12.2 Applied Math: Pounds Formula, Chemical Feed & Flow Conversions

12.2 Applied Math: Pounds Formula, Chemical Feed & Flow Conversions

Mathematical proficiency is indispensable for certified water and wastewater operators. Process calculations dictate chemical feed rates, ensure regulatory compliance limits are maintained, verify pump performance, and prevent toxic chemical overdosing or underdosing. Mastering dimensional analysis and standard conversion factors allows an operator to systematically solve any applied plant calculation.

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Fundamental Unit Equivalencies & Hydraulic Conversions

Before executing process control formulas, memorize these foundational conversion constants:

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|                       ESSENTIAL WATERWORKS CONVERSION EQUIVALENCIES                     |
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|  Flow & Volume Equivalencies:                                                           |
|  - 1 MGD (Million Gallons per Day) = 1,000,000 gpd                                      |
|  - 1 MGD = 694.4 gpm (gallons per minute) = 41,666.7 gph (gallons per hour)              |
|  - 1 MGD = 1.547 cfs (cubic feet per second) = approximately 133,690 cu ft/day                         |
|  - 1 cubic foot (cu ft) = 7.48 gallons = 62.4 pounds of water                           |
|  - 1 gallon of water = 8.34 pounds = 3.785 liters = 3,785 milliliters (mL)              |
|  - 1 liter = 1,000 mL = 0.264 gallons                                                   |
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|  Pressure & Head Equivalencies:                                                         |
|  - 1 psi (pound per square inch) = 2.31 feet of water column head                       |
|  - 1 foot of water column head = 0.433 psi                                              |
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|  Concentration & Mass Equivalencies:                                                    |
|  - 1 mg/L (milligram per liter) = 1 ppm (part per million)                              |
|  - 1 mg/L = 8.34 pounds per Million Gallons (lbs/MG)                                    |
|  - 1% Concentration = 10,000 mg/L (or 10,000 ppm)                                       |
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Dimensional Flow Conversions: Step-by-Step

Example: Converting GPM to MGD

To convert a treatment plant flow of $2,500\text{ gpm}$ into Million Gallons per Day ($\text{MGD}$): $\text{Flow (MGD)} = \frac{2,500\text{ gal}}{\text{min}} \times \frac{1,440\text{ min}}{\text{day}} \times \frac{1\text{ MG}}{1,000,000\text{ gal}} = \frac{3,600,000}{1,000,000} = 3.60\text{ MGD}$ (Shortcut: $\text{MGD} = \frac{\text{gpm}}{694.4}$)

Example: Converting CFS to MGD

To convert an open channel flow of $4.64\text{ cfs}$ into $\text{MGD}$: $\text{Flow (MGD)} = 4.64\text{ cfs} \times \frac{1\text{ MGD}}{1.547\text{ cfs}} = 3.00\text{ MGD}$

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The Universal Pounds Formula

The Universal Pounds Formula is the single most important mathematical equation on water and wastewater certification examinations. It calculates the mass in pounds of any chemical or constituent added to or present in a volume of water over a 24-hour period:

$\text{Pounds per Day (lbs/day)} = \text{Flow (MGD)} \times \text{Dosage or Concentration (mg/L)} \times 8.34\text{ lbs/gal}$

Algebraic Rearrangements

Depending on the unknown variable, the formula is rearranged algebraically:

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

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

Worked Example 1: Chlorine Gas Feed Calculation

Problem: A municipal water plant treats a continuous flow of $4.2\text{ MGD}$. The operator must maintain a chlorine dosage of $3.5\text{ mg/L}$ to achieve required disinfection contact time. How many pounds of $100%$ pure chlorine gas must be fed per day?

• Step 1: Identify given variables
  • $\text{Flow} = 4.2\text{ MGD}$
  • $\text{Dosage} = 3.5\text{ mg/L}$
  • $\text{Density Constant} = 8.34\text{ lbs/gal}$
• Step 2: Apply the Universal Pounds Formula $\text{Feed (lbs/day)} = 4.2\text{ MGD} \times 3.5\text{ mg/L} \times 8.34\text{ lbs/gal} = 122.598\text{ lbs/day}$
• Step 3: Round appropriately $\text{Chlorine Feed} = 122.6\text{ lbs/day}$

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Chemical Feed Calculations for Dry / Non-100% Pure Chemicals

Many dry chemicals used in waterworks (e.g., Calcium hypochlorite $\text{Ca(OCl)}_2$ granules at $65%$ available chlorine, Hydrated lime $\text{Ca(OH)}_2$ at $90%$ purity, or dry Alum at $85%$ active strength) are not $100%$ pure active chemical. Because less than $100%$ of the commercial product is active ingredient, more total commercial product must be fed:

$\text{Commercial Chemical Feed (lbs/day)} = \frac{\text{Pure Chemical Required (lbs/day)}}{\text{Purity Decimal Fraction}} = \frac{\text{Flow (MGD)} \times \text{Dosage (mg/L)} \times 8.34}{\text{\% Purity} / 100}$

Worked Example 2: Calcium Hypochlorite Feed

Problem: A wastewater facility treating $1.8\text{ MGD}$ uses dry calcium hypochlorite ($65%$ available chlorine) to disinfect its final effluent. The required chlorine dose is $4.0\text{ mg/L}$. How many pounds per day of commercial calcium hypochlorite granules must be fed?

• Step 1: Calculate pure active chlorine required $\text{Pure } \text{Cl}_2\text{ (lbs/day)} = 1.8\text{ MGD} \times 4.0\text{ mg/L} \times 8.34\text{ lbs/gal} = 60.048\text{ lbs/day}$
• Step 2: Adjust for chemical purity ($65% = 0.65$) $\text{Commercial Feed (lbs/day)} = \frac{60.048\text{ lbs/day}}{0.65} = 92.38\text{ lbs/day}$
• Step 3: Final answer $\text{Commercial Feed} = 92.4\text{ lbs/day of Calcium Hypochlorite}$

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Liquid Chemical Feed Calculations & Specific Gravity

Liquid chemicals (such as liquid sodium hypochlorite $\text{NaOCl}$ at $12.5%$, liquid alum at $48%$, liquid ferric chloride at $40%$, or liquid caustic soda $\text{NaOH}$ at $50%$) are metered in volumetric units (gallons per day, gallons per hour, or milliliters per minute). Liquid chemical calculations require accounting for two distinct physical properties:

1. Specific Gravity (SG): The ratio of the liquid chemical's density to the density of pure water ($8.34\text{ lbs/gal}$). Total weight of one gallon of solution $= \text{SG} \times 8.34\text{ lbs/gal}$.
2. Percent Active Strength (% Active): The fraction of the total solution weight consisting of active chemical.

$\text{Active Chemical per Gallon (lbs/gal)} = \text{Specific Gravity} \times 8.34\text{ lbs/gal} \times \left(\frac{\%\text{ Active Strength}}{100}\right)$ $\text{Liquid Feed Rate (gal/day)} = \frac{\text{Required Pure Chemical (lbs/day)}}{\text{Active Chemical per Gallon (lbs/gal)}} = \frac{\text{Flow (MGD)} \times \text{Dosage (mg/L)} \times 8.34}{\text{Specific Gravity} \times 8.34 \times (\%\text{ Active} / 100)}$

Notice that the $8.34$ constant cancels out in the numerator and denominator, simplifying to: $\text{Liquid Feed Rate (gal/day)} = \frac{\text{Flow (MGD)} \times \text{Dosage (mg/L)}}{\text{Specific Gravity} \times (\%\text{ Active} / 100)}$

Worked Example 3: Sodium Hypochlorite Volumetric Feed

Problem: A water system treats $3.0\text{ MGD}$ with a target chlorine dosage of $2.5\text{ mg/L}$. The plant feeds commercial liquid sodium hypochlorite containing $12.5%$ available chlorine with a specific gravity of $1.20$. Calculate:

1. Required pure chlorine in lbs/day;
2. Total weight of one gallon of hypochlorite solution;
3. Active chlorine pounds per gallon;
4. Daily liquid feed rate in gallons per day (gpd);
5. Hourly feed rate in gallons per hour (gph);
6. Feed rate in milliliters per minute (mL/min).

• Step 1: Calculate required pure chlorine mass $\text{Pure } \text{Cl}_2\text{ (lbs/day)} = 3.0\text{ MGD} \times 2.5\text{ mg/L} \times 8.34 = 62.55\text{ lbs/day}$
• Step 2: Determine total weight of 1 gallon of solution $\text{Weight/gal} = 1.20\text{ (SG)} \times 8.34\text{ lbs/gal} = 10.008\text{ lbs/gal}$
• Step 3: Calculate active chlorine pounds per gallon $\text{Active } \text{Cl}_2\text{/gal} = 10.008\text{ lbs/gal} \times 0.125\text{ (purity)} = 1.251\text{ lbs pure } \text{Cl}_2\text{/gal}$
• Step 4: Calculate liquid feed rate (gal/day) $\text{Liquid Feed (gal/day)} = \frac{62.55\text{ lbs/day}}{1.251\text{ lbs/gal}} = 50.0\text{ gal/day}$
• Step 5: Convert to gallons per hour (gph) $\text{Feed Rate (gph)} = \frac{50.0\text{ gal/day}}{24\text{ hr/day}} = 2.08\text{ gph}$
• Step 6: Convert to milliliters per minute (mL/min) $\text{Feed Rate (mL/min)} = \frac{50.0\text{ gal/day} \times 3,785\text{ mL/gal}}{1,440\text{ min/day}} = \frac{189,250}{1,440} = 131.4\text{ mL/min}$

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Chemical Metering Pump Calibration (Draw-Down Test)

To verify that chemical metering pumps are delivering accurate volumetric dosages, operators perform a draw-down test using a graduated calibration cylinder on the pump suction line. By isolating the bulk chemical storage tank and timing the drawdown in milliliters over a measured time interval:

$\text{Pump Delivery Rate (gal/day)} = \frac{\text{Volume Pumped (mL)} \times 1,440\text{ min/day}}{\text{Time Elapsed (min)} \times 3,785\text{ mL/gal}}$

Worked Example 4: Pump Draw-Down Verification

Problem: During a calibration check on a chemical feed pump, the liquid level in a graduated calibration cylinder drops $420\text{ mL}$ in exactly $2.0\text{ minutes}$ ($120\text{ seconds}$). What is the actual pumping rate in gallons per day (gpd)?

• Step 1: Calculate pumping rate in mL/min $\text{Flow (mL/min)} = \frac{420\text{ mL}}{2.0\text{ min}} = 210.0\text{ mL/min}$
• Step 2: Convert to gallons per day $\text{Pump Rate (gal/day)} = \frac{210.0\text{ mL/min} \times 1,440\text{ min/day}}{3,785\text{ mL/gal}} = \frac{302,400}{3,785} = 79.89\text{ gal/day}$
• Step 3: Final answer $\text{Calibrated Pump Rate} = 79.9\text{ gpd}$

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Hydraulic Detention Time (DT)

Hydraulic Detention Time (DT) (or retention time) represents the theoretical average time that a parcel of water or wastewater remains within a clarifier, flocculation basin, contact chamber, or sedimentation tank:

$\text{Detention Time (hours)} = \frac{\text{Basin Volume (gallons)}}{\text{Influent Flow Rate (gal/day)}} \times 24\text{ hours/day}$ $\text{Detention Time (minutes)} = \frac{\text{Basin Volume (gallons)}}{\text{Influent Flow Rate (gpm)}} = \frac{\text{Basin Volume (gallons)}}{\text{Flow Rate (gal/day)}} \times 1,440\text{ min/day}$ $\text{Detention Time (days)} = \frac{\text{Basin Volume (Million Gallons)}}{\text{Influent Flow Rate (MGD)}} = \frac{\text{Volume (MG)}}{\text{Flow (MGD)}}$

Tank Volume Calculations

1. Rectangular Basin: $\text{Volume (gal)} = \text{Length (ft)} \times \text{Width (ft)} \times \text{Depth (ft)} \times 7.48\text{ gal/cu ft}$
2. Circular Basin: $\text{Volume (gal)} = \pi \times r^2 \times \text{Depth (ft)} \times 7.48 = 0.7854 \times \text{Diameter}^2\text{ (ft}^2\text{)} \times \text{Depth (ft)} \times 7.48\text{ gal/cu ft}$

Worked Example 5: Rectangular Clarifier Detention Time

Problem: A rectangular sedimentation basin measures $90\text{ feet}$ long, $30\text{ feet}$ wide, and has a water depth of $12\text{ feet}$. If the plant flow rate is $2.4\text{ MGD}$, what is the hydraulic detention time in hours?

• Step 1: Calculate tank volume in cubic feet $\text{Volume (cu ft)} = 90\text{ ft} \times 30\text{ ft} \times 12\text{ ft} = 32,400\text{ cu ft}$
• Step 2: Convert cubic feet to gallons $\text{Volume (gal)} = 32,400\text{ cu ft} \times 7.48\text{ gal/cu ft} = 242,352\text{ gallons}$
• Step 3: Convert flow to gallons per day $\text{Flow (gpd)} = 2.4\text{ MGD} = 2,400,000\text{ gpd}$
• Step 4: Calculate detention time in hours $\text{DT (hours)} = \frac{242,352\text{ gal}}{2,400,000\text{ gpd}} \times 24\text{ hr/day} = 0.10098 \times 24 = 2.4235\text{ hours}$
• Step 5: Final answer $\text{Detention Time} = 2.42\text{ hours} \text{ (or } 145.4\text{ minutes)}$