NCEES PE Environmental

USA · National Council of Examiners for Engineering and Surveying · licensing

5practice questions written to this syllabus
4subjects

Format

Computer-based, 80 questions (multiple choice and alternative item types), 8 hours of exam time within a 9-hour appointment with a 50-minute break.

Sections

Over the years

The official PE Environmental page was read for format. The knowledge-area list is from general knowledge; the detailed specification is a separate PDF that was not read. Taken after the FE and qualifying experience.

Official source: https://ncees.org/exams/pe-exam/environmental/

Subjects this exam draws on

Hydrology & Water Resources for Environmental Scientists1 questionsWater Quality & Agricultural Non-point Source Pollution2 questionsWater & Wastewater Treatment1 questionsEnvironmental & Agro-ecosystem Modelling1 questions

Five practice questions, worked

Original questions written to this exam's published syllabus and pattern. Not past papers.

1. In an unconfined aquifer with K = 15 m/day, saturated thicknesses are 18 m at 20 m and 20 m at 200 m from a pumped well at steady state. What is the pumping rate (Dupuit-Thiem)?

  1. 14,817 m3/day
  2. 1,555 m3/day
  3. 3,581 m3/day
  4. 82 m3/day
Answer and working

B. 1,555 m3/day

Q = pi K (h2^2 - h1^2) / ln(r2/r1) = pi x 15 x 76 / 2.303 = 1,555. Using (h2 - h1) as in confined flow, or log10, is wrong.

Hydrology & Water Resources for Environmental Scientists · Darcy's law and steady-state well hydraulics

2. A dairy effluent has BOD5 = 200 mg/L with k = 0.23 per day (base e). What is its ultimate BOD?

  1. 632 mg/L
  2. 137 mg/L
  3. 293 mg/L
  4. 1000 mg/L
Answer and working

C. 293 mg/L

L0 = BOD5 / (1 - e^-1.15) = 200/0.683 = 293. Multiplying gives 137; the others misuse the equation.

Water Quality & Agricultural Non-point Source Pollution · River pollution and the oxygen sag curve

3. What retention time does a CSTR need to achieve 90% removal for k = 0.5 per day (first order)?

  1. 4.6 days
  2. 18 days
  3. 20 days
  4. 1.8 days
Answer and working

B. 18 days

C/C0 = 0.1 = 1/(1 + k theta) -> k theta = 9 -> theta = 18 d. A PFR would need 4.6 d.

Water & Wastewater Treatment · Reactor kinetics: batch, CSTR and plug flow

4. For a well-mixed lake with Q/V = 0.01 per day and first-order decay k = 0.04 per day, about how long after a step change in input does the concentration reach 95% of its new steady state?

  1. 20 days
  2. 100 days
  3. 300 days
  4. 60 days
Answer and working

D. 60 days

Response time = 1/(Q/V + k) = 1/0.05 = 20 days; 95% is reached after about three response times, 60 days. 100 days is the residence time V/Q alone; 20 days gives only 63%.

Environmental & Agro-ecosystem Modelling · Mass-balance and box models

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