18.1 The Exam Conversion Sheet, Units & Dimensional Analysis

You Get a Formula Sheet

The State Water Board publishes an Exam Conversion Sheet and provides it during the drinking water treatment and distribution examinations. That single fact reframes what preparation means: you are not being tested on whether you memorized 3,960 - you are being tested on whether you can identify what is being asked, choose the right relationship, get the units right, and interpret the answer.

The sheet's core conversions, worth knowing cold anyway because you will use them daily:

And the geometry:

$\text{Area of a circle} = 0.785 \times D^{2} \qquad \text{Circumference} = 3.14 \times D$ $\text{Cylinder volume (gal)} = 0.785 \times D^{2} \times \text{Height (ft)} \times 7.48$ $\text{Rectangular basin (gal)} = L \times W \times H \times 7.48$

[!IMPORTANT] On the wastewater side, 23 CCR 3701(b) requires that examinations at every grade include "mathematical problems related to process control and evaluation." There is no grade at which you can avoid the math, and the difficulty scales: Grade I problems are conversions and simple dosage, Grade III adds "limitations, controls, and performance calculations for primary and secondary treatment and sludge-handling processes," and Grade IV adds the same for tertiary processes plus budget development and control.

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Dimensional Analysis: The Method That Prevents Errors

Write every quantity with its units, arrange the expression so unwanted units cancel, and confirm that what remains is the unit you were asked for. If the units do not come out right, the setup is wrong and no amount of careful arithmetic will fix it.

Example. Convert 2.4 MGD to gpm using only cancellation:

$2.4 \frac{\text{MG}}{\text{day}} \times \frac{1{,}000{,}000 \text{ gal}}{1 \text{ MG}} \times \frac{1 \text{ day}}{1{,}440 \text{ min}} = \textbf{1,667 gpm}$

MG cancels, day cancels, gal/min remains. (Checking against the sheet: 2.4 x 694.4 = 1,667. Same answer, two routes.)

Example. A basin is 60 ft x 22 ft with a water depth of 13 ft. How many gallons?

$60 \text{ ft} \times 22 \text{ ft} \times 13 \text{ ft} = 17{,}160 \text{ ft}^3 \times \frac{7.48 \text{ gal}}{\text{ft}^3} = \textbf{128,357 gal}$

Example. A well casing is 10 inches in diameter with 300 ft of standing water.

$0.785 \times (10/12 \text{ ft})^2 \times 300 \text{ ft} \times 7.48 = 0.785 \times 0.694 \times 300 \times 7.48 = \textbf{1,223 gal}$

Or with the shortcut gal/ft = 0.0408 × D²(in): 0.0408 x 100 x 300 = 1,224 gal. The small difference is rounding; both are acceptable.

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The Pounds Formula: One Relationship, Many Questions

$\text{lb/day} = \text{Flow (MGD)} \times \text{Concentration (mg/L)} \times 8.34$

This single expression answers questions about chemical feed, organic loading, solids loading, sludge mass, and removal - it is the most-used formula in the trade. 8.34 is the weight of a gallon of water in pounds, so the formula is really "gallons times concentration times weight per gallon."

Rearranged:

$\text{mg/L} = \frac{\text{lb/day}}{\text{Flow (MGD)} \times 8.34} \qquad \text{Flow (MGD)} = \frac{\text{lb/day}}{\text{mg/L} \times 8.34}$

Example, three ways. A plant treats 3.6 MGD with an influent BOD of 215 mg/L. Load = 3.6 x 215 x 8.34 = 6,455 lb/day

The same plant feeds 48 lb/day of a coagulant. Dose = 48 / (3.6 x 8.34) = 48 / 30.0 = 1.6 mg/L

The plant must deliver 6,455 lb/day at 180 mg/L. Flow = 6,455 / (180 x 8.34) = 6,455 / 1,501 = 4.30 MGD

When 8.34 Is Not Right

The 8.34 factor assumes water at a specific gravity of 1.0. For a chemical solution, use its actual weight per gallon:

$\text{lb/gal} = \frac{\text{Solution \%}}{100} \times 8.34 \times \text{Specific Gravity}$

For sludge, the specific gravity is typically taken as 1.0 for thin sludges and slightly higher for thickened sludge; for lime slurry, ferric chloride, or 50 percent caustic, the weight per gallon is substantially more than 8.34 and must be used.

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Flow, Velocity, and Detention Time

$Q = A \times V \qquad V = \frac{Q}{A} \qquad A = \frac{Q}{V}$

In operator units:

$Q_{\text{gpm}} = 2.448 \times D^{2}_{(\text{in})} \times V_{(\text{ft/s})}$

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

Example. A 14-inch main carries 1,500 gpm. V = 1,500 / (2.448 x 196) = 1,500 / 479.8 = 3.13 ft/s

Example. A 420,000-gallon basin at 2.8 MGD. Flow = 2.8 x 694.4 = 1,944 gpm DT = 420,000 / 1,944 = 216 minutes = 3.6 hours

[!WARNING] Detention time and contact time are not the same number. Detention time is volume divided by flow. T10 contact time for a CT calculation is detention time multiplied by the baffling factor (0.1 unbaffled, 0.3 poor, 0.5 average, 0.7 superior). Using theoretical detention time in a CT calculation overstates disinfection - it is one of the most consequential errors an operator can make, and it appears on exams constantly.

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The Five Recurring Setup Errors

Sanity Checks

Before writing the answer down, ask three questions:

1. Is the order of magnitude plausible? A chemical dose of 4,000 mg/L is not a dose, it is a spill. A detention time of 0.02 minutes is not a clarifier.
2. Do the units match what was asked? lb/day, mg/L, gpm, gal, ft/s, hours - the question named one.
3. Does it move the right direction? More flow at the same dose means more pounds per day. A lower C-factor means more head loss. Higher MLSS at the same food means a lower F/M.

[!TIP] On a multiple-choice exam, the distractors are built from the classic errors. If a dosage problem offers an answer that is exactly your result multiplied by 0.65, you forgot to divide by purity. If it offers an answer 1.44 times yours, you mixed up per-minute and per-day. Recognizing that a distractor is an error you almost made is a legitimate and fast way to confirm your work.