10.4 Essential Water & Wastewater Math for Operators

Foundational Conversion Factors & The Pounds Formula

Operator mathematics translates analytical laboratory concentrations and flow meter telemetry into physical chemical feed rates, tank sizing dimensions, and process control actions.

┌────────────────────────────────────────────────────────────────────────┐
│                     Essential Operator Conversion Constants            │
├───────────────────────────────────┬────────────────────────────────────┤
│ 1 gallon of pure water            │ = 8.34 pounds (lbs)                │
│ 1 cubic foot (cu ft) of water     │ = 7.48 gallons = 62.4 pounds (lbs) │
│ 1 Million Gallons per Day (MGD)   │ = 1,000,000 gpd = 694.4 gpm        │
│ 1 cubic foot per second (cfs)     │ = 448.8 gpm = 0.646 MGD            │
│ 1 part per million (ppm)          │ = 1 milligram per liter (mg/L)     │
│ 1 foot of water column (head)     │ = 0.433 psi (1 psi = 2.31 ft head) │
│ 1 horsepower (HP)                 │ = 0.746 kilowatts (kW) = 33,000 ft-│
│                                   │   lbs/min                          │
└───────────────────────────────────┴────────────────────────────────────┘

1. The Pounds Formula (Mass Loading & Feed Equation)

The Pounds Formula calculates the daily mass of a constituent entering a process or the neat chemical mass required to achieve a target concentration:

• Physical Derivation: $1\text{ mg/L} = 1\text{ ppm} = \frac{1\text{ lb analyte}}{1,000,000\text{ lbs water}}$. Because $1\text{ gallon of water} = 8.34\text{ lbs}$, $1.0\text{ Million Gallons (MG)} = 8,340,000\text{ lbs}$. Therefore, multiplying $1.0\text{ MGD} \times 1.0\text{ mg/L} \times 8.34\text{ lbs/gal} = 8.34\text{ lbs/day}$.

Worked Example: Plant Organic Loading

Problem: A wastewater facility receives a daily influent flow of $3.60\text{ MGD}$ with a raw $BOD_5$ concentration of $210\text{ mg/L}$ and $TSS$ of $240\text{ mg/L}$. Calculate the daily organic $BOD_5$ loading in pounds per day.

Chemical Feed Rate & Solution Purity Mathematics

Commercial water treatment chemicals are rarely supplied as $100\%$ pure neat compounds. Liquid coagulants, sodium hypochlorite, and caustic soda are delivered as aqueous solutions characterized by specific gravities and active ingredient percentages.

┌────────────────────────────────────────────────────────────────────────┐
│                     Chemical Purity Adjustment Pathways                │
├────────────────────────────────────────────────────────────────────────┤
│ 1. Commercial Dry Chemical:                                            │
│    Chemical Dose (lbs/day) = (MGD × mg/L × 8.34) / (% Purity / 100)    │
├────────────────────────────────────────────────────────────────────────┤
│ 2. Liquid Chemical Solution Feed Rate (gpd):                           │
│    Liquid Feed (gpd) = [Neat Chemical lbs/day] / [Solution lbs/gal]    │
│    Where: Solution lbs/gal = 8.34 × Specific Gravity (SG) × (% Purity) │
└────────────────────────────────────────────────────────────────────────┘

Step-by-Step Worked Example: Sodium Hypochlorite Feed Rate

Problem: A water treatment plant treats a flow of $4.50\text{ MGD}$ with a target chlorine dosage of $3.20\text{ mg/L}$. The plant feeds commercial $12.5\%$ Sodium Hypochlorite ($NaOCl$) solution with a Specific Gravity ($SG$) of $1.20$ ($10.0\text{ lbs/gal}$). Calculate:

1. The neat chlorine required in $\text{lbs/day}$.
2. The liquid hypochlorite feed rate in gallons per day (gpd).
3. The chemical metering pump calibration rate in milliliters per minute (mL/min).

Solution Steps:

1. Calculate Neat Chlorine Demand: $$\text{Neat } Cl_2\text{ (lbs/day)} = 4.50\text{ MGD} \times 3.20\text{ mg/L} \times 8.34 = \mathbf{120.10\text{ lbs/day}}$$
2. Calculate Active Chlorine per Gallon of Solution: $$\text{Active } Cl_2\text{ per Gallon} = 8.34\text{ lbs/gal} \times 1.20\text{ (SG)} \times 0.125\text{ (Purity)} = 10.0\text{ lbs/gal} \times 0.125 = \mathbf{1.25\text{ lbs active } Cl_2\text{/gal}}$$
3. Calculate Liquid Feed Rate (gpd): $$\text{Liquid Feed Rate (gpd)} = \frac{120.10\text{ lbs/day}}{1.25\text{ lbs/gal}} = \mathbf{96.08\text{ gpd}}$$
4. Convert to Metering Pump Delivery Rate (mL/min): $$\text{Feed Rate (mL/min)} = \frac{96.08\text{ gal/day} \times 3,785.4\text{ mL/gal}}{1,440\text{ min/day}} = \frac{363,701\text{ mL/day}}{1,440\text{ min/day}} = \mathbf{252.57\text{ mL/min}}$$

Clarifier Hydraulics: Retention Time, Surface Loading & Weir Overflow Rates

Clarification basins (sedimentation tanks, secondary clarifiers, DAF units) depend upon precise hydraulic loading rates to ensure effective gravity separation of flocs and biological solids.

┌────────────────────────────────────────────────────────────────────────┐
│                     Clarifier Hydraulic Formulas                       │
├──────────────────────────┬─────────────────────────────────────────────┤
│ Hydraulic Retention Time │ DT (hours) = [Tank Volume (gal) × 24] / Flow│
│ (Detention Time - DT)    │ DT (minutes) = [Tank Volume (gal) × 1440] / │
│                          │                Flow (gpd)                   │
├──────────────────────────┼─────────────────────────────────────────────┤
│ Surface Overflow Rate    │ SOR (gpd/ft²) = Flow (gpd) / Surface Area   │
│ (Surface Loading - SOR)  │                 (ft²)                       │
├──────────────────────────┼─────────────────────────────────────────────┤
│ Weir Overflow Rate (WOR) │ WOR (gpd/ft) = Flow (gpd) / Total Weir      │
│                          │                Length (ft)                  │
└──────────────────────────┴─────────────────────────────────────────────┘

Worked Example: Circular Secondary Clarifier Evaluation

Problem: A circular secondary clarifier with a diameter of $70.0\text{ feet}$, a side water depth of $14.0\text{ feet}$, and an exterior peripheral effluent weir receives a treated wastewater flow of $2.80\text{ MGD}$ ($2,800,000\text{ gpd}$). Calculate:

1. Hydraulic Detention Time ($DT$) in hours.
2. Surface Overflow Rate ($SOR$) in $\text{gpd/ft}^2$.
3. Weir Overflow Rate ($WOR$) in $\text{gpd/linear foot}$.

Solution Steps:

1. Calculate Surface Area and Volume: $$\text{Surface Area } (A) = 0.7854 \times D^2 = 0.7854 \times (70\text{ ft})^2 = 0.7854 \times 4,900 = \mathbf{3,848.46\text{ ft}^2}$$ $$\text{Tank Volume } (V) = \text{Area} \times \text{Depth} = 3,848.46\text{ ft}^2 \times 14.0\text{ ft} = 53,878.44\text{ cu ft}$$ $$\text{Volume in Gallons} = 53,878.44\text{ cu ft} \times 7.48\text{ gal/cu ft} = \mathbf{403,010.7\text{ gallons}}$$
2. Calculate Detention Time (DT): $$DT\text{ (hours)} = \frac{403,010.7\text{ gal} \times 24\text{ hr/day}}{2,800,000\text{ gpd}} = \frac{9,672,257}{2,800,000} = \mathbf{3.45\text{ hours}}$$
3. Calculate Surface Overflow Rate (SOR): $$SOR = \frac{2,800,000\text{ gpd}}{3,848.46\text{ ft}^2} = \mathbf{727.56\text{ gpd/ft}^2} \quad (\text{Within standard design: } 400\text{–}800\text{ gpd/ft}^2)$$
4. Calculate Weir Overflow Rate (WOR): $$\text{Weir Length (Circumference)} = \pi \times D = 3.1416 \times 70.0\text{ ft} = \mathbf{219.91\text{ linear feet}}$$ $$WOR = \frac{2,800,000\text{ gpd}}{219.91\text{ ft}} = \mathbf{12,732.5\text{ gpd/linear ft}}$$

Unit Process Efficiency & Percent Removal Calculations

Evaluating the operational efficiency of screens, primary settling basins, biological aeration tanks, and filters requires calculating Percent Removal Efficiency:

Worked Example: Primary Treatment Removal

Problem: Raw wastewater entering a primary clarifier has a $TSS$ concentration of $250\text{ mg/L}$. Clarifier effluent $TSS$ is measured at $85\text{ mg/L}$. Calculate the TSS percent removal efficiency.

Activated Sludge Process Control Math: SVI & MCRT

Biological process optimization requires strict mathematical monitoring of activated sludge settling characteristics and microbial residence time.

┌────────────────────────────────────────────────────────────────────────┐
│                     Activated Sludge Process Formulations              │
├──────────────────────────┬─────────────────────────────────────────────┤
│ Sludge Volume Index      │ SVI (mL/g) = [30-Min Settled Sludge (mL/L) ×│
│ (SVI)                    │               1,000] / MLSS (mg/L)          │
├──────────────────────────┼─────────────────────────────────────────────┤
│ Mean Cell Residence Time │ MCRT (days) = [Aeration Basin Biomass (lbs)]│
│ (MCRT / Sludge Age)      │             / [Total Biomass Lost (lbs/day)]│
│                          │ MCRT = (V_aer × MLSS) / [(Q_WAS × WAS_SS) + │
│                          │        (Q_eff × Eff_SS)]                    │
└──────────────────────────┴─────────────────────────────────────────────┘

1. Sludge Volume Index (SVI)

SVI defines the physical volume in milliliters occupied by $1.0\text{ gram}$ of mixed liquor suspended solids after $30\text{ minutes}$ of quiescent settling in a $1,000\text{-mL}$ settleometer:

• Diagnostic Interpretation:
  • $SVI < 80\text{ mL/g}$: Dense, granular, fast-settling floc; leads to pinpoint floc and turbid effluent ("ashing").
  • $SVI = 80\text{ to }150\text{ mL/g}$: Ideal settling characteristics; rapid compaction with crystal-clear supernatant.
  • $SVI > 150\text{ to }200+\text{ mL/g}$: Filamentous bulking; slow-settling sludge blanket with severe risk of solids carryover over effluent weirs.

2. Mean Cell Residence Time (MCRT / Sludge Age)

MCRT quantifies the average number of days biological microorganisms remain active inside the aeration system before being wasted:

(Note: Constant $8.34$ cancels out in both numerator and denominator).

Worked Example: MCRT Calculation

Problem: An activated sludge facility operates an aeration basin with a volume of $2.50\text{ MG}$ and an $MLSS$ of $2,400\text{ mg/L}$. Daily Waste Activated Sludge ($WAS$) pumping is $0.060\text{ MGD}$ ($60,000\text{ gpd}$) with a $WAS$ solids concentration of $6,800\text{ mg/L}$. Secondary plant effluent flow is $4.20\text{ MGD}$ with an effluent $TSS$ of $12.0\text{ mg/L}$. Calculate the system MCRT in days.

Solution Steps:

1. Calculate Aeration Basin Solids Inventory (Numerator): $$\text{Aerator Biomass (lbs)} = 2.50\text{ MG} \times 2,400\text{ mg/L} \times 8.34 = \mathbf{50,040\text{ lbs}}$$
2. Calculate Daily WAS Solids Wasted: $$\text{WAS Loss (lbs/day)} = 0.060\text{ MGD} \times 6,800\text{ mg/L} \times 8.34 = \mathbf{3,402.72\text{ lbs/day}}$$
3. Calculate Daily Effluent Solids Escaping: $$\text{Effluent Loss (lbs/day)} = 4.20\text{ MGD} \times 12.0\text{ mg/L} \times 8.34 = \mathbf{420.34\text{ lbs/day}}$$
4. Calculate Total Daily Biomass Output (Denominator): $$\text{Total Solids Lost} = 3,402.72 + 420.34 = \mathbf{3,823.06\text{ lbs/day}}$$
5. Calculate MCRT: $$MCRT = \frac{50,040\text{ lbs}}{3,823.06\text{ lbs/day}} = \mathbf{13.09\text{ days}}$$

Pump Horsepower, Electrical Power & Efficiency Calculations

Water transmission and wastewater pumping require calculating mechanical power requirements and total electrical energy operating costs.

                                ENERGY FLOW IN PUMPING SYSTEMS
┌────────────────────────────────────────────────────────────────────────┐
│ [ Electrical Grid Power ] ──► [ Electric Motor ] ──► [ Centrifugal Pump]│
│                                       │                       │        │
│                                Motor Efficiency        Pump Efficiency │
│                                   (η_m ≈ 90%)             (η_p ≈ 80%)  │
│                                       │                       │        │
│                                       ▼                       ▼        │
│                              MOTOR HORSEPOWER         BRAKE HORSEPOWER │
│                                   (MHP)                    (BHP)       │
│                                                               │        │
│                                                               ▼        │
│                                                       WATER HORSEPOWER │
│                                                            (WHP)       │
└────────────────────────────────────────────────────────────────────────┘

The Three Horsepower Equations

(Constant $3,960 = \frac{33,000\text{ ft-lbs/min per HP}}{8.34\text{ lbs/gal}}$).

Worked Example: Booster Station Power & Cost Analysis

Problem: A treated water distribution booster pump delivers $1,500\text{ gpm}$ against a Total Dynamic Head of $180\text{ feet}$. The pump efficiency is $82\%$ ($0.82$), and the electric motor efficiency is $92\%$ ($0.92$). Electricity costs $0.16\text{ per kWh}$. Calculate:

1. Water Horsepower ($WHP$).
2. Brake Horsepower ($BHP$).
3. Motor Horsepower ($MHP$) and wire-to-water efficiency.
4. Daily electrical operating cost assuming 18 hours of operation per day.

Solution Steps:

1. Calculate WHP: $$WHP = \frac{1,500\text{ gpm} \times 180\text{ ft}}{3,960} = \frac{270,000}{3,960} = \mathbf{68.18\text{ WHP}}$$
2. Calculate BHP: $$BHP = \frac{68.18\text{ WHP}}{0.82} = \mathbf{83.15\text{ BHP}}$$
3. Calculate MHP & Total Efficiency: $$MHP = \frac{83.15\text{ BHP}}{0.92} = \mathbf{90.38\text{ MHP}} \quad (\text{Select standard 100 HP motor})$$ $$\text{Wire-to-Water Efficiency} = 0.82 \times 0.92 = 0.7544 = \mathbf{75.44\%}$$
4. Calculate Daily Electrical Operating Cost: $$\text{Power Demand (kW)} = 90.38\text{ MHP} \times 0.746\text{ kW/HP} = \mathbf{67.42\text{ kW}}$$ $$\text{Daily Energy Consumption} = 67.42\text{ kW} \times 18\text{ hr/day} = \mathbf{1,213.6\text{ kWh/day}}$$ $$\text{Daily Electricity Cost} = 1,213.6\text{ kWh} \times \$0.16\text{/kWh} = \mathbf{\$194.18\text{ per day}}$$

Geometric Vessel & Pipeline Volume Formulas

Calculating tank volumes, chemical contact times, and pipe flushing velocities requires geometric volumetric equations:

┌────────────────────────────────────────────────────────────────────────┐
│                     Geometric Tank & Pipe Volume Formulas              │
├──────────────────────────┬─────────────────────────────────────────────┤
│ Rectangular Basin        │ Vol (gal) = Length (ft) × Width (ft) ×      │
│                          │             Depth (ft) × 7.48 gal/cu ft     │
├──────────────────────────┼─────────────────────────────────────────────┤
│ Cylindrical Tank         │ Vol (gal) = 0.7854 × Diameter² (ft) ×       │
│ (Circular Clarifier)     │             Height (ft) × 7.48 gal/cu ft    │
├──────────────────────────┼─────────────────────────────────────────────┤
│ Circular Pipeline Conduits│ Vol (gal) = 0.7854 × (D in / 12)² × Length  │
│                          │             (ft) × 7.48 gal/cu ft           │
└──────────────────────────┴─────────────────────────────────────────────┘

Worked Example: Pipeline Disinfection Volume

Problem: A newly installed $16\text{-inch}$ ductile iron water main measuring $2,500\text{ feet}$ in length must be filled with water and disinfected with chlorine. Calculate the total water volume of the pipeline in gallons.

1. Convert pipe diameter to feet: $D = \frac{16\text{ in}}{12\text{ in/ft}} = 1.333\text{ ft}$.
2. Cross-sectional area: $A = 0.7854 \times (1.333)^2 = 1.396\text{ ft}^2$.
3. Pipe volume in cubic feet: $V = 1.396\text{ ft}^2 \times 2,500\text{ ft} = 3,490.66\text{ cu ft}$.
4. Volume in gallons: $\text{Gallons} = 3,490.66\text{ cu ft} \times 7.48\text{ gal/cu ft} = \mathbf{26,110\text{ gallons}}$.