5.3 Hydraulic, Dosage & Pounds Formula Math

Hydraulic, Dosage & Pounds Formula Math

Precise mathematical competency is essential for water and wastewater operators. Operators utilize applied mathematical calculations daily to determine chemical feeder setpoints, basin retention times, surface loading rates, filtration backwash velocities, and pumping horsepower requirements.

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1. Dimensional Analysis & Essential Conversion Factors

All applied utility calculations rely on a core set of fundamental conversion constants. Committing these constants to memory and tracking units through dimensional cancellation prevents calculation errors.

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2. Tank Geometry, Volume & Pipe Flow Continuity ($Q = A \cdot V$)

Rectangular Basin Volume Calculations

$\text{Volume (cu ft)} = \text{Length (ft)} \times \text{Width (ft)} \times \text{Depth (ft)}$ $\text{Volume (Gallons)} = \text{Volume (cu ft)} \times 7.48\text{ gal/cu ft}$ $\text{Volume (Million Gallons, MG)} = \frac{\text{Volume (Gallons)}}{1,000,000}$

Worked Example 1: Rectangular Flocculation Basin Capacity

A conventional water plant operates a rectangular flocculation basin with dimensions $90\text{ ft}$ long, $30\text{ ft}$ wide, and an effective water depth of $14\text{ ft}$. Calculate the total holding capacity in gallons and million gallons (MG).

1. Calculate volume in cubic feet: $\text{Volume} = 90\text{ ft} \times 30\text{ ft} \times 14\text{ ft} = 37,800\text{ ft}^3$
2. Convert cubic feet to gallons: $\text{Capacity} = 37,800\text{ ft}^3 \times 7.48\text{ gal/ft}^3 = 282,744\text{ gallons}$
3. Convert gallons to million gallons: $\text{Capacity (MG)} = \frac{282,744}{1,000,000} = 0.2827\text{ MG}$

Circular Tank Volume Calculations

$\text{Area (sq ft)} = \pi \times r^2 = 0.7854 \times \text{Diameter (ft)}^2$ $\text{Volume (cu ft)} = 0.7854 \times \text{Diameter (ft)}^2 \times \text{Depth (ft)}$ $\text{Volume (Gallons)} = 0.7854 \times \text{Diameter (ft)}^2 \times \text{Depth (ft)} \times 7.48\text{ gal/cu ft}$

Worked Example 2: Circular Clarifier Volume

A circular secondary clarifier has a diameter of $75\text{ ft}$ and a side water depth (SWD) of $12\text{ ft}$. Calculate the volume in gallons.

1. Calculate surface area: $\text{Area} = 0.7854 \times (75\text{ ft})^2 = 0.7854 \times 5,625 = 4,417.88\text{ ft}^2$
2. Calculate volume in cubic feet: $\text{Volume} = 4,417.88\text{ ft}^2 \times 12\text{ ft} = 53,014.5\text{ ft}^3$
3. Convert to gallons: $\text{Gallons} = 53,014.5\text{ ft}^3 \times 7.48\text{ gal/ft}^3 = 396,548.5\text{ gallons}$ (or $0.3965\text{ MG}$)

Pipe Flow Velocity ($Q = A \cdot V$)

The continuity equation governs flow through closed pressurized pipes and open channels: $Q = A \times V \implies V = \frac{Q}{A}$ Where $Q = \text{Flow (cfs)}$, $A = \text{Cross-sectional Area (sq ft)} = 0.7854 \times D^2$, and $V = \text{Velocity (ft/sec)}$.

Worked Example 3: Velocity in a Transmission Main

A $16\text{-inch}$ diameter ductile iron water main carries a treated water flow rate of $3.5\text{ MGD}$. Calculate the velocity of water in the pipe in feet per second (ft/sec).

1. Convert pipe diameter from inches to feet: $D = \frac{16\text{ in}}{12\text{ in/ft}} = 1.333\text{ ft}$
2. Calculate pipe cross-sectional area: $A = 0.7854 \times (1.333\text{ ft})^2 = 0.7854 \times 1.777 = 1.396\text{ ft}^2$
3. Convert daily flow rate (MGD) to cubic feet per second (cfs): $Q = 3.5\text{ MGD} \times 1.547\text{ cfs/MGD} = 5.4145\text{ cfs}$
4. Calculate flow velocity ($V$): $V = \frac{Q}{A} = \frac{5.4145\text{ cfs}}{1.396\text{ ft}^2} = 3.88\text{ ft/sec}$

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3. Hydraulic Detention Time Calculations

Hydraulic Detention Time (DT) represents the theoretical average time an incremental parcel of water resides within a treatment vessel:

$\text{Detention Time} = \frac{\text{Basin Volume}}{\text{Flow Rate}}$

                    DETENTION TIME TEMPORAL CONVERSIONS

   To get DT in DAYS:    Basin Volume (Gallons) / Flow Rate (Gallons per Day, gpd)
   To get DT in HOURS:   [Basin Volume (Gallons) / Flow Rate (gpd)] x 24 Hours/Day
   To get DT in MINUTES: [Basin Volume (Gallons) / Flow Rate (gpm)]

Worked Example 4: Sedimentation Basin Detention Time in Hours

A water treatment plant treats a flow rate of $4.8\text{ MGD}$. The sedimentation basin has a volume of $800,000\text{ gallons}$. Calculate the hydraulic detention time in hours.

1. Determine hourly flow rate: $\text{Flow Rate (gal/hr)} = \frac{4,800,000\text{ gpd}}{24\text{ hr/day}} = 200,000\text{ gal/hr}$
2. Calculate detention time: $\text{Detention Time} = \frac{800,000\text{ gallons}}{200,000\text{ gal/hr}} = 4.0\text{ hours}$

Worked Example 5: Rapid Mix Basin Detention Time in Seconds

A rapid mix tank holds $1,500\text{ gallons}$. The plant treats a flow of $3.6\text{ MGD}$. Calculate the rapid mix detention time in seconds.

1. Convert plant flow to gallons per second (gps): $\text{Flow Rate (gal/sec)} = \frac{3,600,000\text{ gpd}}{86,400\text{ sec/day}} = 41.67\text{ gal/sec}$
2. Calculate detention time: $\text{Detention Time} = \frac{1,500\text{ gallons}}{41.67\text{ gal/sec}} = 36.0\text{ seconds}$

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4. The Universal Pounds Formula & Active Chemical Feed Math

The Universal Pounds Formula is the single most vital equation on operator certification exams. It establishes the mathematical link between volumetric flow rate, chemical concentration, and mass loading.

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

                          THE POUNDS FORMULA TRIANGLE

                                  /  POUNDS  \
                                 /   lbs/day  \
                                /--------------\
                               / Flow   Dosage  \
                              /  MGD  x  mg/L x 8.34 \
                             -------------------------

Worked Example 6: Pure Chemical Feed (100% Chlorine Gas)

A water treatment plant filters $6.5\text{ MGD}$ and maintains a finished water chlorine dosage of $2.4\text{ mg/L}$. How many pounds of $100%$ chlorine gas must be fed per day?

$\text{Feed (lbs/day)} = 6.5\text{ MGD} \times 2.4\text{ mg/L} \times 8.34\text{ lbs/gal} = 130.10\text{ lbs/day}$

Adjusting for Chemical Purity / Active Strength

Commercial chemicals are rarely $100%$ pure active compound. Dry chemicals (such as calcium hypochlorite, alum, soda ash, or hydrated lime) contain inert binders or carrier salts. The operator must feed more commercial product to deliver the required active mass:

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

Worked Example 7: Granular Calcium Hypochlorite Feed Rate

A well pump delivers $1.2\text{ MGD}$. The required chlorine dosage is $3.0\text{ mg/L}$. Disinfection is provided using granular calcium hypochlorite containing $65%$ available chlorine by weight. How many pounds of calcium hypochlorite powder must be fed daily?

1. Calculate required active chlorine mass: $\text{Active } Cl_2\text{ (lbs/day)} = 1.2\text{ MGD} \times 3.0\text{ mg/L} \times 8.34\text{ lbs/gal} = 30.024\text{ lbs/day}$
2. Divide by purity decimal ($0.65$): $\text{Commercial Hypochlorite (lbs/day)} = \frac{30.024\text{ lbs/day}}{0.65} = 46.19\text{ lbs/day}$

Liquid Chemical Feed Rates & Specific Gravity (SG)

Liquid chemicals (such as sodium hypochlorite $12.5%$, liquid alum $48.5%$, or ferric chloride $40%$) are metered in gallons or milliliters. Calculations must account for the solution's Specific Gravity (SG) (the ratio of the liquid's density to water at $8.34\text{ lbs/gal}$):

$\text{Solution Weight (lbs/gal)} = 8.34\text{ lbs/gal} \times \text{Specific Gravity (SG)}$ $\text{Active Chemical per Gallon (lbs/gal)} = 8.34 \times \text{SG} \times \text{Active Decimal Fraction}$ $\mathbf{\text{Liquid Feed Rate (gpd)} = \frac{\text{Active Chemical Required (lbs/day)}}{\text{Active Chemical per Gallon (lbs/gal)}}}$

                   LIQUID CHEMICAL FEED CONVERSION STEPS

   1. Calculate Active lbs/day   = MGD x mg/L x 8.34
   2. Calculate Solution lbs/gal  = 8.34 x Specific Gravity
   3. Calculate Active lbs/gal    = Solution lbs/gal x Decimal Concentration
   4. Calculate Liquid gpd        = Active lbs/day / Active lbs/gal
   5. Convert gpd to mL/min       = (Liquid gpd x 3,785 mL/gal) / 1,440 min/day

Worked Example 8: Liquid Sodium Hypochlorite Feed in GPD & mL/min

A waterworks plant treats $2.5\text{ MGD}$ with a target sodium hypochlorite dose of $2.0\text{ mg/L}$. The plant uses commercial $12.5%$ sodium hypochlorite solution with a specific gravity of $1.20$. Calculate the feed rate in gallons per day (gpd) and milliliters per minute (mL/min).

1. Calculate active chlorine required per day: $\text{Active } Cl_2\text{ (lbs/day)} = 2.5\text{ MGD} \times 2.0\text{ mg/L} \times 8.34\text{ lbs/gal} = 41.70\text{ lbs/day}$
2. Calculate the weight of one gallon of hypochlorite solution: $\text{Solution Weight} = 8.34\text{ lbs/gal} \times 1.20 = 10.008\text{ lbs/gal}$
3. Calculate the active chlorine contained in one gallon of solution: $\text{Active } Cl_2\text{ per gallon} = 10.008\text{ lbs/gal} \times 0.125 = 1.251\text{ lbs active } Cl_2\text{/gal}$
4. Calculate liquid chemical feed rate in gallons per day: $\text{Feed Rate (gpd)} = \frac{41.70\text{ lbs/day}}{1.251\text{ lbs/gal}} = 33.33\text{ gallons per day (gpd)}$
5. Convert gpd to milliliters per minute (mL/min) for metering pump calibration: $\text{Feed Rate (mL/min)} = \frac{33.33\text{ gal/day} \times 3,785\text{ mL/gal}}{1,440\text{ min/day}} = \frac{126,154}{1,440} = 87.61\text{ mL/min}$

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5. Filtration Hydraulics, Backwash & Surface Loading Rates

Filtration Hydraulic Loading Rate

$\text{Filtration Loading Rate (gpm/sq ft)} = \frac{\text{Influent Flow Rate (gpm)}}{\text{Filter Bed Surface Area (sq ft)}}$

Worked Example 9: Dual-Media Filter Loading Rate

A water treatment plant operates four identical rapid gravity dual-media filters. Each filter bed measures $20\text{ ft}$ long by $15\text{ ft}$ wide. Total plant flow is $6.0\text{ MGD}$ evenly divided across all four filters. Calculate the hydraulic loading rate on each active filter in $\text{gpm/ft}^2$.

1. Calculate flow to each filter: $\text{Flow per filter (MGD)} = \frac{6.0\text{ MGD}}{4} = 1.5\text{ MGD}$
2. Convert MGD to gallons per minute (gpm): $\text{Flow per filter (gpm)} = \frac{1,500,000\text{ gpd}}{1,440\text{ min/day}} = 1,041.67\text{ gpm}$
3. Calculate filter bed surface area: $\text{Surface Area} = 20\text{ ft} \times 15\text{ ft} = 300\text{ sq ft}$
4. Calculate hydraulic loading rate: $\text{Loading Rate} = \frac{1,041.67\text{ gpm}}{300\text{ sq ft}} = 3.47\text{ gpm/sq ft}$

Backwash Rise Rate Calculations

Backwash flow is measured either as a surface wash rate ($\text{gpm/ft}^2$) or as an upward vertical velocity (inches per minute rise rate):

$\mathbf{\text{Rise Rate (in/min)} = \text{Backwash Loading Rate (gpm/ft}^2) \times 1.604\text{ (in/min)}/(\text{gpm/ft}^2)}$ Derivation: $\frac{1\text{ gal}}{\text{ft}^2} = \frac{0.1337\text{ ft}^3}{\text{ft}^2} = 0.1337\text{ ft} \times 12\text{ in/ft} = 1.604\text{ inches}$.

Worked Example 10: Filter Backwash Rise Rate

A filter bed measuring $20\text{ ft} \times 15\text{ ft}$ ($300\text{ ft}^2$) is backwashed at a pump discharge rate of $4,800\text{ gpm}$. Calculate the backwash loading rate in $\text{gpm/ft}^2$ and the vertical rise rate in inches per minute (in/min).

1. Calculate backwash loading rate: $\text{Loading Rate} = \frac{4,800\text{ gpm}}{300\text{ ft}^2} = 16.0\text{ gpm/ft}^2$
2. Convert to rise rate in inches per minute: $\text{Rise Rate (in/min)} = 16.0\text{ gpm/ft}^2 \times 1.604 = 25.66\text{ in/min}$

Clarifier Surface Overflow Rate (SOR) & Weir Overflow Rate (WOR)

$\text{Surface Overflow Rate (SOR, gpd/sq ft)} = \frac{\text{Flow Rate (gpd)}}{\text{Clarifier Surface Area (sq ft)}}$ $\text{Weir Overflow Rate (WOR, gpd/linear ft)} = \frac{\text{Flow Rate (gpd)}}{\text{Total Effluent Weir Length (ft)}}$

Worked Example 11: Clarifier SOR and WOR

A circular clarifier has a diameter of $80\text{ ft}$ and an interior peripheral effluent weir running along its full circumference. The daily flow rate is $3.6\text{ MGD}$. Calculate the SOR ($\text{gpd/ft}^2$) and the WOR ($\text{gpd/ft}$).

1. Calculate clarifier surface area: $\text{Area} = 0.7854 \times (80\text{ ft})^2 = 5,026.55\text{ sq ft}$
2. Calculate Surface Overflow Rate (SOR): $\text{SOR} = \frac{3,600,000\text{ gpd}}{5,026.55\text{ sq ft}} = 716.20\text{ gpd/sq ft}$
3. Calculate total peripheral weir length ($C = \pi \cdot D$): $\text{Weir Length} = 3.1416 \times 80\text{ ft} = 251.33\text{ linear feet}$
4. Calculate Weir Overflow Rate (WOR): $\text{WOR} = \frac{3,600,000\text{ gpd}}{251.33\text{ ft}} = 14,323.8\text{ gpd/linear ft}$

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6. Pumping Hydraulics & Horsepower Calculations

Pumping calculations translate flow and head into mechanical and electrical power requirements.

                      PUMPING HORSEPOWER PROGRESSION

   WATER HORSEPOWER (WHP)   = (gpm x TDH in feet) / 3,960
              |
              v   [Divide by Pump Efficiency Decimal, e.g., 0.80]
   BRAKE HORSEPOWER (BHP)   = WHP / Pump Efficiency
              |
              v   [Divide by Motor Efficiency Decimal, e.g., 0.90]
   MOTOR HORSEPOWER (MHP)   = BHP / Motor Efficiency = WHP / (Pump Eff x Motor Eff)

Derivation of the Pumping Constant ($3,960$)

$1\text{ Horsepower (HP)} = 33,000\text{ foot-pounds per minute (ft-lbs/min)}$ $\text{Power (ft-lbs/min)} = \text{Flow (gpm)} \times 8.34\text{ lbs/gal} \times \text{TDH (ft)}$ $\text{WHP} = \frac{\text{gpm} \times 8.34 \times \text{TDH}}{33,000} = \frac{\text{gpm} \times \text{TDH}}{\frac{33,000}{8.34}} = \frac{\text{gpm} \times \text{TDH}}{3,960}$

Worked Example 12: Comprehensive Pumping Horsepower

A high-service centrifugal pump delivers $1,500\text{ gpm}$ against a Total Dynamic Head (TDH) of $185\text{ feet}$. The pump manufacturer lists a pump mechanical efficiency of $82%$ ($0.82$) and the electric motor has an efficiency of $92%$ ($0.92$). Calculate the Water Horsepower (WHP), Brake Horsepower (BHP), and total Motor Horsepower (MHP).

1. Calculate Water Horsepower (WHP): $\text{WHP} = \frac{1,500\text{ gpm} \times 185\text{ ft}}{3,960} = \frac{277,500}{3,960} = 70.08\text{ WHP}$
2. Calculate Brake Horsepower (BHP) delivered to the pump shaft: $\text{BHP} = \frac{\text{WHP}}{\text{Pump Efficiency}} = \frac{70.08}{0.82} = 85.46\text{ BHP}$
3. Calculate Motor Horsepower (MHP) drawn by the electric motor: $\text{MHP} = \frac{\text{BHP}}{\text{Motor Efficiency}} = \frac{85.46}{0.92} = 92.89\text{ MHP}$ (Operational Decision: The utility would specify a standard commercial $100\text{ HP}$ continuous-duty electric motor to avoid motor overload).