NCEES FE Other Disciplines · USA · NCEES
Original practice questions written to the published syllabus · not past papersNCEES FE Other Disciplines is a licensing and registration exam run by NCEES, for candidates in USA. Computer-based, 110 questions, 6-hour appointment (5 h 20 min exam).
Source: ncees.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Safety, health and environment and chemistry items: BOD and COD calculation, oxygen demand stoichiometry, ppm to mg/m3 conversion, hazardous gases such as hydrogen sulphide, and basic treatment process definitions.
Handling residues, manure and wastewater so that farms and plants do not pollute soil, water and air.
Our 20 NCEES FE Other Disciplines questions in this subject: 11 easy, 8 medium, 1 hard, across 6 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to NCEES FE Other Disciplines and are not shown on any other exam's page.
For a dairy wastewater, the measured COD is greater than the BOD5. The main reason is that:
C. COD also oxidises organic matter that microbes degrade slowly or not at all within five days
The dichromate oxidises nearly all organic matter, including slowly biodegradable and refractory compounds, while BOD5 counts only what microbes oxidise in five days. Chloride is an interference in the COD test (it is masked with mercury sulphate), not something BOD counts; dilution and bottle leaks are not the reason.
A wastewater has an ultimate carbonaceous BOD (L0) of 300 mg/L and a first-order rate constant k = 0.23 per day to base e (BODt = L0[1 − e^(−kt)]). Its BOD5 is:
B. 205 mg/L
BOD5 = 300 × (1 − e^(−0.23×5)) = 300 × (1 − 0.317) = 205 mg/L. 279 mg/L treats 0.23 as a base-10 constant; 95 mg/L is the BOD still remaining after 5 days (L0e^(−kt)), not the BOD exerted; 345 mg/L assumes linear exertion and exceeds the ultimate BOD, which is impossible.
A BOD rate constant is reported as K = 0.10 per day to base 10. The equivalent base-e constant k is:
C. 0.230 per day
Because e^(−kt) = 10^(−Kt), k = 2.303 K = 2.303 × 0.10 = 0.230 per day. 0.043 divides by 2.303 instead of multiplying; 0.100 ignores the change of base, the error that gives wrong BOD5 values in many answers. 1.000 multiplies by 10 instead of 2.303.
A dairy plant discharges 500 m³/day of wastewater with a BOD5 of 1500 mg/L. The daily BOD5 load is:
C. 750 kg/day
Load (kg/d) = flow (m³/d) × concentration (mg/L = g/m³) / 1000 = 500 × 1500/1000 = 750 kg/day. 750000 forgets that 1 g/m³ = 1 mg/L and leaves the answer in grams; 0.75 divides by 10⁶ instead of 10³; 75 is a decimal slip.
When only the carbonaceous BOD of a nitrified effluent is wanted, a nitrification inhibitor is added to the BOD bottle because:
A. Nitrifying bacteria would otherwise consume oxygen oxidising ammonia
Oxidation of ammonia to nitrite and nitrate by nitrifiers uses oxygen and adds a nitrogenous BOD, which inflates the result when only carbonaceous BOD is wanted. The inhibitor blocks nitrifiers; it does not supply nutrients, and nitrate does not release oxygen in the bottle.
In an ideal (Camp) sedimentation basin, the removal of discrete particles of a given settling velocity depends on:
B. The surface overflow rate, so on plan area and not on depth
Every particle whose settling velocity is at least Q/A is removed completely, independent of depth; deeper tanks lengthen detention time but the particle must also fall further. Inlet velocity and detention time alone do not set removal in the ideal model.
Dairy manure contains 80% moisture on a wet basis. Its moisture on a dry basis is:
D. 400%
Per 100 kg: 80 kg water, 20 kg dry matter. Dry basis = 80/20 = 400%. 20% is the dry-matter fraction; 125% is 100/80.
Ammonia is measured at 25 ppm (by volume) in a pig house. Its concentration in mg/m³ at 25 °C and 1 atm (molar volume 24.45 L/mol; NH3 = 17.03 g/mol) is:
D. 17.4 mg/m³
mg/m³ = ppm × MW/24.45 = 25 × 17.03/24.45 = 17.4 mg/m³. 19.0 uses 22.41 L/mol, the molar volume at 0 °C, not the conditions stated; 35.9 inverts the ratio; 426 omits the molar volume.
Hydrogen sulphide (34.08 g/mol) at 10 ppm by volume, at 25 °C and 1 atm (24.45 L/mol), corresponds to:
C. 13.9 mg/m³
10 × 34.08/24.45 = 13.9 mg/m³. 15.2 uses the 0 °C molar volume; 7.2 inverts the ratio; 340.8 omits the molar volume.
Hydrogen sulphide is especially dangerous in manure pits because, at high concentrations, it:
C. Paralyses the sense of smell so the danger is not noticed
H2S smells of rotten eggs at very low levels, but at higher levels it deadens the sense of smell, so a rising, lethal concentration can go unnoticed. It is heavier than air (not water) and collects in low places; it kills people and animals.
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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