14.2 Chemical Feed Rate & Dosage Calculations

10.2 Chemical Feed Rate & Dosage Calculations

Precise chemical addition is fundamental to municipal water treatment (coagulation, disinfection, fluoridation, corrosion control) and wastewater treatment (dechlorination, phosphorus removal, odor control). Operators must calculate exact feed rates to satisfy chemical demand without exceeding maximum contaminant levels (MCLs) or wasting operational budgets.

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1. The Core Pounds Formula

The Pounds Formula (often visualized as the "Pounds Wheel" or "Pounds Pie") is the single most frequently utilized equation in environmental mathematics. It links liquid flow volume, chemical concentration (dose), and chemical mass.

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

Derivation of the Constant 8.34

Why does multiplying MGD by mg/L and 8.34 yield pounds per day? Dimensional analysis reveals the unit cancellation: $\text{Lbs/day} = \left(\frac{1,000,000\text{ gal}}{\text{day}}\right) \times \left(\frac{\text{Lbs Chemical}}{1,000,000\text{ lbs Water}}\right) \times \left(\frac{8.34\text{ lbs Water}}{\text{gal Water}}\right)$ Because 1 mg/L equals 1 part per million (ppm), it represents 1 lb of active chemical per 1,000,000 lbs of water. The factor 1,000,000 cancels out, leaving 8.34 lbs/gal as the converting multiplier.

Variations of the Pounds Formula

Depending on the given parameters, the formula can be rearranged to solve for Dosage or Flow: $\text{Dosage (mg/L)} = \frac{\text{Chemical Feed (lbs/day)}}{\text{Flow (MGD)} \times 8.34\text{ lbs/gal}}$ $\text{Flow (MGD)} = \frac{\text{Chemical Feed (lbs/day)}}{\text{Dosage (mg/L)} \times 8.34\text{ lbs/gal}}$

Worked Example 1: Chlorine Gas Feed Calculation

Problem: A surface water treatment plant processes a flow of 4.5 MGD. The operator wants to maintain a chlorine dose of 3.2 mg/L using 100% pure chlorine gas. How many pounds of chlorine gas must be fed per day?

Solution Step-by-Step: $\text{Lbs/day} = 4.5\text{ MGD} \times 3.2\text{ mg/L} \times 8.34\text{ lbs/gal}$ $\text{Lbs/day} = 14.4 \times 8.34 = 120.096\text{ lbs/day}$ Result: The gas chlorinator rotameter should be set to 120.1 lbs/day.

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2. Liquid Chemical Feed Rates (Accounting for Purity & Specific Gravity)

Most modern treatment facilities utilize liquid chemicals—such as commercial sodium hypochlorite (NaOCl, 12.5% active bleach), liquid alum (48% active), or liquid caustic soda (NaOH, 50% active)—rather than 100% pure dry or gaseous chemicals. Liquid chemicals are less dense or more dense than pure water, which requires accounting for Specific Gravity (SG) and percent active purity.

Specific Gravity (SG)

Specific Gravity is the ratio of the density of a liquid chemical solution to the density of pure water (8.34 lbs/gal): $\text{Weight of Liquid Chemical (lbs/gal)} = \text{Specific Gravity (SG)} \times 8.34\text{ lbs/gal}$

Liquid Feed Rate Formulas

To find the gross solution feed rate in gallons per day (gpd): $\text{Active Lbs/day Required} = \text{Flow (MGD)} \times \text{Dose (mg/L)} \times 8.34$ $\text{Liquid Feed (gal/day)} = \frac{\text{Active Lbs/day Required}}{\text{SG} \times 8.34\text{ lbs/gal} \times \text{Purity (decimal)}}$ $\text{Liquid Feed (gal/day)} = \frac{\text{Flow (MGD)} \times \text{Dose (mg/L)}}{\text{SG} \times \text{Purity (decimal)}}$

Worked Example 2: Liquid Sodium Hypochlorite Feed

Problem: An operator treats a flow rate of 2.0 MGD with liquid sodium hypochlorite to achieve a chlorine dosage of 2.5 mg/L. The sodium hypochlorite solution has a concentration of 12.5% trade purity and a Specific Gravity of 1.20. Calculate the required solution feed rate in gallons per day (gpd).

Solution Step-by-Step:

1. Calculate active chlorine mass required: $\text{Active Lbs/day} = 2.0\text{ MGD} \times 2.5\text{ mg/L} \times 8.34 = 41.7\text{ lbs active Cl}_2\text{/day}$
2. Calculate weight of 1 gallon of liquid hypochlorite solution: $\text{Weight of Solution} = 1.20 \times 8.34\text{ lbs/gal} = 10.008\text{ lbs/gal}$
3. Calculate active chlorine mass contained in 1 gallon of solution: $\text{Active Lbs Cl}_2\text{/gal} = 10.008\text{ lbs/gal} \times 0.125\text{ (purity)} = 1.251\text{ active lbs/gal}$
4. Calculate required liquid solution feed rate in gal/day: $\text{Feed Rate (gal/day)} = \frac{41.7\text{ lbs active/day}}{1.251\text{ lbs active/gal}} = 33.33\text{ gallons/day}$

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3. Dry Chemical Feeder Calibration & Operational Conversions

Operators must frequently convert daily chemical mass rates (lbs/day) into smaller operational units such as pounds per hour (lbs/hr) or grams per minute (g/min) for volumetric drawdown calibrations or bench-scale testing.

Time Conversions for Chemical Feeders:

$\text{Feed Rate (lbs/hr)} = \frac{\text{Feed Rate (lbs/day)}}{24\text{ hours/day}}$ $\text{Feed Rate (g/min)} = \frac{\text{Feed Rate (lbs/day)} \times 453.6\text{ grams/lb}}{1,440\text{ minutes/day}}$ $\text{Feed Rate (mL/min)} = \frac{\text{Feed Rate (gal/day)} \times 3,785\text{ mL/gal}}{1,440\text{ minutes/day}}$

Worked Example 3: Dry Feeder Calibration in Grams per Minute

Problem: A dry lime feeder must supply 180 lbs/day of hydrated lime to maintain raw water pH. During a 1-minute calibration catch-and-weigh test, how many grams of lime should the operator collect?

Solution Step-by-Step: $\text{Feed Rate (g/min)} = \frac{180\text{ lbs/day} \times 453.6\text{ g/lb}}{1,440\text{ min/day}} = \frac{81,648\text{ g/day}}{1,440\text{ min/day}} = 56.7\text{ grams/minute}$

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4. Chlorine Dosage, Demand, and Residual Relationships

When chlorine gas or hypochlorite is added to water, a portion reacts instantly with inorganic reducing agents (hydrogen sulfide, ferrous iron, manganese) and organic matter. This consumed fraction is the Chlorine Demand. The remaining unreacted chlorine available for pathogen inactivation is the Chlorine Residual.

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

Worked Example 4: Chlorine Demand Determination

Problem: A chlorinator feeds 4.2 mg/L of chlorine into treated water. After a 30-minute contact period, the free available chlorine residual in the clearwell effluent is measured at 1.5 mg/L. What is the chlorine demand of the water?

Solution Step-by-Step: $\text{Demand} = \text{Dosage} - \text{Residual} = 4.2\text{ mg/L} - 1.5\text{ mg/L} = 2.7\text{ mg/L}$

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5. Chemical Solution Dilution Mathematics ($C_1 V_1 = C_2 V_2$)

To prepare secondary day-tank solutions or dilute concentrated stock chemicals, operators apply the volumetric mass balance equation:

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

• $C_1$ = Concentration of starting (stock) solution
• $V_1$ = Volume of starting (stock) solution needed
• $C_2$ = Desired concentration of diluted solution
• $V_2$ = Total volume of diluted solution desired

Worked Example 5: Preparing a Polymer Day Tank Solution

Problem: An operator needs to prepare 500 gallons of a 0.2% diluted polymer solution using a concentrated liquid polymer stock solution of 5.0%. How many gallons of stock polymer and how many gallons of dilution water are required?

Solution Step-by-Step:

1. Apply $C_1 V_1 = C_2 V_2$: $5.0\% \times V_1 = 0.2\% \times 500\text{ gallons}$ $V_1 = \frac{0.2 \times 500}{5.0} = \frac{100}{5.0} = 20\text{ gallons of stock polymer}$
2. Calculate required dilution water volume: $\text{Dilution Water} = V_2 - V_1 = 500\text{ gal} - 20\text{ gal} = 480\text{ gallons of water}$