Free FE Environmental practice problems

Ten real questions from our FE Environmental bank, drawn from 10 of its 15 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

5 with figures10 topicsFull worked solutions

1
medium
A. Time value of money & equivalence
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
Figure — drag or pinch to move around
1. $8,593
2. $13,333
3. $20,593
4. $25,333
Show the answer & worked solution
Answer: C.$20,593
The capital recovery factor (FE Reference Handbook 10.6 p. 235) converts a present amount into equal end-of-year payments: $(A/P,\ 6\%,\ 15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=\dfrac{0.06\,(1.06)^{15}}{(1.06)^{15}-1}=0.10296$ , printed as 0.1030 in the handbook's 6% factor table. The annual payment is $200,000 × 0.10296 = $20,593. Distractors: $P/n$ , spreading the principal evenly and ignoring interest ($13,333); the sinking-fund factor $(A/F)$ read in place of $(A/P)$ ($8,593), which treats the loan as a future sum; and simple interest on the full principal for all 15 years, spread evenly, $P(1+in)/n$ ($25,333), which charges interest on money already repaid.
2
medium
A. Fluid statics & dynamics
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 150$ mm. The upstream velocity is $V_1 = 1.5$ m/s. The velocity in the smaller section is most nearly:
Figure — drag or pinch to move around
1. 0.844 m/s
2. 2 m/s
3. 2.67 m/s
4. 4.74 m/s
Show the answer & worked solution
Answer: C.2.67 m/s
Continuity (FE Reference Handbook 10.6 p. 185): the same flow $Q=Av$ passes both sections, so $A_1V_1=A_2V_2$ . With $A=\pi D^2/4$ the $\pi/4$ cancels, leaving $V_2=V_1\left(\dfrac{D_1}{D_2}\right)^2=1.5\left(\dfrac{200}{150}\right)^2=2.67$ m/s, faster in the smaller section, as it must be. Distractors: the diameter ratio used un-squared, $V_1D_1/D_2$ (2 m/s), forgetting that area goes as $D^2$ ; the area ratio inverted, $V_1(D_2/D_1)^2$ (0.844 m/s), which would have the water slow down in the smaller section; and the ratio squared twice, $V_1(D_1/D_2)^4$ (4.74 m/s), the area-ratio squaring applied a second time.
3
medium
A. Central tendency, dispersion & distributions
The life of a carbide cutting insert turning steel at a fixed cutting speed is normally distributed with mean $\mu=50$ min and standard deviation $\sigma=8$ min. The probability that a randomly selected insert lasts between 46 min and 62 min is most nearly:
Figure — drag or pinch to move around
1. 0.2417
2. 0.4332
3. 0.6247
4. 0.6915
Show the answer & worked solution
Answer: C.0.6247
Standardize each limit (Normal Distribution, $z=(x-\mu)/\sigma$ ): $z_a=(46-50)/8=-0.50$ and $z_b=(62-50)/8=1.50$ . From the Unit Normal Distribution table, $F(1.50)=0.9332$ and $F(-0.50)=1-F(0.50)=1-0.6915=0.3085$ (the table lists only positive $x$ ). So $P(46<X<62)=F(z_b)-F(z_a)=0.9332-0.3085=0.6247$ . Distractors: the minus sign on $z_a=-0.50$ dropped, so the table value $F(0.50)=0.6915$ was subtracted in place of $F(-0.50)=0.3085$ (0.2417); only the area from the mean to 62 min counted, $F(1.50)-0.5$ , leaving out the part between 46 min and the mean (0.4332); and the upper limit ignored, reporting $P(X>46)=1-F(-0.50)$ (0.6915).
4
easy
A. Dose-response, exposure & occupational health
The figure shows the dose-response curve from a chronic animal study of an industrial solvent, a noncarcinogen. Which statement about using this curve to set a safe human exposure is correct?
Figure — drag or pinch to move around
1. The slope of the rising part of the curve is the reference dose, and the risk equals the chronic daily intake multiplied by that slope.
2. The reference dose is the NOAEL multiplied by an uncertainty factor, so it lies above the threshold, where the curve begins to rise.
3. No adverse effect is expected below the NOAEL, and the reference dose is the NOAEL divided by an uncertainty factor, so it lies below the threshold.
4. A hazard index (CDI/RfD) greater than 1.0 still means no adverse effect is expected, because the reference dose already carries a safety margin.
Show the answer & worked solution
Answer: C.No adverse effect is expected below the NOAEL, and the reference dose is the NOAEL divided by an uncertainty factor, so it lies below the threshold.
A noncarcinogen's curve is flat (no response) up to a threshold. The handbook defines the NOAEL as the dose below which there are no harmful effects, and derives the reference dose from it: $\text{RfD} = \text{NOAEL}/\text{UF}$ , where the uncertainty factor $\text{UF} > 1$ allows for animal-to-human and person-to-person differences, so the RfD sits below the threshold (Safety — Noncarcinogens and Reference Dose). Distractors: the uncertainty factor applied the wrong way, which would put the "safe" dose on the rising part of the curve; the cancer slope factor's role given to the RfD (only a carcinogen's no-threshold line has a slope used as $\text{Risk} = \text{CDI}\times\text{CSF}$ ); and the hazard index read as harmless above 1.0 (EPA treats $\text{HI} = \text{CDI}/\text{RfD} > 1.0$ as the possibility of an adverse effect, whatever margin the RfD carries).
5
easy
A. Runoff & water budget
An urban catchment of 12 ha has a runoff coefficient of 0.70. The design rainfall intensity, for a storm lasting the catchment's time of concentration, is 90 mm/h. The peak runoff rate is most nearly:
Figure — drag or pinch to move around
1. 0.756 m³/s
2. 0.843 m³/s
3. 2.10 m³/s
4. 3.00 m³/s
Show the answer & worked solution
Answer: C.2.10 m³/s
The handbook's Rational Formula $Q = CIA$ works in any consistent units. With $i = 90\text{ mm/h} = \dfrac{90/1000\text{ m}}{3600\text{ s}} = 2.50\times10^{-5}$ m/s and $A = 12\text{ ha} = 120{,}000$ m², $Q = (0.70)(2.50\times10^{-5})(120{,}000) = 2.10$ m³/s. (The same thing in one step: $Q = CiA/360$ with $i$ in mm/h and $A$ in ha.) Distractors: the runoff coefficient $C$ left out (treated as 1); only the millimeters converted to meters ( $CiA/1000$ ), with the hectares and the hours left unconverted; and the handbook's U.S. form used with $i = 90/25.4 = 3.54$ in./hr but the 12 ha entered as if they were acres (1 ha = 2.47 acres), then the cfs converted to m³/s.
6
medium
A. Codes of ethics, liability & licensure
An engineer preparing the switchgear specification for a hospital project owns a 20 percent interest in one of the three manufacturers likely to bid on the equipment. Under the NCEES Model Rules of Professional Conduct, the engineer should:
1. sell the ownership interest quietly after the contract is awarded to eliminate the conflict
2. write the specification around generic performance requirements so the ownership has no practical effect
3. disclose the ownership interest to the client in writing before continuing with the specification
4. continue normally, since the bids will be evaluated against objective criteria the client controls
Show the answer & worked solution
Answer: C.disclose the ownership interest to the client in writing before continuing with the specification
Model Rule B.6 (FE Reference Handbook 10.6 p. 5, Obligations of Licensees to Employer and Clients) requires a licensee to disclose to the client all known or potential conflicts of interest that could influence or appear to influence their judgment. A stake in a likely bidder is exactly that, and the specification is where it could act, so the client must hear of it before the work goes further; putting it in writing makes the disclosure a record, and the informed client then decides how to proceed. Distractors: drafting around the interest manages the conflict privately, and the client still does not know of it; objective bid criteria do not cure it, because the stake can shape the specification itself before any bid is scored; selling the interest after award keeps the conflict hidden through the very period in which it could operate.
7
medium
A. Analytic geometry, algebra & trigonometry
A ball is thrown upward from the edge of a bridge deck 20 m above a river. Its height above the water is $y=20+15t-4.905t^2$ m, with $t$ in seconds after release. The time at which the ball reaches the water is most nearly:
1. 1.00 s
2. 1.53 s
3. 3.06 s
4. 4.06 s
Show the answer & worked solution
Answer: D.4.06 s
At the water $y=0$ , so $4.905t^2-15t-20=0$ , with $a=4.905$ , $b=-15$ and $c=-20$ . The quadratic formula (Mathematics — Quadratic Equation) gives $t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{15\pm\sqrt{225+392.4}}{9.81}=\dfrac{15\pm24.85}{9.81}$ , so $t=4.06$ s or $t=-1.00$ s. A negative time is before the release, so the ball reaches the water at $t\approx4.06$ s. Distractors: the sign of $b$ lost, so $-b$ was taken as $-15$ instead of $+15$ , which turns the rejected negative root into a positive 1.00 s; the $\pm\sqrt{b^2-4ac}$ term dropped, leaving $-b/2a=15/9.81$ , which is the time at the TOP of the ball's flight (1.53 s); and the 20 m deck height left out, so $15t-4.905t^2=0$ gives the time to fall back to DECK level, not to the water (3.06 s).
8
medium
A. Laws, energy cycles & ideal gases
A rigid compressed-air tank has an internal volume of $0.8\,\text{m}^3$ . Its pressure gauge reads 600 kPa, the local atmospheric pressure is 101 kPa, and the air in the tank is at 27 °C. Treating air as an ideal gas with $R=0.287\,\text{kJ/(kg·K)}$ , the mass of air in the tank is most nearly:
1. 5.57 kg
2. 6.51 kg
3. 8.14 kg
4. 72.4 kg
Show the answer & worked solution
Answer: B.6.51 kg
The ideal-gas law (Thermodynamics — Ideal Gas, $PV=mRT$ ) takes absolute pressure and temperature. Absolute pressure is atmospheric plus gauge: $P=101+600=701$ kPa, and $T=27+273=300$ K. Then $m=\dfrac{PV}{RT}=\dfrac{(701\ \text{kPa})(0.8\ \text{m}^3)}{(0.287\ \text{kJ/(kg·K)})(300\ \text{K})}=\dfrac{560.8}{86.1}=6.513$ kg (a kPa·m³ is a kJ). Distractors: the gauge reading used as the absolute pressure, $P=600$ kPa (5.57 kg); the volume left out: $P/RT$ is the air's density in kg/m³, not the tank's mass (8.14 kg); and the temperature left in °C, $T=27$ instead of 300 K (72.4 kg).
9
easy
A. Air quality, emissions & mass/energy balances
An urban air monitor reports 0.05 ppm of SO₂ (molar mass 64 g/mol) at 25 °C and 1 atm. Expressed as a mass concentration, this is most nearly:
1. 19.1 µg/m³
2. 131 µg/m³
3. 143 µg/m³
4. 1560 µg/m³
Show the answer & worked solution
Answer: B.131 µg/m³
The handbook's air-pollution conversion gives µg/m³ $= \text{ppb}\times\dfrac{P\,MW}{RT}$ , so first $0.05$ ppm $= 50$ ppb. At 25 °C and 1 atm, $\dfrac{RT}{P} = \dfrac{(0.0821)(298.15)}{1} \approx 24.45$ L/mol, giving $C = \dfrac{50\times64}{24.45} = 130.9$ µg/m³. Distractors: the molar mass and the molar volume swapped ( $\text{ppm}\times24.45\times1000/MW$ ); the 0 °C molar volume 22.4 L/mol used instead of 24.45 L/mol (about 9% high); and the temperature entered in °C (25) instead of kelvin (298.15 K) in $V_m = RT/P$ , giving $V_m = 2.05$ L/mol.
10
easy
A. Stoichiometry, equilibrium & acid-base
A mountain stream sample has a hydrogen-ion concentration of $[\text{H}^+] = 2.0\times10^{-5}$ mol/L. Its pH is most nearly:
1. 4.70
2. 5.00
3. 9.30
4. 10.82
Show the answer & worked solution
Answer: A.4.70
By definition (handbook, Acids, Bases, and pH), $\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2.0\times10^{-5}) = 5 - \log_{10}2.0 = 5 - 0.30 = 4.70$ . Distractors: the power of ten read off alone with the mantissa ignored (pH taken as 5); the pOH computed in place of the pH ( $14-\text{pH}$ ); and the natural logarithm used in place of the base-10 logarithm ( $-\ln[\text{H}^+]$ ).

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