19.1 Flow, Concentration, and Load Workbook

Key Takeaways

  • Identify the basis (flow, concentration, mass rate, volume, area, or time) before opening the NCEES PE Civil Reference Handbook.
  • Memorize the load constant: lb/day = flow (MGD) x concentration (mg/L) x 8.34, where 8.34 lb/gal is water density.
  • 1 MGD = 694.4 gpm = 1.547 cfs; 1 cfs = 448.8 gpm; 1 acre-ft = 325,851 gal = 43,560 ft3.
  • Detention time = volume / flow; cubic feet over cfs yields seconds, gallons over MGD yields days.
  • Run a proportional magnitude check: doubling concentration at fixed flow must double load.
Last updated: June 2026

Start with the basis, not the formula

WRE calculation questions look different but share one unit engine. A wastewater load, a chlorine dose, a tank volume, a detention time, and a pollutant allocation all require the same first decision: what is being measured? The answer is a flow rate, concentration, mass rate, volume, surface loading rate, or time. Identify the basis before opening the NCEES PE Civil Reference Handbook (the April 2024 edition supplied on-screen) and the formula becomes obvious. The exam gives roughly 6 minutes per item across 80 questions in a 9-hour appointment, so a disciplined setup buys time.

The conversion constants you must memorize

FromToMultiply by
1 MGDgpm694.4
1 MGDcfs1.547
1 cfsgpm448.8
1 cfsMGD0.6463
1 acre-ftgallons325,851
1 ft3gallons7.481
Water densitylb/gal8.34

The master load equation

The single most-used WRE relationship is mass load = flow x concentration x conversion. In US units:

lb/day = Q (MGD) x C (mg/L) x 8.34

The 8.34 collapses (mg/L)(MG/day)(8.34 lb/gal per mg/L-MG). Worked example: a plant treats 2.5 MGD at 30 mg/L BOD5. Load = 2.5 x 30 x 8.34 = 625.5 lb/day of BOD. If the question instead gives a permit load of 625 lb/day at 2.5 MGD and asks for concentration, rearrange: C = 625 / (2.5 x 8.34) = 30 mg/L. Never start by guessing a treatment formula.

Detention, storage, and loading logic

Detention time

Detention time t = V / Q, but V and Q often arrive on different time bases. A basin of 0.5 MG at 2.5 MGD gives t = 0.5/2.5 = 0.2 day = 4.8 hr. A channel of 1,800 ft3 at 6 cfs gives 1,800/6 = 300 s = 5 min directly, because ft3 over ft3/s is seconds. A stormwater pond of 5 acre-ft (217,800 ft3) at an inflow of 10 cfs gives 21,780 s = 6.05 hr only after converting acre-ft to ft3.

Surface and hydraulic loading

Surface overflow rate (SOR) = Q / plan area (gpd/ft2), used for clarifiers. Hydraulic loading rate = Q / filter area (gpm/ft2), used for filters. The units reveal the concept; a classic distractor substitutes tank volume for plan area. Example: 2.0 MGD over a 40-ft-diameter clarifier (area = pi/4 x 40^2 = 1,257 ft2) gives SOR = 2,000,000/1,257 = 1,591 gpd/ft2, which sits above the typical 600-1,200 gpd/ft2 design range for secondary clarifiers and therefore flags this unit as hydraulically overloaded -- the reasonableness check is the real teaching point.

Workbook checks (run every problem)

  1. State whether the answer should rise or fall as flow rises.
  2. Keep concentration and load separate until the final line.
  3. Convert time once; never switch day to hour mid-equation.
  4. Ask whether the magnitude fits a pipe, plant, or watershed scale.
  5. Re-read the requested unit before selecting an answer.

The four-row load drill

Run this until automatic. Row 1: convert plant flow to the day basis. Row 2: write concentration as mass per volume. Row 3: multiply with the 8.34 constant. Row 4: compare to the requested unit and expected direction. If concentration doubles at fixed flow, load doubles. If flow is halved at fixed concentration, load halves. This proportional check kills wrong choices before detailed arithmetic finishes, and it is the single highest-yield habit for the flow-and-load cluster of the WRE blueprint.

Population, per-capita, and SI cross-checks

Domestic loading often arrives per capita. A useful anchor: typical raw domestic wastewater is roughly 0.17-0.20 lb BOD per capita-day and design flow near 100 gpd per capita. A 50,000-person service area at 100 gpd/capita gives 5.0 MGD; at 0.20 lb BOD/capita-day the influent BOD load is 10,000 lb/day, which back-checks to a concentration of 10,000/(5.0 x 8.34) approximately 240 mg/L -- consistent with medium-strength domestic wastewater. When a problem mixes SI and US units, convert early: 1 mg/L = 1 g/m3, 1 m3/s = 22.8 MGD, and load in kg/day = Q (m3/s) x C (mg/L) x 86.4.

Carrying these anchors lets you reject a choice that is off by an order of magnitude before you finish the arithmetic, which is exactly where the WRE flow-and-load distractors live.

Worked mass-balance example

Mass balance is the unifying tool: in = out + accumulation, and at steady state accumulation is zero. Consider a stream at 8 cfs carrying 4 mg/L of a conservative pollutant that receives a discharge of 1.5 cfs at 60 mg/L. The mixed concentration is the flow-weighted average: C = (8 x 4 + 1.5 x 60) / (8 + 1.5) = (32 + 90) / 9.5 = 12.8 mg/L. Notice the answer must fall between 4 and 60 mg/L and lean toward the larger flow's value -- a sanity check that immediately eliminates any choice outside that band or above 60.

The same flow-weighted blending governs reservoir mixing, recycle streams, and split-flow treatment, so practicing it as a standalone setup pays off across the entire water-quality and treatment portion of the blueprint.

Test Your Knowledge

A plant treats 4.0 MGD of wastewater with an influent BOD of 220 mg/L. What is the BOD mass loading in lb/day?

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Test Your Knowledge

A basin volume is given in cubic feet and flow is given in cfs. What time unit does volume divided by flow produce before any conversion?

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