FE Environmental solid and hazardous waste, and risk
Solid and hazardous waste carries 7–11 questions and health hazards and risk assessment 4–6 more. Both sit on this page, because a waste site is where exposure gets assessed.
Solid waste disposal: landfills, leachate and gas collection.
Hazardous waste compatibility.
Site characterization: sampling, monitoring, remedial investigation.
Hazardous and radioactive waste treatment and disposal: physical, chemical, thermal, biological.
20 waste and risk problems
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Question 1 of 20 · Solid and Hazardous Waste · 7–11 items
Under the U.S. waste management hierarchy established by the Pollution Prevention Act of 1990, which strategy is most preferred for reducing hazardous waste from a chemical processing facility?
Solution
U.S. waste management hierarchy (Pollution Prevention Act of 1990): source reduction → recycling → treatment → disposal.
End-of-pipe treatment ranks below source reduction and recycling, because the waste is still produced.
Recycling ranks second; it is preferred to treatment, but it comes after the waste has already been generated.
More storage only holds the same waste longer and adds accumulation-time compliance burdens; it reduces nothing.
The hierarchy set by the Pollution Prevention Act puts source reduction first: waste never generated needs no handling, treatment or disposal.
Question 2 of 20 · Health Hazards and Risk Assessment · 4–6 items
A subchronic study of a new industrial solvent identified a Lowest Observed Adverse Effect Level (LOAEL) of 50 mg/kg-day; no NOAEL was established. Using default uncertainty factors of 10 each for interspecies variability, intraspecies variability, subchronic-to-chronic extrapolation, and use of a LOAEL instead of a NOAEL, calculate the reference dose (RfD).
Given
LOAEL = 50 mg/kg-day
Interspecies UF = 10
Intraspecies UF = 10
Subchronic-to-chronic UF = 10
LOAEL-to-NOAEL UF = 10
Find
Reference dose RfD (mg/kg-day)
Solution
RfD = (NOAEL or LOAEL) ÷ (product of uncertainty factors) (Health hazards: reference dose).
UF = 10 × 10 × 10 × 10 = 10,000
RfD = 50 mg/kg-day ÷ 10,000 = 0.005 mg/kg-day
Four factors of 10 apply (interspecies, intraspecies, subchronic-to-chronic and LOAEL-to-NOAEL): RfD = 50/10,000 = 0.005 mg/kg-day.
0.05 mg/kg-day leaves out the factor for using a LOAEL instead of a NOAEL.
0.5 mg/kg-day applies only two of the four uncertainty factors.
5 mg/kg-day applies a single factor of 10.
50 mg/kg-day is the LOAEL itself, with no uncertainty factors.
Question 3 of 20 · Health Hazards and Risk Assessment · 4–6 items
A 70-kg worker is exposed to a carcinogen at a concentration of 0.05 mg/m3 for 20 years. The worker inhales 20 m3/day and works 250 days per year. For carcinogenic risk assessment, what is the Lifetime Average Daily Dose (LADD)?
Given
Body weight = 70 kg
Air concentration C = 0.05 mg/m3
Exposure duration = 20 years
Inhalation rate = 20 m3/day
Exposure frequency = 250 days/year
Find
Lifetime average daily dose LADD (mg/kg·day)
Solution
Lifetime average daily dose for a carcinogen (Health hazards: exposure assessment).
4.1 × 10−3 assumes exposure 365 days a year; the worker is exposed 250 days.
1.0 mg/day is the daily inhaled amount (0.05 × 20) on a workday, not normalized to body weight or averaged over a lifetime.
Question 4 of 20 · Health Hazards and Risk Assessment · 4–6 items
A regulatory agency sets a maximum allowable cancer risk of 1 × 10−6. The contaminant has a slope factor of 0.02 (mg/kg-day)−1. For an adult weighing 80 kg with an inhalation rate of 20 m3/day, calculate the maximum allowable concentration in mg/m3.
Given
Target cancer risk = 1 × 10−6
Slope factor = 0.02 (mg/(kg·d))−1
Body weight = 80 kg
Inhalation rate = 20 m3/d
Find
Maximum allowable air concentration (mg/m3)
Solution
Back-calculate a risk-based concentration from the slope factor (Health hazards: risk assessment).
2.0 × 10−1 is the right number in µg/m3 labeled as mg/m3.
Risk = SF × C × IR/BW, so C = Risk × BW/(SF × IR) = 10−6 × 80/(0.02 × 20) = 2.0 × 10−4 mg/m3.
1.3 × 10−5 swaps body weight and inhalation rate (IR/BW instead of BW/IR).
2.5 × 10−6 leaves out the 80 kg body weight.
8.0 × 10−8 multiplies by the slope factor instead of dividing.
Question 5 of 20 · Solid and Hazardous Waste · 7–11 items
A composting facility processes 1000 kg waste at 60% moisture (wet basis). Biological degradation removes 20% of the initial dry solids. Final compost has 40% moisture (wet basis). What is the final mass?
Given
Waste mass = 1000 kg
Initial moisture = 60% (wet basis)
Dry solids degraded = 20%
Final moisture = 40% (wet basis)
Find
Final compost mass (kg)
Solution
Dry-solids mass balance through composting (Solid waste: composting).
solids = 1,000 × 0.40 = 400 kg → after 20% degradation 320 kg
final mass = 320/(1 − 0.40) = 533.3 kg
400.0 kg is the initial dry solids, not the final compost mass.
Dry solids 400 kg, 20% degraded → 320 kg; at 40% moisture solids are 60% of the mass, so 320/0.60 = 533.3 kg.
666.7 kg ignores the 20% of solids lost to degradation (400/0.60).
800.0 kg divides by the moisture fraction 0.40 instead of the solids fraction 0.60.
920.0 kg subtracts the 80 kg of degraded solids but ignores the water lost in drying.
Question 6 of 20 · Solid and Hazardous Waste · 7–11 items
A landfill cell accepts 20,000 Mg of waste in Year 1 and 30,000 Mg in Year 2. With L0 = 100 m3/Mg and k = 0.05 yr−1, use G = Σ kL0Mie−kti (ti = years since placement) to estimate methane generation in Year 3.
Given
Year 1 waste = 20,000 Mg
Year 2 waste = 30,000 Mg
Methane potential L0 = 100 m3/Mg
Decay rate k = 0.05 yr−1
Find
Methane generation in Year 3 (m3)
Solution
First-order decay summed over the two annual cohorts (Solid waste: landfill gas).
9.05 × 104 counts only the Year 1 waste; the Year 2 waste is generating too.
1.43 × 105 counts only the Year 2 waste.
In Year 3 the cohorts are 2 and 1 years old: 0.05 × 100 × 20,000e−0.10 + 0.05 × 100 × 30,000e−0.05 = 90,500 + 142,700 = 2.33 × 105 m3.
2.50 × 105 omits the decay factor e−kt.
4.66 × 106 leaves out the rate constant k in front of L0M.
Question 7 of 20 · Solid and Hazardous Waste · 7–11 items
A leachate collection system pumps to an equalization basin. During a design storm event, the peak influent flow rate is 2,500 L/min for a duration of 4 hours. The transfer pump to the treatment plant has a fixed capacity of 500 L/min and runs continuously. What is the minimum required equalization volume to contain this event without overflow?
V = (2,500 − 500) L/min × 4 h × 60 min/h = 480,000 L
8,000 L multiplies by 4 hours instead of 240 minutes while the rates are per minute.
120,000 L is the volume pumped out during the event, not the volume that must be stored.
Storage = (2,500 − 500) L/min × 240 min = 480,000 L.
600,000 L is the total inflow; the pump removes 500 L/min while the event lasts.
720,000 L adds the pump rate to the inflow instead of subtracting it.
Question 8 of 20 · Solid and Hazardous Waste · 7–11 items
A Waste-to-Energy (WTE) plant processes 500 Mg/day of MSW with a Lower Heating Value (LHV) of 13 MJ/kg. The gross electrical efficiency is 25% and the plant consumes 10% of the gross power for parasitic loads. The resulting ash is 20% of the feed mass with a density of 2,000 kg m−3. Determine the net electrical power exported and the daily volume of ash produced.
Given
MSW feed = 500 Mg/day
Lower heating value LHV = 13 MJ/kg
Gross electrical efficiency = 25%
Parasitic load = 10% of gross power
Ash = 20% of feed mass
Ash density = 2,000 kg/m3
Find
Net electrical power (MW) and daily ash volume (m3)
Solution
Waste-to-energy: heat input × efficiency × (1 − parasitic fraction); ash volume = ash mass ÷ density (Solid waste: combustion).
75.2 MW is the thermal heat input, before the 25% conversion efficiency.
250 m3 divides the full 500 Mg feed by the ash density instead of the 100 Mg of ash.
18.8 MW is the gross output; the plant uses 10% of it for parasitic loads.
20.9 MW divides by 0.90 instead of multiplying, adding the parasitic load rather than removing it.
Heat input = 500,000 kg × 13 MJ/kg ÷ 86,400 s = 75.2 MW; net = 75.2 × 0.25 × 0.90 = 16.9 MW; ash = 100,000 kg ÷ 2,000 kg/m3 = 50 m3.
Question 9 of 20 · Solid and Hazardous Waste · 7–11 items
A community generates 1,000 kg of waste, 60% of which is organic. A composting program is implemented with a 75% resident participation rate. Participating residents successfully divert 90% of their organic waste to the compost bin. If the composting process results in a 50% mass loss due to degradation, what is the final mass of compost produced?
Given
Waste generated = 1,000 kg
Organic fraction = 60 %
Participation rate = 75 %
Diversion by participants = 90 %
Composting mass loss = 50 %
Find
Final compost mass (kg)
Solution
Multiply the fractions along the diversion chain (Solid waste: recycling and composting).
M = 1,000 × 0.60 × 0.75 × 0.90 × (1 − 0.50) = 202.5 ≈ 203 kg
1,000 × 0.60 organic × 0.75 participating × 0.90 captured = 405 kg collected; 50% lost in composting leaves 202.5 ≈ 203 kg.
225 kg assumes participants divert all their organics (no 90% capture factor).
270 kg assumes every resident participates (no 75% factor).
300 kg applies only the organic fraction and the 50% process loss.
405 kg is the organic material collected, before the 50% mass loss during composting.
Question 10 of 20 · Solid and Hazardous Waste · 7–11 items
A landfill generates 200 m3/hr of methane. Initially, the gas collection efficiency is 75% and the cover oxidation factor is 0.10. An impermeable geomembrane cap is installed, increasing collection efficiency to 90% and reducing the oxidation factor to 0.00. What is the percent reduction in reported fugitive methane emissions?
60% compares uncollected gas (50 → 20 m3/hr) and ignores the 10% cover oxidation that applied before the cap.
125% divides the 25 m3/hr drop by the new emissions instead of the original 45 m3/hr.
Question 11 of 20 · Solid and Hazardous Waste · 7–11 items
A landfill design used the HELP model with annual average precipitation data to size the leachate collection system. However, the system overflowed during the first year of operation despite total annual rainfall being normal. Which parameter assumption most likely caused this failure?
Solution
Leachate collection is sized for peak inflow; using annual-average precipitation smooths out storm intensity (Solid waste: leachate).
Annual ET affects the yearly water budget, which was normal; it does not explain peaks exceeding capacity.
Fresh waste below field capacity absorbs water and reduces first-year leachate; it would not cause overflow.
Waste density changes storage capacity and settlement, not the peak flow reaching the collection pipes.
A more permeable liner would leak more and send less to the collection system, not more.
Annual averages hide storm peaks; leachate collection must carry the short-term infiltration from intense events, which is what overflowed.
Question 12 of 20 · Solid and Hazardous Waste · 7–11 items
A city with a fixed landfill capacity increases its waste diversion rate from 0% to 20%. Later, it increases the diversion rate from 20% to 40%. Assuming constant waste generation, how does the additional landfill lifespan gained in the second step compare to the lifespan gained in the first step?
Solution
Landfill life L = capacity/(disposal rate); with R = G(1 − d), equal increments of diversion give increasing gains.
ΔL1 = 1/0.8 − 1/1.0 = 0.25; ΔL2 = 1/0.6 − 1/0.8 = 0.42 (in units of the base life)
The gains grow, not shrink, because life is inversely proportional to the disposal rate.
Life varies with 1/R, not linearly with diversion, so equal diversion steps do not give equal gains.
Density sets the absolute life, but both steps scale with the same density; the ratio of gains does not depend on it.
Life ∝ 1/(disposal rate): 1/0.8 − 1 = 0.25 of the base life for the first step, 1/0.6 − 1/0.8 = 0.42 for the second, so the second gain is larger.
Equal 20-point cuts remove a larger share of the remaining disposal in the second step, so the second gain is the larger one.
Question 13 of 20 · Solid and Hazardous Waste · 7–11 items
An environmental engineer is designing a sampling plan to delineate the unknown horizontal boundaries of a chromium plume in a heterogeneous fill site. The objective is to define the extent of contamination with uniform spatial coverage. Which sampling strategy is most appropriate for this data quality objective?
Solution
Delineation with uniform coverage → systematic (grid) sampling (Solid and hazardous waste: site characterization).
A regular grid gives uniform spatial coverage and a known chance of finding contamination of a given size, which suits delineating an unknown boundary.
Simple random points can cluster and leave gaps, so coverage is not uniform.
Judgmental samples near the source confirm contamination but leave the outer extent undefined.
Sampling along a few transects covers only those lines, not the whole area.
Stratified random sampling needs known strata to divide the site; here the boundaries are what is unknown.
Question 14 of 20 · Solid and Hazardous Waste · 7–11 items
A site is being characterized using a Hydraulic Profiling Tool (HPT). The resulting log shows high injection pressure and low flow from 0–12 meters below ground surface (bgs). From 14–16 meters bgs, the injection pressure drops to near zero and flow rate maximizes. Which depth interval represents the primary preferential flow path for contaminant transport?
Solution
HPT: injection pressure is inversely related to permeability; low pressure + high flow = permeable zone (Site characterization: direct-push tools).
2–4 m lies in the 0–12 m zone of high injection pressure and low flow, which marks tight, low-permeability soil.
4–6 m is inside the tight 0–12 m interval, not a flow path.
8–10 m still shows high pressure and low flow, so it is low permeability.
10–12 m is the bottom of the tight zone; the permeable zone starts below it.
Near-zero injection pressure with maximum flow means water enters the formation easily: 14–16 m is the permeable, preferential flow path.
Question 15 of 20 · Solid and Hazardous Waste · 7–11 items
Groundwater sampling results show toluene concentrations of 35 μg/L in the site samples. The associated trip blank is non-detect (ND), but the field blank contains 12 μg/L of toluene. What is the most likely source of the contamination found in the field blank?
Solution
Blank comparison: trip blank (transport, containers, lab) clean + field blank contaminated → contamination introduced at the sampling site (Site characterization: QA/QC).
Trip blanks are prepared with the same lab-supplied water and preservative; their non-detect result clears it.
A field blank is exposed to site conditions while samples are taken; with a clean trip blank, the toluene must have entered on site.
Trip blanks use the same containers and are non-detect, so the bottles are not the source.
Both blanks are analyzed by the same lab; a calibration problem would show up in the trip blank too.
The trip blank travels with the samples and is non-detect, which rules out contamination in transit.
Question 16 of 20 · Solid and Hazardous Waste · 7–11 items
A proposed hazardous waste landfill cell has a liner area of 10,000 m2. The composite liner includes a compacted clay layer with a thickness of 0.9 m and a hydraulic conductivity of 1 × 10−7 cm/s. The design leachate collection system maintains a maximum hydraulic head of 0.3 m above the liner. If the regulatory performance standard limits leakage to 1,200 L/day, does the design meet the standard, and what is the estimated leakage rate?
Given
Liner area = 10,000 m2
Clay thickness = 0.9 m
Clay hydraulic conductivity = 1 × 10−7 cm/s
Leachate head = 0.3 m
Leakage limit = 1,200 L/d
Find
Leakage rate (L/d) and compliance
Solution
Darcy leakage through the clay liner with the gradient across its full thickness (Solid waste: liner design).
115,200 L/day uses 10−7 m/s; 10−7 cm/s is 10−9 m/s.
3,456 L/day inverts the gradient to (h + t)/h = 4.
864 L/day assumes a unit gradient (i = 1), ignoring the 0.3 m of ponded head.
i = (0.3 + 0.9)/0.9 = 1.33; Q = kiA = 10−9 m/s × 1.33 × 10,000 m2 = 1.33 × 10−5 m3/s = 1,152 L/day, below the 1,200 L/day limit.
288 L/day uses i = h/t = 0.33; the gradient across the clay is (h + t)/t because the head drives flow through the full thickness.
Question 17 of 20 · Solid and Hazardous Waste · 7–11 items
A hazardous waste incinerator treats 10 kg/min of a principal organic hazardous constituent (POHC). The system is required to achieve a Destruction and Removal Efficiency (DRE) of 99.99%. The air intake is 500 m3/min at 25°C, and the stack gas discharges at 1,000°C. Assuming the molar flow rate of gas remains constant through the system, calculate the maximum allowable concentration of the POHC in the stack gas.
Given
POHC feed = 10 kg/min
Required DRE = 99.99 %
Air intake = 500 m3/min at 25 °C
Stack gas temperature = 1,000 °C
Find
Maximum allowable POHC stack concentration (mg/m3)
Solution
Allowed POHC emission from the DRE, divided by the stack gas volume at stack temperature (Hazardous waste: incineration).
2.0 mg/m3 divides by the 500 m3/min intake flow at 25 °C; the hot stack gas occupies 4.3 times that volume.
4.7 mg/m3 uses 99.9% DRE; 99.99% allows only one ten-thousandth of the feed to escape.
8.5 mg/m3 corrects the gas volume in the wrong direction (× 298/1,273), shrinking the hot stack flow to 117 m3/min.
20 mg/m3 uses 99.9% DRE and the cold intake flow, two errors at once.
Question 18 of 20 · Solid and Hazardous Waste · 7–11 items
A sludge sample undergoing stabilization has a total mass of 1.0 kg, a moisture content of 20% by weight (wet basis), and a total Cadmium concentration of 50 mg/kg (wet weight). The solid-water distribution coefficient (Kd) for the binder-sludge mixture is 20 L/kg. Estimate the equilibrium aqueous phase concentration of Cadmium in the pore water.
13 mg/L multiplies Kd by the water volume instead of the solids mass (50/(0.20 × 20)).
250 mg/L assumes all the cadmium is dissolved in the 0.20 L of pore water, ignoring sorption.
Question 19 of 20 · Solid and Hazardous Waste · 7–11 items
A facility is reorganizing its chemical waste storage area. Which of the following waste pairs represents an incompatible combination that must be segregated to prevent a potential fire or explosion hazard?
Solution
Segregate oxidizers (nitric acid) from organics and fuels (acetic acid) (Hazardous waste: storage and compatibility).
Phosphoric acid is not an oxidizer, so storing it with acetic acid is not a fire or explosion risk.
Both are bases; they do not react dangerously with each other.
Ethanol and acetone are both flammable solvents and are commonly bulked together as flammable waste.
Nitric acid is a strong oxidizer and acetic acid is an organic fuel; together they can react violently, so they must be stored apart.
Two non-oxidizing mineral acids (at the usual concentrations) can share an acid cabinet.
Question 20 of 20 · Health Hazards and Risk Assessment · 4–6 items
A worker's exposure was monitored for 6 hours (360 minutes). The sampling pump operated at 2.0 L/min for 120 minutes and 2.5 L/min for 240 minutes, collecting 10.64 mg of dust. Assuming zero exposure for the remaining 2 hours, calculate the 8-hour TWA.
Given
Sampling time = 360 min
Pump rate = 2.0 L/min for 120 min
Pump rate = 2.5 L/min for 240 min
Dust collected = 10.64 mg
Unsampled time = 2 h at zero exposure
Find
8-hour TWA (mg/m3)
Solution
8-hour TWA with an unsampled period at zero (Health hazards: occupational exposure).
V = 2.0 L/min × 120 min + 2.5 L/min × 240 min = 840 L = 0.84 m3
C = 10.64 mg/0.84 m3 = 12.7 mg/m3; TWA = 12.7 × 360/480 = 9.50 mg/m3
12.7 mg/m3 is the average over the 6 sampled hours; the 8-hour TWA spreads it over 480 minutes.
11.1 mg/m3 uses 2.0 L/min for the whole 360 minutes, understating the air volume.
10.2 mg/m3 swaps the two flow rates (2.5 L/min for 120 min, 2.0 L/min for 240 min), understating the air volume at 780 L.
9.85 mg/m3 averages the two flow rates (2.25 L/min) instead of weighting them by time.
Air sampled = 2.0 × 120 + 2.5 × 240 = 840 L; C = 10.64 mg/0.84 m3 = 12.7 mg/m3; TWA = 12.7 × 360/480 = 9.50 mg/m3.
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Common traps
Wet basis or dry basis. Moisture in refuse and compost is usually given on a wet basis: water over total mass. Run the mass balance on dry solids and add the water back at the end.
Loose or compacted. Landfill volume uses the compacted density, and daily cover takes space too; collection vehicle volume uses the loose density.
Higher or lower heating value. The lower value leaves out the heat in the water vapor that leaves the stack. Use whichever the problem states for energy recovery, and do not mix the two in one balance.
Health Hazards and Risk Assessment
The specification's risk area (4–6 questions) covers dose-response for carcinogens and noncarcinogens, exposure routes and pathways, and occupational health: PPE, noise, safety screening. Intake is the same chronic daily intake for both kinds of toxicity:
CDI = (C × IR × ED × EF) / (AT × BW)
Carcinogen: risk = CDI × SF, and the averaging time AT is a lifetime.
Noncarcinogen: hazard quotient HQ = CDI / RfD, above 1 is a concern, and AT is the exposure duration.
Noise adds logarithmically: two equal sources are 3 dB louder than one, not twice the decibels.
Is risk assessment tested on the FE Environmental exam?
Yes. Health hazards and risk assessment is a separate area of the specification, 4–6 questions, covered in the risk section of this page.
What is the difference between RCRA and CERCLA?
RCRA governs waste that is being generated, stored, treated and disposed of now; CERCLA (Superfund) pays for and assigns liability for cleaning up sites that were abandoned or never controlled.
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