16.1 Math Fundamentals, Conversions & Formula Sheets

16.1 Math Fundamentals, Conversions & Formula Sheets

Quick Answer: Operator math rests on a small set of unit conversions, area/volume formulas, and density constants used every day: 7.48 gal/ft³, 8.34 lb/gal, and mg/L ≈ ppm in dilute water. FDEP provides formula sheets at the exam—study them in advance so you recognize every symbol and do not waste clock time re-deriving geometry. Master order of operations, cancel units, and watch common traps (diameter vs radius, MG vs gal, wet vs dry basis).

Florida water, wastewater, and distribution exams put heavy weight on calculation items. Class C treatment exams (100 questions, 3 hours) and shorter Class D / distribution exams all draw from the same core toolbox. This section builds the foundation used in dosage, detention, loadings, and pump problems later in the chapter.

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1. Why Formula Sheets Matter on the FDEP Exam

FDEP’s Operator Certification Program historically provides formula sheets (and related reference material) at the examination. That does not mean you can skip math practice:

Study strategy: Print or obtain the current FDEP/PSI formula reference used for water/wastewater exams, work every example in this chapter with that sheet open, and mark the lines you actually use. On test day you should open the sheet only to confirm a constant or recall a rarely used formula—not to learn geometry for the first time.

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2. Core Unit Conversions Operators Use Daily

Density anchors (nearly pure water at ordinary temperature):

• 1 gallon of water ≈ 8.34 lb
• 1 ft³ of water ≈ 62.4 lb (because 7.48 × 8.34 ≈ 62.4)
• 1 mg/L in 1 MG of water = 8.34 lb of that substance (basis of the famous “pounds formula”)

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3. Area Formulas

Trap: If the problem gives diameter, do not square the radius by mistake without converting. If it gives radius, do not use (0.785 D^2) with (D = r).

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4. Volume Formulas (Tanks)

Depth of water (H) is the liquid depth, not always the tank wall height. A tank 20 ft tall filled only 12 ft uses (H = 12) ft for volume of water.

Worked Example A — Rectangular Clearwell

A clearwell is 40 ft long × 25 ft wide × 12 ft water depth.

Step 1 — Volume in ft³

[ V = 40 \times 25 \times 12 = 12{,}000\ \text{ft}^3 ]

Step 2 — Convert to gallons

[ V = 12{,}000 \times 7.48 = 89{,}760\ \text{gal} ]

Step 3 — Express as MG (if needed for detention later)

[ V = 89{,}760 / 1{,}000{,}000 = 0.08976\ \text{MG} ]

Worked Example B — Cylindrical Tank

A vertical cylindrical tank has diameter 30 ft and water depth 18 ft.

Step 1 — Area of base

[ A = 0.785 \times D^2 = 0.785 \times 30^2 = 0.785 \times 900 = 706.5\ \text{ft}^2 ]

Step 2 — Volume ft³

[ V = 706.5 \times 18 = 12{,}717\ \text{ft}^3 ]

Step 3 — Gallons

[ V = 12{,}717 \times 7.48 \approx 95{,}123\ \text{gal} ]

(Using more digits of (\pi) changes the answer slightly; exam choices are usually spaced to allow the 0.785 convention on the formula sheet.)

Worked Example C — Conical Volume

A cone-bottom section has D = 12 ft and cone height H = 9 ft. Volume of the cone only:

[ V = \frac{1}{3} \times 0.785 \times 12^2 \times 9 = \frac{1}{3} \times 0.785 \times 144 \times 9 ]

[ 0.785 \times 144 = 113.04;\quad 113.04 \times 9 = 1{,}017.36;\quad \div 3 = 339.12\ \text{ft}^3 ]

[ V_{\text{gal}} = 339.12 \times 7.48 \approx 2{,}537\ \text{gal} ]

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5. mg/L, ppm, and Percent Solutions

In dilute aqueous solutions typical of water and wastewater:

[ 1\ \text{mg/L} \approx 1\ \text{ppm (by mass)} ]

Percent by weight relates approximately as:

[ 1% \approx 10{,}000\ \text{mg/L} ]

Exam wording: “12% available chlorine” means 12 lb of available chlorine per 100 lb of solution (weight basis)—feed-rate problems use the decimal fraction 0.12, not “12 mg/L.”

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6. Order of Operations & Dimensional Analysis

Always:

1. Write the formula (from memory or sheet).
2. List given values with units.
3. Convert units before multiplying (gpm → MGD, gal → MG, inches → feet).
4. Multiply/divide left to right after parentheses and powers.
5. Sanity-check magnitude (a 1 MGD plant does not need 10,000 lb/day chlorine at 1 mg/L).

Common conversion you will use constantly:

[ \text{MGD} = \frac{\text{gpm} \times 1{,}440\ \text{min/day}}{1{,}000{,}000} = \frac{\text{gpm}}{694.4} \approx \frac{\text{gpm}}{700}\ \text{(rough check)} ]

Exactly: (\text{MGD} = \text{gpm} \times 0.00144).

Example: 500 gpm = (500 \times 0.00144 = 0.72) MGD.

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7. Common Traps That Steal Points

Worked Example D — Mixed Unit Conversion

Flow is 2,500 gpm. Express as MGD and as ft³/s (cfs) roughly.

MGD:

[ 2{,}500 \times 0.00144 = 3.6\ \text{MGD} ]

ft³/s: First gal/s = (2500 / 60 ≈ 41.67) gal/s; ft³/s = (41.67 / 7.48 ≈ 5.57) cfs.

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8. Practice Setup Before the Next Sections

Every later formula is built from this section:

• Dosage lb/day needs MGD, mg/L, 8.34
• Detention time needs volume (gal or ft³) and consistent flow units
• Loadings reuse the pounds formula on BOD, SS, etc.
• Pump gpm from tank drawdown reuses cylinder volume × 7.48

If conversions feel slow, drill ten volume problems and ten gpm↔MGD problems before attempting full dosage sets.

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Florida Exam Notes

• Formula sheets are a tool, not a substitute for practice.
• Expect geometry + conversion items even on distribution-focused forms.
• Water and wastewater stems share the same constants; only the process story changes.
• Keep a personal “error log” of missed unit conversions—those patterns repeat on PSI forms.