FE Environmental Health Hazards and Risk Assessment Practice Problems

The current FE Environmental specification assigns 4-6 of the 110 exam questions to Health Hazards and Risk Assessment. Form A contains 5 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A private well serving a residence contains tetrachloroethylene at 35 µg/L. An adult weighing 180 lb ingests 0.60 gal/day of this water 350 days per year for 26 years. The endpoint of interest is non-carcinogenic, so the averaging time equals the exposure duration. The chronic daily intake is most nearly:

(A) $1.1 \times 1 0^{- 4}$ mg/(kg-day)
(B) $2.5 \times 1 0^{- 4}$ mg/(kg-day)
(C) $3.5 \times 1 0^{- 4}$ mg/(kg-day)
(D) $9.3 \times 1 0^{- 4}$ mg/(kg-day)

Show worked solution

Answer: (D)

Because the averaging time equals the exposure duration, the two cancel and the intake reduces to a daily dose scaled by exposure frequency, about $9.3 \times 1 0^{- 4}$ mg/(kg-day).

Intake equations are written on a body-mass basis, so the pound and gallon values must become kilograms and litres before anything else.

180 lb \times 0.4536 kg/lb = 81.65 kg

0.60 gal/day \times 3.785 L/gal = 2.271 L/day

Averaging time for a threshold endpoint is the exposure duration itself, which fixes the denominator at 26 years expressed in days.

A T = 26 yr \times 365 d/yr = 9, 490 d

C D I = \frac{C \cdot I R \cdot E F \cdot E D}{B W \cdot A T}

The exposure frequency of 350 d/yr leaves roughly 96 percent of the averaging period as exposed time, so the intake stays close to the daily dose during exposure.

C D I = \frac{( 0.035 ) ( 2.271 ) ( 350 ) ( 26 )}{( 81.65 ) ( 9 , 490 )} = 9.3 \times 1 0^{- 4} mg/(kg \cdot d)

FE Reference Handbook — Environmental Engineering: Risk Assessment, Chronic Daily Intake Equation

Why the other choices appear

(A)Treats the given 180 and 0.60 as kilograms and litres with no conversion: $(0.035) (0.60) (350) (26) / [(180) (9, 490)] = 1.1 \times 1 0^{- 4}$ mg/(kg-day).
(B)Converts body weight to 81.65 kg but leaves the ingestion rate at 0.60 L/day: $191.1/774, 840 = 2.5 \times 1 0^{- 4}$ mg/(kg-day).
(C)Averages over a 70-year lifetime (25,550 d) although the endpoint is non-carcinogenic: $723.3/ [(81.65) (25, 550)] = 3.5 \times 1 0^{- 4}$ mg/(kg-day).