

$$\text{Demand (mg/L)} = \text{Dosage} - \text{Residual} = 3.84\text{ mg/L} - 0.90\text{ mg/L} = 2.94\text{ mg/L}$$

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6. Two-Normal Dilution & Solution Mixing Calculations

When preparing chemical solutions in day tanks or diluting stock reagents, operators apply the Two-Normal Dilution Equation. Because the absolute mass of active chemical solute remains unchanged when water is added, the product of initial concentration and volume equals the product of final concentration and volume:

$$C_1 \times V_1 = C_2 \times V_2$$

Where:

• $C_1 = \text{Concentration of stock / initial solution (\%, mg/L, or normality)}$
• $V_1 = \text{Volume of stock solution required (gallons, liters, or mL)}$
• $C_2 = \text{Concentration of target diluted solution (same units as } C_1)$
• $V_2 = \text{Total volume of target diluted solution (same units as } V_1)$

Worked Example: Diluting Sodium Hypochlorite

Problem Statement: An operator must prepare 150 gallons of a 2.0% sodium hypochlorite feed solution in a chemical day tank from a 10.0% bulk hypochlorite stock. How many gallons of the 10.0% solution must be transferred to the day tank, and how many gallons of dilution water must be added?

Step 1: Identify known values.

• $C_1 = 10.0\%$
• $V_1 = ?$
• $C_2 = 2.0\%$
• $V_2 = 150\text{ gallons}$

Step 2: Solve for $V_1$.

$$V_1 = \frac{C_2 \times V_2}{C_1} = \frac{2.0\% \times 150\text{ gallons}}{10.0\%} = \frac{300}{10.0} = 30.0\text{ gallons}$$

Step 3: Calculate the volume of dilution water needed.

$$\text{Dilution Water Volume} = V_2 - V_1 = 150\text{ gallons} - 30\text{ gallons} = 120.0\text{ gallons}$$

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7. Common Mathematical Pitfalls in Chemical Feed

1. Multiplying by Purity Instead of Dividing: Multiplying by a decimal percentage reduces the chemical amount, which would severely underdose the treatment process. Always divide pure pounds by decimal purity: $\text{Commercial lbs} = \text{Pure lbs} / \text{Purity}$.
2. Omitting Specific Gravity in Liquid Calculations: Assuming liquid chemicals weigh 8.34 lbs/gal like pure water. A dense liquid chemical ($SG = 1.20$) weighs $10.0\text{ lbs/gal}$. Neglecting specific gravity introduces a 15% to 30% dosage error.
3. Entering Gallons per Day into the Pounds Formula: The pounds formula requires flow in Million Gallons per Day (MGD). Entering $2,000,000\text{ gpd}$ instead of $2.0\text{ MGD}$ inflates the calculated chemical requirement by a factor of one million.
4. Confusing Chlorine Demand with Dosage: Dosage is what leaves the chemical feeder; demand is what is consumed by water impurities; residual is what is measured at the tap or outfall. Remember that demand can never exceed dosage unless the residual is zero.